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7th National Certification Exam Energy Managers & Auditors NOVEMBER 2008 Paper 4

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Paper 4 Set A, Energy Auditor Key Regn No: _________________ Name: ___________________ (To be written by the candidates) 7th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY AUDITORS Nov., 2008 PAPER 4: Energy Performance Assessment for Equipment and Utility Systems Date: 23.11.2008 Timings: 1400-1600 HRS Duration: 2 HRS Section - I: SHORT DESCRIPTIVE QUESTIONS Max. Marks: 100 Marks: 10 x 1 = 10 (i) Answer all Ten questions (ii) Each question carries One mark (iii) Answer should not exceed 50 words S-1 What will be the synchronous speed of a VFD driven 4-pole induction motor operating at 38 Hz ? Ans. Ns = 120 x f/P = 120 x 38/4 = 1140 RPM S-2 If the power consumed by a refrigeration compressor is 2 kW per ton of refrigeration, what is the energy efficiency ratio? Ans. EER = 12000 Btu 2000 W = 6 S-3 Explain why heat rate of back pressure turbine is greater than that of a condensing turbine. Ans. As it does not take into account of the heat content of the exhaust steam used in the process. S-4 Why line current method used for estimating loading of a motor is not applicable for motor loading less than 75%. Ans. At lower loadings, power factor of a motor degrades significantly and ampere-load curve becomes nonlinear S-5 Explain why actual air delivered is always converted to (FAD) while measuring delivered air volume flow rates in an air compressor. Ans. As air is compressible, its volume flow rate will vary with pressure on delivery side and hence for comparison purposes the volume flow rates are always converted to their value at standard atmospheric pressure. _________________________ Bureau of Energy Efficiency 1 Paper 4 Set A, Energy Auditor Key S-6 What is the minimum wind speed which is acceptable for viable power generation from a wind turbine? Ans. 15 kmph S-7 Ans. If the dry bulb temperature of air is 35oC and the wet bulb temperature is 35oC what will be the relative humidity %. 100 % S-8 For which fuel the difference between GCV and NCV will be smaller, Coal or Natural Gas? Ans. Coal S-9 What is the conversion efficiency range of a biomass gasifier ? Ans. 60 70 % S-10 How many units of energy will be generated by a wind turbine of 250 kW operating at a capacity factor 0.25 in 8760 hours ? Ans. 250 x 0.25 x 8760 = 5,47,500 kWh -------- End of Section - I --------Section - II: LONG DESCRIPTIVE QUESTIONS Marks: 2 x 5 = 10 (i) Answer all Two questions (ii) Each question carries Five marks L-1 A trial for finding out the actual capacity of a reciprocating instrument air compressor of nominal capacity of 900 Nm^3/Hr was done. The following observations were made : Atmospheric pressure : 1.033 kg/sq.cm Ambient temperature : 30 deg. C Receiver capacity : 12 m^3 Additional hold-up volume : 10% of receiver volume Initial pressure (after bleeding) : 0.2 kg/sq.cm g Final pressure (after pump-up) : 7.0 kg/sq.cm g Pump-up time : 5 min:30sec Motor power(avg) : 105 kW (as per power analyzer) Discharge temperature : 45 deg. C Calculate: i) The actual compressor capacity ii) The specific power consumption in kW/nm3/hr Ans. i) Actual capacity,FAD Q= ( P2 P1) P0 _________________________ Bureau of Energy Efficiency *V T * (273+t1) Nm^3/mte (273+t2) 2 Paper 4 Set A, Energy Auditor Key where , P2 = final pressure at receiver after pump-up ,kg/sq.cm a P1 = initial pressure at receiver after bleeding, kg/sq.cm a P0 = atmospheric pressure, kg./sq.cm a V = total storage volume , m^3 T = pump-up time ,mte Q = (8.033-1.233) * 12*1.1/5.5 *303/318 1.033 = 15.05 Nm3/min, say 15 Nm3/min ii) specific power consumption:= 105 kW/Nm3 15*60 = 0.117 kW/Nm3 L-2 The following parameters were observed during the performance testing of pump. Flow rate of fluid :900m3/hr. Density of fluid :950kg/m3 Discharge pressure : 5.0kg/cm2(a) Suction head :5 metre above the pump centerline. Measured power :180kW Motor efficiency :90% Calculate the pump efficiency. Ans. Hydraulic power = (900/3600) x 45 x 950 x 9.81/ 1000 = 104.7 kW Pump shaft power= 180 x 0.9 = 162 kW Pump efficiency = 104.7/162 = 64.6 % -------- End of Section - II --------- Section - III: Numerical Questions Marks: 4 x 20 = 80 (i) Answer all Four questions (ii) Each question carries Twenty marks _________________________ Bureau of Energy Efficiency 3 Paper 4 Set A, Energy Auditor Key N-1 A furnace oil fired boiler is generating steam 20 t/hr @10 kg/cm2 ( enthalpy 650 kcal/kg & feed water temp-80 0C) The evaporation ratio of the oil fired boiler is 14. The GCV of the fuel is 10,200 kCal/kg. Due to high furnace oil cost the management wants to covert from oil firing to Agro residue briquettes firing with a GCV of 3200 kcal/kg. The expected efficiency of the new Briquette fired boiler is 75%. The cost of furnace oil is Rs.28000/t and briquette cost is Rs.4000/t. The annual operating hrs of the boiler is 7000 hrs. The emission factor for furnace oil is 3 t CO2/ton. a. Find out the annual savings for the company by shifting to Briquettes. b. In addition the management wants to claim carbon credits for fuel switch. Calculate the estimated carbon credits for this measure. Ans. Parameter Unit Steam Generation Steam Enthalpy Feed water temp Evap. Ratio Efficiency GCV T/hr kcal/kg o C t/t % kcal/kg Fuel Consumption t/hr fuel cost Cost of operation Rs/T Rs./hr F.Oil Briquettes 20 650 80 14 10200 0.75 3,200 20(650-80) (20/14) /(3200*0.75) 1.43 4.75 28000.00 4000 (1.43 x 28000) (4.75x 4000) 40000 19000 Energy Cost Savings annual operating hrs annual cost Savings 20 650 80 (40000-19000) 21000 hrs 7000 Rs. Lakh/hr 7000 (21000 x 7000) /100000= 1470 Carbon Credits Emission Factor annual F.Oil savings Expected Carbon Credits _________________________ Bureau of Energy Efficiency t CO2/Ton Ton/year CERs 3 (1.43x 7000) 10000 10000 x 3 30000 4 Paper 4 Set A, Energy Auditor Key N-2 The following are the operating parameters of rerolling mill furnace Weight of input material - 10 T/hr Furnace oil consumption - 600 litres/hr Specific gravity of oil - 0.92 Final material temperature - 1200oC Initial material temperature - 40oC Outlet flue gas temperature - 650oC Specific heat of the material 0.12 kCal/kg/oC GCV of oil - 10,000 kCal/kg Percentage yield - 92 % a. Calculate furnace efficiency by direct method b. Calculate Specific fuel consumption on finished product basis The management installed a recuperator to preheat combustion air from 40oC to 300oC resulted in following benefits: Increase in material input by 10 % Reduction in fuel consumption by 13 % Yield improvement from 92% to 96% c. Calculate the furnace efficiency after the modifications d. Reduction in specific fuel consumption after installing the waste heat recovery Ans. a) Furnace efficiency by direct method Heat input Heat output Efficiency 600 lit/hr x 0.92 x 10000 55,20,000 kCal/hr 10,000 x 0.12 x (1200 40) 1,39,2000 kcal/hr 1,39,2000 /55,20,000 25.21 % b) Specific fuel consumption on finished product basis Weight of finished products Furnace oil consumption Specific fuel consumption 10 x 0.92 9.2 T/hr 600 litres/hr 600/9.2 65.2 litres/ton c) Furnace efficiency with 10 % increase in input material Fuel consumption after 600 x 0.87 modification 522 litres/hr Production after modification 10 + 10 x 0.1 _________________________ Bureau of Energy Efficiency 5 Paper 4 Set A, Energy Auditor Key Heat input Heat output Efficiency 11 T/hr 522 lit/hr x 0.92 x 10000 48,02,400 kCal/hr 11,000 x 0.12 x (1200 40) 15,31,200 kcal/hr 15,31,200/48,02,400 31.9 % d) Reduction in Specific fuel consumption Yield of finished product 11 x 0.96 10.56 T/hr Fuel consumption 522 litres Specific fuel consumption 522/10.56 49.43 litres/T Original specific fuel consumption 65.2 litres/T 65.2 49.43 15.77 litres/T N-3 For a double extraction cum condensing turbine with data as given in the following diagram, evaluate a. Power generated if the efficiency of the turbine is 90 % b. Cooling water flow rate circulation in the condenser if the range is 7oC Ans. a. Power generated if the efficiency of the turbine is 90 % Input heat to turbine = 72000 x 3439.6 = 2.477 x 108 kJ/hr = 2.477 x 108 / 3600 = 68792 kW _________________________ Bureau of Energy Efficiency 6 Paper 4 Set A, Energy Auditor Key Output heat of different streams 1st extraction = 5000 x 2957.2 = 0.148 x 108 kJ/hr = 4107 kW 2nd t extraction = 50,000 x 2768.8 = 1.38 x 108 kJ/hr = 38456 kW Condenser input heat load = 17,000 x 2633.4 = 0.448 x 108 kJ/hr = 12436 kW Total heat leaving the turbine = 4107 + 38456 + 12436 = 54999 kW Heat available for power generation = 68792 54999 = 13793 kW Power generation at 0.9 turbine efficiency = 13793 x 0.9 = 12414 kW b. Cooling water flow rate circulation in the condenser if the range is 7oC Condenser heat load = 17,000 x (2633.4 191.8) = 2442 kJ/hr = 2442 x 4.18 = 10208000 kCal/hr At a range of 7oC cooling water flow rate = 10208000 / 7 = 1458 m3/hr N-4 A steam radiator is used for heating air with steam. Saturated steam enters the radiator at a temperature of 133oC. Air enters the radiator at 30oC and leaves at 85oC. The heat transfer area is 794 m2. The heat duty of the radiator is 14,50,000 kCal/hr. If the correction factor is 0.95 calculate the overall heat transfer coefficient in kW/m2 K. Ans. _________________________ Bureau of Energy Efficiency 7 Paper 4 Set A, Energy Auditor Key LMTD, counter flow [(133-30) (133-85)]/ln [(133-30)/133-85)] 72oC Corrected LMTD 72 x 0.95 68.4oC Heat duty, Q 14,50,000 kCal/hr 1683 kW Area Overall heat transfer coefficient 794 m2 Q/ A x corrected LMTD 1683 / (794 x 68.4) 0.031 kW/m2 K -------- End of Section - III --------- _________________________ Bureau of Energy Efficiency 8

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