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BEE Energy Solved Question Paper 2020 : Energy : energy

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Paper 3 Code : Pink 20th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS September, 2019 PAPER 3: ENERGY EFFICIENCY IN ELECTRICAL UTILITIES Section I: OBJECTIVE TYPE Marks: 50 x 1 = 50 i) Answer all 50 questions ii) Each question carries one mark iii) Please hatch the appropriate oval in the OMR answer sheet with HB pencil only, as per instructions 1. In a pumping system, if then___________________. the temperature of the liquid handled increases, a) NPSHa increases b) NPSHa decreases c) NPSHa remains constant d) NPSHa and NPSHr are independent of temperature 2. Which of the following component has maximum effect on cooling tower performance? a) Fill media 3. b) drift c) louvers d) casing In a vapour compression refrigeration system, the quantum of energy transferred at condenser is more than the energy transferred at ___________. a) Compressor b) Expansion Valve c) Evaporator d) All of the above 4. Demand side Management helps ____________ a) to reduce the energy losses c) to promote energy efficiency among users. 5. b) to reduce system peak demand d) All of the above Which one of the following is true to estimate the range of cooling tower? a) Range = Cooling water inlet temperature Wet bulb temperature b) Range = Cooling water outlet temperature Wet bulb temperature c) Range = Wet Bulb Temperature Cooling Water Inlet Temperature d) None of the above 6. Modest flow variation between 80% to 100%, in a centrifugal fan is achieved more efficiently with ___________________. a) Inlet damper b) Outlet damper c) Inlet guide vanes d) Impeller Change 7. ____________is used as refrigerant both in vapour compression and vapour absorption systems a) Lithium Bromide _______________________ Bureau of Energy Efficiency b) Water c) HFC 134A d) Ammonia 1 Paper 3 Code : Pink 8. In electrical distribution ________________. a) Meter Reading system, commercial b) Metering loss covers discrepancies c) Collection Efficiency due d) All of the above 9. Which of the following parameters is not required for evaluating volumetric efficiency of reciprocating air compressor? a) Power input b) FAD c) Cylinder Stroke d) Cylinder bore 10. _____________ is not used for speed control. a) b) c) d) Variable Frequency drive Soft starter Hydraulic coupling Eddy current drives 11. When compared to standard motors, energy efficient motors will have ____________. a) Higher slip b) Higher starting torque c) Lower No load current d) All the above 12. For a given air requirement, providing higher volume air receiver will ________________. a) b) c) d) 13. Increase energy consumption Reduce energy consumption Reduce Unload Power Reduce Pressure fluctuations Harmonics generation will be more in _________________ a. Inverter drives 14. b. LED Lamps c. Transformers d. Resistance heaters Thermal Power Plant efficiency is low due to ____________________. a) b) c) d) 15. to Higher steam Pressure Higher superheat temperature Low GCV coal Higher Heat loss in condenser Among the following, _____________ has highest design efficiency. a) High tension motors b) Power transformers c) Alternators d) Electric melting furnaces 16. The difference between wet bulb temperature and cooling water inlet temperature in a cooling tower is called _______________. a)Approach 17. b) Range d) None of the above Technical loss in a distribution system can be reduced by _________________. a) Maintaining low HT/LT ratio c) High voltage supply to consumers 18. c) Effectiveness b) Accurate meter reading d) Improving Collection Efficiency Pressure drop can be reduced in a compressed air distribution line by providing ______________. _______________________ Bureau of Energy Efficiency 2 Paper 3 Code : Pink a) After Coolers b) Small diameter distribution pipes c) High pressure air flow d) Large Diameter Distribution pipes. 19. Power consumption is very high for ___________ type of compressed air dryers. a) Refrigeration type b) Blower reactivated type c) Heat of compression type d) Heatless purge type 20. A DC excitation is used to vary the speed of _____________. a) b) c) d) 21. Eddy Current Coupling fluid coupling variable frequency drive None of the above The isothermal power of 500 CFM air compressor is 72 kW and the efficiency is 76 %. The actual power drawn by the compressor will be ________ a) 56 kW 22. b) 94.7 kW b) Increase input power to the motor d) None of the above A 500-kVA transformer is designed for No load loss of 750 watts and load loss of 5700 Watts. The calculated total transformer loss is 1662 watts. What will be the percentage loading of the transformer? a) 54.8 % 24. d) 72 kW Power factor improvement of a 75-kW compressor motor will ___________ a) Reduce input power to the motor c) Reduce the compressor motor shaft power 23. c) 89 kW b) 29 % c) 40 % d) 25.7 % Rating of PF correction capacitors for Induction Motors terminal should be a) 100 % kVAr of the induction motor b) 20 % of Motor Rating c) 25 % of Motor rating d) 90 % of the no-load kVAr induction motor 25. LLF in lighting calculation refers to a) b) c) d) 26. Light Load factor Light lumen factor Light Lux factor Light loss factor A medium voltage end consumer receives 83 million units with a transmission and distribution cascade efficiency of 82%. The million units generated will be ____________. a) 101.2 27. b) 68.1 c) 83 d) None of the above A 1000 kW Gas engine is designed for 38 % efficiency. The operating load of the engine is 825 kW. If the GCV of gas is 8700 kcal/m 3, the hourly gas consumption will be ____________m3/hr a) 214.6 _______________________ Bureau of Energy Efficiency b) 260.13 c)188.89 d) 272.74 3 Paper 3 Code : Pink 28. In an electrical power system, transmission efficiency increases as _______________. a) b) c) d) 29. both voltage and power factor increases both voltage and power factor decrease voltage increases but power factor decreases Voltage decreases but power factor increases. Which of the following is expressed in terms of percentage? a) b) c) d) 30. Absolute humidity Relative humidity Specific Gravity All of the above Which among the following is one of the parameters used to classify fans, blowers & Compressors? a) b) c) d) 31. Volume flow rate Mass flow rate Specific ratio None of the above What is the function of drift eliminators in cooling towers? a) b) c) d) 32. Which of the following statements is not true regarding centrifugal pumps? a) b) c) d) 33. The CRI is expressed in a relative scale ranging from 0 -100. CRI indicates, how perceived colors match with actual colors. LED lamps are having comparatively higher CRI than Incandescent Lamps. The higher the color rendering index, the less color shift or distortion occurs Flow control with ___________in a fan system will not change the fan characteristic curve. a) b) c) d) 35. Flow is zero at shut off head Maximum efficiency will be at design rated flow of the pump Head decreases with increase in flow Power increases with throttling Which of the following is not true with respect to Color Rendering Index (CRI)? a) b) c) d) 34. maximize water and air contact capture water droplets escaping with air stream enables entry of air to the cooling tower eliminates uneven distribution of water into the cooling tower Inlet guide vane speed change with variable frequency drive speed change with hydraulic coupling discharge damper The primary purpose of inter-cooling in a multistage compressor is to _____________. a) remove the moisture in the air b) reduce the work of compression c) separate moisture and oil vapour _______________________ Bureau of Energy Efficiency 4 Paper 3 Code : Pink d) none of the above 36. Illuminance of a surface is expressed in _____________ a) radians 37. b) lux c) lumens d) LPD A pump discharge has to be reduced from 120 m 3/hr to 110 m3/hr by trimming the impeller. What should be the percentage reduction in impeller size? a)10.52 % b) 8.34% c) 9.7 1% d)17.1% 38. Which of the following power plants has the highest efficiency? a) Open cycle Gas Turbine b) Diesel Engine c) Combined cycle gas turbine d) Conventional coal plants 39. COP of a single effect absorption refrigeration system is likely to be in the range of____________ a) 0.6 to 0.7 40. d) 10 TR refrigeration capacity increases refrigeration capacity decreases specific power consumption remains same condenser load increases A 4 pole 50 Hz induction motor is running at 1470 rpm. What is the slip value? b) 0.02 c) 0.04 d) 0.4 The basic function of an air dryer in an air compressor is to a) b) c) d) 45. c) 1TR Reduce technical loss in distribution system Reduce commercial loss in distribution system Reduce capital investment Reduce energy bill for the end consumer a) 0.2 44. b) 3.024TR When evaporator temperature is reduced, ________ a) b) c) d) 43. d) 3.0 to 4.0 HVDS (High Voltage Distribution System) is preferred to___________________ a) b) c) d) 42. c) 1.5 to 2 If 30240 kcal of heat is removed from a room every hour then the refrigeration tonnage will be nearly equal to____________. a) 30.24TR 41. b) 1to 1.2 Prevent dust from entering the compressor Remove moisture before the intercooler Remove moisture in compressor suction Remove moisture in air supplied to the plants Power factor is highest in the case of_________________ a) Sodium vapour lamps b) Induction lamps c) LED Lamps d) Incandescent lamps _______________________ Bureau of Energy Efficiency 5 Paper 3 Code : Pink 46. If the COP of a vapour compression system is 3.5 and the motor draws a power of 10.8 kW at 90% motor efficiency, the cooling effect of vapour compression system will be_________. a) 34 kW 47. b) 42 kW b) 5.3 d ) 48 b)15.5% c)16.6% d)24.7% Energy performance index (EPI) kWh/m2 /yr is the ratio of total building annual energy consumption to __________ a) b) c) d) 50. c) 8 The percentage reduction in distribution losses when tail end power factor is raised from 0.8 to 0.95 is__________ a) 29.4% 49. d) 3.4 kW The blow down requirement in m3/hr of a cooling tower with evaporation rate of 16 m3/hr and CoC of 3 is ____________. a) 4 48. c) 2.8 kW Built up area Carpet area Roof Area Window and Wall Area Which of the following is not a climate zone as per ECBC classification? a) hot-dry b) warm-humid c) Cold-humid d) cold . . End of Section I . . _______________________ Bureau of Energy Efficiency 6 Paper 3 Code : Pink Section II: SHORT DESCRIPTIVE QUESTIONS (i) (ii) S-1 Marks: 8 x 5 = 40 Answer all Eight questions Each question carries Five marks One of the Machining centres has installed 2 No s of 270 cfm compressors for pneumatic operation and also for cleaning operation of components after machining. The compressors are operated at 7 kg/cm2(g) and are on-load for 80 % of the time. The load Power and the un-load Power of each 270 cfm compressor is, 40 kW and 15 kW respectively. The energy audit estimated that cleaning air requirement is 60% of the air generated. Calculate the daily energy consumption for cleaning air alone, assuming continuous operation of the compressor. Ans : Compressor capacity % Loading Air Delivered by 2 compressors Loading Power drawn by the compressors Un-Loading power drawn by the compressors = 270 cfm = 80 % = (270 X 0.80 x 2) = 432 cfm = (40 + 40) = 80 kW = (15 + 15) = 30 kW Average kW drawn by the compressors = [(80 x (0.8 x24))+ (30x (0.2 x 24))]/(24) = 70 kW SEC of compressor = (70/432) = 0.162 kW/cfm Cleaning air consumption at 7 Kg/cm2 = (60 % of generation) = (0.60 x 432) = 259 cfm Energy requirement for Cleaning air per day = (259 x 0.162 x 24) = 1007 kWh/day (or) Alternate Solution = (Load Power x load time) + (Unload Power x Unload time) = (40 x 0.8) + (15 x 0.2) = 32+3 = 35 KW Average KW drawn by the compressors = 35 x 2 = 70 KW Energy requirement for Cleaning air per day = (70 kW x 0.6) x 24 =1008 kWh/day S-2 In a pharmaceutical industry a centrifugal pump is pumping 80 m3/hr of water into a pressurized _______________________ Bureau of Energy Efficiency 7 Paper 3 Code : Pink container. The container pressure is 3 kg/cm2(g). The discharge head of the pump is 5 kg/cm 2(g) and water level is 5 meters below the pump central line. If the power drawn by the motor is 22 kW, find out the pump efficiency. Assume motor efficiency as 90% and the water density as 1000 kg/m3. S-2Sol Ans: Sl. No. S3 Parameter Process Value 1 Water Flow Rate (m3/hr) given 80 2 Discharge Head (meters) given 50 3 Suction Head (meter) given -5 4 Power input to Motor (kW) given 22 5 Motor Efficiency given 90% 6 7 Power Input to Pump (kW) Liquid kW 8 Pump Efficiency Sl. 4* Sl. 5 (Sl. 1/3600)*((Sl. 2*10) Sl. 3)*9.81 =22 x 0.9 = 19.8 = (80/3600) x (50 (-5) x 9.81=11.98 Sl. 7 / Sl. 6 60.56% A refrigeration system designed with 10 TR AHU is operating at 8.25 TR. The measured air parameters are given below: Inlet enthalpy Outlet enthalpy Specific volume of air = 10.26 kcal/kg = 7.26 kcal/kg. = 0.83 m3/kg Calculate the volume of air in m3/hr handled by AHU. Ans : Cooling delivered (TR = (Difference in enthalpy) x (Volume of air / sp. volume x 3024) = (Hi Ho) x V / (v x 3024) Volume of air handled by AHU = (TR x v x 3024 / (Hi Ho)) = ((8.25 x 0.83 x 3024) / (10.26-7.26)) = 6903 m3/hr S4 A fan is designed for 1300 m3/hr, 50 Hz and drawing 3 kW. If the fan is operated with VFD at 37 Hz for 6000 hours, calculate the velocity of air, when air is supplied through 150 mm diameter duct and the annual energy savings. Ans : Power Drawn at 50 HZ Operating frequency Flow at 37 Hz Diameter of the duct Area of the duct _______________________ Bureau of Energy Efficiency = 3 kW = 37 Hz = 1300 x (37 / 50) = 962 m3/hr = 150 mm = 0.0177 m2 8 Paper 3 Code : Pink Velocity of the air in the duct Power consumption with 37 Hz Annual Energy Savings for 6000 hours operation S5 = [(962 / 3600)] / [(0.0177)] = 15.09 m/s = (37/50)3 x 3 = 1.22 kW = 6000 x (3 -1.22) = 10,680 kWh A foundry unit draws power to the tune of 2500 kW. The demand observed during furnace operation is given below: 5 minutes 7 minutes 3 minutes : 2940 kVA : 2550 kVA : 2777 kVA If the billing meter is monitoring demand every 15 minutes, calculate the maximum demand registered and also the average PF, during the demand interval. Ans : Maximum demand registered PF 5 minutes: 2940 KVA 7 minutes 2550 KVA 3 minutes 2777 kVA. Average PF S6 = [ 2940 * (5/15) + 2550 * (7/15) + 2777 * (3/15)] = [ 980 + 1190 + 555.4] = 2725.4 kVA = (2500 / 2940) = 0.85 = (2500 / 2550) = 0.98 = (2500 / 2777) = 0.90 = [ 0.85 *(5/15) + 0.98* (7/15) + 0.9 * (3/15) ] = 0.92 A process plant has installed 4-cell cooling tower, with 45 kW CT fans for each cell and operating at 40 kW at 1450 rpm. As a part of the energy conservation program, the existing fan motors are replaced with two speed motors which would operate at 1450 rpm and 740 rpm. The cooling towers are operated at high speed mode for 5300 hours and at low speed mode for 1800 hours, in a year. Estimate the annual energy savings when compared to operation of fans continuously at a fixed speed of 1450 rpm. Ans : Present energy consumption of all 4 fans Energy consumption for fans at 1450 rpm for 5300 hours Energy consumption for fans at 740 rpm for 1800 hours Annual savings S7 Write short notes on any two of the following: = (4 x 40 x (5300 + 1800)) = 11,36,000 kWh = (4 x 40 x 5300) = 8,48,000 kWh = [ (740/1450)3x 40 x 4 x 1800] = 38281 kWh = [ 11,36,000 - (8,48,000+38,281) ] = 2,49,719 kWh ( Each 2.5 Marks) 1. Integrated Part Load Value (IPLV) for chillers 2. Evaporative Cooling _______________________ Bureau of Energy Efficiency 9 Paper 3 Code : Pink 3. Heat Pump Ans : 1. Integrated Part Load Value (IPLV) for chillers 2. Evaporative Cooling 3. Heat Pump S8 (Page No. 126) (Page No. 136) (Page No. 133) Write short notes on any two of the following: (Each 2.5 Marks) 1. Solar Heat Gain Coefficient (SHGC) 2. Visible Light Transmittance (VLT) 3. Cool Roof Ans : 1. Solar Heat Gain Coefficient (SHGC), 2. Visible Light Transmittance (VLT), 3. Cool Roof, (Page No. 272) (Page No. 272) (Page No. 271) .. . End of Section - II . . _______________________ Bureau of Energy Efficiency 10 Paper 3 Code : Pink Section III: LONG DESCRIPTIVE QUESTIONS (i) (ii) L1 Marks: 6 x 10 = 60 Answer all Six questions Each question carries Ten marks A. For each one of the following, mention whether they belong to Prescriptive Method or Whole Building Performance Method . ( 5 Marks) 1. Compliance by meeting or exceeding specific levels for each individual element of building 2. Allows Trade-off option for building envelope 3. Allows use of energy simulation software 4. Computer model of the proposed design (energy consumption) is compared with Standard Design 5. Compliance if energy use in proposed design is less than energy use in standard design B. Match the Following: ( 5 Marks) 1. 2. 3. Building envelope Passive solar design strategy Visual Light Transmittance 4. Weather stripping 5. Cool roof a) Day lighting of building b) Exfiltration and Infiltration of air c) Roof, walls, windows, skylights, doors and other openings d) Property of high solar reflectance and emittance e) Cross ventilation Ans : A. 1. Prescriptive Method 2. Prescriptive Method 3. Whole Building Performance Method 4. Whole Building Performance Method 5. Whole Building Performance Method B. 1 4 Passive solar design strategy Visual Light Transmittance Weather stripping 5 Cool roof 2 3 L2 Building envelope C Roof, walls, windows, skylights, doors and other openings E Cross-ventilation A Day lighting of building B Exfiltration and Infiltration of air D Property of high solar reflectance and emittance An energy audit was conducted in a large machine shop and the audit report suggested replacing 30 machine motors with energy efficient motors. The loading details of old and new motors are given below: _______________________ Bureau of Energy Efficiency 11 Paper 3 Code : Pink Motor Rating in kW 7.5 11.5 15 Operating Load % 75 85 70 Old Motor Efficiency% New Motor efficiency% No of motors 86 88 89 89 91 92 12 7 11 Assuming motor loading in both cases remains same, calculate the annual energy savings, for 4000 hours operation per year. Ans : Motor Rating in KW Operating Load % 7.5 75 11.5 85 15 70 Actual Old Motor Load In kW 7.5/0.86=8.72 =8.72x 0.75=6.54 11.5/0.88=13.07 =13.07 x 0.85= 11.11 15/0.89=16.85 =16.85x 0.7= 11.79 Actual New Motor Load In kw 7.5/0.89=8.43 =8.43x 0.75= 6.32 11.5/0.91=12.64 =12.64 x 0.85= 10.74 15/0.92=16.30 =16.30 x 0.7 11.41 Annual Savings for 7.5 KW Motors, 12 numbers, operating 4000 hours Annual Savings for 11 KW Motors, 7 numbers, operating 4000 hours Annual Savings for 15 KW Motors, 11 numbers operating 4000 hours Total annual savings for 30 high efficiency motors L3 Old Motor efficiency New Motor efficiency No of motors 86 89 12 88 91 7 89 92 11 = [ 4,000 (6.54-6.32) x 12 ] = 10,560 kWh = [ 4000 (11.11 -10.74) x 7 ] = 10,360 kWh = [ 4,000 (11.79-11.41) x11 ] = 16,720 kWh = 37,640 kWh A 10 MW co-generation plant is operating at a daily load factor of 85 %. Power is generated at 11 KV. 35 % of the power generated, is exported to grid, through a 7.5 MVA Transformer with 99 % efficiency. 32 % power generated, is supplied to mill motors, at 600 Volts, through a 5 MVA step down transformer, with 98 % efficiency. The balance power generated is supplied to other LT Loads and auxiliaries, at 415 Volts, through a 2 MVA transformer, with 98 % efficiency. _______________________ Bureau of Energy Efficiency 12 Paper 3 Code : Pink Calculate the following: 1) Daily energy exported to grid at 33 KV. 2) Daily mill motors consumption at 600 V. 3) Daily LT loads and auxiliary consumption at 415 V. 4) Daily transformers losses in kWh and % transformers losses (Each 2.5 Marks) Ans : 1. Daily generation Daily energy generation for export purpose = (10,000 x 0.85 x 24) = 2,04,000 kWh = (2,04,000 x 0.35) = 71,400 KWh 7.5 MVA transformer loss = [ 71,400 - (71,400 x 0.99) ] = (71,400 70,686) = 714 kWh Net energy export to the Grid at 33 KV level = (71,400 kWh - 714 kWh) =70,686 KWh 2. Daily energy generation for mill motor consumption 5 MVA Transformer loss Net mill Consumption = (2,04,000 x 0.32) = 65,280 kWh = [ 65,280 - (65,280 x 0.98 ) ] = (65,280 - 63,974.4) = 1,306 kWh = 63,974 KWh 3. Daily generation for LT loads & Auxiliary consumption 2MVA Transformer loss Net LT loads & Auxiliary Consumption = (2,04,000 x 0.33) = 67,320 kWh = [ 67320 - (67320 x 0.98) ] = 67,320 - 65,974 = 1,346 kWh = 65,974 kWh 4. Transformers losses % transformers losses = (714 +1306 +1346) = 3,366 kWh day = (3,366 / 2,04,000) x 100 = 1.65 % (Or) To meet the plant LT loads and co-gen auxiliary load, the transformer capacity should be more than 2 MVA. _______________________ Bureau of Energy Efficiency 13 Paper 3 Code : Pink L4 A small machine shop has installed 220 cfm screw compressor to meet air requirement for various operation. The operating details are given below: Shift reference (8 hrs/ Shift) I II III Load Power Un-load power Load time in sec Un-Load time in sec 60 45 25 10 25 45 = 37 KW = 11 KW Calculate the following: 1. Energy loss per day ( 4 Marks) 2. Shift wise average air requirement in cfm ( 2 Marks) 3. The plant has proposed to install a VFD for the compressor. Calculate the energy savings after installing the VFD operated compressor, if the VFD loss is 3 % of load power. ( 4 Marks) Ans : Ist shift consumption = ((60 / 70) x 37) + (10 / 70) x 11) x 8) = (31.71+1.57) x 8 = 266.24 kWh IInd shift consumption = ((0.64 x 37 + 0.36 x 11) x 8 ) = (23.68 + 3.96) x 8 = 221.12 kWh IIIrd shift consumption = ((0.36 x 37 + 0.64 x 11) x 8) = (13.32 +7.04) x 8) = 162.88 kWh Daily Total Energy consumption = (266.24 + 221.12 + 162.88) = 650.24 kWh Daily Energy loss due to unloading = (1.57 +3.96 +7.04) x 8 = 100.56 kWh Daily load cycle Energy consumption = (650.24 100.56) = 549.68 kWh Daily energy consumption with VFD = (549.68 / 0.97) = 566.68 kWh _______________________ Bureau of Energy Efficiency 14 Paper 3 Code : Pink Daily Energy loss due to VFD = (566.68 549.68) = 17 kWh Daily Net Energy savings with VFD compressor = (100.56 17) = 83.56 kWh Ist shift air requirement = (0.86 x 220) = 189.2 cfm IInd shift air requirement = (0.64 x 220) = 140.8 cfm IIIrd shift air requirement = (0.36 x 220) = 79.2 cfm L5 (a) What is L/G ratio and how it is useful in operation of a cooling tower ? (3 Marks) (b) What are the functions of fill media in a cooling tower? (3 Marks) (c) Calculate the L/G ratio for the cooling tower given the following: (4 Marks) Water Flow Approach Air entering enthalpy at 26.67 oC Air leaving enthalpy at 37.8 oC Hot water temperature Cold water temperature Ans : = 4540 m3/hour = 4.45 oC = 24.17 kcal/kg = 39.67 Kcal/kg = 47.77 oC = 31.11oC a) Page 205 (b) Page 209 c) L/G L (47.77 - 31.11) L / G Ratio _______________________ Bureau of Energy Efficiency = (h2 -h1) / (T1 - T2) = G (39.67 - 24.17) = (39.67 - 24.17)/ (47.77 - 31.11) = 0.93 15 Paper 3 Code : Pink L6 a) In an energy audit of a fan, it was observed that the fan was delivering 24,000 Nm 3/hr of air. Suction static pressure was recorded as -15 mm WC and discharge static pressure as 35 mmWC. The power measurement of the motor using power analyzer was recorded as 7 kW. The motor operating efficiency taken from motor performance curve was 90%. What is the static efficiency of the fan? b) Match the Following 1. Heat Pump NPSHR 2. Compressor Static Head 3. Pumping Pressure Static Pressure 4. Fan Compressor 5. Pump Free air delivery test Soln : a) Q = 24.000 Nm3 / hr. = 6.67 m3/sec Static pressure rise = 35 (-15) = 50 mmWC s =? Power input to motor = 7 kW Power input to fan shaft = 7 x 0.90 = 6.3 kW Fan static = Volume in m3/sec x Pst in mmWc 102 x Power input to shaft = (6.67 x 50) / (102 x 6.3) = 0.519 (or) = 51.9 % b) Match the Following 1. Heat Pump NPSHR (5) 2. Compressor Static Head (3) 3. Pumping Pressure Static Pressure (4) 4. 5. Fan Pump Compressor (1) Free air delivery test (2) -------- End of Section - III ----- _______________________ Bureau of Energy Efficiency 16

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