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# MEM 310 Thermodynamic Analysis I Final Exam Example and Answer MEM 238

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This 11 page Study Guide was uploaded by Haikal Fouzi on Monday January 11, 2016. The Study Guide belongs to MEM 238 at Drexel University taught by Dr. Sorin Siegler in Winter 2016. Since its upload, it has received 102 views. For similar materials see Dynamics in Engineering and Tech at Drexel University.

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Date Created: 01/11/16

MEM 310 Dr. Ying Sun Practice Final Exam (120 mins) Thermodynamic Analysis I – MEM 310 Spring 2015 Permitted Aids: textbook, calculator, one handwritten 8.5” x 11” sheet of paper 1. (25 pts) An ideal Otto cycle with air as the working fluid has a compression ratio of 9. The minimum and maximum temperatures in the cycle are 280 K and 1160 K. Using air-standard assumptions, accounting for the variation of specific heats with temperature, and without interpolation (use the nearest table value), determine (a, 10 pts) the amount of heat transferred to the air during the heat-addition process, (b, 10 pts) the thermal efficiency, and (c, 5 pts) the thermal efficiency of a Carnot cycle operating between the same temperature limits. Page 1 of 4 MEM 310 - Midterm 2. (30 pts) A refrigerator uses refrigerant-134a as . QH the working fluid and operates on the ideal vapor-compression refrigeration cycle except 80°C for the compression process. The refrigerant Condenser 3 2 enters the evaporator at 160 kPa with a quality Expansion . of 0.2007 and leaves the compressor at 80°C. If valve W in the compressor consumes 520W of power, Compressor determine 1 (a, 10 pts) the condenser pressure, 4 (b, 10pts) the mass flow rate of the refrigerant, Evaporator (c, 10 pts) the COP of the refrigerator. 160 kPa . x=0.2007 QL Page 2 of 4 MEM 310 - Midterm 3. (25 pts) The volumetric analysis of a mixture of gases is 20 percent argon, 35 percent nitrogen, 15 percent carbon dioxide, and 30 percent methane. This mixture is heated from 10°C to 280°C while flowing through a tube in which the pressure is maintained at 150 kPa. Determine the (a) the constant-pressure specific heat of the mixture and (b) heat transfer of the mixture per unit mass of the mixture. Page 3 of 4 MEM 310 - Midterm 4. (20 points) The adiabatic compressor compresses saturated water vapor at 50°C to 800 kPa and 1000°C. What is the isentropic efficiency of the compressor? Page 4 of 4 MEM 310 Dr. Ying Sun Practice Final Exam (120 mins) Thermodynamic Analysis I – MEM 310 Spring 2015 Permitted Aids: textbook, calculator, one handwritten 8.5” x 11” sheet of paper 1. (25 pts) An ideal Otto cycle with air as the working fluid has a compression ratio of 9. The minimum and maximum temperatures in the cycle are 280 K and 1160 K. Using air-standard assumptions, accounting for the variation of specific heats with temperature, and without interpolation (use the nearest table value), determine (a, 10 pts) the amount of heat transferred to the air during the heat-addition process, (b, 10 pts) the thermal efficiency, and (c, 5 pts) the thermal efficiency of a Carnot cycle operating between the same temperature limits. Page 1 of 7 MEM 310 - Midterm An ideal Otto cycle with air as the working fluid has a compression ratio of 9. The amount of heat transferred to the air during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are P 1160K negligible. 3 Air is an ideal gas with variable specific 3 heats. Properties The properties of air are given in Table A-17. qin 4 Analysis (a) Process 1-2: isentropic compression. 2 q out 280K u1=199.75kJ/kg 1 T1= 280K ! !→ v vr1= 738.0 v v 2 1 1 v r2= v 2 = v r2= (738.0 = 82.0 !!→ u = 281kJ /kg v 1 r 9 Process 2-3: v = constant heat addition. u3= 897.9kJ /kg T3=1160K " "→ v =16.06 r3 qin u 3u = 297.9− 481.0 = 416.9kJ /kg (b) Process 3-4: isentropic expansion. v4 vr4 v 3 rv =39 ( )(6 =144)54 !! → u4= 381.8 kJ/kg v3 Process 4-1: v = constant heat rejection. qout= u4−u =1381.8−199.8 =182kJ /kg q out 182 η th− q =1− 416.9 = 56.3% in (c) The thermal efficiency of a Carnot cycle operating between the same temperature limits is TL 280K η th,C1− =1− = 75.9% T H 1160K Page 2 of 7 MEM 310 - Midterm 2. (30 pts) A refrigerator uses refrigerant-134a as . QH the working fluid and operates on the ideal vapor-compression refrigeration cycle except 80°C for the compression process. The refrigerant Condenser 3 2 enters the evaporator at 160 kPa with a quality Expansion . of 0.2007 and leaves the compressor at 80°C. If valve W in the compressor consumes 520W of power, Compressor determine 1 (a, 10 pts) the condenser pressure, 4 (b, 10pts) the mass flow rate of the refrigerant, Evaporator (c, 10 pts) the COP of the refrigerator. 160 kPa . x=0.2007 QL Page 3 of 7 MEM 310 - Midterm A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor- compression refrigeration cycle except for the compression process. The mass flow rate of the refrigerant, the condenser pressure, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) (b) From the refrigerant-134a tables (Tables A-11 through A-13) P =160kPa# 4 h = 73.19 kJ/kg x4= 0.20 ! 4 h3= h 4 h3= 73.33kJ/kg# " P3= 500kPa x3= 0(sat. liq.)! P = P 2 3 P = 500kPa # 2 "h2= 320.80kJ/kg T2= 80°C ! P1= P 4160kPa# "h1= 241.11kJ/kg x1=1(sat. vap.) ! The mass flow rate of the refrigerant is determined from m = W in = 0.52kW = 0.00653kg/s h −h (320.80-241.11)kJ/kg 2 1 (c) The refrigeration load and the COP are Q L m(h −1h ) 4 = (0.00653kg/s)(241.11−73.19)kJ/kg =1.097kW Q 1.097kW COP = L = = 2.11 Win 0.52kW Page 4 of 7 MEM 310 - Midterm 3. (25 pts) The volumetric analysis of a mixture of gases is 20 percent argon, 35 percent nitrogen, 15 percent carbon dioxide, and 30 percent methane. This mixture is heated from 10°C to 280°C while flowing through a tube in which the pressure is maintained at 150 kPa. Determine the (a) the constant-pressure specific heat of the mixture and (b) heat transfer of the mixture per unit mass of the mixture. Page 5 of 7 MEM 310 - Midterm Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of O , N , CO , and CH are 32.0, 28.0, 44.0, and 16.0 kg/kmol, 2 2 2 4 respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 0.918, 1.039, 0.846, and 2.2537 kJ/kg⋅K, respectively (Table A-2a). Analysis We consider 100 kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are m = NArMAr =(20 kmol)(40 kg/kmol)=800 kg Ar m N2N MN2 N2=(35 kmol)(28 kg/kmol)= 980 kg m CO2= NCO2M CO2=(15 kmol)(44 kg/kmol)= 660 kg m CH4= NCH4M CH4=(30 kmol)(16 kg/kmol)= 480 kg The total mass is m m mAr + m N2+ mCO2 + mCH4 = 960+1120+ 440+320 = 2920kg Then the mass fractions are m 800kg mf Ar= Ar= = 0.274 m m 2920kg m N2 980kg mf N2= = = 0.336 mm 2920kg mf = m CO2= 660kg = 0.226 CO2 m 2920kg m m 480kg mf CH4= CH4= = 0.164 m m 2920kg The constant-pressure specific heat of the mixture is determined from cp= mf Ar p,Ar mf N2 p,N2+ mfCO2cp,CO2 mf CH4 p,CH4 = 0.274×0.520+0.336×1.039+0.226×0.846+0.164×2.2537 =1.05kJ/kg⋅K An energy balance on the tube gives qin c pT 2T )1= (1.05kJ/kg⋅K)(280−10)K = 284kJ/kg Page 6 of 7 MEM 310 - Midterm 4. (20 points) The adiabatic compressor compresses saturated water vapor at 50°C to 800 kPa and 1000°C. What is the isentropic efficiency of the compressor? Page 7 of 7

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