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BEE Energy Sample / Model Paper 2026 : Energy

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Makarand Deole
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PAPER-1 COLOR C0DE : GREEN 24th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER, 2024 PAPER - 1 : GENERAL ASPECTS OF ENERGY MANAGEMENT & ENERGY AUDIT Date : 28.09.2024 Timings : 09:30-12:30 HRS Duration : 3 HRS Section I: OBJECTIVE TYPE Max. Marks : 150 Marks: 50 x 1 = 50 1. Calculate the reduction in CO2 emissions if energy efficiency measures save 1000 kWh, assuming 0.8 kg CO2/kWh. a) 800 kg b) 1250 kg c) 625 kg d) 1000 kg 2. Which of the following is not true, equivalent to 1 atm pressure? a) 1 atm = 101.3 kPa b) 1 atm = 10332 mmWC c) 1 atm = 14.7 psi d) 1 atm = 0.98 kg/cm2 3. Redwood Seconds is measure of ____________ a) Density b) Viscosity c) Specific Gravity d) Flash Point 4. For calculating plant energy performance which of the following data is not required a) Current year production b) Capacity Utilization c) Reference year production d) Reference year Energy use 5. The internal rate of return is discount rate for which NPV is a) Positive b) Zero c) Negative d) All of the above 6. Transit time method is used in which of the instrument a) Lux Meter b) Ultrasonic Flow Meter c) Pitot Tube d) Fyrite 7. Which of the following GHG has the longest atmospheric life time a) Carbon dioxide (CO2) b) Sulphur Hexafluoride (SF6) c) Chloroflurocarbons (CFC) d) Perfluorocarbons (PFC) 8. What is the heat content of 500 liters of water at 6oC in terms of the basic unit of energy in kilojoules? a) 12000 b) 3000 c) 500 d) None of the above 9. The number of moles of water contained in 72 grams of water is a) 2 b) 3 c) 4 d) 5 10. Which of the following is not objective of Demand Side Management? a) Managing Demand by DISCOM to reduce peak demand b) Increasing Load of Generator to meet Peak demand c) Reducing Capital need for Power Capacity Expansion d) None of the above 11. Which of the following has the lowest energy content in terms of MJ/kg a) LPG b) Diesel c) Bagasse d) Furnace Oil 12. Heat transfer in an air-cooled condenser occur predominately by a) Conduction b) Convection c) Radiation d) All of the above 13. The force field analysis in energy action planning considers a) Positive forces only b) Negative Forces only c) No forces d) Both Positive and Negative forces 14. To arrive at the relative humidity at a point we need to know ___________ of air a) DBT b) WBT c) Dew point d) Both A and B 15. Which statement is false regarding Critical path a) CP is longest duration path b) It identifies minimum time to Complete the project c) Activities lies on it cannot be delay d) It is maximum time required to complete the project 16. The roto axis is aligned with wind direction in windmill by ____________control? a) Yaw b) Pitch c) Disc Break d) Both A and B 17. For activity in project, Latest start time is 8 weeks and Latest Finish time is 12 Weeks. If the earliest finish time is 9 Weeks, Slack time for the activity is a) 1 Week b) 3 Weeks c) 4 Weeks d) 7 Weeks 18. Which techniques takes care of time value of money in evaluation a) Payback Period b) IRR c) NPV d) Both B and C 19. If asset depreciation is considered, then the net operating cash inflow will be a) Lower b) Higher c) No effect d) None of the above 20. One Silicon cell in PV modules typically produces a) 0.5 V b) 1.0 V c) 1.5 V d) 2.0 V 21. When the evaporation of water from wet substance is zero, the relative humidity of air is likely to be a) 0% b) 50% c) 100% d) Unpredictable 22. The energy conversion efficiency of solar cell does not depend on a) Solar Energy Insolation b) Inverter c) Area of the Solar Cell d) Maximum Power Output 23. Energy Intensity is ratio of a) Fuel Consumption/GDP b) GDP/Fuel Consumption c) GDP/Energy Consumption d) Energy Consumption/GDP 24. In inductive and resistive combination circuit, the resultant power factor under AC supply will be a) Less than Unity b) More than Unity c) Zero d) Unity 25. If wind speed triples, the energy output from wind turbine will be a) 3 Times b) 6 Times c) 9 Times d) None of the above 26. If we heat air without changing absolute humidity, % relative humidity will a) Increase b) Decrease c) No change d) Can't Say 27. Among which of the following fuel the difference between the GCV and NCV is maximum a) Coal b) Furnace Oil c) Natural Gas d) Rice Husk 28. Which among the following factors is most appropriate for adopting EnMS? a) To improve their energy efficiency b) To reduce cost c) To increase productivity d) Systematically manage their energy use 29. The Ozone layer in stratosphere act as an efficient filter for a) b) c) d) UV-B Rays UV-C Rays X- Ray Gamma Rays 30. An induction motor with 30 kW rating and efficiency of 85% in its name plate means a) It will draw 35.29 kW at full load b) it will always draw 30 kWat full load c) it will draw 25.5 kW at full load d) it will draw 28.23 kW at full load 31. Which of the following macro factors is used in the sensitivity analysis of project finance a) Change in Tax rate b) Change in O & M Cost c) Change in Debt : equity Ratio d) Change in forms of Financing 32. An indication of Sensible heat content in air-water vapour mixture is a) Wet bulb temperature b) Dry bulb temperature c) dew point temperature d) density of air 33. Energy content in 2500 kgs of coal with a calorific value of 4000 kcal/ kg in terms of toe would be a) 1 toe b) 10 toe c) 100 toe d) 1000 toe 34. Reserve per production (R/P) is estimated as a) Reserves remaining at end of year X production in the year b) Reserves remaining at end of year / production in the year c) production in year / Reserves remaining at end of the year d) None of the above 35. In a fuel cell, ___ combines with ____ to generate electricity and ____ comes out as a byproduct. a) Hydrogen, Oxygen, water b) Hydrogen, Nitrogen, nitrous oxide c) Carbon, hydrogen, methane d) Carbon, oxygen, carbon dioxide 36. A manufacturing plant consumes 5 tonnes of coal (CV = 4000 kCal/ kg) to produce 25 tonnes of cement. The Specific Energy Consumption (SEC) of the plant shall be a) 100 kcal/ kg of cement b) 200 kcal/ kg of cement c) 400 kcal/ kg of cement d) 800 kcal/ kg of cement 37. A solution of common salt in water is prepared by adding 25 kg of salt to 100 kg of water. The concentration of salt in this solution as a weight fraction shall be a) 10% b) 15% c) 20% d) 25% 38. An electric iron of power 2000 watts is used for a total of 120 minutes per month. Compute its monthly electricity consumption a) 2.0 kWh b) 2.4 kWh c) 4.0 kWh d) 24.0 kWh 39. The dryness fraction (x) of superheated steam will be a) x = 0.8 b) x = 0.9 c) x = 1 d) x = 0 40. Which entity is responsible for implementing the Energy Conservation Act 2001? a) Ministry of Renewable Energy b) Bureau of Energy Efficiency (BEE) c) Central Pollution Control Board d) National Productivity Council 41. Which of the following is a measure included in the Energy Conservation Act 2001? a) Energy audits b) Energy-saving certificates c) Standards and labelling d) All of the above 42. Calculate the energy consumed by a 200-watt appliance used for 5 hours a day over 30 days. a) 30 kWh b) 27000 kCal c) 6500 kJ d) 30 kJ/h 43. Which of the following is a non-renewable energy source? a) Solar b) Wind c) Biomass d) Coal 44. How is energy efficiency typically improved in industrial processes? a) Reducing production rates b) Optimizing equipment performance c) Increasing labour d) None of the above 45. Which of the following is a typical step in an energy audit? a) Data collection b) Analysis c) Reporting d) All of the above 46. Which tool is commonly used for measuring power factor ? a) Thermometer b) Hygrometer c) Anemometer d) None of the above 47. What is the payback period in energy management? a ) Time taken to identify savings b) Time taken to report savings c) Time taken to recover the investment through savings d) All of the above 48. Which of the following is a method for financing energy efficiency projects? a) Loans b) Leasing c) Performance contracting d) All of the above 49. What is the primary financial metric used to evaluate energy projects? a) Gross margin b) Net present value (NPV) c) Revenue d) Operating income 50. Which principle is energy monitoring and targeting based on? a) Energy consumption is constant b) You can t manage what you don t measure c) Energy consumption is unpredictable d) Production rate has no effect .................. End of Section I .................. Section II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40 (i) Answer all Eight questions (ii) Each question carries Five marks S1 A 10 HP rated induction motor, with nameplate details indicating 415V, 12 amps, and a power factor (PF) of 0.9, is being audited. During the audit, the monitoring equipment displays a reactive power of 2 kVAr and a power factor of 0.758. Calculate the percentage loading of the motor at the time of the test. Solution: PF = kW/KVA (KVA)2 = (kVAr)2 + (kW)2 Given kVAr = 2 and (1) (2) PF=0.758 Solve for kW in eqn (2) using eqn (1) we get, kW measured = 2.32kW Motor rated input kW= 1.732VIcos = 1.732*0.415*12*0.9=7.76 kW Percentage loading of motor= kW measured /rated input kW * 100 = 2.32/7.76 * 100 =29.88% S2 A renovation and modernization (R&M) program of a 1 MW coal-fired thermal power plant was carried out to enhance the operating efficiency from 28% to 32%. The specific coal consumption was 0.7 kg/kWh before R&M. For 7000 hours of operation per year and assuming coal quality remains the same, calculate a) The coal saving per year in tonnes b) The expected avoidance of CO2 into the atmosphere in Tons/year if the emission factor is 1.3 kg CO2/kg coal. Solution: a) Coal consumption per kWh with 32 % efficiency = 28 x 0.7 / 32 = 0.61 kg/kWh Saving in coal = (7000 x 1000 x (0.7 0.61) = 630000 kg b) Expected Avoidance = 63000 x 1.3 = 819 Tons/year S3 A drilling machine drawing continuously 5 kW of input power and with an efficiency of 50%, is used in drilling a bore in an aluminum block of 5 kg of mass. A portion of energy imparted to the block is lost to surroundings and the balance is absorbed by the block in its uniform heating. A 45 oC rise in temperature of the block was observed at the end of 100 seconds and the specific heat of aluminum block is 900 J/kgK. What percentage of drilling machine output power is lost to the surroundings? Solution: Power input to the drilling machine = 5kW Power output of drilling machine = 5 x 0.5 = 2.5 kW Energy used in drilling of bore, Q = 2.5x1000x100 = 250000 J Effective energy absorbed by block,Q = m x c p x (temp rise)= 5*900*45 = 202500J Percentage of energy utilized for heating = 202500/250000= 81% Percentage of energy lost to surroundings = 100-81= 19% S4 Calculate the investment of the project having IRR of 16% and having respective annual savings of Rs 15,000, Rs. 18,000 and Rs. 20,000 at the end of the first, second and third year. Solution: NPV = (15000/1.16)+18000/(1.16x1.16)+(20000/(1.16x1.16x1.16)) = 12931 + 13377 + 12813 = 39,121/S5 In a heat exchanger steam is used to heat 5 kL/ hour of furnace oil from 300C to 90o C. Specific heat of furnace oil is 0.22 kcal/ kg/OC and the specific gravity of furnace oil is 0.95. a) How much steam per hour is required, if steam used is having latent heat of 510 kcal/kg? b) If steam cost is Rs.3.40/kg and electrical energy cost is Rs.6/kWh, which type of heating would be more economical in this particular case? Solution: a) Total heat required = m Cp T = (5 x1000x 0.95) * 0.22 * (90-30) = 62,700 kcal/hr Latent heat of steam = 510 kcal/kg Amount of steam required = 62700/510 = 123 kg/hr b) Steam cost = 123 x Rs.3.40 = Rs.417.9/hr Amount of electricity required = 62700/860 = 72.9 kWh Cost of electricity = 72.9 x Rs. 6 = Rs.437.4/ hr Steam heating will be more economical S6 a) Lower energy intensity of a country need not necessarily mean higher energy efficiency. Explain? b) Why is energy intensity expressed taking into account purchase power parity? Solution: a) Page No 17 b) Page No 18 S7 a) Differentiate between commercial and non-commercial energy with an example for each. b) Differentiate between renewable and non-renewable energy with an example for each. Solution: a) Page No 2 b) Page No 3 S8 a) Write down the parameters, which can be measured by the following instruments. Stroboscope Sling Psychrometer Fyrite Pitot Tube b) An electric resistive heater consumes 3.6 MJ when connected to a 200 V supply for one hour. Find the rating of the heater and the current drawn from the supply. Solution: a) Stroboscope Non Contact Speed measurement Sling Psychrometer Dry and wet bulb temperature Fyrite To measure O2 and CO2 Pitot Tube To measure pressure in gas ducts b) Energy=power x time Power=Energy / time = 3.6 x 106 J/(60 x 60) = 1kW Current = Power / Voltage = 1000 W/200V = 5 Ampere .................. End of Section II .................. Section III: LONG DESCRIPTIVE QUESTIONS (i) (ii) Answer all Six questions Each question carries Ten marks Marks: 6 x 10 = 60 L1 Using the details given below, construct the CUSUM table and calculate the annual savings in MTOE, considering 10,000 kcal/kg of fuel. Energy, MkCal Y X 7.69 Y = 0.176 X Production, T Solution: Month Electrical Power in kWh Apr 90981 May 94993 Jun 88010 Jul 85374 Aug 88741 Sep 88450 Oct 90780 Nov 82216 Dec 90612 Jan 85672 Feb 74939 Mar 83823 Production (T) 493 335 297 493 381 479 585 440 318 234 239 239 Month Electrical Power in kwhr Total Energy in mkcal (y) Eact Apr 90981 78.24 May 94993 81.69 Jun 88010 75.69 Jul 85374 73.42 Aug 88741 76.32 Sep 88450 76.07 Oct 90780 78.07 Nov 82216 70.71 Dec 90612 77.93 Jan 85672 73.68 Feb 74939 64.45 Mar 83823 72.09 Ecal (0.176 X Production +7.69=Y) (x) in mkcalBase Line 493 335 297 493 381 479 585 440 318 234 239 239 Eact-Ecal in mkcal Cusum in mkcal 94 -16 -16 67 15 -1 60 16 15 94 -21 -6 75 2 -5 92 -16 -21 111 -33 -53 85 -14 -68 64 14 -54 49 25 -29 50 15 -14 50 22 8 Savings in MTOe = 8 x 10^6 / 10^7 = 0.8 MTOe L2 You are evaluating a multi-phase investment project with the following cash flows in over a 5-year period. The project includes an initial investment of Rs.20 Lakhs, an additional investment of Rs.5 Lakhs in Year 3, and a salvage value of Rs.3 Lakhs at the end of Year 5. The yearly savings are given below: Year Cash Flow (Rs. Lakhs) 1 6 2 7 3 4 4 9 5 12 Calculate the Internal Rate of Return (IRR) for the project. Solution: The cash flows for each year are as follows: Year Cash Flow (Rs. Lakhs) 0 -20 1 6 2 7 3 -1 (4-5) 4 9 5 15 (12+3) The IRR is the discount rate that makes the Net Present Value (NPV) of these cash flows equal to zero. Let's start with, say r=15% or 0.15. By interpolation Method NPV at 20% = -0.351 NPV at 19% = 0.172 IRR = Lower rate + NPV at lower rate x (Higher rate-lower rate) (NPV at lower rate NPV at higher rate) IRR = 19 + 0.172 x (20-19) (0.172 (-0.351) IRR = 19 + 0.172 0.523 IRR = 19 + 0.32 = 19.32% L3 a) A cement plant is planning for ISO 50001 certification. Write a goal, objective and target for meeting the requirements of energy management system b) For an energy efficiency project, define (i) net operating cash inflows (ii) Economic life (iii) Salvage value c) Compare between NPV and IRR Solution: a) Book 1, Page 157 b) Book 1, Page 173 c) Book 1, Page 172 L4 a) You are part of the team responsible for evaluating the total energy consumption of a manufacturing plant. This plant operates around the clock and has substantial heating and cooling needs due to its production processes. It sources energy from electricity purchased from the grid, furnace oil for thermic fluid heaters, coal for steam boilers, High-Speed Diesel for diesel generators, and Liquefied Petroleum Gas for ovens. To determine if the plant qualifies as a designated consumer under EC Act, list down the data required for assessing the MTOe. b) An energy manager in a factory has gathered following data to arrive at the plant energy performance. Reference year (2022) energy use was 20 million kcal and production factor (PF) for the current year (2023) is 0.9. While the current year s energy use is 19 million kcal. What is the plant energy performance of the factory for the year 2023? State the inference. Solution: a) Energy Source Description Unit of Measurement Electricity Purchased from grid kWh Furnace Oil Used for thermic fluid heater Liters Coal Used for steam boiler Metric tons HSD (High-Speed Diesel) Used for diesel generators Liters LPG (Liquefied Petroleum Gas) Used for ovens Kilograms b) Given: Reference year energy use (2022) = 20 million kcal Production factor (PF) for the current year (2023) = 0.9 Reference year energy equivalent = Reference year energy use Production factor =20 million kcal X 0.9 =18 million kcal To calculate the plant energy performance for the year 2023: Plant Energy Performance = Reference Year Equivalent Current Years Energy use Reference Year Equivalent = 18 19 18 100 = -5.56% 100 Inference: The plant energy performance has decreased by 5.56% in 2023 compared to the reference year energy use in 2022. This indicates that the factory's energy efficiency has worsened, as it consumed 5.56% more energy than expected based on the production factor. L5 a) Draw PERT chart for the following task, dependency and duration. 5 Marks b) Find the critical path 2 Marks c) Calculate expected project duration 3 Marks Task Predecessors Tasks (Dependencies) A B C D E F G H I J A B C E F D G-H Expected Time as Calculated (Weeks) 3 5 7 8 5 5 4 5 6 4 Solution: a) b) The critical path is through activities C, F, H, J c) The expected project duration is 21 weeks (7+5+5+4) L6 An evaporator is to be fed with 5000 kg/hr of a solution having 0.5 % solids. The feed is at 38 oC, and is to be concentrated to 1% solids. Steam is entering at a total enthalpy of 640 kcal/kg and the condensate leaves at 100 oC. Enthalpies of feed are 38.1 kCal/kg, product solution is 100.8 kCal/kg and that of the vapour is 640 kCal/kg. Find the mass of vapour formed per hour and the mass of steam used per hour. Solution: Mass of vapour Feed= 5000 kg/hr @ 0.5 solids Solids = 5000 x 0.5/100 = 25 kg/hr Massout x 1/100 = 25 Massout = 2500 kg/hr Vapour formed = 5000-2500 = 2500 kg/hr Thick liquor = 2500 kg/hr Steam Consumption Enthalpy of Feed = 5000 x 38.1 = 190500 kcal Enthalpy of the thick liquor = 100.8 x 2500 = 252000 kcal Enthalpy of vapour = 640 x 2500 = 16,00,000 kcal Heat Balance Heat input by steam + Heat in Feed = Heat out in vapour + Heat out in thick liquor [M x (640-100) + 38.1 x 5000] = 1600000 + 252000) M x 540 = 1661500 M= 3076.8 kg/hr ___________END_____________ PAPER-1 COLOR C0DE : PINK 25th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS- SEPTEMBER 2025 PAPER-1: GENERAL ASPECTS OF ENERGY MANAGEMENT & ENERGY AUDIT Date:27.09.2025 Timings:09:30-12:30HRS Duration: 3 HRS Max.Marks:150 General instructions: o Please check that this question paper contains 64 questions o The question paper is divided into three sections o All questions in all three sections are compulsory o All parts of a question should be answered at one place Section I: OBJECTIVE TYPE Marks:50x1=50 (i) Answer all 50 questions (ii) Each question carries one mark 1. Select the correct statement about the Critical Path Method (CPM): a) CPM is a deterministic model that does not take into account variation in completion time b) CPM is a probabilistic model that takes into account variation in completion time c) CPM is a probabilistic model that does not take into account variation in completion time d) CPM is a deterministic model that takes into account variation in completion time 2. Acceptable delay time (slack time/float) is equal to: a) Time between Earliest Finish and Latest Finish b) Time between Earliest Start and Latest Start c) Both a and b d) None of the above 3. The objectives of Standards & Labeling (S&L) programme aim to: a) Set sulphur standards for coal-fired power plants b) Provide informed choice about energy saving c) Enforce penalties on renewable obligation non-compliance d) Fix tariff slabs for power-intensive industries 4. Is the activity critical, given ES = 8 days and LS = 10 days? a) Yes b) No c) More details required d) Next activity details required 5. The relation between gauge pressure (pg), system pressure (ps), and atmospheric pressure (pa) is: a) pg = ps + pa b) pg = ps pa c) ps = pg pa d) pa = ps + pg 6. Availability Based Tariff (ABT) was introduced in India to: a) Encourage solar roof-top for industries b) Reduce dependence on oil imports c) Subsidise rural electrification d) Improve grid discipline and frequency control 7. In a Guaranteed Savings ESCO project, the ESCO company would not be involved in: a) Project design b) Project finance c) Project implementation d) Verifying energy savings 8. The Reserves-to-Production (R/P) ratio of coal in India is high compared to oil and gas. This implies: a) Coal reserves can provide secure supply for decades b) India has surplus oil reserves to meet its demand c) Natural gas is India s most secure long-term option d) India s coal imports will vanish completely 9. The use of Purchasing Power Parities (PPPs) in energy intensity calculations ensures that: a) GDP comparisons reflect only exchange rate fluctuations BUREAUOFENERGYEFFICIENCY 1 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. PAPER-1 COLOR C0DE : PINK b) GDP of all countries is valued at a uniform price level, showing only differences in real economic volume c) GDP is measured exclusively in domestic currency terms d) GDP comparisons ignore differences in goods and services consumed Which statement best describes the relationship between energy conservation and energy efficiency?: a) Energy conservation and energy efficiency are identical and interchangeable terms b) Energy efficiency refers to reducing energy intensity per unit of output, while energy conservation refers to reducing overall consumption. c) Energy efficiency requires lowering comfort levels, while energy conservation does not d) Energy conservation excludes energy efficiency measures from its scope Designated Consumers under EC Act are classified mainly because: a) They are exempted from energy audits b) They focus only on renewable generation c) They represent small artisan industries d) They are users of energy in an energy intensive industry Sankey diagrams help energy managers by: a) Prioritizing improvements based on visualized energy losses b) Reducing the need for energy audits c) Replacing thermodynamic calculations d) Eliminating the use of performance indicators Which instrument measures power factor directly? a) Ammeter b) Wattmeter c) Lux meter d) Power analyzer What is the mission of the Bureau of Energy Efficiency (BEE) under the Energy Conservation Act 2001? a) To regulate electricity tariffs at the national level b) To promote renewable energy by providing capital subsidies c) To develop policies and strategies that reduce the energy intensity of the Indian economy d) To license only energy auditors and energy managers What is the main purpose of the Energy Conservation Building Code (ECBC)? a) To set minimum energy efficiency standards for commercial buildings b) To fix electricity tariffs for buildings c) To mandate use of only renewable energy in construction d) To regulate real estate prices Daylight harvesting in lighting systems means: a) Collecting solar energy for night lighting b) Using flat plate collectors for heating c) Adjusting artificial lighting based on natural daylight d) Storing energy in battery banks In a cumulative sum chart, a horizontal graph indicates: a) Nothing can be said b) Energy consumption is reduced c) Specific energy consumption is increasing d) Actual and calculated energy consumption are the same Which of the following is non-commercial energy? a) Lignite b) LPG c) Solar energy for water heating d) Hydro power Power rating of an electrical heater consuming 12,000 J/min is: a) 12 W b) 100 W c) 200 W d) 12,000 W Moles of water in 54 grams: a) 3 b) 4 c) 5 d) 6 BUREAUOFENERGYEFFICIENCY 2 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. PAPER-1 COLOR C0DE : PINK The main purpose of Performance Measurement and Verification (PMV) is to: a) Establish new project costs b) Ensure that guaranteed savings have been achieved c) Increase the baseline consumption d) Eliminate the need for utility bills ROI for an investment of Rs.1,00,000 with an annual return of Rs 20,000 per year is_______ a) 1% b) 10% c) 20% d) 200% Biomass gasifier using 1 kg wood (4,000 kCal/kg) producing 2 m gas (1,000 kCal/m ). What would be the efficiency? a) 25% b) 50% c) 75% d) 100% Mean molecular weight of air (77% N2, 23% O2 by weight) is ___________grams. a) 26.8 b) 27.8 c) 28.8 d) 29.8 Energy saving through DSM is treated as equivalent to: a) A reduction in electricity tariff b) New additions on the supply side in MWs c) Import of cheaper electricity d) Government subsidies What is the main aim of the Accelerated Power Development and Reform Programme (APDRP)? a) To eliminate subsidies for agricultural consumers b) To privatize all power plants in India c) To promote only renewable energy in the power sector d) To cut AT&C losses by audits and system improvements Resistance of 250 V incandescent lamp drawing 0.5 A: a) 5,000 b) 500 c) 50 d) 5 A process receives 1000 kg/hr of raw material. The hourly outputs are 700 kg of product, 200 kg of waste, and 50 kg stored. What is the unaccounted loss? a) 100 kg/hr b) 150 kg/hr c) 200 kg/hr d) 50 kg/hr A boiler receives 100 MJ of fuel energy. The steam output is 70 MJ, the flue gas loss is 20 MJ and the radiation plus unaccounted loss is 10 MJ. What is the boiler efficiency? a) 65% b) 60% c) 70% d) 75% 1 tonne of oil equivalent =: a) 41,868 MJ b) 1,000 kcal c) 1,000 kWh d) 1,000 BTU Maximum specific heat among the following: a) Water b) Lead c) Mercury d) Iron Ozone depletion is mainly due to: a) Oxygen b) Methane c) Chlorofluorocarbons d) Carbon dioxide The first step in an energy action plan is: a) Recognition of achievements BUREAUOFENERGYEFFICIENCY 3 PAPER-1 COLOR C0DE : PINK 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. b) Designing monitoring reports c) Selecting new technologies d) Top management commitment Heat required for Cooling 2000 kg of water for T of 10 C ____________ a) 2,000 kcal b) 20,000 kcal c) 200 kcal d) 2 10 kcal Calculate the quantity of water evaporated when 100 kg of feed containing 6% solids is concentrated to 30% solids. a) 600 kg b) 180 kg c) 80 kg d) 800 kg In force field analysis, which approach is usually more effective for achieving a goal? a) Strengthening forces that are already positive b) Minimising negative forces that act as barriers c) Ignoring external factors and focusing only on internal ones d) Changing the organisational goal Which of the following is NOT a conventional financing option? a) Debt financing b) Performance contracting c) Retained earnings d) Stock buyback Two projects: X (IRR=40%, NPV= `50,000/-) and Y (IRR=30%, NPV= `1,20,000/-) having same life, no finance limit. Choose the best project. a) X b) Y c) Cannot decide d) Question invalid Term for asset value decrease over time: a) Discounting b) Inflation c) Depreciation d) Compounding In Total Productive Maintenance (TPM), which of the following is not one of the six big losses that lower equipment efficiency? a) Breakdowns b) Idling and minor stoppages c) Reduced speed d) Excessive overtime hours Fixed energy consumption can be determined from: a) Bar chart b) Vertical line chart c) Pie chart d) XY coordinate system Carbon capture from point sources and storage is called: a) Carbon sequestration b) Carbon sink c) Carbon capture d) Carbon adsorption Why is an energy baseline established in Monitoring and Targeting (M&T)? a) To record only monthly electricity bills b) To fix a reference point for measuring energy performance improvements c) To eliminate the need for energy performance indicators d) To avoid sharing information with managers and stakeholders Life-cycle costing is better than simple purchase cost because it: a) Includes operation, maintenance and energy costs over life b) Ignores maintenance costs c) Forces single-supplier bidding d) Cuts down procurement cycle time BUREAUOFENERGYEFFICIENCY 4 PAPER-1 COLOR C0DE : PINK 45. Producer gas consists of: a) CO, H , CH b) CO, CH c) CO, H d) Only CH 46. Work Breakdown Structure (WBS) is mainly used for: a) Combining small tasks into one large project b) Dividing complex projects into simpler, manageable tasks c) Preparing cost estimation only d) Eliminating tasks from the project 47. What is a major limitation of the Gantt chart in project management? a) It does not show the duration of activities b) It does not clearly show logical dependencies between activities c) It cannot be used for construction projects d) It requires advanced statistical methods for preparation 48. In a project network diagram, why is a dummy activity used? a) To represent an activity with very small duration b) To show logical dependency between activities with the same start and end nodes c) To reduce the total project duration d) To allocate additional resources to critical activities 49. Solar radiation consists of: a) X-rays, Gamma rays, and Microwaves b) Ultra-violet, Visible, and Infra-red radiation c) Visible, Infra-red, and Radio waves d) Ultra-violet, X-rays, and Cosmic rays 50. Which of the following are basic objectives of sustainable development? a) Economic security and prosperity b) Social development and advancement c) Environmental sustainability d) All of the above ..................End of Section I.................. Section II: SHORT DESCRIPTIVE QUESTIONS (i) (ii) Marks:8x5=40 Answer all Eight questions Each question carries Five marks S-1 The facility has a connected load of 500 kW and currently has a contract demand of 500 kVA. The monthly maximum demand recorded is consistently around 350 kW at 0.85 power factor. The utility imposes a penalty of 350 per excess kVA/month, if recorded demand exceeds contract demand. The demand charge is 300 per kVA/month. a) Determine current demand in kVA. 1 Mark b) The minimum billing demand is 80% of contract demand. Calculate excess demand charges paid above minimum billing demand per month. 2 Marks c) Calculate minimum power factor required to avoid payment of excess demand charges a) over minimum billing demand. 2 Marks S-1 a)Calculate actual kVA demand Ans Actual Demand (kVA)=actual kW/Power Factor=350/0.85 =411.76 kVA b)Excess demand Charges per month Minimum billed demand = 500 0.8 = 400 kVA Excess demand Charges paid= (411.76-400) *300 = Rs.3528.00/month c)Minimum Power Factor Improvement Required To avoid excess demand charges = 350/400 = 0.875 P1P S-2 A food processing unit uses the following per day: LPG consumption: 200 kg/day (Calorific Value = 11,000 kcal/kg, rate 90/kg) DG backup: 100 kWh/day when the grid fails, using diesel at 95/litre with a specific fuel consumption of 260 ml/kWh. (Calorific value = 10,000 kcal/litre) Electrical energy: 1,200 kWh/day at 7.5/kWh BUREAUOFENERGYEFFICIENCY 5 PAPER-1 COLOR C0DE : PINK Each 1 Mark a) Convert the LPG energy to kWh equivalent. b) Calculate the thermal energy input (in kcal) required by the DG to produce 100 kWh. c) Calculate the daily energy cost from all 3 sources. d) Calculate the percentage contribution of each energy source to the total energy input (in kWh equivalent). e) Determine the cost share of each energy source in the total energy cost and identify the most economic source among grid power, LPG and DG power. S-2 a) LPG to kWh Thermal energy = 200 11,000 = 2,200,000 kcal/day Ans Convert to kWh: 2,200,000/860= 2,558.14 kWh b) Diesel Energy Input for DG DG Output = 100 kWh Specific fuel consumption = 260 ml/kWh Input =100*260 /1000 = 26 litres Convert to kcal: 26 10000 = 260,000 kcal /day c)Daily Energy Cost Electricity = 1,200 7.5 = 9,000 LPG = 90/kg 200 = 18,000 Diesel = 26 95 = 2,470 Total Cost= 9,000+ 18,000+ 2,470= 29,470/day d)Percentage contribution of each energy source (in kWh) Source kWh equivalent % Share kWh Grid power LPG DG output Total 1,200 2,558.14 100 3,858.14 kWh =1200/3858.14 *100 = 31.1% =2558.14/3858.14 *100 = 66.3% =100/3858.14 *100 = 2.6% 100% e) Percentage cost share of each energy source Source Grid power LPG DG output Total Daily energy Cost 9,000 18,000 2,470 29,470/day % Share Cost =9000/29470 *100 = 30.53% =18000/29470 *100 = 61.07% =2470/29470 *100 = 8.4% 100% LPG is more economic. S-3 An energy audit conducted in a rubber processing unit identifies the following: A centrifugal pump (motor rating 30 kW) runs continuously for 16 hours/day, 300 days/year. Measured motor loading = 65%, Motor efficiency = 88%, with no flow control. A VFD retrofit is proposed, which is expected to reduce energy consumption by 10% due to optimized flow control. Power cost = 7.0/kWh VFD installation cost = 1,50,000 a) b) c) d) Calculate the current annual energy consumption of the motor. Estimate the expected annual energy savings from the VFD. Calculate the annual cost saving in . Determine the simple payback period for the investment. BUREAUOFENERGYEFFICIENCY 2 Marks 1 Mark 1 Mark 1 Mark 6 Item Annual energy use Energy saved Annual savings Payback period S-3 Ans = = = = Calculation (30x 0.65 /0.88) x16x300 106,364 * 0.1 10,636.4 x7 1,50,000/74454.8 PAPER-1 COLOR C0DE : PINK Value 106,364 kWh 10,636.4 kWh 74,454.8 2.01 years (~24.17 months) A food dryer processes 1,000 kg/hr of wet material with an initial moisture content of 55% S-4 (wet basis) and dries it to a final moisture content of 10% (wet basis). a) b) c) d) e) Steam Flow: 2,500 kg/hr at 3.5 bar (enthalpy = 660 kcal/kg) Latent heat of water vaporization = 540 kcal/kg Specific heat of dry material = 0.45 kcal/kg C Drying temperature rise = 60 C Ignore heat loss and assume 100% steam use for moisture removal and solid heating Each 1 Mark Calculate the mass of bone-dry solid in the feed Calculate the mass of water removed per hour Estimate the energy required to evaporate the moisture Estimate the energy required to heat the dry solids Calculate the total energy input from steam a) Mass of bone-dry solid in the feed S-4 Ans Moisture content (wet basis) = 55% Dry matter fraction = 1 0.55 = 0.45 Dry solid mass = 1000 0.45 = 450 kg/hr b) Mass of water removed per hour Initial water = 1000 450 = 550 kg/hr Final moisture content = 10% (wet basis) Let final product mass = M. Moisture = 0.10M, Dry solid = 0.90M. Dry solid remains constant at 450 kg. 0.90M = 450 M = 500 kg/hr Final water = 500 450 = 50 kg/hr Water removed = 550 50 = 500 kg/hr c) Energy required to evaporate the moisture Qevap = mass of water removed latent heat of vaporization Qevap = 500 540 = 270,000 kcal/hr d) Energy required to heat the dry solids Qsolid = mass of dry solid specific heat temperature rise Qsolid = 450 0.45 60 = 12,150 kcal/hr e) Total energy input from steam Qtotal = Qevap + Qsolid Qtotal = 270,000 + 12,150 = 282,150 kcal/hr Energy available from steam = mass of steam enthalpy = 2500 660 = 1,650,000 kcal/hr Total energy demand = 282,150 kcal/hr Total energy supply = 1,650,000 kcal/hr A medium-sized factory installs an energy-efficient air compressor system costing 6,00,000. S-5 An audit estimates that it will save 1,80,000 per year in energy bills for the next 3 years. Annual maintenance is expected to cost 10,000 starting from second year onward. Assume: Discount rate (cost of capital) is 10% and salvage value at the end of third year is 50,000. BUREAUOFENERGYEFFICIENCY 7 PAPER-1 COLOR C0DE : PINK a) Calculate the net annual cash flow from Year 2 onward b) Compute the Net Present Value (NPV) of the investment c) Based on NPV, assess whether the project is economically acceptable 2 Marks 2 Marks 1 Mark S-5 a) Net Annual Cash Flow from Year 2 Onward Ans Annual saving = 1,80,000 Annual maintenance (from Year 2) = 10,000 Net = 1,80,000 10,000 = 1,70,000 Therefore, the net annual cash flow from Year 2 onward is 1,70,000. b) Cash Flow Table with Discounting (10% discount rate) Year Savings Maintenance Net Cash Salvage ( ) ( ) Flow ( ) Value ( ) 0 - - - - 1 2 3 1,80,000 1,80,000 1,80,000 0 10,000 10,000 1,80,000 1,70,000 1,70,000 0 0 50,000 Total Cash Flow ( ) 6,00,000 1,80,000 1,70,000 2,20,000 PV Factor @10% 1.000 Present Value ( ) 0.909 0.826 0.751 1,63,636 1,40,420 1,65,220 -6,00,000 Total PV of inflows = 1,63,636 + 1,40,420 + 1,65,220 = 4,69,276 NPV = 4,69,276 6,00,000 = 1,30,724 c) Economic Acceptability The NPV is negative ( 1,30,724), meaning the project will not recover its investment cost within 3 years at a 10% discount rate. Therefore, the project is not economically acceptable under these conditions. S-6 A continuous centrifuge separates 36,000 kg of whole milk containing 4% fat in 6-hour period into skim milk with 0.40% fat and cream with 40 % fat. Find out the flow rates of whole milk, cream and skim milk using mass balance. S-6 Mass inlet: Ans Total mass flow of whole milk = 36000/6 = 6000 kg per hour. Fat per hour = 6000 x 0.04 = 240 kg/hr. Therefore, Water plus solids other than fat = (6000-240) kg per hr. = 5760 kg per hr. Mass outlet: Let the mass of cream be X kg then its total fat content is 0.40X. The mass of skim milk is (6000- X) and its total fat content is 0.0040 (6000 - X) Material balance on fat: Fat in = Fat out 6000 x 0.04 = 0.0040(6000 - X) + 0.40X; solving this, X = 545 kg/hr So that the flow of cream is 545 kg/hr and skim milk (6000- 545) = 5455 kg/hr. a) Define energy intensity. Explain what low and high energy intensity indicate about a S-7 country s economy. 3 Marks b) Country A consumes 2000 toe of energy and has a GDP of US$ 100 million. Country B consumes 2500 toe of energy and has a GDP of US$ 140 million. Calculate the Energy Intensity of both countries in toe per US$ million GDP and indicate which country is more efficient in its use of energy? 2 Marks a) Refer guide book S-7 b)Energy Intensity (EI) = Energy Consumption (toe) / GDP (million US $) Ans EI for Country A = 2000 / 100 = 20 EI for Country B = 2500 / 140 = 17.85 Country B consumes less energy per $. This, it is better in terms of energy efficiency. What are the benefits of the Critical Path Method (CPM)? Also explain how the Program S-8 Evaluation and Review Technique (PERT) differs from CPM. BUREAUOFENERGYEFFICIENCY 8 PAPER-1 COLOR C0DE : PINK S-8 Ans Refer guide book ..................End of Section II.................. Section III: LONG DESCRIPTIVE QUESTIONS (i) (ii) Marks:6x10=60 Answer all Six questions Each question carries Ten marks L-1 In a chemical company, variable consumption was measured 2.2 times of the production, and the non-production consumption (fixed energy consumption) was observed 10000 kWh/Month. A company has implemented several energy saving initiatives during the previous financial year. a) Calculate energy saving by preparing a CUSUM chart. The actual production and energy consumption observed during the current financial year for the first two quarters is as follows: (8 Marks) Month Production (kg) Actual Energy Consumption (Kwh) April 75000 170000 May 78000 172000 June 85000 185000 July 72000 155000 Aug 71000 153000 Sept 76000 163000 b)Also mention four names of different financing options for industry. L-1 Ans --- 2 Marks Month Production (kg) Actual Energy Consumption ,kWh (EA) Predicted Energy consumption, kWh (2.2P+10000 ) (EP) Ea-Ep CUSUM April 75000 170000 175000 -5000 -5000 May 78000 172000 181600 -9600 -14600 June 85000 185000 197000 -12000 -26600 July 72000 155000 168400 -13400 -40000 Aug 71000 153000 166200 -13200 -53200 Sept 76000 163000 177200 -14200 -67400 Cumulative saving in six month is 67400 kWh by implementing the energy saving measures. b) Different financing options with the organization i) Debt financing ii) Equity financing iii) Retained earning iv) Capital lease v) True Lease BUREAUOFENERGYEFFICIENCY 9 PAPER-1 COLOR C0DE : PINK vi) Performance contract L-2 An industry is exploring two project development options as part of its pursuing energy efficiency strategy. Using the NPV concept, find out the better option. Consider 10% as the discount rate and 5 year as the project life. Description Capital cost Year 1 2 3 4 5 Project A 80,000 Net Annual Savings ( ) + 25,000 + 25,000 + 25,000 + 25,000 + 25,000 Project B 100,000 Net Annual Savings ( ) + 35,000 + 35,000 + 35,000 + 35,000 + 35,000 0 1 2 3 L-2 NPV = - CF0/(1+r) + CF1/(1+r) + CF2/(1+r) + CF3/(1+r) + . Ans Project A NPV = - 80,000/ (1+0.10)0 + 25,000/ (1+0.10)1 +25,000/ (1+0.10)2 + 25,000/ (1+0.10)3 + 25,000/ (1+0.10)4 + 25,000/ (1+0.10)5 = -80,000 + 22,727 + 20,661 + 18,783 + 17,075 + 15,522 = Rs 14,768 Project B NPV = - 100,000/ (1+0.10)0 + 35,000/ (1+0.10)1 +35,000/ (1+0.10)2 + 35,000/ (1+0.10)3 + 35,000/ (1+0.10)4 + 35,000/ (1+0.10)5 = -100,000 + 31,818 + 28,926 + 26,296 + 23,905 + 21,732 = Rs 32,677 Project B shall be preferable due to higher NPV. L-3 In a chemical company, Natural Gas (NG) is being used to heat 15 kl/hr of water by 15 C. The company is planning to switch this heating process to steam, which is available from neighbouring industries. a) Work out the feasibility of this option, considering annual operating hours of 6000 hrs. The effective heat of NG is 8500 kCal/m , the NG rate is 55/m , and the density of NG is 0.717 kg/m . The latent heat of steam is 540 kCal/kg, and the steam rate is 2.2/kg. 6 Marks b) Also, calculate the tonnes of CO2 emission for both the options, if 0.2 kg of CO2 is emitted per kg of steam consumed and the percentage of carbon in NG is 74%. 4 Marks a)Total heat requirement = 15000X15 = 225000 Kcal/Hr L-3 Current Option : NG Consumption = 225000/8500=26.47 m3/Hr Ans NG Consumption = 26.47 X 6000 = 158820 m3/Year Energy Cost = 158820 X 55/100000=87.35 Lakh/Year Alternative Option : Steam consumption = 225000/540=416.67 Kg/Hr Steam Consumption = 416.67x 6000=2500020 Kg/Year Energy Cost = 2500020 x 2.2/100000=55 Lakh/Year It is suggested to go with an alternative option considering saving of INR 32.35 Lakh per year. b) CO2 emission for using steam = 2500020 X 0.2/1000 =500 Tonne of CO2 emission per annum CO2 emission for using NG = (158820 x 0.717 x 0.74 x (44/12))/1000 = 309 Tonne of CO2 emission per annum BUREAUOFENERGYEFFICIENCY 10 PAPER-1 COLOR C0DE : PINK 4 Marks L-4 (a) Construct a CPM diagram for the activities below: Activity Precedence Duration in weeks A Start 3 B A 4 C B 1 D C 3 E Start 2 F B 1 Finish D, E, F -- (b) Compute the earliest start, earliest finish, latest start & latest finish of all activities. 3 Marks (c) Identify the critical path and its duration. 3 Marks L-4 a) Network diagram (4 marks) Ans b) Early start (ES), Early Finish (EF), Latest start (LS), Latest finish (LF)-3 marks S.no Activity Duration ES EF LS LF 1 A 3 0 3 0 3 2 B 4 3 7 3 7 3 C 1 7 8 7 8 4 D 3 8 11 8 11 5 E 2 0 2 9 11 6 F 1 7 8 10 11 c)Critical Path (3 marks) Total time on critical path: 11 weeks BUREAUOFENERGYEFFICIENCY 11 PAPER-1 COLOR C0DE : PINK L-5 A foundry operates an induction furnace with a capacity of 5 t/hr, having a specific electrical energy consumption of 620 kWh/t of liquid metal produced. The overall casting yield of the foundry is 60%. After melting and casting, the products are heat treated in an oil-fired furnace which consumes 75 kg of fuel oil per tonne of castings. The gross calorific value of fuel oil is 10,000 kCal/kg. Additional information provided includes auxiliary connected electrical load of 50 kW, transformer efficiency of 98%, oil density of 0.88 kg/liter and average oil price of 80 per kg. Average castings produced is 45 tonnes per day and the plant is in continuous operation. Using the above data, a) Calculate the total energy consumption per tonne of finished product, expressed as oil equivalent (kg of oil per tonne of finished casting). 7 Marks b) The foundry is receiving additional order to produce 30 tonnes of casting per day. Assess whether the plant can handle additional demand. 3 Marks L-5 a) Specific electric energy = 620 kWh per tonne of liquid metal. Casting Yield = 60% Ans Therefore, to produce 1 tonne of finished castings Luid metal required=1/0.60=1.67 Energy (electrical) per tonne of finished product: 620 kWh/t (liquid /0.6 = 1033.3 kWh/ t (finished) Equivalent energy in kCal = 1033 x 860 = 8,88,380 kCal/T Equivalent energy of auxiliary loads = (50/45)x24x860 = 22,933 kCal/T (Assuming auxiliaries operating at full load as not mentioned in the question) Total Equivalent electrical energy consumption = 8,88,380 + 22, 933 = 9,11,313 kCal/T Total Equivalent electrical energy with transformer losses = 9,11,313 /0.98 kCal/T = 9,29,911 kCal/T Heat required for heat treatment furnace = 75 x 10,000 = 7,50,000 Kcal/T Total kCals required per tonne of finished casting = 9,29,911 + 7,50,000 = 16,79,911 Kcals/T In terms of oil equivalent = 16,79,911 / 10,000 = 167.99 kg of oil/T of finished product b) Actual liquid metal production by induction furnace = (45/24)/0.6 = 3.125 TPH Addition castings production required Additional Liquid metal production required = 30/24 =1.25 TPH = 1.25/0.6 = 2.083 TPH Total Liquid metal production = 3.125 + 2.083 = 5.208 TPH, which is beyond the induction furnace capacity, therefore the plant will be unable to take-up additional production requirement. L-6 Answer the following questions: Each 2 Marks a) Briefly explain the working principle of a solar PV system. b) If a 1 kW PV system in Chennai operates at an average of 5 peak sun hours/day with 15% efficiency, calculate the daily energy output. c) List the factors that affect the performance of a wind turbine. d) A wind turbine with rotor area 200 m is installed in an area with average wind speed of 8 m/s. If air density is 1.2 kg/m , calculate the wind power available in the air stream. e) Briefly explain how biomass is used for electricity generation. L-6 a) Solar PV works on the principle of the photovoltaic effect in which photons create BUREAUOFENERGYEFFICIENCY 12 PAPER-1 COLOR C0DE : PINK Ans electron-hole pairs at a p n junction, generating DC electricity (or) Refer Guidebook b) Energy Output = 1 kW 5 h = 5 kWh/day. c) Factors: wind speed, air density, rotor area, blade design, hub height, and site conditions. (or) Refer Guidebook d) Power = 0.5 1.2 200 (8 ) = 61,440 W 61.44 kW. e) Biomass is combusted/gasified to generate heat or syngas, which drives turbines/engines to produce power (or) Refer Guidebook .................End of Section III.................. BUREAUOFENERGYEFFICIENCY 13 PAPER-1 COLOR C0DE : GREEN 25th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS- SEPTEMBER 2025 PAPER-1: GENERAL ASPECTS OF ENERGY MANAGEMENT & ENERGY AUDIT Date:27.09.2025 Timings:09:30-12:30HRS Duration: 3 HRS Max.Marks:150 General instructions: o Please check that this question paper contains 64 questions o The question paper is divided into three sections o All questions in all three sections are compulsory o All parts of a question should be answered at one place Section I: OBJECTIVE TYPE Marks:50x1=50 (i) Answer all 50 questions (ii) Each question carries one mark 1. Energy saving through DSM is treated as equivalent to: a) A reduction in electricity tariff b) New additions on the supply side in MWs c) Import of cheaper electricity d) Government subsidies 2. What is the main aim of the Accelerated Power Development and Reform Programme (APDRP)? a) To eliminate subsidies for agricultural consumers b) To privatize all power plants in India c) To promote only renewable energy in the power sector d) To cut AT&C losses by audits and system improvements 3. Resistance of 250 V incandescent lamp drawing 0.5 A: a) 5,000 b) 500 c) 50 d) 5 4. A process receives 1000 kg/hr of raw material. The hourly outputs are 700 kg of product, 200 kg of waste, and 50 kg stored. What is the unaccounted loss? a) 100 kg/hr b) 150 kg/hr c) 200 kg/hr d) 50 kg/hr 5. A boiler receives 100 MJ of fuel energy. The steam output is 70 MJ, the flue gas loss is 20 MJ and the radiation plus unaccounted loss is 10 MJ. What is the boiler efficiency? a) 65% b) 60% c) 70% d) 75% 6. 1 tonne of oil equivalent =: a) 41,868 MJ b) 1,000 kcal c) 1,000 kWh d) 1,000 BTU 7. Maximum specific heat among the following: a) Water b) Lead c) Mercury d) Iron 8. Ozone depletion is mainly due to: a) Oxygen b) Methane c) Chlorofluorocarbons d) Carbon dioxide 9. The first step in an energy action plan is: a) Recognition of achievements b) Designing monitoring reports c) Selecting new technologies d) Top management commitment 10. Heat required for Cooling 2000 kg of water for T of 10 C ____________ a) 2,000 kcal BUREAUOFENERGYEFFICIENCY 1 PAPER-1 COLOR C0DE : GREEN 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. b) 20,000 kcal c) 200 kcal d) 2 10 kcal Calculate the quantity of water evaporated when 100 kg of feed containing 6% solids is concentrated to 30% solids. a) 600 kg b) 180 kg c) 80 kg d) 800 kg In force field analysis, which approach is usually more effective for achieving a goal? a) Strengthening forces that are already positive b) Minimising negative forces that act as barriers c) Ignoring external factors and focusing only on internal ones d) Changing the organisational goal Which of the following is NOT a conventional financing option? a) Debt financing b) Performance contracting c) Retained earnings d) Stock buyback Two projects: X (IRR=40%, NPV= Rs 50,000/-) and Y (IRR=30%, NPV= Rs 1,20,000/-) having same life, no finance limit. Choose the best project. a) X b) Y c) Cannot decide d) Question invalid Term for asset value decrease over time: a) Discounting b) Inflation c) Depreciation d) Compounding In Total Productive Maintenance (TPM), which of the following is not one of the six big losses that lower equipment efficiency? a) Breakdowns b) Idling and minor stoppages c) Reduced speed d) Excessive overtime hours Fixed energy consumption can be determined from: a) Bar chart b) Vertical line chart c) Pie chart d) XY coordinate system Carbon capture from point sources and storage is called: a) Carbon sequestration b) Carbon sink c) Carbon capture d) Carbon adsorption Why is an energy baseline established in Monitoring and Targeting (M&T)? a) To record only monthly electricity bills b) To fix a reference point for measuring energy performance improvements c) To eliminate the need for energy performance indicators d) To avoid sharing information with managers and stakeholders Life-cycle costing is better than simple purchase cost because it: a) Includes operation, maintenance and energy costs over life b) Ignores maintenance costs c) Forces single-supplier bidding d) Cuts down procurement cycle time Producer gas consists of: a) CO, H , CH b) CO, CH c) CO, H BUREAUOFENERGYEFFICIENCY 2 PAPER-1 COLOR C0DE : GREEN d) Only CH 22. Work Breakdown Structure (WBS) is mainly used for: a) Combining small tasks into one large project b) Dividing complex projects into simpler, manageable tasks c) Preparing cost estimation only d) Eliminating tasks from the project 23. What is a major limitation of the Gantt chart in project management? a) It does not show the duration of activities b) It does not clearly show logical dependencies between activities c) It cannot be used for construction projects d) It requires advanced statistical methods for preparation 24. In a project network diagram, why is a dummy activity used? a) To represent an activity with very small duration b) To show logical dependency between activities with the same start and end nodes c) To reduce the total project duration d) To allocate additional resources to critical activities 25. Solar radiation consists of: a) X-rays, Gamma rays, and Microwaves b) Ultra-violet, Visible, and Infra-red radiation c) Visible, Infra-red, and Radio waves d) Ultra-violet, X-rays, and Cosmic rays 26. Which of the following are basic objectives of sustainable development? a) Economic security and prosperity b) Social development and advancement c) Environmental sustainability d) All of the above 27. Select the correct statement about the Critical Path Method (CPM): a) CPM is a deterministic model that does not take into account variation in completion time b) CPM is a probabilistic model that takes into account variation in completion time c) CPM is a probabilistic model that does not take into account variation in completion time d) CPM is a deterministic model that takes into account variation in completion time 28. Acceptable delay time (slack time/float) is equal to: a) Time between Earliest Finish and Latest Finish b) Time between Earliest Start and Latest Start c) Both a and b d) None of the above 29. The objectives of Standards & Labeling (S&L) programme aim to: a) Set sulphur standards for coal-fired power plants b) Provide informed choice about energy saving c) Enforce penalties on renewable obligation non-compliance d) Fix tariff slabs for power-intensive industries 30. Is the activity critical, given ES = 8 days and LS = 10 days? a) Yes b) No c) More details required d) Next activity details required 31. The relation between gauge pressure (pg), system pressure (ps), and atmospheric pressure (pa) is: a) pg = ps + pa b) pg = ps pa c) ps = pg pa d) pa = ps + pg 32. Availability Based Tariff (ABT) was introduced in India to: a) Encourage solar roof-top for industries b) Reduce dependence on oil imports c) Subsidise rural electrification d) Improve grid discipline and frequency control 33. In a Guaranteed Savings ESCO project, the ESCO company would not be involved in: a) Project design b) Project finance BUREAUOFENERGYEFFICIENCY 3 PAPER-1 COLOR C0DE : GREEN c) Project implementation d) Verifying energy savings 34. The Reserves-to-Production (R/P) ratio of coal in India is high compared to oil and gas. This implies: a) Coal reserves can provide secure supply for decades b) India has surplus oil reserves to meet its demand c) Natural gas is India s most secure long-term option d) India s coal imports will vanish completely 35. The use of Purchasing Power Parities (PPPs) in energy intensity calculations ensures that: a) GDP comparisons reflect only exchange rate fluctuations b) GDP of all countries is valued at a uniform price level, showing only differences in real economic volume c) GDP is measured exclusively in domestic currency terms d) GDP comparisons ignore differences in goods and services consumed 36. Which statement best describes the relationship between energy conservation and energy efficiency?: a) Energy conservation and energy efficiency are identical and interchangeable terms b) Energy efficiency refers to reducing energy intensity per unit of output, while energy conservation refers to reducing overall consumption. c) Energy efficiency requires lowering comfort levels, while energy conservation does not d) Energy conservation excludes energy efficiency measures from its scope 37. Designated Consumers under EC Act are classified mainly because: a) They are exempted from energy audits b) They focus only on renewable generation c) They represent small artisan industries d) They are users of energy in an energy intensive industry 38. Sankey diagrams help energy managers by: a) Prioritizing improvements based on visualized energy losses b) Reducing the need for energy audits c) Replacing thermodynamic calculations d) Eliminating the use of performance indicators 39. Which instrument measures power factor directly? a) Ammeter b) Wattmeter c) Lux meter d) Power analyzer 40. What is the mission of the Bureau of Energy Efficiency (BEE) under the Energy Conservation Act 2001? a) To regulate electricity tariffs at the national level b) To promote renewable energy by providing capital subsidies c) To develop policies and strategies that reduce the energy intensity of the Indian economy d) To license only energy auditors and energy managers 41. What is the main purpose of the Energy Conservation Building Code (ECBC)? a) To set minimum energy efficiency standards for commercial buildings b) To fix electricity tariffs for buildings c) To mandate use of only renewable energy in construction d) To regulate real estate prices 42. Daylight harvesting in lighting systems means: a) Collecting solar energy for night lighting b) Using flat plate collectors for heating c) Adjusting artificial lighting based on natural daylight d) Storing energy in battery banks 43. In a cumulative sum chart, a horizontal graph indicates: a) Nothing can be said b) Energy consumption is reduced c) Specific energy consumption is increasing d) Actual and calculated energy consumption are the same 44. Which of the following is non-commercial energy? a) Lignite b) LPG BUREAUOFENERGYEFFICIENCY 4 PAPER-1 COLOR C0DE : GREEN c) Solar energy for water heating d) Hydro power 45. Power rating of an electrical heater consuming 12,000 J/min is: a) 12 W b) 100 W c) 200 W d) 12,000 W 46. Moles of water in 54 grams: a) 3 b) 4 c) 5 d) 6 47. The main purpose of Performance Measurement and Verification (PMV) is to: a) Establish new project costs b) Ensure that guaranteed savings have been achieved c) Increase the baseline consumption d) Eliminate the need for utility bills 48. ROI for an investment of Rs 1,00,000 with an annual return of Rs 20,000 per year is_______ a) 1% b) 10% c) 20% d) 200% 49. Biomass gasifier using 1 kg wood (4,000 kCal/kg) producing 2 m gas (1,000 kCal/m ). What would be the efficiency? a) 25% b) 50% c) 75% d) 100% 50. Mean molecular weight of air (77% N2, 23% O2 by weight) is ___________grams. a) 26.8 b) 27.8 c) 28.8 d) 29.8 ..................End of Section I.................. Section II: SHORT DESCRIPTIVE QUESTIONS P1P (i) (ii) Marks:8x5=40 Answer all Eight questions Each question carries Five marks S-1 A continuous centrifuge separates 36,000 kg of whole milk containing 4% fat in 6-hour period into skim milk with 0.40% fat and cream with 40 % fat. Find out the flow rates of whole milk, cream and skim milk using mass balance. S-1 Mass inlet: Ans Total mass flow of whole milk = 36000/6 = 6000 kg per hour. Fat per hour = 6000 x 0.04 = 240 kg/hr. Therefore, Water plus solids other than fat = (6000-240) kg per hr. = 5760 kg per hr. Mass outlet: Let the mass of cream be X kg then its total fat content is 0.40X. The mass of skim milk is (6000- X) and its total fat content is 0.0040 (6000 - X) Material balance on fat: Fat in = Fat out 6000 x 0.04 = 0.0040(6000 - X) + 0.40X; solving this, X = 545 kg/hr So that the flow of cream is 545 kg/hr and skim milk (6000- 545) = 5455 kg/hr. a) Define energy intensity. Explain what low and high energy intensity indicate about a S-2 country s economy. 3 Marks b) Country A consumes 2000 toe of energy and has a GDP of US$ 100 million. Country B consumes 2500 toe of energy and has a GDP of US$ 140 million. Calculate the Energy BUREAUOFENERGYEFFICIENCY 5 PAPER-1 COLOR C0DE : GREEN Intensity of both countries in toe per US$ million GDP and indicate which country is more efficient in its use of energy? 2 Marks a) Refer guide book S-2 b)Energy Intensity (EI) = Energy Consumption (toe) / GDP (million US $) Ans EI for Country A = 2000 / 100 = 20 EI for Country B = 2500 / 140 = 17.85 Country B consumes less energy per $. This, it is better in terms of energy efficiency. What are the benefits of the Critical Path Method (CPM)? Also explain how the Program S-3 Evaluation and Review Technique (PERT) differs from CPM. Refer guide book S-3 Ans S-4 The facility has a connected load of 500 kW and currently has a contract demand of 500 kVA. The monthly maximum demand recorded is consistently around 350 kW at 0.85 power factor. The utility imposes a penalty of 350 per excess kVA/month, if recorded demand exceeds contract demand. The demand charge is 300 per kVA/month. a) Determine current demand in kVA. 1 Mark b) The minimum billing demand is 80% of contract demand. Calculate excess demand charges paid above minimum billing demand per month. 2 Marks c) Calculate minimum power factor required to avoid payment of excess demand charges a) over minimum billing demand. 2 Marks S-4 a)Calculate actual kVA demand (1 marks) Ans Actual Demand (kVA)=actual kW/Power Factor=350/0.85 =411.76 kVA b)Excess demand Charges per month (2 marks) Minimum billed demand = 500 0.8 = 400 kVA Excess demand Charges paid= (411.76-400) *300 = Rs.3528.00/month c)Minimum Power Factor Improvement Required (2 marks) To avoid excess demand charges = 350/400 = 0.875 S-5 A food processing unit uses the following per day: LPG consumption: 200 kg/day (Calorific Value = 11,000 kcal/kg, rate 90/kg) DG backup: 100 kWh/day when the grid fails, using diesel at 95/litre with a specific fuel consumption of 260 ml/kWh. (Calorific value = 10,000 kcal/litre) Electrical energy: 1,200 kWh/day at 7.5/kWh Calculate the following: Each 1 Mark a) Convert the LPG energy to kWh equivalent. b) Calculate the thermal energy input (in kcal) required by the DG to produce 100 kWh. c) Calculate the daily energy cost from all 3 sources. d) Calculate the percentage contribution of each energy source to the total energy input (in kWh equivalent). e) Determine the cost share of each energy source in the total energy cost and identify the most economic source among grid power, LPG and DG power. BUREAUOFENERGYEFFICIENCY 6 PAPER-1 COLOR C0DE : GREEN S-5 a) LPG to kWh Thermal energy = 200 11,000 = 2,200,000 kcal/day Ans Convert to kWh: 2,200,000/860= 2,558.14 kWh b) Diesel Energy Input for DG DG Output = 100 kWh Specific fuel consumption = 260 ml/kWh Input =100*260 /1000 = 26 litres Convert to kcal: 26 10000 = 260,000 kcal /day c)Daily Energy Cost Electricity = 1,200 7.5 = 9,000 LPG = 90/kg 200 = 18,000 Diesel = 26 95 = 2,470 Total Cost= 9,000+ 18,000+ 2,470= 29,470/day d)Percentage contribution of each energy source (in kWh) Source kWh equivalent % Share kWh Grid power 1,200 LPG 2,558.14 DG output 100 Total 3,858.14 kWh =1200/3858.14 *100 = 31.1% =2558.14/3858.14 *100 = 66.3% =100/3858.14 *100 = 2.6% 100% e) Percentage cost share of each energy source Source Grid power Daily energy Cost 9,000 LPG 18,000 DG output Total 2,470 29,470/day % Share Cost =9000/29470 *100 = 30.53% =18000/29470 *100 = 61.07% =2470/29470 *100 = 8.4% 100% LPG is more economic. S-6 An energy audit conducted in a rubber processing unit identifies the following: A centrifugal pump (motor rating 30 kW) runs continuously for 16 hours/day, 300 days/year. Measured motor loading = 65%, Motor efficiency = 88%, with no flow control. A VFD retrofit is proposed, which is expected to reduce energy consumption by 10% due to optimized flow control. Power cost = 7.0/kWh VFD installation cost = 1,50,000 a) b) c) d) S-6 Ans Calculate the current annual energy consumption of the motor. Estimate the expected annual energy savings from the VFD. Calculate the annual cost saving in . Determine the simple payback period for the investment. Item Annual energy use Energy saved Annual savings Payback period = = = = Calculation (30x 0.65 /0.88)x16x300 106,364 * 0.1 10,636.4 x7 1,50,000/74454.8 2 Marks 1 Mark 1 Mark 1 Mark Value 106,364 kWh 10,636.4 kWh 74,454.8 2.01 years (~24.17 months) A food dryer processes 1,000 kg/hr of wet material with an initial moisture content of 55% S-7 (wet basis) and dries it to a final moisture content of 10% (wet basis). Steam Flow: 2,500 kg/hr at 3.5 bar (enthalpy = 660 kcal/kg) Latent heat of water vaporization = 540 kcal/kg Specific heat of dry material = 0.45 kcal/kg C BUREAUOFENERGYEFFICIENCY 7 PAPER-1 COLOR C0DE : GREEN a) b) c) d) e) S-7 Ans Drying temperature rise = 60 C Ignore heat loss and assume 100% steam use for moisture removal and solid heating Each 1 Mark Calculate the mass of bone-dry solid in the feed Calculate the mass of water removed per hour Estimate the energy required to evaporate the moisture Estimate the energy required to heat the dry solids Calculate the total energy input from steam a) Mass of bone-dry solid in the feed Moisture content (wet basis) = 55% Dry matter fraction = 1 0.55 = 0.45 Dry solid mass = 1000 0.45 = 450 kg/hr b) Mass of water removed per hour Initial water = 1000 450 = 550 kg/hr Final moisture content = 10% (wet basis) Let final product mass = M. Moisture = 0.10M, Dry solid = 0.90M. Dry solid remains constant at 450 kg. 0.90M = 450 M = 500 kg/hr Final water = 500 450 = 50 kg/hr Water removed = 550 50 = 500 kg/hr c) Energy required to evaporate the moisture Qevap = mass of water removed latent heat of vaporization Qevap = 500 540 = 270,000 kcal/hr d) Energy required to heat the dry solids Qsolid = mass of dry solid specific heat temperature rise Qsolid = 450 0.45 60 = 12,150 kcal/hr e) Total energy input from steam Qtotal = Qevap + Qsolid Qtotal = 270,000 + 12,150 = 282,150 kcal/hr Energy available from steam = mass of steam enthalpy = 2500 660 = 1,650,000 kcal/hr Total energy demand = 282,150 kcal/hr Total energy supply = 1,650,000 kcal/hr A medium-sized factory installs an energy-efficient air compressor system costing 6,00,000. S-8 An audit estimates that it will save 1,80,000 per year in energy bills for the next 3 years. Annual maintenance is expected to cost 10,000 starting from second year onward. Assume: Discount rate (cost of capital) is 10% and salvage value at the end of third year is 50,000. a) Calculate the net annual cash flow from Year 2 onward b) Compute the Net Present Value (NPV) of the investment c) Based on NPV, assess whether the project is economically acceptable 2 Marks 2 Marks 1 Mark S-8 a) Net Annual Cash Flow from Year 2 Onward Ans Annual saving = 1,80,000 Annual maintenance (from Year 2) = 10,000 Net = 1,80,000 10,000 = 1,70,000 Therefore, the net annual cash flow from Year 2 onward is 1,70,000. b) Cash Flow Table with Discounting (10% discount rate) BUREAUOFENERGYEFFICIENCY 8 Year Savings ( ) Maintenance ( ) Net Cash Flow ( ) Salvage Value ( ) 0 - - - - 1 2 3 1,80,000 1,80,000 1,80,000 0 10,000 10,000 1,80,000 1,70,000 1,70,000 0 0 50,000 PAPER-1 COLOR Total PV Cash Factor Flow ( ) @10% 1.000 6,00,000 1,80,000 0.909 1,70,000 0.826 2,20,000 0.751 C0DE : GREEN Present Value ( ) -6,00,000 1,63,636 1,40,420 1,65,220 Total PV of inflows = 1,63,636 + 1,40,420 + 1,65,220 = 4,69,276 NPV = 4,69,276 6,00,000 = 1,30,724 c) Economic Acceptability The NPV is negative ( 1,30,724), meaning the project will not recover its investment cost within 3 years at a 10% discount rate. Therefore, the project is not economically acceptable under these conditions. ..................End of Section II.................. Section III: LONG DESCRIPTIVE QUESTIONS (i) (ii) Marks:6x10=60 Answer all Six questions Each question carries Ten marks L-1 (a) Construct a CPM diagram for the activities below: 4 Marks Activity Precedence Duration in weeks A Start 3 B A 4 C B 1 D C 3 E Start 2 F B 1 Finish D, E, F -- (b) Compute the earliest start, earliest finish, latest start & latest finish of all activities. 3 Marks (c) Identify the critical path and its duration. 3 Marks BUREAUOFENERGYEFFICIENCY 9 PAPER-1 COLOR C0DE : GREEN L-1 a)Network diagram (4 marks) Ans b)Early start (ES), Early Finish (EF), Latest start (LS), Latest finish (LF)-3 marks S.no Activity Duration ES EF LS LF 1 A 3 0 3 0 3 2 B 4 3 7 3 7 3 C 1 7 8 7 8 4 D 3 8 11 8 11 5 E 2 0 2 9 11 6 F 1 7 8 10 11 c)Critical Path (3 marks) Total time on critical path: 11 weeks L-2 A foundry operates an induction furnace with a capacity of 5 t/hr, having a specific electrical energy consumption of 620 kWh/t of liquid metal produced. The overall casting yield of the foundry is 60%. After melting and casting, the products are heat treated in an oil-fired furnace which consumes 75 kg of fuel oil per tonne of castings. The gross calorific value of fuel oil is 10,000 kCal/kg. Additional information provided includes auxiliary connected electrical load of 50 kW, transformer efficiency of 98%, oil density of 0.88 kg/liter and average oil price of 80 per kg. Average castings produced is 45 tonnes per day and the plant is in continuous operation. Using the above data, a) Calculate the total energy consumption per tonne of finished product, expressed as oil equivalent (kg of oil per tonne of finished casting). 7 Marks b) The foundry is receiving additional order to produce 30 tonnes of casting per day. Assess whether the plant can handle additional demand. 3 Marks a) L-2 Specific electric energy = 620 kWh per tonne of liquid metal. Casting Yield = 60% Ans Therefore, to produce 1 tonne of finished castings Luid metal required=1/0.60=1.67 Energy (electrical) per tonne of finished product: 620 kWh/t (liquid /0.6 = 1033.3 kWh/ t (finished) Equivalent energy in kCal = 1033 x 860 = 8,88,380 kCal/T Equivalent energy of auxiliary loads = (50/45)x24x860 = 22,933 kCal/T BUREAUOFENERGYEFFICIENCY 10 PAPER-1 COLOR C0DE : GREEN (Assuming auxiliaries operating at full load as not mentioned in the question) Total Equivalent electrical energy consumption = 8,88,380 + 22, 933 = 9,11,313 kCal/T Total Equivalent electrical energy with transformer losses = 9,11,313 /0.98 kCal/T = 9,29,911 kCal/T Heat required for heat treatment furnace = 75 x 10,000 = 7,50,000 Kcal/T Total kCals required per tonne of finished casting = 9,29,911 + 7,50,000 = 16,79,911 Kcals/T In terms of oil equivalent = 16,79,911 / 10,000 = 167.99 kg of oil/T of finished product b) Actual liquid metal production by induction furnace = (45/24)/0.6 = 3.125 TPH Addition castings production required Additional Liquid metal production required = 30/24 =1.25 TPH = 1.25/0.6 = 2.083 TPH Total Liquid metal production = 3.125 + 2.083 = 5.208 TPH, which is beyond the induction furnace capacity, therefore the plant will be unable to take-up additional production requirement. L-3 Answer the following questions: Each 2 Marks a) Briefly explain the working principle of a solar PV system. b) If a 1 kW PV system in Chennai operates at an average of 5 peak sun hours/day with 15% efficiency, calculate the daily energy output. c) List the factors that affect the performance of a wind turbine. d) A wind turbine with rotor area 200 m is installed in an area with average wind speed of 8 m/s. If air density is 1.2 kg/m , calculate the wind power available in the air stream. e) Briefly explain how biomass is used for electricity generation. L-3 a) Solar PV works on the principle of the photovoltaic effect in which photons create electron-hole pairs at a p n junction, generating DC electricity (or) Refer Guidebook Ans b) Energy Output = 1 kW 5 h = 5 kWh/day. c) Factors: wind speed, air density, rotor area, blade design, hub height, and site conditions. (or) Refer Guidebook d) Power = 0.5 1.2 200 (8 ) = 61,440 W 61.44 kW. e) Biomass is combusted/gasified to generate heat or syngas, which drives turbines/engines to produce power (or) Refer Guidebook BUREAUOFENERGYEFFICIENCY 11 PAPER-1 COLOR C0DE : GREEN L-4 In a chemical company, variable consumption was measured 2.2 times of the production, and the non-production consumption (fixed energy consumption) was observed 10000 kWh/Month. A company has implemented several energy saving initiatives during the previous financial year. a) Calculate energy saving by preparing a CUSUM chart. The actual production and energy consumption observed during the current financial year for the first two quarters is as follows: (8 Marks) Month Production (kg) Actual Energy Consumption (Kwh) April 75000 170000 May 78000 172000 June 85000 185000 July 72000 155000 Aug 71000 153000 Sept 76000 163000 b)Also mention four names of different financing options for industry. L-4 Ans --- 2 Marks Month Production (kg) Actual Energy Consumption ,kWh (EA) Predicted Energy consumption, kWh (2.2P+10000 ) (EP) Ea-Ep CUSUM April 75000 170000 175000 -5000 -5000 May 78000 172000 181600 -9600 -14600 June 85000 185000 197000 -12000 -26600 July 72000 155000 168400 -13400 -40000 Aug 71000 153000 166200 -13200 -53200 Sept 76000 163000 177200 -14200 -67400 Cumulative saving in six month is 67400 Kwh by implementing the energy saving measures. b) Different financing options with the organization i) Debt financing ii) Equity financing iii) Retained earning iv) Capital lease v) True Lease vi) Performance contract BUREAUOFENERGYEFFICIENCY 12 PAPER-1 COLOR C0DE : GREEN L-5 An industry is exploring two project development options as part of its pursuing energy efficiency strategy. Using the NPV concept, find out the better option. Consider 10% as the discount rate and 5 year as the project life. Description Capital cost Year 1 2 3 4 5 Project A 80,000 Net Annual Savings ( ) + 25,000 + 25,000 + 25,000 + 25,000 + 25,000 Project B 100,000 Net Annual Savings ( ) + 35,000 + 35,000 + 35,000 + 35,000 + 35,000 0 1 2 3 L-5 NPV = - CF0/(1+r) + CF1/(1+r) + CF2/(1+r) + CF3/(1+r) + . Ans Project A NPV = - 80,000/ (1+0.10)0 + 25,000/ (1+0.10)1 +25,000/ (1+0.10)2 + 25,000/ (1+0.10)3 + 25,000/ (1+0.10)4 + 25,000/ (1+0.10)5 = -80,000 + 22,727 + 20,661 + 18,783 + 17,075 + 15,522 = Rs 14,768 Project B NPV = - 100,000/ (1+0.10)0 + 35,000/ (1+0.10)1 +35,000/ (1+0.10)2 + 35,000/ (1+0.10)3 + 35,000/ (1+0.10)4 + 35,000/ (1+0.10)5 = -100,000 + 31,818 + 28,926 + 26,296 + 23,905 + 21,732 = Rs 32,677 Project B shall be preferable due to higher NPV. L-6 In a chemical company, Natural Gas (NG) is being used to heat 15 kl/hr of water by 15 C. The company is planning to switch this heating process to steam, which is available from neighbouring industries. a) Work out the feasibility of this option, considering annual operating hours of 6000 hrs. The effective heat of NG is 8500 kCal/m , the NG rate is 55/m , and the density of NG is 0.717 kg/m . The latent heat of steam is 540 kCal/kg, and the steam rate is 2.2/kg. 6 Marks b) Also, calculate the tonnes of CO2 emission for both the options, if 0.2 kg of CO2 is emitted per kg of steam consumed and the percentage of carbon in NG is 74%. 4 Marks a)Total heat requirement = 15000X15 = 225000 Kcal/Hr L-6 Current Option : NG Consumption = 225000/8500=26.47 m 3/Hr Ans NG Consumption = 26.47 X 6000 = 158820 m3/Year Energy Cost = 158820 X 55/100000=87.35 Lakh/Year Alternative Option : Steam consumption = 225000/540=416.67 kg/Hr Steam Consumption = 416.67x 6000=2500020 Kg/Year Energy Cost = 2500020 x 2.2/100000=55 Lakh/Year It is suggested to go with an alternative option considering saving of INR 32.35 Lakh per year. b) CO2 emission for using steam = 2500020 X 0.2/1000 =500 Tonne of CO2 emission per annum CO2 emission for using NG = (158820 x 0.717 x 0.74 x (44/12))/1000 = 309 Tonne of CO2 emission per annum .................End of Section III.................. BUREAUOFENERGYEFFICIENCY 13 PAPER-2 COLOR C0DE : GREEN 24th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER, 2024 PAPER - 2 : ENERGY EFFICIENCY IN THERMAL UTILITIES Section I : OBJECTIVE TYPE Marks: 50 x 1 = 50 1. Select the wrong statement with respect to steam traps: a) Discharges condensate as soon as it is formed b) Does not allow steam to escape c) Capable of discharging air and other incondensable gases d) Does not allow condensate to escape 2. The efficiency of a reheating furnace operating at 10 tonne per hour consuming furnace oil of 230 kg/hour for reheating the material from 40 C-1100 C (consider specific heat of material is 0.13 kcal/kg C and calorific value of furnace oil is 10000 kcal/kg) is: a) 55 c) 60 b) 65 d) 70 3. Which fuel requires maximum air for stoichiometric combustion? a) Butane b) Propane c) Hydrogen d) Coal 4. The device that upgrades a low temperature heat source to a high temperature heat sink is called: a) Heat pipe b) Heat pump c) Plate heat exchanger d) Economizer 5. Flash steam can be recovered from: a) Superheated steam b) Saturated steam c) High pressure condensate d) Condensate at atmospheric pressure 6. The minimum capacity of any closed vessel which generates steam under pressure as covered under Indian Boilers Regulation Act is: a) 22.75 litres b) 25 litres c) 15 kilolitres d) 25 kilolitres 7. Major heat loss in an oil-fired boiler is accounted by: a) Blowdown loss b) Un-burnt carbon loss c) Surface radiation loss d) Stack loss 8. The equipment having the highest efficiency in case of coal-fired cogeneration plant is: a) Electric generator b) Boiler feed water pump c) Steam turbine d) Boiler 9. When pure hydrogen is burnt with stoichiometric air percentage CO2 on volume basis in flue gas on dry basis will be: a) 79% b) 21% c) 0% d) 100% 10. Oxygen percentage (by volume) measurement in flue gas can be done by using: a) Ultrasonic tester b) Potassium oxide probe c) Pitot tube d) Zirconium oxide probe 11. In a combustion process the theoretical air required for complete combustion of 1 kg of carbon is: a) 12.5 kg b) 11.6 kg c) 10.5 kg d) None of the above 12. The flue gas analysis of a combustion process indicates a high level of CO2. What does this imply about the combustion process? a) Efficient combustion b) Incomplete combustion c) High excess air supply d) High moisture content in fuel 13. A boiler operates at 85% efficiency. If the energy required for the process is 1000000 kJ calculate the actual energy input needed: a) 850000 kJ b) 1176471 kJ c) 1000000 kJ d) 1250000 kJ 14. In a steam boiler what is the function of the economizer? a) b) c) d) To preheat the feed water To remove impurities from the water To increase steam pressure To cool the flue gases 15. The boiler efficiency can be increased by: a) Increasing the blowdown rate b) Increasing the flue gas temperature c) Using an economizer d) Reducing the feed water temperature 16. In a boiler system the main purpose of a deaerator is to: a) Remove dissolved gases from feed water b) Preheat the combustion air c) Increase steam temperature d) Reduce water hardness 17. Which component is essential for maintaining the quality of steam in a steam distribution system? a) Condenser b) Economizer c) Steam trap d) Superheater 18. Calculate the amount of steam required to heat 2000 kg of water from 30 C to 90 C given that the latent heat of steam is 2260 kJ/kg and specific heat of water is 4.18 kJ/kg C: a) 116 kg b) 132 kg c) 221.9 kg d) 121.9 kg 19. Furnace efficiency can be improved by: a) Reducing the furnace temperature b) Preheating combustion air c) Increasing the fuel flow rate d) Reducing the air supply 20. What is the primary function of a ceramic fiber blanket in high-temperature applications? a) Provide thermal insulation b) Increase structural strength c) Reduce noise levels d) Improve aesthetic appearance 21. Which of the following is an advantage of fluidized bed combustion systems? a) High combustion efficiency b) Lower NOx emissions c) Fuel flexibility d) All of the above 22. In fluidized bed combustion the primary purpose of adding limestone to the bed material is to: a) Reduce SO2 emissions b) Increase combustion temperature c) Improve bed fluidization d) Reduce NOx emissions 23. Which of the following is a common device used for waste heat recovery in industrial furnaces? a) Recuperator b) Heat wheel c) Heat pump d) All of the above 24. What is the primary function of a baffle in a shell and tube heat exchanger? a) Direct the flow of fluid b) Increase heat transfer area c) Reduce pressure drop d) Prevent fouling 25. Calculate the log mean temperature difference (LMTD) for a counter flow heat exchanger with inlet temperatures of 150 C and 50 C for the hot and cold fluids respectively and outlet temperatures of 100 C and 70 C: a) 75 C b) 80 C c) 85 C d) 64 C 26. In a cogeneration system, the efficiency can be maximized by: a) Increasing the temperature of the exhaust gases b) Using high-pressure steam c) Utilizing waste heat d) Increasing fuel consumption 27. A power plant has a gross heat rate of 2,100 kCal/kWh and a net heat rate of 2282 kCal/kWh. if the plant operates 500 MWh of electrical energy, calculate the percentage of auxiliary power consumption. a) 8.6 % b) 8.0 % c) 10 % d) None of the above 28. Heat to Power Ratio is defined as ratio of _______________ to ______________ required by energy consuming facility. a) thermal energy , electricity c) electricity , thermal energy b) thermal enegy , mechanical energy d) chemical energy , mechanical energy 29. Which of the following characteristics of steam make it popular and useful in industries a) Highest specific heat and latent heat b) Highest heat transfer coefficient c) Inert d) All of the above 30. The average temp difference between cold fluid and hot fluid in heat exchanger is described by LMTD. What does LMTD stands for ? a) Log Mean Temperature difference b) Long Mean Temperature Difference c) Log Mean temperature Depth d) Long Mean Temperature depth 31. Water hammer in a steam line can be stopped by a continuous _____________ in flow direction and adequate number of _______________ points at regular intervals: a) bend , vent b) reducer , drain c) slope , drain d) expander , vent 32. What are the three modes of heat transfer ? a) Radiation, Conduction, Convection c) Radiation, Conduction, Compression b) Radiation, Distillation, Convection d) Radiation, Condensation ,Compression 33. Boiler Evaporation ratio means: kilogram of _____________ per kilogram of fuel consumed. a) Heat generated c) Flue gas generated b) Steam generated d) Ash generated 34. The supercritical steam exists when pressure and temp is above the critical point. At the critical point the pressure and temp values are: a) 101.2 bar and 374.15 Deg C c) 101.2 bar and 274.15 Deg C b) 221.2 bar and 274.15 Deg C d) 221.2 bar and 374.15 Deg C 35. Economic Thickness of Insulation (ETI) is that insulation thickness at which: a) Sum of cost of energy loss and cost of insulation is maximum b) Sum of cost of energy loss and cost of insulation is minimum c) Cost of energy loss is more than cost of insulation d) Cost of energy loss is less than cost of insulation 36. In which of the following equipment heat is added in a power plant?: a) Deaerator c) Economizer b) Air preheater d) All of the above 37. A topping cycle of cogeneration the fuel supplied is used for producing: a) Power primarily followed by byproduct heat output b) Heat primarily followed by product power output c) Power, heat, and refrigeration simultaneously d) None of the above 38. Which of the following is not a befitting choice of bottoming cycle of Cogeneration: a) Cement plant requiring thermal energy at 14500C b) Sugar mill needing thermal energy at 1200C c) Blast furnace in Steel plant requiring heat at 12000C d) All the above 39. Which of the following is not a loss from Industrial Heating Furnaces: a) Flue gas loss c) Radiation loss from openings b) Wall loss d) Blowdown loss 40. Which of the following basic type of Fluidised Bed Combustion(FBC) boiler is also called Bubbling bed boiler : a) PFBC b) CFBC c) AFBC d) All the above 41. Steam Turbine cylinder Efficiency is defined as the ratio of ______________ to _______________. a) Actual entropy drop to isentropic entropy drop b) Actual enthalpy drop to isentropic enthalpy drop c) Actual enthalpy drop to isobaric enthalpy drop d) Actual entropy drop to isochoric entropy drop 42. The heat required to change water to steam at boiling point is called as _______________ heat. a) Super c) Specific b) Sensible d) Latent 43. The excess air level of a packaged boiler operating at 5% O2 level is: a) 23.8% c) 21.25% b) 21.25% d) 31.25% 44. The Chemical De-aeration or chemical oxygen scavenging is done by dosing ______________ in condensate water. a) Phosphate c) Alum b) Limestone d) Hydrazine 45. Difference in Gross calorific Value and Net Calorific Value is accounted to the presence of: a) Water vapour and/or moisture b) Sulphur c) Ash d) Flue gas 46. What is the % Flash steam available after the following flashing process: Sensible heat at Higher pressure condensate = 165 kCal/kg; Sensible heat at lower pressure = 120 kCal/kg; Latent heat of flash steam(at lower pressure) = 526 kCal/kg a) 4.6% c) 8.6% b) 6.8 % d) 4.8% 47. Proximate analysis of coal will give: a) % Hydrogen b) % Volatile Matter c) % Carbon d) None of the above 48. In Cross Flow type of Heat Exchanger as regards direction of flow of the cold fluid and hot fluid: a) Hot fluid and cold fluid flow parallel to each other in the same direction b) Hot fluid and cold fluid flow parallel to each other in opposite direction c) Hot fluid and cold fluid flow at perpendicular direction with respect to each other d) None of the above 49. Typical Boiler Specification contains: a) Make b) Maximum Continuous Rating c) Fuel Fired d) All of the above 50. An increase in the steam pressure from 3 bar to 10 bar will result in a decrease of: a) Sensible heat b) Enthalpy of steam c) Specific volume d) Saturation temperature Section - II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40 S1 Match the following wrt to FBC boilers: 1. Complete combustion of volatile matter NOx formation Turbulence and burns residual volatile matter Instant combustion as fuel enters the furnace Prevents channeling and ensures consistent reaction rates 2. 3. 4. 5. a) Secondary Air b) Tertiary Air c) Air Distributor d) Primary Air e) Combustion temperature Solution: instant combustion as fuel Primary Air enters the furnace complete combustion of 2. Secondary Air volatile matter turbulence and burns 3. Tertiary Air residual volatile matter 4. NOx formation Combustion temperature Prevents channeling and 5. ensures consistent reaction Air Distributor rates For heating application, a drier requires 80 m3/min of air at 92 C, which is heated by wood fired thermic fluid heater. The density of air is 1.2 kg/m3 and the specific heat of air is 0.24 kCal/kg C. The inlet air temperature to the drier is 32 C and the drier is operating for 8 hrs per day. The efficiency of the wood fired heater and its distribution piping system is 50 %. The gross calorific value and the cost of purchased wood are 2000 kCal/kg and Rs. 5000 per ton. The auxiliary power consumption for operating the thermic fluid heater is 10 kW. 1. S2 The plant is planning to replace the existing drying system with a 110 kW infrared electric heater drier. The efficiency of the electric heater is 90%. The investment in the new drier is Rs. 10 Lakhs. If the cost of electricity is Rs. 7/kWh, comment on the cost economics of the proposal. Solution: Cost of wood fired thermic fluid heater operation: Air flow rate (volume) = 80 m3/min x 60 = 4800 m3/hr Air flow rate (mass) = 4800 x 1.2 = 5760 kg/hr Sensible heat of air = m x Cp x T = 5760 x 0.24 x (92-32) = 82944 kcal/hr Efficiency of wood fired heater =50% Wood consumption = 82944/ (2000 x0.5) per hr =83 kg per hr Cost of wood per day = 83 x Rs 5 x 8 hour = Rs 3320 per day Cost of Auxiliary electricity = 10 kW x 8 hrs x 7 = Rs.560 per day Total cost of operation = Rs.3880 per day Cost of Infra-red heater operation: Electric heater power consumption= 82944/0.9/860 =107.2 kw Electricity consumption per day= 107.2 kW x 8 hr = 857.3 kwh per day Cost of operation per Day= 857.3 x Rs 7= Rs 6001.00 per day Since the per day expenses are higher in case of electric heater, it is not economical. S3 A textile plant has an extensive steam network and the condensate is not being recovered. The plant management is planning to recover the flash steam from the high-pressure condensate and return the remaining condensate back to the boiler. The following are the parameters observed: Condensate quantity : 1000 kg/hr Condensate Pressure :10 bar Cost of steam :Rs 1100/ T Annual operating hours :6000 Low pressure process steam (flash steam) pressure 2 bar Sensible heat of condensate at 10 bar 188 kCal/kg Sensible heat of condensate at 2 bar 135 kCal/kg Latent heat of steam at 2 bar 518 kCal/kg Boiler Efficiency 80 % GCV of fuel oil 10,200 kCal/kg Specific Gravity of fuel oil 0.92 Condensate temperature when recovered 95 oC Make up water temperature 35 oC Calculate the quantity of flash steam which can be recovered, and the annual fuel oil savings on account of the flas recovery. Solution: Flash steam available % = S1- S2/(L2) % of Flash steam recoverable = (188 135)/518 = 10.2 % a) Quantity of flash steam recovered from condensate = 1000 x 0.102 = 102 kg/hr Balance Condensate available for recovery after flash steam = 1000 102 = 898 kg/hr Heat recovered = Latent Heat of Flash Steam + Sensible Heat of Condensate = (102*518) + (898 x (95 35)) = 106716 kCal/hr b) Oil saved = 106716 x 6000 / (0.80 x 10200) = 78.5 tons/yr S4 Calculate the effectiveness of a parallel flow heat exchanger, where the hot fluid enters at 200 C and leaves at 140 C, and the cold fluid enters at 50 C and leaves at 120 C. The mass flow rates of the hot and cold fluids are 3000 kg/hr and 2500 kg/hr, respectively, with specific heat capacities of 2.5 kJ/kg C and 4.18 kJ/kg C, respectively. Solution: Cmax=mcold Cp,cold=2500kg/hr 4.18kJ/kg C=10450kJ/hr C Cmin=mhot Cp,hot=3000kg/hr 2.5kJ/kg C=7500kJ/hr C Effectiveness( )=Maximum possible heat transfer / Actual heat transfer Qactual=mcold Cp,cold (Tcold,out Tcold,in) Qactual=2500kg/hr 4.18kJ/kg C (120 50) C=2500 4.18 70=731,500kJ/hr Qmax=Cmin (Thot,in Tcold,in) Qmax=7500kJ/hr C (200 50) C=7500 150=1125000 kJ/hr =731,500 /1125000 = 0.65 S5 An oil-fired furnace is operating at 1380 0C with ambient temperature of 30 0C and average fuel consumption of 370 litres/hr. The Calorific value of oil is 10,000 kCal/kg. Specific gravity of oil is 0.93. Weight of the billet heated is 8000 kg/hr and Specific heat of billet is 0.13 kCal/kg/0C. Calculate the following: a)Efficiency of furnace by direct Method. b) Heat loss in furnace in terms of Fuel loss in lit/hr? Solution: Heat Input = 370*0.93* 10000 = 3441000 kCal/hr Heat Output = mCpDeltaT = 8000 *0.13 * (1380-30) = 1404000kCal/ hr Efficiency = Heat Output / Heat Input = 1404000 / 3441000 = 40.8% Energy loss from furnace = 3441000 *(1-0.408) kCal/hr =2,037,072 kCal/hr Fuel Loss / hour = 2,037,072/10000 = 203.7 kg /hr = 203.7/0.93 = 219 lit/h S6 In the chart given below for determining the economic thickness of insulation, what do the following represent: Curve A, Curve B, Curve C, X Axis and Y Axis. Solution: A Combined Costs B Depreciation cost of insulation C Fuel cost due to loss X Axis Insulation thickness Y axis Annual cost S7 How do multiple effect evaporators save energy? Multiple effect evaporators are used to reduce energy consumption by using the vapor produced from one effect to heat the product in the subsequent lower-pressure effect. This allows for efficient use of energy by minimizing steam usage, making the evaporation process more economical. Solution: Page-247 S8 In a cogeneration system using a gas turbine, if the heat rate is 3050 kcal/kWh and the calorific value of the natural gas is 9500 kCal/sm , calculate the fuel consumption for generating 3000 kW of power. Solution: For a gas turbine with a heat rate of 3050 kcal/kWh and the calorific value of natural gas being 9500 kcal/sm , the fuel consumption for generating 3000 kW of power is: Fuel Consumption = 3000 9500 / 3 3050 = 9150000 9500 / 3 = 963.16 3 Section - III: LONG DESCRIPTIVE QUESTIONS L1 Marks: 6 x 10 = 60 A steam pipeline with a diameter of 100 mm is insulated with 30 mm of mineral wool material. The pipeline carries steam at 6 ata and has a length of 150 meters. Due to increased fuel costs, the Energy Auditor has recommended increasing the insulation thickness by an additional 20 mm. Calculate the economic benefits of this recommendation, given the following details: The plant operates for 6,500 hours per year. The existing surface temperature of the insulation is 90 C. The expected surface temperature after additional insulation is 50 C. Ambient temperature is 25 C. Boiler efficiency on NCV (Net Calorific Value) basis is 82%. Cost of fuel oil is Rs. 50,000 per tonne. Net Calorific Value of fuel is 9,500 kCal/kg. The cost of insulation is Rs. 2000/meter. Calculate the following: 1.Heat Loss for Existing Insulation , 2.Heat Loss for Modified Insulation, 3.Fuel Savings, 4.Simple Payback Period, Solution: 1. Calculate Heat Loss for Existing Insulation Given Data: Pipe diameter: 100 mm (0.1 m) Existing insulation thickness: 30 mm (0.03 m) Surface temperature with existing insulation: 90 C Ambient temperature: 25 C Heat Loss Formula: S=[10+20(Ts Ta)] (Ts Ta) S=[10+20(90 25)] (90 25) S=13.25 65=861.25 kCal/hr/m2 2. Calculate Heat Loss for Modified Insulation Given Data: New insulation thickness: 50 mm (0.05 m) New surface temperature: 50 C Heat Loss Formula: S=[10+20(Ts Ta)] (Ts Ta) S=[10+20(50 25)] (50 25) S=11.25 25=281.25 kCal/hr/m2 3. Calculate Total Heat Loss Existing Insulation: Length of pipe = 150 m Diameter of pipe = 100 mm Outer diameter after existing insulation = 0.1 m + (0.03x2) m = 0.16 m Surface area (existing) = 0.16 150 = 75.4 m2 Total Heat Loss (Existing): Total Heat Loss existing=861.25 kCal/hr/m2 75.4 m2= 64938.25 kCal/hr Modified Insulation: New thickness of insulation = 50 mm (0.05 m) New diameter = 0.1+ (0.05*2) = 0.2m Surface area (Amodified) = 0.2 150= 94.23 m2 Total Heat Loss (Modified): Total Heat Loss modified=281.25 kCal/hr/m2 94.23 m2= 26502.2 kCal/hr Reduction in Heat Loss: Reduction=Total Heat Loss existing Total Heat Loss modified= 64938.25 26502.2 = 38436 kCal/hr Annual Heat Savings: Annual Heat Savings=38436 kCal/hr 6,500 hr/year= 24,98,34,325 kCal/year Fuel Savings: Fuel Oil Data: Net Calorific Value (NCV) = 9,500 kCal/kg Boiler Efficiency = 82% Convert heat savings to fuel required: Fuel Required=NCV Efficiency/Annual Heat Savings Substitute the values: Fuel Required=(249834325/9,500/0.82)/1000 = 32 tonnes Monetary Savings: Monetary Savings=Fuel Required Cost per Tonne= 32 tonnes 50,000 Rs/tonne =Rs 16 lakhs 4. Calculate Simple Payback Period Cost of Insulation: Cost = Rs. 2000 x 150 = Rs. 3,00,000/Simple Payback Period: Payback Period= Cost of Insulation / Monetary Savings = 3,00,000 / 16,00,000 = 0.1875 Year or 2.25 Months L2 a) Explain the significance of bulk density in refractory materials. b) Explain the benefits of using monolithic refractories. c) For combustion of 500 liters per hour of furnace oil, estimate the quantity of combustion air required per hour with 20% excess air. The specific gravity of furnace oil is 0.95. The fuel analysis is as follows: Carbon (C) - 84%, Hydrogen (H ) - 12%, Sulphur (S) - 3%, Oxygen (O ) - 1%. Solution: a) Refer Page no 156 or Bulk density impacts a refractory material's strength and heat flow. High bulk density means fewer pores, making the material stronger and more durable but less insulating. Low bulk density means better insulation but less strength. Choosing the right bulk density helps balance strength and insulation based on application needs. b) Refer Page no 161 or Monolithic refractories are seamless and joint-free, reducing weak spots. They are easy to install and repair, saving time and costs. Their design improves heat efficiency and durability, making them suitable for various industrial uses. c) Basis: 1kg of fuel Oxygen requirement for Carbon= (0.84*32/12) = 2.24 kg of O2 Oxygen requirement for Hydrogen= (0.12*16/2) = 0.96 kg of O2 Oxygen requirement for Sulphur= (0.03/1/1) = 0.03 kg O2 Total Oxygen required to be supplied = 2.24+0.96+0.03-0.01 = 3.22 kg of O2 Therefore, Air required to be supplied = 3.22/0.23 = 14 kg of air per kg of fuel Now, Mass of Fuel being supplied = 500*0.95 = 475 kg per hr Quantity of Air required with excess air = 475*14*1.2 = 7980 kg/hr L3 Fill in the following blanks : 1. If the steam generation in a boiler is 24 tonnes in 3 hours and the fuel consumption in the same period is 2 tonnes, the evaporation ratio is _______. 2. A boiler generates 10 TPH of steam. If the feed water temperature is 80 C and the enthalpy of steam at 10 kg/cm pressure is 665 kcal/kg, the total heat output per hour is _______ kcal/hr. 3. If the GCV of coal is 4000 kcal/kg, boiler efficiency is 75%, enthalpy of steam is 665 kcal/kg and enthalpy of feed water is 80 kcal/kg. The heat input required to generate 8000 kg/hr of steam is _______ kcal/hr. 4. If the boiler feed water temperature increases from 30 C to 80 C, the heat added per kg of water is _______ kcal. 5. The stochiometric air required for combustion of 30 kg of carbon is __________. 6. A boiler has a blowdown rate of 2% and produces 20 TPH of steam. The amount of blowdown water per hour is _______ kg/hr. 7. If the specific heat of flue gas is 0.24 kcal/kg C and the flue gas temperature drops from 200 C to 150 C, the heat recovered per kg of flue gas is _______ kcal. 8. The ____________removes oxygen and carbon di-oxide from the boiler feed water on heating. 9. The process of periodically removing a portion of water from the boiler to remove accumulated impurities is called____________. 10. The heat required to convert water into steam at constant pressure and temperature is known as_______. Solution: L4 1. 12 2. 5850000 kCal/hr 3. 6240000 kCal/hr 4. 50 kcal/kg 5. 348 kg 6. 400 kg/hr 7. 12 kCal/kg 8. De-aerator 9. Blowdown 10. latent heat of vaporization Briefly explain the following waste heat recovery devices. L5 Recuperators, Refer Guide Book, Page 219 Regenerators, Refer Guide Book, Page 222 Heat Wheels, Refer Guide Book, Page 222 Heat Pipes, Refer Guide Book, Page 223 Plate Heat Exchangers, Refer Guide Book, Page 226 A textile plant utilizes steam for various processes, including dyeing and finishing. The plant operates a boiler at a pressure of 9 kg/cm (g). For a specific batch process, the plant requires superheated steam at 250 C. The steam saturation temperature at this pressure is 180 C. Given the specific heat of superheated steam is 0.45 kcal/kg C. a) Calculate the total heat content of the superheated steam used in the process. b) Explain, why superheated steam should not be used for process heating? Solution: Superheat temperature: 250 C Saturation temperature: 180 C Specific heat of superheated steam (Cp): 0.45 kcal/kg C Total heat content of dry saturated steam at 9 kg/cm (g) (h): 663 kcal/kg Calculations: 1.Total Heat Content of Superheated Steam (hsuperheat): Hsuperheat = h + Cp x T = 663kcal/kg+0.45kcal/kg Cx70 C = 663kcal/kg+31.5kcal/kg = 694.5kcal/kg 2.Why Superheated Steam Should Not Be Used for Process Heating: Temperature Control: Superheated steam does not condense immediately upon releasing heat, making it difficult to control temperature precisely. For processes requiring precise temperature control, this can lead to overheating or underheating. Heat Transfer Efficiency: The heat transfer coefficient of superheated steam is lower than that of saturated steam. This means that superheated steam is less efficient in transferring heat to the process materials, potentially leading to longer processing times and higher energy consumption. Equipment Wear and Tear: The higher temperature of superheated steam can lead to increased wear and tear on process equipment, reducing its lifespan and increasing maintenance costs. L6 A back pressure cogeneration plant is designed to generate both power and process heat. The electrical power generated is 25 MW. The cogeneration boiler is having coal feed rate of 80 TPH and GCV of 18,800 kJ/kg. The turbine mechanical efficiency is 98%, Gear box efficiency is 97% and alternator efficiency is 98%. Assume steady operating conditions and no steam loss in entire process. The steam parameters are as follows: Main Steam Parameters: Steam Parameters at Turbine inlet: High-pressure steam at 9 MPa, 400 C and enthalpy of 3118.8 kJ/kg Exhaust Steam Parameters: Exhaust of steam of turbine being sent for process heating has an enthalpy of 2815.8 kJ/kg The return Condensate from the process is saturated liquid at 10 kPa having enthalpy of 191.81 kJ/kg. Calculate the following: a) Energy Utilization Factor , b) Heat to Power Ratio, Solution: Net enthalpies drop per kg of Steam: Work done by the turbine per kg of steam: h1 h2=3118.8 kJ/kg 2815.8 kJ/kg=303 kJ/kg Total Mass Flow Rate of Steam: To achieve a net power output of 25 MW Turbine input = 25/(0.98*0.97*0.98) =26.84 MW Steam Required = 26.84 * 1000 *860*4.18 /303/1000 = 318.4 TPH Total Thermal Energy Output: Thermal energy output to the process heater (enthalpy difference): process=h2 hf=2815.8 kJ/kg 191.81 kJ/kg=2624 kJ/kg Total thermal energy output = 318.4 TPH * 1000 * 2624 kJ/kg /4.18/860 /1000= 232.4 MW Energy Utilization Factor (EUF): Total fuel energy input = m fuel GCV=80000 kg/hr 18800 kJ/kg /4.18/860 /1000 =418 MW Total useful energy output = 25 MW + 232.4 MW= 257.4 MW Energy Utilization Factor = 257 MW / 418 MW = 61% Heat to Power Ratio: 232.4/25 = 9.29 Paper-2 Code : Green 25th NATIONAL EXAMINATION FOR CERTIFICATION OF ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER, 2025 PAPER - 2 : ENERGY EFFICIENCY IN THERMAL UNTILITIES Date : 27.09.2025 Timings : 14:00-17:00 HRS Duration : 3 HRS Max. Marks : 150 General instructions : o o o o o Please check that this question paper contains 8 printed pages Please check that this question paper contains 64 questions The question paper is divided into three sections All questions in all three sections are compulsory All parts of a question should be answered at one place Section I: OBJECTIVE TYPE Marks: 50 x 1 = 50 (i) Answer all 50 questions (ii) Each question carries one mark (iii) Please s h a d e t h e a p p r o p r i a t e oval in SECTION-I of MAIN ANSWER BOOKLET with BLUE/BLACK ball point pen 1. 2. 3. 4. 5. 6. The main advantages of a cogeneration system include: A) Higher overall efficiency B) Simultaneous production of power and heat C) Reduced fuel consumption D) All of the above The relationship between heat rate and plant efficiency is such that: A) Higher heat rate means higher efficiency B) Lower heat rate means higher efficiency C) Heat rate and efficiency are unrelated D) Both increase together A metallic radiation recuperator is primarily used to: A) Recover heat from flue gases to preheat combustion air B) Convert radiant heat into electricity C) Cool furnace walls to prevent overheating D) Measure furnace temperature using radiation Select the correct statement about the benefits of waste heat recovery: A) It increases fuel consumption B) It reduces overall plant efficiency C) It saves energy and reduces operating costs D) It always requires high-grade heat sources Terminal temperature difference (TTD) in a steam to water heat exchanger is: A) Inlet temp. difference B) Outlet temp. difference C) Difference of Steam Saturation temperature and outlet water temperature. D) Avg. temperature difference Select the correct statement about flash steam: A) B) C) D) It is produced by cooling steam below saturation temperature It is produced by cooling steam above saturation temperature It forms when hot condensate is released to a lower pressure It forms when hot condensate is released to a higher pressure BUREAU OF ENERGY EFFICIENCY 1 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. Paper-2 Code : Green The specific gravity of a fuel is the ratio of: A) Weight of fuel to its volume B) Density of fuel to the density of water C) Volume of fuel to volume of water D) Mass of fuel to its calorific value Select the correct statement about volatile matter in coal: A) It reduces flame stability in combustion B) It consists of gases released when coal is heated C) It is the same as fixed carbon D) It increases the ash content of coal The Net Calorific Value (NCV) of a fuel is obtained by: A) Subtracting the heat of vaporization of water from the Gross Calorific Value B) Adding the heat of vaporization of water to the Gross Calorific Value C) Measuring only the sensible heat of combustion products D) Ignoring latent heat losses in the fuel Which of the following agro fuels typically has the highest moisture content? A) Saw dust B) Paddy husk C) De-oiled bran D) Coconut shells Select the correct statement about excess air in combustion: A) Excess air always increases boiler efficiency B) Excess air is needed to ensure complete combustion of fuel C) Excess air reduces flue gas losses D) Excess air is less than the theoretical air In a pulverised fuel boiler, the fuel is burned in : A) lump form B) fluidised bed form C) fine powder form D) None of the above Select the correct statement about a supercritical boiler: A) It generates steam at pressures above the critical point B) It always requires a steam drum for separation of steam and water C) It operates only with natural circulation D) It produces steam with high moisture content A thermic fluid heater is primarily used to: A) Generate steam at high pressure B) Heat a mineral oil C) Produce hot water D) Produce hot air A cupola furnace is primarily used for: A) Producing cast iron B) Producing steel ingots C) Heating non-ferrous metals D) Producing coke Select the correct statement about flue gas losses in a reheating furnace: A) Flue gas losses decrease with higher flue gas temperature B) Flue gas losses increase with excess air and higher exhaust temperature C) Flue gas losses are unaffected by excess air D) All of the above The principle of cogeneration is based on: A) Using separate systems to produce heat and power B) Sequential use of energy to produce both electricity and useful heat from the same fuel BUREAU OF ENERGY EFFICIENCY 2 Paper-2 Code : Green C) Converting all heat into electricity D) Using waste heat only for cooling 18. Trigeneration refers to the simultaneous production of: A) Power, heat, and cold B) Power, heat, and steam C) Power, steam, and compressed air D) Heat, cold, and compressed air 19. The flash point of an oil is the: A) Temperature at which the oil ignites spontaneously B) Lowest temperature at which the oil vapour momentarily ignites on application of a flame C) Temperature at which the oil starts boiling D) Temperature at which the oil burns continuously 20. Which type of fuel requires lowest amount of excess air for combustion? A)Furnace oil B) LDO C) Propane D) Rice Husk 21. Which among the following does not release any energy during combustion? A) Carbon B) Sulphur C) Nitrogen D) Hydrogen 22. Which is the best suited pump for pumping viscous liquid fuel? A) Centrifugal pump B) Gear pump C) Vertical turbine pump D) None of the above 23. In a water tube boiler, what is the primary function of the boiler drum? a) To superheat the steam leaving the economizer b) To separate steam from water and provide storage for steam and water mixture c) To maintain draft inside the furnace by balancing air and flue gas pressure d) To preheat the feedwater before entering the boiler tubes 24. Dearation in boilers is primarily carried out to remove: a) Dissolved solids b) Dissolved gases c) Suspended particles d) Boiler scale 25. Which of the following best describes the critical point of steam? A) The point where steam condenses to water at standard atmospheric pressure B) The highest temperature and pressure at which liquid water and steam can coexist in equilibrium C) The temperature at which steam becomes superheated D) The point where latent heat of vaporization is maximum 26. Which of the following is preferable for a heating process? a) Wet steam b) flash steam c) High pressure steam d) Dry saturated steam 27. A float steam trap is an example of which type of steam trap? A) Thermodynamic B) Mechanical C) Thermostatic D) Hydraulic 28. Which of the following is commonly used as a low-temperature insulating material? A) Calcium silicate B) Polyurethane C) Magnesia D) Asbestos 29. The economic thickness of insulation is the thickness at which: A) Heat loss is zero B) Cost of insulation is minimum C) Combined cost (heat loss cost and insulation cost) is minimum D) Heat transfer rate is maximum 30. Which of the following is a key advantage of Fluidized Bed Combustion (FBC) over conventional combustion systems? a) Lower power requirement for fans b) Reduced emissions of SO and NOx c) Higher excess air requirement d) Lower combustion efficiency BUREAU OF ENERGY EFFICIENCY 3 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. Paper-2 Code : Green In a fluidized bed combustion system, the fuel is burned in a bed of: A) Moving metal plates B) Sand, ash, or other granular material suspended by air flow C) Rotating drums D) Water-cooled tubes only In a combined cycle power plant, the waste heat from the gas turbine is used to: A) Preheat the incoming air to the gas turbine B) Generate steam for a steam turbine C) Cool the exhaust gases directly to the atmosphere D) Operate a diesel generator The effectiveness of a heat exchanger is defined as the ratio of: A) Actual heat transfer to the maximum possible heat transfer B) Heat loss to the surroundings to the total heat input C) Outlet temperature difference to inlet temperature difference D) Actual heat transfer to the total mass flow rate In pinch analysis, the "pinch point" represents: A) The location in the heat exchanger where fouling is maximum B) The point of minimum temperature difference between hot and cold streams C) The point where heat transfer rate is maximum D) The location where the heat exchanger pressure drop is minimum The presence of high sulphur in fuels mainly contributes to: A) Reduction in NOx emissions B) Corrosion and air pollution C) Decrease in net calorific value D) Prevention of slag formation Why is viscosity of liquid fuels important in combustion systems? A) It determines the fuel s sulphur content B) It affects atomization and burner performance C) It indicates the ash fusion temperature D) It measures the fuel s calorific value In boilers, natural draft is produced by: A) A steam ejector B) The height and temperature difference in the chimney C) A forced draft fan D) Vacuum pump As per Indian Boiler Regulations (IBR), a pipe is defined as a steam pipe if it carries steam at a pressure exceeding 1 Mark Awarded to all candidates who attempted this question Which of the following is an example of internal water treatment in boilers? A) Deaeration B) Coagulation and filtration C) Sodium Phosphate dosing D) Clarification In boiler water treatment, Reverse Osmosis (RO) is primarily used to: A) Remove only suspended solids from water B) Remove dissolved salts and impurities C) Increase the alkalinity of feedwater D) Convert hard water into soft water Which of the following is not a benefit of condensate recovery in a steam system? A) Energy savings B) Reduced water consumption C) Lower water treatment costs D) None of the above 1 Mark Awarded to all candidates who attempted this question When steam passes through a pressure reducing valve (PRV), its enthalpy: BUREAU OF ENERGY EFFICIENCY 4 Paper-2 Code : Green A) Increases B) Decreases C) Remains the same D) Becomes zero 43. A thermocompressor in a steam system is primarily used to: A) Convert wet steam into dry saturated steam B) Recompress low-pressure steam with high-pressure steam to obtain medium-pressure steam C) Increase boiler steam generation rate D) Reduce steam temperature without affecting pressure 44. Higher excess air in an oil-fired furnace leads to: A) Higher efficiency B) Lower flue gas heat loss C) Zero stack losses D) None of the above 45. The main benefit of using preheated air for combustion is: A) Reduced furnace temperature B) Increased fuel savings C) Increased excess air requirement D) Higher flue gas temperature 46. Choose the correct statement about insulating materials: A) Insulating materials have high thermal conductivity B) Insulating materials reduce heat loss C) Insulating materials increase heat transfer rate D) All of the above 47. Choose the incorrect statement about refractories: A) Refractories are used to withstand high temperatures in furnaces B) Refractories should have low thermal conductivity for insulation purposes C) Refractories should have low melting points for easy shaping D) Refractories must be resistant to thermal shock 48. Which of the following defines the Gross Calorific Value (GCV) of a fuel? a) Heat liberated after complete combustion including latent heat of water vapor b) Heat available after combustion excluding latent heat of water vapor c) Heat required to raise 1 kg of fuel by 1 C d) Ratio of heat released to air-fuel ratio 49. Which of the following statements about a fluidized bed boiler is correct? A) It can efficiently burn low-grade fuels with high ash content B) It requires very high combustion temperatures (above 1600 C) C) It operates without any bed material D) It cannot be used for biomass fuels 50. An extraction condensing turbine is designed to: A) Exhaust all steam to the process at high pressure B) Extract some steam for process use and balance steam for power generation C) Operate only as a back pressure turbine D) Exhaust steam directly to atmosphere .................. End of Section I .................. BUREAU OF ENERGY EFFICIENCY 5 Paper-2 Code : Green Section II: SHORT DESCRIPTIVE QUESTIONS (i) (ii) S-1 Marks: 8 x 5 = 40 Answer all Eight questions Each question carries Five marks A batch furnace uses LPG and draws combustion air at 30 C from the ambient condition. An energy auditor proposes fitting a recuperator to use hot flue gas (950 C) to preheat this air to 400 C. a) Find the % fuel savings from preheating the air to 400 C. b) Find the new LPG consumption after retrofit. c) Calculate the flue-gas exit temperature from the recuperator using an energy balance. Given: S-1 EA = 20%, Stoichiometric air = 15.5 kg air/kg LPG, cp (air & flue gas) = 0.24 kcal/kg C. Ans Actual air (AAS) = (1 + EA) 15.5 = 1.2 15.5 = 18.6 kg/kg fuel. Flue gas per kg fuel: mfg = 1 + AAS = 1 + 18.6 = 19.6 kg/kg fuel. 2 Marks 2 Marks 1 Mark (a) % Fuel savings: Recovered heat to air per kg fuel: P = mair cp (Tair,out Tair,in) = 18.6 0.24 (400 30) = 1651.68 kcal/kg fuel Savings = P / GCV = 1651.68 / 11,500 = 0.1436 14.36 % (b) New LPG consumption: mnew = mold (1 Savings) = 50 (1 0.1436) 42.8 kg/h (c) Flue-gas exit temperature from recuperator (energy balance): mfg x cp x (Tfg,in Tfg,out) = mair x cp x (Tair,out Tair,in) 19.6 (950 Tfg,out) = 18.6 370 = 6882 950 Tfg,out = 6882 / 19.6 351.12 Tfg,out 950 351.12 = 599 C S-2 Match the Following: Term 1. Topping Cycle 2. Bottoming Cycle 3. Heat-to-Power Ratio 4. Combined Cycle Plant 5. Trigeneration Description a. Produces power first, then recovers heat for process use b. Produces power, heat, and cooling from the same energy source c. Produces heat first, then uses waste heat to generate power d. Useful thermal energy to electrical energy output e. Uses gas turbine and steam turbine in series S-2 1 a Ans 2 c 3 d 4 e 5 b S-3 Fill in the Blanks: 1. A furnace oil consumption of 250 liters/day, with density 0.96 kg/l and GCV 10,200 kcal/kg, provides a total heat input of __________ kcal/day. 2. If the NCV of a fuel is 9,500 kcal/kg and the mass of fuel burned is 500 kg, the total heat energy released is __________ MWh 3. For complete combustion of 1 kg of carbon, the theoretical air requirement is __________ kg. 4. Calculate % of nitrogen in dry flue gas if 1 kg of hydrogen is completely burned in presence of 34.8 kg BUREAU OF ENERGY EFFICIENCY 6 5. S-3 Ans Paper-2 Code : Green of air. A coal sample with a GCV of 4,500 kcal/kg and moisture content of 10% has an approximate NCV of __________ kcal/kg, assuming latent heat of vaporization of water = 587 kcal/kg. 1. 250 L/day 0.96 kg/L = 240 kg/day, Heat input = 240 10,200 = 2,448,000 kcal/day 2. Energy = 9,500 500 /(860 *1000) = 5.52 MWh 3. 11.6 kg of theoretical air per kg of carbon 4. 100% 5. NCV = GCV (Moisture fraction 587) = 4,500 (0.10 587) = 4,441 kcal/kg S-4 A 10-TPH forced-draft boiler is tested at a load of 8 TPH. Fuel flow is 0.53 TPH and the fuel GCV is 10,000 kcal/kg. Feedwater temperature is 70 C. The boiler delivers saturated steam at 10 bar(g) at 186 C. Take the specific enthalpy of saturated steam has 664 kcal/kg. a) Calculate the direct efficiency assuming dry saturated steam. 3 Marks b) The indirect method efficiency is 87%, which does not match with the direct method. The auditor claims this difference is due to the wetness of steam. Establish the auditor s claim by calculating the dryness fraction. 2 Marks S-4 Answer Ans Fuel flow to boiler = 0.53 TPH Steam flow = 8TPH Fuel GCV = 10000 kcal/kg Steam enthalpy = 664 kcal/kg Feed water enthalpy = 70 kcal/kg Direct efficiency = 8 (664 70) = 89.7% .53 10000 Indirect efficiency = 87% Moisture content/ Steam quality: If the steam is slightly wet, the actual enthalpy is lower, reducing the direct efficiency closer to the indirect efficiency. Total heat content of wet steam = sensible heat + (dryness fraction * latent heat) = 186 + ( x* (664-186)) = 186+478x = 8 [(186 + 478x) 70] = .87 . 53 10000 X= 96.3%. Hence steam is not fully dry in nature. This is the reason for difference in efficiency A boiler has an air preheater whose design and actual conditions are given below: S-5 Design (0C) Actual (0C) Flue gas Inlet temperature to APH 330 330 Flue gas outlet temperature from APH 180 210 Air inlet temperature to APH 40 40 Air outlet temperature from APH 190 190 Assume the following: specific heats of flue gas and air are equal, there is no heat loss to the surroundings, and as per design the mass of flue gas is approximately to the mass of air. Analyze and discuss the deviations between actual operating condition and design condition in air mass flow, flue-gas mass flow and APH heat transfer. S-5 Ans Cp FG = Cp Air m FG = m Air Design (TFGout TFG in) = (330-180) = 150 (Tair out Tair in) = (190-40) = 150 Heat balance exists Actual BUREAU OF ENERGY EFFICIENCY 7 Paper-2 Code : Green (TFGout TFG in) = (330-210) = 120 (Tair out Tair in) = (190-40) = 150 Heat is not balanced. Implications: a) Air mass flow: Likely lower than design (or part of the air is bypassing the APH). When less air is heated, temperature of outlet air can still reach 190 C, but the gas won t cool as per design. b) Flue-gas mass flow: Effectively higher than air by ~25% at the APH (or flue-gas ingress/leakage upstream of APH inflates the gas stream). This drives the exit flue-gas hotter than the design (210 C vs 180 C). c) APH heat transfer: Heat recovery is less than the design as evidenced by the higher gas outlet temperature (less gas-side cooling). If the insulation thickness on a steam pipe is increased from 50 mm to 100 mm, calculate the percentage S-6 reduction in heat loss. The surface temperature of the insulation is 70 C for 50 mm thickness and 60 C for 100 mm thickness. Ambient temperature is 35 C. S = [10 + (Ts Ta)/20] (Ts Ta) S-6 Calculations Ans 1) Heat loss for 50 mm insulation: T1 = Ts1 Ta = 70 35 = 35 C S50 = [10 + (35/20)] 35 = (10 + 1.75) 35 = 11.75 35 = 411.25 kcal/hr/m2 2) Heat loss for 100 mm insulation: T2 = Ts2 Ta = 60 35 = 25 C S100 = [10 + (25/20)] 25 = (10 + 1.25) 25 = 11.25 25 = 281.25 kcal/hr/m2 Note : Since the diameter of the pipe is not mentioned in the question paper. If the candidate has calculated the rate of heat loss for 50 mm and 100 mm insulation full marks may be awarded. S-7 In a food processing industry, steam at 15 kg/cm was used to heat 10 TPH of milk from 40 C to 80 C. An energy auditor suggests installing a PRV to reduce steam pressure from 15 kg/cm to 3 kg/cm . Specific heat of milk is 1.0 kcal/kg/oC. Present Condition (Steam Pressure 15 kg/cm ) Proposed Condition (Steam Pressure 3 kg/cm ) Sensible heat 200.6 kcal/kg 133.287 kcal/kg Latent Heat 465.72 kcal/kg 517.17 kcal/kg 0.9 ? Steam Parameters Dryness Fraction a) Determine the outlet moisture percentage when steam passes through the PRV. b) Calculate the steam savings per hour due to this measure. S-7 Ans 3 Marks 2 Marks a) Total heat content before PRV = Total heat content after PRV 200.6 + (0.9*465.72) =133.287 +(x *517.17) x = 0.94 b) Heat required for water = (10*1000)*1*(80-40) = 400000 kcal/hr Steam requirement at present condition= 400000/ (465.72*0.9) = 954.3 kg/hr Steam requirement at proposed condition = 400000/ (517.17*.94) = 822.8 kg/hr Savings = 131.5 kg/hr of steam S-8 In an oil-to-water counterflow heat exchanger, hot oil enters at 150 C and leaves at 90 C. Cold water enters at 30 C and leaves at an unknown temperature Tc,o. The log mean temperature difference (LMTD) for the exchanger is 60 C. Calculate the exit temperature of the water. S-8 Given: Hot oil inlet temperature, Th,i = 150 C Ans Hot oil outlet temperature, Th,o = 90 C Cold water inlet temperature, Tc,i = 30 C Cold water outlet temperature, Tc,o = ? BUREAU OF ENERGY EFFICIENCY 8 Paper-2 Code : Green LMTD = 60 C For counterflow heat exchangers: T1 = Th,i Tc,o = 150 Tc,o T2 = Th,o Tc,i = 90 30 = 60 LMTD is given by: LMTD = ( T1 T2) / ln( T1 / T2) = 60 Let T1 = A. Then: (A 60) / ln(A / 60) = 60 This equation holds true when A = 60 (equal terminal differences case). Therefore: 150 Tc,o = 60 Tc,o = 90 C Answer The exit temperature of the cold water is Tc,o = 90 C. .................. End of Section II .................. Section III: LONG DESCRIPTIVE QUESTIONS (i) (ii) Marks: 6 x 10 = 60 Answer all Six questions Each question carries Ten marks L-1 A hot liquid waste stream with a flow rate of 4.0 kg/s, an inlet temperature of 80 C, and a specific heat capacity of 4200 J/kg K is utilized in a heat exchanger to recover heat for preheating boiler make-up water. The makeup water enters at 35 C with a flow rate of 3.0 kg/s and the same specific heat capacity of 4200 J/kg K and it is required to leave at 55 C. The heat exchanger has an overall heat transfer coefficient of 850 W/m K and heat losses to the surroundings are assumed negligible. Based on these conditions, determine: a) The rate of heat transfer 2 Marks b) The exit temperature of the waste stream 6 Marks c) The required area of the heat exchanger 2 Marks L-1 1) Heat Transfer Rate (Q): The heat transfer rate can be calculated using the formula: Ans Q = _m * c_m * (T_m,out - T_m,in) Substituting the values: Q = 3 kg/s * 4200 J/kg K * (55 - 35) K Q = 3 * 4200 * 20 = 252,000 J/s = 252 kW Thus, the heat transfer rate is 252 kW. 2) Exit Temperature of the Effluent (Waste Stream): The heat lost by the waste stream is equal to the heat gained by the make-up water. Therefore, we can calculate the exit temperature of the waste stream using the formula: Q = _w * c_w * (T_w,in - T_w,out) Rearranging for T_w,out: T_w,out = T_w,in - (Q / ( _w * c_w)) Substituting the values: T_w,out = 80 C - (252,000 J/s / (4.0 kg/s * 4200 J/kg K)) T_w,out = 80 C - (252,000 / 16,800) T_w,out = 80 C - 15 C T_w,out = 65 C Thus, the exit temperature of the effluent (waste stream) is 65 C. 3) Area of the Heat Exchanger (A): We can use the heat exchanger equation to calculate the area required: Q = U * A * T_m Where T_m is the log mean temperature difference (LMTD). To calculate T_m, we use: T_m = ((T_w,in - T_m,out) - (T_w,out - T_m,in)) / ln((T_w,in - T_m,out) / (T_w,out - T_m,in)) Substitute the values: T_m = ((80 - 55) - (65 - 35)) / ln((80 - 55) / (65 - 35)) T_m = (25 - 30) / ln(25 / 30) T_m = -5 / ln(0.83) T_m = -5 / -0.186 26.88 K Now, we can solve for A: A = Q / (U * T_m) Substituting the values: A = 252,000 / (850 * 26.88) BUREAU OF ENERGY EFFICIENCY 9 A = 252,000 / 22,848 11.03 m Thus, the area of the heat exchanger required is approximately 11.03 m . (Any value between 10.81 m2 to 11.05 m2 full marks shall be awarded) Paper-2 Code : Green L-2 a) An oil-fired boiler is generating 80 TPH of steam at 88% efficiency, operating 300 days in a year. Management has installed a water treatment plant at an investment of Rs. 1.5 crore to reduce the TDS in boiler feed water from 600 ppm to 200 ppm. The maximum permissible limit of TDS in the boiler is 3000 ppm, and the make-up water is 12%. The temperature of blowdown water is 180 C, and the boiler feed water temperature is 50 C. The calorific value of fuel oil is 10,500 Kcal/kg, and the cost of fuel is Rs. 40,000 per ton. Calculate the payback period for the investment in the water treatment plant. 6 Marks b) True or False 4 Marks 1. Saturated steam and dry steam mean the same thing in practical usage. 2. Installing a steam trap upside down has no effect on its operation because condensate is removed due to pressure difference only. 3. If a boiler operates with 8% blowdown at full load, reducing it to 4% will proportionally increase steam generation without additional fuel consumption. 4. A pressure reducing valve (PRV) saves energy by converting high-pressure steam to low-pressure steam with lower enthalpy. L-2 A) Ans Initial Blowdown: Blowdown % = (Feed water TDS * Make-up water % * 100) / (Max permissible TDS in boiler - Feed water TDS) = (600 * 0.12 * 100) / (3000 - 600) = 72 / 2400 = 3% Improved Blowdown: = (Improved Feed water TDS * Make-up water % * 100) / (Max permissible TDS in boiler - Improved Feed water TDS) = (200 * 0.12 * 100) / (3000 - 200) = 24 / 2800 = 0.86% Reduction in Blowdown = Initial Blowdown % - Improved Blowdown % = 3% - 0.86% = 2.14% Reduction in Blowdown = 2.14 * 80 * 1000 / 100 = 1712 kg/hr Heat Savings Calculation: Heat Savings = m * Cp * (T1 - T2) Heat Savings = 1712 * 1 * (180 - 50) = 1712 * 130 = 222,560 kcal/hr Fuel Oil Savings Calculation: Fuel Oil Savings = Heat Savings / (Calorific value of fuel * Boiler Efficiency) Fuel Oil Savings = 222,560 / (10,500 * 0.88) = 222,560 / 9,240 = 24.0866= 24.09 kg/hr Annual Fuel Oil Savings: Annual Fuel Oil Savings = Fuel Oil Savings * 24 * 300 / 1000 Annual Fuel Oil Savings = 24.09 * 24 * 300 / 1000 = 173.44 MT/year Cost Savings Calculation: Fuel Oil Cost Savings = Annual Fuel Oil Savings * Cost per ton of fuel Fuel Oil Cost Savings = 173.44 * 40,000 = Rs.69,37,600= Rs. 69.37 lakh/year Payback Period Calculation: Payback Period = Investment on water treatment plant / Fuel Oil Cost Savings Payback Period = 1,50,00,000 / 69,37,600 = 2.16 years (or 25.9 months) b)True or False 1. Saturated steam and dry steam mean the same thing in practical usage. Answer: False 2. Installing a steam trap upside down has no effect on its operation because condensate is removed due to pressure difference only. Answer: False 3. If a boiler operates with 8% blowdown at full load, reducing it to 4% will proportionally increase steam generation without additional fuel consumption. Answer: False 4. A pressure reducing valve (PRV) saves energy by converting high-pressure steam to low-pressure steam. Answer: False BUREAU OF ENERGY EFFICIENCY 10 Paper-2 Code : Green Fill in the Blanks Each 1 mark L-3 1. In a _________________, heat exchange takes place between the flue gases and the incoming air through metallic or ceramic walls. 2. The ________________stores heat in brickwork during one part of the cycle and releases it to the incoming cold air during the other part of the cycle. 3. A ______________________is a rotating porous disk that transfers heat between two separate air streams. 4. In a ____________, heat transfer occurs via evaporation and condensation of a working fluid inside a sealed container. 5. _____________ in a boiler recovers waste heat from flue gases to preheat the boiler feedwater. 6. A ____________________ uses a series of thin corrugated plates to separate and transfer heat between two fluids. 7. In high-temperature applications where metallic recuperators are unsuitable, __________ tube recuperators can be used to handle gas inlet temperatures up to around 1550 C. 8. The equipment with a direct contact heat exchange principle in a high pressure boiler system is __________________ 9. A ______________ upgrades low-temperature waste heat to a higher temperature using mechanical work. 10. Steam generation from gas turbine waste heat is typically carried out through a __________________ 1. In a recuperator, heat exchange takes place between the flue gases and the incoming air through metallic L-3 or ceramic walls. Ans 2. The regenerator stores heat in brickwork during one part of the cycle and releases it to the incoming cold air during the other part of the cycle. 3. A heat wheel is a rotating porous disk that transfers heat between two separate air streams. 4. In a heat pipe, heat transfer occurs via evaporation and condensation of a working fluid inside a sealed container. 5. Economiser in a boiler recovers waste heat from flue gases to preheat the boiler feedwater. 6. A plate heat exchanger uses a series of thin corrugated plates to separate and transfer heat between two fluids. 7. In high-temperature applications where metallic recuperators are unsuitable, ceramic tube recuperators can be used to handle gas inlet temperatures up to around 1550 C. 8. The equipment with a direct contact heat exchange principle in a high pressure boiler system is deaerator 9. A heat pump upgrades low-temperature waste heat to a higher temperature using mechanical work. 10. Steam generation from gas turbine waste heat is typically carried out through a Heat Recovery Steam Generator (HRSG) A small-scale industry with an old 2-pass gas fired boiler, operating at 65% efficiency is considering replacing L-4 it with a new 3-pass boiler that has an efficiency of 80%. The industry has an average steam load of 8 TPH at 10 kg/cm with steam enthalpy of 665 kcal/kg, and the new boiler is also equipped with an economizer to preheat the feedwater from 35 C to 75 C. The flue gas exit temperature of the old boiler is 160 C, and the new boiler's flue gas exit temperature will be 85 C. The total operating hours for the year are 6,000. The GCV of natural gas is 9,500 kcal/m . Given the above conditions, calculate the following: 1.The annual fuel savings from replacing the old boiler with the new 3-pass boiler 2. Fuel saving due to preheating the feed water. 3. Evaluate the % improvement in boiler evaporation ratio if Natural gas density is 0.68 kg/m 3 6 Marks 2 Marks 2 Marks L-4 1.Annual fuel savings from replacing the old boiler with the new 3-pass boiler: Ans Fuel Consumption Calculation for the Old Boiler Useful Heat Output (Old Boiler): The useful heat output can be calculated using steam flow rate and enthalpy difference: Useful Heat Output = Steam Load (Enthalpy of Steam - Enthalpy of Feed Water) Useful Heat Output = 8,000 kg/h (665 kcal/kg - 35 kcal/kg) = 5,040,000 kcal/h. Heat Input to Old Boiler: The heat input required by the old boiler is given by: Heat Input (Old Boiler) = Useful Heat Output / Efficiency (Old Boiler) Heat Input (Old Boiler) = 5,040,000 / 0.65 = 7,746,153.85 kcal/h. Fuel Consumption (Old Boiler): BUREAU OF ENERGY EFFICIENCY 11 Paper-2 Code : Green To calculate the fuel consumption, we use the GCV of natural gas: Fuel Consumption (Old Boiler) = Heat Input (Old Boiler) / GCV of Natural Gas Fuel Consumption (Old Boiler) = 7,746,153.85 / 9,500 = 815.3 m /h. Fuel Consumption Calculation for the New Boiler Heat Input to New Boiler: The new boiler operates at 80% efficiency, Heat Input (New Boiler) = Useful Heat Output / Efficiency (New Boiler) Useful Heat Output = Steam Load (Enthalpy of Steam - Enthalpy of Feed Water) Useful Heat Output = 8,000 kg/h (665 kcal/kg - 75 kcal/kg) = 4,720,000 kcal/h Heat Input (New Boiler) = 4,720,000 / 0.80 = 5,900,000 kcal/h. Fuel Consumption (New Boiler): Fuel Consumption (New Boiler) = Heat Input (New Boiler) / GCV of Natural Gas Fuel Consumption (New Boiler) = 5,900,000 / 9,500 = 621.1 m3/h Fuel Savings from Replacing the Old Boiler Fuel Savings: Fuel Savings = Fuel Consumption (Old Boiler) - Fuel Consumption (New Boiler) Fuel Savings = 815.3 - 621.1 m3/h= 195.1 m3/h. Annual Fuel Savings: Annual Fuel Savings = Fuel Savings Operating Hours Annual Fuel Savings = 195.1 6,000 = 1,170,600 m3/year. 2. Fuel saving due to preheating the feed water. Temperature difference = 75-35 = 40oC Heat recovery = 8000*1*40 = 3,20,000 kcal/hr Equivalent fuel savings = 3,20,000 /9500 = 33.68 m3/hr 3. % improvement in boiler evaporation ratio if Natural gas density is 0.68 kg/m 3 Evaporation ratio of the old boiler = 8000 / (815.3*0.68) = 14.43 Evaporation ratio of the new boiler = 8000 / (621.1*0.68) = 18.94 % improvement = {(18.94-14.43) / 14.43} *100 = 31.25 % A chemical plant is considering the following two schemes: Scheme 1: A boiler supplies steam for a condensing turbine as well as for the process heat requirement. The condensing turbine produces 1 MW of electric power with an overall efficiency of 33%. The process requires 3 MW of heat, and the boiler efficiency is 80%. Scheme 2: A boiler supplies steam for a back-pressure turbine, which produces 1 MW of electric power with an overall efficiency of 90%. The process heat requirement of 3 MW is met by the back-pressure steam and the boiler efficiency is 75%. Calculate the following: 1.Determine the heat input to the boiler (in MW) and energy utilization factor for both the schemes. 6 Marks 2. The percentage fuel savings achieved by the energy-efficient scheme compared to the other. 4 Marks Scheme 1: L-5 Power produced by condensing power plant = 1MW Ans Heat input to thermal power plant = 1/.33 = 3 MW Process heat requirement = 3 MW Both the thermal requirements come from boiler running at 80% efficiency Hence to produce 6MW of heat, the fuel input to boiler should be 6/.80 = 7.5 MW L-5 Energy Utilization factor = (3 + 1 )/7.5 = 0.53 Scheme 2: Cogeneration plant power production 1 MW BUREAU OF ENERGY EFFICIENCY 12 Paper-2 Code : Green BP overall efficiency is 90%, output of BP turbine is 3 MW Let the boiler output be x, So(x-3)*0.9 = 1MW x= 4.1 MW Boiler runs at 75% efficiency Fuel input = 4.1/ 0.75 = 5.5 MW Energy Utilization factor = (3 + 1 )/5.5 = 0.73 Therefore, scheme -2, would be the efficient scheme. % reduction = (7.5-5.5)/7.5 x100 = 26.66% improvement The operating parameters of the re-heating furnace in a hot rolling mill, both before and after the L-6 improvements, are presented below: Parameter Fuel consumption Furnace oil density Furnace oil is pre-heated from 30 C to 105 C GCV of furnace oil Cost of fuel per ton Exit flue gas temperature after recuperator Specific heat of flue gas Specific heat of steel Billet temperature Present combustion air preheat temperature Average production Average operating hours per day Annual operation days Ambient temperature Oxygen in flue gas Theoretical air requirement Improved Condition: Oxygen in flue gas Combustion air preheat temperature Flue gas temperature after improving recuperator performance Value 2300 litres/hour 0.92 30 C 105 C 10200 kcal/kg Rs.49,000/400 C 0.24 kcal/kg C 0.12 kcal/kg C 1250 C 290 C 650 tonnes/day 12 hours 300 30 C 11 % 14 kg/kg fuel 5% 390 C 340 C Calculate the following: a) Present Specific Energy Consumption (SEC) in litres/ton b) Fuel savings achieved after improvements in recuperator in per tonne of metalc) New Specific Energy Consumption (SEC) in litres/ton d)Annual savings in Rs. Lakhs L-6 Ans Production in TPH SEC (lts/ton) SEC (kg of fuel / ton of metal) Fuel Savings = Equivalent Heat Recovery from recouperator Before Improvements Excess air in Present Condition = 11/(21-11)*100 AAS= 14*(1+1.11) Mass of Flue Gas = AAS+1 Heat Loss= 30.54*0.24*(400-30) 1 Mark 6 Marks 1 Mark 2 Marks 54.17 42.46 39.06 110 29.54 30.54 2711.952 % kg /kg fuel kg /kg fuel kCal/kg Fuel After Improvements Excess air in Present Condition = {5/(21-5)}*100 AAS= 14*(1+0.3125) BUREAU OF ENERGY EFFICIENCY 31.25 18.375 % kg /kg fuel 13 Mass of Flue Gas = AAS+1 Heat Loss = 19.375*0.24*(340-30) Heat Loss Reduction/Heat Recovery % heat recovery = 1270.45/10200 Equivalent Fuel Quantity Savings = 2300*0.92*0.1246 Fuel Savings per Tonne of Metal New Fuel Consumption Paper-2 Code : Green 19.375 kg /kg fuel 1441.5 kCal/kg Fuel 1270.452 12.46 263.56 4.87 1852.44 New SEC (lts/Hr) = 1852.44/0.92/54.17 Annual Savings Difference in fuel consumption Density of furnace oil Furnace oil cost Cost per liter Average production Average operating hours per day Daily fuel savings Annual Savings kCal/kg Fuel % kg/Hr kg/ Ton of Metal kg/Hr 37.17 lts/ton 5.29 0.92 49000.00 45.08 650 12 1.55 465 lts/ton Rs/ton Rs/liter tonnes/day hours Rs.Lakh/Day Lakhs ................. End of Section III .................. BUREAU OF ENERGY EFFICIENCY 14 Paper-2 Code : Pink 25th NATIONAL EXAMINATION FOR CERTIFICATION OF ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER, 2025 PAPER - 2 : ENERGY EFFICIENCY IN THERMAL UNTILITIES Date : 27.09.2025 Timings : 14:00-17:00 HRS Duration : 3 HRS Max. Marks : 150 General instructions : o o o o o Please check that this question paper contains 8 printed pages Please check that this question paper contains 64 questions The question paper is divided into three sections All questions in all three sections are compulsory All parts of a question should be answered at one place Section I: OBJECTIVE TYPE Marks: 50 x 1 = 50 (i) Answer all 50 questions (ii) Each question carries one mark (iii) Please s h a d e t h e a p p r o p r i a t e oval in SECTION-I of MAIN ANSWER BOOKLET with BLUE/BLACK ball point pen 1. 2. The flash point of an oil is the: A) Temperature at which the oil ignites spontaneously B) Lowest temperature at which the oil vapour momentarily ignites on application of a flame C) Temperature at which the oil starts boiling D) Temperature at which the oil burns continuously Which type of fuel requires lowest amount of excess air for combustion? A)Furnace oil B) LDO C) Propane D) Rice Husk 3. Which among the following does not release any energy during combustion? A) Carbon B) Sulphur C) Nitrogen D) Hydrogen 4. Which is the best suited pump for pumping viscous liquid fuel? A) Centrifugal pump B) Gear pump C) Vertical turbine pump D) None of the above 5. In a water tube boiler, what is the primary function of the boiler drum? a) To superheat the steam leaving the economizer b) To separate steam from water and provide storage for steam and water mixture c) To maintain draft inside the furnace by balancing air and flue gas pressure d) To preheat the feedwater before entering the boiler tubes Dearation in boilers is primarily carried out to remove: a) Dissolved solids b) Dissolved gases c) Suspended particles d) Boiler scale Which of the following best describes the critical point of steam? A) The point where steam condenses to water at standard atmospheric pressure B) The highest temperature and pressure at which liquid water and steam can coexist in equilibrium C) The temperature at which steam becomes superheated D) The point where latent heat of vaporization is maximum Which of the following is preferable for a heating process? a) Wet steam b) flash steam c) High pressure steam d) Dry saturated steam 6. 7. 8. 9. A float steam trap is an example of which type of steam trap? A) Thermodynamic B) Mechanical C) Thermostatic D) Hydraulic BUREAU OF ENERGY EFFICIENCY 1 Paper-2 Code : Pink 10. Which of the following is commonly used as a low-temperature insulating material? A) Calcium silicate B) Polyurethane C) Magnesia D) Asbestos 11. The economic thickness of insulation is the thickness at which: A) Heat loss is zero B) Cost of insulation is minimum C) Combined cost (heat loss cost and insulation cost) is minimum D) Heat transfer rate is maximum 12. Which of the following is a key advantage of Fluidized Bed Combustion (FBC) over conventional combustion systems? a) Lower power requirement for fans b) Reduced emissions of SO and NOx c) Higher excess air requirement d) Lower combustion efficiency 13. In a fluidized bed combustion system, the fuel is burned in a bed of: A) Moving metal plates B) Sand, ash, or other granular material suspended by air flow C) Rotating drums D) Water-cooled tubes only 14. In a combined cycle power plant, the waste heat from the gas turbine is used to: A) Preheat the incoming air to the gas turbine B) Generate steam for a steam turbine C) Cool the exhaust gases directly to the atmosphere D) Operate a diesel generator 15. The effectiveness of a heat exchanger is defined as the ratio of: A) Actual heat transfer to the maximum possible heat transfer B) Heat loss to the surroundings to the total heat input C) Outlet temperature difference to inlet temperature difference D) Actual heat transfer to the total mass flow rate 16. In pinch analysis, the "pinch point" represents: A) The location in the heat exchanger where fouling is maximum B) The point of minimum temperature difference between hot and cold streams C) The point where heat transfer rate is maximum D) The location where the heat exchanger pressure drop is minimum 17. The presence of high sulphur in fuels mainly contributes to: A) Reduction in NOx emissions B) Corrosion and air pollution C) Decrease in net calorific value D) Prevention of slag formation 18. Why is viscosity of liquid fuels important in combustion systems? A) It determines the fuel s sulphur content B) It affects atomization and burner performance C) It indicates the ash fusion temperature D) It measures the fuel s calorific value 19. In boilers, natural draft is produced by: A) A steam ejector B) The height and temperature difference in the chimney C) A forced draft fan D) Vacuum pump 20. As per Indian Boiler Regulations (IBR), a pipe is defined as a steam pipe if it carries steam at a pressure exceeding 1 Mark Awarded to all candidates who attempted this question 21. Which of the following is an example of internal water treatment in boilers? A) Deaeration B) Coagulation and filtration C) Sodium Phosphate dosing D) Clarification BUREAU OF ENERGY EFFICIENCY 2 Paper-2 Code : Pink 22. In boiler water treatment, Reverse Osmosis (RO) is primarily used to: A) Remove only suspended solids from water B) Remove dissolved salts and impurities C) Increase the alkalinity of feedwater D) Convert hard water into soft water 23. Which of the following is not a benefit of condensate recovery in a steam system? 1 Mark Awarded to all candidates who attempted this question 24. When steam passes through a pressure reducing valve (PRV), its enthalpy: A) Increases B) Decreases C) Remains the same D) Becomes zero 25. A thermocompressor in a steam system is primarily used to: A) Convert wet steam into dry saturated steam B) Recompress low-pressure steam with high-pressure steam to obtain medium-pressure steam C) Increase boiler steam generation rate D) Reduce steam temperature without affecting pressure 26. Higher excess air in an oil-fired furnace leads to: A) Higher efficiency B) Lower flue gas heat loss C) Zero stack losses D) None of the above 27. The main benefit of using preheated air for combustion is: A) Reduced furnace temperature B) Increased fuel savings C) Increased excess air requirement D) Higher flue gas temperature 28. Choose the correct statement about insulating materials: A) Insulating materials have high thermal conductivity B) Insulating materials reduce heat loss C) Insulating materials increase heat transfer rate D) All of the above 29. Choose the incorrect statement about refractories: A) Refractories are used to withstand high temperatures in furnaces B) Refractories should have low thermal conductivity for insulation purposes C) Refractories should have low melting points for easy shaping D) Refractories must be resistant to thermal shock 30. Which of the following defines the Gross Calorific Value (GCV) of a fuel? a) Heat liberated after complete combustion including latent heat of water vapor b) Heat available after combustion excluding latent heat of water vapor c) Heat required to raise 1 kg of fuel by 1 C d) Ratio of heat released to air-fuel ratio 31. Which of the following statements about a fluidized bed boiler is correct? A) It can efficiently burn low-grade fuels with high ash content B) It requires very high combustion temperatures (above 1600 C) C) It operates without any bed material D) It cannot be used for biomass fuels BUREAU OF ENERGY EFFICIENCY 3 Paper-2 Code : Pink 32. An extraction condensing turbine is designed to: A) Exhaust all steam to the process at high pressure B) Extract some steam for process use and balance steam for power generation C) Operate only as a back pressure turbine D) Exhaust steam directly to atmosphere 33. The main advantages of a cogeneration system include: A) Higher overall efficiency B) Simultaneous production of power and heat C) Reduced fuel consumption D) All of the above 34. The relationship between heat rate and plant efficiency is such that: A) Higher heat rate means higher efficiency B) Lower heat rate means higher efficiency C) Heat rate and efficiency are unrelated D) Both increases together 35. A metallic radiation recuperator is primarily used to: A) Recover heat from flue gases to preheat combustion air B) Convert radiant heat into electricity C) Cool furnace walls to prevent overheating D) Measure furnace temperature using radiation 36. Select the correct statement about the benefits of waste heat recovery: A) It increases fuel consumption B) It reduces overall plant efficiency C) It saves energy and reduces operating costs D) It always requires high-grade heat sources 37. Terminal temperature difference (TTD) in a steam to water heat exchanger is: A) Inlet temp. difference B) Outlet temp. difference C) Difference of Steam Saturation temperature and outlet water temperature. D) Avg. temperature difference 38. Select the correct statement about flash steam: A) It is produced by cooling steam below saturation temperature B) It is produced by cooling steam above saturation temperature C) It forms when hot condensate is released to a lower pressure D) It forms when hot condensate is released to a higher pressure 39. The specific gravity of a fuel is the ratio of: A) Weight of fuel to its volume B) Density of fuel to the density of water C) Volume of fuel to volume of water D) Mass of fuel to its calorific value 40. Select the correct statement about volatile matter in coal: A) It reduces flame stability in combustion B) It consists of gases released when coal is heated C) It is the same as fixed carbon D) It increases the ash content of coal 41. The Net Calorific Value (NCV) of a fuel is obtained by: A) Subtracting the heat of vaporization of water from the Gross Calorific Value B) Adding the heat of vaporization of water to the Gross Calorific Value C) Measuring only the sensible heat of combustion products D) Ignoring latent heat losses in the fuel 42. Which of the following agro fuels typically has the highest moisture content? A) Saw dust B) Paddy husk BUREAU OF ENERGY EFFICIENCY 4 Paper-2 Code : Pink C) De-oiled bran D) Coconut shells 43. Select the correct statement about excess air in combustion: A) Excess air always increases boiler efficiency B) Excess air is needed to ensure complete combustion of fuel C) Excess air reduces flue gas losses D) Excess air is less than the theoretical air 44. In a pulverised fuel boiler, the fuel is burned in : A) lump form B) fluidised bed form C) fine powder form D) None of the above 45. Select the correct statement about a supercritical boiler: A) It generates steam at pressures above the critical point B) It always requires a steam drum for separation of steam and water C) It operates only with natural circulation D) It produces steam with high moisture content 46. A thermic fluid heater is primarily used to: A) Generate steam at high pressure B) Heat a mineral oil C) Produce hot water D) Produce hot air 47. A cupola furnace is primarily used for: A) Producing cast iron B) Producing steel ingots C) Heating non-ferrous metals D) Producing coke 48. Select the correct statement about flue gas losses in a reheating furnace: A) Flue gas losses decrease with higher flue gas temperature B) Flue gas losses increase with excess air and higher exhaust temperature C) Flue gas losses are unaffected by excess air D) All of the above 49. The principle of cogeneration is based on: A) Using separate systems to produce heat and power B) Sequential use of energy to produce both electricity and useful heat from the same fuel C) Converting all heat into electricity D) Using waste heat only for cooling 50. Trigeneration refers to the simultaneous production of: A) Power, heat, and cold B) Power, heat, and steam C) Power, steam, and compressed air D) Heat, cold, and compressed air .................. End of Section I .................. BUREAU OF ENERGY EFFICIENCY 5 Paper-2 Code : Pink Marks: 8 x 5 = 40 Section II: SHORT DESCRIPTIVE QUESTIONS (i) (ii) Answer all Eight questions Each question carries Five marks S-1 A 10-TPH forced-draft boiler is tested at a load of 8 TPH. Fuel flow is 0.53 TPH and the fuel GCV is 10,000 kcal/kg. Feedwater temperature is 70 C. The boiler delivers saturated steam at 10 bar(g) at 186 C. Take the specific enthalpy of saturated steam has 664 kcal/kg. a) Calculate the direct efficiency assuming dry saturated steam. 3 Marks b) The indirect method efficiency is 87%, which does not match with the direct method. The auditor claims this difference is due to the wetness of steam. Establish the auditor s claim by calculating the dryness fraction. 2 Marks S-1 Fuel flow to boiler = 0.53 TPH Steam flow = 8TPH Ans Fuel GCV = 10000 kcal/kg Steam enthalpy = 664 kcal/kg Feed water enthalpy = 70 kcal/kg Direct efficiency = 8 (664 70) = 89.7% .53 10000 Indirect efficiency = 87% Moisture content/ Steam quality: If the steam is slightly wet, the actual enthalpy is lower, reducing the direct efficiency closer to the indirect efficiency. Total heat content of wet steam = sensible heat + (dryness fraction * latent heat) = 186 + ( x* (664-186)) = 186+478x = 8 [(186 + 478x) 70] = .87 . 53 10000 X= 96.3%. Hence steam is not fully dry in nature. This is the reason for difference in efficiency A boiler has an air preheater whose design and actual conditions are given below: S-2 Design (0C) Actual (0C) Flue gas Inlet temperature to APH 330 330 Flue gas outlet temperature from APH 180 210 Air inlet temperature to APH 40 40 Air outlet temperature from APH 190 190 Assume the following: specific heats of flue gas and air are equal, there is no heat loss to the surroundings, and as per design the mass of flue gas is approximately to the mass of air. Analyze and discuss the deviations between actual operating condition and design condition in air mass flow, flue-gas mass flow and APH heat transfer. S-2 Ans Cp FG = Cp Air m FG = m Air Design (TFGout TFG in) = (330-180) = 150 (Tair out Tair in) = (190-40) = 150 Heat balance exists Actual (TFGout TFG in) = (330-210) = 120 (Tair out Tair in) = (190-40) = 150 Heat is not balanced. Implications: a) Air mass flow: Likely lower than design (or part of the air is bypassing the APH). When less air is heated, temperature of outlet air can still reach 190 C, but the gas won t cool as per design. BUREAU OF ENERGY EFFICIENCY 6 Paper-2 Code : Pink b) Flue-gas mass flow: Effectively higher than air by ~25% at the APH (or flue-gas ingress/leakage upstream of APH inflates the gas stream). This drives the exit flue-gas hotter than the design (210 C vs 180 C). c) APH heat transfer: Heat recovery is less than the design as evidenced by the higher gas outlet temperature (less gas-side cooling). If the insulation thickness on a steam pipe is increased from 50 mm to 100 mm, calculate the percentage S-3 reduction in heat loss. The surface temperature of the insulation is 70 C for 50 mm thickness and 60 C for 100 mm thickness. Ambient temperature is 35 C. S = [10 + (Ts Ta)/20] (Ts Ta) S-3 Calculations Ans 1) Heat loss for 50 mm insulation: T1 = Ts1 Ta = 70 35 = 35 C S50 = [10 + (35/20)] 35 = (10 + 1.75) 35 = 11.75 35 = 411.25 kcal/hr/m2 2) Heat loss for 100 mm insulation: T2 = Ts2 Ta = 60 35 = 25 C S100 = [10 + (25/20)] 25 = (10 + 1.25) 25 = 11.25 25 = 281.25 kcal/hr/m2 Note : Since the diameter of the pipe is not mentioned in the question paper. If the candidate has calculated the rate of heat loss for 50 mm and 100 mm insulation full marks may be awarded. S-4 In a food processing industry, steam at 15 kg/cm was used to heat 10 TPH of milk from 40 C to 80 C. An energy auditor suggests installing a PRV to reduce steam pressure from 15 kg/cm to 3 kg/cm . Specific heat of milk is 1.0 kcal/kg/oC. Present Condition (Steam Pressure 15 kg/cm ) Proposed Condition (Steam Pressure 3 kg/cm ) Sensible heat 200.6 kcal/kg 133.287 kcal/kg Latent Heat 465.72 kcal/kg 517.17 kcal/kg 0.9 ? Steam Parameters Dryness Fraction a) Determine the outlet moisture percentage when steam passes through the PRV. b) Calculate the steam savings per hour due to this measure. S-4 Ans 3 Marks 2 Marks a) Total heat content before PRV = Total heat content after PRV 200.6 + (0.9*465.72) =133.287 +(x *517.17) x = 0.94 b) Heat required for water = (10*1000)*1*(80-40) = 400000 kcal/hr Steam requirement at present condition= 400000/ (465.72*0.9) = 954.3 kg/hr Steam requirement at proposed condition = 400000/ (517.17*.94) = 822.8 kg/hr Savings = 131.5 kg/hr of steam S-5 In an oil-to-water counterflow heat exchanger, hot oil enters at 150 C and leaves at 90 C. Cold water enters at 30 C and leaves at an unknown temperature Tc,o. The log mean temperature difference (LMTD) for the exchanger is 60 C. Calculate the exit temperature of the water. S-5 Given: Hot oil inlet temperature, Th,i = 150 C Ans Hot oil outlet temperature, Th,o = 90 C Cold water inlet temperature, Tc,i = 30 C Cold water outlet temperature, Tc,o = ? LMTD = 60 C For counterflow heat exchangers: T1 = Th,i Tc,o = 150 Tc,o T2 = Th,o Tc,i = 90 30 = 60 LMTD is given by: LMTD = ( T1 T2) / ln( T1 / T2) = 60 Let T1 = A. Then: (A 60) / ln(A / 60) = 60 BUREAU OF ENERGY EFFICIENCY 7 This equation holds true when A = 60 (equal terminal differences case). Therefore: 150 Tc,o = 60 Tc,o = 90 C Answer The exit temperature of the cold water is Tc,o = 90 C. S-6 Paper-2 Code : Pink A batch furnace uses LPG and draws combustion air at 30 C from the ambient condition. An energy auditor proposes fitting a recuperator to use hot flue gas (950 C) to preheat this air to 400 C. a) Find the % fuel savings from preheating the air to 400 C. b) Find the new LPG consumption after retrofit. c) Calculate the flue-gas exit temperature from the recuperator using an energy balance. Given: S-6 EA = 20%, Stoichiometric air = 15.5 kg air/kg LPG, cp (air & flue gas) = 0.24 kcal/kg C. Ans Actual air (AAS) = (1 + EA) 15.5 = 1.2 15.5 = 18.6 kg/kg fuel. Flue gas per kg fuel: mfg = 1 + AAS = 1 + 18.6 = 19.6 kg/kg fuel. 2 Marks 2 Marks 1 Mark (a) % Fuel savings: Recovered heat to air per kg fuel: P = mair cp (Tair,out Tair,in) = 18.6 0.24 (400 30) = 1651.68 kcal/kg fuel Savings = P / GCV = 1651.68 / 11,500 = 0.1436 14.36 % (b) New LPG consumption: mnew = mold (1 Savings) = 50 (1 0.1436) 42.8 kg/h (c) Flue-gas exit temperature from recuperator (energy balance): mfg x cp x (Tfg,in Tfg,out) = mair x cp x (Tair,out Tair,in) 19.6 (950 Tfg,out) = 18.6 370 = 6882 950 Tfg,out = 6882 / 19.6 351.12 Tfg,out 950 351.12 = 599 C S-7 Match the Following: Term 1. Topping Cycle 2. Bottoming Cycle 3. Heat-to-Power Ratio 4. Combined Cycle Plant 5. Trigeneration Description a. Produces power first, then recovers heat for process use b. Produces power, heat, and cooling from the same energy source c. Produces heat first, then uses waste heat to generate power d. Useful thermal energy to electrical energy output e. Uses gas turbine and steam turbine in series S-7 1 a Ans 2 c 3 d 4 e 5 b S-8 Fill in the Blanks: 1. A furnace oil consumption of 250 liters/day, with density 0.96 kg/l and GCV 10,200 kcal/kg, provides a total heat input of __________ kcal/day. 2. If the NCV of a fuel is 9,500 kcal/kg and the mass of fuel burned is 500 kg, the total heat energy released is __________ MWh 3. For complete combustion of 1 kg of carbon, the theoretical air requirement is __________ kg. 4. Calculate % of nitrogen in dry flue gas if 1 kg of hydrogen is completely burned in presence of 34.8 kg of air. 5. A coal sample with a GCV of 4,500 kcal/kg and moisture content of 10% has an approximate NCV of __________ kcal/kg, assuming latent heat of vaporization of water = 587 kcal/kg. BUREAU OF ENERGY EFFICIENCY 8 Paper-2 Code : Pink S-8 Ans 1. 250 L/day 0.96 kg/L = 240 kg/day, Heat input = 240 10,200 = 2,448,000 kcal/day 2. Energy = 9,500 500 /(860 *1000) = 5.52 MWh 3. 11.6 kg of theoretical air per kg of carbon 4. 100% 5. NCV = GCV (Moisture fraction 587) = 4,500 (0.10 587) = 4,441 kcal/kg .................. End of Section II .................. Section III: LONG DESCRIPTIVE QUESTIONS (i) (ii) Marks: 6 x 10 = 60 Answer all Six questions Each question carries Ten marks L-1 A small-scale industry with an old 2-pass gas fired boiler, operating at 65% efficiency is considering replacing it with a new 3-pass boiler that has an efficiency of 80%. The industry has an average steam load of 8 TPH at 10 kg/cm with steam enthalpy of 665 kcal/kg, and the new boiler is also equipped with an economizer to preheat the feedwater from 35 C to 75 C. The flue gas exit temperature of the old boiler is 160 C, and the new boiler's flue gas exit temperature will be 85 C. The total operating hours for the year are 6,000. The GCV of natural gas is 9,500 kcal/m . Given the above conditions, calculate the following: 1.The annual fuel savings from replacing the old boiler with the new 3-pass boiler 2. Fuel saving due to preheating the feed water. 3. Evaluate the % improvement in boiler evaporation ratio if Natural gas density is 0.68 kg/m 3 6 Marks 2 Marks 2 Marks L-1 1.Annual fuel savings from replacing the old boiler with the new 3-pass boiler: Ans Fuel Consumption Calculation for the Old Boiler Useful Heat Output (Old Boiler): The useful heat output can be calculated using steam flow rate and enthalpy difference: Useful Heat Output = Steam Load (Enthalpy of Steam - Enthalpy of Feed Water) Useful Heat Output = 8,000 kg/h (665 kcal/kg - 35 kcal/kg) = 5,040,000 kcal/h. Heat Input to Old Boiler: The heat input required by the old boiler is given by: Heat Input (Old Boiler) = Useful Heat Output / Efficiency (Old Boiler) Heat Input (Old Boiler) = 5,040,000 / 0.65 = 7,746,153.85 kcal/h. Fuel Consumption (Old Boiler): To calculate the fuel consumption, we use the GCV of natural gas: Fuel Consumption (Old Boiler) = Heat Input (Old Boiler) / GCV of Natural Gas Fuel Consumption (Old Boiler) = 7,746,153.85 / 9,500 = 815.3 m /h. Fuel Consumption Calculation for the New Boiler Heat Input to New Boiler: The new boiler operates at 80% efficiency, Heat Input (New Boiler) = Useful Heat Output / Efficiency (New Boiler) Useful Heat Output = Steam Load (Enthalpy of Steam - Enthalpy of Feed Water) Useful Heat Output = 8,000 kg/h (665 kcal/kg - 75 kcal/kg) = 4,720,000 kcal/h Heat Input (New Boiler) = 4,720,000 / 0.80 = 5,900,000 kcal/h. Fuel Consumption (New Boiler): Fuel Consumption (New Boiler) = Heat Input (New Boiler) / GCV of Natural Gas Fuel Consumption (New Boiler) = 5,900,000 / 9,500 = 621.1 m3/h BUREAU OF ENERGY EFFICIENCY 9 Paper-2 Code : Pink Fuel Savings from Replacing the Old Boiler Fuel Savings: Fuel Savings = Fuel Consumption (Old Boiler) - Fuel Consumption (New Boiler) Fuel Savings = 815.3 - 621.1 m3/h= 195.1 m3/h. Annual Fuel Savings: Annual Fuel Savings = Fuel Savings Operating Hours Annual Fuel Savings = 195.1 6,000 = 1,170,600 m3/year. 2. Fuel saving due to preheating the feed water. Temperature difference = 75-35 = 40oC Heat recovery = 8000*1*40 = 3,20,000 kcal/hr Equivalent fuel savings = 3,20,000 /9500 = 33.68 m3/hr 3. % improvement in boiler evaporation ratio if Natural gas density is 0.68 kg/m 3 Evaporation ratio of the old boiler = 8000 / (815.3*0.68) = 14.43 Evaporation ratio of the new boiler = 8000 / (621.1*0.68) = 18.94 % improvement = {(18.94-14.43) / 14.43} *100 = 31.25 % A chemical plant is considering the following two schemes: Scheme 1: A boiler supplies steam for a condensing turbine as well as for the process heat requirement. The condensing turbine produces 1 MW of electric power with an overall efficiency of 33%. The process requires 3 MW of heat, and the boiler efficiency is 80%. Scheme 2: A boiler supplies steam for a back-pressure turbine, which produces 1 MW of electric power with an overall efficiency of 90%. The process heat requirement of 3 MW is met by the back-pressure steam and the boiler efficiency is 75%. Calculate the following: 1.Determine the heat input to the boiler (in MW) and energy utilization factor for both the schemes. 6 Marks 2. The percentage fuel savings achieved by the energy-efficient scheme compared to the other. 4 Marks Scheme 1: L-2 Power produced by condensing power plant = 1MW Ans Heat input to thermal power plant = 1/.33 = 3 MW Process heat requirement = 3 MW Both the thermal requirements come from boiler running at 80% efficiency Hence to produce 6MW of heat, the fuel input to boiler should be 6/.80 = 7.5 MW L-2 Energy Utilization factor = (3 + 1 )/7.5 = 0.53 Scheme 2: Cogeneration plant power production 1 MW BP overall efficiency is 90%, output of BP turbine is 3 MW Let the boiler output be x, So(x-3)*0.9 = 1MW x= 4.1 MW Boiler runs at 75% efficiency Fuel input = 4.1/ 0.75 = 5.5 MW Energy Utilization factor = (3 + 1 )/5.5 = 0.73 Therefore, scheme -2, would be the efficient scheme. % reduction = (7.5-5.5)/7.5 x100 = 26.66% improvement BUREAU OF ENERGY EFFICIENCY 10 Paper-2 Code : Pink The operating parameters of the re-heating furnace in a hot rolling mill, both before and after the L-3 improvements, are presented below: Parameter Fuel consumption Furnace oil density Furnace oil is pre-heated from 30 C to 105 C GCV of furnace oil Cost of fuel per ton Exit flue gas temperature after recuperator Specific heat of flue gas Specific heat of steel Billet temperature Present combustion air preheat temperature Average production Average operating hours per day Annual operation days Ambient temperature Oxygen in flue gas Theoretical air requirement Improved Condition: Oxygen in flue gas Combustion air preheat temperature Flue gas temperature after improving recuperator performance Value 2300 litres/hour 0.92 30 C 105 C 10200 kcal/kg Rs.49,000/400 C 0.24 kcal/kg C 0.12 kcal/kg C 1250 C 290 C 650 tonnes/day 12 hours 300 30 C 11 % 14 kg/kg fuel 5% 390 C 340 C Calculate the following: a) Present Specific Energy Consumption (SEC) in litres/ton b) Fuel savings achieved after improvements in recuperator in per tonne of metalc) New Specific Energy Consumption (SEC) in litres/ton d)Annual savings in Rs. Lakhs L-3 Ans Production in TPH SEC (lts/ton) SEC (kg of fuel / ton of metal) Fuel Savings = Equivalent Heat Recovery from recouperator Before Improvements Excess air in Present Condition = 11/(21-11)*100 AAS= 14*(1+1.11) Mass of Flue Gas = AAS+1 Heat Loss= 30.54*0.24*(400-30) 1 Mark 6 Marks 1 Mark 2 Marks 54.17 42.46 39.06 110 29.54 30.54 2711.952 % kg /kg fuel kg /kg fuel kCal/kg Fuel 31.25 18.375 19.375 1441.5 % kg /kg fuel kg /kg fuel kCal/kg Fuel 1270.452 kCal/kg Fuel After Improvements Excess air in Present Condition = {5/(21-5)}*100 AAS= 14*(1+0.3125) Mass of Flue Gas = AAS+1 Heat Loss = 19.375*0.24*(340-30) Heat Loss Reduction/Heat Recovery % heat recovery = 1270.45/10200 Equivalent Fuel Quantity Savings = 2300*0.92*0.1246 Fuel Savings per Tonne of Metal BUREAU OF ENERGY EFFICIENCY 12.46 263.56 4.87 % kg/Hr kg/ Ton of Metal 11 New Fuel Consumption New SEC (lts/Hr) = 1852.44/0.92/54.17 Paper-2 Code : Pink 1852.44 kg/Hr 37.17 lts/ton Annual Savings Difference in fuel consumption 5.29 lts/ton Density of furnace oil 0.92 Furnace oil cost 49000.00 Rs/ton Cost per liter 45.08 Rs/liter Average production 650 tonnes/day Average operating hours per day 12 hours Daily fuel savings 1.55 Rs.Lakh/Day Annual Savings 465 Lakhs L-4 A hot liquid waste stream with a flow rate of 4.0 kg/s, an inlet temperature of 80 C, and a specific heat capacity of 4200 J/kg K is utilized in a heat exchanger to recover heat for preheating boiler make-up water. The makeup water enters at 35 C with a flow rate of 3.0 kg/s and the same specific heat capacity of 4200 J/kg K and it is required to leave at 55 C. The heat exchanger has an overall heat transfer coefficient of 850 W/m K and heat losses to the surroundings are assumed negligible. Based on these conditions, determine: a) The rate of heat transfer 2 Marks b) The exit temperature of the waste stream 6 Marks c) The required area of the heat exchanger 2 Marks L-4 1) Heat Transfer Rate (Q): The heat transfer rate can be calculated using the formula: Ans Q = _m * c_m * (T_m,out - T_m,in) Substituting the values: Q = 3 kg/s * 4200 J/kg K * (55 - 35) K Q = 3 * 4200 * 20 = 252,000 J/s = 252 kW Thus, the heat transfer rate is 252 kW. 2) Exit Temperature of the Effluent (Waste Stream): The heat lost by the waste stream is equal to the heat gained by the make-up water. Therefore, we can calculate the exit temperature of the waste stream using the formula: Q = _w * c_w * (T_w,in - T_w,out) Rearranging for T_w,out: T_w,out = T_w,in - (Q / ( _w * c_w)) Substituting the values: T_w,out = 80 C - (252,000 J/s / (4.0 kg/s * 4200 J/kg K)) T_w,out = 80 C - (252,000 / 16,800) T_w,out = 80 C - 15 C T_w,out = 65 C Thus, the exit temperature of the effluent (waste stream) is 65 C. 3) Area of the Heat Exchanger (A): We can use the heat exchanger equation to calculate the area required: Q = U * A * T_m Where T_m is the log mean temperature difference (LMTD). To calculate T_m, we use: T_m = ((T_w,in - T_m,out) - (T_w,out - T_m,in)) / ln((T_w,in - T_m,out) / (T_w,out - T_m,in)) Substitute the values: T_m = ((80 - 55) - (65 - 35)) / ln((80 - 55) / (65 - 35)) T_m = (25 - 30) / ln(25 / 30) T_m = -5 / ln(0.83) T_m = -5 / -0.186 26.88 K Now, we can solve for A: A = Q / (U * T_m) Substituting the values: A = 252,000 / (850 * 26.88) A = 252,000 / 22,848 11.03 m Thus, the area of the heat exchanger required is approximately 11.03 m . (Any value between 10.81 m2 to 11.05 m2 full marks shall be awarded) BUREAU OF ENERGY EFFICIENCY 12 Paper-2 Code : Pink a) L-5 An oil-fired boiler is generating 80 TPH of steam at 88% efficiency, operating 300 days in a year. Management has installed a water treatment plant at an investment of Rs. 1.5 crore to reduce the TDS in boiler feed water from 600 ppm to 200 ppm. The maximum permissible limit of TDS in the boiler is 3000 ppm, and the make-up water is 12%. The temperature of blowdown water is 180 C, and the boiler feed water temperature is 50 C. The calorific value of fuel oil is 10,500 Kcal/kg, and the cost of fuel is Rs. 40,000 per ton. Calculate the payback period for the investment in the water treatment plant. 6 Marks b) True or False 4 Marks 1. Saturated steam and dry steam mean the same thing in practical usage. 2. Installing a steam trap upside down has no effect on its operation because condensate is removed due to pressure difference only. 3. If a boiler operates with 8% blowdown at full load, reducing it to 4% will proportionally increase steam generation without additional fuel consumption. 4. A pressure reducing valve (PRV) saves energy by converting high-pressure steam to low-pressure steam with lower enthalpy. L-5 A) Ans Initial Blowdown: Blowdown % = (Feed water TDS * Make-up water % * 100) / (Max permissible TDS in boiler - Feed water TDS) = (600 * 0.12 * 100) / (3000 - 600) = 72 / 2400 = 3% Improved Blowdown: = (Improved Feed water TDS * Make-up water % * 100) / (Max permissible TDS in boiler - Improved Feed water TDS) = (200 * 0.12 * 100) / (3000 - 200) = 24 / 2800 = 0.86% Reduction in Blowdown = Initial Blowdown % - Improved Blowdown % = 3% - 0.86% = 2.14% Reduction in Blowdown = 2.14 * 80 * 1000 / 100 = 1712 kg/hr Heat Savings Calculation: Heat Savings = m * Cp * (T1 - T2) Heat Savings = 1712 * 1 * (180 - 50) = 1712 * 130 = 222,560 kcal/hr Fuel Oil Savings Calculation: Fuel Oil Savings = Heat Savings / (Calorific value of fuel * Boiler Efficiency) Fuel Oil Savings = 222,560 / (10,500 * 0.88) = 222,560 / 9,240 = 24.0866= 24.09 kg/hr Annual Fuel Oil Savings: Annual Fuel Oil Savings = Fuel Oil Savings * 24 * 300 / 1000 Annual Fuel Oil Savings = 24.09 * 24 * 300 / 1000 = 173.44 MT/year Cost Savings Calculation: Fuel Oil Cost Savings = Annual Fuel Oil Savings * Cost per ton of fuel Fuel Oil Cost Savings = 173.44 * 40,000 = Rs.69,37,600= Rs. 69.37 lakh/year Payback Period Calculation: Payback Period = Investment on water treatment plant / Fuel Oil Cost Savings Payback Period = 1,50,00,000 / 69,37,600 = 2.16 years (or 25.9 months) b)True or False 1. Saturated steam and dry steam mean the same thing in practical usage. Answer: False 2. Installing a steam trap upside down has no effect on its operation because condensate is removed due to pressure difference only. Answer: False 3. If a boiler operates with 8% blowdown at full load, reducing it to 4% will proportionally increase steam generation without additional fuel consumption. Answer: False 4. A pressure reducing valve (PRV) saves energy by converting high-pressure steam to low-pressure steam. Answer: False Fill in the Blanks Each 1 mark L-6 1. In a _________________, heat exchange takes place between the flue gases and the incoming air through metallic or ceramic walls. 2. The ________________stores heat in brickwork during one part of the cycle and releases it to the incoming cold air during the other part of the cycle. 3. A ______________________is a rotating porous disk that transfers heat between two separate air streams. 4. In a ____________, heat transfer occurs via evaporation and condensation of a working fluid inside a sealed BUREAU OF ENERGY EFFICIENCY 13 5. 6. 7. 8. L-6 Ans Paper-2 Code : Pink container. _____________ in a boiler recovers waste heat from flue gases to preheat the boiler feedwater. A ____________________ uses a series of thin corrugated plates to separate and transfer heat between two fluids. In high-temperature applications where metallic recuperators are unsuitable, __________ tube recuperators can be used to handle gas inlet temperatures up to around 1550 C. The equipment with a direct contact heat exchange principle in a high pressure boiler system is __________________ 9. A ______________ upgrades low-temperature waste heat to a higher temperature using mechanical work. 10. Steam generation from gas turbine waste heat is typically carried out through a __________________ 1. In a recuperator, heat exchange takes place between the flue gases and the incoming air through metallic or ceramic walls. 2. The regenerator stores heat in brickwork during one part of the cycle and releases it to the incoming cold air during the other part of the cycle. 3. A heat wheel is a rotating porous disk that transfers heat between two separate air streams. 4. In a heat pipe, heat transfer occurs via evaporation and condensation of a working fluid inside a sealed container. 5. Economiser in a boiler recovers waste heat from flue gases to preheat the boiler feedwater. 6. A plate heat exchanger uses a series of thin corrugated plates to separate and transfer heat between two fluids. 7. In high-temperature applications where metallic recuperators are unsuitable, ceramic tube recuperators can be used to handle gas inlet temperatures up to around 1550 C. 8. The equipment with a direct contact heat exchange principle in a high pressure boiler system is deaerator 9. A heat pump upgrades low-temperature waste heat to a higher temperature using mechanical work. 10. Steam generation from gas turbine waste heat is typically carried out through a Heat Recovery Steam Generator (HRSG) ................. End of Section III .................. BUREAU OF ENERGY EFFICIENCY 14 Paper-3 Code : Pink 25th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER, 2025 PAPER - 3 : ENERGY EFFICIENCY IN ELECTRICAL UNTILITIES Date : 28.09.2025 Timings: 09:30-12:30 HRS Duration: 3 HRS Max. Marks: 150 General instructions: o o o o o Please check that this question paper contains 8 printed pages Please check that this question paper contains 64 questions The question paper is divided into three sections All questions in all three sections are compulsory All parts of a question should be answered at one place Section I: OBJECTIVE TYPE Marks: 50 x 1 = 50 (i) Answer all 50 questions (ii) Each question carries one mark (iii) Please shade the appropriate oval in SECTION-I of MAIN ANSWER BOOKLET with BLUE/BLACK ball point pen 1 2 3 4 5 6 7 8 9 10 11 Which lamp is most suitable for color-critical applications? a) Halogen lamps b) LED lamps c) CFLs d) Metal halide lamps Iron losses in an electric motor can be reduced by using a) More copper and large conductors b) Use of thinner gauge lower loss core steel c) Use of low loss fan design d) Optimised design and strict quality control A cooling tower has an evaporation loss of 12 m /hr and COC of 2.5. What will be the blowdown loss in m /hr ? a) 5.2 b) 8.0 c) 9.6 d) 10.2 Energy savings by motor replacement can be worked out by: a) KW output ( old new) b) KW output ( new old) c) KW output (1/ old 1/ new) d) KW output (1/ new 1/ old) Stray losses in a motor are mainly caused by: a) Leakage flux induced by load currents b) Hysteresis and eddy currents c) Frictional losses d) Copper winding losses In a vapor compression refrigeration system, enthalpy changes occur across: a) Compressor b) Condenser c) Evaporator d) All of the above The purpose of inter-cooling in a multistage compressor is to: a) Increase final pressure b) Reduce compression work c) Separate oil vapour d) Remove all moisture When air is cooled by evaporation in an air washer: a) Humidity ratio decreases b) Dry bulb temperature decreases c) Dry bulb temperature increases d) Enthalpy increases Amorphous core transformers primarily reduce: a) Load loss b) No-load loss c) Stray loss d) None of the above If a pump delivery valve is throttled to 30% of rated flow, best energy efficiency measure is: a) Replacing the motor b) Installing a larger impeller c) Increasing pump speed d) None of the above Calculate the FAD in CFM for an air compressor with a cylinder displacement of 150 CFM and volumetric efficiency of 90%: BUREAU OF ENERGY EFFICIENCY 1 Paper-3 Code : Pink 12 13 14 15 16 17 18 19 a) 165 b) 135 c) 150 d) None of the above If pump speed is reduced to 2/3rd of its original speed, power consumption will: a) Decrease by half b) Decrease to one-fourth c) Decrease to approx. 30% of original d) Remains same At higher altitudes, for same FAD, air compressors: a) Consume less power b) Consume more power c) Show no difference d) Work without lubrication In a 4-stroke diesel engine, fuel is injected during: a) Induction stroke b) Compression stroke c) Ignition and Power stroke d) Exhaust stroke Voltage unbalance in motors: a) Reduces motor temperature b) Increases motor slip c) Causes excessive heating and reduces life d) Improves torque Soft starters are used to: a) Increase motor speed b) Reduce inrush current c) Convert AC to DC d) Improve efficiency An air compressor is driven by an IE3 premium efficiency motor. Compared to an IE2 motor of the same rating, which of the following statements is most accurate? A) The IE3 motor will always consume less power under all load conditions. B) The IE3 motor achieves higher efficiency mainly by reducing copper and iron losses. C) The IE3 motor has lower inrush current during starting compared to IE2. D) The IE3 motor achieves efficiency by increasing slip. Scale in condenser tubes: a) Increases energy use b) Reduces heat transfer c) Can lead to higher operating pressure d) All of the above Larger diameter ducts in fans: a) Increase system resistance b) Reduce system resistance c) No effect d) Increase static pressure 20 Which of the following is a common symptom indicating that a pump is oversized? a) High discharge pressure b) Throttle valve-controlled systems c) Low suction pressure d) High motor power consumption 21 Reducing the diameter of an impeller in a centrifugal pump will: a) Increase head b) Decrease head c) No effect on head d) Increase flow The main function of fill media in a cooling tower is to: a) Reduce drift losses b) Increase water air contact c) Reduce fan noise d) Filter suspended solids Energy-saving opportunities in cooling towers include: a) Optimizing fan blade angle seasonally b) Maintaining correct water chemistry c) Cleaning fill media regularly d) All of the above The L/G ratio in a cooling tower is: a) Ratio of liquid water mass flow to gas (air) mass flow b) Ratio of drift loss to make-up water c) Ratio of cooling load to fan power d) Ratio of TDS in blowdown to TDS in make-up water Increasing chilled water leaving temperature in a centrifugal chiller: a) Increases efficiency b) Decreases efficiency c) No effect d) Increases refrigerant flow Luminaires are used to: a) Store electrical energy b) Distribute and control light from lamps c) Produce light directly d) Increase lamp wattage The Envelope Performance Factor (EPF) in ECBC is used to: a) Compare energy efficiency of proposed and baseline building designs b) Determine cooling tower sizing c) Calculate lighting power density d) Measure indoor air quality 22 23 24 25 26 27 BUREAU OF ENERGY EFFICIENCY 2 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 Paper-3 Code : Pink In a cooling tower, if any three of the four parameters: heat load, range, approach, and wet-bulb temperature, are kept constant, the required tower size will vary a)Directly with the heat load b)Inversely with the range c)Inversely with the approach d) All the above A system resistance curve of a fan changes with: a) Inlet guide vanes b) Discharge dampers c) Speed change with VFD d) Any of the above A centrifugal pump has BEP efficiency of 65%. At shut-off head, efficiency is: a) 0% b) 65% c) 50% d) 30% A pump with 200 mm impeller delivers 120 m /h. To deliver 100 m /h by trimming, impeller size will be approximately: a) 240 mm b) 167 mm c) 60 mm d) 276 mm If tail-end power factor is improved from 0.80 to 0.95, distribution loss reduction is: a) 13.33% b) 21% c) 29% d) 16% In a vapour compression system, refrigerant changes from vapour to liquid in the: a) Compressor b) Evaporator c) Condenser d) Expansion valve Unity power factor means: a) No reactive power is drawn from the supply b) Current leads voltage c) Current lags voltage d) Reactive power is maximum Examples of lighting controls include: a) Dimmer switches b) Timers c) Photo-sensors d) All of the above A 4-pole, 50 Hz induction motor runs at 1470 rpm. Slip is: a) 0.02 b) 0.20 c) 0.25 d) 0.30 Which type of compressed air dryer consumes the least power for capacities higher than 250 CFM? a) Refrigeration type b) Blower reactivated type c) Heat of compression type d) Heatless purge type A DG set operates at 1250 kVA, 0.8 PF, with specific fuel consumption of 0.23 L/kWh. Quantity of fuel used is_____________. a) 175 L/h b) 230 L/h c) 250 L/h d) 300 L/h In a UPS, DC to AC conversion is carried out by: a) Converter b) Charger c) Battery d) Inverter If a plant receives 96 Million Units (MU) with a T&D efficiency of 80%, generation required is: a) 101.2 MU b) 76.9 MU c) 120 MU d) 68.1 MU Synchronous speed of a motor is inversely proportional to: a) Number of poles b) Frequency c) Voltage d) Temperature If 27,216 kcal of heat is removed per hour, refrigeration tonnage is: a) 10 TR b) 7 TR c) 11 TR d) None of the above The most influential component for cooling tower performance is: a) Fill media b) Drift eliminator c) Casing d) Fan motor An equipment room measures 12 8 3.5 m. Ventilation required for 15 ACH is: a) 5060 m /h b) 5020 m /h c) 5040 m /h d) 5080 m /h Luminous efficacy is: a) Ratio of lumens to watts b) Measured in candela c) Same as luminance d) Measures reflection of light If dew point temperature equals air temperature, relative humidity is: a) 0% b) 45% c) 50% d) 100% In a DG set, the component causing maximum energy loss is: a) Coolant loss b) Alternator loss c) Radiation loss d) Flue gas loss The Energy Conservation Act applies to buildings with connected load: a) All HT connections b) Commercial building having 100 kW c) Residential buildings only d) Government buildings having 100 kW A 750 kVA transformer has 1200 W no-load loss and 7200 W full-load copper loss. At 60% load, total loss is: BUREAU OF ENERGY EFFICIENCY 3 Paper-3 Code : Pink 50 a) 4320 W b)5520 W c) 7632 W d) 3792 W In vapour compression and vapour absorption systems, the common refrigerant is: a) Lithium bromide b) R-134a c) Ammonia d) None of the above .................. End of Section I .................. Section II: SHORT DESCRIPTIVE QUESTIONS (i) (ii) Marks: 8 x 5 = 40 Answer all Eight questions Each question carries Five marks S-1 A steel manufacturing facility is powered by a 3-phase, 6.6 kV, 50 Hz supply and operates the following electrical loads: An electric arc furnace consumes 1.2 MW at a lagging power factor of 0.65, a bank of induction motors for rolling operations consumes 800 kW at a 0.80 lagging power factor, and the lighting and instrumentation systems consume 100 kW at unity power factor. Due to utility regulations, the overall plant power factor must be improved to 0.95 lagging. A capacitor bank will be installed for compensation. As an energy auditor evaluate the following: a. b. c. d. Total active power consumption. (1 Mark) Total initial apparent power drawn by the facility. (1 Mark) The operating power factor. (1 Mark) Determine the total reactive power required to achieve the desired power factor. (2 Mark) S-1 Load Ans Arc furnace All Motors Lighting & instruments Total a. b. c. d. Real Power (KW) 1200 800 100 2100 PF 0.65 0.8 1 KVA = KW/PF 1846 1000 100 2946 Reactive Power (KVAr) = (KVA)2-(KW2) 1402 600 0 2002 Total active Power = 1200+800+100 = 2100 kW Total apparent power drawn: 1846+1000+100 = 2946 kVA Operating Power factor = 2100/2946 = 0.713 KVAR = KW (Tan PF1 -Tan PF2) = 2100 * (Tan (Cos-1(0.713)- Cos-1(0.95)) = 1376 kVAr S-2 Determine the difference in heat rejected in kCal/TR to the cooling tower for two different types of air conditioning system operating at same capacity. Parameter Centrifugal chiller VAM 3 - 180 - 340 13.0 14.6 - 33.5 7.7 9.0 - 39.1 0.6 - Chilled water flow (m /h) 3 Condenser water flow (m /h) Chiller inlet temp ( C) Condenser water inlet temp ( C) Chiller outlet temp ( C) Condense water outlet temp ( C) Specific power consumption (kW/TR) 1 TR = 3024 kcal/h S-2 Centrifugal chiller: Ans Power input = 0.6 860 = 516 kcal/TR Heat rejected = 3024 + 516 = 3540 kcal/TR BUREAU OF ENERGY EFFICIENCY 4 Paper-3 Code : Pink VAM: Chilled water flow = 180 m /h = 180000 kg/h T = 14.6 9.0 = 5.6 C Cooling load = 180000 1 5.6 = 1008000 kcal/h TR = 1008000 / 3024 = 333.33 TR Heat Rejected = 340000 kg/hr x 1 kcal/kg x (39.1-33.5) = 1904000 kcal/hr Heat Rejected per TR = 1904000 / 333.33 = 5712 kcal/TR Difference per TR = 5712 - 3540 = 2172 kcal/TR S-3 An energy audit in an industrial unit revealed a fan directly coupled with motor was operating at 37 Hz through VFD for 500 hours/month and supplying air through a 150 mm diameter duct. The fan is designed to deliver an air flow of 1300 m /h with a rated input power of 3 kW at 50 Hz. Calculate the air velocity and annual energy savings ignoring the motor losses. Given: S-3 Design air flow rate = 1300 m /h at 50 Hz Ans Operating frequency = 37 Hz Power drawn at 50 Hz = 3 kW Operating hours = 500 hours/month Duct diameter = 150 mm = 0.15 m Flow rate at 37 Hz: Q2 = 1300 (37 / 50) = 962 m /h Q2 = 962 / 3600 = 0.2672 m /s Duct area: A = /4 (0.15) = 0.01767 m Air velocity: V = Q / A = 0.2672 / 0.01767 = 15.12 m/s Power at 37 Hz: P2 = 3 (37 / 50) = 3 0.405 = 1.215 kW Monthly energy savings: (3 - 1.215) 500 = 892.5 kWh/month Annual energy savings: 892.5 12 = 10710 kWh/year Answers: Air velocity = 15.12 m/s Annual energy savings = 10710 kWh/year A pump is used to fill a rectangular overhead tank measuring 5 m 3.5 m with a height of 10 m. The inlet S-4 pipe to the tank is positioned at a height of 25 m above ground level. The following additional data is available: The pump draws water from an underground sump situated 4 meters below the pump level and delivers it to a tank whose overflow line is positioned 8 meters above the tank bottom. The motor driving the pump draws 7.5 kW of power. The operating efficiencies of the motor and the pump are 90% and 70% respectively. Calculate the time taken by the pump to fill the tank up to the overflow level. BUREAU OF ENERGY EFFICIENCY 5 Paper-3 Code : Pink S-4 Step 1: Volume of water filled Ans Volume = 5 3.5 8 = 140 m Mass of water = 140 1000 = 140000 kg Step 2: Total head (H) Total head = 25 + 4 = 29 m Step 3: Shaft power (output of motor) Shaft Power = 7.5 0.90 = 6.75 kW Step 4: Water power (hydraulic power output of pump) Water Power = 6.75 0.70 = 4.725 kW = 4725 W Step 5: Time taken Pump efficiency = Mass flow g Head / Shaft Power 0.7 = mass flow x 9.81 x 29 /(6.75x1000) Mass flow = 16.6 kg/sec = 59791 kg/hr = 59.79 m3/hr Time = 140 / 59.79 = 2.34 hrs = 140.5 minutes S-5 A commercial training hall with dimensions 18 m 12 m is being planned. Calculate the number of 18 W LED lamps, each providing 1800 lumens, required to achieve an illuminance level of 300 Lux. The lamps will be installed at a height of 3 meters from the working plane. The utilisation factor (UF) of the system is 0.70, and the light loss factor (LLF) is 0.80. S-5 Area of room (A): 18 12 = 216 m Ans Total lumens required ( _total): _total = E A = 300 216 = 64800 lumens Effective lumens per lamp ( _lamp_effective): _lamp_effective = Lumen Output UF LLF = 1800 0.70 0.80 = 1008 lumens Number of lamps required (N): N = _total / _lamp_effective = 64800 / 1008 = 64.3 65 lamps S-6 A private power distribution company has implemented new digital metering and billing systems to improve efficiency in a residential zone. After six months of operation, the following data was recorded: Input energy to the system = 75 MU Metered billed energy = 56 MU Unmetered average billing = 4 MU Amount billed = 680 million Total amount received = 600 million Arrears collected = 90 million Purchased energy cost = 8.50 per kWh i) Estimate the Aggregate Technical and Commercial (AT&C) loss (%) and the revenue realized per kWh 4 Marks ii) Calculate the revenue loss per kWh to the company due to AT&C loss. 1 Mark BUREAU OF ENERGY EFFICIENCY 6 Paper-3 Code : Pink S-6 Input Energy = 75 MU = 75,000,000 kWh Ans Metered Billed Energy = 56 MU Unmetered Average Billing = 4 MU Total Energy Billed = 56 + 4 = 60 MU Amount Billed = 680 million Arrears Collected = 90 million Amount Received = 600 million Purchased Energy Cost = 8.50/kWh i) AT&C Loss (%) and Revenue Realized ( /kWh) Billing Efficiency = (60 / 75) 100 = 80.0% Collection Efficiency = ((600 - 90) / 680) 100 = (510 / 680) 100 = 75.0% AT&C Loss (%) = 1 - (Billing Efficiency Collection Efficiency) = 1 - (0.80 0.75) 100 = 1 - 0.60 100 = 40.0% Revenue Realized per kWh = (600 - 90) / 75 = 510 / 75 = 6.80/kWh ii) Revenue Loss per kWh Revenue Loss = 8.50 - 6.80 = 1.70/kWh S-7 List five energy saving measures in compressed air system. S-7 Energy-Saving Measures in a Compressed Air System (Headings Only) Ans 1. Fixing Air Leaks 2. Reducing Compressor Discharge Pressure 3. Proper Sizing of Compressors 4. Using Variable Speed Drives (VSDs) 5. Avoiding Artificial Demand 6. Heat Recovery from Compressor 7. Reducing Air Intake temperature 8. Optimizing Piping Layout and Size 9. Using Efficient Air Dryers 10. Eliminating Inappropriate Uses of Compressed Air 11. Using Automatic Drain Traps 12. Implementing Demand-Side Control 13. Operating Compressors at Full Load 14. Performing Regular Maintenance and Monitoring S-8 Match the Following: 1. Solar Heat Gain Coefficient (SHGC) a. Coefficient of Performance (COP) 2. U-value b. Solar Reflectance 3. HVAC System Efficiency c. Fenestration Heat Gain 4. Cool Roof d. Building Insulation e. Conductive path for unwanted heat 5. Thermal Bridging transfer S-8 Answer Key: Ans 1 c. Fenestration Heat Gain 2 d. Building Insulation BUREAU OF ENERGY EFFICIENCY 7 Paper-3 Code : Pink 3 a. Coefficient of Performance (COP) 4 b. Solar Reflectance 5 e. Conductive path for unwanted heat transfer .................. End of Section II .................. Section III: LONG DESCRIPTIVE QUESTIONS (i) (ii) Marks: 6 x 10 = 60 Answer all Six questions Each question carries Ten marks L-1 A distribution company (DISCOM) plans to implement a comprehensive Demand Side Management (DSM) initiative to reduce its peak load and overall energy procurement cost. The program targets two consumer categories Residential consumers and Industrial consumers. For residential consumers, DISCOM has taken LED replacement as a DSM intervention whereas for industrial consumers load shifting strategy has been adopted through peak and off-peak electricity pricing. The DISCOM supplies electricity to 10,000 households, each using 4 CFL bulbs (30 W each). These bulbs are used for 5 hours per day during evening peak hours. The DISCOM replaces each CFL bulb with a 9 W LED bulb. Procurement cost of each LED bulb costs 100, however the DISCOM provides the LED bulbs in place of CFL to consumers at a subsidized rate of 70/bulb. Administrative cost per household for the program Rs.10/-. DISCOM also serves 50 industrial consumers, each with a shiftable evening load of 100 kW, used from 6 PM to 10 PM. DISCOM incentivizes these consumers to shift their load from 10 PM to 2 AM by offering an incentive of 2 per kWh shifted. Eighty percent of industrial consumers agree to take advantage of the tariff incentive scheme. Power purchase cost for DISCOM: o Evening Peak (5 10 PM): 7/kWh o Late Night (10 PM 6 AM): 3/kWh Calculate the following: a) For the residential consumers: i) Calculate the total daily energy savings in kWh/day from the LED replacement program. 2 Marks ii) Determine the daily cost savings for the DISCOM from the LED program. 1 Mark iii) Calculate the total one-time cost to the DISCOM for the LED program, including subsidies and administrative costs. 2 Marks iv) Estimate the simple payback period in days for the LED program. 1 Mark b) For the industrial consumers: i) Calculate the total energy shifted in kWh/day. 1 Mark ii) Compute the net daily savings for DISCOM, considering power cost reduction and incentive payout. 1 Mark c) Estimate the carbon emission avoidance due to above two DSM activity, if emission factor of grid electricity is 0.716 tCO2/MWh. 2 Marks Daily energy savings from LED replacement: (10000*4*5) * (30 9) /1000 = 4200 kWh L-1 Daily cost savings for DISCOM = 4200 *(7) = Rs.29400 Ans One time cost to DISCOM: Subsidy for LED replacement = 10000 * 4 * (100-70) = Rs. 12 Lakh Administrative cost = 10000*10 = Rs. 1.0 Lakh Total One-time cost to DISCOM = 12+ 1.0 = Rs. 13.0 Lakh Simple Payback = 1300000 / (16800*365) = 44 Days Energy Shifted by industrial consumers = 50*100*4*0.8 = 16000 kWh per Day Net daily savings = 1600 *(Purchase cost savings incentive cost) = 16000* (7-3-2) = Rs 32000/ Day Energy Savings achieved only by the LED program, hence emission reduction = 4200*0.716/1000 = 3.0 tCO2 per day. BUREAU OF ENERGY EFFICIENCY 8 L-2 Paper-3 Code : Pink A) A clear water pump with rated flow of 125 m3/hr, head 55 m at rated speed of 1460 rpm and 79% efficiency is being used for supplying clarified water to a residential colony s water treatment facility. The daily water requirement is for 3000 M3. The pump is directly coupled and driven by a three phase 50 Hp, 415 V, 64A, 0.9 pf, 1460 rpm induction motor with 90.5% full load efficiency. During an internal energy audit, it has found that the motor is designed to operate with only 65% loading at pump rated conditions, therefore compromising on motor efficiency. The plant management has considered replacing the standard motor with a 30 kW IE3 motor. The following are operating parameters before and after motor replacement: Parameters Before motor replacement After motor replacement Flow (m3/hr) 130 ? Head (m) 52 51 Supply Voltage (V) 415 415 Current (Amp) 42 39 Power Factor 0.9 0.92 Motor Eff (%) 0.88 0.932 As an external auditor, you observed that although the plant has reduced the size of the induction motor to improve loading and enhance motor efficiency, the slip of the new IE3 motor has decreased by 20 rpm. This raises concerns about the actual energy savings achieved. Validate the savings claimed, calculate the following: i) ii) iii) iv) % Loading of motor after replacement. (1 Mark) Flow after replacing the standard motor with 30 kW IE 3 motor. (1 Mark) Operating Pump Efficiency before and after motor replacement. (2 Marks) Daily energy saving during operation due to motor replacement. (1 Marks) B) Mark the following statements as True/False i) Totally enclosed, fan cooled (TEFC) motors are less efficient than screen-protected, drip-proof (SPDP) motors. ii) Stray loss in induction motors is inversely proportional to load current. iii) As per BIS standard, the motor output should not be affected with voltage variation up to +/- 6%. iv) Motor life doubles for each 100C reduction in operating temperature. v) Starting torque of energy efficient motors is higher than standard motors. L-2 Ans A) i) Loading of the IE3 Motor: = (1.732*0.415*39*0.92*0.932)/30 = 80.12% ii) Flow after replacement = 130 m3/hr * 1480/1460 = 131.8 m3/hr iii) Operating Pump Efficiency: Power Consumption before replacement = 1.732*0.415*42*0.9 = 27.17 kW Power Consumption after replacement = 1.732*0.415*39*0.92 = 25.79 kW Before replacement = [(130 m3/hr /3600) * 52 *9.81] / (27.17*0.88) = 77% After replacement = [(131.8 m3/hr/3600) * 51 *9.81] / (25.79*0.932) = 76.2% Daily Operating Hour before Pump replacement = 3000 /130 = 23.08 Hrs Daily Operating Hour after Pump replacement = 3000 /131.8 = 22.77 Hrs iv) Daily Energy Savings = (23.08*27.17) (25.79*22.77) = 39.9 kWh B) True/False: i) False ii) False iii) True iv) True v) False BUREAU OF ENERGY EFFICIENCY 9 Paper-3 Code : Pink A Commercial Office building accommodates two government departments. The total employees working in L-3 both the departments is 250 out of which 70% average number of employees present at any time. Being a government office, the building is operational 6 days a week with 10 working Hrs a day. The building receives Electricity supply from local electricity Distribution company through a 33KV feeder, and it is distributed after stepping down to 415 V. The building does not have separate parking, lawn, internal roads etc. The building information sheet is as below: Office Building 1. 2. 3. Contract Demand (kW) Installed capacity: Diesel Generating (DG) Set(s) (kVA) a) Annual Electricity Consumption, purchased from Utilities (kWh) b) Annual Electricity Consumption, through DG Set(s) (kWh) 4. a) Annual Cost of Electricity, purchased from Utilities (Rs.) b) Annual Cost of Electricity generated through DG Set(s) (Rs.) 5. 6. 7. 8. 9. 10. Built Up Area (sq.m) Area of the Building Conditioned Area (sq.m) Installed capacity of Chiller of Air Conditioning System (TR) Installed lighting load (kW) Officer Appliances (kW) Other Loads (kW) HSD Consumption in DG (GCV 10800 Kcal/kg and density of 0.85) Annual Data (April 24-March 25) 130 160 105753 2136 1043557 54405 3591.96 2155.18 137.5 8.11 11.0 12.5 585 Litres Calculate the following: a. b. L-3 Ans The total electricity consumed by the building and average electricity unit cost. (2 Mark) EPI of the building considering the reported data for past one year. Also recommend the appropriate rating under BEE star rating program for buildings if the bandwidth of the EPI range between 15050 kWh/sq. m/year. (3 Marks) c. Calculate the design diversity factor of the building, if the design EER of the chiller is 3.5 and recorded maximum demand is 75% of the contract demand. (2 Marks) d. Estimate the overall operating efficiency of the DG set (2 Marks) e. Calculate the lighting power density (1 Marks) a. Total electricity consumed: 105753+2136 =107889 kWh Average unit cost: (1043557+54405)/107889 = Rs.10.18 per kWh b. EPI = 107889/3591.96 = 30 kWh/m2/year, the EPI is below the bandwidth for star rating, hence 5 Star rated. c. Diversity Factor = Maximum Demand / Connected Load Maximum Demand = 130 * 0.75 = 97.5 kW Connected Load = Lighting load + appliances +others+ AC = 8.11+11.0+12.5+ (137.5*3024/860/3.5) = 169.75 kW Diversity Factor = 97.5/169.75 = 0.57 (or) 169.75/97.5 = 1.74 d. Overall, DG Set efficiency= (2136*860) / (585*0.85*10800) = 34.2% e. Lighting Power Density = 8.11*1000/3591.96 = 2.25 W/m2 A 5-star business hotel operates a centralized HVAC system operating round the clock with the following L-4 configuration. Only one chiller operates at a time, while the other is on standby. Two centrifugal chillers, each rated at 250 TR, with EER varying with load as below: No change in EER observed above 85% load, assume chiller motor efficiency of 90% at all loading conditions and the energy consumption by the auxiliary systems is as below: During chiller operation, two pumps run in parallel at an 80% load factor, consuming a total of 19.7 kW. In addition, two cooling tower fans operate continuously with a power consumption of 5.89 kW. Both the BUREAU OF ENERGY EFFICIENCY 10 Paper-3 Code : Pink pumps and fans function 24 hours a day, with overall efficiencies of 75% and 70% respectively. The applicable electricity tariff is 6.5 per kWh. Evaluate the following: a. The total annual energy consumption (in MWh) and cost of the HVAC system, considering part-load EERs and auxiliary loads. (4 Marks) b. Heat removal by condenser in (TR) at different loads. (3 Marks) c. The hotel is planning to use the chiller partially as a heat pump by mounting a plate heat exchanger in series between the compressor and condenser (desuperheater for partial heat recovery) for producing hot water. The heat recovery can be only 20% of the condenser heat discharge. If the hot water requirement is 2000 litters/hr with 100C temperature rise, evaluate whether the hot water requirement can be met at 40% loading conditions. (3 Marks) a. Energy Consumption: L-4 Energy Consumed by Chiller: Ans Loading (%) Cooling Load (TR) EER A B C Cooling Load (KW) D=B*3024/86 0 Input Power (KW) Days Mwh E=D/C F G=E*F*24/100 0 0.85 212.5 5.2 747.21 143.69 180 620.8 0.6 150 4.6 527.44 114.66 120 330.2 0.4 100 3.9 351.63 90.16 65 140.7 Energy Consumed by auxiliaries: (19.7+5.89) *24*365/1000= 224.5 Mwh Total Annual Energy Consumption = 620.8+330.2+140.7+224.5 = 1316 Mwh Annual Cost = 1316000*6.5 = Rs. 85.55 Lakh b. Heat Removal by Condenser (TR) Loading (%) X 0.85 0.6 0.4 Cooling Load (TR) Y 212.5 150 100 Input Power (KW) Z 143.69 114.66 90.16 Power Input to Compressor (kW) P=Z8*0.9 129.32 103.20 81.14 Condenser Heat Load (TR) HL =Y+(P*860/3024) 249.28 179.35 123.08 c) Heat Hoad at 40% loading = 123.08*3024 = 372194 kCal/Hr Recovery potential (20%) = 74439 kCal/Hr Heat requirement for hot water generation = 2000*1*10 = 20000kCal/H Therefore, the heating requirement can easily be met at 40% loading. L-5 a) In a large-scale steel manufacturing facility, a cooling tower is used to reject heat from continuous casting operations. The circulating water flow rate is 2000 m /hr. The cooling tower is currently operating at a Cycles of Concentration (COC) of 3. The evaporation loss is estimated at 1.0% of the circulating flow, and the drift loss is 0.1% of the circulating flow. The facility is planning to improve the COC from 3 to 6 through advanced water treatment. Evaluate the following: i. ii. iii. iv. Calculate the make-up water requirement at the current COC of 3. Calculate the revised make-up water requirement if the COC is increased to 6. Estimate the total water savings per day. Discuss one limitation or risk associated with increasing the COC. b) True or False i. Cooling towers primarily reject heat through evaporative cooling. ii. The approach temperature in a cooling tower is the difference between the hot water temperature and the BUREAU OF ENERGY EFFICIENCY 11 iii. iv. v. vi. L-5 Paper-3 Code : Pink ambient dry bulb temperature. Blowdown in a cooling tower is required to prevent the build-up of dissolved solids. Drift losses in a cooling tower refer to water carried away with the exhaust air. Cooling tower effectiveness improves with higher approach temperatures. Cycles of concentration in a cooling tower relate to how many times the water is reused before discharge. a) Given: Ans Circulating Water Flow (CWF) = 2000 m /hr Initial COC = 3 Final COC = 6 Evaporation Loss (E) = 1% of 2000 = 20 m /hr Drift Loss (D) = 0.1% of 2000 = 2 m /hr i) Current Make-up Water Requirement at COC = 3 B = 20 / (3 - 1) = 10 m /hr Make-up Water = E + D + B = 20 + 2 + 10 = 32 m /hr ii) Revised Make-up Water Requirement at COC = 6 B = 20 / (6 - 1) = 4 m /hr Make-up Water = E + D + B = 20 + 2 + 4 = 26 m /hr iii) Water Savings Hourly Savings = 32 - 26 = 6 m /hr Daily Savings = 6 24 = 144 m /day iv) Limitation of Higher COC Increasing COC can lead to higher concentrations of dissolved solids in the water, which may cause scaling, corrosion, and microbiological fouling in the system. Effective water treatment and frequent monitoring are necessary to avoid operational issues. b) True or False i) ii) iii) iv) v) vi) L-6 True False (It is the difference between the cold-water temperature and the wet bulb temperature.) True True False (Lower approach means better effectiveness.) True a) A manufacturing plant operates a 180 kVA diesel generator set rated at 0.8 lagging PF. The prime mover is a diesel engine rated 240 BHP. The alternator has total losses (including exciter power) of 5.44 kW. Assume no derating for site conditions. The generator is required to supply a mixed industrial load at its full kVA rating. The plant manager wishes to improve system efficiency by operating at a higher power factor. The diesel engine operates at a brake thermal efficiency of 32% when loaded near its rated capacity. The calorific value of the diesel fuel is 10,500 kCal/kg, and the specific gravity of the fuel is 0.85. Calculate the following: i) Maximum power factor that can be maintained at full kVA load without exceeding the engine capacity. (3 Marks) ii) Corresponding diesel fuel consumption (litres per hour) at this maximum power factor. (2 Marks) b) True or False Question: i) Improving power factor of the load on a DG set reduces the apparent power drawn and increases the system's overall fuel efficiency. BUREAU OF ENERGY EFFICIENCY 12 Paper-3 Code : Pink ii) iii) iv) v) L-6 Alternator losses are independent of the load power factor. Turbocharger in a diesel engine helps to reduce engine noise. A diesel generator set must always be operated at unity power factor for maximum efficiency. DG sets are designed to handle unbalanced load between phases to 25% of their capacity. a) i. Convert BHP to kW (shaft power): Ans 240 BHP 0.746 = 179.04 kW Net electrical power available = 179.04 kW - 5.44 kW = 173.6 kW At full load (180 kVA), maximum PF = Real Power / Apparent Power PF = 173.6 / 180 = 0.964 ii. Thermal Input Required = Electrical Output / Efficiency = 179.04 / 0.32 = 559.5 kW Convert kcal to kW: 1 kg diesel = 10,500 kcal = 10,500 4.1868 = 43,961.4 kJ/kg Fuel consumption (kg/hr) = (559.5 3600) / 43961.4 = 45.8 kg/hr In litres per hour = 45.8 / 0.85 = 53.88 L/hr b) True or False Question: i) Improving power factor of the load on a DG set reduces the apparent power drawn and increases the system's overall fuel efficiency. True ii) Alternator losses are independent of the load power factor. False iii) Turbocharger in a diesel engine helps to reduce engine noise. False iv) A diesel generator set must always be operated at unity power factor for maximum efficiency. False v) DG sets are designed to handle unbalanced load between phases to 25% of their capacity. False ................. End of Section III .................. BUREAU OF ENERGY EFFICIENCY 13 PAPER-3 COLOR C0DE : GREEN 24th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER, 2024 PAPER - 3 : ENERGY EFFICIENCY IN ELECTRICAL UTILITIES SECTION I : OBJECTIVE QUESTIONS Marks 50x1=50 1. How does an increase in system resistance affect the operation of a centrifugal fan? a) Increases the airflow b) Reduces the airflow c) Reduces the static pressure d) No effect on fan performance 2. What is the primary purpose of trimming the impeller in a centrifugal pump? a) To increase the pump speed b) To adjust the pump capacity to match system requirements c) To reduce the pump speed d) To increase the NPSH required 3. How does increasing the diameter of the suction pipe affect the NPSHA in a pumping system? a) Reduces NPSHA b) Increases NPSHA c) Decreases NPSHR d) Increases NPSHR 4. A pump has a flow rate of 200 cubic meters per hour and operates against a head of 30 meters. If the pump efficiency is 70%, what is the input power required? a) 60.5 kW b) 23.36 kW c) 95.2 kW d) 100 kW 5. What is the effect of cavitation in pump? a) Increases efficiency b) Reduces noise and vibration c) Causes erosion of impeller surfaces d) Increases NPSH required 6. What is the relationship between pump speed and flow rate in a centrifugal pump according to the Affinity Laws? a) Flow rate is proportional to the pump speed b) Flow rate is proportional to the square of the pump speed c) Flow rate is proportional to the cube of the pump speed d) Flow rate is independent of pump speed 7. What is the impact of using larger diameter pipes on the system resistance in a pumping system? a) Reduces system resistance by lowering friction head losses b) Increases system resistance c) Increases power d) Increases static head 8. A cooling tower reduces the temperature of water from 40 C to 30 C. If the mass flow rate of water is 5 kg/s, what is the heat removed by the cooling tower? a) 500 kW b) 209 kW c) 2 kW d) 5 kW 9. The approach temperature of a cooling tower is 5 C, and the range is 10 C. If the inlet water temperature is 40 C, what is the outlet water temperature? a) 25 C b) 30 C c) 35 C d) 45 C 10. How can the performance of a cooling tower be improved? a) Proper water treatment b) Regular maintenance c) Optimizing air and water flow d) All of the above 11. What is the primary function of a luminaire in a lighting system? a) To generate light b) To store electrical energy c) To distribute light emitted from lamps d) To control the voltage supply 12. Which type of lamp has the highest luminous efficacy among the following? a) Low pressure sodium vapour lamp b) Halogen lamp c) LED lamp d) Compact fluorescent lamp (CFL) 13. How does the use of high-efficiency luminaries contribute to energy conservation? a) By decreasing the power consumption b) By increasing the luminous efficacy c) By improving light distribution characteristics d) All of the above 14. How does altitude affect the performance of a DG set? a) Increases power output b) Reduces fuel consumption c) Reduces power output d) No change in fuel consumption 15. What is the Solar Heat Gain Coefficient (SHGC) used for in building energy analysis? a) To measure light transmittance b) To measure the heat gain through fenestration due to solar radiation c) To measure air leakage through windows d) To measure the thermal emittance of roofing materials 16. What is the primary function of an economizer in an HVAC a) To increase indoor air pollution b) To reduce the cost of heating equipment c) To use outdoor air for cooling when conditions are favorable, saving energy d) To increase the use of mechanical cooling systems system? 17. Energy Performance Index is the ratio of total building annual energy consumption to -----a) Carpet area b) Built up area c) roof area d)Windows and Walls area 18. In a D G set, the generator is generating 1000kVA at 0.7PF. If the specific fuel consumption of this D G set is 0.25 lits per kWh, then how much fuel in litres will be consumed while delivering generated power for one hour? a)230 b) 250 c)175 d) 225 19. The power measured in an Induced Draft (I D) fan operating at 49 Hz is 52 kW. A Variable Frequency Drive (VFD) is installed and the fan was operated at 34 Hz, The estimated Power saving will be_________ a)35.7 kW b)17.3 kW c)34.6 kW d)36 kW 20. A fan is drawing 16 kW at 800 RPM. If its speed is reduced to 600 RPM, the power drawn by the fan will be____________ a) 6.75 kW b) 9 kW c) 12 kW d) None of the above 21. Harmonics generation is more in_____________ a) Inverter Drive b) LED lamp c) Transformer d) Resistance heater 22. A 500 cfm reciprocating compressor has a loading and unloading period of 5 seconds and 20 seconds respectively during a compressor leakage test. The air leakage in the compressor air system will be ________ a)125 cfm b)100 cfm c) 200 cfm d) none of the above 23. What is the efficiency of motor with the following nameplate details 22 kW, 415V, 42 A, 0,8 p.f, 1475 rpm? a)94.5% b) 91% c) 89.9% d) None of the above 24. A hotel building has 14 floors, each of 1000m2 area, If the Lighting Power Density is 10.8 per m2 the interior lighting power allowance for the hotel building is_________ a)110800 W b) 129600 W c)151200 W d) 186600 W 25. A pump with 230mm diameter impeller is delivering a flow of 150 m 3/hr. If the flow is to be reduced to 110m3 /hr by trimming the impeller, what should be the approximate size of the impeller? a)207mm b)175 mm c) 169 mm d)195 mm 26. What is the main advantage of using a rotary screw air compressor over a reciprocating compressor? a) Lower initial cost c) Higher maximum pressure b) Continuous, pulsation-free air delivery d) None of the above 27. What is the primary function of substations in the electrical power supply system? a) To generate electricity b) To communicate over long distances c) To facilitate voltage transformation d) None of the above 28. How does the transmission voltage level affect the efficiency of long-distance power transmission? a) Higher voltage levels reduce transmission losses b) Higher voltage levels increase transmission losses c) Voltage levels do not affect transmission losses d) None of the above 29. What is the primary purpose of using high voltage direct current (HVDC) transmission over long distances? a) To increase the frequency of electricity b) To step down the voltage for distribution c) To minimize transmission losses over long distances d) All of the above 30. What is the impact of voltage imbalance among the three phases in an electrical system? a) Improved motor efficiency b) Increased motor losses and reduced equipment life c) Reduced power consumption d) Enhanced power factor 31. How does adding capacitors to an electrical distribution system improve power factor? a) By providing the reactive power b) By increasing the active power consumption c) By lowering the system voltage d) By increasing the frequency of the system 32. A transformer has a primary voltage of 220V and a secondary voltage of 110V. If the primary current is 5A, what is the secondary current assuming no losses? a) 2.5A b) 5A c) 10A d) 20A 33. A factory consumes 500,000 kWh of electricity per month with a power factor of 0.8. How much is the reactive power (kVAR)? a) 400,000 kVAR b) 375,000 kVAR c) 800,000 kVAR d) 500,000 kVAR 34. Which of the following best describes an induction motor's operation? a) It uses direct current to create mechanical energy. b) It generates a rotating magnetic flux that induces current in the rotor. c) It operates synchronously with the AC supply frequency. d) It requires external excitation to operate. 35. How does operating a motor in star mode affect its performance? a) It reduces the voltage and derates the motor capacity. b) It increases motor speed. c) It improves the power factor at high loads. d) It eliminates the need for external capacitors. 36. In a VFD-controlled motor, what happens when the supply frequency is reduced while maintaining the same voltage? a) The motor speed increases. b) The motor efficiency improves. c) The motor draws higher current and may overheat. d) The motor torque decreases. 37. Which of the following describes the function of a soft starter in a motor system? a) It increases the motor's full-load speed. b) It converts AC power to DC power. c) It reduces the inrush current during motor start-up. d) It improves the motor's efficiency at low speeds. 38. A motor operates at 75% load with an efficiency of 88%. If the motor's rated power is 20 kW, what is the actual output power? a) 15 kW b) 13.2 kW c) 17.6 kW d) 14.4 kW 39. What is the main purpose of an after-cooler a) To remove moisture from the air by cooling it b) To increase the pressure of the compressed air c) To filter out dust and particles d) To lubricate the compressed air in a compressed air system? 40. How does a desiccant air dryer remove moisture from compressed air? a) By cooling the air b) By using adsorbents like silica gel or activated carbon c) By increasing the pressure d) By reducing the air flow rate 41. Which of the following is an efficient method to control the capacity of a centrifugal compressor? a) Automatic on/off control b) Variable inlet guide vanes c) Load and unload control d) Multi-step control 42. What is the effect of increasing the intake air temperature on the efficiency of an air compressor? a) Increases efficiency by reducing power consumption b) Decreases efficiency by increasing power consumption c) No significant effect on efficiency d) Increases the volumetric capacity of the compressor 43. What is the main benefit of using a variable speed drive (VSD) with a screw compressor? a) Increases the maximum pressure capacity b) Reduces the size of the compressor c) Eliminates unloaded running condition d) Simplifies maintenance 44. Which of the following describes the primary function of an air receiver in a compressed air system? a) To increase the air pressure b) To act as a reservoir and dampen pulsations c) To filter out impurities d) To cool the compressed air 45. What is the primary function of the evaporator in a refrigeration cycle? a) To compress the refrigerant b) To absorb heat from the environment c) To condense the refrigerant d) To regulate the flow of refrigerant 46. In which component of an ideal refrigeration system, the refrigeration temperature will increase? a) Compressor b) Condenser c) Evaporator d) Expansion valve 47. Which refrigerant is commonly used in vapor absorption refrigeration systems? a) R-22 b) R-134a c) H2O d) LiBr 48. What is the effect of increasing the chilled water leaving temperature on the efficiency of a centrifugal chiller? a) It increases the efficiency of the chiller b) It decreases the efficiency of the chiller c) It has no effect on efficiency d) It increases the refrigerant flow rate 49. An HVAC system operates with a COP (Coefficient of Performance) of 4. If the system provides 100 kW of cooling, what is the power input to chiller? a) 0.04 kW b) 25 kW c) 400 kW d) None of the above 50. What is the effect of decreasing the RPM of a fan by 10% on its power requirement? a) Decreases the power requirement by 27% b) Decreases the power requirement by 19% c) Increases the power requirement by 10% d) No significant effect SECTION II : SHORT DESCRIPTIVE QUESTIONS (i) (ii) Marks 8x5 = 40 Answer all EIGHT questions Each question carries FIVE marks S1 A process plant continuously operates a furnace oil operated DG set of capacity 3.0 MW to avoid any process safety incident in case of tripping of critical equipment on power failure. Total critical load on DG set is 2.5 MW and exhaust flue gas at 430 deg.C is vented as original design intent was to operate DG set intermittently only during power failure. Since it is being operated continuously, the process team developed a scheme to generate saturated steam at 5 bar(g) using the waste heat boiler. Other operating parameters are given below: Specific heat of flue gas Final stack temperature to avoid Sulphur dewing Flue gas flow Sat. temp. of steam at 5 barg Latent heat at 5 barg Feed water temperature 0.24 210.0 17.5 159.0 498.0 130.0 kcal/kg-Deg.C Deg.C TPH Deg.C kcal/kg Deg.C Calculate the quantity of steam generated from waste heat boiler in TPH. Solution: Heat available for steam generation= 17500*0.24*(430-210) = 924000 Kcal/hr Steam Generation= 924000/( 498+(159-130)) = 1753.3 kg/hr = 1.75 TPH S2 List five energy efficiency measures in Compressed air system. Solution: Book 3, Refer Page 101 S3 Analyse the following data collected for a water pump. If the operating head is 16m explain what will happen to other parameters. Design Parameters Flow (Q) Head (H) Power(P) Efficiency Values 40 lps 20 m 15 kW 51% Solution : 1. If the operating head is 16m instead of 20 m, the operating flow will be higher than the rated flow. 2. Since the operating point has deviated from the BEP, the operating efficiency will be less than design efficiency. 3. Since, the flow has increased and pump efficiency decreased than rated, the operating power demand will be more than the rated power. S4 A 3 phase Induction motor has the following details: Name plate details: 55 kW,415 V,95 A,0.9 p.f, 50 Hz Running load details: 410 V,75 A,0.80 p.f, 48 Hz Calculate the following: a) loading percentage, b) Rated efficiency Ans: Actual power drawn by the motor = 1.732 x 410 x 75 x 0.80 /1000 = 42.6 kW Rated input power = 1.732 x 415 x 95 x 0.90 /1000 = 61.5 kW Percentage loading of motor = 42.6 / 61.5 = 69.3 % Rated efficiency of motor = (55 / 61.5) x 100 = 89.4% S5 Match the following: Column A 1. Envelope Performance Factor (EPF) 2. Luminous Efficacy 3. Economizer 4. Thermal Mass 5. Energy Simulation Software Each 1 mark Column B a. Lighting System Efficiency b. Heat Storage in Building Materials c. ECBC Compliance d. Building Energy Performance Modeling e. Outdoor Air for Free Cooling Solution: Column A 1. Envelope Performance Factor (EPF) 2. Luminous Efficacy 3. Economizer 4. Thermal Mass 5. Energy Simulation Software Column B c. ECBC Compliance a. Lighting System Efficiency e. Outdoor Air for Free Cooling b. Heat Storage in Building Materials d. Building Energy Performance Modeling S6 A cooling water pump has a positive suction head of 5 meters. The discharge pressure is 3.0 kg/cm , and the water flow rate is 150 m /hr. Determine the pump efficiency given that the actual power input of the connected motor is 18.0 kW and the motor operates with an efficiency of 85%. Solution: Flow Rate: 150 m /hr Total Head: 30-5 = 25m Power input to pump = 18*0.85 = 15.3 kW Hydraulic Power = (150/3600)*25*9.81 = 10.2 kW Pump Efficiency = 10.2/15.3 = 66.7% S7 A steel industry has 100 MW of captive power plant with 2 nos. of identical extraction condensing steam turbine. Power demand is 80 MW. The turbine specific condensing load is 3.20 kg/kWh and heat rejection in condenser is 560 kCal/kg. Cold cooling water temperature: 32.0 0C Hot cooling water temperature 39.2 0C Calculate the following: a) The Cooling Water circulation flow (m3/hr) through both condensers, If only one cooling tower supplies water to condenser of both the steam turbines. b) Make-up water flow rate (kg/hr) to basin, assuming blowdown loss is 1.0 % of circulation flow. Solution: Condenser heat load = (3.2*80000*560/1000000) = 143.36 Circulated cooling water flow = (143.36*1000000/7.2/1000) = MkCal/hr 19911m3/hr Evaporation losses = 0.00085*1.8*19911*7.2= 219.3 m3/hr Blowdown loss = 19911*0.01 = 199 m3/hr Water makeup to cooling tower = 219.3+199 = 418.4 m3/hr = 418400 kg/hr S8 A process engineer develops a scheme to put 500 TR absorption-based refrigeration system to bring down process fluid temperature from 34 0C to 26 0C and this will result in higher production by 10%. 5 TPH excess steam is available in the plant and this new scheme utilizes this excess steam. COP of refrigeration system is 0.65 and available latent of steam for refrigeration system is 540 kcal/kg. A)Estimate excess steam utilization for absorption- based refrigeration system in TPH. .. b) Estimate required Cooling water (m3/hr), if available approach in condenser is 10 0 C. .. Solution: Energy required for refrigeration system 500*3024/0.65 = 2326153.8 kcal/hr Steam needed for refrigeration system 2326153.8/540 = 4.3 TPH Steam utilization for VAM = 4.3 TPH Required condenser duty 2326153.8+(500*3024) = 3838153.8 kcal/hr Required Cooling water 3838153.8/10 = 383.8 m3/hr SECTION III : LONG DESCRIPTIVE QUESTIONS Marks 6 x 10 = 60 L1 1. The lumen (lm) is the photometric equivalent of the Watt, weighted to match the eye response of the "standard observer," with blue light receiving the greatest weight False. 2. The CRI of a lamp is 100 if it renders the color of the chips identical to the reference light source, indicating perfect color rendering. - True. 3. A commercial building with a high window-to-wall ratio (WWR) and low SHGC glazing will experience higher cooling loads, as more solar heat will be transmitted through the windows - False. 4. Rotary screw compressors are preferable for fluctuating air demand- False. 5. Operating compressors at lower delivery pressures always results in higher energy efficiency- True. 6. Heat of compression dryers have higher operating costs compared to heatless purge dryers- False 7. Using variable speed drives in compressors can eliminate unloaded running conditions and save energy-True 8. Motor efficiency generally increases as the motor's rated capacity increases. True 9. The power factor of an induction motor improves as the load on the motor decreases. False 10. A decrease in supply voltage by 10% will decrease the torque of the motor by approximately 19% - False L2 A 2-stage reciprocating compressor is supplying nitrogen from low pressure header to high pressure vessel. This high-pressure nitrogen is only used during any process upset. Compressor is cut-off, once vessel pressure reaches 45 barg, and started, when vessel pressure comes down to 35 barg. During energy audit, it was observed that compressor is started at gap of every 36 hrs when there is no intended consumption. Other data is given below: Vol. of high pressure N2 vessel 11.5 m3 Vessel temperature 35.0 Deg.C Initial gas density 50.3 kg/m3 End gas density 39.4 kg/m3 Compressor load kW drawn 30.0 kW Compressor capacity at constant suction pressure 250.0 kg/hr i. ii. Estimate the leak rate (kg/hr). 5 Marks Estimate the energy saving potential (kWh/Annum), if all leaks are attended. Consider operating time of 8760 hrs/annum. 5 Marks Solution: Initial Vessel Pressure End vessel pressure Initial gas density End gas density Change in gas quantity in 36 hrs 45.0 barg 35.0 barg 50.3 kg/m3 39.4 kg/m3 125.7 kg N2 leakage rate 3.5 kg/hr Time needed for compressor run % time of compressor running Running time of compressor per annum 0.51 hrs or 30.6 1.40 % 122.3 hrs min Power consumption per annum due to air leakage 3670.0 kWh/Annum Energy Saving Potential L3 3670.00 kWh/Annum a) During performance guarantee test of an induced draft cooling tower, it was found that design approach of cooling tower is not achieved. As one of the probable causes, the team decided to check the efficiency of cooling tower fan. If design static efficiency is 70%, estimate the operating static efficiency using following parameters: Pitot tube coefficient 0.9 Velocity pressure 49.0 mmWC Air Density at operating condition 1.129 kg/m3 Duct diameter 2.1 m Differential pressure across fan 130.0 mmWC Motor shaft Power 190.0 kW Motor Efficiency 95.0 % Gear Box Efficiency 96.0 % b) A centrifugal fan drawing 54 kW and operating at 1440 rpm is delivering air at 30,000 m3 /hr. The head developed by the fan is 400 mm WC. If the speed is decreased by 200 rpm, calculate the following: 1. Air flow in m3 / hr 2. Static pressure in mm WC 3. Power drawn in kW Solution: a) Air Velocity Duct Area Vol. flowrate Air kW transferred Power input to fan Static Efficiency = 26.3 m/sec = 3.5 m2 = 92.0 m3/sec = 92*130/102 = 117.25 kW = 190*0.95*0.96 = 182.4 kW = 117.25 / 182.4 x100 = 64.28% b) 1. Airflow in m3 / hr = (1240/1440) x 30000 = 25833 m3 /hr 2. Static pressure in mm WC = (1240 /1440)2 x 400 = 296.6 mm WC 3. Power drawn in kW = (1240 /1440)3 x54= 34.48 kW L4 A 20 MW co-generation plant operates at a daily load factor of 85% and 8% auxiliary power consumption. The power is generated at 11 kV. Out of the total energy generated, 45% is exported to the grid through a 15 MVA transformer with 99% efficiency. Additionally, 35% of the generated energy is supplied to mill motors at 600 Volts through an 8 MVA step-down transformer with 98.5% efficiency. The remaining energy is used for other LT loads and auxiliaries at 415 Volts through a 4 MVA transformer with 98.2% efficiency. Calculate the following: 1. Daily energy generation in MWh. 2. Daily energy exported in MWh to the grid at 33 kV. 3. Daily mill motors consumption in MWh at 600 V. 4. Daily LT loads and auxiliary consumption in MWh at 415 V. 5. Daily transformer losses in kWh and % transformer losses. Solution 1. Daily Energy Generation Calculation Gross energy generated=Plant capacity Load factor Hours per day Gross energy generat ed =20 MW 0.85 24 hours Gross energy generated=408 MWh per day Net energy available after auxiliary consumption: Net energy available=Gross energy generated (1 Auxiliary power consumption) =408 MWh 0.92 = 375.36 MWh 2. Energy Exported to the Grid Energy available for exported =Net energy available 45% =375.36 MWh 0.45 =168.912 MWh Energy exported to the grid after transformation loss=168.912 MWh 0.99 =167.223 MWh 3. Daily Mill Motors Consumption Energy available for mill motors =Net energy available 35% =375.36 MWh 0.35 =131.376 MWh Energy supplied to mill motors after transformation loss = 131.376 MWh 0.985 =129.278 MWh 4. Daily LT Loads and Auxiliary Consumption Energy available to LT loads and auxiliaries = =375.36 MWh 0.20 =75.072 MWh Energy supplied to LT loads and auxiliaries after transformation loss = 75.072 0.982 =73.78 MWh 5. Calculate Daily Transformer Losses in kWh and % Transformer Losses Total Transformer Losses: Grid Transformer Losses+Mill Motors Transformer Losses+LT Transformer Losses Total l osses= =(168.912-167.223)+(131.376-129.278)+(75.072-73.78) 1,689 kWh+2,098 kWh+1,292 kWh Total losses=5,079 kWh Percentage of Transformer Losses from Net Generation: 5079/408000 = 1.24% L5. As part of a management initiative to advance green energy in a new process plant, a process engineer is assessing the economic viability of a 650 TR chiller. She is considering proposals for both LiBr-based vapor absorption chillers and vapor compression refrigeration systems. While power is sourced from renewable energy, the steam required is partially generated from excess process heat and additionally from firing furnace oil. COP of advance Vapor Absorption Chiller : 1.3 COP of Vapor Compression Chiller : 4.50 Net steam price including excess steam and from boiler : 1500.00 INR/MT Net Power cost from green source : 7.20 INR/kWh Price of Cooling water : 3.00 INR/M3 Cooling water range : 80C Specific steam heat available for chiller : 490.0 kcal/kg Evaluate both the offers and find out the offer which is economical in terms of operating cost. Solution: Vapor Absorption chiller Chilling capacity 650.0 TR Required heat duty 1965600 kcal/hr Heat equivalent input to VAM = 1965600/1.3 = 1512000 Required Steam flow = 1512000/490 = 3084 kG/hr Condenser heat duty = (650*3024) + (3084* 490) = 3476760 kcal/hr Required Cooling Water = 434.6 M3/hr Operating cost of Vapor Absorption Chiller =(3084*1.5)+(434.6*3) = 5930 INR/hr Vapor Compression chiller Chilling capacity 650.0 TR Required heat duty 1965600 kcal/hr Required Power consumption 508 kW Condenser heat duty = (650*3024)+(508*860)=2402480kcal/hr Required Cooling Water =300 M3/hr Operating cost of Vapor Absorption Chiller = (508*7.2)+(300*3) = 4558 INR/hr Hence, operating vapor compression chiller is economical. L6 a) Match the following: (4 Marks) 1 2 3 4 Prescriptive Approach Whole Building Performance Approach Building envelope Effective Aperture 1. 2. 3. 4. Exterior fa ade Trade-Off option light admitting potential Uses simulation to show compliance for the entire building b. Fill in the following blank statements: 1. The Effective Aperture (EA) or light admitting potential of a glazing system is determined by multiplying the Visible Light Transmittance (VLT) of the glazing by the _____________ of the building. 2. Thermal emittance is the relative ability of a material to __________the absorbed heat. 3. If a window has a SHGC of 0.25 and the total incident solar radiation is 600 W/m , the solar heat gain through the window is _________ 4. The emissivity of a material is the ratio of energy radiated by a particular material to energy radiated by a __________ at the same temperature. 5. As per ECBC the unit of Energy Performance Index (EPI) __________ 6. Fenestration surface having a slope of less than 60 degrees from the horizontal plane is termed _________________ Solution: a) 1. 2. 3. 4. Prescriptive Approach b Whole Building Performance Approach d Building envelope a Effective Aperture c b) 1. The Effective Aperture (EA) or light admitting potential of a glazing system is determined by multiplying the Visible Light Transmittance (VLT) of the glazing by the Window-Wall Ratio (WWR) of the building. 2. Thermal emittance is the relative ability of a material to radiate the absorbed heat. 3. If a window has a SHGC of 0.25 and the total incident solar radiation is 600 W/m , the solar heat gain through the window is 150 W/m . 4. The emissivity of a material is the ratio of energy radiated by a particular material to energy radiated by a black body at the same temperature. 5. As per ECBC the unit of Energy Performance Index (EPI) kWh/Sq.mt/year 6. Fenestration surface having a slope of less than 60 degrees from the horizontal plane is termed Skylight. Paper-3 Code : Green 25th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER, 2025 PAPER - 3 : ENERGY EFFICIENCY IN ELECTRICAL UNTILITIES Date : 28.09.2025 Timings: 09:30-12:30 HRS Duration: 3 HRS Max. Marks: 150 General instructions: o o o o o Please check that this question paper contains 8 printed pages Please check that this question paper contains 64 questions The question paper is divided into three sections All questions in all three sections are compulsory All parts of a question should be answered at one place Section I: OBJECTIVE TYPE Marks: 50 x 1 = 50 (i) Answer all 50 questions (ii) Each question carries one mark (iii) Please shade the appropriate oval in SECTION-I of MAIN ANSWER BOOKLET with BLUE/BLACK ball point pen 1 2 3 4 5 6 7 8 9 10 11 12 The most influential component for cooling tower performance is: a) Fill media b) Drift eliminator c) Casing d) Fan motor An equipment room measures 12 8 3.5 m. Ventilation required for 15 ACH is: a) 5060 m /h b) 5020 m /h c) 5040 m /h d) 5080 m /h Luminous efficacy is: a) Ratio of lumens to watts b) Measured in candela c) Same as luminance d) Measures reflection of light If dew point temperature equals air temperature, relative humidity is: a) 0% b) 45% c) 50% d) 100% In a DG set, the component causing maximum energy loss is: a) Coolant loss b) Alternator loss c) Radiation loss d) Flue gas loss The Energy Conservation Act applies to buildings with connected load: a) All HT connections b) Commercial building having 100 kW c) Residential buildings only d) Government buildings having 100 kW A 750 kVA transformer has 1200 W no-load loss and 7200 W full-load copper loss. At 60% load, total loss is: a) 4320 W b)5520 W c) 7632 W d) 3792 W In vapour compression and vapour absorption systems, the common refrigerant is: a) Lithium bromide b) R-134a c) Ammonia d) None of the above Which lamp is most suitable for color-critical applications? a) Halogen lamps b) LED lamps c) CFLs d) Metal halide lamps Iron losses in an electric motor can be reduced by using a) More copper and large conductors b) Use of thinner gauge lower loss core steel c) Use of low loss fan design d) Optimised design and strict quality control A cooling tower has an evaporation loss of 12 m /hr and COC of 2.5. What will be the blowdown loss in m /hr ? a) 5.2 b) 8.0 c) 9.6 d) 10.2 Energy savings by motor replacement can be worked out by: a) KW output ( old new) b) KW output ( new old) c) KW output (1/ old 1/ new) d) KW output (1/ new 1/ old) BUREAU OF ENERGY EFFICIENCY 1 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 Paper-3 Code : Green Stray losses in a motor are mainly caused by: a) Leakage flux induced by load currents b) Hysteresis and eddy currents c) Frictional losses d) Copper winding losses In a vapor compression refrigeration system, enthalpy changes occur across: a) Compressor b) Condenser c) Evaporator d) All of the above The purpose of inter-cooling in a multistage compressor is to: a) Increase final pressure b) Reduce compression work c) Separate oil vapour d) Remove all moisture When air is cooled by evaporation in an air washer: a) Humidity ratio decreases b) Dry bulb temperature decreases c) Dry bulb temperature increases d) Enthalpy increases Amorphous core transformers primarily reduce: a) Load loss b) No-load loss c) Stray loss d) None of the above If a pump delivery valve is throttled to 30% of rated flow, best energy efficiency measure is: a) Replacing the motor b) Installing a larger impeller c) Increasing pump speed d) None of the above Calculate the FAD in CFM for an air compressor with a cylinder displacement of 150 CFM and volumetric efficiency of 90%: a) 165 b) 135 c) 150 d) None of the above If pump speed is reduced to 2/3rd of its original speed, power consumption will: a) Decrease by half b) Decrease to one-fourth c) Decrease to approx. 30% of original d) Remains same At higher altitudes, for same FAD, air compressors: a) Consume less power b) Consume more power c) Show no difference d) Work without lubrication In a 4-stroke diesel engine, fuel is injected during: a) Induction stroke b) Compression stroke c) Ignition and Power stroke d) Exhaust stroke Voltage unbalance in motors: a) Reduces motor temperature b) Increases motor slip c) Causes excessive heating and reduces life d) Improves torque Soft starters are used to: a) Increase motor speed b) Reduce inrush current c) Convert AC to DC d) Improve efficiency An air compressor is driven by an IE3 premium efficiency motor. Compared to an IE2 motor of the same rating, which of the following statements is most accurate? A) The IE3 motor will always consume less power under all load conditions. B) The IE3 motor achieves higher efficiency mainly by reducing copper and iron losses. C) The IE3 motor has lower inrush current during starting compared to IE2. D) The IE3 motor achieves efficiency by increasing slip. Scale in condenser tubes: a) Increases energy use b) Reduces heat transfer c) Can lead to higher operating pressure d) All of the above Larger diameter ducts in fans: a) Increase system resistance b) Reduce system resistance c) No effect d) Increase static pressure 28 Which of the following is a common symptom indicating that a pump is oversized? a) High discharge pressure b) Throttle valve-controlled systems c) Low suction pressure d) High motor power consumption 29 Reducing the diameter of an impeller in a centrifugal pump will: a) Increase head b) Decrease head c) No effect on head d) Increase flow BUREAU OF ENERGY EFFICIENCY 2 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 Paper-3 Code : Green The main function of fill media in a cooling tower is to: a) Reduce drift losses b) Increase water air contact c) Reduce fan noise d) Filter suspended solids Energy-saving opportunities in cooling towers include: a) Optimizing fan blade angle seasonally b) Maintaining correct water chemistry c) Cleaning fill media regularly d) All of the above The L/G ratio in a cooling tower is: a) Ratio of liquid water mass flow to gas (air) mass flow b) Ratio of drift loss to make-up water c) Ratio of cooling load to fan power d) Ratio of TDS in blowdown to TDS in make-up water Increasing chilled water leaving temperature in a centrifugal chiller: a) Increases efficiency b) Decreases efficiency c) No effect d) Increases refrigerant flow Luminaires are used to: a) Store electrical energy b) Distribute and control light from lamps c) Produce light directly d) Increase lamp wattage The Envelope Performance Factor (EPF) in ECBC is used to: a) Compare energy efficiency of proposed and baseline building designs b) Determine cooling tower sizing c) Calculate lighting power density d) Measure indoor air quality In a cooling tower, if any three of the four parameters: heat load, range, approach, and wet-bulb temperature, are kept constant, the required tower size will vary a)Directly with the heat load b)Inversely with the range c)Inversely with the approach d) All the above A system resistance curve of a fan changes with: a) Inlet guide vanes b) Discharge dampers c) Speed change with VFD d) Any of the above A centrifugal pump has BEP efficiency of 65%. At shut-off head, efficiency is: a) 0% b) 65% c) 50% d) 30% A pump with 200 mm impeller delivers 120 m /h. To deliver 100 m /h by trimming, impeller size will be approximately: a) 240 mm b) 167 mm c) 60 mm d) 276 mm If tail-end power factor is improved from 0.80 to 0.95, distribution loss reduction is: a) 13.33% b) 21% c) 29% d) 16% In a vapour compression system, refrigerant changes from vapour to liquid in the: a) Compressor b) Evaporator c) Condenser d) Expansion valve Unity power factor means: a) No reactive power is drawn from the supply b) Current leads voltage c) Current lags voltage d) Reactive power is maximum Examples of lighting controls include: a) Dimmer switches b) Timers c) Photo-sensors d) All of the above A 4-pole, 50 Hz induction motor runs at 1470 rpm. Slip is: a) 0.02 b) 0.20 c) 0.25 d) 0.30 Which type of compressed air dryer consumes the least power for capacities higher than 250 CFM? a) Refrigeration type b) Blower reactivated type c) Heat of compression type d) Heatless purge type A DG set operates at 1250 kVA, 0.8 PF, with specific fuel consumption of 0.23 L/kWh. Quantity of fuel used is_____________. a) 175 L/h b) 230 L/h c) 250 L/h d) 300 L/h In a UPS, DC to AC conversion is carried out by: a) Converter b) Charger c) Battery d) Inverter BUREAU OF ENERGY EFFICIENCY 3 Paper-3 Code : Green 48 49 50 If a plant receives 96 Million Units (MU) with a T&D efficiency of 80%, generation required is: a) 101.2 MU b) 76.9 MU c) 120 MU d) 68.1 MU Synchronous speed of a motor is inversely proportional to: a) Number of poles b) Frequency c) Voltage d) Temperature If 27,216 kcal of heat is removed per hour, refrigeration tonnage is: a) 10 TR b) 7 TR c) 11 TR d) None of the above .................. End of Section I .................. Section II: SHORT DESCRIPTIVE QUESTIONS (i) (ii) Marks: 8 x 5 = 40 Answer all Eight questions Each question carries Five marks S-1 A commercial training hall with dimensions 18 m 12 m is being planned. Calculate the number of 18 W LED lamps, each providing 1800 lumens, required to achieve an illuminance level of 300 Lux. The lamps will be installed at a height of 3 meters from the working plane. The utilisation factor (UF) of the system is 0.70, and the light loss factor (LLF) is 0.80. S-1 Area of room (A): 18 12 = 216 m Ans Total lumens required ( _total): _total = E A = 300 216 = 64800 lumens Effective lumens per lamp ( _lamp_effective): _lamp_effective = Lumen Output UF LLF = 1800 0.70 0.80 = 1008 lumens Number of lamps required (N): N = _total / _lamp_effective = 64800 / 1008 = 64.3 65 lamps S-2 A private power distribution company has implemented new digital metering and billing systems to improve efficiency in a residential zone. After six months of operation, the following data was recorded: Input energy to the system = 75 MU Metered billed energy = 56 MU Unmetered average billing = 4 MU Amount billed = 680 million Total amount received = 600 million Arrears collected = 90 million Purchased energy cost = 8.50 per kWh i) Estimate the Aggregate Technical and Commercial (AT&C) loss (%) and the revenue realized per kWh 4 Marks ii) Calculate the revenue loss per kWh to the company due to AT&C loss. BUREAU OF ENERGY EFFICIENCY 1 Mark 4 Paper-3 Code : Green S-2 Input Energy = 75 MU = 75,000,000 kWh Ans Metered Billed Energy = 56 MU Unmetered Average Billing = 4 MU Total Energy Billed = 56 + 4 = 60 MU Amount Billed = 680 million Arrears Collected = 90 million Amount Received = 600 million Purchased Energy Cost = 8.50/kWh i) AT&C Loss (%) and Revenue Realized ( /kWh) Billing Efficiency = (60 / 75) 100 = 80.0% Collection Efficiency = ((600 - 90) / 680) 100 = (510 / 680) 100 = 75.0% AT&C Loss (%) = 1 - (Billing Efficiency Collection Efficiency) = 1 - (0.80 0.75) 100 = 1 - 0.60 100 = 40.0% Revenue Realized per kWh = (600 - 90) / 75 = 510 / 75 = 6.80/kWh ii) Revenue Loss per kWh Revenue Loss = 8.50 - 6.80 = 1.70/kWh Final Answers: i) AT&C Loss = 40.0%, Revenue Realized = 6.80/kWh ii) Revenue Loss per kWh = 1.70/kWh S-3 List five energy saving measures in compressed air system. S-3 Energy-Saving Measures in a Compressed Air System (Headings Only) Ans 1. Fixing Air Leaks 2. Reducing Compressor Discharge Pressure 3. Proper Sizing of Compressors 4. Using Variable Speed Drives (VSDs) 5. Avoiding Artificial Demand 6. Heat Recovery from Compressor 7. Reducing Air Intake temperature 8. Optimizing Piping Layout and Size 9. Using Efficient Air Dryers 10. Eliminating Inappropriate Uses of Compressed Air 11. Using Automatic Drain Traps 12. Implementing Demand-Side Control 13. Operating Compressors at Full Load 14. Performing Regular Maintenance and Monitoring S-4 Match the Following: 1. Solar Heat Gain Coefficient (SHGC) a. Coefficient of Performance (COP) 2. U-value b. Solar Reflectance 3. HVAC System Efficiency c. Fenestration Heat Gain 4. Cool Roof d. Building Insulation e. Conductive path for unwanted heat 5. Thermal Bridging transfer BUREAU OF ENERGY EFFICIENCY 5 Paper-3 Code : Green S-4 Answer Key: Ans 1 c. Fenestration Heat Gain 2 d. Building Insulation 3 a. Coefficient of Performance (COP) 4 b. Solar Reflectance 5 e. Conductive path for unwanted heat transfer A steel manufacturing facility is powered by a 3-phase, 6.6 kV, 50 Hz supply and operates the following S-5 electrical loads: An electric arc furnace consumes 1.2 MW at a lagging power factor of 0.65, a bank of induction motors for rolling operations consumes 800 kW at a 0.80 lagging power factor, and the lighting and instrumentation systems consume 100 kW at unity power factor. Due to utility regulations, the overall plant power factor must be improved to 0.95 lagging. A capacitor bank will be installed for compensation. As an energy auditor evaluate the following: a. b. c. d. Total active power consumption. (1 Mark) Total initial apparent power drawn by the facility. (1 Mark) The operating power factor. (1 Mark) Determine the total reactive power required to achieve the desired power factor. (2 Mark) S-5 Load Ans Arc furnace All Motors Lighting & instruments Total a. b. c. d. Real Power (KW) 1200 800 100 2100 PF 0.65 0.8 1 KVA = KW/PF 1846 1000 100 2946 Reactive Power (KVAr) = (KVA)2-(KW2) 1402 600 0 2002 Total active Power = 1200+800+100 = 2100 kW Total apparent power drawn: 1846+1000+100 = 2946 kVA Operating Power factor = 2100/2946 = 0.713 KVAR = KW (Tan PF1 -Tan PF2) = 2100 * (Tan (Cos-1(0.713)- Cos-1(0.95)) = 1376 kVAr S-6 Determine the difference in heat rejected in kCal/TR to the cooling tower for two different types of air conditioning system operating at same capacity. Parameter Centrifugal chiller VAM 3 - 180 - 340 13.0 14.6 - 33.5 7.7 9.0 - 39.1 0.6 - Chilled water flow (m /h) 3 Condenser water flow (m /h) Chiller inlet temp ( C) Condenser water inlet temp ( C) Chiller outlet temp ( C) Condense water outlet temp ( C) Specific power consumption (kW/TR) 1 TR = 3024 kcal/h S-6 Centrifugal chiller: Ans Power input = 0.6 860 = 516 kcal/TR Heat rejected = 3024 + 516 = 3540 kcal/TR VAM: Chilled water flow = 180 m /h = 180000 kg/h BUREAU OF ENERGY EFFICIENCY 6 Paper-3 Code : Green T = 14.6 9.0 = 5.6 C Cooling load = 180000 1 5.6 = 1008000 kcal/h TR = 1008000 / 3024 = 333.33 TR Heat Rejected = 340000 kg/hr x 1 kcal/kg x (39.1-33.5) = 1904000 kcal/hr Heat Rejected per TR = 1904000 / 333.33 = 5712 kcal/TR Difference per TR = 5712 - 3540 = 2172 kcal/TR S-7 An energy audit in an industrial unit revealed a fan directly coupled with motor was operating at 37 Hz through VFD for 500 hours/month and supplying air through a 150 mm diameter duct. The fan is designed to deliver an air flow of 1300 m /h with a rated input power of 3 kW at 50 Hz. Calculate the air velocity and annual energy savings ignoring the motor losses. Given: S-7 Design air flow rate = 1300 m /h at 50 Hz Ans Operating frequency = 37 Hz Power drawn at 50 Hz = 3 kW Operating hours = 500 hours/month Duct diameter = 150 mm = 0.15 m Flow rate at 37 Hz: Q2 = 1300 (37 / 50) = 962 m /h Q2 = 962 / 3600 = 0.2672 m /s Duct area: A = /4 (0.15) = 0.01767 m Air velocity: V = Q / A = 0.2672 / 0.01767 = 15.12 m/s Power at 37 Hz: P2 = 3 (37 / 50) = 3 0.405 = 1.215 kW Monthly energy savings: (3 - 1.215) 500 = 892.5 kWh/month Annual energy savings: 892.5 12 = 10710 kWh/year A pump is used to fill a rectangular overhead tank measuring 5 m 3.5 m with a height of 10 m. The inlet S-8 pipe to the tank is positioned at a height of 25 m above ground level. The following additional data is available: The pump draws water from an underground sump situated 4 meters below the pump level and delivers it to a tank whose overflow line is positioned 8 meters above the tank bottom. The motor driving the pump draws 7.5 kW of power. The operating efficiencies of the motor and the pump are 90% and 70% respectively. Calculate the time taken by the pump to fill the tank up to the overflow level. S-8 Step 1: Volume of water filled Ans Volume = 5 3.5 8 = 140 m Mass of water = 140 1000 = 140000 kg Step 2: Total head (H) Total head = 25 + 4 = 29 m Step 3: Shaft power (output of motor) BUREAU OF ENERGY EFFICIENCY 7 Paper-3 Code : Green Shaft Power = 7.5 0.90 = 6.75 kW Step 4: Water power (hydraulic power output of pump) Water Power = 6.75 0.70 = 4.725 kW = 4725 W Step 5: Time taken Pump efficiency = Mass flow g Head / Shaft Power 0.7 = mass flow x 9.81 x 29 /(6.75x1000) Mass flow = 16.6 kg/sec = 59791 kg/hr = 59.79 m3/hr Time = 140 / 59.79 = 2.34 hrs = 140.5 minutes .................. End of Section II .................. Section III: LONG DESCRIPTIVE QUESTIONS (i) (ii) Marks: 6 x 10 = 60 Answer all Six questions Each question carries Ten marks L-1 a) In a large-scale steel manufacturing facility, a cooling tower is used to reject heat from continuous casting operations. The circulating water flow rate is 2000 m /hr. The cooling tower is currently operating at a Cycles of Concentration (COC) of 3. The evaporation loss is estimated at 1.0% of the circulating flow, and the drift loss is 0.1% of the circulating flow. The facility is planning to improve the COC from 3 to 6 through advanced water treatment. Evaluate the following: i. ii. iii. iv. L-1 Calculate the make-up water requirement at the current COC of 3. Calculate the revised make-up water requirement if the COC is increased to 6. Estimate the total water savings per day. Discuss one limitation or risk associated with increasing the COC. b) True or False i. Cooling towers primarily reject heat through evaporative cooling. ii. The approach temperature in a cooling tower is the difference between the hot water temperature and the ambient dry bulb temperature. iii. Blowdown in a cooling tower is required to prevent the build-up of dissolved solids. iv. Drift losses in a cooling tower refer to water carried away with the exhaust air. v. Cooling tower effectiveness improves with higher approach temperatures. vi. Cycles of concentration in a cooling tower relate to how many times the water is reused before discharge. a) Given: Ans Circulating Water Flow (CWF) = 2000 m /hr Initial COC = 3 Final COC = 6 Evaporation Loss (E) = 1% of 2000 = 20 m /hr Drift Loss (D) = 0.1% of 2000 = 2 m /hr i) Current Make-up Water Requirement at COC = 3 B = 20 / (3 - 1) = 10 m /hr Make-up Water = E + D + B = 20 + 2 + 10 = 32 m /hr ii) Revised Make-up Water Requirement at COC = 6 B = 20 / (6 - 1) = 4 m /hr Make-up Water = E + D + B = 20 + 2 + 4 = 26 m /hr iii) Water Savings Hourly Savings = 32 - 26 = 6 m /hr BUREAU OF ENERGY EFFICIENCY 8 Paper-3 Code : Green Daily Savings = 6 24 = 144 m /day iv) Limitation of Higher COC Increasing COC can lead to higher concentrations of dissolved solids in the water, which may cause scaling, corrosion, and microbiological fouling in the system. Effective water treatment and frequent monitoring are necessary to avoid operational issues. b) True or False i) ii) iii) iv) v) vi) L-2 True False (It is the difference between the cold-water temperature and the wet bulb temperature.) True True False (Lower approach means better effectiveness.) True a) A manufacturing plant operates a 180 kVA diesel generator set rated at 0.8 lagging PF. The prime mover is a diesel engine rated 240 BHP. The alternator has total losses (including exciter power) of 5.44 kW. Assume no derating for site conditions. The generator is required to supply a mixed industrial load at its full kVA rating. The plant manager wishes to improve system efficiency by operating at a higher power factor. The diesel engine operates at a brake thermal efficiency of 32% when loaded near its rated capacity. The calorific value of the diesel fuel is 10,500 kCal/kg, and the specific gravity of the fuel is 0.85. Calculate the following: i) Maximum power factor that can be maintained at full kVA load without exceeding the engine capacity. (3 Marks) ii) Corresponding diesel fuel consumption (litres per hour) at this maximum power factor. (2 Marks) b) True or False Question: i) Improving power factor of the load on a DG set reduces the apparent power drawn and increases the system's overall fuel efficiency. ii) Alternator losses are independent of the load power factor. iii) Turbocharger in a diesel engine helps to reduce engine noise. iv) A diesel generator set must always be operated at unity power factor for maximum efficiency. v) DG sets are designed to handle unbalanced load between phases to 25% of their capacity. BUREAU OF ENERGY EFFICIENCY 9 L-2 Ans a) Paper-3 Code : Green i. Convert BHP to kW (shaft power): 240 BHP 0.746 = 179.04 kW Net electrical power available = 179.04 kW - 5.44 kW = 173.6 kW At full load (180 kVA), maximum PF = Real Power / Apparent Power PF = 173.6 / 180 = 0.964 ii. Thermal Input Required = Electrical Output / Efficiency = 179.04 / 0.32 = 559.5 kW Convert kcal to kW: 1 kg diesel = 10,500 kcal = 10,500 4.1868 = 43,961.4 kJ/kg Fuel consumption (kg/hr) = (559.5 3600) / 43961.4 = 45.8 kg/hr In litres per hour = 45.8 / 0.85 = 53.88 L/hr b) True or False Question: i) Improving power factor of the load on a DG set reduces the apparent power drawn and increases the system's overall fuel efficiency. True ii) Alternator losses are independent of the load power factor. False iii) Turbocharger in a diesel engine helps to reduce engine noise. False iv) A diesel generator set must always be operated at unity power factor for maximum efficiency. False v) DG sets are designed to handle unbalanced load between phases to 25% of their capacity. False L-3 A distribution company (DISCOM) plans to implement a comprehensive Demand Side Management (DSM) initiative to reduce its peak load and overall energy procurement cost. The program targets two consumer categories Residential consumers and Industrial consumers. For residential consumers, DISCOM has taken LED replacement as a DSM intervention whereas for industrial consumers load shifting strategy has been adopted through peak and off-peak electricity pricing. The DISCOM supplies electricity to 10,000 households, each using 4 CFL bulbs (30 W each). These bulbs are used for 5 hours per day during evening peak hours. The DISCOM replaces each CFL bulb with a 9 W LED bulb. Procurement cost of each LED bulb costs 100, however the DISCOM provides the LED bulbs in place of CFL to consumers at a subsidized rate of 70/bulb. Administrative cost per household for the program Rs.10/-. DISCOM also serves 50 industrial consumers, each with a shiftable evening load of 100 kW, used from 6 PM to 10 PM. DISCOM incentivizes these consumers to shift their load from 10 PM to 2 AM by offering an incentive of 2 per kWh shifted. Eighty percent of industrial consumers agree to take advantage of the tariff incentive scheme. Power purchase cost for DISCOM: o Evening Peak (5 10 PM): 7/kWh o Late Night (10 PM 6 AM): 3/kWh Calculate the following: a) For the residential consumers: i) Calculate the total daily energy savings in kWh/day from the LED replacement program. 2 Marks ii) Determine the daily cost savings for the DISCOM from the LED program. 1 Mark iii) Calculate the total one-time cost to the DISCOM for the LED program, including subsidies and administrative costs. 2 Marks iv) Estimate the simple payback period in days for the LED program. 1 Mark b) For the industrial consumers: i) Calculate the total energy shifted in kWh/day. 1 Mark ii) Compute the net daily savings for DISCOM, considering power cost reduction and incentive payout. 1 Mark c) Estimate the carbon emission avoidance due to above two DSM activity, if emission factor of grid electricity is 0.716 tCO2/MWh. 2 Marks BUREAU OF ENERGY EFFICIENCY 10 Paper-3 Code : Green Daily energy savings from LED replacement: (10000*4*5) * (30 9) /1000 = 4200 kWh L-3 Daily cost savings for DISCOM = 4200 *(7) = Rs.29400 Ans One time cost to DISCOM: Subsidy for LED replacement = 10000 * 4 * (100-70) = Rs. 12 Lakh Administrative cost = 10000*10 = Rs. 1.0 Lakh Total One-time cost to DISCOM = 12+ 1.0 = Rs. 13.0 Lakh Simple Payback = 1300000 / (16800*365) = 44 Days Energy Shifted by industrial consumers = 50*100*4*0.8 = 16000 kWh per Day Net daily savings = 1600 *(Purchase cost savings incentive cost) = 16000* (7-3-2) = Rs 32000/ Day Energy Savings achieved only by the LED program, hence emission reduction = 4200*0.716/1000 = 3.0 tCO2 per day. A) A clear water pump with rated flow of 125 m3/hr, head 55 m at rated speed of 1460 rpm and 79% L-4 efficiency is being used for supplying clarified water to a residential colony s water treatment facility. The daily water requirement is for 3000 M3. The pump is directly coupled and driven by a three phase 50 Hp, 415 V, 64A, 0.9 pf, 1460 rpm induction motor with 90.5% full load efficiency. During an internal energy audit, it has found that the motor is designed to operate with only 65% loading at pump rated conditions, therefore compromising on motor efficiency. The plant management has considered replacing the standard motor with a 30 kW IE3 motor. The following are operating parameters before and after motor replacement: Parameters Before motor replacement After motor replacement Flow (m3/hr) 130 ? Head (m) 52 51 Supply Voltage (V) 415 415 Current (Amp) 42 39 Power Factor 0.9 0.92 Motor Eff (%) 0.88 0.932 As an external auditor, you observed that although the plant has reduced the size of the induction motor to improve loading and enhance motor efficiency, the slip of the new IE3 motor has decreased by 20 rpm. This raises concerns about the actual energy savings achieved. Validate the savings claimed, calculate the following: i) ii) iii) iv) % Loading of motor after replacement. (1 Mark) Flow after replacing the standard motor with 30 kW IE 3 motor. (1 Mark) Operating Pump Efficiency before and after motor replacement. (2 Marks) Daily energy saving during operation due to motor replacement. (1 Mark) B) Mark the following statements as True/False i) Totally enclosed, fan cooled (TEFC) motors are less efficient than screen-protected, drip-proof (SPDP) motors. ii) Stray loss in induction motors is inversely proportional to load current. iii) As per BIS standard, the motor output should not be affected with voltage variation up to +/- 6%. iv) Motor life doubles for each 100C reduction in operating temperature. v) Starting torque of energy efficient motors is higher than standard motors. BUREAU OF ENERGY EFFICIENCY 11 Paper-3 Code : Green L-4 A) i) Loading of the IE3 Motor: = (1.732*0.415*39*0.92*0.932)/30 = 80.12% ii) Flow after replacement = 130 m3/hr * 1480/1460 = 131.8 m3/hr iii) Operating Pump Efficiency: Power Consumption before replacement = 1.732*0.415*42*0.9 = 27.17 kW Power Consumption after replacement = 1.732*0.415*39*0.92 = 25.79 kW Ans Before replacement = [(130 m3/hr /3600) * 52 *9.81] / (27.17*0.88) = 77% After replacement = [(131.8 m3/hr/3600) * 51 *9.81] / (25.79*0.932) = 76.2% Daily Operating Hour before Pump replacement = 3000 /130 = 23.08 Hrs Daily Operating Hour after Pump replacement = 3000 /131.8 = 22.77 Hrs iv) Daily Energy Savings = (23.08*27.17) (25.79*22.77) = 39.9 kWh B) True/False: i) False ii) False iii) True iv) True v) False A Commercial Office building accommodates two government departments. The total employees working in L-5 both the departments is 250 out of which 70% average number of employees present at any time. Being a government office, the building is operational 6 days a week with 10 working Hrs a day. The building receives Electricity supply from local electricity Distribution company through a 33KV feeder, and it is distributed after stepping down to 415 V. The building does not have separate parking, lawn, internal roads etc. The building information sheet is as below: Office Building 1. 2. 3. Contract Demand (kW) Installed capacity: Diesel Generating (DG) Set(s) (kVA) a) Annual Electricity Consumption, purchased from Utilities (kWh) b) Annual Electricity Consumption, through DG Set(s) (kWh) 4. a) Annual Cost of Electricity, purchased from Utilities (Rs.) b) Annual Cost of Electricity generated through DG Set(s) (Rs.) 5. 6. 7. 8. 9. 10. Built Up Area (sq.m) Area of the Building Conditioned Area (sq.m) Installed capacity of Chiller of Air Conditioning System (TR) Installed lighting load (kW) Officer Appliances (kW) Other Loads (kW) HSD Consumption in DG (GCV 10800 Kcal/kg and density of 0.85) Annual Data (April 24-March 25) 130 160 105753 2136 1043557 54405 3591.96 2155.18 137.5 8.11 11.0 12.5 585 Litres Calculate the following: a. b. c. L-5 Ans d. e. a. b. c. The total electricity consumed by the building and average electricity unit cost. (2 Mark) EPI of the building considering the reported data for past one year. Also recommend the appropriate rating under BEE star rating program for buildings if the bandwidth of the EPI range between 15050 kWh/sq. m/year. (3 Marks) Calculate the design diversity factor of the building, if the design EER of the chiller is 3.5 and recorded maximum demand is 75% of the contract demand. (2 Marks) Estimate the overall operating efficiency of the DG set (2 Marks) Calculate the lighting power density (1 Marks) Total electricity consumed: 105753+2136 =107889 kWh Average unit cost: (1043557+54405)/107889 = Rs.10.18 per kWh EPI = 107889/3591.96 = 30 kWh/m2/year, the EPI is below the bandwidth for star rating, hence 5 Star rated. Diversity Factor = Maximum Demand / Connected Load Maximum Demand = 130 * 0.75 = 97.5 kW BUREAU OF ENERGY EFFICIENCY 12 Paper-3 Code : Green Connected Load = Lighting load + appliances +others+ AC = 8.11+11.0+12.5+ (137.5*3024/860/3.5) = 169.75 kW Diversity Factor = 97.5/169.75 = 0.57 (or) 169.75/97.5 = 1.74 d. Overall, DG Set efficiency= (2136*860) / (585*0.85*10800) = 34.2% e. Lighting Power Density = 8.11*1000/3591.96 = 2.25 W/m2 A 5-star business hotel operates a centralized HVAC system operating round the clock with the following L-6 configuration. Only one chiller operates at a time, while the other is on standby. Two centrifugal chillers, each rated at 250 TR, with EER varying with load as below: No change in EER observed above 85% load, assume chiller motor efficiency of 90% at all loading conditions and the energy consumption by the auxiliary systems is as below: During chiller operation, two pumps run in parallel at an 80% load factor, consuming a total of 19.7 kW. In addition, two cooling tower fans operate continuously with a power consumption of 5.89 kW. Both the pumps and fans function 24 hours a day, with overall efficiencies of 75% and 70% respectively. The applicable electricity tariff is 6.5 per kWh. Evaluate the following: a. The total annual energy consumption (in MWh) and cost of the HVAC system, considering part-load EERs and auxiliary loads. (4 Marks) b. Heat removal by condenser in (TR) at different loads. (3 Marks) c. The hotel is planning to use the chiller partially as a heat pump by mounting a plate heat exchanger in series between the compressor and condenser (desuperheater for partial heat recovery) for producing hot water. The heat recovery can be only 20% of the condenser heat discharge. If the hot water requirement is 2000 litters/hr with 100C temperature rise, evaluate whether the hot water requirement can be met at 40% loading conditions. (3 Marks) BUREAU OF ENERGY EFFICIENCY 13 Paper-3 Code : Green L-6 Ans a. Energy Consumption: Energy Consumed by Chiller: Loading (%) Cooling Load (TR) EER A B C Cooling Load (KW) D=B*3024/86 0 Input Power (KW) Days Mwh E=D/C F G=E*F*24/100 0 0.85 212.5 5.2 747.21 143.69 180 620.8 0.6 150 4.6 527.44 114.66 120 330.2 0.4 100 3.9 351.63 90.16 65 140.7 Energy Consumed by auxiliaries: (19.7+5.89) *24*365/1000= 224.5 Mwh Total Annual Energy Consumption = 620.8+330.2+140.7+224.5 = 1316 Mwh Annual Cost = 1316000*6.5 = Rs. 85.55 Lakh b. Heat Removal by Condenser (TR) Loading (%) X 0.85 0.6 0.4 Cooling Load (TR) Y 212.5 150 100 Input Power (KW) Z 143.69 114.66 90.16 Power Input to Compressor (kW) P=Z8*0.9 129.32 103.20 81.14 Condenser Heat Load (TR) HL =Y+(P*860/3024) 249.28 179.35 123.08 c) Heat Hoad at 40% loading = 123.08*3024 = 372194 kCal/Hr Recovery potential (20%) = 74439 kCal/Hr Heat requirement for hot water generation = 2000*1*10 = 20000kCal/H Therefore, the heating requirement can easily be met at 40% loading. ................. End of Section III .................. BUREAU OF ENERGY EFFICIENCY 14 Paper-4 Code: Green 25TH NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER 2025 PAPER 4: ENERGY PERFORMANCE ASSESSMENT FOR EQUIPMENT AND UTILITY SYSTEMS Date: 28.09.2025 Timings: 14:00-16:00 HRS Duration: 2 HRS Max. Marks: 100 General instructions: o o o o o o Please check that this question paper contains 6 printed pages Please check that this question paper contains 16 questions The question paper is divided into three sections All questions in all three sections are compulsory All parts of a question should be answered in one place Time management guidance: Section - I & II: 10 Minutes each, Section - III: 100 Minutes Section - I: BRIEF QUESTIONS (i) (ii) Marks: 10 x 1 = 10 Answer all Ten questions Each question carries One mark 1. Evaporation ratio is based on actual performance data (steam output and fuel input) and does not depend on whether efficiency is expressed on GCV or NCV basis. True 2. Isentropic efficiency of a back-pressure turbine will be more if extraction temperature is higher than steam temperature at back-pressure condition. False 3. Profitability Index (PI) will be greater than 1 for projects with positive NPV, but its value varies with cash flow patterns and is not always higher for all positive NPV projects. In an integrated iron & steel plant, all the rolling mills consume more energy compared to energy consumed for iron making. False 5. In indirect method of boiler efficiency calculations, blowdown losses are also considered. False 6. Lower the TTD (Terminal Temperature Difference) and DCA (Drain Cooler Approach) for feedwater heaters, the higher will be the efficiency of the cycle. True 7. If wet steam is generated, the high Evaporation ratio indicates high efficiency of boiler. False 8. The only reason for installing condensate recovery systems is to reduce makeup water. False 9. The heat rate of a thermal power plant can be improved by decreasing the condenser cooling water temperature. True 10. In a cement rotary kiln, the highest heat loss occurs through clinker discharge. False 4. True . End of Section - I . BUREAU OF ENERGY EFFICIENCY 1 Paper-4 Code : Green Section - II: SHORT NUMERICAL QUESTIONS Marks: 2 x 5 = 10 (i) Answer all the Two questions (ii) Each question carries Five marks L-1 In a petrochemical industry, both the Low-Pressure (LP) boiler and the High-Pressure (HP) boiler operate with the same evaporation ratio of 14, using the same fuel oil. The operating details are provided below: Particulars LP Boiler HP Boiler Pressure 10 kg/cm a 32 kg/cm a Temperature Saturated steam 400 C Enthalpy of steam 665 kcal/kg 732 kcal/kg Temperature of feed water 80 C 105 C Evaporation Ratio 14 14 If the efficiency of the LP boiler is 82%, calculate the efficiency of the HP boiler. L-1 Effy = ER. (hg hf) / GCV Ans EffyL.P 1 = 0.82 = 14 x(665 80) / GCV EffyH.P 2 = 14 x(732 105) / GCV EffyH.P 2 / EffyL.P 1 = (732 105)0.82 / (665 80) = 0. 8789 =87.89% Or EffyL.P 1= 0.82 = 14 x(665 80) / GCV GCV= 14x(665-80) / 0.82 = 9987.8 kcal/kg EffyH.P 2 = 14 x(732 105) / GCV = 14 x(732 105) / 9987.8 = 0.8789 = 87.89 % L-2 During the assessment year 2024 25, one of the thermal power plants reported a gross heat rate of 2300 kCal/kWh and an auxiliary power consumption of 9%, while the baseline net heat rate was 2600 kCal/kWh. If the baseline generation is 5000 MU and considering that 1 kg of oil equivalent = 10,000 kCal and 1 e-Certificate = 1 ton of oil equivalent (TOE), calculate the reduction in net heat rate compared to the baseline and determine the expected number of e-Certificates. L-2 Ans Assessment Year Net Heat Rate = 2300/ (100-9)/100 = 2527.47 kCal/kWh Reduction in net heat rate = Baseline Assessment year Net Heat Rate = 2600 2527.47 = 72.53 kCal/kWh Expected E-certs = 5000 x 10^6 x 72.53 / 10000 x 1000 = 36,263.74 e-Certificates . End of Section - II . Paper-4 Code : Green Section - III: LONG NUMERICAL QUESTIONS Marks: 4 x 20 = 80 (i) Answer all the Four questions (ii) Each question carries Twenty marks N-1 A pharmaceutical manufacturing plant operates a central chilled water system that serves both cleanroom AHUs and a process cooling water loop for tablet coating machines. The process cooling water passes through a counterflow heat exchanger, entering at 20 C and leaving at 14 C with a flow rate of 166 m /h. On the other side, the chilled water enters at 7 C and leaves at 12 C, with a flow rate of 200 m /h. The heat exchanger has an overall heat transfer coefficient of 2.8 kW/m C. The chilled and condenser cooling water loops are each served by centrifugal pumps with operating efficiencies of 78 % and 80 %, respectively and with a head of 18 m. The cooling tower is fitted with an induced-draft axial fan delivering 28 m /s of air at 42.83 mmWC total pressure, with a fan efficiency of 62 %. All pumps and the fan are driven by directly coupled three-phase induction motors with an efficiency of 92 %. The plant s HVAC system operates for 300 days annually, running 18 hours per day, to handle an average cooling load of 850 kW with an ISEER value of 4.4 Calculate the following: a) Estimate the required heat exchanger surface area. (6 marks) b) Determine the combined electrical load in kW of both pumps and the cooling tower fan, if condenser cooling water flow is 600 m /h. (6 marks) c) Determine the Specific Energy Consumption (SEC) of the chiller in kWh/TR and calculate the overall SEC. (4 marks) d) Calculate the annual energy consumption in kWh and the annual operating cost in . Lakhs if electricity is charged at 7.5 per kWh. (4 marks) N-1 a) Heat Exchanger Surface Area Sol (6 Marks) - Process water flow rate: 166 m /h = 46.1 kg/s - T = 20 - 14 C = 6 C - Heat load (Q): Q = cp T = 1156 kW - LMTD (counterflow): T1 = 8 C, T2 = 7 C, LMTD = 7.49 C - Area: A = Q / (U LMTD) = 1156 / (2.8 7.49) = 55.2 m b) Combined Electrical Load of Pumps and Fan (6 Marks) Chilled water pump (200 m /h, 18 m head): - Hydraulic power = 9.81 kW - Motor input = 13.7 kW Condenser pump (600 m /h, 18 m head): - Hydraulic power = 29.43 kW - Motor input = 40.0 kW Cooling tower fan (28 m /s, 42.83 mmWC): - Air power = 11.76 kW - Motor input = 20.6 kW Total auxiliary load = 13.7 + 40.0 + 20.6 = 74.3 kW c) Specific Energy Consumption (SEC) - Cooling load = 850 kW = 242 TR - Chiller input = 193.2 kW (ISEER = 4.4) (4 Marks) Paper-4 Code : Green Chiller SEC = 193.2 / 242 = 0.79 kWh/TR System SEC = (193.2 + 74.3) / 242 = 1.106 kWh/TR d) Annual Energy & Cost (4 Marks) Total power demand = 193.2+74.3 = 267.46 kW - Annual energy consumption = 267.46 18 x 300 = 1444304.94 kWh/year - Annual cost (@ 7.5/kWh) = 1444304.94 7.5 /100000 = 108.32 lakh/year N-2 A boiler is fired with 200 kg/hr of a hydrogen-enriched hydrocarbon fuel (C H H ) at atmospheric pressure and a temperature of 20 C. The flue gas, leaving the boiler at atmospheric pressure and 300 C, has the following dry composition by weight: CO = 12% O = 3% N = 85% Based on this information, determine: a) The main constituents of the fuel (carbon and hydrogen) 12 Marks b) The percentage composition of each constituent in the fuel 4 Marks c) The total mass flow rate (kg/hr) of the dry flue gas. 4 Marks Assumptions (per 100 kg dry flue gas): Nitrogen -85 kg/hr, Oxygen 3 kg/hr, CO2 12 kg/hr Sol Take air O mass fraction = 0.23 and N mass fraction = 0.77. N-2 Carbon in the fuel (from CO ): CO C: = 12 2 12 144 36 = = kg 44 44 11 =36/11=3.2727 kg/h of carbon/ per 100 kg/hr of flue gas Actual air supplied: N2 Air mass = N2 mass fraction in air = 85 0.77 = 110.39 kg/hr of air / 100 kg/hr of flur gas O supplied with that air: O2supplied=0.23 110.39 =25.39 kg/hr / 100 kg/hr of flue gas O consumed (used in combustion) = supplied residual in flue gas consumed O2 = 25.39 3 = 22.39 kg/hr /100 kg/hr of flue gas O used to oxidize carbon (to form CO ): 32 8 Stoichiometry: C + O2 CO2 mass factor = = . 12 O2 for C = 3 8 8 36 288 = = = 8.73 g/hr/100kg/hr of flue gas 3 3 11 33 O available for hydrogen: consumed O2 for H = O2 O2 for C = 22.39 8.73 = 13.66 kg/hr/100kg/hr of flue gas Each kg H requires 8 kg O to form H O, so H mass: = 13.66 = 1.70779 kg/hr 8 Fuel mass & composition (per 100 kg dry flue gas): Total fuel mass (C + H): fuel = + = 3.2727 + 1.70779 = 4.9805kg/hr Mass fractions: 3.2727 100 = 65.732663% ( 65.73%) 4.9805 1.7077 % = 100 = 34.267337% ( 34.27%) 4.9805 % = If fuel flow = 200 kg/h: Carbon: (200) = 200 0.6573 = 131.465326 kg/h Paper-4 Code : Green = 131.46 kg/hr Hydrogen: (200) = 200 0.3426 = 68.534674 kg/h = 68.54 kg /hr c) The total Mass Flow rate of Dry Flue gas in kg/hr: Carbon (C) = 131.46 kg/h Hydrogen (H) = 68.54 kg/h Mass of dry flue gas = Mass of wet flue gas Mass of H2O = 200 + Actual Air 9* 68.54 = Actual Air 416.86 Actual Air = Ta +Ea Ta = (2.67 x131.46 + 8x68.54) / 23% = 3910 kg/hr Ea = E% x Ta E% = O2%/ (21% - O2%) O2%= 3/32 / (12/44 + 3/32 + 85/28) = 2.75% E% = 2.75% / (21%- 2.75%) = 15.07% Ea = 15.07% x 3910 kg/hr = 589.24 kg/hr Therefore, Actual Air = Ta +Ea = 3910 kg/hr + 589.24 kg /hr = 4499.24 kg/hr Mass of Dry flue gas = 4499.24 416.86 = 4082.38 kg/hr N-3 i) A small Topping Cycle Gas Turbine cogeneration plant has the following operating parameters. Estimate the power generation in kW and the steam supplied from the HRSG in TPH. The Energy Auditor has suggested that the HRSG exit flue gas temperature can be maintained at 95 C by recovering more heat from the HRSG. If the HRSG exit flue gas temperature is maintained at 95 C, estimate the additional steam generation and also the EUF with improved steam generation. The operating parameters are given below: 12 Marks Parameter Natural Gas Fuel Firing Rate Lower Heating Value (LHV) Exhaust Gas Flow Rate Exhaust Gas Temperature Mean Specific Heat of Gas (Cp) Specific Power Generation HRSG Inlet Temperature HRSG Exit Temperature Steam Pressure Saturated Steam Temperature Steam Enthalpy Feedwater Temperature HRSG Efficiency Value 1500 Sm /hr 9600 kcal/Sm 16.35 kg/s 525 C 0.265 kcal/kg C 3.044 kWh/Sm 520 C 135 C 10 kg/cm 179 C 663 kcal/kg 105 C 80 % ii) A coal-fired boiler operates with the following parameters. Parameter Hours of Operation Feed Water Temperature Steam Enthalpy Value 24 hours 140 C 805 kcal/kg 8 Marks Paper-4 Code : Green GCV of Coal Evaporation Ratio Steam Flow Rate 4200 kcal/kg 5.7 265 TPH Calculate the Boiler Efficiency and Coal Consumption per hour. If the boiler efficiency is improved by 2% relative to the existing efficiency, then estimate the coal savings per day. N-3 a) Ans Power Generation Power = Fuel rate Specific power generation = 1500 3.044 = 4566 kW Steam Generation at Baseline (Exit 135 C) Heat recovered = 48,04,153.20 kcal/hr Steam generated = 8609.6 kg/hr = 8.61 TPH Steam Generation at Improved Condition (Exit 95 C) Heat recovered = 53,03,286 kcal/hr Steam generated = 9504 kg/hr = 9.5 TPH Additional steam = 9504 8609.6 = 894.5 kg/hr = 0.89 TPH Energy Utilization Factor (EUF) Fuel input = 16,747.2 kW Useful output = Power (4566) + Steam thermal (5587.23) = 10,153.23 kW EUF = 10,153.23 / 16,747.2 = 60.63 % Determination of EUF at improved Condition (Exit 95 C) New Useful output = Power (4566) + Steam thermal (6167.72) = 10,733.72 kW EUF = 10,733.72 / 16,747.2 = 64.09 % b) Step 1: Data Given Steam flow = 265 TPH = 265,000 kg/hr Steam enthalpy = 805 kcal/kg Feedwater enthalpy 140 kcal/kg GCV of coal = 4200 kcal/kg Evaporation ratio = 5.7 Step 2: Heat Output Net enthalpy gain = 805 140 = 665 kcal/kg Q_out = 265,000 665 = 17,62,25,000 kcal/hr Step 3: Coal Consumption (from ER) Coal/hr = 265,000 5.7 = 46,491 kg/hr = 46.491 TPH Step 4: Heat Input Q_in = 46,491 4200 = 19,52,62,200 kcal/hr Step 5: Boiler Efficiency = Q_out Q_in 100 = (17,62,25,000 19,52,62,200) 100 = 90.25% Step 6: Improved Efficiency (relative 2%) _new = 90.25 1.02 = 92.055% Step 7: New Coal Consumption Coal_new = Q_out ( _new GCV) Coal_new = 17,62,25,000 (0.92055 4200) = 45,580.70 kg/hr = 45.58 TPH Step 8: Coal Savings Coal/hr = 46.491 45.580 = 910 .53 kg/hr Daily Saving = 910.53 24 = 21,852.77 kg/day 21.85 TPD Answer any ONE of the following among four questions given below: Paper-4 Code : Green N-4 A 10,000 TPD cement plant purchases power from the grid, operates an 18 MW Captive Power (A) Plant (CPP) and also has a Waste Heat Recovery (WHR) system with a 9 MW turbine. Electricity is used for cement production and also supplied to the colony and other utilities. The plant also exports excess energy to the grid. The annual energy and production data are given below: Parameter Value Annual operating hours 8,300 hrs Energy imported 60,210,000 kWh Energy exported 3,150,000 kWh Energy supplied to colony & others 3,500,000 kWh CPP average generation 18 MW CPP heat rate 3100 kcal/kWh WHR turbine average generation 9 MW WHR turbine heat rate 3600 kcal/kWh Indian coal consumption 96,000 MT (GCV 4500 kcal/kg) Pet coke consumption 80,000 MT (GCV 7500 kcal/kg) Imported coal consumption 198,000 MT (GCV 7200 kcal/kg) Biomass consumption 10,000 MT (GCV 2850 kcal/kg) Clinker produced 2,700,000 MT Clinker-to-cement ratio 1.4 Calculate the following: a) Specific Electrical Energy Consumption (SEEC) in kWh/ton of cement. (8 Marks) b) Estimate the Specific Thermal Energy Consumption (STEC) in kcal/kg of clinker (7 Marks) c) If the CPP is operated using only Indian coal and Imported coal, calculate the coal blending ratio by weight required to achieve a blended coal GCV of 6000 kcal/kg. (5 Marks) N-4 Solution (A) Specific Electrical Energy Consumption (SEEC) 8 Marks Sol Energy Input CPP generation = 18 8300 1000 = 149,400,000 kWh WHR generation = 9 8300 1000 = 74,700,000 kWh Grid import = 60,210,000 kWh Total Input = 284,310,000 kWh Deductions Export to grid = 3,150,000 kWh Colony & others = 3,500,000 kWh Net for cement production = 284,310,000 3,150,000 3,500,000 = 277,660,000 kWh Cement Production Clinker produced = 2,700,000 MT Cement produced = 2,700,000 1.4 = 37,80,000 MT SEEC SEEC = 277,660,000 37,80,000 = 73.46 kWh/ton cement Specific Thermal Energy Consumption (STEC) 7 Marks Fuel Heat Input Indian coal = 96,000 4500 = 432 10 kcal Pet coke = 80,000 7500 = 600 10 kcal Imported coal = 198,000 7200 = 1425.6 10 kcal Biomass = 10,000 2850 = 28.5 10 kcal Total Heat Input = 2486.1 10 kcal Heat to Kiln As per PAT convention (no CPP deduction applied) Heat to kiln = 2486.1 10 kcal STEC Clinker produced = 2,700,000 MT = 2,700,000,000 kg STEC = 2486.1 10 2,700,000,000 = 920.78 kcal/kg clinker Coal Blending Ratio 5 Marks Define ratio Let x = fraction of Indian coal, (1 x) = fraction of Imported coal Equation 6000 = 4500x + 7200(1 x) 6000 = 4500x + 7200 7200x 6000 = 7200 2700x Paper-4 Code : Green 2700x = 1200 x = 0.444 (44.4%) Step 4: Ratio (1 Mark) Indian coal = 44.4%, Imported coal = 55.6% Coal Blending Ratio = 44 : 56 Or N-4 The following data was collected from a 500 MW turbine unit during an energy audit. The power (B) plant operates with a main steam (MS) flow of 1561 TPH at a pressure of 166 kg/cm and temperature of 529 C. The hot reheat (HRH) flow is 1413 TPH, with steam conditions of 42.4 kg/cm and 540 C, while the cold reheat (CRH) section records a pressure of 44.3 kg/cm and temperature of 341 C. The feed water enters at 246 C, whereas the MS, CRH, and HRH enthalpies are 806.47 kCal/kg C, 730.71 kCal/kg C, and 844.28 kCal/kg C, respectively. These optimized steam cycle parameters enable the generator to deliver a substantial power output of 501 MW with a boiler efficiency of 88%. The turbine cycle heater operating parameters are as below: Steam Heater Reference LP Heater 1 LP Heater 2 LP Heater 3 HP Heater 5 HP Heater 6 Feed Water in Feed Water out Design Values Temp ( C) Pressure (kg/cm ) Saturation Temp ( C) Drain Temp ( C) Temp ( C) Pressure (kg/cm ) Temp ( C) Pressure (kg/cm ) TTD DCA 92.5 140 209 416 335 -0.23 0.49 1.97 17.4 43 93.07 111.23 132.9 207.33 254.94 64.2 70.4 110 171 212 47.2 170 - 13.7 202 - 63.6 105 130 210 255 12.6 11.5 10.4 199 197 2.88 2.95 2.95 0 0.1 4.8 4.95 4.95 5 5 Neglect temperature loss in the feedwater line between heaters and calculate the following: a. Calculate the Turbine Heat Rate and the Unit Heat Rate. 8 Marks b. Determine the loss or gain in the Turbine Heat Rate due to deviations of the TTD (Terminal Temperature Difference) and DCA (Drain Cooler Approach) of the LP/HP Heater systems from their design values. 12 Marks Consider the following criteria: For every 0.56 C increase or decrease in TTD from the design value, the Heat Rate will increase or decrease by 0.014%. For every 0.56 C increase or decrease in DCA from the design value, the Heat Rate will increase or decrease by 0.005%. N-4 Turbine Heat Rate: (B) Turbine: HR: {(1561000 X (806.47-246))+ (1413000 X (844.27 730.714))}/ (501000) = 2066.57 kCal/kWh Ans Unit Heat Rate: = 2066.57 / 88% = 2348.38 kCal/kWh From the Above data, the following heater data can be inferred Heater Ref. Feed water In- Feed water In- Steam Inlet let Temp OC let Temp OC temp OC LP Heater -1 LP Heater-2 LP Heater -3 HP heater-5 HP Heater -6 47.2 63.6 105 170 210 63.6 105 130 210 255 92.5 140 209 416 335 Inlet steam Saturation Temp OC 93.07 111.23 132.9 207.33 254.94 Drain temp OC 64.2 70.4 110 171 212 So TTD (Terminal Temperature Difference) = Inlet Steam Saturation Temp OC Feed Outlet Temp OC DCA (Drain Cooler Approach = Drain temperature oC Feed Water Inlet Temperature oC Heater Ref. TTD oC DCA OC TTD oC DCA OC (Design) (Design) (Calculated) (Calculated ) 2.88 4.8 LP Heater -1 29.47 17 Paper-4 Code : Green LP Heater-2 LP Heater -3 HP heater-5 HP Heater -6 2.95 2.95 0 0.1 4.95 4.95 5 5 6.23 2.9 -2.67 -0.06 6.8 5 1 2 Difference Between design values and Operating values of TTD and DCA of Heaters. Heater Ref. LP Heater -1 LP Heater-2 LP Heater -3 HP heater-5 HP Heater -6 Total Difference TTDOperating TTDDesign 26.59 3.28 -0.05 -2.67 -0.16 26.66 DCAOperating DCADesign 12.2 1.85 0.05 -4 -3 7.1 Change in Heat rate because of deviation in TTD = (Net Change in TTD for All heaters X 0.014%/0.56oC ) o Since given, for every 0.56 C change in TTD HR will increase by 0.014% So Increase in HR because of TTD deviation= 26.66oC X 0.014 / 0.56 = 0.67475 % Change in Heat rate because of deviation in DCA = (Net Change in DCA for All heaters X 0.005%/0.56oC ) o Since given, for every 0.56 C change in TTD HR will increase by 0.005% So Increase in HR because of DCA deviation 7.1 X 0.005 / 0.56 = 0.064 % Total % Increase in Turbine HR because of deviation in operation of TTD and DCA of Heaters from Design Values = 0.7381 %. =0.7381 X 2066.57 = 15.25 kCal/kWh Or N-4 A DRI-route steel plant operates a DRI unit and a Steel Melting Shop (SMS). The plant also has a (C) coal-based captive power plant (CPP). Any shortfall in electrical energy is met by imported grid power. On average, the plant imports 1,20,000 kWh/day and the operational parameters are given below: Description Parameter Value Rated capacity 500 TPD Capacity utilization 70% DRI Unit Specific coal consumption 1.25 t coal / t sponge iron Specific power consumption 95 kWh / t sponge iron Coal GCV 5000 kcal/kg Yield 85% SMS Specific power consumption 830 kWh / t liquid steel Gross efficiency 27% CPP Auxiliary power consumption 8% of gross generation Coal GCV 5000 kcal/kg Grid Grid electricity heat rate 2700 kCal/ kWh Calculate the following: a) The daily production of sponge iron and liquid steel in TPD. (2 marks) b) DRI coal consumption in TPD and its thermal input in Million kcal/day. (2 marks) c) Total daily electrical energy demand of DRI and SMS in kWh/day. (3 marks) d) The CPP gross generation in kWh/day, CPP heat rate in kCal/kWh, CPP thermal input in Million kcal/day and the CPP coal consumption in TPD. (6 marks) e) The overall specific energy consumption (SEC) in Million kcal per tonne of liquid steel. (4 marks) f) Compare your SEC with a benchmark of 6.5 Million kcal/t of liquid steel, and comment briefly on performance. (3 marks) Paper-4 Code : Green N-4 1) Daily Production (2 marks) (C) Sponge iron = 500 70% = 350 TPD Ans Liquid steel = 350 0.85 = 297.5 TPD 2) DRI Coal & Heat Input (2marks) Coal = 1.25 350 = 437.5 TPD Thermal input = 437.5 5,000 1000 = 2,187.5 Million kcal/day 3) Electrical Demand (3 marks) DRI = 95 350 = 33,250 kWh/day SMS = 830 297.5 = 246925 kWh/day Total = 280,175 kWh/day 4) CPP Generation & Coal (6 marks) CPP net = 280175 120,000 = 160175 kWh/day CPP gross = 160175/ 0.92 = 174103.26 kWh/day Heat rate = 860 / 0.27 = 3,185.2 kcal/kWh Thermal input = 174103.26 3,185.2 = 554.55 Million kcal/day CPP coal = 554.55 10^6 / 5000 = 110.91 TPD 5) Overall SEC (4 marks) Total heat input = Thermal input to DRI + Thermal input to CPP + Thermal Input of Import Imported electricity thermal eq. = 120,000 2700 = 324.0 Million kcal/day Total heat input = 2,187.5 + 554.55 + 324.0 = 3066.05 Million kcal/day SEC = 3066.05 / 297.5 = 10.31 Million kcal/t 6) Benchmark Comparison & Comment (3 marks) Benchmark = 6.5 Million kcal/t; Actual = 10.31 Million kcal/t (~58.6% higher) Comments: Need reduction in DRI coal use, improved SMS SPC, better CPP efficiency, adoption of WHR, and auxiliary load reduction. Or N-4 (D) A composite textile mill uses stenters for drying and heat-setting applications, currently the stenter system is running on a coal-fired boiler, and there is a proposal to modify the system to a biomass-fired thermic fluid heater. The relevant data is given below: Parameter Cloth inlet temperature Cloth outlet temperature Cloth inlet moisture Cloth outlet moisture Stenter output Stenter Efficiency Latent heat of inlet steam to stenter at 10 bar Sensible heat of inlet steam to stenter at 10 bar Dryness fraction of inlet steam Condensate temperature Boiler efficiency (coal-fired) Distribution line losses Boiler Cost of coal GCV of Coal Operating hours per annum Thermic fluid heater efficiency Distribution line losses Thermic Fluid Heater Biomass cost GCV of Biomass Value 32 C 78 C 65 % 6% 1250 kg/hr 48% 477 kcal/kg 184 kcal/kg 0.95 87 C 72 % 5% Rs. 7000/ton 4200 kCal/kg 7200 hours 70 % 6% Rs. 4000/ton 3800 kCal/kg Calculate the following: a) Steam and coal required for the current coal-fired boiler. 8 Marks b) If the system is converted to a biomass-fired thermic fluid heater, calculate the biomass required and its associated cost per hour. 8 Marks Paper-4 Code : Green c) Estimate the difference in annual fuel cost savings. N-4 a) Steam and Coal Required for the Coal-Fired Boiler (D) Dry cloth mass: Ans m_dry = 1,250 (1 0.06) = 1,175 kg/h Wet inlet mass and moisture evaporated: m_inlet = 1,175 / 0.35 = 3,357 kg/h Moisture evaporated = 3,357 1,250 = 2,107 kg/h Heat required to evaporate moisture: Q_evap = 2,107x[ 540 +(78-32)] = 2,107 586 = 12,34,702 kcal/h Heat input to stenter (48% efficiency): Q_stenter = 12,34,702 / 0.48 = 25,72,296 kcal/h Steam required: h_steam = 0.95 477 + 184 = 637.15 kcal/kg Also condensate leaves at 87oC = 87 kCal/kg Required Steam = 2,567,490 / (637.15-87) = 4675 kg/h Boiler output required (5% distribution loss): Q_boiler_out = 25,72,296 / 0.95 = 27,07,680 kcal/h Boiler heat input (72% efficiency): Q_boiler_in = 27,07,680 / 0.72 = 37,60,666 kcal/h Coal required: Coal = 37,60,666 / 4,200 = 895.4 kg/h = 0.895 t/h Coal cost per hour: Cost = 0.895 7,000 = Rs. 6265 Rs/hr b) Biomass Required & Cost per Hour Thermic Fluid Heater (6% Loss) Heater output required (6% distribution loss): Q_heater_out = 25,72,296 / 0.94 = 2,736,485.106383 kcal/h TFH thermal input (70% efficiency): Q_TFH_in = 2,736,485.106383 / 0.70 = 3,909,264.437689 kcal/h Biomass required: Biomass = 3,909,264.437689/ 3,800 1,028.22485255 kg/h = 1.028 t/h Biomass cost per hour: Cost = 1.028 4,000 = Rs. 4112 Rs/hr P4P c) Annual Fuel Cost Savings Hourly Fuel Saving: Saving = 6265 4112 = Rs. 2153 Rs/hr Annual Saving (7,200 h/year): Annual Saving = 2153 7,200 = Rs. 1,55,01,600 Rs. 1.55 crore/year . End of Section - III . 4 Marks Paper-4 Code: PINK 25TH NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - SEPTEMBER 2025 PAPER 4: ENERGY PERFORMANCE ASSESSMENT FOR EQUIPMENT AND UTILITY SYSTEMS Date : 28.09.2025 Timings: 14:00-16:00 HRS Duration: 2 HRS Max. Marks: 100 General instructions: o o o o o o Please check that this question paper contains 6 printed pages Please check that this question paper contains 16 questions The question paper is divided into three sections All questions in all three sections are compulsory All parts of a question should be answered in one place Time management guidance: Section - I & II: 10 Minutes each, Section - III: 100 Minutes Section - I: BRIEF QUESTIONS (i) (ii) Marks: 10 x 1 = 10 Answer all Ten questions Each question carries One mark 1. Lower the TTD (Terminal Temperature Difference) and DCA (Drain Cooler Approach) for feedwater heaters, the higher will be the efficiency of the cycle. True 2. If wet steam is generated, the high Evaporation ratio indicates high efficiency of boiler. False 3. The only reason for installing condensate recovery systems is to reduce makeup water. False 4. The heat rate of a thermal power plant can be improved by decreasing the condenser cooling water temperature. True 5. In a cement rotary kiln, the highest heat loss occurs through clinker discharge. False 6. Evaporation ratio is based on actual performance data (steam output and fuel input) and does not depend on whether efficiency is expressed on GCV or NCV basis. True 7. Isentropic efficiency of a back-pressure turbine will be more if extraction temperature is higher than steam temperature at back-pressure condition. False 8. Profitability Index (PI) will be greater than 1 for projects with positive NPV, but its value varies with cash flow patterns and is not always higher for all positive NPV projects. In an integrated iron & steel plant, all the rolling mills consumes more energy compared to energy consumed for iron making. False In indirect method of boiler efficiency calculations, blowdown losses are also considered. False 9. 10. True . End of Section - I . BUREAU OF ENERGY EFFICIENCY 1 Paper-4 Code : PINK Section - II: SHORT NUMERICAL QUESTIONS Marks: 2 x 5 = 10 (i) Answer all the Two questions (ii) Each question carries Five marks L-1 During the assessment year 2024 25, one of the thermal power plants reported a gross heat rate of 2300 kCal/kWh and an auxiliary power consumption of 8%, while the baseline net heat rate was 2600 kCal/kWh. If the baseline generation is 5000 MU and considering that 1 kg of oil equivalent = 10,000 kCal and 1 e-Certificate = 1 ton of oil equivalent (TOE), calculate the reduction in net heat rate compared to the baseline and determine the expected number of e-Certificates. L-1 Ans Assessment Year Net Heat Rate = 2300/ (100-8)/100 = 2500 kCal/kWh Reduction in net heat rate = Baseline Assessment year Net Heat Rate = 2600 2500 = 100 kCal/kWh Expected E-certs = 5000 x 10^6 x 100 / 10000 x 1000 = 50,000 e-Certificates L-2 In a petrochemical industry, both the Low-Pressure (LP) boiler and the High-Pressure (HP) boiler operate with the same evaporation ratio of 14, using the same fuel oil. The operating details are provided below: Particulars LP Boiler HP Boiler Pressure 10 kg/cm a 32 kg/cm a Temperature Saturated steam 400 C Enthalpy of steam 665 kcal/kg 732 kcal/kg Temperature of feed water 80 C 105 C Evaporation Ratio 14 14 If the efficiency of the LP boiler is 80%, calculate the efficiency of the HP boiler. L-2 Effy = ER. (hg hf) / GCV Ans EffyL.P 1 = 0.8 = 14 x(665 80) / GCV EffyH.P 2 = 14 x(732 105) / GCV EffyH.P 2 / EffyL.P 1 = (732 105)0.8 / (665 80) = 0. 8574 =85.74% Or EffyL.P 1= 0.8 = 14 x(665 80) / GCV GCV= 14x(665-80) / 0.8 = 10237.5kcal/kg EffyH.P 2 = 14 x(732 105) / GCV = 14 x(732 105) / 10237.5 = 0.8574 = 85.74% . End of Section - II . Paper-4 Code : PINK Section - III: LONG NUMERICAL QUESTIONS Marks: 4 x 20 = 80 (i) Answer all the Four questions (ii) Each question carries Twenty marks N-1 A boiler is fired with 200 kg/hr of a hydrogen-enriched hydrocarbon fuel (C H H ) at atmospheric pressure and a temperature of 20 C. The flue gas, leaving the boiler at atmospheric pressure and 300 C, has the following dry composition by weight: CO = 12% O = 3% N = 85% Based on this information, determine: a) The main constituents of the fuel (carbon and hydrogen) 12 Marks b) The percentage composition of each constituent in the fuel 4 Marks c) The total mass flow rate (kg/hr) of the dry flue gas. 4 Marks Assumptions (per 100 kg dry flue gas): Nitrogen -85 kg/hr, Oxygen 3 kg/hr, CO2 12 kg/hr Sol Take air O mass fraction = 0.23 and N mass fraction = 0.77. N-1 Carbon in the fuel (from CO ): CO C: = 12 2 12 144 36 = = kg 44 44 11 =36/11=3.2727 kg/h of carbon/ per 100 kg/hr of flue gas Actual air supplied: N2 Air mass = N2 mass fraction in air = 85 0.77 = 110.39 kg/hr of air / 100 kg/hr of flur gas O supplied with that air: O2supplied=0.23 110.39 =25.39 kg/hr / 100 kg/hr of flue gas O consumed (used in combustion) = supplied residual in flue gas consumed O2 = 25.39 3 = 22.39 kg/hr /100 kg/hr of flue gas O used to oxidize carbon (to form CO ): 32 8 Stoichiometry: C + O2 CO2 mass factor = = . 12 O2 for C = 3 8 8 36 288 = = = 8.73 g/hr/100kg/hr of flue gas 3 3 11 33 O available for hydrogen: consumed O2 for H = O2 O2 for C = 22.39 8.73 = 13.66 kg/hr/100kg/hr of flue gas Each kg H requires 8 kg O to form H O, so H mass: = 13.66 = 1.70779 kg/hr 8 Fuel mass & composition (per 100 kg dry flue gas): Total fuel mass (C + H): fuel = + = 3.2727 + 1.70779 = 4.9805kg/hr Mass fractions: 3.2727 100 = 65.732663% ( 65.73%) 4.9805 1.7077 % = 100 = 34.267337% ( 34.27%) 4.9805 % = If fuel flow = 200 kg/h: Carbon: (200) = 200 0.6573 = 131.465326 kg/h = 131.46 kg/hr Hydrogen: (200) = 200 0.3426 = 68.534674 kg/h Paper-4 Code : PINK = 68.54 kg /hr c) The total Mass Flow rate of Dry Flue gas in kg/hr: Carbon (C) = 131.46 kg/h Hydrogen (H) = 68.54 kg/h Mass of dry flue gas = Mass of wet flue gas Mass of H2O = 200 + Actual Air 9* 68.54 = Actual Air 416.86 Actual Air = Ta +Ea Ta = (2.67 x131.46 + 8x68.54) / 23% = 3910 kg/hr Ea = E% x Ta E% = O2%/ (21% - O2%) O2%= 3/32 / (12/44 + 3/32 + 85/28) = 2.75% E% = 2.75% / (21%- 2.75%) = 15.07% Ea = 15.07% x 3910 kg/hr = 589.24 kg/hr Therefore, Actual Air = Ta +Ea = 3910 kg/hr + 589.24 kg /hr = 4499.24 kg/hr Mass of Dry flue gas = 4499.24 416.86 = 4082.38 kg/hr N-2 i) A small Topping Cycle Gas Turbine cogeneration plant has the following operating parameters. Estimate the power generation in kW and the steam supplied from the HRSG in TPH. The Energy Auditor has suggested that the HRSG exit flue gas temperature can be maintained at 95 C by recovering more heat from the HRSG. If the HRSG exit flue gas temperature is maintained at 95 C, estimate the additional steam generation and also the EUF with improved steam generation. The operating parameters are given below: 12 Marks Parameter Natural Gas Fuel Firing Rate Lower Heating Value (LHV) Exhaust Gas Flow Rate Exhaust Gas Temperature Mean Specific Heat of Gas (Cp) Specific Power Generation HRSG Inlet Temperature HRSG Exit Temperature Steam Pressure Saturated Steam Temperature Steam Enthalpy Feedwater Temperature HRSG Efficiency Value 1500 Sm /hr 9600 kcal/Sm 16.35 kg/s 525 C 0.265 kcal/kg C 3.044 kWh/Sm 520 C 135 C 10 kg/cm 179 C 663 kcal/kg 105 C 79% ii) A coal-fired boiler operates with the following parameters. Parameter Hours of Operation Feed Water Temperature Steam Enthalpy GCV of Coal Evaporation Ratio Steam Flow Rate Value 24 hours 145 C 805 kcal/kg 4200 kcal/kg 5.7 265 TPH 8 Marks Paper-4 Code : PINK Calculate the Boiler Efficiency and Coal Consumption per hour. If the boiler efficiency is improved by 2% relative to the existing efficiency, then estimate the coal savings per day. N-2 a) Ans Power Generation Power = Fuel rate Specific power generation = 1500 3.044 = 4566 kW Steam Generation at Baseline (Exit 135 C) Heat recovered = 4,744,101 kcal/hr Steam generated = 8502 kg/hr = 8.50 TPH Steam Generation at Improved Condition (Exit 95 C) Heat recovered = 5,236,995 kcal/hr Steam generated = 9385 kg/hr = 9.39 TPH Additional steam = 9385 8502 = 883 kg/hr = 0.88 TPH Energy Utilization Factor (EUF) Fuel input = 16,744 kW Useful output = Power (4566) + Steam thermal (6086) = 10,652 kW EUF = 10,652 / 16,744 = 63.60 % b) Heat Output Net enthalpy gain = 805 145 = 660 kcal/kg Q_out = 265,000 660 = 174,900,000 kcal/hr Coal Consumption (from ER) Coal/hr = 265,000 5.7 = 46,491.22807 kg/hr = 46.49 TPH Heat Input Q_in = 46,491.22807 4200 = 195,263,157.9 kcal/hr Boiler Efficiency = Q_out Q_in 100 = (174,900,000 195,263,157.9) 100 = 89.57% Improved Efficiency (relative 2%) _new = 89.57 1.02 = 91.36% New Coal Consumption Coal_new = Q_out ( _new GCV) Coal_new = 174,900,000 (0.9136 4200) = 45,584.085 kg/hr = 45.58 TPH Coal Savings Coal/hr = 46,491.228 45,584.085= 907.143 kg/hr Daily Saving = 907.143 24 = 21,771.43 kg/day 21.77 TPD N-3 A pharmaceutical manufacturing plant operates a central chilled water system that serves both cleanroom AHUs and a process cooling water loop for tablet coating machines. The process cooling water passes through a counterflow heat exchanger, entering at 20 C and leaving at 14 C with a flow rate of 166 m /h. On the other side, the chilled water enters at 7 C and leaves at 12 C, with a flow rate of 200 m /h. The heat exchanger has an overall heat transfer coefficient of 2.8 kW/m C. The chilled and condenser cooling water loops are each served by centrifugal pumps with operating efficiencies of 78 % and 80 %, respectively and with a head of 18 m. The cooling tower is fitted with an induced-draft axial fan delivering 28 m /s of air at 42.83 mmWC total pressure, with a fan efficiency of 62 %. All pumps and the fan are driven by directly coupled three-phase induction motors with an efficiency of 92 %. The plant s HVAC system operates for 300 days annually, running 18 hours per day, to handle an average cooling load of 850 kW with an ISEER value of 4.5. Calculate the following: Paper-4 Code : PINK a) Estimate the required heat exchanger surface area. (6 marks) b) Determine the combined electrical load in kW of both pumps and the cooling tower fan, if condenser cooling water flow is 600 m /h. (6 marks) c) Determine the Specific Energy Consumption (SEC) of the chiller in kWh/TR and calculate the overall SEC. (4 marks) d) Calculate the annual energy consumption in kWh and the annual operating cost in Lakhs if electricity is charged at 7.5 per kWh. (4 marks) N-3 a) Heat Exchanger Surface Area Sol - Process water flow rate: 166 m /h = 46.1 kg/s - T = 20 14 C = 6 C - Heat load (Q): Q = cp T = 1156 kW - LMTD (counterflow): T1 = 8 C, T2 = 7 C, LMTD = 7.49 C - Area: A = Q / (U LMTD) = 1156 / (2.8 7.49) = 55.2 m b) Combined Electrical Load of Pumps and Fan Chilled water pump (200 m /h, 18 m head): - Hydraulic power = 9.81 kW - Motor input = 13.7 kW Condenser pump (600 m /h, 18 m head): - Hydraulic power = 29.43 kW - Motor input = 40.0 kW Cooling tower fan (28 m /s, 42.83 mmWC): - Air power = 11.76 kW - Motor input = 20.6 kW Total auxiliary load = 13.7 + 40.0 + 20.6 = 74.3 kW c) Specific Energy Consumption (SEC) - Cooling load = 850 kW = 242 TR - Chiller input = 189 kW (ISEER = 4.5) Chiller SEC = 189 / 242 = 0.78 kWh/TR System SEC = (189 + 74.3) / 242 = 1.09 kWh/TR d) Annual Energy & Cost Total power demand = 189+74.3 = 263.3 kW - Annual energy consumption = 263.3 18 x 300 = 15,64,002 kWh/year - Annual cost (@ 7.5/kWh) = 15,64,002 7.5 /100000 = 117.3 lakh/year Answer any ONE of the following among four questions given below: N-4 The following data was collected from a 500 MW turbine unit during an energy audit. The power (A) plant operates with a main steam (MS) flow of 1561 TPH at a pressure of 166 kg/cm and temperature of 529 C. The hot reheat (HRH) flow is 1413 TPH, with steam conditions of 42.4 kg/cm and 540 C, while the cold reheat (CRH) section records a pressure of 44.3 kg/cm and temperature of 341 C. The feed water enters at 246 C, whereas the MS, CRH, and HRH enthalpies are 806.47 kCal/kg C, 730.71 kCal/kg C, and 844.28 kCal/kg C, respectively. These optimized steam cycle parameters enable the generator to deliver a substantial power output of 501.7 MW with a boiler efficiency of 88%. The turbine cycle heater operating parameters are as below: Steam Heater Reference Temp ( C) Pressure (kg/cm ) Saturation Temp ( C) Feed Water in Drain Temp ( C) Temp ( C) Pressure (kg/cm ) Feed Water out Temp ( C) Pressure (kg/cm ) Design Values TTD DCA Paper-4 Code : PINK LP Heater 1 LP Heater 2 LP Heater 3 HP Heater 5 HP Heater 6 92.5 140 209 416 335 -0.23 0.49 1.97 17.4 43 93.07 111.23 132.9 207.33 254.94 64.2 70.4 110 171 212 47.2 170 - 13.7 202 - 63.6 105 130 210 255 12.6 11.5 10.4 199 197 2.88 2.95 2.95 0 0.1 4.8 4.95 4.95 5 5 Neglect temperature loss in the feedwater line between heaters and calculate the following: a. Calculate the Turbine Heat Rate and the Unit Heat Rate. 8 Marks b. Determine the loss or gain in the Turbine Heat Rate due to deviations of the TTD (Terminal Temperature Difference) and DCA (Drain Cooler Approach) of the LP/HP Heater systems from their design values. 12 Marks Consider the following criteria: For every 0.56 C increase or decrease in TTD from the design value, the Heat Rate will increase or decrease by 0.014%. For every 0.56 C increase or decrease in DCA from the design value, the Heat Rate will increase or decrease by 0.005%. N-4 Turbine Heat Rate: (A) Turbine: HR: {(1561000 X (806.47-246))+ (1413000 X (844.27 730.714))}/ (501700) = 2063.72kCal/kWh Ans Unit Heat Rate: = 2063.72 / 88% = 2345.131 kCal/kWh From the Above data, the following heater data can be inferred Heater Ref. Feed water In- Feed water In- Steam Inlet let Temp OC let Temp OC temp OC LP Heater -1 LP Heater-2 LP Heater -3 HP heater-5 HP Heater -6 47.2 63.6 105 170 210 63.6 105 130 210 255 92.5 140 209 416 335 Inlet steam Saturation Temp OC 93.07 111.23 132.9 207.33 254.94 Drain temp OC 64.2 70.4 110 171 212 So TTD (Terminal Temperature Difference) = Inlet Steam Saturation Temp OC Feed Outlet Temp OC DCA (Drain Cooler Approach = Drain temperature oC Feed Water Inlet Temperature oC Heater Ref. TTD oC DCA OC TTD oC DCA OC (Design) (Design) (Calculated) (Calculated ) LP Heater -1 2.88 4.8 29.47 17 LP Heater-2 2.95 4.95 6.23 6.8 LP Heater -3 2.95 4.95 2.9 5 HP heater-5 0 5 -2.67 1 HP Heater -6 0.1 5 -0.06 2 Difference Between design values and Operating values of TTD and DCA of Heaters. Heater Ref. LP Heater -1 LP Heater-2 LP Heater -3 HP heater-5 HP Heater -6 Total Difference TTDOperating TTDDesign 26.59 3.28 -0.05 -2.67 -0.16 26.66 DCAOperating DCADesign 12.2 1.85 0.05 -4 -3 7.1 Change in Heat rate because of deviation in TTD = (Net Change in TTD for All heaters X 0.014%/0.56oC ) Since given, for every 0.56oC change in TTD HR will increase by 0.014% So Increase in HR because of TTD deviation= 26.66oC X 0.014 / 0.56 = 0.67475 % Paper-4 Code : PINK Change in Heat rate because of deviation in DCA = (Net Change in DCA for All heaters X 0.005%/0.56oC ) o Since given, for every 0.56 C change in TTD HR will increase by 0.005% So Increase in HR because of DCA deviation 7.1 X 0.005 / 0.56 = 0.064 % Total % Increase in Turbine HR because of deviation in operation of TTD and DCA of Heaters from Design Values = 0.7381 %. =0.7381 X 2063.72 = 15.23 kCal/kWh Or N-4 A DRI-route steel plant operates a DRI unit and a Steel Melting Shop (SMS). The plant also has a (B) coal-based captive power plant (CPP). Any shortfall in electrical energy is met by imported grid power. On average, the plant imports 1,20,000 kWh/day and the operational parameters are given below: Description Parameter Value Rated capacity 500 TPD Capacity utilization 70% DRI Unit Specific coal consumption 1.25 t coal / t sponge iron Specific power consumption 95 kWh / t sponge iron Coal GCV 5000 kcal/kg Yield 87% SMS Specific power consumption 830 kWh / t liquid steel Gross efficiency 27% CPP Auxiliary power consumption 8% of gross generation Coal GCV 5000 kcal/kg Grid Grid electricity heat rate 2700 kCal/ kWh Calculate the following: a) The daily production of sponge iron and liquid steel in TPD. (2 marks) b) DRI coal consumption in TPD and its thermal input in Million kcal/day. (2marks) c) Total daily electrical energy demand of DRI and SMS in kWh/day. (3 marks) d) The CPP gross generation in kWh/day, CPP heat rate in kCal/kWh, CPP thermal input in Million kcal/day and the CPP coal consumption in TPD. (6 marks) e) The overall specific energy consumption (SEC) in Million kcal per tonne of liquid steel. (4 marks) f) Compare your SEC with a benchmark of 6.5 Million kcal/t of liquid steel, and comment briefly on performance. (3 marks) 1) Daily Production N-4 (B) Sponge iron = 500 70% = 350 TPD Ans Liquid steel = 350 0.87 = 304.5 TPD 2) DRI Coal & Heat Input Coal = 1.25 350 = 437.5 TPD Thermal input = 437.5 5,000 1000 = 2,187.5 Million kcal/day 3) Electrical Demand DRI = 95 350 = 33,250 kWh/day SMS = 830 304.5 = 252,735 kWh/day Total = 285,985 kWh/day 4) CPP Generation & Coal CPP net = 285,985 120,000 = 165,985 kWh/day CPP gross = 165,985 / 0.92 = 180,418.5 kWh/day Heat rate = 860 / 0.27 = 3,185.2 kcal/kWh Thermal input = 180,418.5 3,185.2 = 574.67 Million kcal/day CPP coal = 574.67 10^6 / 5000 = 114.93 TPD 5) Overall SEC Total heat input = Thermal input to DRI + Thermal input to CPP + Thermal Input of Import Imported electricity thermal eq. = 120,000 2700 = 324.0 Million kcal/day Total heat input = 2,187.5 + 574.67 + 324.0 = 3,086.17 Million kcal/day SEC = 3,086.17 / 304.5 = 10.14 Million kcal/t 6) Benchmark Comparison & Comment Paper-4 Code : PINK Benchmark = 6.5 Million kcal/t; Actual = 10.14 Million kcal/t (~56% higher) Comments: Need reduction in DRI coal use, improved SMS SPC, better CPP efficiency, adoption of WHR, and auxiliary load reduction. Or N-4 (C) A composite textile mill uses stenters for drying and heat-setting applications, currently the stenter system is running on a coal-fired boiler, and there is a proposal to modify the system to a biomass-fired thermic fluid heater. The relevant data is given below: Parameter Cloth inlet temperature Cloth outlet temperature Cloth inlet moisture Cloth outlet moisture Stenter output Stenter Efficiency Latent heat of inlet steam to stenter at 10 bar Sensible heat of inlet steam to stenter at 10 bar Dryness fraction of inlet steam Condensate temperature Boiler efficiency (coal-fired) Distribution line losses Boiler Cost of coal GCV of Coal Operating hours per annum Thermic fluid heater efficiency Distribution line losses Thermic Fluid Heater Biomass cost GCV of Biomass Value 32 C 78 C 65 % 6% 1250 kg/hr 48% 477 kcal/kg 184 kcal/kg 0.95 87 C 72 % 5% Rs. 7000/ton 4200 kCal/kg 7200 hours 70 % 7% Rs. 4000/ton 3800 kCal/kg Calculate the following: a) Steam and coal required for the current coal-fired boiler. 8 Marks b) If the system is converted to a biomass-fired thermic fluid heater, calculate the biomass required and its associated cost per hour. 8 Marks c) Estimate the difference in annual fuel cost savings. 4 Marks N-4 a) Steam and Coal Required for the Coal-Fired Boiler (C) Dry cloth mass: Ans m_dry = 1,250 (1 0.06) = 1,175 kg/h Wet inlet mass and moisture evaporated: m_inlet = 1,175 / 0.35 = 3,357 kg/h Moisture evaporated = 3,357 1,250 = 2,107 kg/h Heat required to evaporate moisture: Q_evap = 2,107x[ 540 +(78-32)] = 2,107 586 = 12,34,702 kcal/h Heat input to stenter (48% efficiency): Q_stenter = 12,34,702 / 0.48 = 25,72,296 kcal/h Steam required: h_steam = 0.95 477 + 184 = 637.15 kcal/kg Also condensate leaves at 87oC = 87 kCal/kg Required Steam = 2,567,490 / (637.15-87) = 4675 kg/h Paper-4 Code : PINK Boiler output required (5% distribution loss): Q_boiler_out = 25,72,296 / 0.95 = 27,07,680 kcal/h Boiler heat input (72% efficiency): Q_boiler_in = 27,07,680 / 0.72 = 37,60,666 kcal/h Coal required: Coal = 37,60,666 / 4,200 = 895.4 kg/h = 0.895 t/h Coal cost per hour: Cost = 0.895 7,000 = Rs. 6265 Rs/hr b) Biomass Required & Cost per Hour Thermic Fluid Heater (7% Loss) Heater output required (7% distribution loss): Q_heater_out = 25,72,296 / 0.93 = 2,765,909.677419355kcal/h TFH thermal input (70% efficiency): Q_TFH_in = 2,765,909.677419355/ 0.70 = 3,951,299.539170507 kcal/h Biomass required: Biomass = 3,951,299.539170507/ 3,800 1,039.289352409603 kg/h = 1.039 t/h Biomass cost per hour: Cost = 1.039 4,000 = Rs. 4157 Rs/hr c) Annual Fuel Cost Savings Hourly Fuel Saving: Saving = 6265 4157 = Rs. 2108 Rs/hr Annual Saving (7,200 h/year): Annual Saving = 2108 7,200 = Rs. 1,51,77,600 Rs. 1.51 crore/year Or N-4 A 10,000 TPD cement plant purchases power from the grid, operates an 18 MW Captive Power (D) Plant (CPP) and also has a Waste Heat Recovery (WHR) system with a 9 MW turbine. Electricity is used for cement production and also supplied to the colony and other utilities. The plant also exports excess energy to the grid. The annual energy and production data are given below: Parameter Value Annual operating hours 8,300 hrs Energy imported 60,210,000 kWh Energy exported 3,150,000 kWh Energy supplied to colony & others 3,500,000 kWh CPP average generation 18 MW CPP heat rate 3100 kcal/kWh WHR turbine average generation 9 MW WHR turbine heat rate 3600 kcal/kWh Indian coal consumption 96,000 MT (GCV 4500 kcal/kg) Pet coke consumption 80,000 MT (GCV 7500 kcal/kg) Imported coal consumption 198,000 MT (GCV 7200 kcal/kg) Biomass consumption 10,000 MT (GCV 2850 kcal/kg) Clinker produced 2,700,000 MT Clinker-to-cement ratio 1.375 Calculate the following: a) Specific Electrical Energy Consumption (SEEC) in kWh/ton of cement. (8 Marks) b) Estimate the Specific Thermal Energy Consumption (STEC) in kcal/kg of clinker (7 Marks) c) If the CPP is operated using only Indian coal and Imported coal, calculate the coal blending ratio by weight required to achieve a blended coal GCV of 6000 kcal/kg. (5 Marks) N-4 Solution (D) Specific Electrical Energy Consumption (SEEC) 8 Marks Energy Input Paper-4 Code : PINK Sol CPP generation = 18 8300 1000 = 149,400,000 kWh WHR generation = 9 8300 1000 = 74,700,000 kWh Grid import = 60,210,000 kWh Total Input = 284,310,000 kWh Deductions Export to grid = 3,150,000 kWh Colony & others = 3,500,000 kWh Net for cement production = 284,310,000 3,150,000 3,500,000 = 277,660,000 kWh Cement Production Clinker produced = 2,700,000 MT Cement produced = 2,700,000 1.375 = 3,712,500 MT SEEC SEEC = 277,660,000 3,712,500 = 74.7 kWh/ton cement Specific Thermal Energy Consumption (STEC) 7 Marks Fuel Heat Input Indian coal = 96,000 4500 = 432 10 kcal Pet coke = 80,000 7500 = 600 10 kcal Imported coal = 198,000 7200 = 1425.6 10 kcal Biomass = 10,000 2850 = 28.5 10 kcal Total Heat Input = 2486.1 10 kcal Heat to Kiln As per PAT convention (no CPP deduction applied) Heat to kiln = 2486.1 10 kcal STEC Clinker produced = 2,700,000 MT = 2,700,000,000 kg STEC = 2486.1 10 2,700,000,000 = 920.04 kcal/kg clinker Coal Blending Ratio Define ratio Let x = fraction of Indian coal, (1 x) = fraction of Imported coal Equation 6000 = 4500x + 7200(1 x) 6000 = 4500x + 7200 7200x 6000 = 7200 2700x 2700x = 1200 x = 0.444 (44.4%) Ratio Indian coal = 44.4%, Imported coal = 55.6% Coal Blending Ratio = 44 : 56 P4P . End of Section - III . PAPER-4 COLOUR CODE: GREEN 24th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS AND ENERGY AUDITORS SEPTEMBER, 2024 PAPER -4 ENERGY PERFORMANCE ASSESSMEN T FOR EQUIPMENT AND UTILITY SYSTEMS SECTION I : BRIEF QUESTIONS Marks 10x1=10 STATE TRUE OR FALSE (EACH QUESTION CARRIES 1 MARK) 1. The air velocity is uniform across the cross section of a circular duct. False 2. A pump operating with VFD at a low speed to accommodate a high static head does not adhere to the affinity laws - True 3. Lower CO percentage in flue gas indicates better combustion efficiency - True 4. Higher power factor results in higher power loss in the power systemFalse 5. Increasing the condenser vacuum will increase the heat rate of a thermal power plant.- False 6. The excess air in cement kiln combustion can be assessed by measuring CO2 % in kiln exhaust- False 7. A higher solar cooling load factor results in a lower air-conditioning load.- False 8. For a 20 MW co-generation plant, a backpressure turbine system configuration will have less steam rate (kg/kWh) compared to an extraction condensing turbine - False 9. The heat transfer coefficient of a shell and tube heat exchanger is reduced solely due to changes in the temperatures of the cold and hot fluids. - False 10. For a given motor rating, stray losses as a percentage decrease with a change in the output kW of the motor. - False ****** End of Section -I ****** PAPER-4 COLOUR CODE: GREEN SECTION II : SHORT NUMERICAL QUESTIONS (i) (ii) Marks: 2 x 5 = 10 Answer all the TWO questions Each question carries Five marks L-1 In a chemical plant, a cooling water pump supplies cooled water to both the process and the refrigeration system. During performance testing, the following operating parameters were recorded: Measured Data: Rated Flow Rated Head Running Pump Flow (Q) Motor Input Parameters Motor losses Suction Head Delivery Head : 2124 m3/hr : 70 m : 1700 m3/hr : V = 440 V, I = 480 A, Power Factor = 0.89, : 19.5 kW :8m : 55 m Calculate the following: (a) Pump Efficiency (Hydraulic Efficiency) (b) Overall Pump Set (Pump + Motor) Efficiency Solution: Flow Delivered by pump = 1700 m3/hr = 0.47 m3/s Head Differential (h) = Delivery Head Suction Head (m) = 55 8 = 47m Hydraulic Power = = 1000 x 9.81 x 0.47 x 47/1000 = 216.70 kW Motor Input = 1.732 x V x I x Cos (phi) = 1.732 x 440 x 480 x 0.89/1000 = 325.5 kW Motor Output = Motor Input Motor Losses = 325.5 19.5 = 306 kW Pump Input = Motor Output (a) Pump Efficiency Hydraulic Efficiency = Pump Output/Pump Input x 100 Hydraulic Efficiency = 216.70/306 x 100 = 70.82% (b) Overall Efficiency Overall Efficiency = Pump Output/Motor Input x 100 Overall Efficiency = 216.70/325 x 100 = 66.67% PAPER-4 COLOUR CODE: GREEN L-2 A medium-sized re-rolling plant has installed a batch-type reheating furnace. The furnace operating details are as follows: Average Furnace Oil Consumption : 1565 LPH Average Production : 36 TPH Cost of Furnace Oil : 55/kg Cost of Electricity : Rs 9.25/kWh Furnace Oil Density : 0.93 kg/liter GCV of Furnace Oil : 10,200 kCal/kg Billet Loading Temperature : 35 C Billet Reheat Temperature : 1250 C Mean Specific Heat of Billet : 0.13 kCal/kg C The plant management has proposed replacing the reheating furnace with a 95% efficient electric furnace. Calculate the following: a) Oil fired furnace efficiency. b) The specific energy consumption for both the cases and comment on the plant management proposal based on the cost benefits. Solution: Heat to Billet = m cp t Production Production Sp Heat of Billet Billet Final Temperature Billet Loading temperature Heat required for the Billet(mcp T) Furnace oil Consumption Density Furnace oil Consumption GCV of Furnace Oil Heat Input Oil fired Furnace Efficiency 36 36000 0.13 1250 35 5686200 1565 0.93 1455.45 10200 14845590 38.3 TPH Kg/T kcal/kg Deg c Deg c kCal/hr LPH SEC Oil 43.5 40.43 =40.43x 55 Ltrs/T Kg/Ton of billet Rs/ton Cost/ton with FO Kg/hr Kcal/Kg Kcal/Hr % PAPER-4 COLOUR CODE: GREEN Electrical furnace Efficiency Heat to billet Electricity Heat Equivalent kWh required kWh at 95 % Furnace Efficiency SEC Electrical Cost/ton with electrical furnace =2223.6 95 5686200 860 6611.86 6960 193.3 1788 % Kcal/hr Kcal/kwh Kwh Kwh Kwh/Ton Rs Conversion to Electrical is Economical. ****** End of Section -II ****** SECTION III: LONG NUMERICAL QUESTIONS (i) (ii) Marks 4 x 20 = 80 Answer all the Four questions Each question carries Twenty marks N-1 A medium-sized textile processing plant has installed a 20 TPH traveling grate coal-fired boiler. As part of a green energy initiative and to optimize steam costs, the plant management has proposed replacing the coal-fired boiler with a paddy husk boiler. The ultimate analysis of paddy husk and other boiler operating parameters are provided below. Average Monthly Steam Demand : 10800 Tonnes Operating hours per month : 720 Hours Coal-Fired Boiler Efficiency : 67 % Steam Generation Pressure : 12 kg/cm Steam Enthalpy : 665 kcal/kg Ambient temperature :32 c Feed Water temperature : 84 C Exit Flue Gas temperature after APH : 225 C Oxygen % In Flue gas before APH with Coal as Fuel : 8% Oxygen % In Flue gas before APH with Paddy husk as Fuel: 6% Radiation loss accounted for Husk Boiler :1.6% Humidity factor : 0.025 Kg/Kg dry air GCV of Coal : 4200 kcal/kg GCV of Paddy Husk : 3500 kcal/Kg PAPER-4 COLOUR CODE: GREEN Cost of Coal Cost of Rice Husk Auxiliary Cost for coal fired boiler Auxiliary Cost for paddy husk boiler Ultimate Analysis of Paddy Husk (%): : Rs 12000/Tonne : Rs 6700/ Tonne : Rs 750/tonne : Rs 550/tonne Moisture Mineral Matter Carbon Hydrogen Nitrogen Sulphur Oxygen Calculate the following: :10.79 :16.73 :33.95 :5.01 :0.91 :0.09 :32.52 a) Evaporation ratio of the coal-fired boiler b) Steam cost (fuel cost + auxiliary cost) of the coal-fired boiler in Rs/tonne c) Efficiency of the paddy husk boiler using the indirect method d) Evaporation ratio of the paddy husk boiler e) Steam cost (fuel cost + auxiliary cost) of the paddy husk boiler in Rs/tonne Solution for N2 Coal Fired Boiler Efficiency = 67 % GCV of Coal = 4200 Kcal/kg Steam Generation Pressure = 12 kg/Cm Steam Enthalpy = 665 kcal/kg Feed Water temperature = 84 C Evaporation Ratio = (4200 x 0.67)/ (665-84) Cost of Coal Cost of Coal Fired Boiler Steam Auxiliary cost Total cost of steam from Coal Boiler = 4.84 = 12000/Tonne =12000/4.84 = Rs2479/Tonne = Rs 750/Tonne = Rs( 2479+750) = Rs 3,229/Tonne Efficiency of Paddy Husk Boiler: Theoretical air required for complete combustion Paddy Husk PAPER-4 COLOUR CODE: GREEN = {11.6. C + [34.8 (H2 O2/8)] + 4.35 S} / 100 = {11.6 x 33.52 [34.8 (5.01 32.53/8)] + 4.35 x 0.09} / 100 = 4.27 Kg/ Kg of Husk % O2 in fuel gas = 6 % Excess air = [%O2 / (21 - % O2)] x 100 = [6 / (21 6)] x 100 = 40% Actual Air Supplied (ASS) = (1 + 0.4) x 4.27 = 5.98 Kg/Kg fuel Mass of dry flue gas = mdfg Mass of dry flue gas = mass of combustion gases due to presence C, S, O2, N2 + mass of N2 in air supplied Mdfg = 0. 3395x (44 / 12) + 0.0009 x (64 / 32) + [(5.98 4.27) x (23 / 100)] + 5.98 x (77/100) Mdfg = 6.165 Kg/Kg fuel Alternatively, Mdfg. = (AAS+1) (9xH2) Mmoist = (5.98+1) (9 x 0.0501) - 0.1079 = 6.34 kg/kg fuel % heat loss in dry flue gas = mdfg x Cpf x (Tg Ta) / GCV of fuel Tg = flue gas temperature = 225 c Ta = ambient temperature = 32 c Cp = SP ht of flue gas = 0.24 Kcal/KgC (Value referred from the guidebook) GCV = Gross Calorific Value of Paddy Husk =3500 Kcal/kg L1 = % heat loss in dry flue gases = [(6.34 x 0.24 x (225-32))/3500] X 100 = 8.4 % Heat loss due to evaporation of water due to H2 in fuel = {9 x H2 [584 + CPS (Tg Ta)]} / GCV CPS = Specific heat of superheated steam = 0.43 Kcal/Kg (Value referred from the guidebook) L2 = {9 x 0.0501 [ 584 + 0.43 (225 32)] / 3500} x 100 = 8.59% L3 = % heat loss due to moisture in fuel == M x[584 + CPS (Tg Ta)/GCV L3=0.1079 x (584+0.43(225-32)/3500=2.06 % L4 = AAS x humidity factor x CPS x (Tg Ta) / GCV Humidity factor = 0.025 Kg/Kg dry air L4 = {[5.98 x 0.025 x 0.43 (225-32)] / 3500} x 100 = 0.35% L5 = Radiation and convection loss from the boiler = 1.6% (given data) Total losses in the boiler in %= L1 + L2 + L3 + L4 + L5 = 8.4+8.59+2.06+0.35+1.6=21 Efficiency of boiler by indirect method = 100 21% = 79 % Evaporation Ratio of Paddy Husk Boiler = 3500 X 0.79/ (665-84) = 4.76 PAPER-4 COLOUR CODE: GREEN Cost of Husk = Rs 6700 Cost of steam from paddy Husk Boiler = 6700/4.76 = Rs1407.6/Tonne Auxiliary Cost = Rs 550/tonne Total Steam Cost with Paddy husk Boiler = Rs 1957.6/Tonne N-2 One of the sugar plant, operating at 4,000 TCD (Tonnes of Crushing per day), has installed a 13 MW back pressure turbine co-generation system. The co-generation plant comprises of one old bagasse-fired boilers with a capacity of 50 TPH and another with 35 TPH. This co-generation plant meets both the steam and power demands of the sugar plant and exports power to the grid during the crushing season. The power required for the sugar plant is 29 kW per tonne of crushing. The schematic of the present system is given below: Main Steam Header 77 TPH PRDS & Gland Seal 5 TPH 43 TPH Boiler 50 TPH (Rated) Boiler 35 TPH (Rated) T G I/L 72 TPH ? (Rated) L P Process Steam72 TPH The plant management has conducted an energy audit of the sugar plant and the co-generation system. The energy auditor has recommended replacing the existing co-generation system with a high-pressure boiler of 80 TPH and a back pressure steam turbine to increase power export. The operating details for both the existing and proposed systems are presented in the table below: Description Bagasse GCV Boiler steam generation pressure Boiler super heat steam temperature Main steam pressure enthalpy Feed water temperature Units Present Proposed Co-Gen System Co-Gen System kCal/kg 2270 2270 kg/cm (g) 42 85 C 485 520 kCal/kg 815 823 C 95 105 PAPER-4 COLOUR CODE: GREEN Turbine inlet steam pressure kg/cm (g) 42 85 C 480 520 TPH 72 72 kCal/kg 812 823 kg/cm (g) 1.5 1.5 kCal/kg 676 676 Back pressure steam temperature C 178 178 Boilers operating efficiency % 67 (Avg) 74 Turbine efficiency % 90 92 Alternator efficiency % 96 96 Rs./kWh 4.25 4.25 % 6 6 TPH 5 5 Turbine inlet steam temperature Turbine inlet steam flow Inlet steam enthalpy Turbine back pressure Back pressure steam enthalpy Cost realised for exported power Auxiliary power consumption for the co-generation plant Steam consumption for PRDS & Gland Seal Calculate the following: a) Power generation in the present and proposed system. b) Power export with the present and proposed system. c) Bagasse consumption in the present and proposed system. d) Savings in bagasse with the proposed system. e) Heat-to-power ratio (kWth/kWe) of the present and proposed co-generation system. f) Heat-to-power ratio (kWth/kWe) of the sugar plant. g) Steam rate (kg of steam per kW) for the present and proposed system. h) Energy Utilization Factor (EUF) in the present and proposed system. Solution: Daily Crushing Load = 4000TCD PAPER-4 COLOUR CODE: GREEN Crushing Load/Hour = 4000/24 =166.6 TPH Power required = 29 kW /Tonne = 29x166.66= 4833 Kw Power Generation Stem inlet to turbine = 72 TPH Inlet Enthalpy = 812 Kcal/kg Backpressure Enthalpy = 676 Kcal/kg Power Generation potential in kW = 72 x1000 x (812-676)/860= 11386 Kw Turbine and alternator efficiency = 11386 x 0.9 x0.96 = 9837 Kw (Gross Gen) Auxiliary Power Consumption in Co-Gen: 6% of gross generation Sugar Plant = 4833 kW Power Export in Present System = (0.94*9837)-4833 = 4413.78 kW Steam Rate Kg/kw = 72000/9837= 7.32 kg/kW Present boiler Efficiency = 67 % Bagasse Required = 77 x1000 (815-95)/2270 x 0.67 = 36.45 TPH Heat to power ratio of cogenator = kW Thermal = 72 x1000 x (676-95)/860 = 48641.86 kW kW Electrical = 9837 Kwth/Kwe = 48641.86/9837= 4.94 Energy Utilisation Factor in Present system= (Q Elect+ Q ther)/ Q Fuel = (9837x860+72000x(676-95))/ (36.35x1000 x2270) x100 = 60.95% Heat to power ratio of Sugar Plant Kw Thermal = (72 x1000 x (676-95))/860=48641.86 Kw PAPER-4 COLOUR CODE: GREEN Kw Electrical = 4833 Kwth/Kwe = 48641.86/4833 = 10.1 Power generation with High Pressure Cogeneration System = (72000 x(823-676))/860= 12307 Kw Applying Turbine and Alternator Eff Actual Power Generation = 0.92 x0.96 x 12307= 10869 kw Steam Rate Kg/kw = 72000/ 10869 = 6.62 kg/kW Export with High Pressure Cogeneration=(0.94*10869)-4833- Kw= 5383.86 kW Additional Export with High Pressure Cogeneration Plant = 5383.86-4413.78= 970.08 Kw Bagasse required for High pressure Cogeneration system= 77000*(823-105)/(2270 x0.74) = 32.91 TPH Savings in Bagasse =36.45-32.91 = 3.54 TPH Energy utilisation with High Pressure Cogeneration= =(Q Elect+ Q therm)/ Q Fuel = (10869x860+72000x(676-105))/ (32.91x1000 x2270) x100 = 67.5% N-3 A steam turbine power plant utilizes a Circulating Water (CW) system with an Induced Draft Cooling Tower (IDCT) for condenser cooling. The steam flow rate to the condenser is 440 T/hr. It is observed that 45 m of CW is needed to condense 1 T of steam. To compensate for the water loss from the IDCT sump, two makeup water pumps each with a capacity of 220 m /hr are available in parallel. During operation, it was observed that a single pump delivers 200 m /hr and both pumps operating in parallel deliver 350 m /hr together. One pump operates continuously and the second pump switches on when the IDCT sump level falls below 90% of its full capacity. The second pump is switched off once the sump is full. The TDS of circulating water is 2,000 ppm and the TDS of the makeup water is 500 ppm. The cooling tower operates with an effectiveness of 63% and an approach of 4 C. The size of the IDCT sump is 100 m x 15 m x 40 m and to maintain proper water chemistry, a continuous blowdown from the sump is given. During steady state operation, calculate the following: PAPER-4 COLOUR CODE: GREEN a) Evaporative loss in m3/hr, blow down in m3/hr and draw mass flow diagram of the system with all the flow rates and losses. 10 Marks b) What is the time period in hours for which only one IDCT makeup pump is in service? 5 Marks c) What is the time period in hours for which both the IDCT makeup pumps are in service? 5 Marks Solution: = 19800 m3/hr Total CW flow = 440 * 45 Effectiveness = Range / (Range + Approach ) Range = (4*0.63)/(1-0.63) = 6.810C Evaporative loss = 0.00085*1.8*19800*6.81 COC = 2000/500 Blow down from CT sump = Evaporation Loss/ (COC-1) = 206.3 m3/hr =4 = 68.76 m3/hr Mass flow diagram of the system: Evaporative Loss 206.3 m3/hr Hot CW 19800 m3/hr Cold CW 19800 m3/hr IDCT Sump Make up (200+350/2) = 275 m /hr 3 CT make up (one pump) Blow down 68.76 m3/hr = 200 m3/hr a) Calculation of time period for which only one pump is in service Total volume of sump = 100*15*40 = 60000m3 Volume change needed before 2nd make up pump is stared = (100-90)*60000 = 6000 m3 Volume change in sump from mass flow diagram = 200-206.3-68.76 = -75 m3/hr PAPER-4 COLOUR CODE: GREEN Time period for which only one pump is in service = 6000/75= 80 hrs b) Calculation of time period for which both pumps are in service When two pumps are in service, total make up flow will become 350m3/hr, All other flows remain same Volume change needed before 2nd make up pump is stopped= (100-90)*60000 = 6000 m3 Volume change in sump from mass flow diagram = 350-206.3-68.76 Time period for which both pumps are in service = 6000/75= 80hrs = 75 m3/hr Answer any ONE of the following among four questions given below: N4- (A) i) A 500 MW power plant is operated with the turbine back pressure of 0.14 ata and after improving the condenser cooling system the turbine back pressure is maintained at 0.11 ata. The design heat rate of the turbine at backpressure of 0.14 ata is 2040 kcal/kwh and at 0.11 Back pressure will be 2000 kcal/kWh. The operating turbine efficiency is 93 % and the alternator efficiency is 96 % and the boiler efficiency is 87 %. The coal used in Power plant is having GCV of 5500 kCal/kg. Calculate the following: 1. 2. 3. 4. Condenser vacuum in mmHg at 0.14 ata turbine Back pressure Condenser vacuum in mmHg at 0.11 ata turbine Back pressure Improvement in gross heat Rate At 74 % Power Plant Loading what will be the coal savings per day ii) The operating Details of a 1000 MW thermal power plant are given below: Plant Load Factor : 76 % Annual Operating Hours : 7200 hours Annual Coal Consumption : 40,45,400 MT Annual Furnace oil Consumption (Support Fuel) : 3,500 MT Annual HSD Consumption (Earthmoving Equipment) : 150 MT 2 Marks 2 Marks 4 Marks 4 Marks PAPER-4 COLOUR CODE: GREEN Coal GCV : 5,500 kCal/ kg Furnace Oil GCV : 10,200 kCal/kg HSD GCV : 10,500 kCal/kg Calculate the following: 1. Annual power generation in Million kWh. 2 Marks 2. Gross Heat Rate kCal/kWh, 3 Marks 3. Net Heat Rate kCal/kWh, if auxiliary power consumption is 8% of the running Load. 3 Marks Solution: i) 1. Condenser vacuum in mmHg = 0.14 x 1.0332 kg/cm =0.145 kg/cm = 109.93 mm Hg = 760-109.93 = 650 mmHg 2.Condenser vacuum in mmHg = 0.11 x 1.0332 kg/cm =0.114 Kg/cm = 86.37 mm Hg =760-86.37= 674mmHg 3.Improvement in gross heat Rate Gross Heat Rate with turbine Pressure O.14 ata = 2040/(0.87)= 2345 kCal/kW Gross Heat Rate with O.11 ata = 2000/(0.87)= 2299 kCal/kW Improvement = 2345-2299 = 46 Kcal/kWh 4.Heat savings at 74 % Loading = 500 X1000x 0.74 x (46) Kcal/Hr = 17020000 kCal/hr Coal savings = 17020000/5500 = 3094.55 kg/Hr Daily Coal savings= 3094.55 kg x 24 hrs= 74.27 Ton/ Day Solution: ii) Operating Load = 1000 x0.76 = 760 MW PAPER-4 COLOUR CODE: GREEN Annual Units Generated Gross Heat Rate = (760 x 1000 x7200)/1000000= 5472 million Units =[((4045400*1000)x 5500) + (3500*1000x10200))]/ 5472 million units = 4073.64 kCal/kWh Net Heat Rate = 4073.64/ (1-0.08) = 4427.87 kCal/kg (Or) N4- (B) A textile processing unit currently operates a 5-chamber stenter with a dryer efficiency of 40%. The average amount of dry cloth processed by this stenter is 1,500 kg/hr, containing 4% moisture. The incoming cloth to the stenter has a moisture content of 42%, with a feed temperature of 37 C, and exits at 89 C. The management plans to upgrade to an 8-chamber stenter, which has a dryer efficiency of 58%. The heat required for both stenters is supplied by a thermic fluid heating system powered by furnace oil, which operates at an efficiency of 80%. The furnace oil has a gross calorific value (GCV) of 10,000 kcal/kg and a density of 0.92 kg/liter. Calculate the following: 1. The dryer heat input (kcal/kg) for both the 5-chamber and 8-chamber stenters. 12 Marks 2. The amount of furnace oil required (in liters per hour) for the thermic fluid heater when using the 5-chamber stenter and when using the 8-chamber stenter. 8 Marks Solution: Bone dry cloth weight Inlet cloth weight with Moisture Inlet Moisture/kg dry cloth =1500*0.96 =1440/(1-0.42) =2483*0.42/1440 Outlet Moisture/kg dry cloth =1500*0.04/1440 Mass of moisture evaporated =1440*(0.724-0.042) 1440 kg/hr 2483 Kg/hr 0.724 Kg moisture/Kg cloth 0.042 Kg moisture/Kg cloth 982 kg/hr dry dry PAPER-4 COLOUR CODE: GREEN Heat Load to stenter Five Chamber stenter drying Efficiency 1a Input heat to stenter Thermopack efficiency Heat Input to Thermopack Furnace Oil GCV Furnace oil qty Furnace oil density 2a Furnace oil required Eight Chamber stenter drying efficiency 1b Input heat required for 8 chamber stenter Thermopack Efficiency =982*((89-37)+540) Heat Input at Thermopack Furnace Oil GCV Furnace oil qty Furnace oil density 2b Furnace oil required =1002317.24/0.8 581344 kCal/hr 40 % =581344/0.4 1453360 80 1816700 10000 181.67 0.92 197 58 =1453360/0.8 =1816700/10000 =181.67/0.92 =581344/0.58 kCal/hr % kCal/hr Kcal/kg Kg/hr Kg/lit LPH % 1002317.24 kCal/hr 80 % =1252896.55/10000 =125.29/0.92 1252896.55 10000 125.29 0.92 136.18 kCal/hr Kcl/kg Kg/hr Kg/lit LPH (Or) N4 (C) The management of a cement plant with a capacity of 7200 TPD has decided to install waste heat recovery boilers (WHRB) to generate steam from preheater gas and clinker cooler gas for power generation. The relevant data is given below: Raw meal feed Rate Clinker Output Preheater Outlet temp Clinker Cooler Outlet temp WHRB Exit temperature Preheater Gas heat availability Cooler gas heat availability Enthalpy of Main Steam Feed Water temperature Power plant Condensate return Temperature Heat recovery potential in WHRB Turbine Cycle Efficiency 7200 62 325 310 160 152 120 815 95 46 75 36 TPD % of feed c c c Kcal/kg of clinker Kcal/kg of clinker Kcal/kg c c % % PAPER-4 COLOUR CODE: GREEN Gear Box Efficiency Alternator Efficiency 95 96 % % The chemical analysis of clinker (Loss-free basis) is given below: Constituents % SiO 22.68 Fe O3 5.92 Al O3 5.29 CaO 63.00 MgO 1.25 Calculate the following: 1. The power output from the co-generation plant in MW. (15 Marks) 2. The heat of formation of the clinker. (5 Marks) Solution : Clinker Output Heat in Pre Heater-Gas Heat In cooler gas Steam Enthalpy Feed Water temperature Power plant Condensate return Temp Heat recovery potential in WHRB Steam from Preheater Steam From Cooler gas Total Steam Generation Turbine Cycle Efficiency Gear Box Efficiency Alternator Efficiency Power generation Heat of Formation of clinker = (7200x0.62/ 24hrs) = 152x186 x1000 = 120 x186x1000 given given given given =28272000x0.75/(815-95)/1000 =22320000x0.75/(815-95)/1000 =29.48 + 23.25 given given given = 52.7x1000x(815-95)x0.36x0.95x0.96/860/1000 186 28272000 22320000 815 95 46 75 29.48 23.25 52.7 36 95 96 14.48 TPH kcal/kg/hr kcal/kg/hr Kcal/Kg c c % TPH TPH TPH % % % MW PAPER-4 COLOUR CODE: GREEN = 2.22 X Al O3+6.48 MgO +7.646 Cao-5.116x SiO -0.59x Fe O3 = 2.22 x5.29+6.48 x1.25 +7.646 x63 5.116 x22.68 0.59 x5.92 = 382 kCal/kg of clinker (Or) N4 (D) The production data for a steel plant using the Direct Reduced Iron (DRI) route is outlined below. The DRI unit has a daily production capacity of 500 tonnes of sponge iron but operates at 60% of this capacity. The produced sponge iron is transported to the Steel Melting Shop (SMS) where it is processed into ingots, the final product. The plant also operates a captive power station to fulfill its energy requirements. The operational parameters for both the baseline year and the assessment year are provided as follows: Parameter Sponge Iron Full Production Capacity Plant operating Capacity Specific Coal Consumption of DRI Specific Power Consumption of DRI Yield of Steel Melting Shop SEC of Steel Melting Shop Captive Power Station Efficiency GCV of Coal Unit Base Year (2022) Assessment Year (2023) T/Day % T/T kWh/T % kWh/Ton % kCal/kg 500 60 1.3 110 85 850 26.06 6000 500 60 1.15 95 88 830 27.74 6200 Calculate the following: (1) Specific Energy Consumption of the plant in Million kCal/Tonne of Finished Product for Base Year. 8 Marks (2) Specific Energy Consumption of the plant in Million kCal/Tonne of finished product for assessment year. 8 Marks (3) Reduction in coal consumption considering both DRI and captive power plant in tonnes per day for the assessment year. 4 Marks Solution : 1. Base Year Performance Specific Energy Consumption = 1300 kg x 6000 + 110 kWh x 3300 = 8.163 million kCal/Tonne of SI PAPER-4 COLOUR CODE: GREEN Plant Capacity Plant Actual Running Capacity Total Energy Consumption of Sponge Iron /day Total production of Ingots from Sponge Iron considering Heat Rate of the Captive Power Station Specific Energy Consumption for Ingot = 500 T/day = 60% = 0.6 x 500 = 300 Tonnes = 300 x 8.163 = 2448.9 million kCal =300 x 0.85 = 255 Tonnes/Day =860/0.2606 = 3300 kCal/kWh =850 x 3300 = 2.805 million kCal/Tonne of Ingot Total Specific Energy Consumption for = 2.805 x 255 = 715.275 million kcal Ingot Production per year Plant Specific Energy Consumption for = (2448.9+ 715.275)/255 production of finished product (ingot) = 12.41 million kcal/tonne during base year 2. Assessment Year Performance Revised Parameter Specific Power Consumption Yield The Specific Energy Consumption of SMS Plant Heat Rate GCV of Coal = (1 0.1363) x 110 = 95 kWh/Tonne = (1 + 0.0352) x 85 = 88 % = (1-0.0235) x 850 = 830 kwh/tonne =3300 200 = 3100 kCal/kWh =6200 kCal/kg Specific Energy Consumption = 1150 kg x 6200 + 95 kWh x 3100 =7.425 million kCal/Tonne of SI Plant Capacity Plant Actual Running Capacity Total Energy Consumption of Sponge Iron /day Total production of Ingots from Sponge Iron considering Heat Rate of the Captive Power Station during assessment year Specific Energy Consumption for Ingot = 500 T/day = 60% = 0.6 x 500 = 300 Tonnes = 300 x 7.425 = 2227.5 million kCal = 300 x 0.88 = 264 Tonnes/Day = 860/0.2774 = 3100 kCal/kWh = 830 x 3100 = 2.573 million kCal/Tonne of Ingot Total Specific Energy Consumption for = 2.573 x 264 = 679.27 million kcal Ingot Production per year Plant Specific Energy Consumption for = (2227.5+679.27)/264 production of finished product (ingot) = 11.01 million kcal/tonne during base year PAPER-4 COLOUR CODE: GREEN 3. Reduction in Coal Consumption Energy Saving in Sponge Iron Plant Energy Saving in Steel Melting Plant Total Energy Saving Equivalent Coal Reduction (Saving) = (8.163 7.425) x 300 = 221.4 million kCal/day = (2.805 x 255 2.573 x 264) = 38.07 million kCal/day = 221.4 + 38.07 = 259.47 million kCal = 259.47 x 106/ (6200 x 103) = 41.85 Tonnes per Day Comment: The coal consumption is reduced by 41.85 Tonnes / Day ****** End of Section -III ******

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