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20th National Certification Exam Energy Managers & Auditors SEPTEMBER 2019 Paper 2

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Paper-2 Code: Pink 20th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS& ENERGY AUDITORS- September, 2019 PAPER 2: Energy Efficiency in Thermal Utilities 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 (i) (ii) (iii) Marks: 50 x 1 = 50 Answer all 50 questions Each question carries one mark Please hatch the appropriate oval in the OMR answer sheet with Black Pen or HB pencil, as per instructions 1. Arrange the following fuels in decreasing order of their GCV s - (p) Bagasse, (q) Furnace Oil, (r) Coal, (s) Hydrogen a) s-q-r-p b) p-q-r-s c) r-s-q-p d) q-r-s-p 2. Which of the following contributes to spluttering of flame at burner tip during combustion of fuel oil? a) ash content b) water content c) Sulphur content d) ambient air humidity and temperature 3. Which trap is preferred in condensate removal from steam main lines? a) Float trap b) Thermodynamic trap c) Thermostatic trap d) All of the above 4. In an FBC boiler, with low ash fusion coal, if the bed temperature exceeds 950 C, the result is: a) Boiler explosion b) clinker formation c) Melting of lime stone d) Ash carry over 5. Water logging of 3 m lift of condensate, at trap discharge, will result in back pressure of _________ a) 0.03 kg/cm2 b) 0.3 kg/cm2 c) 3 kg/cm2 d) 30 kg/cm2 6. When pure hydrogen is burnt with stoichiometric air, percentage CO2 on volume basis, in dry flue gas, will be _____ a) 79% b) 21% c) 0% d) 100% 7. Heat transfer rate for drying application will be low if we heat with _________ a) Saturated steam b) Dry steam c) Superheated steam d) None of the above 1 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink 8. The viscosity of furnace oil will be maximum at which of the following temperatures? a) 40 oC b) 60 oC c) 90 oC d) 105 oC 9. Carpet loss in the context of coal consumers is related to : a) Short receipt b) Accounting mistakes c) Ash handling 10. Thermo-compressor is commonly used for : a) compressing hot air c) converting saturated steam to super-heated steam 11. Latent heat of any vapour at its critical point will be : a) highest b) above zero c) zero d) coal storage b) upgrading low pressure steam d) reverse compression of CO2 d) less than zero 12. The temperature at which, refractory will deform under its own weight, is it s softening temperature, indicated by : a) melting point b) Pyrometric Cone Equivalent c) reform temperature d) critical point 13. Which type of the following co-generation system has high heat-to-power ratio? a) gas turbine b) back pressure turbine c) extraction condensing turbine d) reciprocating engine 14. Drain pockets are provided in a steam line for : a) effective removal of line condensate c) removal of dirt 15. Capillary wick is a part of: a) heat pump c) heat pipe b) effective removal of steam d) checking of steam line b) heat wheel d) regenerator 16. Scale losses in reheating furnaces will: a) increase with CO in combustion gases c) decrease with excess air b) increase with excess air d) have no relation with excess air 17. Tuyeres is a terminology associated with: a) induction furnace b) pusher type furnace c) arc furnace d) cupola 18. The low combustion temperature in FBC boilers results in minimal formation of : a) CO b) SOx c) NOx d) CO2 19. Turn down ratio of a burner is the ratio of a) maximum to minimum fuel input without affecting optimum excess air levels b) minimum to maximum fuel input without affecting optimum excess air levels c) maximum to average fuel input d) average to minimum fuel input 20. Dolomite is a ________ type of refractory: a) acidic b) basic c) neutral d) none of the above 21. Which of the following is not true of condensate recovery? a) reduces water charges b) reduces fuel costs c) increases boiler output d) increases boiler blow down 22. Calcium and magnesium bicarbonates present in feedwater fed to a boiler would form : a) acidic solution b) alkaline solution c ) neutral solution d) none of the above 2 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink 23. In steam systems, the purpose of venting air is because, air is ________. a) a good conductor b) an inert substance c) an oxidizing agent d) an insulator 24. A bottoming cycle is one in which fuel is used for producing : a) power primarily followed by byproduct heat output b) heat primarily followed by byproduct power output c) power, heat and refrigeration simultaneously d) none of the above 25. A supercritical boiler has parameters beyond critical point, which refers to : a) 221.2 bar (a) pressure and 374.15 0C temperature b) 246.1 bar (a) pressure and 538.44 0C temperature c) 306.5 bar (a) pressure and 538.82 0C temperature d) 170.0 bar (a) pressure and 374.18 0C temperature 26. Of the total volume of natural gas, the main constituent is : a) methane b) iso-octane c) propane d) hexane 27. For optimum combustion of fuel oil, O2 percentage in flue gases should be maintained at : a) 2-3 % b) 14-15 % c) 23 % d) 21% 28. The draft caused solely by the difference in weight between the column of hot gas inside the chimney and column of outside air is known as : a) balanced draft b) induced draft c) forced draft d) natural draft 29. The cogeneration system which has high overall system efficiency is : a) back pressure steam turbine b) combined cycle c) extraction condensing steam turbine d) reciprocating engine 30. When 10 kg of fuel, with 60% carbon, is burnt with theoretical air, the mass of CO2 released will be : a) 32 kg b) 440 kg c) 450 kg d) 22 kg 31. F & A (from and at) rating of the boiler is the amount of steam generated from : a) water at 0 C to saturated steam at 100 C b) water at feed water temperature to saturated steam at 100 C c) water at 100 C to saturated steam at 100 C d) water at ambient temperature to saturated steam at 100 C 32. Steam generated in a boiler is 36 tonnes in 3 hours. Fuel consumption in the same period is 1 tonne per hour. Continuous blow down is 8% of feed water input. The boiler evaporation ratio is ? a) 12 b) 11.7 c) 36 d) 24 33. Which of the following statement is false? a) LPG vapour is twice as light as air c) LPG is a gas at normal atmospheric pressure b) LPG is a mixture of propane and butane d) LPG is required to be odorized 34. The inverted bucket operates on the principle of _______ difference between water and steam: a) pressure b) density c) temperature d) velocity 35. Which of the following is not measured in proximate analysis ? a) volatile matter b) fixed carbon c) sulphur d) ash 36. Reduction of steam pressure, in a process heating application will : a) reduce the steam temperature c) increase the enthalpy of evaporation b) reduce the sensible heat d) all of the above 3 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink 37. The TDS level in boiler water, in the context of boiler blow down, can be determined by measuring : a) alkalinity of water c) electrical conductivity of water b) thermal conductivity of water d) turbidity of water 38. De-aeration of boiler feed water helps in combating : a) corrosion b) TDS c) silica d) hardness 39. In stoichiometric combustion of furnace oil, which of the following will be absent in flue gas ? a) nitrogen b) carbon dioxide c) oxygen d) Sulphur dioxide 40. Furnace wall heat loss does not depend on : a) temperatures of external wall surfaces c) thermal conductivity of wall brick b) velocity of air around the furnace d) material of stock to be heated 41. In determining the optimal economic insulation thickness for a steam pipeline, which of the following factors need not be considered ? a) annual hours of operation b) calorific value c) pipe material d) cost of fuel c) high heat storage d) thermal shock resistant 42. Which is not a property of Ceramic fiber insulation? a) low thermal conductivity b) light weight 43. Which property is the most important, for an insulating brick ? a) Mechanical strength c) Compact strength b) Chemical resistance d) Porosity 44. Quality of waste heat in flue gas refers to : a) dust concentration in flue gas c) moisture in flue gas b) Temperature of flue gas d) corrosive gases in flue gas 45. In a low temperature waste heat recovery system, which of the following, is the most suitable ? a) Economizer b) Heat Pipe c) regenerator d) ceramic recuperator 46. Which of the following heat recovery equipment, requires a compressor for its operation? a) thermo-compressor b) heat wheel c) Heat pump d) heat pipe 47. Pinch analysis of process streams, depicts the plot of : a) temperature vs entropy c) temperature vs specific heat b) temperature vs. area d) temperature vs enthalpy 48. Which of the following is true for a process heating requiring direct injection of steam? a) Thermodynamic trap is required c) Inverted bucket trap is required b) Thermostatic trap is required d) None of the above 49. If a vapor-liquid combination of 1 kg at 120 C is supplied with 50 kcal of heat without change in state and at constant pressure conditions; its temperature will be? a) 220 C b) 190 C c) 170 C d) 120 C 50. Which of the following constituent in flue gas is used for determining excess air? a) % nitrogen b) % Sulphur Dioxide c) % Carbon dioxide d) % 4 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink Moisture . . End of Section I .. . Section - II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40 (i) (ii) Answer all Eight questions Each question carries Five marks S1 In an industry, an electrical oven consuming 1100 kWh/batch, is proposed for replacement, by a FO fuel fired oven. Calculate the simple payback period, given the following data: Number of batches / years Efficiency of electric oven Efficiency of FO fired oven Cost of FO GCV of FO Electricity cost Investment for FO fired oven = 4000 = 82% = 55% = Rs.35,000/Tonne = 10,200 kcal/kg = Rs.6.0/kWh = Rs. 125 Lakhs Ans : Useful heat, required per batch = (1100 x 860 x 0.82) = 7,75,720 kcal/batch FO input per batch = (7,75,720 /(0.55 x 10,200)) = 138.27 kg FO/batch FO cost per batch = (138.27 kg FO/batch x Rs.35/kg FO) = Rs.4,839.45 Electricity cost per batch = (1,100 kWh/batch x Rs.6.0/kWh) = Rs.6,600 Cost savings per batch on account of replacement = (Rs.6,600 Rs.4,839.45) = Rs.1,760.55 Annual cost savings at 4000 batches per year = (1,760.55 x 4000) = Rs.70,42,200 (Or) = Rs.70.422 lakhs Investment = Rs.125 lakhs Simple payback period = (125/70.422) = 1.78 years S2 In a process plant, 30 TPH of steam, after pressure reduction to 20 kg/cm2(g), through a pressure reducing valve, gets superheated. The temperature of superheated steam is 350 oC. The management desires to install a de-super heater to convert the superheated steam into useful saturated steam at 20 kg/cm2(g) for process use. The saturated steam temperature is 210oC. Calculate the quantity of water required to be injected at 30 oC, in the de-super heater, in order to obtain the desired saturated steam, using the following data: Specific heat of superheated steam Latent heat of steam at 20 kg/cm2(g) = 0.45 kcal/kgoC = 450 kcal/kg 5 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink Ans : Quantity of heat available above saturation By Heat & Mass balance: Quantity of water (Q) required to be added in de-super heater = (30,000 x 0.45 x (350-210)) = 18,90,000 kcal/hr Q x{1 x (210-30) + 450} = 18,90,000 = 18,90,000/{1 x (210-30) + 450} = 18,90,000/630 = 3000 kg/hr S3 In an industry the process equipment needs 5000 kg/hr of saturated steam at 10 kg/ cm2(g). For a steam velocity of 25 m/sec, what will be the diameter of the steam pipe in mm , given that the specific volume of steam at 10 kg/ cm2(g) is 0.1802 m3/kg. Ans : Specific volume of steam at 10 kg/cm2(g) = 0.1802 m3/kg Flow rate = 25m/sec Mass flow rate = 5000 kg/hr = 1.389 kg/sec Volume flow rate = 1.389 x 0.1802 = 0.25 m3/sec Volume flow rate is also = ( /4 x D2) x25 Therefore, ( /4 x D2) x 25 = 0.25 Hence, diameter of steam pipe line D = [(0.25/(( /4) x 25)]0.5 = 0.1128m or 112.8 mm S4 An economizer was installed in an oil fired boiler. The following data was obtained after commissioning the economizer. Air to fuel ratio = 18 Evaporation ratio of the boiler = 12.5 Specific heat of flue gas = 0.25 kcal/kg C. Condensate recovery in the plant = Nil. Calculate the rise in temperature of feed water across the economizer, corresponding to a drop in flue gas temperature from 280 C to 190 C. Ans : Steam generated per kg of fuel, (from evaporation ratio) Required combustion air per kg of fuel, (from air to fuel ratio) Flue gas generated per kg of fuel Heat balance across the Economizer : Rise in temperature of water T = 12.5 kg = 18 kg combustion air/kg fuel oil = (18 +1) = 19 kg flue gas/kg fuel oil Heat given by flue gas = Heat received by water ((19 x 0.25 x (280-190)) = (12.5 kg x 1kcal/kgoC x T) = 34.2 C S5 Compute the heat loss in percentage, due to unburnt in fly ash and bottom ash, for an AFBC Boiler, using Indian coal, with: GCV % Ash in coal Ratio of bottom ash to fly ash GCV of fly ash GCV of bottom ash = 4200 kcal/kg. = 38.8 = 15: 85 = 452.5 kcal/kg = 800 kcal/kg 6 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink Ans: Unburnt in fly ash Amount of fly ash in 1 kg of coal GCV of fly ash Heat loss in fly ash % Heat loss in fly ash Unburnt in bottom ash Amount of bottom ash in 1 kg of coal GCV of bottom ash Heat loss in bottom ash % Heat loss in bottom ash = (0.85 x0.388) = 0.3298 kg fly ash/kg coal = 452.5 kcal/kg fly ash = (0.3298 x 452.5 kcal per kg fly ash) = 149.23 kcal/kg coal = (149.23 x 100 /4200 ) = 3.55 % = 0.15 x 0.388 = 0.0582 kg bottom ash/kg coal = 800 Kcal/kg bottom ash = (0.0582 x 800 kcal per kg bottom ash) = 46.56 kcal/kg coal = (46.56 x 100 /4200) = 1.11 % S6 List five main parameters considered for the selection of refractories? Ans : (Each 1 Mark) (Page No:166, Sec 5.11) S7 What is the significance of volatile matter, in case of solid fuels? Ans : S8 (Page No: 9) (i) List three functions of a steam trap. (ii) Explain the working principle of thermodynamic trap. (3 Marks) (2 Marks) Ans : (i) List three functions of a steam trap. (Page 82) (ii) Explain the working principle of thermodynamic trap. (Page 86-87) . . End of Section II .. . Section III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60 (i) Answer all six questions (ii) Each question carries ten marks L-1 A process industry consuming 10 TPH of saturated steam at 10 kg/sq.cm(g) pressure has been using coal as fuel in boiler. Typical ultimate analysis of the coal: Carbon Hydrogen : 41.11% : 2.76 % 7 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink Nitrogen Oxygen Sulphur Ash Water : 1.22 % : 9.89 % : 0.41% : 38.63 : 5.89 Flue gas temperature Ambient temperature Enthalpy of steam Feed water temperature Specific heat of flue gases Boiler efficiency with Indian coal GCV of coal Oxygen content in dry flue gases Annual Hours of operation = 200 C = 30 C = 668 kcal/kg = 80 C = 0.23 kcal/kgoC =72 % = 4,000 kCal/kg = 10% = 8000 hrs. Determine: (i) Quantity of annual coal requirement in tonnes/year (ii) Calculate % dry flue gas losses (5 Marks) (5 Marks) Solution: a) Coal requirement Q = Steam (q) x (hg hf)/( Efficiency x GCV) = 10 x (668-80) / (0.72 x 4000) = 2.042 T/Hr = 2.042 x 8000 hrs = 16336 Tonnes/year Theoretical air requirement for coal = [(11.6 x C%) + {34.8 x(H2% - O2%/8)} + (4.35 x S%)] kg / kg of coal 100 = [(11.6 x 41.11) + {34.8 x(2.76 9.89/8)} + (4.35 x 0.41)] 100 = 5.31 kg / kg of coal (Or) ================================================================================================= = C+ O2 = CO2 2H2+ O2 = 2H2O S+ O2 = SO2 12+32= 44 4+32=36 32+32=64 Total oxygen required (C%*32)/12 (H%*32)/4 (S%*32)/32 = (41.11* 32/12) + (2.76*32/4) + (0.41*32/32) = (109.63) + (22.08) + (0.41) = 132.1 kg/ 100 kg fuel 8 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink Oxygen already present in 100 kg fuel = 9.89 kg/ 100 kg fuel Additional oxygen required Quantity of dry air required (Air contains 23% O2 by weight) = 132.1 9.89 kg/ 100 kg fuel = 122.21 kg/ 100 kg fuel = 122.1/ 0.23 = 531.35 kg/ 100 kg fuel Theoretical air required = 531.35/100 = 5.31 kg air/ kg fuel ================================================================================================= Excess air = O2x100/(21-O2) Excess air = 10 x 100/ (21 10) = 90.9% Actual air= 5.31 * (100+90.9)/100 =10.137 kg air/kg coal Heat loss in dry flue gas = m x CP (Tf Ta) x 100 /GCV = (10.137+ 1) x 0.23 x (200 30) x 100 4000 = 10.89 % L-2 a) In a double pipe heat exchanger, flow rates of the hot and the cold-water streams flowing through a heat exchanger are 10 and 25 kg/min, respectively. Hot and cold-water stream inlet temperatures are 70 C and 27 C, respectively. The exit temperature of the hot stream is required to be 50 C. The specific heat of water is 4.179 kJ/ kg K. The overall heat transfer coefficient is 900 W/m2 K. Neglecting the effect of fouling, calculate the heat transfer area for a) Parallel-flow b) Counter-flow. ..7 marks b) Write a brief note on the operation and application of plate heat exchangers in process industries. ..3 marks Ans: a) Rate of heat transfer, Q (watts) Q= m x Cp x 1000 x (T2-T1) =(10/60) 4.179 x 1000 (70 50) = 13930 W Cold water exit temperature, T2 T2 = [Q/(mx Cp x 1000)]+ T1 = (13930/((25/60)* (4.179*1000)))+27 = 35 Terminal temperature differences for parallel = (70-27) & (50 35); i.e., flow heat exchangers 43 C and 15 C respectively. LMTD (43 15)/ln(43/15) = 26.59 C Overall heat transfer coefficient U 900 W/m2 K Heat transfer area required for parallel flow A = Q/ (U*LMTD) = [13930/ (900 26.59)] = 0.582 m2 9 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink Terminal temperature differences for counter flow heat exchangers LMTD Overall heat transfer coefficient U Heat transfer area required for counter flow (70 35) and (50 27) 0C i.e., 35 C and 23 C respectively. (35 23)/ln(35/23) = 28.58 900 W/m2 K A = Q/ (U*LMTD) =[13930/ (900 28.58)] = 0.542 m2 B) Plate heat exchangers consist of a stack of parallel thin plates that lie between heavy end plates. Each fluid stream passes alternately between adjoining plates in the stack, exchanging heat through the plates. The plates are corrugated for strength and to enhance heat transfer by directing the flow and increasing turbulence. These exchangers have high heat-transfer coefficients and area, the pressure drop is also typically low, and they often provide very high effectiveness. However, they have relatively low-pressure capability. The biggest advantage of the plate and frame heat exchanger, and a situation where it is most often used, is when the heat transfer application calls for the cold side fluid to exit the exchanger at a temperature significantly higher than the hot side fluid exit temperature i.e. temperature cross . This would require several shell and tube exchangers in series due to the lack of purely counter-current flow. The overall heat transfer coefficient of plate heat exchangers under favorable circumstances can be as high as 8,000 W/m2 C. With traditional shell and tube heat exchangers, the U-value will be below 2,500 W/m2 C. L-3 a) In a fruit processing plant, 105 TPD of syrup at 33% concentration is dried to 50% concentration. The existing single effect evaporator, where steam input for water removal ratio is 1.0 kg/kg is proposed to be replaced by a triple effect evaporator where the ratio of steam input for water removal is 0.4 kg/kg. Calculate the annual fuel cost savings for 300 days of operation at an evaporation ratio of 13.5 in the oil-fired boiler and at a furnace oil cost of Rs. 35,000/tonne. ..7 marks b) Why steam is recommended to be used at the lowest practicable pressure for indirect process heating? ..3 marks Ans.: a) Bone Dry material = (105 TPD x 0.33) = 34.65 TPD Product at 50 % concentrate = (34.65 / 0.5) = 69.3 Water removed/ day = (105 69.3) = 35.7 TPD Initial steam consumption with single effect evaporator at 1 kg/kg Steam consumption with triple effect evaporator at 0.4 kg/kg Steam savings per day = (35.7 TPD x 1.0 kg/kg) = 35.7 TPD = (35.7 TPD x 0.4 kg/kg) = 14.28 TPD = (35.7 TPD 14.28 TPD) = 21.42 TPD 10 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink FO savings per day at evaporation ratio of 13.5 Rupee savings per day at Rs. 35,000/MT Annual monetary savings at 300 working days per year = (21.42 TPD / 13.5 Tonne steam per Tonne FO) = 1.5867 TPD = (1.5867 TPD FO X Rs. 35,000/MT FO) = Rs. 55,535 = (Rs. 55,535 X 300 Days) = Rs.166.6 Lakhs b) The latent heat in steam reduces as the steam pressure increases. It is only the latent heat of steam, which takes part in the heating process when applied to an indirect heating system. Thus, it is important that its value be kept as high as possible. This can only be achieved if we go in for lower steam pressures. However, lower the steam pressure, the lower will be its temperature. Since temperature is the driving force for the transfer of heat at lower steam pressures, the rate of heat transfer will be slower and the processing time greater. In equipment where fixed losses are high (e.g. big drying cylinders), there may even be an increase in steam consumption at lower pressures due to increased processing time. There are however, several equipment s in certain industries where one can profitably go in for lower pressures and realize economy in steam consumption, without materially affecting production time. Therefore, there is a limit to the reduction of steam pressure. Depending on the equipment design, the lowest possible steam pressure with which the equipment can work should be selected without sacrificing either on production time or on steam consumption. L-4 a) An oil fired reheating furnace has an operating temperature of around 1000 C. Average furnace oil consumption is 330 litres/hour. Flue gas exit temperature after the air preheater is 820 C. Combustion air is preheated from ambient temperature of 35 C to 215 C through the air preheater. The other data are as given below. Specific gravity of oil = 0.92 Calorific value of oil = 10,200 kcal/kg Average O2 percentage in flue gas = 13.5 % Theoretical air required = 14 kg of air per kg of oil Specific heat of air = 0.23 kcal/kg C Specific heat of flue gas = 0.25 kcal/kg C Find out : The sensible heat carried away by the exhaust flue gases in kcals/hr and as a percentage of the energy input. ..4 marks The heat recovered by the combustion air in kcal/hr and as a percentage of the energy input. ..3 marks b) Explain the concept and the advantage of a self-recuperative burner? ..3 marks Ans: a) Fuel input Energy Input Excess air = (330 litres/hr x 0.92 kg/litre) = 303.6 kg/hr = (303.6 kg oil/hr x 10,200 kcals/kg oil) = 30,96,720 kcal/hr = [O2 x 100/(21-O2)] = (13.5 x 100)/(21 - 13.5) 11 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink Theoretical air required Actual mass of air required Mass of flue gas (m) Specific heat of flue gas (Cp) Sensible heat loss in the flue gas Sensible heat loss in the flue gas as % heat loss to input energy Heat recovered by combustion air Heat recovered by combustion air as % of input energy = 180 % = 14 kg of air/kg of oil = 14 x (1 + 180 /100) = 39.2 kg air/kg of oil = (39.2 + 1) = 40.2 kg flue gas/kg oil = 0.25 kcal/kg.oC = (m x Cp x T)flue gas = (40.2 x 0.25 x (820-35)) = 7889.3 kcal/kg of oil (Or) = (7889.3 kcal/kg of oil x 303.6 kg oil/hr) = 2395176.3 kcal/hr = (2395176.3 / 30,96,720) x 100 = 77.35 % = (39.2 x 0.23 x (215-35)) = 1622.88 kcal/kg of oil = (1622.88 kcal/kg oil x 303.6 kg oil/hr) = 492706.37 kcal/hr = (492706.37 kcal/hr/30,96,720 kcal/hr) x100 = 15.91 % b) Self-recuperative burner (SRB) is based on traditional heat recovery techniques, in that, the products of combustion are drawn, through a concentric tube recuperator, around the burner body and used to pre-heat the combustion air. A major advantage of this type of system is that, it can be retro-fitted to an existing furnace structure, to increase production capability, without having to alter the existing exhaust gas ducting arrangements. SRBs are generally more suited to Heat-treatment furnaces, where exhaust gas temperatures are lower and there are no stock recuperation facilities. L-5 a) An open cycle gas turbine was running with naphtha as fuel. The following are the data collected during the gas turbine operation: Fuel (Naphtha) consumption = 300 kg/hr GCV of naphtha fuel = 11,500 kcal/kg Overall Efficiency of gas turbine (which includes air compressor and alternator) = 22% Cost of naphtha fuel = Rs.40,000/Tonne a) Find out the output power and cost of fuel for generating one unit of electricity. ..6 marks b) The management has decided to install a waste heat boiler, to generate 2 TPH of saturated steam, at 4 kg/cm2(g), with an enthalpy of 656 kcal/kg. Assuming that, 50% of the input heat is available in the turbine exhaust gases, how much steam can be generated if the feed water temperature is 30 oC. ..4 marks 12 _____________________ Bureau of Energy Efficiency Paper-2 Code: Pink Ans: a) Heat input to turbine Efficiency of gas turbine Gas turbine output power Cost of generating 882.56 units of electricity Cost of One unit of Electricity generation b) Waste heat potential in existing gas turbine Heat required for raising 1 kg of steam (feed water temp 30 oC) Steam generation potential L6 Explain any two of the following: = (300 kg Naptha/hr x 11,500 kcal/kg) = 34,50,000 kcal/hr = 22% = ((34,50,000 kcal/hr x 0.22)/ 860) = 882.56 kW = (300 kg Naptha/hr x Rs.40/kg Naptha) = Rs.12000/hr = (Rs.12000 per hour/882.56 kWh per hour) = Rs.13.6/kWh = (0.5 x 34,50,000 kcal/hr) = 17,25,000 kcal/hr. = (656-30) kcal/kg steam = 626 kcal/kg steam = (17,25,000 kcal per hour/626 kcal per kg steam) = 2755.6 kg steam/hr = 2.7556 TPH (Each 5 Marks) 1. Regenerator (Page 222) 2. Heat Pipe (Page 223) 3. Gas Turbine cogeneration system (Page 192) 13 _____________________ Bureau of Energy Efficiency

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