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17th National Certification Exam Energy Managers & Auditors SEPTEMBER 2016 Paper 4

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Regular set A Regn No: __________________ Name : __________________ (To be written by the candidate) 17th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS September, 2016 PAPER 4:Energy Performance Assessment for Equipment and Utility Systems Date: 25.09.2016 Timings: 14:00-16:00 HRS Duration: 2 HRS General instructions: o o o o o Please check that this question paper contains 6 printed pages Please check that this question paper contains 16 questions The question paper is divided into three sections All questions in all three sections are compulsory All parts of a question should be answered at one place Section - I: S-1 Ans S-2 BRIEF QUESTIONS Ans S-9 An air washer cools the water and a cooling tower cools the air. True or False. False. A 11 kW induction motor has an efficiency of 90% what will be its maximum delivered output? 11 kW. The COP of a vapour absorption refrigeration system is lower than the COP of a vapour compression refrigeration system-True /false. True. An industrial electrical system is operating at unity power factor. Addition of further capacitors will reduce the maximum demand (kVA). True or False. False. Which parameter in the proximate analysis of coal is an index of ease of ignition? Volatile matter. The major source of heat loss in a coal fired thermal power plant is through flue gas losses in the boiler. True or false. False. With evaporative cooling, it is possible to attain water temperatures below the atmospheric wet bulb temperature. True or False False A pump is retrofitted with a VFD and operated at full speed. Will the power consumption increase or decrease or remain the same? Increase De-aeration in boiler refers to removal of dissolved gases. True or false Ans True Ans S-3 Ans S-4 Ans S-5 Ans S-6 Ans S-7 Ans S-8 Regular set A S-10 In a compressed air system, the function of the after cooler is to reduce the work of compression. True or False Ans False . End of Section - I . Section - II: L-1 Ans SHORT NUMERICAL QUESTIONS In a petrochemical industry the LP & HP boilers have the same evaporation ratio of 14 using the same fuel oil. The operating details of LP & HP boiler are given below: Particulars LP Boiler HP Boiler Pressure 10 Kg./cm2a 32 Kg./cm2a Temperature Saturated Steam 400oC Enthalpy of steam 665 Kcal/kg 732 Kcal/kg Enthalpy of feed water 80oC 105oC Evaporation Ratio 14 14 Find out the efficiency of HP boiler if the LP boiler efficiency is 80%. Effy = ER. (hg hf) / GCV EffyL.P 1 = 0.8 = 14 x(665 80) / GCV EffyH.P 2 = 14 x(732 105) / GCV EffyH.P 2 / EffyL.P 1 = (732 105)0.8 / (665 80) = 0. 8574 =85.74% Or EffyL.P 1= 0.8 = 14 x(665 80) / GCV GCV= 14x(665-80) / 0.8 = 10237.5kcal/kg EffyH.P 2 = 14 x(732 105) / GCV = 14 x(732 105) / 10237.5 = 0.8574 = 85.74% L-2 While carrying out an energy audit of a pumping system, the treated water flow (in open channel) was measured by the tracer method. 20% salt solution was used as the tracer which was dosed @ 2 lts/min. The water analysis about 500 mtrs away revealed salt concentration of 0.5%. Assuming complete mixing and no losses, calculate the water flow rate. Ans 20% salt solution 0.5% salt solution Dosing rate Salt added in water Total flow Water flow rate = = = = = = 200 gms of salt in 1 Litre of water 5 gms of salt in 1 litre of water 2 lts/min 2 x 200 = 400 gms/min 400/5 = 80 lts/min 80 2 = 78 lts/min Or C1V1 = C2V2 V2 = C1V1/C2 = 0.2x2/0.005 =80 lts/min Actual flow = total flow dosage flow = 80-2 =78 lts/min . End of Section - II . Regular set A Section - III: N-1 LONG NUMERICAL QUESTIONS In a chemical plant, a 3000 Million Cal/hr cooling tower with one CW pump caters to the cooling water requirements. The management had decided to refurbish the cooling tower as its performance is felt to be low. The operating parameters of the CW system before and after refurbishment are presented below. S.No Parameter 1 2 CW inlet temp to CT Atmospheric air conditions COC Suction head of CW pump Discharge pressure of CW pump Efficiency CW Pump CW Pump motor CT fan CT fan motor 3 4 5 6 7 8 9 10 Pressure developed by CT fan Effectiveness of CT L/G ratio Density of air Before refurbishment 35oC WbT -25 oC, DbT - 38 oC 3.5 -1m After refurbishment 35oC WbT -25 oC, DbT - 38 oC 5 -1m 4kg/cm2(g) 4kg/cm2(g) 54% 89% 55% 90% 53% 89% 54% 90% 20mmwc 20mmwc 60 % 1.5 1.29kg/m3 70% 1.5 1.29kg/m3 As a result of cooling tower refurbishment the effectiveness has increased from 60% to 70 %. Also with improved water treatment the COC has increased to 5. Find out 1. Reduction in power consumption of pump and fan due to improvements in cooling tower. 2. Reduction in make up water consumption (ignoring drift losses) in KL/day Regular set A Ans Paramet er Effectiven ess Equation / formulae = (TCWi-TCW0)/(TCWiWbT) CW flow rate Q = heat load/( TCWiTCW0) Evaporati on loss =1.8*.00085*CW flow x Range Blow down loss Total water loss Make-up water = Evaporation Loss/(COC-1) = Eva loss+ Blow down loss Total head H Pump LKW = discharge headsuction head = ((Q*1000/3600)*(H* 9.81))/1000 Pump input Motor input Air flow in CT fan Qf = Pump LKW/Eff.Pump = Pump input/motor eff =[( CW flow)x1000]/ [((L/G)]*1.29) Hf Pressure developed by fan Hf = [(Qf in m3/h)*(Hf in mmWC)]/(3600*102 ) =Air KW/(FanEffi x Motor Eff) Air KW Fan motor input = Total water loss x 24hrs Before refurbishment 0.6=(35- TCWO)/(3525) TCWO = 29 oC =(3000x106/103) / ( 35- 29) = 500000 kg/h = 500 m3/h 1.8x0.00085x500x (35-29) = 4.59 m3/h = 4.59/(3.5-1) = 1.84 m3/h = 4.59+1.84 =6.43 m3/h After refurbishment 0.7=(35- TCWO)/(3525) TCWO = 28 oC =(3000x106/103)/(3528) = 428571 kg/h = 429 m3/hr 1.8x.00085x429x(3528) = 4.59 m3/h 4.59/(5-1) = 1.15 m3/h =4.59+1.15 =5.74 m3/h = 6.43 x 24 = 154.2 m3/day =154.2KL/day = 40-(-1) = 41 mWC = 5.74 x 24 = 137.76m3/day =137.76 KL/day = 40-(-1) = 41 mWC = (500*1000/3600)*(41 *9.81)/1000 = 55.86KW =55.86/0.54 =103.4 kW = 103.4/0.89 =116.2 kW = (500x1000/(1.5x1.29) = 258398 m3/h = 20mmWC = (429*1000/3600)*(41 *9.81)/1000 = 47.9 kW = 47.9/0.53 =90.4 kW =90.4/0.89 = 101.6kW = (429x1000/(1.5x1.29 ) = 221705m3/h = 20mmWC =(258398*20)/(3600* 102) =14.07 kW =(221705*20)/(3600* 102) = 12.08 kW =14.07/(0.55*0.9) = 28.43kW =12.058/(0.54*0.9) = 24.9 kW (1) Reduction in power of pump and motor = (116.2+28.43) - (101.6+24.9) = 18.13 kW (2) Reduction in makeup water = 154.2-137.76 = 16.44 or 16.5 KL/day Regular N-2 set A In a beverages industry the product stream (liquid) flowing at a rate of 5000 kgs/hr at 90 oC is first cooled in counter type cooling water (CW) heat exchanger to 55 oC and then by a chilled water (ChW) heat exchanger, to reduce temperature of the product to 11oC. The specific heat of the product is 0.9 kCal/kgoC. The other operating data and parameters are: Cooling Water heat exchanger Chilled Water heat exchanger Inlet temp Outlet temp Inlet temp Outlet temp Product 90oC 55 oC Product 55oC 11 oC Cooling Water 25 oC 32 oC Chilled water 7 oC 12 oC The chilled water is supplied by a reciprocating chiller, whose motor is drawing 60 KW with a motor efficiency of 87%. The management decides to upgrade cooling water heat exchanger by providing additional heat exchanger area to further enhance heat recovery i.e. to reduce the temperature of product at its outlet to 40oC. A. Depict the heat exchanger in existing and upgraded (improved) heat recovery case in a simple block diagram B. Calculate i. The additional heat exchanger area (as a % of the existing area) for cooling water heat exchanger, assuming there is no change in cooling water circulation rate and the overall heat transfer coefficient. ii. The COP of the chiller. iii. Reduction in refrigeration /chiller load and yearly energy savings at 600 hours per month operation, assuming energy consumption is proportional to load delivered. Regular Ans set A (A) Existing mh = 5000 Kg/hr, 90oC product 55oC Cooling water heat exchanger area (A1) T1 11oC Chilled Water heat exchanger T2 cw.out 32oC ch.w out 12oC cw.in 25oC Heat load on CW heat exchanger Heat rejected by product stream To chiller ch.w in 7oC From chiller = heat gained by CW Q1 = mc X 1 X ( Tc) Q1 = = 5000 X 0.9 X (90 55) 1,57,500 Kcal/hr mc = Cooling Water flow rate = (5000X0.9 X 35)/(32 25) = 22,500 Kg/hr. 5000 X 0.9 X (90 40) = 2,25,000 Kcal/hr 5000 X 0.9 X (90 40) ------------------------------ = 10oC 22,500 Heat exchanger duty with increased heat recovery Q2 Cooling water temp. Rise with increased heat exchanger duty for cooling product stream to 40oC = = Cooling water outlet temp with the above case = 25 + 10 = 35oC (A) Upgraded mh = 5000 Kg/hr, 90oC product T1 40oC Upgraded Cooling water heat exchanger area (A2) T2 cw.out 35oC 11oC Chilled Water heat exchanger ch.w out 12oC cw.in 25oC To chiller ch.w in 7oC From chiller ..2.5 marks LMTD1 with existing case T1 - T2 = ---------------------Ln T1 / T2 (90 32) (55 25) = -------------------------------- = 42.49oC Ln (90 32) / (55 25) Regular set A LMTD2 with additional heat recovery Q2 = U X A2 X LMTD2 Q2 = 2,25,000 = U X A2 X 30.77 (90 35) (40 25) = -------------------------------- = 30.77oC Ln (90 35) / (40 25) Q1 = U X A1 X LMTD1 Q1 = 1,57,500 = U X A1 X 42.49 A2 / A1 = (2,25,000 /1,57,500) X (42.49 / 30.77) = 1.973 Additional area required = Refrigeration load in existing case = 97.3% of existing heat exchanger area of CW heat exchanger 5000 X 0.9 X (55 11) = = = = = = = = = = 1,98,000 Kcal/hr 1,98,000 /3024 65.476 TR 60 KW 87% 198000 /( 60 X 0.87 X 860) 4.41 60 / 65.476 = 0.916 5000 X 0.9 X (55 40) / 3024 22.32 TR = 22.32 X 0.916 X 600 X 12 1,47,204.86 Kwh Motor input power Motor eff. C.O.P. of refrigeration chiller Input KW / TR Reduction in refrigeration load due to lower input temperature of the product to chilled water heat exchanger Yearly energy savings at 600 hrs. operation per month N-3 = In a continuous process industry Steam and Power are supplied through a cogeneration plant interconnected with grid. The design and actual operating parameters of the cogeneration plant as represented in the schematic are given in the table below. Double Extraction Condensing Steam Turbine Cogeneration System Regular set A Design B- Boiler TG- Steam Turbine Generator actual 68.75tph, 75tph,64kg/cm2(a), 450oC @82% 64kg/cm2(a), 450oC efficiency @81% efficiency Double Extraction Condensing type 10MW 7.2MW Stream Steam flow location Ref 1 2 3 4 Steam input to turbine First extraction Second extraction Condenser in Condenser out Steam Flow (tph) Steam Pressure (kg/cm2) Steam Temp (oC) 68.75 64 450 Steam enthal py (kCal/k g) 745 18.75 31.25 18.75 18.75 17 9 0.1 - 270 200 - 697 673 550 46 The industry is installing a 1200 TR double effect absorption chiller to meet the refrigeration load due to product diversification. Additional steam will be generated by the boiler, which will go into the turbine and be extracted at 9kg/cm 2(a) to meet the VAM requirement. The additional power thus generated will reduce the imported grid power. The following additional data has been provided: Maximum allowable steam flow the extraction at 9 Kg/cm2a Minimum allowable steam to condenser 40 TPH 9 TPH Critical power requirement of the plant 3800 KW Power import from grid 500 KW Cost of grid power Rs.4.25 / Kwh G.C.V. of coal 4000 Kcal/Kg. Cost of coal Rs. 4000/ton Feed Water temperature 105oC Feed Water enthalpy 105 Kcal/Kg. Combined efficiency of gear box and generator 96% Steam requirement for double effect absorption chiller 4.5 Kg./TR hr at 9 Kg/cm2a Annual hours of operation. 8000 hrs/y Steam rate at 9 Kg/cm2a at 2nd extraction for 1 KW 12 (Kg/hr)/kW turbine output Ignore auxiliary power consumption and also return condensate from extracted steam to process. Calculate (i) The Energy Utilization Factor (EUF) for the existing operating case (ii) The net additional annual operating cost, after installation of VAM. (iii) The Energy Utilization Factor (EUF) after installation of VAM. Regular Ans set A (i) Energy Utilization Factor (EUF) = (before VAM installation) Q thermal + P electrical ----------------------------------Fuel Consumption X G.C.V. Q thermal = m2 h2 + m3h3 + m4h 4 Q in = m (h1 hf) ------------ X G.C.V. Q thermal = = = = 18750 X 697 + 31250 X 673 + 18750 X 46 (18.75 X 697 + 31.25 X 673 + 18.75 X 46) X 103 Kcal/hr (13068 X 21031 + 862.5) X 103 Kcal/hr 34962.5 X 103 Kcal/hr Pe = = 7200 X 860 6192 X 103 Kcal/hr Fuel Consumption = (745 105) X 68.750x1000 -------------------------0.81X 4000 X 1000 = 13.58 TPH = 34962.5 X 103 + 6192 X 103 ---------------------------------- -x 100 13.58 X 103 X 4000 = 75.76% = 1200 TR EUF (ii) Refrigeration Load 1TR requires 4.5 Kg./hr steam at 9 Kg./cm2a Steam consumption in double effect absorption chiller Increase in steam extraction at 9 Kg./ cm2a = = = 1200 X 4.5 5400 Kg./hr. 5400Kg/hr Every 12 Kg./hr extraction at 9 kg/ cm2a gives 1 KW output at turbine , efficiency of generator and gear box = 0.96 Additional power recovery due to increase in extraction = = (5400 / 12) X 0.96 432 KW Additional coal consumption due to increase in extraction = (745 105)X 5400/(0.81x4000) = 1066 kg/h Additional cost of coal = 4000x1.066 = Rs 4266.6 /hr ..2 marks Monetary realisation by reducing import cost of purchased electricity = 4.25 Rs./unit = 432 X 4.25 = 1836 Rs./hr Regular set A Net additional annual operating cost after VAM installation (iii) Stream Steam flow Ref location 1 Steam input to turbine First extraction Second extraction 2 3 Condenser in Condenser out 4 Steam Temp (oC) 450 Steam enthalpy (kCal/kg) 745 17 9 270 200 697 673 0.1 - - 550 46 Q thermal + P electrical ----------------------------------Fuel Consumption X G.C.V. Q thermal = m2 h2 + m3h3 + m4h 4 Q in = m (h1 hf) ------------ X G.C.V. Q thermal = 18750 X 697 + 36650 X 673 + 18750 X 46 =(18.75 X 697 + 36.65 X 673 + 18.75 X 46) X 103 Kcal/hr = (13068 X 21031 + 862.5) X 103 Kcal/hr = 38596.7 X 103 Kcal/hr Pe = = (7200+432) X 860 6563.5 X 103 Kcal/hr = 13.58 TPH +1.066 TPH = 14.646TPH = 38596.7 X 103 + 6563.5 X 103 ----------------------------------------x100 = 14.646 X 103 X 4000 EUF A) Steam Pressure (kg/cm2) 64 Energy Utilization Factor (EUF)= (after VAM installation) Fuel Consumption N-4 Steam Flow (tph) 68.75+5.4 =74.15 18.75 31.25+5.4 =36.65 18.75 18.75 = (4266.6-1836)*8000 = Rs 1.94 crore/y 77.08% Answer ANY ONE OF THE FOLLOWING among A, B, C and D The operating parameters observed w.r.t. design in a 110 MW power generation unit are given below: Regular set A Parameters Generator output Steam generator temperature outlet super Design Operation 110 MW 540oC 110 MW 525oC 140 Kg/cm2a 135oC 87.5% 3800 0.09 Kg./cm2a 0.87 130 Kg/cm2a 135oC 87.5% 3800 0.11 Kg./cm2a 0.89 2362.5 Kcal /Kwh ------ 96% 4420 KW 96% 4420 KW 810 Kcal/Kg. 550 Kcal/Kg heat Steam generator outlet pressure Feed water inlet temperature Boiler GCV of Coal Turbine exhaust pressure Dryness fraction of exhaust steam Turbine heat rate Efficiency generator Energy loss in gear box Enthalpy of steam at 520oC, 130 Kg/cm2a, Enthalpy of steam at 0.11 Kg./cm2a Calculate the I. Actual steam flow to the turbine II. Specific steam consumption of turbine III. % increase in gross unit heat rate compared to design IV. Increase in monthly (720 hours/month) coal consumption due to deviation in operation w.r.t. design at a plant load factor of 80% Ans Generator output = of generator 110 MW 96% Generator input = = 110 / 0.96 114.58 MW Energy loss in gear box = = 4420 KW 4.42 MW Turbine output = = = Total input at gear box + energy loss in gear box 114.58 + 4.42 119 MW = ms (810 550) X (1 / 860) Turbine out put Turbine out put Regular set A ms (810 550) X (1 / 860) Steam flow rate through the turbine Specific steam = = = consumption = Boiler = = Coal consumption mf Specific coal consumption Actual unit heat rate 119 X 1000 3,93,615 Kg /hr 393.615 Tonne/hr (393.615 X 1000) / (110 X 1000) 3.58 Kg./kW Ms (hg hw) --------------- X 100 GCV X mf 393615 (810 135) ------------------------0.875 X 3800 79906.8 Kg./hr = = 79906.8 / (110 X 1000) 0.726 Kg./Kwh = = 0.726 X 3800 2758.8 Kcal/Kwh Design turbine heat rate of steam generator or boiler 2362.5 Kcal / KW 87.5% Design unit heat rate = 2362.5 / 0.875 = 2700 Kcal/Kwh % increase in heat rate w.r.t. = [(2758.8 2700) / 2700] X 100 design = 2.17 % Specific coal consumption for = 2700/3800 design heat rate = 0.71kg/kwh = (0.726 0.71) X 110 X 1000 X 0.8 X 720 -----------------------------------------------Additional coal consumption per 1000 month with a PLF of 80% 1013.76 tonnes = Or B) In a textile process house the production from the stenter machine is 72000 mtrs per day. The effective operation of stenter is 20 hours per day. The percentage moisture in the dried cloth (output) is 6% and its temperature is 75oC and wet cloth inlet is at 25oC . The stenter is heated by steam at 8 kg/cm2a and the daily steam consumption for the stenter is 16.5 tonnes. The efficiency of the stenter dryer is 47%. Calculate the (i) Linear speed of the stenter machine (ii) Inlet moisture (iii) Feed rate of the stenter. The following data have been provided Weight of 10 meter of dried cloth Enthalpy of the steam to the stenter = 1 kg. = 665 kcal/kg. Regular set A Enthalpy of condensate at the exit of stenter Ignore losses in start-up and stoppage. Ans = 130 kcal/kg. Production per day = Actual hours of operation = Linear speed of the stenter = 72000 meters 20 hours/ day 72000 / (20x60) = 60 meters per min Dried cloth output = 72000 / (20x10) = 360 kg/hr. Moisture in dry cloth Bone dry cloth = = 6% 360 x 0.94 = 338.4 kg/hr Moisture in outlet cloth mo = (360 338.4) / 338.4 0.0638 Kg./Kg. bone dry cloth = Steam consumption per day = = 16.5 tonnes 16500 / 20 = 825 Kg./hr. Heat load on the dryer =Energy input in steam x Dryer Efficiency = Steam flow rate x (Enthalpy steam Enthalpy condensate) x Efficiency Dryer = 825 x (665 130) x 0.47 = 207446.3 Kcal/hr. Further Heat load on the dryer = w x (mi mo) X [(Tout Tin) + 540] Kcal/hr. w =weight of bone dry cloth rate kg/hr mi = weight of cloth inlet moisture Kg./Kg. bone dry cloth Tout = dried cloth outlet temperature= 75oC Tin = wet cloth inlet temperature = 25oC 338.4 x (mi 0.0638) X [(75 25) + 540] = 207446.3 Kcal/hr mi = 1.1028 Kg./Kg. bone dry cloth(1.1028) / (1.1028+1)x100 % inlet moisture in wet cloth = 52.44 % total moisture in inlet cloth feed rate(inlet cloth rate), = 1.1028x338.4= 373.2 kg/hr = total inlet moisture/hr +bone dry cloth/hr = 373.2+338.4 = 711.6 Kg./hr. or C) The preheater exhaust gas from a cement kiln has the following composition on dry basis :CO2 23.9%, O2 5.9%, CO 0.2%, remaining is N2. The static pressure and temperature measured in the duct are -730 mmWC and 3500C respectively. The velocity pressure measured with a pitot tube is 19 mmWC in a duct of 2800 mm diameter ( Pitot tube constant = 0.89 ). The atmospheric pressure at the site is 10350 mmWC and universal gas constant is 847.84 mmWCm3/kg mol k. The specific heat capacity of preheater exhaust gas is 0.25 kcals/kg0C. Regular set A The static pressure developed by PH exhaust fan is 630mmWC and power drawn is 1582 kW. Calculate the efficiency of fan given that the motor efficiency is 92%. Ans The management had decided to install a 1.3 MW power plant with a cycle efficiency of 15% by using this preheater exhaust gas. Calculate the exhaust gas temperature at the outlet of waste heat recovery boiler of the power plant. Molecular weight exhaust gas (dry basis) M = %CO2xMCO2 + %O2xMO2 + %COxMCO + %N2 x MN2 = {(23.9 x 44) + (5.9 x 32) + (0.2 x 28) + (70 x 28)}/100 = 32.06 kg/kg mole Exhaust Gas density at operating temperature = = [ PM / RT ] = [ (10350 730) x 32.06 ) / { 847.84 x (273+350) } = 0.584 kg/m3 = 3.14 x( 2.8/2) 2= 6.15 m2 Duct Area Volume flow rate =A Cp (2 x g x P / )1/2 = 6.15 x 0.89 (2 x 9.81x 19 /0.584)1/2 = 138.3 m3/s Volume flow rate= 497880 m3/ h Fan efficiency = volumetric flow rate x pressure developed (102 x power drawn x motor eff) = 138.3 x 630 (102x1582x0.92) x100 = 58.69% Mass flow rate of preheater exhaust gas = Volume flow rate x density = 497880*0.584 = 2,90,762 kg/hr Heat equivalent of power generated from power plant =1.3MW =1300 x 860 = 1118000 kCals/hr Heat given up to power plant by exhaust gas = 290762 x0.25 x(350-To)x0.15 To = 350 - (1118000/(290945x0.25x0.15)) = 247.50C or D) For a commercial building, using the following data, (i) Determine the building cooling load in TR (ii) Calculate the supply air quantity to the cooling space in m3/s Outdoor conditions : DBT = 40 C, WBT = 28 C, Humidity = 19 g of water / kg of dry air Desired indoor conditions : DBT = 25 C, RH = 60 %, Humidity = 12 g of water / kg of dry air Regular set A Total area of wall = 324 m2, out of which 50% is window area. U Factor ( Wall ) = 0.33 W/m2K U Factor ( Roof ) = 0.323 W/m2K U factor [ fixed windows with aluminium frames and a thermal break ] = 3.56 W/m2K Other data: Ans 20 m x 25 m roof constructed of 100 mm concrete with 90 mm insulation & steel decking. CLTD at 17:00 hr : Details : Wall = 12 C; Roof = 44 C; Glass Window = 7 C SCL at 17 : 00 hr : Details : Glass Window = 605 W/ m2 Shading coefficient of Window = 0.74 Space is occupied from 8:00 to 17:00 hr by 30 people doing moderately active work. Sensible heat gain / person = 75 W ; Latent heat gain / person = 55 W ; CLF for people = 0.9 Fluorescent light in space = 21.5 W/m2 ; CLF 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 Supply air dry bulb temperature is 150C (i) Cooling Load Determination: I. External Heat Gain (i) Conduction heat gain through the wall = U factor x net area of wall x CLTD = 0.33 x (324*0.5) x 12 ] = 641.5 W (ii) Conduction heat gain through the roof = U factor x net area of roof x CLTD = 0.323 x ( 20 x 25 ) x 44 = 7106 W (iii) Conduction heat gain through the windows = U factor x net area of windows x CLTD = (3.56 x 162 x 7) = 4037 W (iv) Solar radiation through glass = Surface area x Shading coefficient x SCL = (162 x 0.74 x 605) = 72527 W II. Internal Heat Gain (i) Heat gain from people =Sensible heat gain + Latent heat gain Regular set A Sensible heat gain = (No. of people x Sensible heat gain / person x CLF) =(30 x 75 x 0.9) = 2025 W Latent heat gain = No. of people x Latent heat gain / person = (30 x 55 ) = 1650 W Therefore, Heat gain from people = (2025 + 1650 ) = 3675 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 (20 x 25) = 10750 W Therefore, heat gain from lighting = (10750 x 1.2 x 0.9) =11610 W (iii) Heat generated by equipment : Sensible heat generated by coffee maker Latent heat generated by coffee maker Sensible heat gain by computers and office equipment Therefore, Heat generated by equipment =1050 W = 450 W = 5.4 x 500 = 2700 W = 4200 W (iv)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 = { (20 x 25 x 3.6 ) x 0.3 } / 3600 = 0.15 m3 / s Therefore, sensible heat gain =1210 x 0.15 x ( 40 25 ) =2722.5 W Latent heat gain =3010 x 0.15 x ( 19 12 ) =3160.5 W No. 1. 2. 3. 4. 5. 6. 7. 8. Space Load Components Sensible Heat Load (W) Conduction through exterior wall 641.5 Conduction through roof 7106 Conduction through windows 4037.0 Solar radiation through windows 72527 Heat gained from people 2025 Heat gained from lighting 11610 Heat gained from equipment 3750 Heat gained by air infiltration 2722.5 Total space cooling load 104419 Total Cooling Load = 109679.5W/3516 =31.2 TR Latent Heat Load (W) --------1650 --450 3160.5 5260.5 (ii) Supply Air Quantity Calculation: Supply air flow = Sensible heat gain / {1210 * (Room dry bulb temperature Supply dry bulb temperature)} = 104419 W / {1210 J/m30K*(25 15)0C} = 8.63 m3/s Regular set A -------- End of Section - III ---------

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