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19th National Certification Exam Energy Managers & Auditors SEPTEMBER 2018 Paper 4

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Paper 4 Set A 19th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS - September, 2018 PAPER 4: ENERGY PERFORMANCE ASSESSMENT FOR EQUIPMENT AND UTILITY SYSTEMS Section - I: BRIEF QUESTIONS S-1 (i) Section I contains Ten questions (ii) Each question carries One mark Marks: 10 x 1 = 10 The capacity of a screw compressor cannot be controlled by discharge throttling. True or False Ans True S-2 A package air conditioner of 5 TR capacity delivers a cooling effect of 4 TR. If Energy Efficiency Ratio (W/W) is 2.90, the power in kW drawn by compressor would be ___________ Ans =(4*3024)/860 = 14.065/2.90 = 4.85 S-3 If the steam generation in a boiler is reduced to 45%, the radiation loss from the surface of the boiler will reduce by the same ratio. True or False Ans False S-4 If the coal Gross Calorific Value is 4200 kcal/kg and specific coal consumption is 0.6 kg/kWh, what is the power station gross efficiency ? Ans (860 /(4200 x 0.6)) x100 = 34.12% S-5 If the measured input power of a 90 kW motor is 45 kW, then the calculated loading of the motor is 50 %. True or False Ans False S-6 The speed of an energy efficient motor will be more than the standard motor of same capacity because ________________ decreases. Ans Slip S-7 O2 % in flue gas is required in the direct method efficiency evaluation of a boiler. True or False Ans False S-8 To minimize scale losses in a reheating furnace, the furnace should be operated at a negative pressure. True or False Ans False S-9 The heat rate of a power plant will reduce when there is an increase in the inlet cooling water temperature to the condenser. True or False Ans False S-10 A typical co-generation system in a cement plant will come under the category of topping cycle. True or False Ans False . End of Section - I . Section - II: SHORT NUMERICAL QUESTIONS Marks: 2 x 5 = 10 (i) Section II containsTwo questions (ii) Each question carries Five marks L-1 A coal based power plant A is having a Gross Unit Heat Rate of 2400 kCal/kWh with Auxiliary powerconsumption of 8 % whereas Plant B of same size and make, has an operating Net Heat Rate of 2500 kCal/kWh. In your opinion, which plant is more efficient and why? Ans L-2 Gross Heat Rate of Plant A 2400 kcal/kwh Auxiliary Power Consumption 8% Net Heat Rate of Plant A = Gross Heat Rate/(100- APC) = 2400/(1-0.08) = 2608.70 kcal/kwh Therefore, Plant B is more efficient with a lower Net Heat Rate of 2500 kcal/kwh than that of Plant A (2608.70 kcal/kwh). Milk is flowing in a pipe cooler at a rate of 0.85 kg/sec. Initial temperature of the milk is 55 C and it is cooled to 18 C using a stirred water bath with a constant temperature of 10 C around the pipe. Specific heat of milk is 3.86 kJ/kg C. Calculate the heat transfer rate (kCal/hr) and also LogrithmicMeanTemprature Difference (LMTD) of the exchanger. Heat transfer in cooling milk = 0.85 * 3.86 * (55 18) = 121.4 kJ/sec = (121.4* 3600) = (437040 kJ/hr) / (4.18) Ans = 104555 kcal/hr LMTD: DT1 = 55 10 = 45 C DT2 = 18 10 = 8 C LMTD of the heat exchanger = (45 8)/ ln (45 / 8) = 21.4 C . End of Section - II Section - III: LONG NUMERICAL QUESTIONS Marks: 4 x 20 = 80 (i) Section III containsFour questions (ii) Each question carries Twenty marks N-1 The schematic and operating data of a steam turbine cogeneration plant with a back pressure turbine is given below. Enthalpy of steam at 180 bar, 550 0C 3420 kJ/kg Exhaust steam enthalpy at isentropic expansion from 180 bar to 2 bar 2430 kJ/kg Enthalpy of boiler feed water 504.7 kJ/kg Efficiency of boiler - 80 % Calorific value of coal 4500 kcal/kg Steam flow rate into the Turbine - 91 TPH Turbine isentropic efficiency - 90 % Generator efficiency - 98 % Gear box efficiency - 97 % Calculate: Ans (each carries 5 Marks) a) Electrical output from the generator in MW b) Fuel consumption in Boiler in TPH c) Energy Utilization factor of the cogeneration plant d) Heat to power ratio of the cogeneration plant, kCal/kW a) Electrical output from the generator in MW Actual exhaust steam enthalpy = [3420 - (0.9 x (3420 - 2430))] = 2529 kJ/kg Turbine power output = 1 x (1000/3600)) x (3420 - 2529)]/1000 = 22.52 MW Electrical output = (22.52 x 0.97 x 0.98) = 21.4 MW b) Fuel consumption in Boiler in TPH Fuel consumption in Boiler = (91,000 x (3420 504.7)) / (4.18 x 4500 x 0.80) = 17.6 TPH c) Energy Utilization factor of the cogeneration plant = [(21,400 x 860) + (91000 x ((2529 504.7)/ 4.18))] / (17,600 x 4500) = [(1,84,04,000 + 44069689) / (7,92,00,000)] = 0.79 d) Heat to power ratio of the cogeneration plant, kCal/kW Heat to power ratio, kcal/kW = (91000 x (2529 - 504.7)/4.18) / 21400 = 2059 kCal/kWh (or) = 2.39 kWthermal/kWelectrical N-2 In a process plant, the hot effluent having a flow rate of 63450 kg/hr at 80 0C from the process is sent to a finned tube air cooled heat exchanger for cooling. The outlet temperature of theeffluent from the heat exchanger is 38 0C. Air at a temperature of 30 0C enters the heat exchanger and leaves at 60 0C. The fan develops a static pressure of 30 mmWC. The operating efficiency of the fan is 65 % and fan motor efficiency is 90 %. The plant operates for 5000 hours per year. The management decided to replace the existing air-cooled heat exchanger with water-cooled Plate Heat Exchanger (PHE). Following are the relevant data: Existing: Specific heat of air : 0.24 kcal/kg 0C Specific heat of hot effluent : same as water Density of air : 1.29 kg/m3 : 75 % Proposed: Cooling water pump efficiency Pump motor efficiency : 90 % Effectiveness of water cooled heat exchanger : 0.4 Cooling water inlet temperature : 25 0C Total head developed by the cooling water pump : 30 m Over all heat transfer coefficient of PHE : 22300 kcal/hrm2 0C Calculate the following: Annual energy savings due to replacement of existing air cooled plate heat exchanger by water cooled counter flow plate heat exchanger. (15 Marks) Area of the proposed water cooled plate heat exchanger. Ans Heat duty in hot fluid = M x Cp hot x (Ti - To) = 63450 x 1 x (80 - 38) = 2664900 kcal / hr In a heat exchanger, Heat duty in hot fluid = Heat duty in cold Air Mass of the cold air = 2664900 / (0.24 X (60-30)) = 370125 kg/hr Existing System: Fan Shaft Power = Volume,m3/s X Static Press mmWc 102 X Fan Efficiency factor = (370125/( 3600x 1.29)) x 30 102 X 0.65 = 36.06 kW Motor Input Power = 36.06/ 0.9 = 40.07 kW Proposed System: Effectiveness of water cooled heat exchanger= 0.4 Cold Water outlet temperature = TWO Cold water inlet temperature = T Wi Hot effluent inlet temperature = TEff.in Hot effluent outlet temperature = TEff.out TWO TWi Effectiveness = TEff.in TWi Cold Water Outlet = (0.4 x (80 25)) + 25 = 47 OC Heat duty in hot fluid Mass flow rate of cooling water (M)= Cp x (TWO TWi) = 2664900 1 x (47 25) x 1000 = 121.13 m3 /hr (5 Marks) Hydraulic Power Requirement for one Cooling Water Pump: = (Flow in m3/hr x Head in m x Density in kg/m3 x g in m/s2) (1000 x 3600) = (121.13 x 30 x 1000 x 9.81) (1000X3600) = 9.9 kW Pump input Power Requirement = 9.9 kW / 0.75 2 kW Pump Motor Input Power = 13.2 / 0.9 = 14.67 kW Thus savings = Power consumption by fans Water Pumping Power = 40.07 14.67 = 25.4 kW Annual energy savings in kWh = 25.4 kW x 5000 hrs = 127000 kWh/annum Calculations for LMTD for Proposed counter flow PHE: LMTD for counter flow in PHE= {(80-47) (38-25)} / ln {(80-47) / (38-25)} = 21.5 OC Considering overall heat transfer coefficient (U) = 22300 kCal/hrm20C Heat transfer Area = Q / (U x Tlmtd) = 2664900 /(22300 x 21.5) = 5.56 m2 N3 In a process industry, the wet products are to be dried in a drier. The plant has a pressurizedhot water boiler which supplies hot water at 145 0C to the heating coils in the drier. The return water to the boiler is at a temperature of 110 0C. The boiler is fired by saw dustbriquettes. The other relevant data are given below. Fuel firing rate = 375 kg/hr O2 in flue gas = 12.2 % CO in flue gas = 189 ppm CO2 in flue gas = 8.5 % Avg. exit flue gas temperature = 235 0C Ambient temperature = 31 0C Humidity in ambient air = 0.0204 kg / kg dry air Gross Calorific Value of ash = 800 kCal/kg Radiation & other unaccounted losses = 0.5 % Specific heat of flue gas = 0.23 kCal/kg0C Ash = 8.0 Moisture = 7.5 Carbon = 45.3 Hydrogen = 4.4 Nitrogen = 1.4 Oxygen = 33.3 Sulphur = 0.1 Gross Calorific Value of saw dust briquette = 3300 kCal/kg Fuel (briquettes) Ultimate Analysis (in %) Calculate the hot water circulation rate (m3/hr) in the boiler. An 1. Theoretical air required for complete combustion s = [(11.6 x C ) + {34.8 x ( H 2 O2 / 8)}+ (4.35 x S )] / 100 kg/kg of coal = [(11.6 x 45.3) + {34.8 x (4.4 33.3/8)} + (4.35 x 0.1)] / 100 = 5.34 kg / kg of briquette. 2. Excess air supplied Actual O2 measured in flue gas = 12.2 % % Excess air supplied (EA) = O2 % 100 21 O2 % = 12.2% 100 21 12.2% = 138.6 % 3. Actual mass of air supplied = {1 + EA/100} x theoretical air = {1 + 138.6/100} x 5.34 = 12.74 kg/kg of briquette 4. To find actual mass of dry flue gas Mass of dry flue gas = Mass of CO2 +Mass of N2 content in the fuel+ Mass of N2 in the combustion air supplied + Mass of oxygen in flue gas + Mass of SO2in flue gas = 0.453 x 44 12.74 x 77 (12.74 5.34) x 23 0.001x64 + 0.014 + + + 12 100 100 32 = 13.19 kg / kg of briquette 5. To find all losses a) % Heat loss in dry flue gas (L1) = = m x C p x (T f Ta ) GCV of fuel x 100 13.19 x 0.23 x (235 31) x 100 3300 = 18.75 % b) % Heat loss due to formation of water from H2 in fuel (L2) = = 9 x H 2 x {584 + C p (Tf - Ta )} GCV of fuel x 100 9 x 0.044 x {584 + 0.45 (235 - 31 )} x 100 3300 = 8.1 % c) % Heat loss due to moisture in fuel (L3) = = = 1.54 % M x {584 + C p (Tf - Ta )} GCV of fuel x 100 0.075 x {584 + 0.45 (235 - 31 )} x 100 3300 d) % Heat loss due to moisture in air (L4) = = AAS x humidity x C p x (T f Ta ) GCV of fuel x 100 12.74 x 0.0204 x 0.45 x (235 31) x 100 3300 = 0.723 % e) % Heat loss due to partial conversion of C to CO (L5) = 0.0189 x 0.453 5654 x x 100 0.0189 + 8.5 3300 = = f) %CO x C 5654 x x 100 %CO + %CO2 GCV of fuel 0.172 % % Heat loss due to Ash (L6) Gross Calorific Value of Ash =800 kCal/kg Amount of Ash in 1 kg of coal = 0.08 kg/kg coal Heat loss in bottom ash 0.08 x 800 = =64 kcal/kg of coal % Heat loss in bottom ash = (64x 100) / (3300) = 1.94 % g) % Heat loss due to radiation & other unaccounted losses (L7) = 0.5% (given) HWG efficiency by indirect method = 100- (L1+ L2+ L3+ L4+ L5+ L6+ L7) =100-(18.75 + 8.1+ 1.54+ 0.723+ 0.172 +1.94+0.5) = 68.28 % Hot water circulation rate in m3/hr: HWG efficiency % = Mass of hot water = (Mass of hot water x Cp x T) Mass of fuel x GCV of fuel 375 x 3300 x 0.6828 (145 - 110) x 1 =24,141.86kg/hr =24.142 m3/hr x 100 N-4 A) Answer any ONE of the following A 60 MW captive power plant (CPP) of a chemical plant has a coal fired Boiler, condensing steam Turbine and Generator. The CPP after meeting its auxiliary power consumption is exporting power to the chemical plant. The operating data of CPP is as follows: Generator output : 60 MW Auxiliary power consumption : 6 MW Steam flow to the turbine : 231 Tons/hr Steam inlet pressure and temperature : 105 kg/cm2 (a) and 480 0C Enthalpy of inlet steam at operating pressure and temperature : 793 kCal/kg Enthalpy of feed water to boiler : 130 kCal/kg Condenser exhaust steam pressure and temperature : 0.1 kg/cm2(a) and 45.5 0C Enthalpy of water at operating pressure and temperature of condenser : 45.5 kCal/kg Latent heat of vaporisation of steam at operating pressure and temperature of condenser : 571.6 kCal/kg Enthalpy of exhaust steam : 554 kCal/kg GCV of coal used : 4240 kCal/kg Efficiency of the boiler : 86.5 % Based on the above data, calculate the following parameters of the power plant: Ans a) Gross Heat Rate (8 Marks) b) Net Heat Rate (3 Marks) c) Dryness fraction of exhaust steam (2 Marks) d) Condenser heat load (3 Marks) e) Specific coal consumption (2 Marks) f) Overall efficiency (2 Marks) (a) Gross heat rate Gross heat rate= Coal consumption (kg/hr) x GCV of coal (kcal/kg) -----------------------------------------------------------Generator output (kw) Given : --------------(1) Coal consumption=? GCV of coal=4240 Kcal/kg Generator output= 60 MW Boiler efficiency= Q (H-h)/ (q x GCV) Where, Q= Quantity of steam generation (kg/hr)=231x1000 H= Enthalpy of steam (Kcal/kg) =793 h=Enthalpy of boiler feed water (kcal/kg) =130 q=Coal consumption (kg/hr) =? Boiler efficiency=0.865 Substituting the given values in equation (2) we get, 0.865=( 231 x 1000 x (793 130) )/q x 4240 q= 41758 kg/hr --------------(2) Substituting the calculated value of q in equation (1) we get, Gross heat rate= (41758 x 4240) / (60x1000) =2950.9 kcal /kWh (b) Net heat rate Net heat rate = Gross heat rate -----------------------------------------------------1 ( % Auxiliary Power consumption/100) Auxiliary consumption =6 MW Generation= 60 MW % Auxiliary consumption=( 6/60) x 100 = 10% Substituting the values in the equation (3) we get, Net heat rate= 2950.9/( 1 10/100)= 3278.8 kCal/kWh ----------------------(3) (C) Dryness fraction of exhaust steam Enthalpy of exhaust steam = Enthalpy of feed water + Dryness fraction of steam x L.H. of vaporisation of steam Substituting the given values in the above, we get 554= 45.5+ dryness fraction of steam x 571.6 Dryness fraction of steam= (554 45.5)/571.6 = 0.889 (d) Condenser heat load Heat load on condenser= Steam flow rate x L.H of vaporisation of steam x dryness fraction of steam = 231x 1000 x 571.6 x 0.889 =117383.2 MCal/hr (e) Calculation of specific coalconsumption Specific coal consumption = Total coal consumption/Gross generation = 41758 kg/hr / (60 x 1000) kW = 0.696 kg/kWh (f) Calculation of overall efficiency of plant Overall efficiency= 860/Gross heat rate, kCal/kWh -----------------------(4) Substituting the values we get, 860/2950.9 =29.14% (OR) Overall efficiency = (Generator Output, kW x 860 kCal/kWh) /(Mass flow rate of coal kg/hr x GCV of coal,kCal/kg) = (60 x 1000 x 860)/(41758 x 4240 ) =29.14% OR B) In the energy audit of a 6-stage Preheater (PH) section of a 4000 TPD (clinker) Cement kiln operating at full load, the following were the field measurements taken. S.No. Ans: Description Value 1. Reference temperature (0C) 0 2. Reference pressure and the Barometric pressure (mmWG) 3. Average Dynamic Pressure (mmWC) 17.1 4. Static Pressure at Fan Inlet (mmWC) -860 5. Static Pressure at Fan outlet (mmWC) -16 6. Temperature (0C) 328 7. Density of the PH Gas (NM3/kg), at reference condition 1.422 8. Pitot Tube constant 0.854 9. Diameter of PH Duct (m) 3.2 10. Cp of PH Gas (kcal/kg0C) 0.245 11. Power Input to the PH fan motor (kW) 1812 12. PH fan Motor Efficiency (%) 13. GCV of coal (kCal/kg) 5600 14. Annual Operating Hours 7300 15. Cost of Coal (Rs./Ton) 4836 10323 95 a) Estimate the specific heat losses (kCal/kg clinker) carried away by PH gases. (5 Marks) b) Estimate the PH fan Efficiency. c) Estimate the envisaged specific fuel savings (kCal/kg clinker), annual fuel savings and annual monetary savings by reduction in PH gas temperature to 290 0C by appropriate modification in the PH cyclones. (5 Marks) d) Estimate energy savings in fan power consumption in the proposed case where PH exit temperature is reduced to 290 0C. Also consider the static pressure at the fan inlet will reduce by 8 % from the present level due to PH modification (Fan and motor efficiency in both the cases are same). (5 Marks) (5 Marks) a) Specific heat losses by PH gases(kCal/kg clinker) Density of gas at Present operating Conditions (kg/m3) Velocity of PH Gas (m/s) Area of the PH Duct (m2) t , p = stp 10323 * PS 273 kg/m3 10323 273 + t e 0.59 = 1.422*(273/(273+328))*((10323 -860)/10323) kg/m3 Velocity =Pt (2g( Prms)avg / t , p ) m/sec 20.36 =0.854 * ((2 * 9.81 * 17.1/0.59)) = Pi/ 4 * D2 = 3.14 * (3.22)/4 0.5 8.04 Flow rate of the PH Gas (m3/hr) Flow rate of the PH Gas (Nm3/hr) Specific PH Gas generation (Nm3PH gas /kg Clinker) Specific PH Gas generation (kgPH gas /kg Clinker) Specific Heat Loss in existing case (kcal/kg Clinker) = Area * Velocity = 8.04 * 20.36 *3600 m3/hr = Flow rate of the PH Gas (m3/hr) x (0.59/1.422) = 589299.8 X (0.59/1.422) = Flow rate of Ph gas (Nm3/hr) / Clinker Production (kg/hr) = 244505.5/ ((4000x1000)/24) = Specific PH Gas generation (Nm3PH gas /kg Clinker) x 1.422 kg/Nm3 =1.467 x 1.422 = m Cp (Tph-Tref.) =2.086*0.245*(328-0) 589299.8 244505.5 1.467 2.086 167.6 b) PH fan efficiency Air Power (kW) Fan Efficiency (%) = ((Q (m3/hr)/3600)x (Pst(mmWC)))/102 = (589299.8/3600)* (-16+860)/102 = Air power*100/(motor power *motor effi) = (1354.5*100)/(1812*0.95) c) Envisaged fuel and monetary savings Specific Heat Loss in the = M Cp (Tph-new-Tref.) proposed case (Kcal/Kg Clinker) =2.086 * 0.245 * (290-0) =old heat loss/kg cli- new heat loss/kg cli Fuel Savings (Kcal/Kg Cli =167.6-148.21 =Clinker prod. * run hrs/yr* heat saving/kg cli /coal gcv Annual Fuel Savings =166.67*7300*19.39/5600 Annual Monetary savings (Rs. =fuel savings in tons * fuel cost in (Rs. Ton) Lakhs/yr) = (4212.79 *4836)/100000 d) Fan energy savings Envisaged static pressure at Fan Inlet after PH modification (mmWC) = 92% of Original Static pressure at Fan inlet (reduction in Friction loss due to temperature reduction- given) = 92% * 860 Envisaged Fan Flow after PH modification = Flow (Nm3/hr) * ((273+Tph-new)/273) * (10323/(10323+Pst-new)) Fan efficiency (%) Already estimated above (considering the same) Fan motor Input power in the proposed case (kW) = ((Q (m3/hr)/3600)x (Pst(mmWC)))/102 = ((546091.5/3600)* (-16+791.2))/(102*.787*.95) =Fan power (old-new) =1812 - 1541 =(power saved * Annual operating hrs)/105 = ( 271 * 7300) / 10^5 Fan Power saving (kW) Annual Energy saving (Lakh kWh/yr) 1354.5 78.7 148.21 19.39 4212.79 203.73 -791.2 546091.5 = 244505.5* ((273+290)/273)*(10323/(10323-791.2)) OR 78.7 1541 271 19.783 C) A building is currently using Vapour Compression Refrigeration (VCR) chillers for meeting its cooling requirements. The following are the existing data pertaining to the building. Existing System: Total Power drawn from grid for the whole building including chiller loads : 1300 kW Grid Power required for VCR : 300 kW Building cooling load : 7,56,000 kCal/hr Cost of Grid Power : Rs.8 /kWh The management proposes to install a natural gas engine with a Waste Heat Recovery Boiler (WHRB), which will generate power as well as steam for an operating Vapour Absorption Machine (VAM). A part of the total chilling load and power requirement of the building is proposed to be met by this cogeneration system. The following are the data for the proposed system. Proposed System: Total power generated from gas engine co-gen plant : 1000 kW Gas engine efficiency : 40 % Heat absorbed for steam generation in WHRB (as a % of heat input to gas engine) : 21 % Specific steam consumption for VAM : 5 kg/TR Calorific value of Natural Gas : 8500 kcal/sm3 Cost of Natural Gas : Rs.40/sm3 Annual operating hours : 4000 Total enthalpy of steam : 660 kCal/kg Feed water temperature to WHRB : 60 0C Calculate the following: Cost of generating one unit of electricity from the gas engine ? (5 marks) TR generated from Vapour Absorption Chiller driven by WHRB generated steam? (5 Marks) Total energy cost of existing & proposed system and state whether the proposed scheme is viable ? Ans 1. Cost of generating one unit of electricity from gas engine? Fuel Consumption = 1000 kW X 860 / ( 0.4 X 8500) = 252.94 sm3/hr Cost per unit of electricity from gas engine = (252.94 sm3/hr X 40 Rs./ sm3) / 1000 kW (10 marks) = Rs.10.12/ kWh 2. TR generated from VAM driven by WHRB generated steam? Heat absorbed by WHRB for Steam generation = 21% x (252.94 x 8500) = 451497.9 kcal /hr Amount of steam generated = 451497.9/(660-60) = 752.49 kg/hr TR generated by VAM = 752.49/5 = 150.49 TR 3. Techno-economic viability of the proposed scheme? Present cost of Electricity (Grid) = 1300 x 8 = 10,400 Rs./hr Proposed Scheme Cost of NG for Electricity = 252.94 sm3/hr X 40 Rs./ sm3 = 10,118 Rs. /hr TR required by the building = 756000/3024 = 250 TR Energy performance of chiller (VCR) Cost of Electricity from Grid to meet the balance chiller load = (250-150.49) x 1.2 x 8 = 955.30 Rs./ hr Total energy cost with proposed system = 300/250 = 1.2 kW/TR = 10118 +955.30 = 11,073 Rs./hr Proposed project is not viable, because total cost is more in proposed scheme than in present scheme. OR D) In a textile process unit, a five chamber stenter is installed for drying the cloth. The inlet and outlet conditions of the cloth are shown in the figure below. The production output of the stenter is 10,000 kgs/day. The heat input to the stenter is provided by a thermic fluid heater fired by fire wood as fuel. Gross Calorific Value (GCV) of Fire Wood is 3500 kcal/kg. The efficiency of the thermic fluid heater is 70% and distribution loss in the thermic fluid system is 20,000 kcal/hr. The average fire wood consumption rate is 427 kg/hr. Calculate the following: (each carries 10 Marks) a) Drier efficiency b) Fuel savings in thermic fluid heater if the inlet moisture is reduced from 60 % to 50 % by mechanical squeezing. (Assume that drier efficiency does not change) Ans Calculation of Stenter Efficiency Stenter Efficiency, % = W x ( min mout) x {( Tout - Tin ) + 540} Qin x 100 OR Stenter Efficiency, % = (moisture removed from fabric) x {( Tout - Tin ) + 540} Qin x 100 Where, W =Bone dry weight of the fabric, kg/hr min = kg moisture / kg bone dry fabric at inlet mout = kg moisture / kg bone dry fabric at outlet Qin = Thermal energy input to the stenter (kCal/hr) (a) Drier Efficiency Amount of moisture removed: Bone dry weight of fabric Hence, Total weight of inlet fabric Inlet Moisture weight = 14,250 kg/day = (10,000 x (95%)) = 9500 kg/day = (9500)/(0.4/1.0) = 23,750 kg/day = (23,750 9500) OR min = (14,250 kg/day) / (9500 kg/day) = 1.50 kg/kg dry fabric Outlet Moisture weight = 10000 - 9500 = 500 kg/day OR mout = (500 kg/day) / (9500 kg/day) = 0.053 kg/kg dry fabric Moisture removed from fabric in Stenter min -mout = 1.5 0.053 = 1.447 kg/kg dry fabric W x (min -mout) = 9500 x 1.447 =13747 kg/day = 573 kg/hr OR Moisture at I/L moisture at O/L = (14,250-500) = 13,750 kg/day = 573 kg/hr Heat required for removing the moisture = 573 x {(80-28) + 540} = 3,39,216 kcal/hr Heat input to the Thermic Fluid = (Firewood consumption rate x Calorific value x thermic fluid heater efficiency) = (427 kg/hr x 3500 kcal/kg x 70%) = 10,46,150 kcal/hr Heat input to the Drier Stenter Efficiency (%) stenter) = 10,46,150 20,000 = 10,26,150 kcal/hr = (Heat reqd. for removing moisture / Heat input to the = (3,39,216 / 10,26,150) x 100 = 33% (b) Fuel savings if the inlet moisture reduced from 60 to 50% Moisture removed in Drier with 50% input moisture (At 50 % moisture : Bone dry weight = moisture weight = 9500 kg) Moisture removed from the fabric in the Stenter = Inlet moisture Outlet moisture = (9500-500) = 9000 kg/day = 375 kg/hr Drier efficiency, 33 % = [375 kg/hr x {(80-28) 0C + 540 kCal/kg} ] / [(Fuel consumption kg/hr x 0.70 x 3500 kCal/kg) 20000kCal/hr] Therefore, Fuel consumption, kg/hr (for reduction of inlet moisture) = [{(375 kg/hr x {(80-28) 0C + 540 kCal/kg}) /0.33} + 20000] / (0.70 x 3500) = (672727 + 20000)/ (0.70 x 3500) = 283 kg/hr Fuel Savings = 427 283 = 144 kg/hr = 3456 kg/day . End of Section - III

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