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13th National Certification Exam Energy Managers & Auditors SEPTEMBER 2012 Paper 4

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Click Here & Upgrade PDF Complete Expanded Features Unlimited Pages Documents Paper 4 SET A KEY 13th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY AUDITORS September, 2012 PAPER 4: Systems Energy Performance Assessment for Equipment and Utility Date: 16.9.2012 Section - I: Timings: 14:00-16:00 HRS Duration: 2 HRS Max. Marks: 100 BRIEF QUESTIONS Marks: 10 x 1 = 10 (i) Answer all Ten questions (ii) Each question carries One mark S-1 In a vapour compression refrigeration system, why the heat rejected in the condenser is more than the heat absorbed in the evaporator ? Ans Because heat of compression is also added to it S-2 If the unit heat rate is 3120 kcal/kWh and the turbine heat rate is 2808 kCal/kWh what is the boiler efficiency ? Ans (2808/3120) x 100 = 90 % S-3 A rise in conductivity of boiler feed water indicates ____ . Ans Rise in the TDS level of feed water S-4 Why is it preferable to measure the flow at the inlet side of the fan? Ans Less turbulence S-5 The critical point of steam occurs at ____bar and _______ oC Ans 221.2 bar and 374.15oC S-6 In a heat exchanger _______ is the ratio of actual heat transfer rate to the maximum heat transfer rate. 1 _______________________ Bureau of Energy Efficiency Click Here & Upgrade PDF Complete Expanded Features Unlimited Pages Documents Paper 4 SET A KEY Ans Effectiveness S-7 In an integrated steel plant pig iron is produced from _________furnace? Ans Blast furnace S-8 If the PLF of a 210 MW power plant is 80% , what is the annual gross generation in MWh Ans 1,471,680 MWH S-9 A pump operates on water with a total head of 12 m. If water is replaced by brine with a specific gravity of 1.2 what will be the total head developed by the pump ? Ans 12 m or same S-10 A draft system in a boiler which uses both FD and ID fan is called . Ans Balanced Draft . 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 Ans Calculate pressure drop in meters when pipe diameter is increased from 250 mm to 300 mm for a length of 600 meters. Water velocity is 2 m/s in the 250 mm diameter pipe and friction factor is 0.005. Pressure drop = 4fLV2 -------------2gD Velocity of water in pipe of 300 mm diameter = (0.25 x 0.25 x 2) /(0.3 x 0.3) = 1.39 m/s Pressure drop with 300 mm = 4 x 0.005 x 600 x 1.392 / (2 x 9.81 x 0.300) = 3.94 m 2 _______________________ Bureau of Energy Efficiency Click Here & Upgrade PDF Complete Expanded Features Unlimited Pages Documents L-2 Paper 4 SET A KEY A three phase 37 kW four pole induction motor operating at 49.8 Hz is rated for 415 V, 50 Hz and 1440 RPM. The actual measured speed is 1460 RPM. Find out the percentage loading of the motor if the voltage applied is 410 V. Ans % Loading = Slip x 100% (Ss Sr) x (Vr / V)2 Synchronous speed = 120 x 49.8 / 4 = 1494 rpm Slip = Synchronous Speed Measured speed in rpm. = 1494 1460 = 34 rpm. % Loading = 34 x 100% = 61.45% (1494 - 1440) x (415/410)2 . End of Section - II . 3 _______________________ Bureau of Energy Efficiency Click Here & Upgrade Expanded Features Unlimited Pages PDF Complete Documents Section - III: Paper 4 SET A KEY LONG NUMERICAL QUESTIONS Marks: 4 x 20 = 80 (i) Answer all Four questions Refer Original question paper for questions N1 Key a) Theoretical air required for complete combustion O2 (11.6 x C ) + 34.8 x + (4.35 x S ) H 2 8 = 100 = 32.52 (11.6 x 33.95) + 34.8 x 5.01 8 + (4.35 x 0.09) 100 Moles of N2 = % CO2 theoretical = = 4.27 kg / kg of paddy husk 77 4.27 100 0.0091 + = 0.1178 28 28 Moles of C Moles of N 2 + Moles of C + Moles of S 0.3395 12 = 0.3395 0.0009 0.1178 + + 12 32 Max theoretical ( CO2 )t = 19.36 % Actual CO2 measured in flue gas = 14.0% b) % Excess air supplied = 7900 x [( CO 2 ) t (CO 2 ) a (CO 2 ) a x [100 ( CO 2 ) t ] c) Actual mass of air supplied = = = = 37.5 % {1 + EA/100} x theoretical air {1 + 37.5/100} x 4.27 5.87 kg/kg of coal 4 _______________________ Bureau of Energy Efficiency Click Here & Upgrade PDF Complete Expanded Features Unlimited Pages Documents Mass of dry flue gas = Paper 4 SET A KEY 0 .3395 44 5 .87 77 (5 .87 4 .27 ) 23 + 0 .0091 + + 12 100 100 = 6.15 kg / kg of coal (or) (actual mass of air supplied + 1) mass of H20 (5.87 + 1) (9H + M) = 6.87 (9x.05 + 0.1079)= 6.87 0.5579 = 6.31 kg/kg of coal % Heat loss in dry flue gas = = = Loss due to CO m x C P x ( Tf Ta ) x 100 GCV of fuel 6.15 x 0.23 x (160 32 ) x 100 3568 5.07 % = = L2 0.35 x 0.3395 x 5654 (0.35+14) x 3568 = 1.31 % Heat Loss in ash % heat loss due to unburnt flyash % ash in paddy husk Ratio of bottom ash to flyash GCV of flyash Amount of flyash in 1 kg of husk Heat loss in flyash GCV of bottom ash = 16.73 = 10:90 = 450 kcal/kg = 0.9 x 0.1673 = 0.15 kg = 0.15 x 450 = 67.5 kcal/kg of husk = 800 kcal/kg Amount of bottom ash in 1 kg of husk = 0.1 x 0.1673 = 0.01673 kg Heat loss in bottom ash = 0.01673 x 800 = 13.4 kcal/kg of husk 5 _______________________ Bureau of Energy Efficiency Click Here & Upgrade Expanded Features Unlimited Pages PDF Complete Documents Paper 4 SET A KEY Total heat loss in ash = 67.5 + 13.4 = 80.9 kcal/kg = 80.9/3568 = 2.26 % % loss in ash Total losses = 100 (5.07 + 1.31 +2.26) (15.4) Boiler efficiency = N-2 100 8.64 15.4 = 75.96 % KEY Hot Water use per day : 20,000 L/day Water in = 200C Water out = 600C Temp. diff. = 400C Total Heat required = mCpdt = 20000 x 1 x 40 = 8,00,000 kcal/day 1. Energy Requirement for 20KL/day of water for a temperature differential of 40 deg.C in an Electric Boiler/Geyser Energy Requirement (for 20 KL/day) = Total heat required (800000) 860 kcal/kWh x 0.99 ( efficiency of electric heating)) = 939.6 kWh/day 2. For 20 KL/day, of water flow with 400C Temperature Diff. Energy to be drawn by Heat Pump = 8,00,000 860x0.95x2.5 Energy drawn by circulation pump Energy drawn by evaporator fan = 391.68 Kwh/day = 3.74 x 24 hr = 1.4 kW x 16 hr = 89.76 kWh/day = 22.4 kWh/day Total Energy drawn by heat pump system = 391.68 +89.76+22.4 = 503.8 kWh /day SAVINGS IN COMPARISON TO ELECTRIC WATER HEATER = 939.6 503.8 = 435.75 Kwh/day = 1,52,516 kWh/year ( @ 350 days/year) = 12.20 lakhs ( @ Rs8.0 per kWh) 6 _______________________ Bureau of Energy Efficiency Click Here & Upgrade Expanded Features Unlimited Pages PDF Complete Documents Paper 4 SET A KEY 3. SIMPLE PAY BACK PERIOD = Rs.16.0 LAKHS Investment/ Rs.12.20 lakhs per year savings = 1.30 years or 16 months KEY N-3 Ans Power generation from cogen plant = 5000X 0.9 X 8000 = 360 lac Kwh/yr Auxiliary power Net power generation = = 1% 0.99 X 360 = 356.4 lac Kwh Natural gas requirement for power generation = 360 X 3050 / 9500 = 115.57 lac sm3 Cost of fuel per annum = 115.57 X 8 = Rs.924.56 lacs Annual expenditure for interest, depreciation and O&M = 500 + 200 = 700 lacs Total cost of generation = Rs.1624.56 lacs. Cost of cogeneration power = 1624.56 X 105 / 356.4 X 105 = Rs.4.56 / Kwh. Gas consumption in existing gas fired boiler = = = [10000 (665 85) / (0.86 X 9500)] 710 Sm3/hr 710 x 24 = 17040 sm3/day Cost of steam from existing boiler = = 710*Rs. 8 x8000 Rs. 454.4 Lacs /yr Cost of power generation after giving = credit for steam generation 1624.56 454.4 = Rs.1170.16 lacs Cost of power generation after accounting = 1170.16 X 105 / 356.4 X 105 for steam cost = Rs. 3.28 / Kwh Grid power cost = Rs. 4.5 / Kwh Cost advantage for cogen plant generation = 4.5 3.28 = Rs.1.22 / Kwh Daily gas requirement for operating GT cogen plant = 5000 X 0.9 X 3050 X 24 9500 34673.68 Sm3 / day = 7 _______________________ Bureau of Energy Efficiency Click Here & Upgrade Expanded Features Unlimited Pages PDF Complete Documents Paper 4 SET A KEY Additional gas requirement for co-gen plant N-4 = To attempt ANY ONE OF THE FOLLOWING among A, B, C and D N4 A Ans 34673.68 17040 = 17633.68 Sm3/day KEY i) Turbine power output kW = Steam flow to turbine kg/hr x enthalpy drop across the turbine kcal/kg ------------------------------------------------------------------------------------------------------860 Inlet enthalpy of steam =794.4 kcal/kg Enthalpy of exhaust steam is calculated as given below exhaust steam dryness fraction = 90% enthalpy of exhaust steam = (45.9 + 0.9 x 572.5) = 561 kcal/kg turbine out put = ((120 x 1000 kg/hr x (794.4 561) kcal/kg) /860 turbine output = 32567.4 kW ii) generator output kW = turbine output x combined efficiency of mechanical, gear transmission & generator = 32567.4 x 0.92 =29962 kW iii) turbine heat rate = heat input in to the turbine/ generator out put =q x (h1 hw)/generator out put Where q = steam inflow to turbine kg/hr h1= enthalpy of turbine inlet steam =794.4 kcal/kg hw= enthalpy of feed water to boiler = 100 kcal/kg Turbine heat rate = ((120 x 1000 kg/hr) x (794.4 100) kcal/kg))/ 29962 kw = 2781 kcal/kwh iv) unit heat rate = turbine heat rate /boiler efficiency = 2781 / 0.88 = 3160 kcal/ kwh v) turbine cycle efficiency = (860 / turbine heat rate) x 100 = 860 /2781 =0.309 =0.309 x 100 = 30.9% vi) condenser heat load = m x cp x dt Where m = cooling water flow through condenser, kg/hr 8 _______________________ Bureau of Energy Efficiency Click Here & Upgrade Expanded Features Unlimited Pages PDF Complete Documents Paper 4 SET A KEY note: density of water is given as 0.95 g /cubic centimetre = 950 kg/ cubic meter cp = specific heat of cooling water, kcal/ kg. oC = 0.98 kcal /kg. oC dt = cooling water temperature rise, oC = 10 Condenser heat load =6318 x 950 x 0.98 x 10 = 5,88,20,580 kcal /hr vii) specific steam consumption of turbine = 860 / (enthalpy drop x combined efficiency) = 860/ ((794.4 561) x 0.92)) =860 / (233.4 x 0.92) =4.0 kg/kwh = 4.0 kg / kwh N4-B Ans KEY Volumetric flow rate of PH gas at NTP = 1.47 x 125 x 1000 = 183750 [Nm3/hr] Mass flow rate of PH gas [kg/hr] = 183750 x 1.42 = 260925 Calculation for 4 stage pre-heater kiln Heat loss in PH Gas = m x cp x T = 260925 x 0.244 x 370 = 23556309 Equivalent coal wasted = 23556309 5540 1000 = 4.252 [kcal/hr] [kcal/hr] [tons of coal/hr] Electrical Energy consumption of PH Fan Volumetric flow rate of PH Gas at 370 oC temperature and -400 mm WC static pressure: = 183750 ( 273 + 370) 10333 = 450216 V or [m3/hr] 273 (10333 400 ) = 450216/3600 = 125 V Pressure difference across PH fan Power consumption of PH fan P= [m3/sec] = 50 (- 400) = 450 125 450 = 806.24 102 0.72 0 .95 [mm WC] [kW] Calculation for 6 stage pre-heater kiln Heat loss in PH Gas = m x cp x T = 260925 x 0.244 x 295 = 18781381 Equivalent coal wasted = 18781381 5540 1000 = 3.39 [kcal/hr] [kcal/hr] [tons of coal/hr] Electrical Energy consumption of PH Fan Volumetric flow rate of PH Gas at 295 oC temperature and -600 mm WC static pressure: V = 183750 ( 273 + 295) 10333 = 405875 [m3/hr] 273 (10333 600) Or V = 405875/3600 = 112.75 [m3/sec] 9 _______________________ Bureau of Energy Efficiency Click Here & Upgrade Expanded Features Unlimited Pages PDF Complete Documents Paper 4 SET A KEY Pressure difference across PH fan = 50 (- 600) = 650 Power consumption of PH fan P = 112.75 650 = 1050.4 [mm WC] [kW] 102 0.72 0.95 The above kilns can be compared as follows: Item PH Gas heat loss (kcal/hr) Equivalent coal wasted (tons of coal) Power consumption in PH Gas (kW) 6 Stage PH Kiln 18781381 3.39 1050 4 stage PH kiln 23556309 4.252 806.24 Calculation for annual Monetary savings Coal savings in 6 stage PH Kiln = 4.252 3.39 = 0.862 Annual monetary savings (Thermal) [ton of coal/hr] = 0.862 x 8000 x 6150 = 4,24,10,400 [Rs.] Additional Electrical energy requirement for 6 stage PH Kiln = 1050.4 806.24 = 244.16 [kW] Annual additional electrical cost [Rs.] = 244.16 x 8000 x 5 = 97,66,400 It is obvious that in monetary terms, thermal energy saving in 6 stage pre-heater kiln is higher than the additional electrical energy cost in 4 stage kiln. Therefore, 6 stage pre-heater kiln is better option than 4 stage pre-heater kiln. So the net annual monetary saving in case of 6 stage pre-heater kiln is = 4,24,10,400 97,66,400 = 3,26,44,000 N4-C Ans [Rs.] KEY a) Before insulation Surface heat loss, S = [ 10 + (TS-Ta)/20] x (Ts Ta) Total heat Loss =SxA where A= Surface area, m2 Surface heat loss , S = [ 10 + ( 110-25)/20] x (110-25) = 1211.25 K.Cal/m2/hr Total heat loss = 1211.25 x 20 m2 = 24225 kCal/hr After insulation Surface heat loss ,S = [ 10 + (55-25)/20] x (55-25) =345 K.Cal/m2/hr Total heat loss = 345 x 20 m2 = 6900 kCal/hr Heat reduction per hour after proper insulation = 24225- 6900 = 17325 kCal/hr Annual heat loss reduction = 17325 x 8000 = 138600000 = 138.6 million kCal/year Steam distribution loss Heat loss Boiler efficiency = 20% = 138.6 million kCal/ 0.8 = 173.25 million kcal/year = 70% Equivalent coal consumption reduction = 173.25 x106 /0.7 x 4800 = 51.56 Ton /year 10 _______________________ Bureau of Energy Efficiency Click Here & Upgrade Expanded Features Unlimited Pages PDF Complete Documents Paper 4 SET A KEY Monetary Cost savings per year Investment @ Rs 1000 per M2 = 51.5 x 5000 = Rs 2.575 lacs = 20 x 1000 = Rs 20000 Condensate recovery Reduction in coal consumption through heat recovered from condensate return Annual coal savings Annual savings = 2000 x 1 x (80 40) / 0.7 x 4800 = 23.8 kg of coal per hour = 23.8 x 8000/1000 = 190.4 ton / year = 23.8 x 8000 x Rs.5/kg coal = Rs. 9.52 lacs b)Simple payback period Total savings from both the measures = 2.575 + 9.52 = 12.1 lakhs Total investment = Rs. 20,000 + Rs 2 lakhs = Rs.2.2 lakhs Simple payback period (combined) = 2.2/12.1= 2.2 months c)GHG emission reduction Carbon content in the coal Total Coal saving /year CO2 reduction = 40% by weight = 51.5 + 190.4 = 241.9 Ton per year = 241.9 x 0.4 x 44/12 = 355 Ton of CO2/year N4-D Ans KEY Theoretical air required for complete combustion =[(11.6x85.9)x(34.8x(12-0.7/8))+4.35x0.5]/100 =996.44+414.12+2.175/100 =14.1 kg/kg of oil Existing oxygen % in flue gas =6% % excess air supplied Actual mass of air supplied =6 x100/(21-6) =40% =(1+Excess air/100)x Theoretical air =(1+40/100)x 14.1 =19.74 kg/kg of oil After modification, oxygen % in flue gas =3% % excess air supplied =3 x100/(21-3) =16.67% Actual mass of air supplied =(1+Excess air/100)x Theoretical air =(1+16.67/100)x 14.1 =16.45 kg/kg of oil a) Heat loss reduction through actual mass of air supplied 11 _______________________ Bureau of Energy Efficiency Click Here & Upgrade PDF Complete Expanded Features Unlimited Pages Documents Paper 4 SET A KEY Actual mass of air supplied before WHR =19.74 kg/kg of oil Actual mass of air supplied AFTER WHR =16.45 kg/kg of oil Existing oil consumption per hour = 25 ton/hr x 60kg/ton = 1500 kg of oil /hr Flue gas loss before WHR = [1500 kg oil + (1500 x 19.74 kg air)] x 0.24 x (600-30) = 4255848 kcal/hr Flue gas loss after WHR = [1500 kg oil + (1500 x 16.45 kg air)] x 0.24 x (300-30) = 1696140 kcal/hr Flue gas heat loss reduction after WHR implementation = 4255848-1696140 = 2559708 kcal/hr Reduction in fuel oil consumption after installing Waste heat recovery and reduction in excess air Furnace efficiency after WHR = 256 kg/hr = 25000 x 0.12 x (1200-40) x 100 [(1500-256) x10000)] = 28 % b) Calculate fuel oil reduction after charging hot ingot in reheating furnace Ingot charging temperature is increased from 40 oC to 500 oC Fuel oil reduction due to increased charge temperature = = 25 x 1000 x 0.12 x(500-40)/0.28 x10,000 = 492.86 kg/hr = 493 kg/hr c ) Specific oil and power consumption after implementing both the above measure Fuel oil reduction after implementation of both measures = 256 + 493 = 749 kg oil/hr Fuel oil consumption after implementation of both measures = 1500 749 = 751 kg oil/hr Yield improvement Production after implementation of both measures = 3% = 25 x 1.03 = 25.75 ton/hr Specific oil consumption = 751/ 25.75 = 29.2 kg/Ton Specific power consumption = 25x90 / 25.75 = 87.37 kWh/ton -------- End of Section - III --------12 _______________________ Bureau of Energy Efficiency

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