9-13The three processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the back work ratio and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air are given as R = 0.287 kJ/kg.K, c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4.Analysis (a) The P -v and T -s diagrams of the cycle are shown in the figures. (b) The temperature at state 2 is K 2100kP a100kP a 700K) 300(1212===P P T TK 210023==T TDuring process 1-3, we havekJ/kg516.600)K 21K)(300kJ/kg 287.0()()(3131113,13=-⋅-=--=--=-=⎰-T T R P Pd w in V V vDuring process 2-3, we havekJ/kg8.1172n7K)(2100)Kl kJ/kg 287.0(7ln 7ln ln22233232,32=⋅======⎰⎰-RT RT RT d RTPd w out V VV V v Vv The back work ratio is then0.440===--kJ/kg8.1172kJ/kg6.516,32,13outin bw w w rHeat input is determined from an energybalance on the cycle during process 1-3,kJ/kg2465kJ/kg 1172.8300)K)(2100kJ/kg 718.0()(,3213,3231,3131,32,31=+-⋅=+-=+∆=-∆=--------outv outin out in w T T c w u q u w qThe net work output issvkJ/kg 2.6566.5168.1172,13,32=-=-=--in out net w w w(c) The thermal efficiency is then26.6%====266.0kJ2465kJ656.2in net th q w η9-21An air-standard cycle executed in a piston-cylinder system is composed of threespecified processes. The cycle is to be sketcehed on the P -v and T -s diagrams; the heat and work interactions and the thermal efficiency of the cycle are to bedetermined; and an expression for thermal efficiency as functions of compression ratio and specific heat ratio is to be obtained.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air are given as R = 0.3 kJ/kg·K and c v = 0.3 kJ/kg·K. Analysis (a) The P -v and T -s diagrams of the cycle are shown in the figures. (b) Noting that1.4297.00.1KkJ/kg 0.13.07.0===⋅=+=+=vv c c k R c c p pProcess 1-2: Isentropic compressionK 4.584)5)(K 293(429.01112112===⎪⎪⎭⎫ ⎝⎛=--k k r T T T vvkJ/kg 204.0=-⋅=-=-K )2934.584)(K kJ/kg 7.0()(12in 2,1T T c w v0=-21qFrom ideal gas relation,2922)5)(4.584(3212323==−→−===T r T T v v v v Process 2-3: Constant pressure heat additionkJ/kg701.3=-⋅=-=-==⎰-K )4.5842922)(K kJ/kg 3.0()()(2323232out 3,2T T R P Pd w v v vskJ/kg2338=-⋅=-=∆=∆+=----K )4.5842922)(K kJ/kg 1()(233232,32in 3,2T T c h u w q p outProcess 3-1: Constant volume heat rejectionkJ/kg 1840.3=⋅=-=∆=--K 293)-K)(2922kJ/kg 7.0()(1331out 1,3T T c u q v0=-13w(c) Net work isK kJ/kg 3.4970.2043.701in 2,1out 3,2net ⋅=-=-=--w w wThe thermal efficiency is then21.3%====213.0kJ2338kJ497.3in net th q w η9-32The two isentropic processes in an Otto cycle are replaced with polytropic processes.The heat added to and rejected from this cycle, and the cycle’s thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa·m 3/kg·K, c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The temperature at the end of the compression isK 4.537K)(8) 288(13.11112112===⎪⎪⎭⎫ ⎝⎛=---n n r T T T vvAnd the temperature at the end of the expansion isK 4.78981K) 1473(113.11314334=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=---n n r T T T vvThe integral of the work expression for the polytropic compression giveskJ/kg 6.238)18(13.1K) K)(288kJ/kg 287.0(1113.1121121=--⋅=⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛-=---n n RT w vvSimilarly, the work produced during the expansion iskJ/kg 0.65418113.1K) K)(1473kJ/kg 287.0(1113.1143343=⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎭⎫⎝⎛-⋅-=⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛--=---n n RT w vv Application of the first law to each of the four processes giveskJ/kg 53.59K )2884.537)(K kJ/kg 718.0(kJ/kg 6.238)(122121=-⋅-=--=--T T c w q v kJ/kg 8.671K )4.5371473)(K kJ/kg 718.0()(2332=-⋅=-=-T T c q vkJ/kg 2.163K )4.7891473)(K kJ/kg 718.0(kJ/kg 0.654)(434343=-⋅-=--=--T T c w q vkJ/kg 0.360K )2884.789)(K kJ/kg 718.0()(1414=-⋅=-=-T T c q vThe head added and rejected from the cycle arekJ/kg419.5kJ/kg 835.0=+=+==+=+=----0.36053.592.1638.6711421out 4332in q q q q q qThe thermal efficiency of this cycle is then0.498=-=-=0.8355.41911in out th q q η9-37An ideal Otto cycle with air as the working fluid has a compression ratio of 8. Theamount 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 negligible. 3 Air is an ideal gas with variable specific heats.Properties The properties of air are given in Table A-17E. Analysis (a) Process 1-2: isentropic compression.32.144Btu/lbm92.04R 540111==−→−=r u T v()Btu/lbm 11.28204.1832.144811212222=−→−====u r r r r v v v v v Process 2-3: v = constant heat addition.Btu/lbm241.42=-=-===−→−=28.21170.452419.2Btu/lbm452.70R 240023333u u q u T in r vvP(b) Process 3-4: isentropic expansion.()()Btu/lbm 205.5435.19419.28434334=−→−====u r r r r v v v v v Process 4-1: v = constant heat rejection.Btu/lbm 50.11304.9254.20514out =-=-=u u q53.0%=-=-=Btu/lbm241.42Btu/lbm113.5011in out th q q η (c) The thermal efficiency of a Carnot cycle operating between the same temperature limits is 77.5%=-=-=R2400R54011C th,H L T T η9-40The expressions for the maximum gas temperature and pressure of an ideal Otto cycleare to be determined when the compression ratio is doubled.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis The temperature at the end of the compression varies with the compression ratio as1112112--=⎪⎪⎭⎫⎝⎛=k k r T T T v vsince T 1 is fixed. The temperature rise during thecombustion remains constant since the amount of heat addition is fixed. Then, the maximum cycle temperature is given by11in 2in 3//-+=+=k r T c q T c q T v vThe smallest gas specific volume during the cycle isr13v v =When this is combined with the maximum temperature, the maximum pressure is given by ()11in 1333/-+==k r T c qRrRT P v v v9-47An ideal diesel cycle has a compression ratio of 20 and a cutoff ratio of 1.3. The maximum temperature of the air and the rate of heat addition are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K, c v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis()K 6.95420K) 288(14.11112112===⎪⎪⎭⎫ ⎝⎛=---k k r T T T vvK 1241===⎪⎪⎭⎫ ⎝⎛=K)(1.3) 6.954(22323c r T T T vv Combining the first law as applied to the various processes with the process equations gives6812.0)13.1(4.113.12011)1(1114.111.41th =---=---=--c k c k r k r r ηAccording to the definition of the thermal efficiency,kW 367===0.6812kW 250th net inηW Q9-59An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 1.4. The net work,heat addition, and the thermal efficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.3704 psia·ft 3/lbm.R (Table A-1E), c p = 0.240 Btu/lbm·R, c v = 0.171 Btu/lbm·R, and k = 1.4 (Table A-2Ea).Analysis Working around the cycle, the germane properties at the various states are()R 158015R) 535(14.11112112===⎪⎪⎭⎫ ⎝⎛=---k k r T T T vvout()psia 2.62915psia) 2.14(4.112112===⎪⎪⎭⎫ ⎝⎛=k kr P P P vvpsia 1.692psia) 2.629)(1.1(23====P r P P p xR 1738psia 629.2psia 692.1R) 1580(22=⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=PP T T xxR 2433R)(1.4) 1738(33===⎪⎪⎭⎫⎝⎛=c x xx r T T T vvR 2.942151.4R) 2433(14.11314334=⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=---k c k rr T T T vvApplying the first law to each of the processes givesBtu/lbm 7.178R )5351580)(R Btu/lbm 171.0()(1221=-⋅=-=-T T c w v Btu/lbm 02.27R )15801738)(R Btu/lbm 171.0()(22=-⋅=-=-T T c q x x vBtu/lbm 8.166R )17382433)(R Btu/lbm 240.0()(33=-⋅=-=-x p x T T c qB t u /l b 96.47R )17382433)(R Btu/lbm 171.0(Btu/lbm 8.166)(333=-⋅-=--=--x x x T T c q w vBtu/lbm 9.254R )2.9422433)(R Btu/lbm 171.0()(4343=-⋅=-=-T T c w vThe net work of the cycle isBtu/lbm 124.2=-+=-+=---7.17896.479.25421343net w w w w x and the net heat addition isBtu/lbm 193.8=+=+=--8.16602.2732in x x q q q Hence, the thermal efficiency is0.641===Btu/lbm193.8Btu/lbm124.2in net th q w η9-61An expression for cutoff ratio of an ideal diesel cycle is to be developed.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potentialenergy changes are negligible. 3 Air is an ideal gas with constant specific heats. Analysis Employing the isentropic process equations,112-=k rT Toutwhile the ideal gas law gives1123T r r r T T k c c -==When the first law and the closed system work integral is applied to the constant pressure heat addition, the result is)()(111123in T r T r r c T T c q k k c p p ---=-=When this is solved for cutoff ratio, the result is11in1T r c q r k p c -+=9-81A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.Properties The properties of air are given in Table A-17E. Analysis (a ) Noting that process 1-2 is isentropic,T h P r 11112147=−→−==520R124.27Btu /lbm .()()Btu/lbm 240.11 147.122147.110221212==−→−===h T P P P P r r R 996.5(b ) Process 3-4 is isentropic, and thus()Btu/lbm38.88283.26571.504Btu/lbm115.8427.12411.240Btu/lbm 265.834.170.1741010.174Btu/lbm 504.71R 200043out T,12inC,43433343=-=-==-=-==−→−=⎪⎭⎫⎝⎛====−→−=h h w h h w h P P P P P h T r r rThen the back-work ratio becomess200052048.5%===Btu/lbm238.88Btu/lbm115.84outT,in C,bw w w r(c ) 46.5%====-=-==-=-=Btu/lbm264.60Btu/lbm123.04Btu/lbm123.0484.11588.238Btu/lbm264.6011.24071.504inout net,th in C,out T,out net,23in q w w w w h h q η9-87A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10.The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.Properties The properties of air are given in Table A-17E. Analysis (a ) Noting that process 1-2 is isentropic,T h P r 11112147=−→−==520R124.27Btu /lbm .()()Btu/lbm 240.11 147.122147.110221212==−→−===h T P P P P r r R 996.5(b ) Process 3-4 is isentropic, and thus()Btu/lbm38.88283.26571.504Btu/lbm115.8427.12411.240Btu/lbm 265.834.170.1741010.174Btu/lbm 504.71R 200043out T,12inC,43433343=-=-==-=-==−→−=⎪⎭⎫⎝⎛====−→−=h h w h h w h P P P P P h T r r rThen the back-work ratio becomes48.5%===Btu/lbm238.88Btu/lbm115.84outT,in C,bw w w rs2000520(c ) 46.5%====-=-==-=-=Btu/lbm264.60Btu/lbm123.04Btu/lbm123.0484.11588.238Btu/lbm264.6011.24071.504inout net,th in C,out T,out net,23in q w w w w h h q η(d) The expression for the cycle thermal efficiency is obtained as follows:⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛-=⎪⎭⎫⎝⎛---=⎪⎪⎭⎫ ⎝⎛---=-⎪⎪⎭⎫ ⎝⎛--=---=----=-==-----------1111111111111111111231223in in 2,1out 3,2in net th 11)1(11111)1(11)1(1)1(1)()()()()(k k p k p k p k k v p k k p k v p p v r r k k r r k c R r T T r k c R r r T c r T T r T c c R r T r rT c T r T c c RT T c T T c T T R q w w q w η since 111kc c c c c c R p v p v p p -=-=-=。
