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BEE Energy Sample / Model Paper 2015 : supplementary paper 3

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Paper 3 Set A SUPPLIMENTARY MODEL SOLUTIONS 16th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS September, 2015 PAPER 3: Energy Efficiency in Electrical Utilities Date: 20.09.2015 Marks: 150 Timings: 0930-1230 HRS Duration: 3 HRS Section I: OBJECTIVE TYPE Max. 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 Black Pen or HB pencil 1. A 100 kVAr capacitor rated for 415 Volts is connected across a load with 390 Volts normal supply. The effective kVAr will be a)100 b) 93.98 c) 88.31 d) none of the above 2. A 22 kW, 415 kV, 45A, 0.8 pf, 1475 rpm, 4 pole 3 phase induction motor operating at 420 V, 40 A and 0.8 pf. What will be the rated efficiency a) 85.0% b) 94.5% c) 89.9% d) 88.2% 1 Paper 3 Set A SUPPLIMENTARY 3. A 4 pole 50 Hz 3 phase induction motor has a full load slip of 5 %. The full load speed is: a) 750 b) 1425 c) 1500 d) 1475 4. A 50 hp motor with a full load efficiency rating of 90 percent was metered and found to be operating at 25 kW. The percent Motor Load is a) 75% b) 50% c) 60% d) 25% 5. A DG set consumes 70 litres per hour of diesel oil. If the specific fuel consumption of this DG set is 0.33 litres/ kWh at that load, then what is the kVA loading of the set at 0.8 PF? a) 212 kVA b) 265 kVA c) 170 kVA d) none of the above 6. A hotel building has four floors each of 1000 m2 area. If the interior lighting power allowance for the hotel building is 43000W, the Lighting Power Density (LPD) is: a) 10.75 b) 0.09 c) 43 d) data insufficient 7. A spark ignition engine is used for firing which type of fuels: a) high speed diesel furnace oil b) light diesel oil c) natural gas d) 8. Calculate the density of air at 11400 mmWC absolute pressure and 650C. (Molecular weight of air: 28.92 kg/kg mole and Gas constant:847.84 mmWC m3/kg mole K) a) 1.2 kg/m3 above b) 1.5 kg/m3 c) 1.15 kg/m3 d) none of the 9. FAD of a compressor means discharge at a) standard temperature conditions at suction conditions at suction c) standard temperature conditions at delivery conditions at delivery b) actual temperature d) actual temperature 10. Flow control by damper operation in fan system will a) increase energy consumption b) reduce energy 2 Paper 3 Set A SUPPLIMENTARY consumption c) reduce system resistance d) none of the above 11. Friction losses in a pumping system is a) inversely proportional to flow of flow c) proportional to square of flow flow b) inversely proportional to cube d) inversely proportional square of 12. Higher COP can be achieved with_____. a) lower evaporator temperature and higher condenser temperature b) higher evaporator temperature and Lower condenser temperature c) higher evaporator temperature and higher condenser temperature d) lower evaporator temperature and Lower condenser temperature 13. Humidification involves a) reducing wet bulb temperature and specific humidity b) reducing dry bulb temperature and specific humidity c) increasing wet bulb temperature and decreasing specific humidity d) reducing dry bulb temperature and increasing specific humidity 14. If EER of One Ton Split AC is 3.5, what is its power rating? a) 1.0 k W b) 1.5 kW c) 0.8 kW d) None of the above 15. If the COP of a vapour compression system is 3.5 and the motor draws power of 10.8 kW at 90% motor efficiency, the cooling effect of vapour compression system will be: a) 34 kW above b) 37.8 kW c) 0.36 kW d) none of the 16. If the observed temperature in air receiver is higher than ambient air temperature the correction factor for free air delivery will be: a) less than one b) greater than one c) equal to one d)equal to zero 3 Paper 3 Set A SUPPLIMENTARY 17. If the speed of a reciprocating pump is reduced by 50 %, the head a) is reduced by 50% c) remains same b) is reduced by 12.5% d) none of the above 18. Ignitors are used for starting a) FTL above b) CFL c) Metal Halide d) none of the 19. Illuminance of a surface is expressed in a) radians b) lux c) lumens d) LPD 20. In a large compressed air system, about 70% to 80% of moisture in the compressed air is removed at the a) air dryer b) after cooler c) air receiver d) inter cooler 21. In a no load test of a poly-phase induction motor, the measured power by the wattmeter consists of: a) core loss c) core loss, windage & friction loss windage & friction loss b) copper loss d) stator copper loss, iron loss, 22. In a rolling mill, the loading on transformer was 1200 kVA with the power factor of 0.86. The plant improved the power factor to 0.98 by adding capacitors. What is the reduction in kVA a) 1053 b) 147 c) 24 d) nil 23. In three phase transformer, the load current is 139.1A, and secondary voltage is 415V. The rating of the transformer would be ___________. a) 50 kVA b) 57.72 kVA c) 100 kVA d) 173 kVA 24. Installation of Variable frequency drives (VFD) allows the motor to be operated with 4 Paper 3 Set A SUPPLIMENTARY a) constant current d)none of the above b) lower start-up current c) higher voltage 25. Installing larger diameter pipe in pumping system results in a) increase in static head c) increase in frictional head b) decrease in static head d) decrease in frictional head 26. L / G ratio in cooling tower is the ratio of a) length and girth b) length and gradient of temperature c) Water mass flow rate and air mass flow rate d) Water volume flow rate and air volume flow rate 27. Lower power factor of a DG set demands a) lower excitation currents c) higher excitation currents b) no change in excitation currents d) none of the above 28. Occupancy linked Control can be achieved using a) ultrasonic sensors b) acoustic sensors c) infrared sensors d) all the above 29. Power factor Improvement will result in a) reduction in active power c) reduction in reactive power b) reduction in active current d) all the above 30. Power factor is highest in case of a) sodium vapour lamps incandescent lamps b) mercury vapour lamps c) fluorescent lamps d) 31. Pumps of different sizes can run in parallel provided their .. are similar a) static head d) none of these b) discharge head c) closed valve heads 32. Slip power recovery system is applicable in case of a) squirrel cage induction motor. b) wound rotor motor c) synchronous motor d) DC shunt motor 5 Paper 3 Set A SUPPLIMENTARY 33. Speed of the rotor of an induction motor is always a) greater than its synchronous speed c) equal to its synchronous speed speed b) less than its synchronous speed d)integer multiple of its synchronous 34. Star delta starter of an induction motor a) reduces voltage by inserting resistance in rotor circuit b) reduces voltage by inserting resistance in stator circuit c) reduces voltage through a transformer d) reduces the supply voltage due to change in connection configuration 35. The percentage imbalance when line-line voltages are 415 V, 418 V and 408 V is a) 1.047% b) 0.32% c) 1.44% d) none of the above 36. The actual measured load of 1000 k VA transformer is 400 k VA . Find out the total transformer loss corresponding to this load if no load loss is 1500 Watts and full load Copper Loss is 12,000 Watts a) 1920 watts watts b) 1500 watts c) 3420 watt d) 13500 37. The adsorption material used in an adsorption air dryer for compressed air is a) calcium chloride chloride b) magnesium chloride c) activated alumina d) potassium 38. The capacitor size in kVAr required to improve power factor from 0.90 to unity for 900 kW Load will be a) 720 kVAr above b) 436 kVAr c) 485 kVAr d) none of the 39. The cooling tower size is _____________ with the entering WBT when heat load, range and approach are constant. a) directly proportional none of above b) inversely proportional c) constant 6 d) Paper 3 Set A SUPPLIMENTARY 40. The fan system resistance is predominately due to a) more bends used in the duct c) volume of air handled b) more equipments in the system d) density of air 41. The illuminance of a lamp at one meter distance is 10 Lm/m2. What will be the corresponding value at 0.7 meter distance a)14.28 b) 20.41 c) 10 d) None of these 42. The indicator of cooling tower performance is best assessed by a) wet bulb temperature approach b) dry bulb temperature c) range d) 43. The inexpensive way to improving energy efficiency of a motor which operates consistently at below 40% of rated capacity is by _____. a) operating in star mode c) operating in delta mode b) replacing with correct sized motor d) none 44. The nomenclature T2,T5,T8 and T12 for fluorescent lamps are categorized based on a) diameter of the tube b) length of the tube c) both diameter and length of the tube d) power consumption 45. The percentage reduction in distribution loses when tail end power factor raised from 0.85 to 0.95 is a) 10.1% b)19.9% c)71% d)84% 46. The unit of specific humidity of air is: a) grams moisture/kg of dry air c) grams moisture/kg of air b) moisture percentage in air d) percentage 47. Time of the Day metering (TOD) is a way to a) reduce the peak demand of the distribution company b) increase the revenue of the distribution company 7 Paper 3 Set A SUPPLIMENTARY c) increase the peak demand demand in a industry d) increase the maximum 48. Trivector meter measures three vectors representing a) active, reactive and maximum demand apparent power c) active, harmonics and maximum demand power b) active, power factor and d)active, reactive and apparent 49. What is the function of drift eliminators in cooling towers a) maximize water and air contact b) capture water droplets escaping with air stream c) enables entry of air to the cooling tower d) Eliminates uneven distribution of water into the cooling tower 50. Which loss in a distribution transformer is dominating; if the transformer is loaded at 68% of its rated capacity a) core loss b) copper loss c) hysteresis loss d)magnetic field loss 8 Paper 3 Set A SUPPLIMENTARY Section II: SHORT DESCRIPTIVE QUESTIONS x 5 = 40 Marks: 8 (i) Answer all Eight questions (ii) Each question carries Five marks S-1 The total system resistance of a piping loop is 50 meters and the static head is 15 meters at designed water flow. Calculate the system resistance offered at 75%, 50% and 25% of water flow Solution: Total System Resistance of piping loop: 50m Static Head :15 m So, Dynamic Head at designed water flow: 35m Sl. No. Flow % 1 2 3 75.0% 50.0% 25.0% Dynamic Head (m) = 35 x (%flow)2 19.68 8.75 2.19 Static Head Total (m) Resistance (m) 15 15 15 34.68 23.75 17.19 S-2 The input power to a fan is 30kW for a 2500 Nm3/hr fluid flow. The fan pulley diameter is 300mm. If the flow to be reduced by 15% by changing the fan pulley, what should be the diameter of fan pulley and power input to fan. 9 Paper 3 Set A SUPPLIMENTARY Solution: 1 Input power to fan kW 2 Fluid flow Nm3/hr 3 Diameter of Fan pulley (mm) 4 Governing Equation is N1D1=N2D2 5 N2 = 0.85N1 From Eqn-1 D2=(D1)*(N1/N2); = 6 (300)*(N1/0.85N1) (KW1/KW2) = (N13/N23); Hence, KW2=(N2/N1)3*(KW1 ) =(0.85 7 N1/N1)3*(30) So Power requirement for fan will be 18.4 kW. Fan pulley to be changed to 352 mm diameter. 30 2500 300 Eqn-1 given 352mm 18.42KW S-3 List five energy saving measures for air conditioning system. Solution: Insulate all cold lines / vessels using economic insulation thickness to minimize heat gains. Optimize air conditioning volumes by measures such as use of false ceiling and segregation of critical areas for air conditioning by air curtains. Minimize the air conditioning loads by measures such as roof cooling, roof painting, efficient lighting, pre-cooling of fresh air by air- to-air heat exchangers Variable volume air system Optimal thermo-static setting of temperature of air conditioned spaces Sun film application Minimize the process heat loads by measures involving TR reduction and 10 Paper 3 Set A SUPPLIMENTARY refrigeration temperature level reduction Flow optimization Heat transfer area increase to accept higher temperature coolant Avoid wastages by heat gains Avoid wastages by loss of chilled water Avoid wastages by ideal flows. Frequent cleaning and descaling of all heat exchangers At refrigeration and AC plant area Minimize part load operations by matching loads and plant capacity on line; adopt variable speed drives for varying load. Ensure regular maintenance of all AC plant component as per manufacturers guide lines. Ensure adequate quantity of chilled water flows, cooling water flows Avoid by pass flow by closing valves of ideal equipments Adopt VAR system where economics permit as non CFC solutions Make efforts to continuously optimize condenser and evaporator parameters for minimizing specific energy consumption and maximizing capacity S-4 In a DG set, the generator is rated at 1000 kVA, 415V, 1390 A, 0.8 PF, 1500 RPM. The full load specific energy consumption of this DG set as measured by the energy auditor is 4.0 kWh per liter of fuel and air drawn by the DG set is 14 kg/kg of fuel. The energy auditor has recommended a waste heat recovery (WHR) system. Also the auditor indicated that the waste heat recovery potential is 2.6x105 kCal/hr at the existing engine exhaust gas temperature of 583oC. Estimate the exhaust temperature to chimney after installation of proposed WHR system. The specific gravity of fuel oil is 0.86 and specific heat of flue gas is 0.25 kCal/kg 0C. Solution: 1 Rated kVA of Diesel Generator (given) 1000 11 Paper 3 Set A SUPPLIMENTARY 2 3 4 5 Rated kW @ 0.8 pf (1*0.8pf) Specific fuel consumption (kwh/lts) (given) Specific gravity of fuel oil (given) Oil consumption at full load in kg/hr((2*4)/3) 6 Air supplied per kg of fuel (kg) (given) 7 Mass of flue gas (14+1) per kg of fuel 8 Mass of flue gas kg per hour (7*5) 9 waste heat recovery potential kCal/hr (given) 10 Delta T across waste heat recovery system (Heat kCal/hr)/(mass of flue gas kg/hr*specific heat, kcal/kg0C) (260000/2580x0.25) 11 Present Flue gas temp. or temp. before waste heat recovery system (given) 12 Exit flue gas temp. after waste heat recovery system(583 delta T) 800 4 0.86 172 14 15 2580 260000 403 583 180 S-5 The following single line diagram depicts the location of a 100 kW heater load and a 200 kW motor (which is 200 metres away from the 415V, LT bus). The main incoming line power factor of the system is 0.85 lag. Calculate the rating of capacitors to improve PF of main incoming line to 0.9 lag. Ans: 12 Paper 3 Set A SUPPLIMENTARY Total Inductive load requiring PF compensation=200kW (since the other 100 kW is a resistive load) Operating PF cos 1= 0.85 lag. Desired PF cos 2= 0.90lag -1 -1 kVAr required=kW((tan(cos 1)-tan(cos 2)) -1 -1 = 200(tan (cos 0.85)-tan (cos 0.90)) = 200( tan(31.78)-tan(25.84)) = 200(0.619-0.484) = 200(0.135) = 27 kVAr S-6 Define Lux and Luminous efficacy Ans Lux (lx) is the illuminance produced by a luminous flux of one lumen, uniformly distributed over a surface area of one square meter. It is also defined as the International System unit of illumination, equal to one lumen per square meter. Luminous efficacy is defined as the ratio of luminous flux emitted by a lamp to the power consumed by the lamp. Efficacy is energy efficiency of conversion from electricity to light form. S-7 State five possible methods of electrical demand management in a plant electrical system to minimize maximum demand. Ans i. Rescheduling operation time period of loads 13 Paper 3 Set A SUPPLIMENTARY ii. Storage of products, in process material or utilities such as refrigeration iii. Shedding of non essential loads iv. Reactive Power Compensation v. Operation of Captive Power Generator S-8 Match the following with respect to energy efficient compressed air system. A1 After Cooler As cool as possible for maximum energy A2 efficiency A3 Refrigerant Drier A4 Air receiver Compressed air line A5 drain traps Lower the dew point of the B1 compressed air discharge Locate at the bottom of lines at B2 required intervals Remove heat of compression B3 from last stage of compression B4 Compressor Air Intake Dampen Flow pulsations and helps compressed air peak B5 demand Ans A1-B3; A2-B4; A3-B1; A4-B5; A5-B2 .. . End of Section - II . . 14 Paper 3 Set A SUPPLIMENTARY Section III: LONG DESCRIPTIVE QUESTIONS 60 Marks: 6 x 10 = (i) Answer all Six questions (ii) Each question carries Ten marks L-1 a) Calculate the ventilation rate for an engine room of 20m length, 10.5m width and 15m height; if the recommended Air Changes per Hour (ACH) is 20. b) Air at 25,200 m3/hr and at 1.2 kg/m3 density is flowing into an air handling unit of an inspection room. The enthalpy difference between the inlet and outlet air is 2.38 kcal/kg. If the motor draws 22 kW with an efficiency of 90%, find out the kW/TR of the refrigeration system. (1 cal = 4.18 J). Solution: a) Ventilation Rate: Room Length (m) Room Height (m) Room Width (m) Air Changes per Hr (ACH) ventilation rate (m3/Hr) = Length (m) * Height (m) * Width (m) * ACH 20 15 10.5 20 63000 b) Refrigeration tonnes Q x x (h2-h1) 25200 x 1.2 x (2.38) kcal/kg 71,971 kcal/hr TR 71,971 /3024 23.8 TR Power input to the compressor 22 x 0.9= 19.8 kW 15 Paper 3 Set A SUPPLIMENTARY kW/TR L-2 19.8/23.8 = 0.83 In a diary plant 3 numbers of cooling water pumps, identical in characteristics are installed in parallel to supply cooling. During normal operation two of the pumps are operational while one pump is on standby. All pump combinations develop a discharge pressure of 3.4 kg/cm2 (a). The installed water flow meter at the common header during an energy audit reads the following: Operating Pump No. Pump No 1 & 2 Pump No 2 & 3 Pump No 3 & 1 Flow Rate (m3/hr) 545 535 550 The power drawn by motors of cooling water pump 1, 2 & 3 are 33 kW, 31.5 kW & 32.5 kW respectively. While the operating motor efficiency for pump no. 1 & 2 is 92% the motor efficiency for pump no. 3 is 91.5%. If the water level in suction of all pumps is 3 meter below pump central line calculate the following: i) Individual pump efficiencies ii) Specific energy consumption (kWh/m3) iii) Which is the best operating pump combination Solution: Let flow of pump 1,2 &3 be X, Y and Z respectively. From given: X + Y = 545 ---------(1) Y + Z = 535 ---------(2) X + Z = 550 ---------(3) Subtracting eqn (2) from eqn (1): 16 Paper 3 Set A SUPPLIMENTARY X Z = 10 --------(4) Adding eqn (3) and eqn (4): 2X = 560 X= 280 Putting X value in eqn (1) and (2): Y = 265 and Z = 270 Therefore, individual pump flow rates are: 280 m3/hr, 265 m3/hr and 270 m3/hr respectively. Pump Ref: A) Flow Rate (M3/hr) (calculated) B) Discharge Head (m) =3.4 kg/cm2 (a) = 2.4 kg/cm2 (g) =24 m (given) C) Suction Head (m) (g) (given) D)Total Head (Discharge Head - Suction Head)* (B-C) E) Liquid kW [flow (m2/s)*total head (m)*density (1000 kg./m3) * 9.81 (m/s2)/1000] F) Power Drawn by motor kW (given) G) Motor eff. % (given) H) Pump input power kW (FxG) I) Pump eff. % (E/H) J) Specific Energy Consumption (kWh/M3) (F/A) 1 280 2 265 3 270 24 -3 24 -3 24 -3 27 27 27 20.60 20.22 19.87 33 31.5 32.5 92.0% 92.0% 91.5% 30.36 28.98 29.74 67.9% 69.8% 66.8% 0.118 0.119 0.120 Pump No. 1 & 2 are the best performing operating combination. Note: *The total head has been calculated subtracting Discharge Gauge pressure from suction gauge pressure. The candidates can also calculate total head as difference of absolute pressures as follows: 17 Paper 3 Set A SUPPLIMENTARY Discharge Head= 3.4 kg/cm2 (a) Suction Head = 1- 0.3 kg/cm2 = 0.7 kg/cm2 Total Head Developed = 3.4 0.7 = 2.7 kg/cm2 = 27 m L-3 a) In a chemical industry, cooling water of 9000 m3/hr and 6000 m3/hr from two independent heat exchangers with temperature of 410C and 520C respectively are fed to one cooling tower after proper mixing at top basin. If measured heat rejection by the cooling tower is 45,000TR, calculate effectiveness and evaporation loss of the cooling tower at 310C WBT. b) In an air conditioning duct 0.5 m x 0.5 m, the average velocity of air measured by vane anemometer is 28 m/s. The static pressure at suction of the fan is -20 mmWC and at the discharge is 30 mmWC. The three phase induction motor draws 10.8 A at 415 V with a power factor of 0.9. Find out efficiency of the fan if motor efficiency = 88% (neglect density correction) Solution: a) Sl. No. 1 2 3 4 5 6 7 Particulars Flow Rate (m3/hr) (given) 0 Temp. C (given) 3 Mix. Flow Rate (m /Hr) (Sl.1 +2) Mix. Hot Water Temp. 0C [(Flow1 * Temp. 1) + (Flow 2 *Temp. 2)]/ (Flow1 + Flow 2) Heat Rejection (TR) (given) Range of Cooling Tower 0C: (Heat Rejection TR * 3024) / (Flow M3/hr * 1000) WBT 0C (given) Stream Stream 1 2 9000 6000 41 52 15000 45.4 45000 9.072 31 18 Paper 3 Set A SUPPLIMENTARY Cold Water Temp. 0C (Mix. Hot Water Temp. Range) Approach 0C (Cold Water Temp. WBT) Effectiveness (Range/ (Range + Approach)) Evaporation Loss (m3/hr) = 0.00085*1.8*Mix. Flow m3/hr*Range 8 9 10 11 36.328 5.328 63 208.2 b) 1 2 3 4 5 6 7 8 9 L-4 Area of the Duct: (0.5*0.5) m2 Avg. velocity (m/s) (given) 3 Air Flow (m /s) (Sl. 1* Sl. 2) Suction static Pr. (mmWC) (given) Discharge Static Pr. (mmWC) (given) Power drawn by the motor (kW): (1.732 * 415*10.8*0.8/1000) Air Power kW: = Flow (m3/s)* (Dis. Pr Suc. Pr.) mmWC /102 Power to fan Shaft kW (Motor Drawn power * Motor eff. Of 88%) Fan Static Eff. (%) = Air Power*100%/Shaft Input 0.25 28 7 -20 30 6.99 3.43 6.15 55.76 One of the process industries has installed 18 MW cogeneration plant. The Cogeneration plant maximum condenser load is 7 MW and the extraction steam of 57 TPH is used for process and also for vapour absorption machine. The condenser heat load is 550 Kcal/kg of steam and the steam rate is 5 kg/KW for condenser power. The heat load of VAM in 127 Kcal/min/TR and the capacity of VAM is 1100 TR. Estimate cooling tower heat load in Kcal/hr. If the tower is designed for 6oC range, 19 SUPPLIMENTARY Paper 3 Set A calculate the water flow in cooling tower. The design approach temperature of the CT is 5oC. Ans. Condenser load = 7 MW Steam rate for condenser = 5 kg/KW Total steam required for condenser power = 7000 X 5 = 35000 Kg/hr. Condenser heat load = 35000 x 550 = 19250000 Kcal/hr. Heat load of VAM = 1100 x 127 x 60 = 8382000 Kcal/hr. Total heat load = 19250000 + 8382000 = 27632000 Kcal/hr. Range of tower = 6 deg C Cooling water Flow required=27632000/6=4605333 lts or 4605 m3. L-5 a) List the design improvements that have been incorporated in Energy Efficient motors to increase operating efficiency over standard motors? b) The power input to a three phase induction motor is 45 kW. If the induction motor is operating at a slip of 1% and with total stator losses of 1.80 kW, find the total mechanical power developed Solution: a) Following design modifications have been done in EEMs : a. Use of lower loss silicon steel b. A longer core c. Thicker wire d. Thinner lamination 20 Paper 3 Set A SUPPLIMENTARY e. f. g. h. Smaller air gap between rotor and stator Use of copper Superior bearings Smaller fan b) Stator input: 45kW Stator losses: 1.80 kW Stator output: 45-1.80= 43.2kW OR Rotor Input= 43.2 kW Slip= 1% Mechanical Power Output = ( 1-S) x Rotor Input = (0.99 x 43.2) = 42.77 kW L-6 Compare the performance of centrifugal chiller with vapour absorption chiller using the data given below: Sl. No. 1 2 3 4 5 6 7 8 9 Parameter Chilled water flow (m3/h) Condenser water flow (m3/h) Chiller inlet water temperature (0C) Condenser water inlet temperature (0C) Chiller outlet water temperature (0C) Condenser water outlet temperature (0C) Chilled water pump consumption (kW) Condenser water pump consumption (kW) Cooling tower fan consumption (kW) Centrifugal Chiller 192 245 13 VAM 183 360 14.5 28 32 7.8 9.2 36.2 40.7 32 31 38 52 9 22 If the compressor of centrifugal chiller consumes 205 kW, the steam consumption for VAM is 1620 kg/Hr. calculate the following: 21 SUPPLIMENTARY Paper 3 Set A i) Refrigeration load delivered (TR) for both systems? ii) Condenser Heat load (TR) for both systems? iii) Compare auxiliary power consumption for both systems, give reason? iv) If electricity cost is Rs.4.0/kWh and steam cost is Rs.0.45/kg compare the operating cost for both systems. 22 SUPPLIMENTARY Paper 3 Set A Solution: Compression Chiller vs VAM Sl. Centrifug Parameter VAM No. al Chiller 1 Refrigeration load delivered (TR) = Mass of Chilled water flow x 330.16 320.73 Specific heat * Delta T of Chilled water 3 3 Condenser delivered = Sl. No.1 mheat /hr *load 1000kg/m *1 0 2 (TR) kcal/kg C * (Sl. No. 3. Sl. No. 1035.71 = Mass 5) of condenser water flow x 664.35 Specific heat * Delta T of condenser water 3 Auxiliary Power Consumption 79 105 = Sl. No.2 m3/hr * 1000kg/m3 *1 (kW) = 0 kcal/kg C * (Sl. No. 6 Sl. No. in case of VAM system 4 The auxiliary (Sl. No. 7 + Sl. power No. 8 +consumption Sl. No. 9) is 4) higher because heat rejection in VAM condenser is 5 6 comparatively higher than centrifugal chiller with Total Energy Consumption: 284 kW Auxiliary approximate similar cooling load. Power of (Auxiliary 105 kW Power of and Steam 79kW consumpti and on of 1620 Chiller kg/hr consump Operating Energy Cost per Hour Rs. Rs 1149/tion of of Operation 1136/205 kW) (105 * 4 = Rs. 420/plus 1620 (284 * 4 = * 0.45 = Rs. Rs. 729/-) -------- End of Section - III1136/-) ----- 23

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