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15th National Certification Exam Energy Managers & Auditors AUGUST 2014 Paper 4

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Paper 4 SET A 15th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY AUDITORS August, 2014 PAPER 4: Energy Performance Assessment for Equipment and Utility Systems Date: 24.8.2013 Timings: 14:00-16:00 Hrs Section - I: BRIEF QUESTIONS Duration: 2 Hrs Max. Marks: 100 Marks: 10 x 1 = 10 (i) Answer all Ten questions (ii) Each question carries One mark S-1 Which loss is not considered while evaluating boiler efficiency by Indirect Method ? Ans Blow down loss S-2 What will be the synchronous speed of a VFD driven 4-pole induction motor operating at 40 Hz ? Ans S-3 Ns = 120 x f/P = 120 x 40/4= 1200 RPM What is the refrigerant used in a vapour absorption system with lithium bromide as an absorbent? Ans Water S-4 Other than rated kW of motor and the actual power drawn, what other parameter is required to determine the percentage loading of the motor ? Ans Motor Efficiency or rated motor efficiency S-5 Ans S-6 Ans S-7 Ans S-8 Inclined tube manometer is used for measuring gas flow in a duct when the air velocity is very high: True or False? False. A pump will cavitate if the NPSHrequired is _________ than the NPSHavailable. More To determine the effectiveness of the cooling tower, it is required to measure cooling water inlet, outlet and _____temperatures. Ambient Wet bulb The ratio of actual heat transfer to the heat that could be transferred by heat exchanger of infinite size is termed as 1 _______________________ Bureau of Energy Efficiency Paper 4 SET A Ans Effectiveness S-9 If the unit heat rate of a power plant is 3070 kcal/kWh ,what is the power plant efficiency ? Ans (860/3070) x 100 = 28 % S-10 Ans The difference between GCV and NCV of hydrogen fuel is Zero: True or False False . End of Section - I . Section - II: SHORT NUMERICAL QUESTIONS Marks: 2 x 5 = 10 (i) Answer all Two questions (ii) Each question carries Five marks L-1 Hot water at 80 OC is used for room heating in a 5 Star hotel for 4 months in a year. About 200 litres per minute of hot water is maintained in circulation with the return temperature at 50 OC. The hot water is generated using a hot waste stream , through a Plate Heat Exchanger (PHE). The hot stream enters the PHE in counterflow direction at 95 OC and leaves at 60 OC. The area of the heat exchanger is 20 m2. Calculate the LMTD and the overall heat transfer coefficient. Ans Heat load, Q = 200 * 60 * (80 50) = 360000 Kcals/hr (or) 418.7 kW LMTD (for counter flow) (95 80)/(60 50) = ------------------------ = 3.7 OC ln (15/10) Overall Heat Transfer Coefficient, U = Q/( AxLMTD) = 418.7/(20 x 3.7) = 5.66 kW/m2.OC (OR) = 4864.8 kcal/hr.m2.OC L-2 A gas turbine generator is delivering an output of 20 MW in an open cycle with a heat rate of 3440 kcal/kWh. It is converted to combined cycle plant by adding heat recovery steam generator and a steam turbine raising the power generation output to 28 MW. However, with this retrofitting and increased auxiliary consumption, the fuel consumption increases by 5% in the gas turbine. 2 _______________________ Bureau of Energy Efficiency Paper 4 SET A Calculate the combined cycle gross heat rate and efficiency. Ans Gas turbine output = 20 MW Combined cycle output = 28 MW Heat rate in GT open cycle for 20 MW = 3440 kcal/kwh Increase in fuel consumption in combined cycle operation = 5% Combined cycle heat rate = (3440 X 1.05) X( 20 / 28) = 2580 kcal/kwh Combined cycle plant efficiency = (860 / 2580) X 100 = 33.33% . End of Section - II . Section - III: LONG NUMERICAL QUESTIONS Marks: 4 x 20 = 80 (i) Answer all Four questions N-1 The steam requirement of an export oriented unit is met by a 6 TPH oil fired package boiler generating steam at 10 kg/cm2. The monthly steam consumption of the unit is 3000 tonnes. Other data are given below: Fuel oil composition: Carbon = 86%; Hydrogen = 12%; Oxygen= 0.5%; Sulphur =1.5% Specific heat of flue gases, Cp = 0.27 kcal/kgoC G.C.V. of fuel oil = 10,000 kcal/kg Sp.heat of super heated water vapour = 0.45 kcal/kgoC Enthalpy of steam at 10 kg/cm2 = 665kcal/kg Feed water temperature = 85 oC % O2 in dry flue gas = 6% Flue gas temperature at boiler outlet = 240 oC Ambient temperature = 30oC Cost of fuel oil = Rs.43 per kg. Radiation and other unaccounted losses = 2.45% The export oriented unit is costing its steam cost based on the fuel consumption cost with additional 15% to account for the auxiliary and consumables. A neighbouring continuous process plant now offers to supply the required steam at 10 3 _______________________ Bureau of Energy Efficiency Paper 4 SET A kg/cm2 to the export oriented unit at a cost of Rs 3300 per tonne with a condition that all the condensate will be returned back. Calculate the following: a) Boiler efficiency b) Cost advantage per tonne of availing steam from neighbouring plant in place of in-house generation and also monthly monetary saving. Ans First calculate the efficiency of Boiler (in EOU) Theoretical air required = = 11.6 C + 34.8 (H O/8) + 4.35 S = [11.6 X 86 + 34.8 (12 0.5/8) + 4.35 X 1.5] x 1/100 = 14.195 = Say 14.2 % Excess Air = = [% O2 / (21 - % O2)] X 100 [6 / (21 6)] X 100 = 40% AAS = Actual amount of air supplied = 14.2 X 1.4 = 19.88 kg per kg. of fuel oil Mass of dry flue gas mdfg = Mass of combustion gases due to presence of C,H,S +Mass of N2 supplied = (0.86 X 44/12) + (0.015 X 64 / 32) + [(19.88 14.2) X 23 / 100] + (19.88 X 77/100) = 19.797 Mass dry flue gas ,say = 19.8 Kg / kg fuel Or Alternatively mass of dry flue gas = (AAS + 1) 9 H = (19.88 + 1) 9 X 0.12 = 19.8 Kg./Kg. fuel L1 = % heat loss in dry flue gas = [mdfg x Cp x (Tq Ta) / GCV] x 100 L1 19.8 X 0.27 x (240 30) = --------------------------------- x 100 10,000 = 11.23% L2 L2 = Loss due to presence of hydrogen forming water vapour 9 x H [584 + Cps (Tg Ta)] = ------------------------------------------ x 100 GCV 9 X 0.12 [584 + 0.45 (240 30)] = ------------------------------------------- x 100 10000 = 7.33% L3 = Radiation and other unaccounted losses = 2.45% Total losses = L1 + L2 + L3 4 _______________________ Bureau of Energy Efficiency Paper 4 SET A = 11.23 + 7.33 + 2.45 = 21.05 % Efficiency of the EOU boiler by indirect method = 100 21.05 =78.99 % = Say 79 % Secondly calculate the cost of steam in the EOU plant Evaporation Ratio = [(n X GCV) / (hg hf)] X 100 = [(0.79 X 10000) / (665 85)] X 100 = 13.62 kg Steam / kg. Fuel Fuel oil consumption Fuel oil consumption = 1000 / 13.62 kg. per tonne of steam = 73.42 kg./tonne of steam gen Cost of fuel oil. Cost of steam in EOU = = = = Say = Rs. 43 per kg Fuel cost + 15% fuel cost 73.42 x 1.15 x 43 Rs.3,599 per tonne Rs 3600 per tonne Selling cost of steam from neighboring plant = Rs 3300 per tonne Cost advantage = 3600 3300 = Rs.300 per tonne Annual Savings = Rs.300 per tonne x 3000tonne/month X 12 month = Rs.108 Lacs N-2 a) The operating parameters of a Vapour Compression Refrigeration system are indicated below. Parameter Water Flow (m3/hr) Inlet Temperature (OC) Outlet Temperature (OC) Density (kg/m3) Chiller side 89 10.1 6.8 1000 Condenser side 87 32.3 36.6 990 Find the COP of the Refrigeration system ignoring heat losses. b) A 6 pole, 415 volt, 3 , 50 Hz induction motor delivers 22 kW power at rotor shaft at a speed of 950 rpm with PF of 0.88. The total loss in the stator including core, copper and other losses, is 2 kW. Calculate the following. i) ii) iii) iv) v) i) Slip ii) Rotor Copper Loss iii) Total Input to motor ivi) Line current at 415 V and motor pf of 0.88 v) Motor operating efficiency a) Refrigeration Effect = 89 x 1000 x (10.1 6.8) 5 _______________________ Bureau of Energy Efficiency Paper 4 SET A Ans Condenser load Compressor work C.O.P. b) Synchronous Speed Motor Speed Slip Power input to rotor = (120 x 50 / 6 ) = 1000 rpm = 950 rpm = (1000 950 ) / 1000 = 5 % = { ( 22 / ( 1 0.05 ) } = 23.16 kW (ii) Rotor Copper Loss Or = ( 0.05 x 23.16 ) = 1 . 1 5 8 k W = 23.16-22 =1.16 kW (iii) Total Input to motor = ( 23.16 + 2 ) (iv) Line Current = ( 25.16 x 1000) /( 3 x 415 x0.88 ) = 39.75 Amps (v) Motor Efficiency = (i) N-3 = 293700 kcal/hr = 87 x 990 x (36.6 32.3) = 370359 kcal/hr = Condenser load Refrigeration effect = 370359 293700 = 76659 Kcal/hr = Refrigeration Effect/ Compressor work = 293700/76659 = 3.83 = 25.16 kW ( 22 / 25.16 ) = 87.44 % A common plant facility is installed to provide steam and power to textile and paper plant with a co-generation system. The details and operating parameters are given below: 60 T/hr.80 bar at 5000C Enthalpy- 810 kcal/kg Boiler To paper industry Turbine G Coal Alternator Boiler efficiency- 80% GCV of Coal-5000 kcal/kg Boiler feed water temp.- 800C Steam 20 T/hr To textile industry To textile industry Steam 60T/hr. at 10 bar Enthalpy- 660 kcal/kg Steam Header Steam 40 T/hr To paper industry Other data: - Turbine, alternator and other losses = 8% - Specific steam consumption in paper industry= 5 Tons/Ton of paper - Specific power consumption in paper industry= 600 kWh/Ton of paper Calculate: 6 _______________________ Bureau of Energy Efficiency Paper 4 SET A Ans i. i. Coal consumption in boiler per hour or per day. ii. Power generation from co-generation plant iii. If 10% is auxiliary power consumption in co-generation plant, how much power is consumed by the textile industry per hour? iv. What is the gross heat rate of turbine? i) Boiler efficiency = Steam production ( steam enthalpy- Feed water enthalpy) / Quantity of coal x G.C.V. of coal Quantity of coal = 60,000 (810-80)/ 0.8 x 5000 = 10.95 tons/hr. ii) Gross power generation from co-generation plant Total enthalpy input to turbine = 60,000 x 810 = 48.6 Million kcal. Total enthalpy out put through back pressure= 60,000* 660 = 39.6 Million kcal Enthalpy difference = 48.6- 39.6 = 9 Million kcal/hr Turbine,alternator and other losses =8% or 9x0.08 = 0.72 Million kcal/hr Useful energy for power generation = 9- 0.72 = 8.28 Million kcal/hr 6 Power generation from co-generation plant = 8.28 x 10 /860 = 9628kWh iii) If 10% is auxiliary power consumption in co-generation plant, power consumed by textile industry 10% of total power generation = 9628 x 0.10 = 962.8kWh Total power consumed by industries = 9628 962.8 = 8665.2kWh Total steam consumption in paper plant 40 tons/hr. and specific steam consumption 5 ton/ton of paper. So Paper production per hour is 8 tons. Specific power consumption = 600kWh/ton. Total power consumption in paper industry = 8 x 600 = 4800kWh Total power consumption in textile industry = 8665.2- 4800 = 3865.2 kWh iv) Gross heat rate= Input enthalpy output enthalpy/ gross generation =( 48.6- 39.6) 106/ 9628 = 934.7 kCal/kWh N-4 A To attempt ANY ONE OF THE FOLLOWING among A, B, C and D A captive thermal plant is delivering an output of 29 MW at the generator terminal. The generator efficiency is 96%. The steam generated in a utility boiler with an efficiency of 86% at 105 ata and 485oC is fed to the turbine. The turbine exhausts steam to condenser maintained at 0.1 ata and 45.5oC. The feed water temperature at inlet to the boiler is 105oC. 7 _______________________ Bureau of Energy Efficiency Paper 4 SET A The other data pertaining to captive power plant are, Enthalpy of steam at 105 ata 485oC = 795 kcal/kg. Dryness fraction of steam at inlet to condenser = 0.9 Enthalpy of dry saturated steam at 0.1ata =618 kcal/Kg. o Enthalpy of water at 0.1 ata & at 45.5 C = 45.5 kcal/Kg. Loss in the gear box connecting turbine and generator = 1100 kW Enthalpy of feed water at inlet to the boiler , = 105 kcal/Kg. Based on the above data determine: i. Output of the steam turbine in kW ii. Steam flow through the turbine iii. Turbine heat rate iv. Unit heat rate Ans Enthalpy of steam at turbine exhaust i.e. h3 i) = 45.5 +0.9 (618 45.5) = 560.75 Kcal/Kg. Generator electric output Generator input = 29000 KW = 29000 / 0.96 = 30208.33 KW Loss in gear box Output of steam turbine = 1100 KW = Generator input + Gear box loss = 30208.33 + 1100 = 31308.33 KW Output of the steam turbine Say = 31308 KW ms = Steam flow through turbine 8 _______________________ Bureau of Energy Efficiency Paper 4 SET A = (Turbine output x 860) (h2 h3) Turbine enthalpy drop h2 = Enthalpy at turbine inlet h3 = Enthalpy at turbine exhaust ms = 31308 x 860 / 795 560.75) ii) Steam flow through the turbine iii) Turbine heat rate = 795 kcal/kg = 560.75 kcal/Kg. = 114940.78 kg/Hr. = 114.94 TPH Say = 115 TPH = Heat input to turbine / Generator output = [ms (h2 h1)] / 29000 = 115000 (795 105) / 29000 = 2736.2 kcal/ kWh iv)Unit heat rate = Turbine heat rate / Efficiency of boiler Unit heat rate = 2736.2 / 0.86 = 3181.63 kcal/ kWh Or B In a textile unit a stenter is delivering 80 meters/min of dried cloth at 5% moisture. The moisture of wet cloth at inlet is 50%. The stenter is heated by steam at 7 kg/cm2 with inlet enthalpy of 660 kcal/kg. and condensate exits the stenter at 135 kcal/kg. Other data Latent heat of water evaporated from the wet cloth Weight of 10 meters of dried cloth Inlet temperature of wet cloth Outlet temperature of dried cloth at stenter outlet i) ii) = 540 kcal/kg = 1 kg = 27oC = 80oC. Estimate the steam consumption in the stenter considering a dryer efficiency of 48%. Determine the specific steam consumption kg/kg of dried cloth Ans Output of stenter = 80 mts/min. = 80 x 60 /10 = 480 Kg/hr. Moisture in the dried output cloth = 5% Wt of bone dry cloth = 480 X (1 0.05) i.e. W = 456 Kg/hr. mo =moisture in outlet cloth =(480 456)/456 =0.0526 Kg./Kg. of bone dried cloth Inlet moisture = 50% Wt of inlet cloth = 456 / (1 0.50) = 912 Kg./hr. mi = moisture in inlet cloth = 912 X 0.5 / 456 = 1.00 Kg./Kg. bone dried cloth Inlet temperature of cloth = 27oC 9 _______________________ Bureau of Energy Efficiency Paper 4 SET A Final temperature of cloth = 80oC Heat load on the dryer . Heat load on the dryer = w x (mi mo) X [(Tout Tin) + 540] Kcal/hr. = = 456 (1 0.0526) X [(80 27) + 540] 2,56,184.5 Kcal/hr Efficiency of the dryer = 48% Heat input to the stenter = 2,56,184.5 / 0.48 = 5,33,717.71 Kcal/hr Steam consumption in the stenter = = 5,33,717.71 / (660 135) 1016.61 Kg/hr Steam consumption per Kg. of dried at stenter outlet cloth = 1016.61 / 480 = 2.12 Kg./Kg. dried cloth Or C Determine the cooling load of a commercial building for the following given data. Outdoor conditions : DBT = 35 C ; WBT = 25 C; Humidity = 18 g of water / kg of dry air Desired indoor conditions : DBT = 25.6 C ; RH = 50 %; Humidity = 10 g of water / kg of dry air Total area of wall = 40 m2 Total area of window = 20m2 U Factor ( Wall ) = 0.33 W / m2K U Factor ( Roof ) = 0.323 W / m2K U factor [ fixed windows with aluminum frames and a thermal break ] = 3.56 W / m2K 15 m x 25 m roof constructed of 100 mm concrete with 90 mm insulation & steel decking. CLTD at 17:00 h :Details : Wall = 12 C Roof = 44 C Glass Window = 7 C SCL at 17 : 00 h :Details : Glass Window = 605 W/ m2 Shading coefficient of Window = 0.74 Space is occupied from 8:00 to 17:00 h by 25 people doing moderately active work. Sensible heat gain / person = 75 W ; Latent heat gain / person = 55 W ; CLF for people = 0.9 10 _______________________ Bureau of Energy Efficiency Paper 4 SET A Fluorescent light in space = 21.5 W/m2 FLF for lighting = 0.9 Ballast factor details = 1.2 for fluorescent lights & 1.0 for incandescent lights Computers and office equipment in space produces 5.4 W/m2 of sensible heat One coffee maker produces 1050 W of sensible heat and 450 W of latent heat. Air changes / hr of infiltration = 0.3 Height of building = 3.6 m Ans I External Heat Gain (i) Conduction heat gain through the wall =U factor x net area of wall x CLTD =[ 0.33 x 40 x 12 ] = 158.4 W (ii) Conduction heat gain through the roof =U factor x net area of roof x CLTD =0.323 x ( 15 x 25 ) x 44 = 5 329.5 W (iii) Conduction heat gain through the windows =U factor x net area of windows x CLTD = (3.56 x 20 x 7) = 498.4 W (i) Solar radiation through glass = Surface area c Shading coefficient x SCL =(20 x 0.74 x 605) = 8 954 W II Internal Heat Gain (i) Heat gain from people =Sensible heat gain + Latent heat gain Sensible heat gain =(No.of people x Sensible heat gain / person x CLF) =(25 x 75 x 0.9) = 1 687.5 W Latent heat gain =No.of people x Latent heat gain / person =(25 x 55 ) = 1 375 W Therefore, Heat gain from people=(1687.5 + 1375 ) = 3 062.5 W (ii) Heat gain from lighting Energy input =( Energy input x Ballast factor x CLF ) =( Amount of lighting in space / unit area ) x Floor area =21.5 x ( 15 x 25 ) =8 062.5 W Therefore, heat gain from lighting =( 8062.5 x 1.2 x 0.9 ) =8 707.5 W 11 _______________________ Bureau of Energy Efficiency Paper 4 SET A (iii) Heat generated by equipment : Sensible heat generated by coffee maker =1050 W Latent heat generated by coffee maker =450 W Sensible heat gain by computers and office equipment = 5.4 x 375 = 2025 W Therefore, Heat generated by equipment (iv) = 3 525 h Heat gain through air infiltration =( Sensible heat gain + Latent heat gain ) Sensible heat gain Airflow =(1210 x airflow x T ) =( Volume of space x air change rate ) / 3600 ={ (15 x 25 x 3.6 ) x 0.3 } / 3600 =0.1125 m3 / s Therefore, sensible heat gain Latent heat gain No Space Load Components 1 Conduction through exterior wall 2 Conduction through roof 3 =1210 x 0.1125 x ( 35 25.6 ) = 1 279.58 W =3010 x 0.1125 x ( 18 10 ) = 2 709 W Sensible Heat Load (W) Latent Heat Load (W) 158.4 ------ 5 329.5 ------ Conduction through windows 498.4 ------ 4 Solar radiation through windows 8954 ------ 5 Heat gained from people 1 687.5 1 375 6 Heat gained from lighting 8 707.5 ------ 7 Heat gained from equipment 3 075 450 8 Heat gained by air infiltration 1 279.58 2 709 29 689.88 4 534 Total space cooling load Or D During heat balance of a 5 stage preheater Kiln in a cement plant, the following data was measured at Preheater (PH) Fan Inlet and clinker cooler vent air fan inlet: 12 _______________________ Bureau of Energy Efficiency Paper 4 SET A Parameter measured Temperature Static Pressure Specific heat 316 (Ps) mm WC -650 Avg. Dynamic Pressure (Pd) mm WC 28.6 268 -56 9.7 0.24 o Unit C PH Exit Gas at PH fan Inlet Clinker cooler vent air at cooler Stack Fan Inlet kcal/kg o C 0.248 Gas Density at STP kg/m3 Duct Area 1.4 2.27 1.29 2.01 m2 Note: take Pitot tube constant as 0.85, reference temperature 20 oC and atmospheric pressure 9908 mm WC. Other Data Clinker Production TPH 45.16 Designed specific volume of PH gas Nm3/kg clinker 1.75 NCV of Coal kcal/kg 5500 Cost of coal Rs./ton 6500 Annual Operation hrs 8000 Calculate the following: Ans i. Specific volume of PH gas as well as cooler vent air (Nm3/kg clinker) ii. Heat loss in pre-heater exit gas (kcal/kg clinker) iii. Heat loss in cooler vent air (kcal/kg clinker) iv. If the measured specific volume of PH gas (Nm3/kg clinker) exceeds the design value, calculate the heat loss (kcal/kg clinker) and annual monetary loss due to excessive specific volume of PH gas. i. Density of Pre-heater gas at PH Fan Inlet at prevailing temp., pressure conditions: 273 (273 273 1.40 T,P (273 Velocity of PH gas 2 g Pd v Pt T,P STP (9908 PS ) T ) 10334 (9908 650) = 0.581 kg/m3 316) 10334 T,P 13 _______________________ Bureau of Energy Efficiency Paper 4 SET A 2 9.8 28.6 = 26.4 m/sec 0.581 Volumetric flow rate of PH gas = velocity X duct cross-sectional area = 26.4 X 2.27 = 59.9 m3/sec = 59.9 X 3600 = 215640 m3/hr Specific volume of PH gas = 215640 X 0.58/1.4 = 89491 Nm3/hr = 89491/45160 = 1.98 Nm3/kg clinker v 0.85 Similarly density of cooler vent air at cooler vent air fan Inlet at prevailing temp., pressure conditions: 273 (9908 PS ) T,P STP (273 T ) 10334 273 (9908 56) 1.29 = 0.62 kg/m3 T,P (273 268) 10334 Velocity of cooler vent air in the fan inlet duct 2 g Pd v Pt T,P 2 9.8 9.7 = 14.88 m/sec 0.62 Volumetric flow rate of PH gas = velocity X duct cross-sectional area = 14.88 X 2.01 = 29.9 m3/sec = 29.9 X 3600 = 107640 m3/hr Specific volume of cooler vent air = 107640 X 0.62/1.29 = 51734 Nm3/hr = 51734/45160 = 1.15 Nm3/kg clinker v 0.85 ii)Heat loss in PH exit gas Q1 = m ph c p T (Cp of PH gas = 0.248 kcal/kg oC) Q1= 1.98 X 1.4 X 0.248 x (316-20) = 203.5 kcal/kg clinker iii) Heat loss in cooler vent air Q2 = mCA c p T (Cp of cooler vent air = 0.24 kcal/kg oC) Q2= 1.15 x 1.29 x 0.24 x (268-20) = 88.3 kcal/kg clinker 14 _______________________ Bureau of Energy Efficiency Paper 4 SET A iv) Heat Loss due to excess specific volume of PH gas Vexcess =1.98 1.75 = 0.23 Nm3/kg clinker Heat loss Q = 0.23 x 1.4 x 0.248 x (316-20) = 23.6 kcal/kg clinker Equivalent coal saving = 23.6/5500 = 0.0043 kg coal/kg clinker or ton of coal/ton of clinker Coal saving in one hour = 0.0043 x 45.16 = 0.194 TPH Annual Coal Saving = 0.194 x 8000= 1552 tons of coal per annum Annual Monitory Saving = 1552 x 6500 = Rs. 100.88 lakhs -------- End of Section - III --------- 15 _______________________ Bureau of Energy Efficiency

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