工程热力学第三版答案【英文】第9章
第九章工程热力学思考题答案
第九章气体动力循环1、从热力学理论瞧为什么混合加热理想循环的热效率随压缩比ε与定容增压比λ的增大而提高,随定压预胀比ρ的增大而降低？答:因为随着压缩比ε与定容增压比λ的增大循环平均吸热温度提高,而循环平均放热温度不变,故混合加热循环的热效率随压缩比ε与定容增压比λ的增大而提高。
混合加热循环的热效率随定压预胀比ρ的增大而减低,这时因为定容线比定压线陡,故加大定压加热份额造成循环平均吸热温度增大不如循环平均放热温度增大快,故热效率反而降低。
2、从内燃机循环的分析、比较发现各种理想循环在加热前都有绝热压缩过程,这就是否就是必然的？答:不就是必然的,例如斯特林循环就没有绝热压缩过程。
对于一般的内燃机来说,工质在气缸内压缩,由于内燃机的转速非常高,压缩过程在极短时间内完成,缸内又没有很好的冷却设备,所以一般都认为缸内进行的就是绝热压缩。
3、卡诺定理指出两个热源之间工作的热机以卡诺机的热效率最高,为什么斯特林循环的热效率可以与卡诺循环的热效率一样？答:卡诺定理的内容就是:在相同温度的高温热源与相同温度的低温热源之间工作的一切可逆循环,其热效率都相同,与可逆循环的种类无关,与采用哪一种工质无关。
定理二:在温度同为T1的热源与同为T2的冷源间工作的一切不可逆循环,其热效率必小于可逆循环。
由这两条定理知,在两个恒温热源间,卡诺循环比一切不可逆循环的效率都高,但就是斯特林循环也可以做到可逆循环,因此斯特林循环的热效率可以与卡诺循环一样高。
4、根据卡诺定理与卡诺循环,热源温度越高,循环热效率越大,燃气轮机装置工作为什么要用二次冷却空气与高温燃气混合,使混合气体降低温度,再进入燃气轮机？答:这就是因为高温燃气的温度过高,燃气轮机的叶片无法承受这么高的温度,所以为了保护燃气轮机要将燃气降低温度后再引入装置工作。
同时加入大量二次空气,大大增加了燃气的流量,这可以增加燃气轮机的做功量。
5、卡诺定理指出热源温度越高循环热效率越高。
工程热力学第三版答案【英文】第4章
4-4The work done during the isothermal process shown in the figure is to be determined. Assumptions The process is quasi-equilibrium. Analysis From the ideal gas equation,vRTP =For an isothermal process,/kg m 0.6kP a200kP a 600/kg)m (0.2331221===P P v v Substituting ideal gas equation andthis result into the boundary work integral produceskJ395.5-=⎪⎪⎭⎫⎝⎛⋅====⎰⎰333312112121out ,m kP a 1kJ 1m 0.6m 0.2ln )m kP a)(0.6 kg)(200 (3lnv v v vvv mP d mRTd P W b The negative sign shows that the work is done on the system.4-9Refrigerant-134a in a cylinder is heated at constant pressure until its temperature risesto a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium.Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-11 through A-13)/kg m 0.052427C 07kP a 005/kg m 0.0008059liquid Sat.kP a 00532223kPa 005@11=⎭⎬⎫︒====⎭⎬⎫=v v v T P P fAnalysiskg 04.62/kgm 0.0008059m 0.053311===v V m andvPkJ1600=⎪⎪⎭⎫⎝⎛⋅-=-=-==⎰33121221out ,m kPa 1kJ1/kg m 0.0008059)427kPa)(0.052 kg)(500 (62.04)()( v v V VV mP P d P W b Discussion The positive sign indicates that work is done by the system (work output).4-23A saturated water mixture contained in a spring-loaded piston-cylinder device is heated until the pressure and temperature rises to specified values. The work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The initial state is saturated mixture at 90︒C. The pressure and the specific volume at this state are (Table A-4),/kgm 23686.0)001036.03593.2)(10.0(001036.0kP a183.70311=-+=+==fgf x P v v vThe final specific volume at 800 kPa and 250°C is (Table A-6)/kg m 29321.032=vSince this is a linear process, the work done is equal to the area under the process line1-2:kJ24.52=⎪⎭⎫ ⎝⎛⋅-+=-+==331221out ,m kPa 1kJ 1)m 23686.01kg)(0.2932 (12)kPa 800(70.183)(2Area v v m P P W b4-29An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and the process is to be shown on a P-v diagram.Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The device is well-insulated and thus heat transfer is negligible. 3The energy stored in the resistance wires, and the heat transferred to the tank itself is negligible.Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as)(V 0)=PE =KE (since )(1212in ,energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in u u m t I Q u u m U W E E E e -=∆=-=∆=∆=-The properties of water are (Tables A-4 through A-6)()[]()k J /k g 2569.7v a p o rs a t ./k g m 0.29065k J /k g980.032052.30.25466.97/km 0.290650.0010531.15940.250.001053k J/k g 3.2052,97.466/kgm 1.1594,001053.025.0kP a 150/kg m 0.29065@2312113113113==⎪⎭⎪⎬⎫===⨯+=+==-⨯+=+=====⎭⎬⎫==g fg f fg f fg f g f u u u x u u x u u x P v v v v v v v Substituting,min60.2==∆⎪⎪⎭⎫⎝⎛-=∆s 33613kJ/s 1VA 1000.03)kJ/kg 980kg)(2569.7 (2)A 8)(V 110(t t4-39A saturated water mixture contained in a spring-loaded piston-cylinder device isheated until the pressure and volume rise to specified values. The heat transfer and the work done are to be determined.Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed asv)(0)=PE =KE (since )(12ou ,in 12ou ,in energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in u u m W Q u u m U W Q E E E t b t b -+=-=∆=-∆=-The initial state is saturated mixture at 75 kPa.The specific volume and internal energy at this state are (Table A-5),kJ/kg30.553)8.2111)(08.0(36.384/kgm 1783.0)001037.02172.2)(08.0(001037.0131=+=+==-+=+=fg f fg f xu u u x v v vThe mass of water iskg 22.11/kgm 1783.0m 23311===v V m The final specific volume is/kg m 4458.0kg22.11m 53322===m V vThe final state is now fixed. The internal energy at this specific volume and 225 kPa pressure is (Table A-6) kJ/kg 4.16502=u Since this is a linear process, the work done is equal to the area under the process line 1-2:kJ 450=⎪⎭⎫⎝⎛⋅-+=-+==331221out ,m kP a 1kJ 1)m 2(52)kP a 225(75)(2Area V V P P W b Substituting into energy balance equation giveskJ 12,750=-+=-+=kJ/kg )30.553kg)(1650.4 22.11(kJ 450)(12out ,in u u m W Q b4-43Two tanks initially separated by a partition contain steam at different states. Now thepartition is removed and they are allowed to mix until equilibrium is established. The temperature and quality of the steam at the final state and the amount of heat lost from the tanks are to be determined.Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions.Analysis (a ) We take the contents of both tanks as the system. This is a closed system since no massenters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as[][]0)=PE =KE (since )()(1212out energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in =-+-=∆+∆=-∆=-W u u m u u m U U Q E E E B A B AThe properties of steam in both tanks at the initial state are (Tables A-4 through A-6)kJ/kg 7.2793/kg m 25799.0C 300kPa 1000,13,1,1,1==⎪⎭⎪⎬⎫︒==A A A A u T P v ()[]()kJ/kg4.15954.19270.50.66631/kg m 0.196790.0010910.392480.500.001091kJ/kg 4.1927,66.631/kgm .392480,001091.050.0C 1501,131,131,1=⨯+=+==-⨯+=+=====⎭⎬⎫=︒=fg f B fg f B fg f g f B u x u u x u u x T v v v v vThe total volume and total mass of the system arekg523m 106.1/kg)m 19679.0kg)( 3(/kg)m 25799.0kg)( 2(333,1,1=+=+==+=+=+=B A B B A A B A m m m m m v v V V VNow, the specific volume at the final state may be determined/kg m 22127.0kg5m 106.1332===m Vvwhich fixes the final state and we can determine other properties()kJ/kg8.12821.19820.3641.11561001073.060582.0001073.022127.0/kg m 22127.0kPa 0032222kPa 300 @sat 2322=⨯+=+==--=--=︒==⎪⎭⎪⎬⎫==fg f fg f u x u u x T T P 0.3641C 133.5v v v v v (b ) Substituting,[][]kJ 3959kJ/kg )4.15958.1282(kg) 3(kJ/kg )7.27938.1282(kg) 2()()(1212out -=-+-=-+-=∆+∆=-BA B A u u m u u m U U QorkJ 3959=out Q4-60The air in a rigid tank is heated until its pressure doubles. The volume of the tank and the amount of heat transfer are to be determined.Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -221︒F and 547 psia. 2 The kinetic and potential energy changes are negligible, ∆∆pe ke ≅≅0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. Properties The gas constant of air is R = 0.3704 psia.ft 3/lbm.R (Table A-1E).Analysis (a3ft 80.0=⋅⋅==psia50R) R)(540/lbm ft psia 4lbm)(0.370 (20311P mRT V(b) We take the air in the tank as our system. The energy balance for this stationary closed system can be expressed as)()(1212in in energiesetc. potential, kinetic, internal,in Change systemmassand work,heat,by nsferenergy tra Net out in T T mc u u m Q UQ E E E -≅-=∆=∆=-vThe final temperature of air isR 1080R) (540211222211=⨯==−→−=T P P T T P T P V V The internal energies are (Table A-17E)u u u u 12====@@540R 1080R 92.04Btu /lbm 186.93Btu /lbmSubstituting, Q in = (20 lbm)(186.93 - 92.04)Btu/lbm = 1898 BtuAlternative solutions The specific heat of air at the average temperature of T avg = (540+1080)/2= 810 R = 350︒F is, from Table A-2Eb, c v ,avg = 0.175 Btu/lbm.R. Substituting,Q in = (20 lbm)( 0.175 Btu/lbm.R)(1080 - 540) R = 1890 BtuDiscussion Both approaches resulted in almost the same solution in this case.4-64A student living in a room turns her 150-W fan on in the morning. The temperature in the room when she comes back 10 h later is to be determined.Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141︒C and 3.77 MPa. 2 The kinetic andQpotential energy changes are negligible, ∆∆ke pe ≅≅0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 4 All the doors and windows are tightly closed, and heat transfer through the walls and the windows is disregarded.Properties The gas constant of air is R = 0.287 kPa.m 3/kg.K (Table A-1). Also, c v = 0.718 kJ/kg.K for air at room temperature (Table A-2).Analysis We take the room as the system. This is a closed system since the doors and the windows are said to be tightly closed, and thus no mass crosses the system boundary during the process. The energy balance for this system can be expressed as)()(1212,,energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net T T mc u u m W UW E E E in e in e out in -≅-=∆=∆=-vThe mass of air iskg174.2K) K)(288/kg m kP a (0.287)m kP a)(144 (100m 14466433113=⋅⋅===⨯⨯=RT P m V VThe electrical work done by the fan isW W t e e==⨯= ∆(0.15kJ /s)(103600s)5400kJ Substituting and using the c v value at room temperature, 5400 kJ = (174.2 kg)(0.718 kJ/kg ⋅︒C)(T 2 - 15)︒CT 2 = 58.2︒CDiscussion Note that a fan actually causes the internal temperature of a confinedspace to rise. In fact, a 100-W fan supplies a room with as much energy as a 100-W resistance heater.4-69Carbon dioxide contained in a spring-loaded piston-cylinder device is heated. The work done and the heat transfer are to be determined.Assumptions 1 CO 2 is an ideal gas since it is at a high temperature relative to its critical temperature of 304.2 K. 2 The kinetic and potential energy changes are negligible, 0pe ke ≅∆≅∆. Properties The properties of CO 2 are R = 0.1889 kJ/kg ⋅K and c v = 0.657 kJ/kg ⋅K (Table A-2a ).PAnalysis We take CO 2 as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as)(12out ,in energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in T T mc U W Q E E E b -=∆=-∆=-vThe initial and final specific volumes are33111m 5629.0k P a 100K) K)(298/kg m kPa kg)(0.1889 (1=⋅⋅==P mRT V 33222m 1082.0k P a1000K)K)(573/kg m kPa kg)(0.1889 (1=⋅⋅==P mRT V Pressure changes linearly with volume and the work done is equal to the area underthe process line 1-2:kJ1.250m kPa 1kJ 1)m 5629.0(0.10822)kPa1000(100)(2Area 331221out ,-=⎪⎪⎭⎫⎝⎛⋅-+=-+==V V P P W b Thus,kJ 250.1=in ,b WUsing the energy balance equation,kJ 4.69K )25K)(300kJ/kg 657.0(kg) 1(kJ 1.250)(12out ,in -=-⋅+-=-+=T T mc W Q b vThus,kJ 69.4=out Q4-76Air at a specified state contained in a piston-cylinder device with a set of stops is heated until a final temperature. The amount of heat transfer is to be determined.Assumptions 1 Air is an ideal gas since it is at a high temperature relative to its critical temperature of 304.2 K. 2 The kinetic and potential energy changes are negligible, 0pe ke ≅∆≅∆. Properties The properties of air are R = 0.287 kJ/kg ⋅K and c v = 0.718 kJ/kg ⋅K (Table A-2a ). Analysis We take air as the system. This is a closedsystem since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as)(12out ,in energiesetc. potential, kinetic, internal,in Change systemmassand work,heat,by nsferenergy tra Net out in T T mc U W Q E E E b -=∆=-∆=-vThe volume will be constant until the pressure is 300 kPa:K 900kP a100kP a 300K) (3001212===P P T T The mass of the air iskg 4646.0K)K)(300/kg m kPa (0.287)m kPa)(0.4 (10033111=⋅⋅==RT P m V The boundary work done during process 2-3 iskJ04900)K -K)(1200/kg m kP a (0.287)kg 4646.0()()(323232out ,=⋅⋅=-=-=T T mR P W b V V Substituting these values into energy balance equation,kJ 340=-⋅+=-+=K )300K)(1200kJ/kg 718.0(kg) 4646.0(kJ 40)(13out ,in T T mc W Q b v4-85An egg is dropped into boiling water. The amount of heat transfer to the egg by the time it is cooked is to be determined.Assumptions 1 The egg is spherical in shape with a radius of r 0 = 2.75 cm. 2 The thermal properties of the egg are constant. 3 Energy absorption or release associated with any chemical and/or phase changes within the egg is negligible. 4 There are no changes in kinetic and potential energies.Properties The density and specific heat of the egg are given to be ρ = 1020 kg/m 3 and c p = 3.32 kJ/kg.︒C.Analysis We take the egg as the system. This is a closes system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as)()(1212egg in energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in T T mc u u m U Q E E E -=-=∆=∆=-Then the mass of the egg and the amount of heat transfer becomeBoilingPV (m 3)kJ 21.2=︒-︒=-=====C )880)(C kJ/kg. 32.3)(kg 0889.0()(kg0889.06m ) 055.0()kg/m 1020(612in 333T T mc Q D m p ππρρV。
工程热力学第三版课后习题答案
工程热力学第三版课后习题答案【篇一：工程热力学课后答案】章）第1章基本概念⒈闭口系与外界无物质交换，系统内质量将保持恒定，那么，系统内质量保持恒定的热力系一定是闭口系统吗? 答：否。
当一个控制质量的质量入流率与质量出流率相等时（如稳态稳流系统），系统内的质量将保持恒定不变。
⒉有人认为，开口系统中系统与外界有物质交换，而物质又与能量不可分割，所以开口系不可能是绝热系。
这种观点对不对，为什么?答：不对。
“绝热系”指的是过程中与外界无热量交换的系统。
热量是指过程中系统与外界间以热的方式交换的能量，是过程量，过程一旦结束就无所谓“热量”。
物质并不“拥有”热量。
一个系统能否绝热与其边界是否对物质流开放无关。
⒊平衡状态与稳定状态有何区别和联系，平衡状态与均匀状态有何区别和联系?答：“平衡状态”与“稳定状态”的概念均指系统的状态不随时间而变化，这是它们的共同点；但平衡状态要求的是在没有外界作用下保持不变；而平衡状态则一般指在外界作用下保持不变，这是它们的区别所在。
⒋倘使容器中气体的压力没有改变，试问安装在该容器上的压力表的读数会改变吗？在绝对压力计算公式p?pb?pe(p?pb); p?pb?pv(p?pb)中，当地大气压是否必定是环境大气压?答：可能会的。
因为压力表上的读数为表压力，是工质真实压力与环境介质压力之差。
环境介质压力，譬如大气压力，是地面以上空气柱的重量所造成的，它随着各地的纬度、高度和气候条件不同而有所变化，因此，即使工质的绝对压力不变，表压力和真空度仍有可能变化。
“当地大气压”并非就是环境大气压。
准确地说，计算式中的pb 应是“当地环境介质”的压力，而不是随便任何其它意义上的“大气压力”，或被视为不变的“环境大气压力”。
⒌温度计测温的基本原理是什么?答：温度计对温度的测量建立在热力学第零定律原理之上。
它利用了“温度是相互热平衡的系统所具有的一种同一热力性质”，这一性质就是“温度”的概念。
⒍经验温标的缺点是什么?为什么?答：由选定的任意一种测温物质的某种物理性质，采用任意一种温度标定规则所得到的温标称为经验温标。
工程热力学(第三版)习题答案全解可打印第九章
π
pV 400 ∴ q m = 400 × 1 es = × R g T1 60
可逆定温功压缩功率为：
Wc ,T = − p1V1 ln =−
p2 p1
400 8 π × 0.1 × 10 6 × 0.786 × × 0.3 2 × 0.2 × ln = 15.4 × 10 3 J = 15.4kW s 60 4 1
n = 3× p1 v1 (π n −1 t1 = 20°C
n −1 n
p4 12.5 =3 =5 p1 0 .1
（1） wc = 3wc , L
Q p1 = 0.1MPa
− 1) ∴v =
1 1.3−1 1.3
R g T1 p1
=
287 × 293.15 0.1× 10
6
= 0.8413 m
3
kg kg
− 1] =
1.4 × 0.1× 10 6 × 140 × [6 1.4 − 1
1.4 −1 1.4
− 1] = 327.9 × 10 5 J
h
327.9 ×10 5 = 9108.3W = 9.11KW 3600 p n p1V1 [( 2 ) n −1 p1
n −1 n
（3）多变压缩
Wt , n = Nn = − 1] = 1 .2 × 0.1×10 6 × 140 × [6 1 .2 − 1
V h = 0.009m 3
π =7
1 n
σ = 0.06
1 n
n = 1.3
1
（1） η v = 1 −
Vc (π Vh
− 1) = 1 − σ (π
− 1) = 1 − 0.06 × (7 1.3 − 1) = 0.792
工程热力学第三版电子教案第9章
第9章气体和蒸汽的流动9．1 基本要求 (85)9．2 本章难点 (85)9．3 例题 (85)9.4 思考及练习题 (93)9．5自检题 (97)9．1 基本要求1．深入理解喷管和扩压管流动中的基本关系式和滞止参数的物理意义，熟练运用热力学理论分析亚音速、超音速和临界流动的特点。
2．对于工质无论是理想气体或蒸汽，都要熟练掌握渐缩、渐缩渐扩喷管的选型和出口参数、流量等的计算。
理解扩压管的流动特点，会进行热力参数的计算。
3．能应用有摩擦流动计算公式，进行喷管的热力计算。
4．熟练掌握绝热节流的特性，参数的变化规律。
9．2 本章难点1．喷管和扩压管截面变化与速度、压力变化的关系。
2．喷管的选型与临界截面的关系。
39．3 例1：压力为降为5bar 解：由初压p 图（图9.1h 1=3350kJ/kgs 1=7.1kJ/(kg ·K)因绝热节流前、后焓相等，故由h 1=h 2及p 2可求节流后的蒸汽状态点2，查得t 2=440℃； s 2=7.49kJ/(kg ·K)因此，节流前后熵变量为Δs =s 2-s 1=7.94－7.1=0.84kJ/(kg ·K)Δs ＞0，可见绝热节流过程是个不可逆过程。
若节流流汽定熵膨胀至0.1bar ，由1h '=2250kJ/kg ，可作技术功为 kJ/kg 11002250335011=-='-h h若节流后的蒸汽定熵膨胀至相同压力0.1bar ，由图查得2h '=2512kJ/kg ，可作技术功为假定气体为理想气体，则：2(211010c T T c h h p =-=-）K111587.11141000089.12180100222110≈=⨯⨯+=+=pc c T T 应用等熵过程参数间的关系式得：11010-⎪⎪⎭⎫⎝⎛=k k T T p pbar 0525.1110011151136.136.111010=⎪⎭⎫ ⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=--k k T T p p喷管出口状态参数也可根据等熵过程参数之间的关系求得：11010-⎪⎪⎭⎫⎝⎛=k k T T p p即：136.136.121115343.00525.1-⎪⎪⎭⎫⎝⎛=T即喷管出口截面处气体的温度为828.67K 。
工程热力学第三版答案【英文】第10章
10-8Heat rejected decreases; everything else increases.The pump work remains the same, the moisture content decreases, everything else increases.The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) the same as the boiler inlet pressure in ideal cycles.10-16A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),Btu/lbm81.11140.240.109Btu/lbm40.2ft psia 5.404Btu 1 psia )3800)(/lbm ft 01630.0()(/lbmft 01630.0Btu/lbm40.109in p,1233121in p,3psia 3 @1psia 3 @1=+=+==⎪⎪⎭⎫⎝⎛⋅-=-=====w h h P P w h h f f v v vBtu/lbm 24.975)8.1012)(8549.0(40.1098549.06849.12009.06413.1 psia 3RBtu/lbm 6413.1Btu/lbm0.1456F 900psia 80044443443333=+=+==-=-=⎭⎬⎫==⋅==⎭⎬⎫︒==fgf fg fh x h h s s s x s s P s h T PKnowing the power output from the turbine the mass flow rate of steam in the cycle is determined fromlbm/s 450.3kJ 1Btu 0.94782)Btu/lbm 24.975(1456.0kJ/s 1750)(43out T,43out T,=⎪⎭⎫ ⎝⎛-=-=−→−-=h h W m h h m WThe rates of heat addition and rejection areBtu/s2987Btu/s 4637=-=-==-=-=Btu/lbm )40.109.24lbm/s)(975 450.3()(Btu/lbm )81.1110.6lbm/s)(145 450.3()(14out 23inh h m Q h h m Qand the thermal efficiency of the cycle is35.6%==-=-=3559.04637298711inout thQ Q η10-24A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heatsource. The mass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output from the turbine, and the thermal efficiency of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a ) We use properties of water for geothermal water (Tables A-4 through A-6)kJ/kg 14.990kP a 500 14.9900C 23022122111=-=⎭⎬⎫====⎭⎬⎫=︒=fg f h h h x h h P h x T The mass flow rate of steam through the turbine is===kg/s) 230)(1661.0(123m x m (b ) Turbine:kJ/kg 7.2344)1.2392)(90.0(81.19190.0kPa 10kJ/kg 3.2160kPa 10K kJ/kg 8207.6kJ/kg1.27481kPa 500444443443333=+=+=⎭⎬⎫===⎭⎬⎫==⋅==⎭⎬⎫==fg f s h x h h x P h s s P s h x P0.686=--=--=3.21601.27487.23441.27484343s T h h h h η (c ) The power output from the turbine iskW 15,410=-=-=kJ/kg )7.23448.1kJ/kg)(274 38.20()(433out T,h h mW (d ) We use saturated liquid state at the standard temperature for dead state enthalpykJ/kg 83.1040C 25000=⎭⎬⎫=︒=h x TkW 622,203kJ/kg )83.104.14kJ/kg)(990 230()(011in=-=-=h h m E7.6%====0.0757622,203410,15inout T,thE W η10-36An ideal reheat Rankine with water as the working fluid is considered. The temperatures at the inlet of both turbines, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),kJ/kg87.19806.781.191kJ/kg06.7m kPa 1kJ 1 kPa )107000)(/kg m 001010.0()(/kgm 001010.0kJ/kg81.191in p,1233121in p,3kPa 10 @1kPa 10 @1=+=+==⎪⎭⎫ ⎝⎛⋅-=-=====w h h P P w h h f f v v vC373.3︒==⎭⎬⎫==⋅=+=+==+=+=⎭⎬⎫==33433444444kJ/kg5.3085kP a 7000K kJ/kg 3385.6)6160.4)(93.0(0457.2kJ/kg0.2625)5.2047)(93.0(87.720 93.0kP a 800T h s s P s x s s h x h h x P fg f fg fC416.2︒==⎭⎬⎫==⋅=+=+==+=+=⎭⎬⎫==55655666666kJ/kg0.3302kP a 800K kJ/kg 6239.7)4996.7)(93.0(6492.0kJ/kg 4.2416)1.2392)(93.0(81.191 90.0kP a 10T h s s P s x s s h x h h x P fg f fg fThus, kJ/kg6.222481.1914.2416kJ/kg 6.35630.26250.330287.1985.3085)()(16out 4523in =-=-==-+-=-+-=h h q h h h h qand37.6%==-=-=3757.06.35636.222411in out th q q η10-38A steam power plant that operates on a reheat Rankine cycle is considered. The condenser pressure, the net power output, and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a()()()()()()3)(Eq. 2.335885.02.33582) (Eq. ?1) (Eq.95.0?K kJ/kg 2815.7kJ/kg2.3358C 450 MP a 2kJ/kg3.30271.29485.347685.05.3476kJ/kg 1.2948MP a 2K kJ/kg 6317.6kJ/kg5.3476C 550 MP a 5.12665566565656666655554334434343443333s s T sT s s T sT s s h h h h h h h h h h s s P h x P s h T P h h h h h h h h h s s P s h T P --=--=−→−--==⎭⎬⎫===⎭⎬⎫==⋅==⎭⎬⎫︒===--=--=→--==⎭⎬⎫==⋅==⎭⎬⎫︒==ηηηη The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EES from the above three equations:P 6 = 9.73 kPa , h 6 = 2463.3 kJ/kg,(b ) Then,()()()()kJ/kg59.20302.1457.189kJ/kg14.020.90/m kP a 1kJ 1kP a 73.912,500/kg m 0.00101//kgm 001010.0kJ/kg57.189in ,123121in ,3kPa 10 @1kPa 73.9 @1=+=+==⎪⎪⎭⎫⎝⎛⋅-=-=====p pp f f w h h P P w h h ηv v v Cycle analysis:()()kW 10,242==-==-=-==-+-=-+-=kg 2273.7)kJ/-.8kg/s)(3603 7.7()(kJ/kg7.227357.1893.2463kJ/kg 8.36033.24632.335859.2035.3476out in net16out 4523in q q m W h h q h h h h q (c ) The thermal efficiency is36.9%==-=-=369.0kJ/kg3603.8kJ/kg2273.711in out th q q η。
工程热力学第9章
1
1
1
1
1
1
1
57
讨论： k
t
b) ; t不变，但wnet
c ) 重负荷（q1 ）时内部热 效率下降，因温度上升使 ，造成热效率下降
58
汽油易爆燃
一般汽油机 5 10
柴油机 14 21
一般柴油机效率高于汽油机的效率 但汽油机小巧
9–4 活塞式内燃机各种理想循环的热力学比较
p Gasoline engine cycle
0
火花塞
Atmosphere
v
Gasoline engine cycle
p
进气阀门打开
0
火花塞
Atmosphere
v
Gasoline engine cycle
p
开始进气
0
火花塞
Atmosphere
v
Gasoline engine cycle
p
进气压力稍低于 大气压
在热力学基本定律的基 础上，分析循环能量转化的 经济性，合理安排循环，提 高热效率，寻求提高经济性 的方向及途径。
汽车发动机P－V图 2
动力循环的分类
按工质
internal combustion engine
气体动力循环：内燃机 空气为主的燃气 按理想气体处理
蒸汽动力循环：外燃机 水蒸气等实际气体
v3
v2
t
1
q2 q1
q2 cV T4 T1
t
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T4 T1
T3 T2
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q1 cp T3 T2
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工程热力学第三版答案【英文】第一章
1-2There is no truth to his claim. It violates the second law of thermodynamics. 1-14A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a relation is to be obtained for the filling time. Assumptions Gasoline is an incompressible substance and the flow rate is constant.Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the unit of time is ‘seconds’. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds. Putting the given information into perspective, we havet [s] [L], and [L/s}It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate. Therefore, the desired relation is Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities.1-25A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains constant is called isochoric.1-38The change in water temperature given in F unit is to be converted to C, K and R units.Analysis Using the conversion relations between the various temperature scales,1-49The pressure in a pressurized water tank is measured by a multi-fluid manometer. The gage pressure of air in the tank is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air-water interface.Properties The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kg/m3, respectively.Analysis Starting with the pressure at point 1 at the air-water interface, and moving along the tube by adding (as we go down) or subtracting (as we go up) th e terms until we reach point 2, and setting the result equal to P atm since the tube is open to the atmosphere givesSolving for P1,or,Noting that P1,gage = P1 - P atm and substituting,Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.1-55The pressure in chamber 1 of the two-piston cylinder shown in the figure is to be determined.Analysis Summing the forces acting on the piston in the vertical direction givesF1F2F3which when solved for P1 givessince the areas of the piston faces are given bythe above equation becomes1-63A gas contained in a vertical piston-cylinder device is pressurized by a spring and by the weight of the piston. The pressure of the gas is to be determined. Analysis Drawing the free body diagram of thepiston and balancing the vertical forces yieldF springP atmPW = mgThus,1-74Fresh and seawater flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer. The pressure difference between the two pipelines is to be determined.Assumptions 1 All the liquids areincompressible. 2 The effect of aircolumn on pressure is negligible.FreshWaterh wh seah airSeaWaterMercuryAirh HgProperties The densities of seawaterand mercury are given to be sea =1035 kg/m3 and Hg = 13,600 kg/m3.We take the density of water to be w=1000 kg/m3.Analysis Starting with the pressurein the fresh water pipe (point 1) andmoving along the tube by adding (aswe go down) or subtracting (as wego up) the terms until we reach thesea water pipe (point 2), and settingthe result equal to P2givesRearranging and neglecting the effect of air column on pressure, Substituting,Therefore, the pressure in the fresh water pipe is 3.39 kPa higher than the pressure in the sea water pipe.Discussion A 0.70-m high air column with a density of 1.2 kg/m3 corresponds to a pressure difference of 0.008 kPa. Therefore, its effect on the pressure difference between the two pipes is negligible.1-83A multi-fluid container is connected to a U-tube. For the given specific gravities and fluid column heights, the gage pressure at A and the height of a mercury column that would create the same pressure at A are to be determined.Assumptions 1 All the liquids areincompressible. 2 The multi-fluidcontainer is open to the atmosphere.A90 cm70 cm30 cm15 cmGlycerinSG=1.26OilSG=0.90Water20 cmProperties The specific gravities aregiven to be 1.26 for glycerin and 0.90for oil. We take the standard density ofwater to be w=1000 kg/m3, and thespecific gravity of mercury to be 13.6.Analysis Starting with the atmosphericpressure on the top surface of thecontainer and moving along the tube byadding (as we go down) or subtracting(as we go up) the terms until we reachpoint A, and setting the result equal toP A giveRearranging and using the definition of specific gravity,orSubstituting,The equivalent mercury column height isDiscussion Note that the high density of mercury makes it a very suitable fluid for measuring high pressures in manometers.1-109The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock. The mass of the petcock is to be determined.Assumptions There is no blockage of the pressure release valve.W = mgP atmPAnalysis Atmospheric pressure is acting on allsurfaces of the petcock, which balances itself out.Therefore, it can be disregarded in calculations ifwe use the gage pressure as the cooker pressure. Aforce balance on the petcock (F y = 0) yields1-119A relation for the air drag exerted on a car is to be obtained in terms of on the drag coefficient, the air density, the car velocity, and the frontal area of the car.Analysis The drag force depends on a dimensionless drag coefficient, the air density, the car velocity, and the frontal area. Also, the unit of force F is newton N, which is equivalent to kgm/s2. Therefore, the independent quantities should be arranged such that we end up with the unit kgm/s2 for the drag force. Putting the given information into perspective, we haveF D [ kgm/s2] C Drag [], A front [m2],[kg/m3], and V [m/s]It is obvious that the only way to end up with the unit “kgm/s2” for drag force is to multiply mass with the square of the velocity and the fontal area, with the drag coefficient serving as the constant of proportionality. Therefore, the desired relation isDiscussion Note that this approach is not sensitive to dimensionless quantities, and thus a strong reasoning is required.。
工程热力学第9章答案
p cr = υ cr p 0 = 0.528 × 2.5 = 1.32MPa <Pb
出口气流温度
(Pb=1.5 MPa)
所以，渐缩喷管的出口处的压力为 P2= Pb=1.5 MPa
k −1 k
⎛ P2 T2 = T0 ⎜ ⎜P ⎝ 0
⎞ ⎟ ⎟ ⎠
⎛ 1.5 ⎞ = 453.15 × ⎜ ⎟ ⎝ 2.5 ⎠
2(h0 − h2 ) = 2 × (3250 − 2890) × 1000 = 848.528m / s
喷管出口处流速 c f 2 ' = ϕc f 2 = 0.95× = 806.102m / s
2
第 9 章 气体和蒸汽的流动
出口处焓 h2 ' = h0 −
1 1 2 c f 2 ' = 3250 − × 806.102 2 / 1000 = 2925.10kJ / kg 2 2
A2 c 2 10 × 10 −4 × 351.51 = = 5.226kg / s v2 0.06726
出口质量流量为 q m =
出口空气温度 t 2 = T2 − 273.15 = 391.615 − 273.15 = 118.465 ℃ 9-6 如果进入喷管的蒸汽状态为 p1=2MPa，t1=400℃，喷管出口处的压力 p2=0.5MPa， 速度系数 ϕ =0.95，入口速度不计。试求喷管出口处蒸汽的速度和比体积。 解：则查水蒸汽的焓熵图，滞止焓 h0 = h1 = 3250kJ / kg 如果气流可逆绝热流动到压力 p2，则查水蒸汽的焓熵图，此时的焓 h2 = 2890kJ / kg 所以 c 2 =
(Pb=0.270 MPa)
1.4
p cr = υ cr P0 = 0.528 × 0.5334 = 0.2816MPa >Pb
工程热力学 课后习题答案 可打印 第三版 第九章
缩功； （4）压缩过程中传出的热量。 解： （1）多变指数
0.32 0.1 = 1.186 ∴n = = V 0 . ln( 1 ) ln 032 V2 0.012 ln(
2
p
Q p1v1 = p 2 v 2
n
n
p
)
ln
1
（2）空气终温
p T2 = T1 ( 2 ) p1
n −1 n
0.32 = (273.15 + 50) × ( ) 0.1
3
温升为 14℃。求（1）空气可能达到的最高压力； （2）压气机必需的功率。 解：
Q w = q m,W c∆t = q m, a 465 × 4.187 × 14 = 7.5715kW 3600 pV 0.1× 10 6 × 250 = 1 1 = = 0.08254 kg s R g T1 3600 × 287 × 293.15
9-4 三台空气压缩机的余隙容积比均为 6%， 进气状态均为 0.1MPa,27°C ， 出口压力为 0.5MPa ，
但压缩过程的指数分别为：n1=1.4、n2=1.25、n3=1,试求各压气机的容积效率（假设膨胀过程的 指数和压缩过程相同） 。 解： 据题意
Vc = 0.06 Vh 0.5 π= =5 0 .1
或
wc = p n RgT1 [( 2 ) p1 n −1
n −1 n
− 1] =
1.2 0.9 × 287 × 293.15 × [( ) 1.2 − 1 0.1
1.2 −1 1.2
− 1] = 2.23 × 10 5 J
kg
N = q m ,a ⋅ wc = 0.0825 × 2.23 × 10 5 = 18.4kW
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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 -=-=-=。