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BEE Energy Solved Question Paper 2024 : Energy

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Kakoli
Bureau of Energy Efficiency (BEE), New Delhi
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Paper 4 Code : GREEN 23rd NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - MARCH, 2023 PAPER 4 : ENERGY PERFORMANCE ASSESSMENT FOR EQUIPMENT AND UTILITY SYSTEMS Date : 26.03.2023 Timings : 14:00-16:00 HRS Duration : 2 HRS Section - I: BRIEF QUESTIONS Max. Marks : 100 Marks: 10 x 1 = 10 (i) Answer all Ten questions (ii) Each question carries One mark 1. A slight negative pressure is maintained in the heating zone of the furnace to maintain uniform heating of the stock. True/False False 2. In a thermal power plant, the boiler efficiency affects the turbine heat rate. True/False False 3. Higher the slip, higher will be the loading of motor. True/False True 4. A roof exhaust fan in an industrial shed belong to the category of axial flow fans. True/False True 5. A screw compressor designed for 7 bar if operated at 5 bar, then its FAD will decrease. True/False False 6. If an electric motor is operated at the rated frequency and voltage without any shaft load, then the power factor will be high. True/False False 7. For the same head and flow, a pump operating with water will consume less power than operating with kerosene. True/False False 8. The higher the TTD (Terminal Temperature Difference) and the DCA (Drain Cooler Approach), the more efficient is the Rankine Cycle. True/False False 9. For a centralized air conditioning system with an air-cooled condenser the kW/TR will be lesser than the one with water cooled condenser. True/False False 10. The direct method of efficiency calculation of boilers does not require steam generation pressure. True/False False . End of Section - I . Section - II: SHORT NUMERICAL QUESTIONS L1 The operating data of Vapor Absorption Chiller is given below: Chilled Water Flow Specific Heat of Chilled Water Density of water Chilled Water Inlet Temperature Chilled Water Outlet Temperature Enthalpy of Steam Steam Flow Rate Condensate Temperature Ans Marks: 2 x 5 = 10 : 0.18 m3/sec : 4.18 kJ/kg/K : 1000 kg/m3 : 9.0 oC : 7.0 oC : 2700 kJ/kg : 1 kg/sec : 100 oC Calculate the following: A) Refrigeration effect in TR B) Thermal input energy in kJ/s C) Coefficient of performance (COP) A) Refrigeration Effect in TR Refrigeration Effect = Mc x x Cp x (tcin tcout) kJ/s = 0.18 x 1000 x 4.18 x (9 7) = 1504.8 kJ/s = 1504.8/3.516 TR = 428 TR B) Thermal Input Energy Input Energy = Mst x (hst hcond) = 1 x (2700 418) = 2282 kJ/s C) Coefficient of Performance (COP) = Refrigeration Effect/Thermal Input Energy = 1504.8/2282 = 0.66 L2 Ans The flue gas analysis of an Industrial Boiler revealed CO : 5000 PPM and CO 2 :14%. The Gross Calorific Value of the fuel is 4200 kCal/kg. The fuel has 36 kg of carbon per 100 kg of fuel. Estimate the percentage loss due incomplete combustion of C to CO. The heat Loss due to Partial Combustion = [(%CO x C)/ (%CO + %CO 2)] x 5654 CO = 5000 PPM = 5000x100/1000000 = 0.5 CO2= 14 = [(0.5 x 0.36)/ (0.5 +14)] x 5654 = 70.19 kCal GCV of Fuel = 4000 kCal % loss due to incomplete combustion = 70.19/4200 x 100 = 1.67% . End of Section - II Section - III: LONG NUMERICAL QUESTIONS (i) Answer all the Four questions (ii) Each question carries Twenty marks Marks: 4 x 20 = 80 N-1 In an engineering company, the compressed air is required for pneumatic equipment s and other processes. To meet the requirement, one reciprocating compressor of 500 CFM is installed. To assess the performance of compressor, Free Air Delivery test was carried out. The test and other data are given below. Receiver Capacity : 10 m3 Interconnecting pipe : 1 m3 Initial Pressure in Receiver : 1.0 kg/cm2 a Atmospheric Pressure : 1.0 kg/cm2 a Final Pressure : 8.5 kg/cm2 a Time Taken to Fill Receiver : 6 Minute Inlet air temperature (Tin) : 30 oC Air temperature in the receiver (T out) : 40 oC Motor RPM : 1400 RPM Motor Efficiency : 93% Motor Pulley Diameter : 300 mm Compressor RPM : 700 RPM Compressor Pulley Diameter : 600 mm Average Loading Time : 65% Average Unloading Time : 35% Power Consumption during loading : 82 kW Power Consumption during unloading : 21 kW Cost of Energy : 10.00 Rs / kWh Answer the followings: A) B) C) D) Estimate the Free Air Deliver of Compressor? [5 Marks] Evaluate the operating energy cost per day (for 24 hours of operation) [2 Marks] Calculate the isothermal power and isothermal efficiency of the compressor. [5 Marks] To reduce the energy loss due to unloading, the maintenance team has decided to reduce the speed of the compressor by reducing the motor pulley size. Evaluate the speed of compressor required for 10 min unloading time and 50 min of loading time and accordingly evaluate the diameter of the pulley of the motor. [5 Marks] E) Estimate the hourly energy consumption, energy saving after replacement of the pulley and payback period. The cost of pulley and belts is Rs. 1.2 Lakhs. The operating hours of the compressor is 8000 in a year. Consider power consumption during loading is 64 kW and power consumption during unloading is 18 kW after motor pulley change. [3 Marks] Ans A) Free Air Delivery (with temperature correction) Q = P2-P1 x V P0 T x (273+t1) (273+t2) = 13.31 m3/min = 13.31 x 35.31 = 470 CFM B) To evaluate the operating energy cost per day (for 24 hours of operation) Reciprocating compressor rated capacity = 500 CFM Percentage with respect to rated capacity = ( 500 470 / 500 ) x 100 =6% Loading energy ( 65% Loading ) = Average loading time x Power consumption during loading = 65 % x 82 kW = 53.3 kWh Unloading energy (65% Loading) = Average unloading time x Power consumption during unloading = 35 % x 21 kW = 7.35 kWh Hourly energy consumption =53.3 + 7.35 = 60.65 kWh Energy cost per day = 60.65 x 10 x 24 = Rs.14556/- C) Calculate the isothermal power and isothermal efficiency of the compressor. To calculate Isothermal Power = P1 x Q1 x loger / 36.7 r = Pressure ratio, P2/P1 P1 = Absolute intake pressure, kg/cm2 P2 = Absolute delivery pressure, kg/cm2 Q1 = Free air delivered, m3/hr = 1 x13.31x 60 x (Ln (8.5/1)/36.7) = 46.57 kW To calculate Isothermal efficiency = Isothermal Power Actual measured input power = 46.57 / (82 x 0.93) = 61.07 % D) Hourly air delivery = 470 CFM x 65%, Average loading time = 305.50 Cu.ft/hr Loading Time Existing (T1) New Loading Time after pully change(T2) = = 65% 83.3% Existing RPM of Compressor (N1) = 700 RPM New RPM of Compressor after Pully Change (N2) = (700x65%)/83.3% = 546 RPM Existing Motor Pully Size New Motor Pully Diameter 234 mm E) New Loading Power Unloading Power = = =300 MM = (300 x 546)/ 700 = 64 18 kW (given) kW (given) Hourly Energy Consumption after pulley Change = (64 x 50/60) + (18 x 10/60) = 56.33 kWh Reduction in Power Consumption = 60.65 56.33 = 4.32 kWh Annual Energy reduction = 4.32 x 8000 = 34533.33 kWh Annual Cost Reduction = 34533.33 x 10 = Rs.3,45,333/Cost of Pully & Belt Change Simple Payback Period N-2 = 120000 Rs. = 120000/345333= 4.17 Months A) A factory having a water requirement of 1000 lit/sec is planning to draw water from river basin located 3 km away from the plant. The following two options are evaluated: Option A: Two separate pumps connected to separate dedicated discharge line of 500 mm dia with 3 km length and each pump delivering 500 l/s. The efficiency of the pump is 80% and motor efficiency is 93%. Option B: Single pump connected to a discharge line of 750 mm diameter with 3 km length and delivering 1000 l/s. The efficiency of the pump is 80% and motor efficiency is 95%. The other data s are follows: System static head : 12 m Pipe friction factor is : 0.006 Cost of energy : 10 /kWh Annual operating hours : 7680 As an energy auditor which option you will recommend considering the annual energy cost. [15 Marks] B) A cooling tower connected to the condenser of a refrigeration system is operating with a heat load of 280 x 103 kCal/hr. The hot water inlet temperature to the cooling tower and approach are 33oC and 3oC respectively. The wet bulb temperature is 25oC. The condenser pump is operating at a total head of 25m. The combined efficiency of pump and motor is 52 %. Calculate the power drawn by the condenser pump. [5 Marks] Ans A) Two separate pumps connected to separate dedicated discharge line of 500 mm dia with 3 km length and each pump delivering 500 l/s. The efficiency of the pump is 80% and motor efficiency is 93%. Single pump connected to a discharge line of 750 mm diameter with 3 km length and delivering 1000 l/s. The efficiency of the pump is 80% and motor efficiency is 95%. Units Discharge flow 500 1000 lps (Q) 0.5 1 m3/s Diameter of pipe 0.5 0.75 m =3.141*0.5*0.5/4 =3.141*0.75*0.75/4 = 0.196 = 0.442 =0.5/0.196 = 1/0.442 = 2.55 = 2.26 Length of Pipe 3000 3000 Pipe friction factor 0.006 0.006 = 4flv2/2gD = 4flv2/2gD =4*0.006*3000*2.552/(2*9.81*0.5) =4*0.006*3000*2.262/(2*9.81*0.75) =47.61 = 25.08 12 12 = 47.61 + 12 = 25.08+12 = 59.61 = 37.08 = 0.8 x 0.93 = 0.8 x 0.95 = 0.744 =0.76 = Q*H* *g/1000 = Q*H* *g/1000 =0.5*59.61*1000*9.81/1000 =1*37.08*1000*9.81/1000 = 292.39 = 363.75 Motor input power No. of pumps = Hydraulic Power / Combined Eff = 292.39/0.744 = 393.00 = Hydraulic Power / Combined Eff = 363.75/0.76 = 478.61 kW 2 1 nos. Total power drawn = Motor input power x No of pumps = Motor input power x No of pumps = 393 x 2 = 786.00 = 478.61x 1 = 478.61 7680 7680 Parameter Area Velocity Friction resistance Static Head m/s Total head Combined Efficiency (Pump & Motor) Hydraulic power, Ph No. of hours of operation m2 M M M M kW kw hours Total units consumed = Total power drawn x no of opr hrs = Total power drawn x no of opr hrs = 786 x 7680 = 6036480 = 478.61 x 7680 = 3675724.8 Energy Cost 10 10 Annual Energy cost kWh = Total units consumed x Cost /kW h = Total units consumed x Cost 6,03,64,800 /yea r 3,67,57,248 2,36,07,552 Cost Reduction Option B is recommended considering the annual energy cost. B) Cooling Tower Heat Rejection 280000 kCal/hr Cold water temperature of basin Range of Cooling tower = 25+3= 28 o = (33- 28) = 5 o C C = (280000/5)/1000 = 56.0 m3/ hr Flow Rate of Cooling Water Head of Pump 25 M Combined Efficiency 52 % = Q*H* *g/1000 = ((56/3600)* 25*1000*9.81)/1000 kW = 3.82 = Hydraulic Power / Combined Eff kW Power Drawn by the Pump = 3.82/52*100 =7.3 N-3 An oil refinery has captive power plant with pet coke fired boiler. The following are the data collected to assess the boiler performance. Pet coke elementary analysis Hydraulic Power Carbon Hydrogen Nitrogen Oxygen Sulphur Moisture GCV of Petcoke Stack O2 Stack Temperature Heat loss due to radiation & convection Loss due to unburnt in fly ash & bottom ash Specific heat of flue gas Specific heat of water vapour Ambient Temperature Humidity in ambient air 88.8 3.6 1.2 1.4 3.6 1.4 : 8340 kCal/kg : 6.0 mol% : 250 oC : 1.0 % : 0.5% : 0.29 kCal/kg oC : 0.45 kCal/kg oC : 30 oC : 0.0204 kg/kg dry air Steam generation at 110 bar g & 520 oC Steam enthalpy at generation pressure and temperature : 816 kCal/kg Feed water temperature : 200 oC Steam drum pressure : 115 bar g A) Calculate Boiler efficiency using indirect method. [12] B) Calculate evaporation ratio. [3] C) Calculate the quantity (kg) flash steam generated per kg of blow down if: [5] - Saturated liquid enthalpy at steam drum pressure: 352 kCal/kg - Total enthalpy flash steam at 2 bar (g) pressure: 646 kCal/kg - Latent heat of evaporation at 2 bar (g) pressure: 526 kCal/kg Ans = {(11.6x 0.888)+[34.8x(0.0360.014/8)]+4.35x0.036} A) Theoretical Air required = 11.65 kg air / kg of petcoke = 100 x 6.0/(21-6.0) % Excess Air supplied = 40 = (1+40/100)x11.65 Actual Air supplied = 16.31 % kg air / kg of petcoke ={(0.888x44/12)+0.012+(16.31x0.77) +(16.31 -11.65)x0.23+(0.036x64/32)} = 16.97 Mass of dry flue gas Or = (16.31+1)-(0.014+(9*0.036)) =16.97 kg air / kg of petcoke = 16.97x0.29x(250-30)/8340x100 Stack losses, L1 = 12.98 Loss due to formation of water vapor from H2 in fuel, L2 = 9x0.036x(584+0.45*(250-30))/8340x100 = 2.65 = 0.014x(584+0.45x(250-30))/8340x100 Loss due to moisture in fuel, L3 = 0.11 Loss due to moisture in Air, L4 % % % = 16.31x0.0204x0.45x(250-30)/8340x100 = 0.39 % Loss due to radiation and convection = 1.0 % Loss due to unburnt in fly ash & bottom ash = 0.5 % Efficiency of boiler using indirect method B) Heat required per unit of steam generation Heat supplied by fuel for steam generation = (100-12.98-2.65-0.11-0.39-1-0.5) = 82.37 = (816-200) = 616.0 = (82.37/100) x 8340 = 6870 = 6870/616 Evaporation Ratio C) = 11.15 Quantity (kg) flash steam generated per kg of blow down S1 = 352 kCal/kg S2 = 646 526 = 120 kCal/kg L2 = 526 kCal kg % Flash Steam quantity from blowdown: (352-120)/526 = 44.1% Quantity of flash steam per kg of blow down: 0.44 kg/ kg of blow down. N4 A Answer any ONE of the following among four questions given below: A) Determine the cooling load of a commercial building based on the following data Outdoor Conditions DBT 360C WBT 260C Humidity 19 gm of water/kg of dry air Desired Indoor Conditions DBT 250C RH 50% Humidity 11 gm of water/kg of dry air Total area of wall 45 m2 Total area of window 15 m2 U - Factor (Wall) 0.35 W/m2 K U - Factor (Roof) 0.363 W/m2 K U - Factor (Windows) 3.00 W/m2 K Other Data % kcal/kg of Steam kcal/kg of petcoke MT Steam/MT of petcoke Roof Area 20 m x 20 m CLTD Wall 120C Roof 420C Window 80C SCL Window 500 W/m2 Shading coefficient (Window) Space occupancy 0.84 30 people Heat Gain/person Sensible 80 W Latent 50 W CLF for people 0.9 Fluorescent light in space CLF for lighting 0.9 Ballast factor 1.2 21 W/m2 Computers and office equipment heat production 6.3 W/m2 of sensible heat Air changes/hr of infiltration 0.25 Height of building 3.9 m Product of density and specific heat of air Latent heat factor 1210 J/m3 0K 3010 J kg dry air/m3 grams of water 14 marks B) The management has decided to give additional insulation to the roof at the cost of 40,000/-. This treatment modifies the U factor of the roof to value 0.300 W/m2K. The other data is The COP of the vapour compression system 3.75 The efficiency of the motor coupled with compressor 90%. The number of hours of operation in a year for the building 6000 hours The power cost is 10 /kWh Find out the simple payback period for this additional insulation. Ans Soln: A) External heat gain a) Conduction heat gain through the wall = U-factor x net area of wall x CLTD = 0.35 x (45-15) x 12 = 126 W b) Conduction heat gain through the roof = U-factor x area of roof x CLTD = 0.363 x (20 x 20) x 42= 6098.4 W c) Conduction heat gain through the windows = U-fact. x net area of window xCLTD = 3 x 15 x 8 = 360 W d) Solar radiation through glass = surface area x shading coeff. x SCL = 15 x 0.84 x 500 = 6300 W 1. Internal heat gain a) Heat gain from people = sensible heat gain + latent heat gain Sensible heat gain = No. of people x sens. heat gain/person x CLF = 30 x 80 x 0.9 = 2160W Latent heat gain = No. of people x latent heat gain/person = 30 x 50 Heat gain from people b) Heat gain from lighting = 1500 W = 2160 + 1500 =3660 W = (Energy Input x ballast factor x CLF) Energy Input = (Amount of lighting in space/ unit area) x floor area = 21 x (20 x 20) = 8400 W Heat gain from lighting = 8400 x 1.2 x 0.9 = 9072 W c) Heat generated by equipment 1. Sensible heat gain by computers and office equipment Total heat generated by equipment = 6.3 x 20 x 20 = 2520 W =2520 d) Heat gain by air infiltration = (Sensible + latent) heat gain Sensible heat gain = 1210 x airflow x T Airflow = (Vol. of space x air change rate)/3600 ={(20 x 20 x 3.9) x 0.25}/3600 = 0.11 m3/s Sensible heat gain = 1210 x 0.11 x (36-25) = 1464.1 W Latent heat gain = 3010 x airflow x h = 3010 x 0.11 x (19-11) = 2649 W Heat Load No . 1 Space Load Component Sensile Conduction through exterior wall 2 Conduction through roof 3 Conduction through windows 126 6098.4 360 Latent 4 Solar radiation through windows 6300 5 Heat gain from people 2160 6 Heat gain from lighting 9072 7 Heat gain from equipment 2520 8 Heat gain by air infiltration 1464.1 2649 28100.5 4149 Total Space Cooling Load 1500 Total Cooling Load = 32249.5 W B) Modified conduction heat gain through the roof = U-factor x net area of roof x CLTD = 0.3 x (20 x 20) x 42 = 5040 W = 6098.4 - 5040 = 1058.4 W 2 marks Savings in cooling load Savings in TR rating = Total savings in cooling load / 3516 = 1058.4/3516 = 0.301 TR COP = (Refrigeration effect kCal/hr)/(Power input kCal/hr) Savings in Power input = Refrigeration effect /COP = 0.301 x 3024 / (3.75 x 860) = 0.282 kW Annual energy savings = Energy savings x hours = 0.282 x 6000 = 1692 kWh Annual cost savings = Energy savings x Energy rate = 1692 x 10 = 16920 /year Simple payback period = 40,000 / 16920 = 2.36 years N4 B The following is the data collected from a 500 MW Turbine unit by an Energy Auditor. Main Steam Flow (TPH) 1561 Hot reheat Flow (TPH) 1413 Main steam pressure (kg/cm2)/ Temperature (oC) 166/529 Cold reheat (CRH) pressure (kg/cm2)/ Temperature (oC) 44.3/341 Hot reheat pressure (HRH) (kg/cm2)/ Temperature (oC) Feed water temperature (oC) LP Heater 1 LP Heater 2 LP Heater 3 HP Heater 5 HP Heater 6 246 Main steam enthalpy (kCal/kgoC) 806.47 Feed water enthalpy (kCal/kgoC) 246 CRH Enthalpy (kCal/kg oC) 730.71 HRH Enthalpy (kCal/kg oC) 844.27 Generator output (MW) 501.7 Boiler Efficiency (%) 87% Steam Heater Reference 42.4/540 Feed water in Design Values Feed water out Tem p (oC) Saturati on Temp (oC) Drain Temp (oC) Temp (oC) Pressure (kg/cm2) Temp (oC) Pressur e (kg/cm2 ) TT D (oC) DCA (oC) 73 70 64.2 47.2 13.7 63.6 12.6 2.88 4.8 140 111.23 70.4 - - 105 11.5 2.95 4.95 209 132.9 110 - - 130 10.4 2.95 4.95 416 207.33 171 170 202 210 199 0 5 335 254.94 212 - - 255 197 0.1 5 Calculate the following: a.) Turbine cycle heat rate and unit gross heat rate. (6 Marks) b.) Net heat rate and efficiency of the power plant if the auxiliary power consumption is 6%. (2 Marks) c.) Loss/gain in the turbine heat rate because of the deviation of the TTD and DCA of the LP/HP Heater systems from the design values. (12 Marks) Consider, for every 0.56oC increase or decrease of TTD from the design value, Heat Rate will increase or decrease by 0.014%. For every 0.56oC increase or decrease of DCA from the design value, Heat Rate will increase or decrease by 0.005%. Ans a.) Turbine Heat Rate: Unit Gross Heat Rate: = {(1561000 X (806.47-246)) + (1413000 X (844.27 730.71))}/ (501700) = 2063.72 kCal/kWh = 2063.72 / 87% = 2372.092 kCal/kWh b.) Net Heat Rate Power plant efficiency = 2372.092/(1-0.06) = 2523.50 kCal/kWh = 860 / 2523.50 * 100= 34.1% c.) From the above data, the following heater data can be inferred. Heater Ref. LP Heater-1 LP Heater-2 LP Heater-3 HP Heater-5 HP Heater-6 Feed water In-let Temp O C 47.2 63.6 105 170 210 Feed water Out-let Temp O C 63.6 105 130 210 255 Steam Inlet temp OC 73 140 209 416 335 Inlet steam Saturation Temp OC 70 111.23 132.9 207.33 254.94 Drain temp O C 64.2 70.4 110 171 212 So TTD (Terminal Temperature Difference) = Inlet Steam Saturation Temp OC Feed Outlet Temp O C DCA (Drain Cooler Approach = Drain temperature oC Feed Water Inlet Temperature oC Heater Ref. LP Heater -1 LP Heater-2 LP Heater -3 HP Heater-5 HP Heater -6 TTD oC (Design) 2.88 2.95 2.95 0 0.1 DCA OC (Design) 4.8 4.95 4.95 5 5 TTD oC (Calculated) 70-63.6 = 6.4 6.23 2.9 -2.67 -0.06 DCA OC (Calculated) 64.2-47.2 =17 6.8 5 1 2 Difference Between design values and Operating values of TTD and DCA of Heaters. Heater Ref. LP Heater -1 LP Heater-2 LP Heater -3 TTDOperating TTDDesign 3.52 3.28 -0.05 DCAOperating DCADesign 12.2 1.85 0.05 HP Heater-5 HP Heater -6 Total Difference -2.67 -0.16 3.92 -4 -3 7.1 Change in Heat rate because of deviation in TTD = (Net Change in TTD for All heaters X 0.014%/0.56oC ) o Since given, for every 0.56 C change in TTD HR will increase by 0.014% Increase in HR because of TTD deviation= 3.92 X 0.014 / 0.56 = 0.098 % Change in Heat rate because of deviation in DCA = (Net Change in DCA for All heaters X 0.005%/0.56oC ) Since given, for every 0.56oC change in TTD HR will increase by 0.005% So Increase in HR because of DCA deviation = 7.1 X 0.005 / 0.56 = 0.063 % N4 C Total % Increase in Turbine HR because of deviation in operation of TTD and DCA of Heaters from Design Values = 0.161 %. =0.161% X 2063.72 = 3.32 kCal/kWh Or A cement company manufactures ordinary clinker and low heat clinker. The chemical composition of the cement clinkers (loss free basis) are given in the following table: Percentage of constituent Constituent Ordinary Clinker Low Heat Clinker SiO2 20 44 Fe2O3 7 15 Al2O3 7 1 CaO 65 39 MgO 1 1 Following heat data is given for both types of manufacturing process. Item Heat output excluding heat of formation Total Heat Input kCal/kg of Ordinary Clinker kCal/kg of Low Heat Clinker 360+6.7mfuel 400+7.1mfuel 27+6000mfuel 27+6000mfuel Calculate the following: 1. Heat of formation of clinker of both the types. 8 marks 2. Total heat output in terms of mfuel 2 marks 3. Amount of fuel needed per kg of clinker for each type of clinker from heat balance principle. 4 marks 4. Which type of clinker consumes more energy? 1 mark 5. If cost of coal is 6,200/- per ton, what is the difference between energy consumption cost per ton of clinker, in Ordinary Clinker and Low Heat Clinker? Ans 5 marks Heat of formation of clinker HR = 2.22 Al2O3+6.48 MgO+7.646 CaO-5.116 SiO2-0.59 Fe2O3. Percentage of constituent in Constituent Ordinary Clinker Heat in kcal/kg of Clinker Multiplie r Low Heat Ordinary Clinker Clinker Low Heat Clinker SiO2 20 44 -5.116 -20x5.116= -102.32 -44x5.116= -225.10 Fe2O3 7 15 -0.59 -7x0.59= -4.13 -15x0.59= -8.85 Al2O3 7 1 2.22 7x2.22= 15.54 1x2.22= 2.22 CaO 65 39 7.646 65x7.646= 496.99 39x7.646= 298.19 MgO 1 1 6.48 1x6.48= 6.48 1x6.48= 6.48 Heat of formation of clinker 412.56 72.94 Additional Heat Output 360+ 6.7mfuel 400+ 7.1mfuel Total Heat Output 772.56+ 6.7mfuel 472.94+ 7.1mfuel Total Heat Input 27+ 6000mfuel 27+ 6000mfuel Balancing the Heat Output with Heat Input 772.56+6.7mfue 472.94+7.1mfuel =27+ =27+ 6000mfuel l 6000mfuel Mass of fuel consumed 0.1244 kg/kg 0.0744 Clinker Clinker kg/kg Fuel Consumed per ton of clinker 0.1244 ton/ton 0.0744 Clinker Clinker ton/ton Fuel cost = 0.1244x6200 =0.0744x6200 = 771.28 /ton = 461.28 /ton Therefore, the Ordinary Clinker consumes more energy. The difference between energy consumption cost of Ordinary clinker and the Low Heat Clinker is = 771.28-461.28 = 310 /ton. N4 D A) Or In a textile mill 150 kg of fabric is dyed per batch in a jigger. The dye liquor is heated from 250 C to 900 C. The specific steam consumption is 0.65 kg/kg (with enthalpy 660 kCal/kg). Calculate the liquor ratio allowing 15% margin for the losses. B) 5 marks In a textile mill a thermic fluid heater of 20 lakh kCal/hour is meeting process heat requirement. Outlet temperature of fluid : 2800C Return temperature of fluid : 2600C Specific Heat of fluid : 0.55 kcal/kg 0C : 840 kg/m3 Density of fluid Current coal consumption : 400 kg/hour GCV of coal : 4100 kcal/kg % loading of the thermic fluid heater : 45% 1) What is the thermic fluid circulation rate? 3 marks 2) What is the existing thermal efficiency? 2 marks 3) By providing an air pre-heater the efficiency of the thermic fluid heater improved to 62%. Calculate the coal savings for 6000 Hrs of annual operation? 5 marks C) The operating details of a stenter in a textile unit are given below: 1. The inlet and outlet conditions of the cloth are shown in the figure below. The feed rate of wet cloth to the stenter is 1000 kgs/hour. 2. The heat input to the stenter is provided by a thermic fluid heater fired by firewood as fuel. 3. Gross Calorific Value (GCV) of Firewood is 3500 kCal/kg. 4. Firewood Consumption is 320 kg/hr. 5. The efficiency of the thermic fluid heater is 65%. Calculate the drying efficiency of the stenter. Ans A)Steam quantity needed per batch = quantity of fabric x specific steam consumption = 150 kg x 0.65 kg/kg = 97.5 kg Heat energy consumed per batch = Quantity of steam x enthalpy of steam (5 Marks) = 97.5 kg x 660 kcal/kg = 64350 kCal/batch Heat energy consumed per batch = textile mass x liq. Ratio x 1.15(incl margin) x T 64350 = 150 x L x 1.15 x (90-25) Liquor Ratio = 5.74 Therefore liquor ratio of the jigger is 1:5.74 .. B) 1) Capacity of thermic fluid heater = 20 lakh kcal/hour Heat Duty = Capacity x % loading =2000000 x 0.45 =900000 kcal/hr. = Circulation rate x density x sp. Heat x T Heat Duty 900000 = Q x 840 x 0.55 x (280-260) Q=97.40 m3/hour Thermic fluid Circulation rate 2) Input Energy = Coal consumption x GCV = 400 x 4100 = 1640000 or 1.64 x 106 kcal/hour Efficiency of thermic fluid heater = 900000/1640000 = 0.5487 i.e. 54.87 % 3) Efficiency of thermic fluid heater: 62% New coal consumption for thermic flid heater = 0.5487 * 400/0.62 = 354 kg/hr Annual Coal Savings = (400-354) * 6000/1000 = 276 T/Hr. C)Calculation of Stenter Efficiency 1. 1000 kg/hr wet cloth contains 60% moisture (water)=1000*0.6 = 600 kg/hr moisture. 2. 1000 kg/Hr wet cloth contains 40% Bone dry cloth= 1000*0.4 = 400 kg/hr Bone dry cloth. 3. Weight of wet cloth at outlet contains 5% moisture = 400/0.95=421 kg/hr 4. Quantity of Moisture at the outlet = 421-400 =21 kg/hr. 5. Moisture evaporated = (600-21)=579 kg/hr 6. Heat energy required to evaporate 579kg/hr of water = m*cp*deltaT + m*540 =579*1*(80-30)+(579*540) =(579*50)+(579*540) =341610 kCal/hr 7. Heat Energy provided by firewood heated thermic fluid at 65% efficiency 8. Stenter Efficiency = 320 kg/hr*3500 kCal/kg*0.65 = 728000 kCal/hr =341610/728000*100 =46.92 % . End of Section - III

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