工程热力学英语习题.
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。
高等工程热力学试卷和答案
1True or false(20pts)1) Mass can cross the system boundary in the Open system .(T)2) Heat and work can not cross in the Closed system. (F)3)Intensive property is a property that is independent of the extent or mass of the system.(T)4)Heat exchangers are devices that transfer energy between fluids at different temperatures.(T)5) Heat engine can have a thermal efficiency of 100%. (F)6)Heat Engines are Devices that are used to convert heat to work.(T)7)The decrease in quality is always accompanied by an increase in entropy.(T)8)Reversible processes actually can occur in nature.F)9)Energy can be neither created or destroyed;it can only change forms.（T）10)Both Heat and Work are associated with a process and a state.(F)11)Once the temperature equality is established,the heat transfer stops.(T)12)The direction of heat transfer is always from the lower temperature body to the higher temperature.body.(F)13)Work is not a property of a system or its surroundings.(T)14)Closed Feedwater Heaters require a separate pump for each heater.(F)15)The reheat process in general can significantly change the cycle efficiency.(T)2Chosing a correct answer(30pts)1) A thermodynamic system generally has a boundary, this boundary （b）a)can be personally defined from your own interestsb)can only be defined following certain rolesc)is depend on the physical boundariesd)is independent on the physical as well as personal wills2) A system is called a closed system when （c）a)there is only heat exchange between the system and its surroundingsb)there is only work interactions between the system and its surroundingsc)there are both heat exchange and work interactoins between the system and its surroundingsd)there are neither heat exchange nor work interactions between the system and its surroundings3)The specific internal energy,u,other than U,of a system is determined by the（c）of the systema) volume b) pressure c) temperature d) mass4) An example of internal property of a system is （c？）a)mass b)total volume c)pressure d)enthalpy5) A steady state process is related to （a ？）a) a closed system with invarible propertiesb) a control volume with its property changesc) an open system keeping its properties constantd) a closed system having its properties changed6) A good approximation for a stream flowing through a porous plug is （a ）a) h in = h out b) P in = P out c)T in = T out d) a and b e) b and c7) Δ T= 0 is true （d ）a) for an adiabatic process. b) if no work is done.c) for an isentropic process. d) for a reversible process.8) Characteristics of Heat Engines include （e ）a) They receive heat from a high-temperature source.1b)They convert part of this heat to work.c)They reject the remaining waste heat to low-temperature sink.d) They operate on a cycle.e)all of above9) Quality （a）a) is defined by M /M + M ) liq liq vapb) is defined by v = x v + (1-x) v f gc)equals zero for a saturated vapor.d)is only defined in the two phase region.e)is the fraction of mass that is liquid.10) A good approximation for a stream flowing through a heat exchanger is （e？）a) h = h in out b) P = P in out c) T = T in outd) a and b e) b and c11)In an adiabatic process:（b）a)temperature cannot change.b)there is no heat flow.c)internal energy cannot change.d) a and be)a,b and c12) Which of the following is an extensive property ? （b）a)temperature b)weight c)compositiond)pressure e)none of the above13) What is the efficiency of an ideal Carnot heat engine cycle operating between 38 C and 482 C ? （c）a) 90 % b) 89 % c) 59 % d) 16 % e) 11 %14) Which of the following does not have units of energy? （d）a) N-m b) kw-hr c) Pa-m3 d) All have units of energy15)Extensive properties（d）a)apply at a point.b) do not depend on the size of the system.c)depend on position within a system.d)are proportional to the mass of a system.e)are printed in the tables in the back of your book.3Provide definition for the following terms(10pts)1)Internal energy:Internal energy is really the kinetic and potential energy of the atoms in the molecules.2)Heat:Energy crossing the system boundary,not associated with mass,due to a temperature difference.3)Work:Work is the energy transfer associated with a force acting through a distance.4) Closed system: No mass crosses the boundary. Heat and work can cross.5) Open system: Mass can cross the system boundary.4Answering and calculations(40pts)1) Feedwater Heater:Inlet 1 T1 = 200 ºC, p1 = 700 kPa, m1 40 kg/s2Inlet 2 T2 = 40 ºC, p2 = 700 kPa,Exit sat. liquid, p3 = 700 kPa,Find m 2 ?, m 3 ? andV 2 ? A2 = 25 cm2 Steady State m i m e m 1 m 2 m 3AVvInlet 2: compressed liquidTable A-4, v2 = 0.001008 m3/kgExit: saturated liquid Table A-5, v3 = 0.001108 m3/kgm 3 (AV)3v 3 0.060.00110854.15 kg/s m 2 m 3 m 1= 54.15 – 40 = 14.15 kg/sV m 2v 2 (14.15)(0.001008) 5.7 m/s2) A rigid tank contains 10 kg of water at 90 ºC. If 8 kg of the water is in the liquid form and the rest is in the vapor form. Determine the pressure in the tank and the volume of the tank . Table A-4, Psat = 70.14 kPa, P = Psat = 70.14 kPaTable A-4, vf = 0.001036 m3/kg, vg = 2.361 m3/kg v = vf + x(vg -vf) = 0.001036 + 0.2(2.361 – 0.001036) = 0.473 m3/kg V = mv = 10(0.473) = 4.73 m33)Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency. W Q H Q L 80 50 30 MWQ H 80 MW, Q L 50 MW0.375Q H 80 m A V 2 A 0.00252(AV)3 0.06 m 3 /sW304) A heat engine receives 600 kJ of heat from a high-temperature source at 1000 K during a cycle. It converts150kJ of this heat to work and rejects the remaining450kJ to a low-temperature sink at300 K. Determine if this heat engine violates the 2nd law of thermodynamics on the basis of(a)the Clausius inequality.(b)the Carnot principle.(a)Clausius inequalityQ QH QL600450T T T10003003H L= - 0.9 kJ/K < 0(b) Carnot principleth 1 Q Q H L 1 0.25T 300ηth < ηrev5)A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power consumption of the heat pump are given. The exterior temperature outside of the house is T L , 20C. The interior temperature e inside of the house is T H ,220C. The rate of heat loss of the house is Q H , 110.000kj/h. It is to be determined if this heat pump can do thejob.Assumptions: The heat pump operates steadily.Analysis: The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined fromCOP HPjev =11（T /T ）LH = 11（2273K ）/(22273K ) 14.75The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to beW net.jev.min= Q H = 110.000kj /h ( 1h ) =2.07kwThis heat pump is powerful enough since 8 kw >2.07 kw.6) 0.5 kg of air undergoes a Carnot cycle with η = 0.5.Given the initial pressure p1 = 700 kPa, initial volume V1 = 0.12 m3 and heat transfer during the isothermal expansion process Q12 = 40 kJ, Find the highest and the lowest temperatures in the cycle.p 1V 1 (700)(0.12)mR (0.5)(0.287)T H T 1 585.4 K H COP P 14.75 3600srev 1 T L 1 10000.7 Hth T T Q T，T L = 292.7 K4 H1 L 1 L Q H T H 1 1 0.5 0.5 L。
工程热力学与传热学习题(英文版)：第五章 水蒸气与湿空气
第五章 水蒸气和湿空气水蒸气 英文习题1. Volume and energy change during evaporationA mass of 200 g of saturated liquid water is completely vaporized at constant pressure of 100 kPa. Determine (a) the volume change and (b) the amount of energy added to the water.2.Pressureand volume of a mixtureA rigid tank contains 10 kg of water at 90℃. If 8 kg of the water is in the liquid form and rest is in thevapor form. Determine (a) the pressure in the tank and (b) the volume of the tank.3. Properties of saturated liquid-vapor mixtureAn 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160 kPa. Determine (a) the temperature of the refrigerant, (b) the quality, (c) the enthalpy of the refrigerant, and (d) the volume occupied by the vapor phase.4.Charging of a rigid tankby steamA rigid insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 1 MPa and 300℃. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed.Determine the final temperatureFIGURE 5-1FIGURE 5-2FIGURE 5-3FIGURE 5-4of the steam in the tank.湿空气 英文习题1. The amount of water vapor in room air100 kPa A 5-m ×5-m×3-m room shown in Fig.5-1 contains air at 25℃ and at a relative humidity of 75 percent. Determine (a) the partial pressure of dry air, (b) the specific humidity, (c) the enthalpy per unit mass of the dry air,and (d) the masses of the dry air and water vapor in the room.2. Fogging of the windows in a houseIn cold weather, condensation frequently occurs on the inner surfaces of the windows due to the lower air temperatures near the window surface. Consider a house, shown in Fig.5-6, that contains air at 20℃ and 75 percent relative humidity. At what window temperature will themoisture in the air start condensing on the inner surfaces of thewindows?3. The specific and relative humidity of airThe dry and the wet-bulb temperatures of atmospheric air at 1 atm (101.325 kPa) pressure are measured with a sing psychrometer and determined to be 25℃ and 15℃, respectively. Determine (a0 the specific humidity, (b0 the relative humidity, and © the enthalpy of the air.4. Heating and humidification of airAn air-conditioning system is to take in outdoor air at 10℃ and 30 percent relative humidity at a steady rate of 45 m 3/min and to condition it to 25℃ and 60 percent relative humidity. The outdoor air is first then heated to 22℃ in the heating section and humidified by the injection of hot steam in the humidifying section. Assuming the entire process takes place at a pressure of 100 kPa, determine (a) the rate of heat supply in the heating section and (b) the mass flow rate of the steamrequired in the humidifying section.5. Cooling and dehumidification of airAir enters a window air conditioner at 1 atm, 30℃, and 80 percent relative humidity at rate of 10 m 2/min, and it leaves as saturated air at 14℃. Part of the moisture in the air that condenses during the process is also removed at 14℃. Determine the rates of heat and moisture removal from the air.工程热力学与传热学第五章 水蒸气与湿空气 习题FIGURE 5-5FIGURE 5-6FIGURE 5-7习题1.热水泵必须安装在热水容器下面距容器有一定高度的地方，而不能安装在热水容器的上面，为什么？2.锅炉产生的水蒸气在定温过程中是否满足q=w的关系？为什么？3.有无0℃或低于0℃的蒸汽存在？有无高于400℃的水存在，为什么？4.25MPa的水，是否也象1MPa的水那样经历汽化过程？为什么？5.dh=c p dT适用于任何工质的定压过程，水蒸气定压汽化过程中dT=0，由此得出结论，水定压汽化时dh=c p dT=0，此结论是否正确，为什么？6.试解释湿空气，湿蒸汽，饱和湿空气。
《工程热力学B(双语)》试题13141
2013-2014第一学期《工程热力学B(双语)》期末试题考试专业：车辆工程、机械工程及自动化、农业机械化、食品科学、交通运输、工业设计一、单项选择题（每小题2分，共10分）1. If the system undergoes an irreversible process, entropy of gas ( )A：unable to determine B：0=dS D：0dS C：0〈dS〉2. Gas enters the nozzle at a pressure of 2000 kPa with a low velocity, and it expands to a pressure of 100 kPa. Then ( ) should be used.A. Converging nozzleB. Diverging nozzleC. Converging-diverging nozzleD. Any one of the above three kinds of nozzles3. 某理想气体经历一个可逆多变过程使温度下降、熵增加，则（）A：压力升高，比体积增大，对外作正功B：压力降低，比体积增大，对外作正功C：压力升高，比体积减小，对外作负功D：压力降低，比体积减小，对外作负功4. 在湿空气的定压加热过程中，湿空气的相对湿度将（）A：不确定B：增大C：不变D：减小5. 下面有可能看成可逆过程的是（）A：温差传热B：混合C：节流D：绝热膨胀二、判断题（每小题1分，共10分，对的画“√”，错的画“×”）1. The specific entropy of ideal gas is the only function of temperature. （）2. Carnot principle can be expressed as: No engine can be more efficient than areversible engine. （）3. 大气压力的变化会引起压力表读数的变化，但不会引起容器内工质绝对压力的改变。
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,/kgm 0.6kPa200kPa 600/kg)m (0.2331221===P P v vSubstituting ideal gas equation and this result into the boundary work integral produceskJ395.5-=⎪⎪⎭⎫⎝⎛⋅====⎰⎰333312112121out ,m kPa 1kJ 1m0.6m 0.2ln)m kPa)(0.6 kg)(200 (3lnv v v vvv mP d mRTd P W bThe 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 rises to 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)/kgm 0.052427C 07kPa 005/kgm 0.0008059liquid Sat.kPa 00532223k Pa005@11=⎭⎬⎫︒====⎭⎬⎫=v vv T P P fAnalysiskg04.62/kgm 0.0008059m0.053311===v V mandvPkJ1600=⎪⎪⎭⎫⎝⎛⋅-=-=-==⎰33121221out ,m kPa 1kJ 1/kg m 0.0008059)427kPa)(0.052 kg)(500 (62.04)()( v v V V V mP P d P W bDiscussion 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(001036.0kPa183.70311=-+=+==fgfx P vvvThe final specific volume at 800 kPa and 250°C is (Table A-6)/kgm 29321.032=vSince this is a linear process, the work done is equal to the area under the process line 1-2: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, k inetic,internal,in Change system massand 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 g2569.7v a p o rs a t ./k g m 0.29065k J /k g980.032052.30.25466.97/k g m 0.290650.0010531.15940.250.001053k J /k g3.2052,97.466/kg m 1.1594,001053.025.0kPa 150/k gm 0.29065@2312113113113==⎪⎭⎪⎬⎫===⨯+=+==-⨯+=+=====⎭⎬⎫==g fgffg f fgf g f u u ux uu x uu x P v v v vv vvSubstituting,min60.2==∆⎪⎪⎭⎫ ⎝⎛-=∆s 33613kJ/s1VA1000.03)kJ/kg 980kg)(2569.7(2)A8)(V 110(t t4-39A saturated water mixture contained in a spring-loaded piston-cylinder device is heated 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, k inetic,internal,in Change system massand 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/kg m 1783.0)001037.02172.2)(08.0(001037.0131=+=+==-+=+=fgffg f xuuu x v vvThe mass of water iskg22.11/kgm 1783.0m23311===v V mThe final specific volume is/kgm 4458.0kg22.11m53322===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:kJ450=⎪⎭⎫ ⎝⎛⋅-+=-+==331221out ,m kPa 1kJ 1)m 2(52)kPa225(75)(2Area V V P P W bSubstituting into energy balance equation giveskJ12,750=-+=-+=kJ/kg )30.553kg)(1650.422.11(kJ 450)(12out ,in u u m W Q b4-43Two tanks initially separated by a partition contain steam at different states. Now the partition 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, k inetic,internal,in Change systemmass and work ,heat,by nsfer energy tra Net out in =-+-=∆+∆=-∆=-W u u m u u m UUQ E E E B A BAThe properties of steam in both tanks at the initial state are (Tables A-4 through A-6)kJ/kg7.2793/kgm 25799.0C 300kPa 1000,13,1,1,1==⎪⎭⎪⎬⎫︒==A A A A u T P v()[]()kJ/kg4.15954.19270.50.66631/kgm 0.196790.0010910.392480.500.001091kJ/kg4.1927,66.631/kgm .392480,001091.050.0C 1501,131,131,1=⨯+=+==-⨯+=+=====⎭⎬⎫=︒=fgfB fg f B fgf gf B ux uu x u u x T v vv vvThe total volume and total mass of the system arekg523m106.1/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/kgm 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 22127.0kPa0032222k Pa 300 @sat 2322=⨯+=+==--=--=︒==⎪⎭⎪⎬⎫==fg f f g f u x u u x T T P 0.3641C133.5v v v v v(b ) Substituting,[][]kJ 3959kJ/kg )4.15958.1282(kg) 3(kJ/kg )7.27938.1282(kg) 2()()(1212out -=-+-=-+-=∆+∆=-BA BAu u m u u m UUQorkJ3959=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 (a3ft80.0=⋅⋅==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, k inetic,internal,in Change system massand work ,heat,by nsfer energy tra Net out in T T mc u u m Q UQ E E E -≅-=∆=∆=-vThe final temperature of air isR1080R) (540211222211=⨯==−→−=T P P T T P T P V VThe 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 inthe 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 atroom 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, k inetic,internal,in Change system massand 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 kPa (0.287)m kPa)(144 (100m14466433113=⋅⋅===⨯⨯=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, k inetic,internal,in Change systemmass and 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 are 33111m5629.0k P a100K)K)(298/kg m kPa kg)(0.1889 (1=⋅⋅==P mRT V33222m1082.0kPa1000K)K)(573/kg m kPa kg)(0.1889 (1=⋅⋅==P mRT VPressure changes linearly with volume and the work done is equal to the area underthe process line 1-2:kJ1.250m kPa 1kJ 1)m5629.0(0.10822)kPa 1000(100)(2Area 331221out ,-=⎪⎪⎭⎫⎝⎛⋅-+=-+==V V P P W b Thus,kJ250.1=in ,b WUsing the energy balance equation,kJ4.69K )25K)(300kJ/kg 657.0(kg) 1(kJ 1.250)(12out ,in -=-⋅+-=-+=T T mc W Q b vThus, kJ 69.4=outQ4-76Air 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, k inetic,internal,in Change systemmassand work ,heat,by nsfer energy 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:K900kPa100kPa 300K)(3001212===P P T TThe mass of the air iskg4646.0K)K)(300/kg m kPa (0.287)m kPa)(0.4 (10033111=⋅⋅==RT P m VThe boundary work done during process 2-3 iskJ04900)K -K)(1200/kg m kPa (0.287)kg 4646.0()()(323232out ,=⋅⋅=-=-=T T mR P W b V VSubstituting these values into energy balance equation,kJ340=-⋅+=-+=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, k inetic,internal,in Change system massand 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 becomeBoiling PV (m 3)kJ 21.2=︒-︒=-=====C )880)(C kJ/kg. 32.3)(kg 0889.0()(kg0889.06m)055.0()kg/m1020(612in 333T T mc Q Dm p ππρρV。
工程热力学与传热学习题(英文版)：第四章 热力学第二定律
FIGURE 4-6
750K.
9. Entropy generation associated with heat transfer
A frictionless piston-cylinder device contains a saturated liquid-vapor mixture of water at 100℃ (Fig. 4-7). During a constant-pressure process, 600 kJ of heat is transferred to the surrounding air at 25℃. result, part of the water vapor contained in the cylinder condenses. Determine (a) the entropy change of the water and (b) the total entropy generation during this transfer process.
efficiency of this Carnot engine and (b) the amount of heat rejected to the sink per cycle.
FIGURE 4-4
FIGURE 4-5
6. Heating a house by a Carnot heat pump
must be obtained or discarded into the river ?
3. Heat rejection by a refrigerator
The food compartment of a refrigerator, shown in Fig 4-2, is maintained at 4℃ by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2 kW, determine (a) the coefficient of performance of the refrigerator and (b) the rate of heat rejection to the room that houses the refrigerator.
工程热力学与传热学习题(英文版)：第一章 热力学基本概念
第一章 热力学基本概念英文习题1. Expressing temperature rise in different unitsDuring a heating process, the temperature of a system rises by 10℃. Express this rise in temperature in K, ℉ and R.2. Absolute pressure of a vacuum chamberA vacuum gage connected to a chamber reads 5.8 psi at location where the atmosphere pressure is 14.5 psi. Determine the absolutepressure in the chamber.3. Measuring pressure with a manometerA manometer is used to measure the pressure in a tank. The fluid used has a specific gravity of 0.85, and the manometer column height is 55 cm, as shown in Fig.1-1. If the local atmospheric pressure is 96kPa,determine the absolute pressure within the tank.4. Measuring pressure with a multi-fluid manometerThe water in a tank is pressurized by air, and the pressure is measured by a multi-fluid manometer as shown in Fig. 1-2. The tank is located on a mountain at an altitude of 1400 m where the atmospheric pressure is 85.6 kPa. Determine the air pressure in the tank if h 1=0.1 m, h 2=0.2 m, and h 3=0.35 m. Take the densities of water, oil, and mercury to be 1000 kg/m 3, 850 kg/m 3, and 13600 kg/m 3 respectively.5. Effect of piston weight on pressure in a cylinderThe piston of a vertical piston-cylinder device containing a gas has a mass of 60 kg and a cross-sectional area of 0.04 m 2, as shown in Fig.1-3. The local atmosphere pressure is 0.97 bar, and the gravitational acceleration is 9.81 m/s 2. (a) Determine the pressure inside the cylinder. (b) If some heat is transferred to the gas and its volume is doubled, do you expect the pressure inside thecylinder to change?6. Burning off lunch caloriesA 90-kg man had two hamburgers, a regular serving of French fries, and a 200-ml Coke for lunch. Determine how long it will take for him to burn the lunch calories off (a) by watching TV and (b) by fast swimming. What would your answers be for a 45-kg man?7. Burning of a candle in an insulated roomA candle is burning in a well-insulated room. Taking the room (the air plus the candle) as the system, determine (a) if there is any heat transferduringthisFIGURE 1-1FIGURE 1-2FIGURE 1-3FIGURE 1-4burning process and (b) if there is any change in the internal energy of the system.8. Boundary work during a constant-volume processA rigid tank contains air at 500 kPa and 150℃. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65℃and 400 kPa, respectively. Determine the boundary work done during the process.FIGURE 1-5FIGURE 1-69. Isothermal compression of an ideal gasA piston-cylinder device initially contains 0.4 m3 of air at 100 kPa and 80℃. The air is now compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process.10. Heat transfer from a personConsider a person standing in a breezy room at 20℃. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 m2and 29℃, respectively, and the convection heat transfer coefficient is 6 W/m2.℃ (Fig.1-7)FIGURE 1-7工程热力学与传热学第一章基本概念习题习题中文习题1.平衡状态与稳定状态有何区别？热力学中为什么要引入平衡状态的概念？2.表压力或真空度能否作为状态参数进行热力计算？若工质的压力不变，问测量其压力的压力表或真空计的读数是否可能变化？3.真空表指示数值越大，被测对象的实际压力愈大还是愈小？4.准平衡过程与可逆过程有何区别？5.不可逆过程是无法回到初态的过程，这种说法是否正确？6.没有盛满开水的热水瓶，其瓶塞有时被自动顶开，有时被自动吸紧，这是什么原理？7.用U形管压力表测定工质的压力时，压力表液柱直径的大小对读数有无影响？8.某容器被一刚性壁分为两部分，在容器不同部位装有3个压力表，如图示，压力表B的读数为1.75bar，压力表A的读数为1.10bar。
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。
工程热力学第三版答案【英文】第9章
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 -=-=-=。
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.。
工程热力学第三版答案【英文】第17章
17-5Air at 320 K is flowing in a duct. The temperature that a stationary probe inserted into the duct will read is to be determined for different air velocities. Assumptions The stagnation process is isentropic.Properties The specific heat of air at room temperature is c p = 1.005 kJ/kg ⋅K. Analysis The air which strikes the probe will be brought to a complete stop, and thus it will undergo a stagnation process. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature, T 0. It is determined frompc V T T 220+=. The results for each case are calculated below:(a ) K 320.0=⎪⎪⎭⎫ ⎝⎛⋅⨯=2s /2m 1000kJ/kg 1K kJ/kg 005.122m /s) (1+K 3200T (b ) K 320.1=⎪⎭⎫⎝⎛⋅⨯=2220s /m 1000kJ/kg 1K kJ/kg 005.12m/s) (10+K 320T (c ) K 325.0=⎪⎭⎫⎝⎛⋅⨯=2220s /m 1000kJ/kg 1K kJ/kg 005.12m/s) (100+K 320T(d ) K 817.5=⎪⎭⎫⎝⎛⋅⨯=2220s /m 1000kJ/kg 1K kJ/kg 005.12m/s) (1000+K 320T Discussion Note that the stagnation temperature is nearly identical to thethermodynamic temperature at low velocities, but the difference between the two is significant at high velocities.17-8Air flows through a device. The stagnation temperature and pressure of air and its velocity are specified. The static pressure and temperature of air are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Air is an ideal gas.Properties The properties of air at an anticipated average temperature of 600 K are c p = 1.051 kJ/kg ⋅K and k = 1.376.Analysis The static temperature and pressure of air are determined fromK 518.6=⎪⎭⎫⎝⎛⋅⨯-=-=22220s /m 1000kJ/kg 1K kJ/kg 051.12m/s) (5702.6732p c V T TVandMPa 0.23=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=--)1376.1/(376.1)1/(022022K 673.2K 518.6MPa) 6.0(k k TT P PDiscussion Note that the stagnation properties can be significantly different thanthermodynamic properties.17-21The Mach number of a passenger plane for specified limiting operating conditions is to be determined.Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4. Analysis From the speed of sound relationm /s 293 kJ/kg1s /m 1000K) 273K)(-60 kJ/kg 287.0)(4.1(22=⎪⎪⎭⎫⎝⎛+⋅==kRT c Thus, the Mach number corresponding to the maximum cruising speed of the plane is0.897===m /s293m /s)6.3/945(Ma max c VDiscussion Note that this is a subsonic flight since Ma < 1. Also, using a k value at-60︒C would give practically the same result.17-25The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.287 kJ/kg·K and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded.Analysis The final temperature of air is determined from the isentropic relation of ideal gases,K 4.228MPa 1.5MPa 0.4K) 2.333(4.1/)14.1(/)1(1212=⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛=--kk PP T TTreating k as a constant, the ratio of the initial to the final speed of sound can be expressed as1.21=====4.2282.333Ratio 21221112T T RT k RT k c cDiscussion Note that the speed of sound is proportional to the square root of thermodynamic temperature.17-40The critical temperature, pressure, and density of air and helium are to be determined at specified conditions.Assumptions Air and Helium are ideal gases with constant specific heats at room temperature.Properties The properties of air at room temperature are R = 0.287 kJ/kg·K, k = 1.4, and c p = 1.005 kJ/kg·K. The properties of helium at room temperature are R = 2.0769 kJ/kg·K, k = 1.667, and c p = 5.1926 kJ/kg·K. Analysis (a ) Before we calculate the critical temperature T *, pressure P *, and density ρ*, we need to determine the stagnation temperature T 0, pressure P 0, and density ρ0.C 1.131s /m 1000 kJ/kg 1C kJ/kg 005.12m/s) (250+1002C 10022220︒=⎪⎭⎫⎝⎛︒⋅⨯=+︒=p c V TkP a 7.264K 373.2K 404.3 kP a)200()14.1/(4.1)1/(00=⎪⎭⎫⎝⎛=⎪⎭⎫⎝⎛=--k k T T P P33000 kg/m 281.2K)K)(404.3/kg m kPa 287.0( kPa7.264=⋅⋅==RT P ρ Thus,K 337=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=1+1.42K) 3.404(12*0k T TkPa 140=⎪⎭⎫⎝⎛=⎪⎭⎫⎝⎛+=--)14.1/(4.1)1/(01+1.42kPa) 7.264(12*k k k P P3kg/m 1.45=⎪⎭⎫⎝⎛=⎪⎭⎫⎝⎛+=--)14.1/(13)1/(101+1.42)kg/m 281.2(12*k k ρρ(b ) For helium, C 48.7s /m 1000 kJ/kg1C kJ/kg 1926.52m/s) (300+40222220︒=⎪⎭⎫⎝⎛︒⋅⨯=+=p c V T TkP a 2.214K 313.2K 321.9 kP a)200()1667.1/(667.1)1/(00=⎪⎭⎫⎝⎛=⎪⎭⎫⎝⎛=--k k T T P P33000 kg/m 320.0K)K)(321.9/kg m kPa 0769.2( kPa2.214=⋅⋅==RT P ρ Thus,K 241=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=1+1.6672K) 9.321(12*0k T TkPa 104.3=⎪⎭⎫⎝⎛=⎪⎭⎫⎝⎛+=--)1667.1/(667.1)1/(01+1.6672kPa) 2.214(12*k k k P P3kg/m 0.208=⎪⎭⎫⎝⎛=⎪⎭⎫⎝⎛+=--)1667.1/(13)1/(101+1.6672)kg/m 320.0(12*k k ρρDiscussion These are the temperature, pressure, and density values that will occur at the throat when the flow past the throat is supersonic.。
工程热力学英语习题.
4. Consider a device with one inlet and one exit.
If the volume flow rates at the inlet and at the
exit are the same, is the flow through the device necessarily steady？ Why?
e) What is the change in entropy of the surroundings?
f) Is this process reversible or irreversible? How do you know? (Note: you do not have to prove your answer mathematically to get full credit. Explain in a sentence how you know whether the process is reversible or irreversible)
[a] -38 ℃ [e]144 ℃
[b]30 ℃
[c]40 ℃
[d] 53 ℃
2.2 A 5 m3 rigid vessel contains steam at 20 MPa and 400
℃.The mass of the steam is
[a] 5.0kg [b] 0.322kg [c] 322kg [e]680kg
b) Is the final internal energy Ufinal of the system (both compartments together) lower, equal or higher than initial internal energy Uinitial of the system? Why? (A sentence or two is expected, perhaps bolstered by some equations.)
工程热力学第三版答案【英文】第二章
2-8A person with his suitcase goes up to the 10th floor in an elevator. The part of the energy of the elevator stored in the suitcase is to be determined. Assumptions 1 The vibrational effects in the elevator are negligible.Analysis The energy stored in the suitcase is stored in the form of potential energy, which is mgz . Therefore,kJ 10.3=⎪⎭⎫⎝⎛=∆=∆=∆222suitcase /s m 1000 kJ/kg 1m) 35)(m/s(9.81) kg (30z mg PE ETherefore, the suitcase on 10th floor has 10.3 kJ more energy compared to an identical suitcase on the lobby level.Discussion Noting that 1 kWh = 3600 kJ, the energy transferred to the suitcase is 10.3/3600 = 0.0029 kWh, which is very small.2-14A river is flowing at a specified velocity, flow rate, and elevation. The total mechanical energy of the river water per unit mass, and the power generation potential of the entire river are to be determined.Assumptions 1 The elevation given is the elevation of the free surface of the river. 2 The velocity given is the average velocity. 3 The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be ρ = 1000 kg/m 3. Analysis Noting that the sum of the flow energy and the potential energy is constant for a given fluid body, we can take theelevation of the entire river water to be the elevation of the free surface, and ignore the flowenergy. Then the total mechanical energy of the river water per unit mass becomesk J /k g 0.887=⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛+=+=+=22222me ch /s m 1000kJ/kg 12)m/s 3(m) 90)(m/s (9.812Vgh ke pe e The power generation potential of the river water is obtained by multiplying the totalmechanical energy by the mass flow rate,kg/s500,000/s)m 00)(5 kg/m 1000(33===V ρmMW444=====kW 000,444kJ/kg) 7kg/s)(0.88 000,500(mech mech max e m E WTherefore, 444 MW of power can be generated from this river as it discharges into thelake if its power potential can be recovered completely.Discussion Note that the kinetic energy of water is negligible compared to the potential energy, and it can be ignored in the analysis. Also, the power output of an actual turbine will be less than 444 MW because of losses and inefficiencies.2-21C(a ) From the perspective of the contents, heat must be removed in order to reduce and maintain the content's temperature. Heat is also being added to the contents from the room air since the room air is hotter than the contents.(b ) Considering the system formed by the refrigerator box when the doors are closed, there are three interactions, electrical work and two heat transfers. There is atransfer of heat from the room air to the refrigerator through its walls. There is also a transfer of heat from the hot portions of the refrigerator (i.e., back of the compressor where condenser is placed) system to the room air. Finally, electrical work is being added to the refrigerator through the refrigeration system.(c ) Heat is transferred through the walls of the room from the warm room air to the cold winter air. Electrical work is being done on the room through the electrical wiring leading into the room.2-27A man is pushing a cart with its contents up a ramp that is inclined at an angle of 10° from the horizontal. The work needed to move along this ramp is to be determined considering (a) the man and (b) the cart and its contents as the system.Analysis (a ) Considering the man as the system, letting l be the displacement along the ramp, and letting θ be the inclination angle of the ramp,Btu6.248ft lbf 4862=⎪⎭⎫⎝⎛⋅⋅=⋅=+==ft lbf 169.778Btu 1ft)lbf 4862(ft)sin(10) 100)(lbf 180100(sin θFl WThis is work that the man must do to raise the weight of the cart and contents, plus hisown weight, a distance of l sin θ.(b ) Applying the same logic to the cart and its contents givesBtu2.231ftlbf 1736=⎪⎭⎫⎝⎛⋅⋅=⋅===ft lbf 169.778Btu 1ft)lbf 1736(ft)sin(10) 100)(lbf 100(sin θFl W2-30The work required to compress a spring is to be determined.Analysis The force at any point during the deflection of the spring is given by F = F 0 + kx , where F 0 is the initial force and x is the deflection as measured from the point where the initial force occurred. From the perspective of the spring, this force acts in the direction opposite to that in which the spring is deflected. Then,[]Btu0.0214=⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛⋅⋅=⋅=-+-=-+-=+==⎰⎰in 12ft 1ft lbf 169.778Btu 1in)lbf 200(inlbf 200in)01(2lbf/in2000)in (1lbf) 100()(2)()(22221221202121x x k x x F dx kx FFds W2-43The classrooms and faculty offices of a university campus are not occupied anaverage of 4 hours a day, but the lights are kept on. The amounts of electricity and money the campus will save per year if the lights are turned off during unoccupied periods are to be determined.Analysis The total electric power consumed by the lights in the classrooms and faculty offices iskW528264264kW 264264,000= W)1106(400=lamps) of (mp)per consumed (Power kW264264,000= W)11012(200=lamps) of (mp)per consumed (Power officeslighting,classroomlighting,totallighting,offices lighting,classroom lighting,=+=+==⨯⨯⨯==⨯⨯⨯=E E E E ENoting that the campus is open 240 days a year, the total number of unoccupiedwork hours per year isUnoccupied hours = (4 hours/day)(240 days/year) = 960 h/yrThen the amount of electrical energy consumed per year during unoccupied work period and its cost are$41,564/yr======.082/kWh)kWh/yr)($0 506,880(energy) of cost nit savings)(U (Energy savings Cost kWh506,880h/yr) kW)(960 (528hours) Unoccupied )((savings Energy totallighting,EDiscussion Note that simple conservation measures can result in significant energyand cost savings.2-47A gasoline pump raises the pressure to a specified value while consuming electric power at a specified rate. The maximum volume flow rate of gasoline is to be determined.Assumptions 1 The gasoline pump operates steadily. 2 The changes in kinetic and potential energies across the pump are negligible.Analysis For a control volume that encloses the pump-motor unit, the energy balance can be written as/energiesetc. potential, k inetic, internal,in change of Rate (steady)0system massand work ,heat,by nsferenergy tra net of Rate out in ==-dtdE E E →outin E E =21in)()(v v P m P m W =+ →P P P m W ∆=-= V v )(12insince /v V =mand the changes in kinetic and potential energies of gasoline are negligible, Solving for volume flow rate and substituting, the maximum flow rate is determined to be/s m 0.5433=⎪⎪⎭⎫⎝⎛⋅=∆=kJ 1mkPa 1kPa 7kJ/s 8.3P 3inmaxWVDiscussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the volume flow rate will be less because of the losses associated with the conversion of electrical-to-mechanical shaft and mechanical shaft-to-flow energy.2-65Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pump-motor unit and the pressure difference between the inlet and the exit of the pump are to be determined.Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictional losses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible.Properties We take the density of water to be ρ = 1000 kg/m 3. Analysis (a ) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2. We also take the lake surface as the reference level (z 1 = 0), and thus the potential energy at points 1and 2 are pe 1 = 0 and pe 2 = gz 2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P 1 = P 2 = P atm ). Further, the kinetic energy at both points is zero (ke 1 = ke 2 = 0) since the water at both locations is essentiallystationary. The mass flow rate of water and its potential energy at point 2 are k g /s70/s)m 070.0)( kg/m 1000(33===V ρmk J /k g 196.0/s m 1000 kJ/kg 1m) 20)(m/s(9.8122222=⎪⎭⎫⎝⎛==gz pe Then the rate of increase of the mechanical energy of water becomeskW13.7kJ/kg) 6kg/s)(0.19 70()0()(22in mech,out mech,fluid mech,===-=-=∆pe m pe m e e m EThe overall efficiency of the combined pump-motor unit is determined from itsdefinition,67.2%or 0.672 kW20.4 kW 7.13inelect,fluid mech,motor-pump==∆=W E η(b ) Now we consider the pump. The change in the mechanical energy of water as itflows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy arenegligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 13.7 kW:P P P me e m E ∆=-=-=∆V ρ12in mech,out mech,fluid mech,)(Solving for ∆P and substituting,kPa 196=⎪⎪⎭⎫⎝⎛⋅=∆=∆ kJ 1m kPa 1/s m 0.070 kJ/s13.733fluidmech,VE PTherefore, the pump must boost the pressure of water by 196 kPa in order to raise its elevation by 20 m.Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor .2-112The work required to compress a gas in a gas spring is to be determined.Assumptions All forces except that generated by the gas spring will be neglected.Analysis When the expression given in the problem statement is substituted into the work integral relation, and advantage is taken of the fact that the force and displacement vectors are collinear, the result is[]Btu0.0228=⎪⎭⎫⎝⎛⋅⋅=⋅=⎪⎭⎫ ⎝⎛--⋅=--===----⎰⎰ft lbf 169.778Btu 1ft)lbf 74.17(ftlbf 74.17in 12ft 1 in)1(in)4(4.11inlbf 200)(1Constant Constant0.40.41.411122121k k kx x k dxxFds W。
工程热力学：英文例题1
(1) Example Boundary Work during a Constant-VolumeA rigid tank contains air at 500kPa and 150℃. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65℃ and 400kPa, respectively. Show this process on the p-V diagram. Determine the boundary work done during this process.SolutionThe boundary work can be determined to be 0d 21==⎰V p WThis is expected since a rigid tank has a constant volume and d V = 0 in the above equation. Therefore, there is no boundary work done during this process. That is, the boundary work done during a constant-volume process is always zero. This is also evident from the p-V diagram of the process (the area under the process curve is zero).(2) Example Cooling of a Hot Fluid in a TankA rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800kJ. During the cooling process, the fluid loses 500kJ of heat, and the paddle wheel does 100kJ of work on the fluid.p , kPa 500 400Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel.Solution Take the contents of the tank as the system.This is a closed system since no mass crosses the boundary during the process. We observe that the volume of a rigid tank is constant and thus there is no boundary work and V 2=V 1 . Also, heat is lost from the system and shaft work is done in the system.Assumption The tank is stationary and thus the kinetic and potential energy changes are zero.0=∆=∆E E P KTherefore, U E ∆=∆ and internal energy is the only form of the system ’s energy that may change during this process. system out in E E E ∆=-12out in pw,U U U Q W -=∆=-100kJ-500kJ = U 2 - 800kJ , U 2 = 400kJTherefore, the final internal energy of the system is 400kJ.(3)(4)。
工程热力学（英语）
1-11 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined. Assumptions The variation of air density with altitude is negligible.Properties The density of air is given to be ρ = 1.18 kg/m 3. The density of mercury is 13,600 kg/m 3.Analysis Atmospheric pressures at the top and at the bottom of the building arekPa100.70N/m 1000kPa1m/s kg 1N1m) )(0.755m/s )(9.807kg/m (13,600)(kPa97.36N/m 1000kPa1m/s kg 1N1m) )(0.730m/s )(9.807kg/m (13,600)(2223bottombottom 2223toptop =⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅===⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅==h g P h g ρP ρTaking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtainkPa 97.36)(100.70N/m 1000kPa 1m/s kg 1N1))(m/s )(9.807kg/m (1.18)(/2223topbottom air topbottom air -=⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅-=-=h P P gh P P A W ρIt yields h = 288.6 mwhich is also the height of the building.1-21 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined.Assumptions The manometer fluid is an incompressible substance.Properties The density of the liquid is given to be ρ = 0.81 kg/L = 810 kg/m 3. Analysis The gage pressure in the duct is determined fromPa636=⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅==-=2223atm abs gage N/m 1Pa1m/s kg 1N1m) )(0.08m/s )(9.81kg/m (810ghP P P ρ The length of the differential fluid column iscm 13.9=︒==35sin /)cm 8(sin /θh L730 mmHgDiscussion Note that the length of the differential fluid column is extended considerably by inclinin g the manometer arm for better readability.2-4 No. This is the case for adiabatic systems only.2-6 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined.Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined fromQ Q Q Q cooling lights people heat gain =++whereQ Q Q lights people heat gain 10100W 1kW40360kJ /h 4kW 15,000kJ /h 4.17kW=⨯==⨯=== Substituting,.Q cooling 9.17kW =++=14417Thus the number of air-conditioning units required is units 2−→−=1.83kW/unit5kW9.172-18 The flow of air through a flow channel is considered. The diameter of the wind channel downstream from the rotor and the power produced by the windmill are to be determined. Analysis The specific volume of the air is/kg m 8409.0kP a 100K) K)(293/kg m kP a 287.0(33=⋅⋅==P RT v The diameter of the wind channel downstream from the rotor ism 7.38===−→−=−→−=m/s9m/s10m) 7()4/()4/(21122221212211V V D D V D V D V A V A ππ The mass flow rate through the wind mill iskg/s 7.457/kg)m 4(0.8409m/s)(10m) 7(3211===πvV A mThe power produced is thenkW 4.35=⎪⎭⎫ ⎝⎛-=-=22222221/s m 1000kJ/kg 12)m/s 9()m/s 10(kg/s) 7.457(2V V m Wcool·2-19 The available head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined.Assumptions 1 The flow is steady. 2 Water levels at the reservoir and the discharge site remain constant. 3 Frictional losses in piping are negligible. Properties We take the density of water to beρ = 1000 kg/m 3 = 1 kg/L.Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy,which is gz per unit mass, and gz m for a given mass flow rate.kJ/kg 177.1/s m 1000kJ/kg 1m ) 120)(m /s (9.81222mech =⎪⎭⎫⎝⎛===gz pe e The mass flow rate iskg/s ,000100/s)m 0)(10kg/m 1000(33===V ρmThen the maximum and actual electric power generation becomeMW 7.117kJ/s 1000MW 1kJ/kg) 7kg/s)(1.17 000,100(mech mech max =⎪⎭⎫ ⎝⎛===e mE WMW 94.2===MW) 7.117(80.0max overall electric W W η Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the efficiency of the turbine –generator unit.3-9 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the initial and final states.Analysis This is a constant volume process. The specific volume is/kg m 0014.0kg10m 014.03321====m Vv vThe initial state is determined to be a mixture, and thus the temperature is the saturation temperature at the given pressure. From Table A-12 by interpolation C 0.61︒==kPa 300 @sat 1T TandR-134a 300 kPa 10 kgvkJ/kg52.54)13.198)(009321.0(67.52009321.0/kgm )0007736.0067978.0(/kg m )0007736.00014.0(113311=+=+==--=-=fg f fgf h x h h x v v vThe total enthalpy is thenkJ 545.2===)kJ/kg 52.54)(kg 10(11mh HThe final state is also saturated mixture. Repeating the calculations at this state,C 21.55︒==kPa 600 @sat 2T TkJ/kg64.84)90.180)(01733.0(51.8101733.0/kgm )0008199.0034295.0(/kg m )0008199.00014.0(223322=+=+==--=-=fg f fgf h x h h x v v vkJ 846.4===)kJ/kg 64.84)(kg 10(22mh H3-22 rigid tank contains an ideal gas at a specified state. The final temperature is to be determined for two different processes.Analysis (a ) The first case is a constant volume process. When half of the gas is withdrawn from the tank, the final temperature may be determined from the ideal gas relation as()K 400=⎪⎭⎫ ⎝⎛==K) 600(kP a 300kP a 1002112212T P P m m T (b ) The second case is a constant volume and constant mass process. The ideal gas relation for this case yieldskPa 200=⎪⎭⎫ ⎝⎛==kP a) 300(K 600K 4001122P T T P3-32 Complete the following table for H 2 O : P , kPa T , ︒C v , m 3 / kgu , kJ/kg Phase description 200 30 0.001004 125.71 Compressed liquid 270.3130--Insufficient information200 400 1.5493 2967.2 Superheated steam 300133.520.5002196.4Saturated mixture, x=0.825500 473.1 0.6858 3084 Superheated steam4-14 Oxygen is heated to experience a specified temperature change. The heat transfer is to be determined for two cases.Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 154.8 K and 5.08 MPa. 2 The kinetic and potential energy changes arenegligible, 0pe ke ≅∆≅∆. 3 Constant specific heats can be used for oxygen.Properties The specific heats of oxygen at the average temperature of (25+300)/2=162.5︒C=436 K are c p = 0.952 kJ/kg ⋅K and c v = 0.692 kJ/kg ⋅K (Table A-2b ).Analysis We take the oxygen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for a constant-volume process can be expressed as)(12in energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in T T mc U Q E E E -=∆=∆=-vThe energy balance during a constant-pressure process (such as in a piston-cylinder device) can be expressed as)(12in out ,in out ,in energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsferenergy tra Net out in T T mc H Q U W Q UW Q E E E p b b -=∆=∆+=∆=-∆=-since ∆U + W b = ∆H during a constant pressure quasi-equilibrium process. Substituting for both cases, kJ 190.3=-⋅=-==K )25K)(300kJ/kg 692.0(kg) 1()(12const in,T T mc Q v VkJ 261.8=-⋅=-==K )25K)(300kJ/kg 952.0(kg) 1()(12const in,T T mc Q p P4-25 A rigid tank filled with air is connected to a cylinder with zero clearance. The valve is opened, and air is allowed to flow into the cylinder. The temperature is maintained at 30︒C at all times. The amount of heat transfer with the surroundings is to be determined.Assumptions 1 Air is an ideal gas. 2 The kinetic and potential energy changes are negligible,∆∆ke pe ≅≅0. 3 There are no work interactions involved other than the boundary work. Properties The gas constant of air is R = 0.287 kPa.m 3/kg.K (Table A-1). Analysis We take the entire air in the tank and the cylinder to be the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this closed system can be expressed asoutb,in 12out b,in energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in 0)(W Q u u m U W Q E E E ==-=∆=-∆=-since u = u (T ) for ideal gases, and thus u 2 = u 1 when T 1 = T 2 . The initial volume of air is33112212222111m 0.80)m (0.41kP a200kP a 400=⨯⨯==−→−=V V V V T T P P T P T PThe pressure at the piston face always remains constant at 200 kPa. Thus the boundary work done during this process iskJ 80m kPa 1kJ 10.4)m kPa)(0.8 (200)( 3312221out b,=⎪⎪⎭⎫⎝⎛⋅-=-==⎰V V V P d P W Therefore, the heat transfer is determined from the energy balance to bekJ 80==in out b,Q W4-27 An insulated cylinder is divided into two parts. One side of the cylinder contains N 2 gas and the other side contains He gas at different states. The final equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely.Assumptions 1 Both N 2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible. Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m 3/kg.K is c v = 0.743 kJ/kg·°C for N 2, and R = 2.0769 kPa.m 3/kg.K is c v = 3.1156 kJ/kg·°C for He (Tables A-1 and A-2)Analysis The mass of each gas in the cylinder is()()()()()()()()kg0.808K 298K /kg m kPa 2.0769m 1kPa 500kg 4.77K 353K /kg m kPa 0.2968m 1kPa 50033He111He33N 111N 22=⋅⋅=⎪⎪⎭⎫ ⎝⎛==⋅⋅=⎪⎪⎭⎫ ⎝⎛=RT P m RT P m V VTaking the entire contents of the cylinder as our system, the 1st law relation can be written as()()He12N 12HeN energiesetc. potential, kinetic,internal,in Change system massand work,heat,by nsferenergy tra Net out in )]([)]([0022T T mc T T mc U U U E E E -+-=∆+∆=∆=∆=-v vSubstituting,()()()()()()0C 25C kJ/kg 3.1156kg 0.808C 80C kJ/kg 0.743kg 4.77=︒-︒⋅+︒-⋅f fT TIt givesT f = 57.2︒Cwhere T f is the final equilibrium temperature in the cylinder.The answer would be the same if the piston were not free to move since it would effect only pressure, and not the specific heats.Discussion Using the relation P V = NR u T , it can be shown that the total number of moles in the cylinder is 0.170 + 0.202 = 0.372 kmol, and the final pressure is 510.6 kPa.6-9 An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer, and work output measurements. The claim is to be evaluated.Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from42%or 42.0K500K29011C th,max th,=-=-==H L T T ηη The actual thermal efficiency of the heat engine in ques tion is42.9%or 0.429kJ700kJ300net th ===H Q W η which is greater than the maximum possible thermal efficiency. Therefore, this heat engine is a PMM2 and the claim is false .6-11 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power consumption of the heat pump are given. It is to be determined if this heat pump can do the job.Assumptions The heat pump operates steadily.Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined from()()()14.75K 27322/K 273211/11COP rev HP,=++-=-=H L T TThe required power input to this reversible heat pump is determined from the definition of the coefficient of performance to bekW 2.07=⎪⎪⎭⎫ ⎝⎛==s 3600h 114.75kJ/h 110,000COP HP minin,net,H Q W This heat pump is powerful enough since 5 kW > 2.07 kW.7-8A reversible heat pump with specified reservoir temperatures is considered. The entropy change of two reservoirs is to be calculated and it is to be determined if this heat pump satisfies the increase in entropy principle.Assumptions The heat pump operates steadily. Analysis Since the heat pump is completely reversible, the combination of the coefficient of performance expression, first Law, and thermodynamic temperature scale gives73.26)K 294/()K 283(11/11COP rev HP,=-=-=HL T T The power required to drive this heat pump, according to the coefficient of performance, is thenkW 741.326.73kW 100COP rev HP,in net,===HQ WAccording to the first law, the rate at which heat is removed from the low-temperature energy reservoir is5 kWnetkW 26.96kW 741.3kW 100in net,=-=-=W Q Q H L The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, iskW/K 0.340===∆K294kW 100H H H T Q Sand that of the low-temperature reservoir iskW/K 0.340-=-==∆K283kW 26.96L L L T Q S The net rate of entropy change of everything in this system iskW/K 0=-=∆+∆=∆340.0340.0total L H S S Sas it must be since the heat pump is completely reversible.7-11 Steam is expanded in an isentropic turbine. The work produced is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The process is isentropic (i.e., reversible-adiabatic).Analysis There is one inlet and two exits. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form asoutin energiesetc. potential, kinetic,internal,in change of Rate (steady) 0systemmassand work,heat,by nsferenergy tra net of Rate out in 0E E E E E==∆=-332211out out332211h m h m h m W W h m h m h m --=++=From a mass balance,kg/s 75.4)kg/s 5)(95.0(95.0kg/s 25.0)kg/s 5)(05.0(05.01312======m mm mNoting that the expansion process is isentropic, the enthalpies at three states are determined as follows:6)-A (Table KkJ/kg 6953.7kJ/kg4.2682 C 100 kPa 503333 ⋅==⎭⎬⎫︒==s h T P 6)-A (Table kJ/kg 3.3979K kJ/kg 6953.7 MPa 41311=⎭⎬⎫⋅===h s s Ps6)-A (Table kJ/kg 1.3309K kJ/kg 6953.7 kPa 7002322=⎭⎬⎫⋅===h s s P Substituting,kW6328=--=--=kJ/kg) .4kg/s)(2682 75.4(kJ/kg) .1kg/s)(3309 25.0(kJ/kg) .3kg/s)(3979 5(332211outh m h m h m W 7-13 The entropy change relations of an ideal gas simplify to∆s = c p ln(T 2/T 1) for a constant pressure processand ∆s = c v ln(T 2/T 1) for a constant volume process.Noting that c p > c v , the entropy change will be larger for a constant pressure process.7-22 Air is compressed in a piston-cylinder device. It is to be determined if this process is possible. Assumptions 1 Changes in the kinetic and potential energies are negligible. 4 Air is an ideal gas with constant specific heats. 3 The compression process is reversible.Properties The properties of air at room temperature are R = 0.287 kPa ⋅m 3/kg ⋅K, c p = 1.005 kJ/kg ⋅K (Table A-2a).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 asin,out 12out in ,12out in ,12out in ,energiesetc. potential, kinetic,internal,in Change systemmassand work,heat,by nsfer energy tra Net out in )(since 0)( )(b b p b b W Q T T Q W T T mc Q W u u m U Q W E E E ===--=--=∆=-∆=-The work input for this isothermal, reversible process iskJ/kg 8.897kP a100kP a250K)ln K)(300kJ/kg 287.0(ln12in =⋅==P P RT w That is,kJ/kg 8.897in out ==w qThe entropy change of air during this isothermal process isK kJ/kg 0.2630kP a100kP a250K)ln kJ/kg 287.0(ln ln ln121212air ⋅-=⋅-=-=-=∆P P R P P R T T c s p The entropy change of the reservoir isK kJ/kg 0.2630K300kJ/kg 89.78R R ⋅===∆R T q s Note that the sign of heat transfer is taken with respect to the reservoir. The total entropy change (i.e., entropy generation) is the sum of the entropy changes of air and the reservoir:K kJ/kg 0⋅=+-=∆+∆=∆2630.02630.0R air total s s sNot only this process is possible but also completely reversible.Heat8-1 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle 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 c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).Analysis (b ) From the ideal gas isentropic relations and energy balance,()()K 579.2kPa 100kPa 1000K 3000.4/1.4/11212=⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛=-kk P P T T()()()K3360==−→−-⋅=-=-=3max 32323in 579.2K kJ/kg 1.005kJ/kg 2800T T T T T c h h q p(c )()K 336K 3360kP a1000kP a1003344444333===−→−=T P P T T P T P v v ()()()()()()()()21.0%=-=-==-⋅+-⋅=-+-=-+-=+=kJ/kg2800kJ/kg 221211kJ/kg2212K 300336K kJ/kg 1.005K 3363360K kJ/kg 0.718in out th14431443out 41,out 34,out q q T T c T T c h h u u q q q p ηvDiscussion The assumption of constant specific heats at room temperature is not realistic in this case the temperature changes involved are too large.8-5 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure 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-2). Analysis (a ) Process 1-2: isentropic compression.()()()()kP a 2338kP a 100K 308K 757.99.5K757.99.5K 3081122121112220.412112=⎪⎪⎭⎫⎝⎛==−→−===⎪⎪⎭⎫ ⎝⎛=-P T T P T P T P T T k v v v v vvProcess 3-4: isentropic expansion.()()K 1969==⎪⎪⎭⎫ ⎝⎛=-0.4134439.5K 800k T T vvvvsProcess 2-3: v = constant heat addition.()kPa 6072=⎪⎪⎭⎫⎝⎛==−→−=kPa 2338K 757.9K 19692233222333P T T P T P T P v v (b ) ()()()()kg 10788.6K 308K /kg m kPa 0.287m 0.0006kPa 100433111-⨯=⋅⋅==RT P m V()()()()()kJ0.590=-⋅⨯=-=-=-K 757.91969K kJ/kg 0.718kg 106.78842323in T T mc u u m Q v(c) Process 4-1: v = constant heat rejection.()()()()kJ0.2 40K 308800K kJ/kg 0.718kg 106.788)(41414out =-⋅⨯-=-=-=-T T mc u u m Q v kJ 0.350240.0590.0out in net =-=-=Q Q W59.4%===kJ0.590kJ0.350inout net,th Q W η(d ) ()()kPa 652=⎪⎪⎭⎫⎝⎛⋅-=-=-===kJm kPa 1/9.51m 0.0006kJ0.350)/11(MEP 331outnet,21outnet,max 2min r W W rV V V V V V8-7 An ideal diesel cycle has a a cutoff ratio of 1.2. The power produced is 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 The specific volume of the air at the start of the compression is/kg m 8701.0kPa95K)288)(K /kg m kPa 287.0(33111=⋅⋅==P RT vThe total air mass taken by all 8 cylinders when they are charged is kg 008665.0/kgm 8701.0m)/4 12.0(m) 10.0()8(4/3212cyl 1cyl===∆=ππv v VS B N N m The rate at which air is processed by the engine is determined fromkg/s 1155.0rev/cycle2rev/s) 1600/60kg/cycle)( (0.008665rev ===N n m msince there are two revolutions per cycle in a four-stroke engine. The compression ratio is2005.01==routAt the end of the compression, the air temperature is()K 6.95420K) 288(14.1112===--k r T TApplication of the first law and work integral to the constant pressure heat addition giveskJ/kg 1325K )6.9542273)(K kJ/kg 005.1()(23in =-⋅=-=T T c q pwhile the thermal efficiency is6867.0)12.1(4.112.12011)1(1114.111.41th =---=---=--c k c k r k r r ηThe power produced by this engine is thenkW105.1====kJ/kg) 67)(1325kg/s)(0.68 (0.1155in th net netq m w m W η8-12 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power 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 constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).Analysis For the isentropic compression process,K .1527K)(10) 273(0.4/1.4/)1(12===-kk p r T TThe heat addition iskJ/kg 500kg/s1kW500in in===m Q qApplying the first law to the heat addition process,K 1025KkJ/kg 1.005kJ/kg500K 1.527)(in 2323in =⋅+=+=-=p p c q T T T T c q The temperature at the exit of the turbine isK 9.530101K) 1025(10.4/1.4/)1(34=⎪⎭⎫⎝⎛=⎪⎪⎭⎫⎝⎛=-kk p r T TApplying the first law to the adiabatic turbine and the compressor producekJ/kg 6.496K )9.5301025)(K kJ/kg 1.005()(43T =-⋅=-=T T c w pskJ/kg 4.255K )2731.527)(K kJ/kg 1.005()(12C =-⋅=-=T T c w pThe net power produced by the engine is thenkW 241.2=-=-=kJ/kg )4.2556kg/s)(496. 1()(C T net w w m W 0.482===kW 500kW 241.2in net th Q W ηFinally the thermal efficiency is。
工程热力学课后答案 英文版 电子工业出版社
1-11 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined. Assumptions The variation of air density with altitude is negligible.Properties The density of air is given to be ρ = 1.18 kg/m 3density of mercury is 13,600 kg/m 3.Analysis building arekPa100.70N/m 1000kPa1m/s kg 1N1m) )(0.755m/s )(9.807kg/m (13,600)(kPa97.36m/s kg 1N1m) )(0.730m/s )(9.807kg/m (13,600)(2223bottombottom 223toptop =⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅===⎪⎪⎭⎫⎝⎛⋅==h g P h g ρP ρTaking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtainkPa 97.36)(100.70N/m 1000kPa 1m/s kg 1N1))(m/s )(9.807kg/m (1.18)(/2223topbottom air topbottom air -=⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅-=-=h P P gh P P A W ρIt yields h = 288.6 mwhich is also the height of the building.1-21 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined.Assumptions The manometer fluid is an incompressible substance.Properties The density of the liquid is given to be ρ = 0.81 kg/L = 810 kg/m 3. Analysis The gage pressure in the duct is determined fromPa636=⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅==-=2223atm abs gage N/m 1Pa1m/s kg 1N1m) )(0.08m/s )(9.81kg/m (810ghP P P ρ The length of the differential fluid column is730 mmHgcm 13.9=︒==35sin /)cm 8(sin /θh LDiscussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability.2-4 No. This is the case for adiabatic systems only.2-6 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined.Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined fromQ Q Q Q cooling lights people heat gain =++whereQ Q Q lights people heat gain 10100W 1kW40360kJ /h 4kW 15,000kJ /h 4.17kW=⨯==⨯=== Substituting,.Q cooling 9.17kW =++=14417Thus the number of air-conditioning units required is units 2−→−=1.83kW/unit 5kW9.172-18 The flow of air through a flow channel is considered. The diameter of the wind channel downstream from the rotor and the power produced by the windmill are to be determined. Analysis The specific volume of the air is/k g m 8409.0k P a100K) K)(293/kg m kP a 287.0(33=⋅⋅==P RT v The diameter of the wind channel downstream from the rotor ism 7.38===−→−=−→−=m/s9m/s10m) 7()4/()4/(21122221212211V V D D V D V D V A V A ππ The mass flow rate through the wind mill iscool·kg/s 7.457/kg)m 4(0.8409m/s)(10m) 7(3211===πvV A mThe power produced is thenkW 4.35=⎪⎭⎫⎝⎛-=-=22222221/s m 1000kJ/kg 12)m/s 9()m/s 10(kg/s)7.457(2V V m W 2-19 The available head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined.Assumptions 1 The flow is steady. 2 Water levels at the reservoir and the discharge site remain constant. 3 Frictional losses in piping are negligible. Properties We take the density of water to beρ = 1000 kg/m 3 = 1 kg/L.Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy,which is gz per unit mass, and gz m for a given mass flow rate.kJ/kg 177.1/s m 1000kJ/kg 1m ) 120)(m /s (9.81222mech =⎪⎭⎫ ⎝⎛===gz pe e The mass flow rate iskg/s ,000100/s)m 0)(10kg/m 1000(33===V ρmThen the maximum and actual electric power generation becomeMW 7.117kJ/s 1000MW 1kJ/kg) 7kg/s)(1.17 000,100(mech mech max =⎪⎭⎫ ⎝⎛===e mE WMW 94.2===MW) 7.117(80.0max overall electric W W η Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the efficiency of the turbine –generator unit.3-9 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the initial and final states.Analysis This is a constant volume process. The specific volume is/kg m 0014.0kg10m 014.03321====m Vv vR-134a 300 kPa 10 kgThe initial state is determined to be a mixture, and thus the temperature is the saturation temperature at the given pressure. From Table A-12 by interpolation C 0.61︒==kPa 300 @sat 1T TandkJ/kg52.54)13.198)(009321.0(67.52009321.0/kgm )0007736.0067978.0(/kg m )0007736.00014.0(113311=+=+==--=-=fg f fgf h x h h x v v vThe total enthalpy is thenkJ 545.2===)kJ/kg 52.54)(kg 10(11mh HThe final state is also saturated mixture. Repeating the calculations at this state,C 21.55︒==kPa 600 @sat 2T TkJ/kg64.84)90.180)(01733.0(51.8101733.0/kgm )0008199.0034295.0(/kg m )0008199.00014.0(223322=+=+==--=-=fg f fgf h x h h x v v vkJ 846.4===)kJ/kg 64.84)(kg 10(22mh H3-22 rigid tank contains an ideal gas at a specified state. The final temperature is to be determined for two different processes.Analysis (a ) The first case is a constant volume process. When half of the gas is withdrawn from the tank, the final temperature may be determined from the ideal gas relation as()K 400=⎪⎭⎫ ⎝⎛==K) 600(kP a 300kP a 1002112212T P P m m T (b ) The second case is a constant volume and constant mass process. The ideal gas relation for this case yields kPa 200=⎪⎭⎫ ⎝⎛==kP a) 300(K 600K 4001122P T T P3-32 Complete the following table for H 2 O :P , kPa T , ︒C v , m 3 / kgu , kJ/kg Phase description 200 30 0.001004 125.71 Compressed liquid 270.3130--Insufficient informationv200 400 1.5493 2967.2 Superheated steam 300133.520.5002196.4Saturated mixture, x=0.825500 473.1 0.6858 3084 Superheated steam4-14 Oxygen is heated to experience a specified temperature change. The heat transfer is to be determined for two cases.Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 154.8 K and 5.08 MPa. 2 The kinetic and potential energy changes are negligible, 0pe ke ≅∆≅∆. 3 Constant specific heats can be used for oxygen.Properties The specific heats of oxygen at the average temperature of (25+300)/2=162.5︒C=436 K are c p = 0.952 kJ/kg ⋅K and c v = 0.692 kJ/kg ⋅K (Table A-2b ).Analysis We take the oxygen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for a constant-volume process can be expressed as)(12in energiesetc. potential, kinetic, internal,in Change system massand work,heat,by nsfer energy tra Net out in T T mc U Q E E E -=∆=∆=-vThe energy balance during a constant-pressure process (such as in a piston-cylinder device) can be expressed as)(12in out ,in out ,in energiesetc. potential, kinetic, internal,in Change system massand work,heat,by nsfer energy tra Net outin T T mc H Q U W Q UW Q E E E p b b -=∆=∆+=∆=-∆=-since ∆U + W b = ∆H during a constant pressure quasi-equilibrium process. Substituting for both cases, kJ 190.3=-⋅=-==K )25K)(300kJ/kg 692.0(kg) 1()(12const in,T T mc Q v VkJ 261.8=-⋅=-==K )25K)(300kJ/kg 952.0(kg) 1()(12const in,T T mc Q p P4-25 A rigid tank filled with air is connected to a cylinder with zero clearance. The valve is opened, and air is allowed to flow into the cylinder. The temperature is maintained at 30︒C at all times. The amount of heat transfer with the surroundings is to be determined.Assumptions 1 Air is an ideal gas. 2 The kinetic and potential energy changes are negligible,∆∆ke pe ≅≅0. 3 There are no work interactions involved other than the boundary work. Properties The gas constant of air is R = 0.287 kPa.m 3/kg.K (Table A-1).Analysis We take the entire air in the tank and the cylinder to be the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this closed system can be expressed asoutb,in 12out b,in energiesetc. potential, kinetic,internal,in Change system massand work,heat,by nsfer energy tra Net out in 0)(W Q u u m U W Q E E E ==-=∆=-∆=-since u = u (T ) for ideal gases, and thus u 2 = u 1 when T 1 = T 2 . The initial volume of air is33112212222111m 0.80)m (0.41kP a200kP a 400=⨯⨯==−→−=V V V V T T P P T P T P The pressure at the piston face always remains constant at 200 kPa. Thus the boundary work done during this process iskJ 80m kPa 1kJ 10.4)m kPa)(0.8 (200)( 312221out b,=⎪⎪⎭⎫⎝⎛⋅-=-==⎰V V V P d P W Therefore, the heat transfer is determined from the energy balance to bekJ 80==in out b,Q W4-27 An insulated cylinder is divided into two parts. One side of the cylinder contains N 2 gas and the other side contains He gas at different states. The final equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely.Assumptions 1 Both N 2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible. Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m 3/kg.K is c v = 0.743 kJ/kg·°C for N 2, and R = 2.0769 kPa.m 3/kg.K is c v = 3.1156 kJ/kg·°C for He (Tables A-1 and A-2)Analysis The mass of each gas in the cylinder is()()()()()()()()kg0.808K 298K /kg m kPa 2.0769m 1kPa 500kg 4.77K 353K /kg m kPa 0.2968m 1kPa 50033He111He33N 111N 22=⋅⋅=⎪⎪⎭⎫ ⎝⎛==⋅⋅=⎪⎪⎭⎫ ⎝⎛=RT P m RT P m V VTaking the entire contents of the cylinder as our system, the 1st law relation can be written as()()He12N 12HeN energiesetc. potential, kinetic, internal,in Change system massand work,heat,by nsfer energy tra Net out in )]([)]([0022T T mc T T mc U U U E E E -+-=∆+∆=∆=∆=-v vSubstituting,()()()()()()0C 25C kJ/kg 3.1156kg 0.808C 80C kJ/kg 0.743kg 4.77=︒-︒⋅+︒-⋅f fT TIt givesT f = 57.2︒Cwhere T f is the final equilibrium temperature in the cylinder.The answer would be the same if the piston were not free to move since it would effect only pressure, and not the specific heats.Discussion Using the relation P V = NR u T , it can be shown that the total number of moles in the cylinder is 0.170 + 0.202 = 0.372 kmol, and the final pressure is 510.6 kPa.6-9 An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer, and work output measurements. The claim is to be evaluated.Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from42%or 42.0K500K 29011C th,max th,=-=-==H L T T ηη The actual thermal efficiency of the heat engine in question is42.9%or 0.429kJ700kJ300net th ===H Q W η which is greater than the maximum possible thermal efficiency. Therefore, this heat engine is a PMM2 and the claim is false .6-11 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power consumption of the heat pump are given. It is to be determined if this heat pump can do the job.Assumptions The heat pump operates steadily.Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined from ()()()14.75K 27322/K 273211/11COP rev HP,=++-=-=H L T TThe required power input to this reversible heat pump is determined from the definition of the coefficient of performance to bekW 2.07=⎪⎪⎭⎫ ⎝⎛==s 3600h 114.75kJ/h 110,000COP HP min in,net,H Q W This heat pump is powerful enough since 5 kW > 2.07 kW.7-8A reversible heat pump with specified reservoir temperatures is considered. The entropy change of two reservoirs is to be calculated and it is to be determined if this heat pump satisfies the increase in entropy principle.Assumptions The heat pump operates steadily. Analysis Since the heat pump is completely reversible, the combination of the coefficient of performance expression, first Law, and thermodynamic temperature scale gives73.26)K 294/()K 283(11/11COP rev HP,=-=-=HL T T The power required to drive this heat pump, according to the coefficient of performance, is thenkW 741.326.73kW 100COP rev HP,in net,===HQ WAccording to the first law, the rate at which heat is removed from the low-temperature energy reservoir iskW 26.96kW 741.3kW 100in net,=-=-=W Q Q H L The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, iskW/K 0.340===∆K294kW 100H HHT Q S and that of the low-temperature reservoir iskW/K 0.340-=-==∆K283kW 26.96L L L T Q S The net rate of entropy change of everything in this system iskW/K 0=-=∆+∆=∆340.0340.0total L H S S Sas it must be since the heat pump is completely reversible.net7-11 Steam is expanded in an isentropic turbine. The work produced is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The process is isentropic (i.e., reversible-adiabatic).Analysis There is one inlet and two exits. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form aso u tin energiesetc. potential, kinetic, internal,in change of Rate (steady) 0systemmassand work,heat,by nsfer energy tra net of Rate out in 0E E E E E==∆=-332211out out332211h m h m h m W W h m h m h m --=++=From a mass balance,kg/s 75.4)kg/s 5)(95.0(95.0kg/s 25.0)kg/s 5)(05.0(05.01312======m mm mNoting that the expansion process is isentropic, the enthalpies at three states are determined as follows:6)-A (Table KkJ/kg 6953.7kJ/kg4.2682C 100 kPa 503333 ⋅==⎭⎬⎫︒==s h T P 6)-A (Table kJ/kg 3.3979K kJ/kg 6953.7 MPa 41311=⎭⎬⎫⋅===h s s P 6)-A (Table kJ/kg 1.3309K kJ/kg 6953.7 kPa 7002322=⎭⎬⎫⋅===h s s P Substituting,kW6328=--=--=kJ/kg) .4kg/s)(2682 75.4(kJ/kg) .1kg/s)(3309 25.0(kJ/kg) .3kg/s)(3979 5(332211outh m h m h m W7-13 The entropy change relations of an ideal gas simplify to∆s = c p ln(T 2/T 1) for a constant pressure processand ∆s = c v ln(T 2/T 1) for a constant volume process.Noting that c p > c v , the entropy change will be larger for a constant pressure process.7-22 Air is compressed in a piston-cylinder device. It is to be determined if this process is possible.Assumptions 1 Changes in the kinetic and potential energies are negligible. 4 Air is an ideal gas with constant specific heats. 3 The compression process is reversible.Properties The properties of air at room temperature are R = 0.287 kPa ⋅m 3/kg ⋅K, c p = 1.005 kJ/kg ⋅K (Table A-2a).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 asin,out 12out in ,12out in ,12out in ,energiesetc. potential, kinetic, internal,in Change system massand work,heat,by nsfer energy tra Net out in )(since 0)( )(b b p b b W Q T T Q W T T mc Q W u u m U Q W E E E ===--=--=∆=-∆=-The work input for this isothermal, reversible process iskJ/kg 8.897kP a100kP a 250K)ln K)(300kJ/kg 287.0(ln12in =⋅==P P RT w That is,kJ/kg 8.897in out ==w qThe entropy change of air during this isothermal process isK kJ/kg 0.2630kP a100kP a250K)ln kJ/kg 287.0(ln ln ln121212air ⋅-=⋅-=-=-=∆P P R P P R T T c s p The entropy change of the reservoir isK kJ/kg 0.2630K300kJ/kg 89.78R R ⋅===∆R T q s Note that the sign of heat transfer is taken with respect to the reservoir. The total entropy change (i.e., entropy generation) is the sum of the entropy changes of air and the reservoir:K kJ/kg 0⋅=+-=∆+∆=∆2630.02630.0R air total s s sNot only this process is possible but also completely reversible.8-1 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle 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 c p = 1.005 kJ/kg.K, c v = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).Analysis (b ) From the ideal gas isentropic relations and energy balance,()()K 579.2k P a 100k P a1000K 3000.4/1.4/11212=⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=-kk P P T THeatv()()()K3360==−→−-⋅=-=-=3max 32323in 579.2K kJ/kg 1.005kJ/kg 2800T T T T T c h h q p(c )()K 336K 3360kP a1000kP a1003344444333===−→−=T P P T T P T P v v ()()()()()()()()21.0%=-=-==-⋅+-⋅=-+-=-+-=+=k J /k g2800k J /k g 221211k J /k g2212K 300336K k J /k g 1.005K 3363360K k J /k g 0.718in out th14431443out 41,out 34,out q q T T c T T c h h u u q q q p ηvDiscussion The assumption of constant specific heats at room temperature is not realistic in this case the temperature changes involved are too large.8-5 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure 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-2). Analysis (a ) Process 1-2: isentropic compression.()()()()kP a 2338kP a 100K 308K 757.99.5K757.99.5K 3081122121112220.412112=⎪⎪⎭⎫⎝⎛==−→−===⎪⎪⎭⎫ ⎝⎛=-P T T P T P T P T T k v v v v vvProcess 3-4: isentropic expansion.()()K 1969==⎪⎪⎭⎫ ⎝⎛=-0.4134439.5K 800k T T vvProcess 2-3: v = constant heat addition.()kPa 6072=⎪⎪⎭⎫⎝⎛==−→−=kPa 2338K 757.9K 19692233222333P T T P T P T P v v (b ) ()()()()kg 10788.6K 308K /kg m kPa 0.287m 0.0006kPa 100433111-⨯=⋅⋅==RT P m V vs()()()()()kJ0.590=-⋅⨯=-=-=-K 757.91969K kJ/kg 0.718kg 106.78842323in T T mc u u m Q v(c) Process 4-1: v = constant heat rejection.()()()()kJ0.2 40K 308800K kJ/kg 0.718kg 106.788)(41414out =-⋅⨯-=-=-=-T T mc u u m Q v kJ 0.350240.0590.0out in net =-=-=Q Q W59.4%===kJ0.590kJ0.350inout net,th Q W η(d ) ()()kPa 652=⎪⎪⎭⎫⎝⎛⋅-=-=-===kJm kPa 1/9.51m 0.0006kJ0.350)/11(MEP 331outnet,21outnet,max 2min r W W rV V V V V V8-7 An ideal diesel cycle has a a cutoff ratio of 1.2. The power produced is 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 The specific volume of the air at the start of the compression is/kg m 8701.0kPa95K)288)(K /kg m kPa 287.0(33111=⋅⋅==P RT vThe total air mass taken by all 8 cylinders when they are charged is kg 008665.0/kgm 8701.0m)/4 12.0(m) 10.0()8(4/3212cyl 1cyl===∆=ππv v VS B N N mThe rate at which air is processed by the engine is determined fromkg/s 1155.0rev/cycle2rev/s) 1600/60kg/cycle)( (0.008665rev ===N nm msince there are two revolutions per cycle in a four-stroke engine. The compression ratio is2005.01==r At the end of the compression, the air temperature is()K 6.95420K) 288(14.1112===--k r T ToutApplication of the first law and work integral to the constant pressure heat addition giveskJ/kg 1325K )6.9542273)(K kJ/kg 005.1()(23in =-⋅=-=T T c q pwhile the thermal efficiency is6867.0)12.1(4.112.12011)1(1114.111.41th =---=---=--c k c k r k r r ηThe power produced by this engine is thenkW105.1====kJ/kg) 67)(1325kg/s)(0.68 (0.1155in th net netq m w m W η8-12 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power 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 constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).Analysis For the isentropic compression process,K .1527K)(10) 273(0.4/1.4/)1(12===-kk p r T TThe heat addition iskJ/kg 500kg/s1kW 500in in===m Q qApplying the first law to the heat addition process,K 1025KkJ/kg 1.005kJ/kg500K 1.527)(in 2323in =⋅+=+=-=p p c q T T T T c q The temperature at the exit of the turbine isK 9.530101K) 1025(10.4/1.4/)1(34=⎪⎭⎫⎝⎛=⎪⎪⎭⎫⎝⎛=-kk p r T TApplying the first law to the adiabatic turbine and the compressor produceskJ/kg 6.496K )9.5301025)(K kJ/kg 1.005()(43T =-⋅=-=T T c w pkJ/kg 4.255K )2731.527)(K kJ/kg 1.005()(12C =-⋅=-=T T c w pThe net power produced by the engine is thenkW 241.2=-=-=kJ/kg )4.2556kg/s)(496. 1()(C T netw w m W Finally the thermal efficiency is0.482===kW 500kW241.2innet thQ W η。
工程热力学英文版9
A + B AB.
(10.3)
Suppose we introduce NA,0 moles of A and NB,0 moles of B into a container with volume V at time t = 0. We allow the system to evolve with no interaction with the surroundings (isolated). As time proceeds, we find that the mole numbers change, due to formation of AB molecules. If we wait long enough, the mole numbers come to steady-state values, which we will denote NA∗ , NB∗ , and NA∗B. Evidently, once steady state is reached, the rate at which AB is formed by reaction (10.1) just balances the rate at which it is destroyed by reaction (10.2).
When this condition is reached, we say the system is in chemical equilibrium. Depending on the nature of the chemical reaction, and also on the temperature
10.3 Reactions in Ideal Gases
If the mixture of A, B, and AB is an ideal gas mixture, then we have already
工程热力学第三版答案【英文】第5章
5-7Air is expanded and is accelerated as it is heated by a hair dryer of constant diameter. The percent increase in the velocity of air as it flows through the drier is to be determined.Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 1.20 kg/m 3 at the inlet, and 1.05 kg/m 3 at the exit. Analysis There is only one inlet and one exit, and thusmm m 12==. Then,)of increase and (or, 1.263kg/m 0.95kg/m 1.2033211222112126.3% =====ρρρρV V AV AV m mTherefore, the air velocity increases 26.3% as it flows through the hair drier.5-14A smoking lounge that can accommodate 15 smokers is considered. The required minimum flow rate of air that needs to be supplied to the lounge and the diameter of the duct are to be determined.Assumptions Infiltration of air into the smoking lounge is negligible.Properties The minimum fresh air requirements for a smoking lounge is given to be 30 L/s per person.Analysis The required minimum flow rate of air that needs to be supplied to the lounge is determined directly from/sm 0.453=L/s 450= persons)person)(15L/s (30= persons)of No.(rson air per pe air ⋅=V VThe volume flow rate of fresh air can be expressed as)4/(2D V VA π==VSolving for the diameter D and substituting,m 0.268===m /s)(8)/s m 45.0(443ππVD VTherefore, the diameter of the fresh air duct should be at least 26.8 cm if the velocity of air is not to exceed 8 m/s.VSmokingLounge15 smokers5-20An air compressor compresses air. The flow work required by the compressor is to be determined.Assumptions 1 Flow through the compressor is steady. 2 Air is an ideal gas.Properties The gas constant of air is R = 0.287 kPa ⋅m 3/kg ⋅K (Table A-1).Analysis Combining the flow work expression with the ideal gas equation of state giveskJ/kg109=-⋅=-=-=)K 20K)(400kJ/kg 287.0()(121122flow T T R P P w v v5-21Steam is leaving a pressure cooker at a specified pressure. The velocity, flow rate, thetotal and flow energies, and the rate of energy transfer by mass are to be determined. Assumptions 1 The flow is steady, and the initial start-up period is disregarded. 2 The kinetic and potential energies are negligible, and thus they are not considered. 3 Saturation conditions exist within the cooker at all times so that steam leaves the cooker as a saturated vapor at 20 psia.Properties The properties of saturated liquid water and water vapor at 20 psia are v f = 0.01683 ft 3/lbm, v g = 20.093 ft 3/lbm, u g = 1081.8 Btu/lbm, and h g = 1156.2 Btu/lbm (Table A-5E).Analysis (a ) Saturation conditions exist in a pressure cooker at all times after the steady operating conditions are established. Therefore, the liquid has the properties of saturated liquid and the exiting steam has the properties of saturated vapor at the operating pressure. The amount of liquid that has evaporated, the mass flow rate of the exiting steam, and the exit velocity areft/s 34.1=⎪⎪⎭⎫⎝⎛⨯===⨯===∆==⎪⎪⎭⎫ ⎝⎛=∆=22233-33liquidft 1in 144in 0.15/lbm)ft 093lbm/s)(20. 10(1.765lbm/min 1059.0min45lbm 766.4lbm 766.4gal 1ft 13368.0/lbm ft 0.01683gal0.6c g c g f A m A m V t m m m v v V ρlbm/s101.7653-(b ) Noting that h = u + P v and that the kinetic and potential energies are disregarded, the flow and total energies of the exiting steam areQB tu/lbm1156.2B tu/lbm74.4=≅++==-=-==h pe ke h u h P e θ8.10812.1156flow vNote that the kinetic energy in this case is ke = V 2/2 = (34.1 ft/s)2 /2 = 581 ft 2/s 2 =0.0232 Btu/lbm, which is very small compared to enthalpy.(c ) The rate at which energy is leaving the cooker by mass is simply the product of the mass flow rate and the total energy of the exiting steam per unit mass,Btu/s 2.04=⨯==-Btu/lbm ) 6.2lbm /s)(115 10765.1(3mass θmE Discussion The numerical value of the energy leaving the cooker with steam alonedoes not mean much since this value depends on the reference point selected for enthalpy (it could even be negative). The significant quantity is the difference between the enthalpies of the exiting vapor and the liquid inside (which is h fg ) since it relates directly to the amount of energy supplied to the cooker.5-30Air is decelerated in an adiabatic diffuser. The velocity at the exit is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5 The diffuser is adiabatic.Properties The specific heat of air at the average temperature of (20+90)/2=55°C =328 K is c p = 1.007 kJ/kg ⋅K (Table A-2b ).Analysis There is only one inlet and one exit, and thus m m m==21. We take diffuser as the system, which is a control volume since mass crosses the boundary. The energybalance for this steady-flow system can be expressed in the rate form asoutin energiesetc. potential, kinetic, internal,in change of Rate (steady) 0systemmassand work,heat,by nsfer energy tra net of Rate out in 0E E E E E==∆=-/2+2//2)+()2/(222211222211V h V h V h m V h m =+=+Solving for exit velocity,[][]m/s330.2=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⋅+=-+=-+=5.02225.021215.021212kJ/kg 1/s m 1000)K90K)(20kJ/kg 007.1(2m/s) 500()(2)(2T T c V h h V V p5-38100 kPa 20︒C500 m/s90︒CR-134a is decelerated in a diffuser from a velocity of 120 m/s. The exit velocity of R-134a and the mass flow rate of the R-134a are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 There are no work interactions. Properties From the R-134a tables (Tables A-11 through A-13) kJ/kg 267.29/kgm 0.025621.kPa 8001311==⎭⎬⎫=h vapor sat P v andkJ/kg 274.17/kgm 0.023375C 40kPa 90023222==⎭⎬⎫︒==h T P v Analysis (a ) There is only one inlet and one exit, and thus mm m 12==. Then the exit velocity of R-134a is determined from the steady-flow mass balance to be()m/s 60.8===−→−=m/s 120/kg)m (0.025621/kg)m (0.0233751.811133121122111222V A A V V A V A v v v v (b ) We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form asoutin energiesetc. potential, kinetic, internal,in change of Rate (steady) 0systemmassand work,heat,by nsfer energy tra net of Rate out in 0E E E E E==∆=-⎪⎪⎭⎫⎝⎛-+-=≅∆≅=++20)pe W (since /2)V +()2/(212212in 222211inV V h h mQ h m V h m QSubstituting, the mass flow rate of the refrigerant is determined to be()⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-+-=2222/s m 1000kJ/kg 12m /s) (120m /s 60.8kg 267.29)kJ/(274.17kJ/s 2m It yieldskg/s 1.308=m5-46Steam expands in a turbine. The change in kinetic energy, the power output, and theturbine inlet area are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time.212Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Properties From the steam tables (Tables A-4 through 6) kJ/kg3178.3/kgm 0.047420C 400MPa 613111==⎭⎬⎫︒==h T P v andkJ/kg 2318.52392.10.9262.31792.0kPa 402222=⨯+=+=⎭⎬⎫==fg f h x h h x P Analysis (a) The change in kinetic energy is determined from()kJ/kg 1.95-=⎪⎪⎭⎫⎝⎛-=-=∆22222122/s m 1000kJ/kg12m /s) (80m /s 502V V ke(b ) There is only one inlet and one exit, and thus mm m 12==. We take the turbine as the system, which is a control volumesince mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form asoutin energiesetc. potential, kinetic, internal,in change of Rate (steady) 0systemmassand work,heat,by nsfer energy tra net of Rate out in 0E E E E E==∆=-⎪⎪⎭⎫⎝⎛-+--=≅∆≅+=+20)pe Q (since /2)+()2/(212212out 222out211V V h h mW V h m W V h mThen the power output of the turbine is determined by substitution to beMW 14.6==---=kW 14,590kJ/kg )1.953178.32318.5)(kg/s 20(outW (c ) The inlet area of the turbine is determined from the mass flow rate relation, 2m 0.0119===−→−=m/s80)/kg m 0.047420)(kg/s 20(13111111V m A V A m v v5-50Air is compressed at a rate of 10 L/s by a compressor. The work required per unitmass and the power required are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time.2P 1 = 6 MPa T 1 = 400︒CV 1P 2 = 40 kPa x 2 = 0.92 V 2 = 50 m/sKinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.Properties The constant pressure specific heat of air at the average temperature of (20+300)/2=160°C=433 K is c p = 1.018 kJ/kg·K (Table A-2b). The gas constant of air is R = 0.287 kPa ⋅m 3/kg ⋅K (Table A-1).Analysis (a ) There is only one inlet and one exit, and thus m m m==21. We take the compressor as the system, which is a control volume since mass crosses the boundary.The energy balance for this steady-flow system can be expressed in the rate form aso u tin energiesetc. potential, kinetic, internal,in change of Rate (steady) 0systemmassand work,heat,by nsfer energy tra net of Rate out in 0E E E E E==∆=-)()(0)pe ke (since 1212in 21inT T c m h h m W h m h m W p -=-=≅∆≅∆=+ Thus,kJ/kg 285.0=-⋅=-=0)K 2K)(300kJ/kg (1.018)(12in T T c w p(b ) The specific volume of air at the inlet and the mass flow rate are/kg m 7008.0kPa120K) 273K)(20/kg m kPa 287.0(33111=+⋅⋅==P RT vkg/s 0.01427/kgm 0.7008/s m 010.03311===v V m Then the power input is determined from the energy balance equation to bekW 4.068=-⋅=-=0)K 2K)(300kJ/kg 8kg/s)(1.01 (0.01427)(12inT T c m W p5-65Steam is throttled by a well-insulated valve. The temperature drop of the steam afterthe expansion is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved. Properties The inlet enthalpy of steam is (Tables A-6),kJ/kg 1.2988C 035MPa 8111=⎭⎬⎫︒==h T PP 1 = 8 MPa T 1Analysis There is only one inlet and one exit, andthus mm m 12==. We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as02121outin (steady) 0systemout in h h h m h m E E E E E ====∆=-since QW ke pe ≅=≅≅∆∆0. Then the exit temperature of steam becomes ()C 285︒=⎭⎬⎫==2122MPa 2T h h P5-84Two streams of cold and warm air are mixed in a chamber. If the ratio of hot to coldair is 1.6, the mixture temperature and the rate of heat gain of the room are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties The gas constant of air is R = 0.287 kPa.m 3/kg.K. The enthalpies of air are obtained from air table (Table A-17) ash 1 = h @280 K = 280.13 kJ/kg h 2 = h @ 307 K = 307.23 kJ/kg h room = h @ 297 K = 297.18 kJ/kg Analysis (a ) We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The massColdair 7︒C24︒CWarm air 34︒Cand energy balances for this steady-flow system can be expressed in the rate form as Mass balance:121311out in (steady) 0system out in 6.1 since 6.26.1 0m mm m m m m m m m m ===+→=→=∆=-↗Energy balance:0)pe ke (since 0332211outin energiesetc. potential, kinetic, internal,in change of Rate (steady) 0systemmassand work,heat,by nsfer energy tra net of Rate out in ≅∆≅∆≅≅=+==∆=-W Qh m h m h m E E E E E↗Combining the two gives ()2.3/2.22.32.2213312111h h h or h m h m h m+==+ Substituting, h 3 = (280.13 +2.2⨯ 307.23)/3.2 = 298.76 kJ/kg From air table at this enthalpy, the mixture temperature isT 3 = T @ h = 298.76 kJ/kg = 298.6 K = 25.6︒C (b ) The mass flow rates are determined as followskg/s.1363kg/s) 9799.0(2.32.3kg/s 9799.0/kg m 0.7654/s m 0.75kg/m 7654.0kPa 105K)273K)(7/kg m kPa (0.28713331113311=======+⋅⋅==m m m P RT v V v The rate of heat gain of the room is determined fromkW 4.93-=-=-=kJ/kg )76.29818.297(kg/s) 136.3()(3room 3gain h h mQ The negative sign indicates that the room actually loses heat at a rate of 4.93 kW.5-102A room is to be heated by an electric resistance heater placed in a duct in the room. The power rating of the electric heater and the temperature rise of air as it passes through the heater are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The heating duct is adiabatic, and thus heat transfer through it is negligible. 5 No air leaks in and out of the room.Properties The gas constant of air is 0.287 kPa.m 3/kg.K (Table A-1). The specific heats of air at room temperature are c p = 1.005 and c v = 0.718 kJ/kg·K (Table A-2). Analysis (a ) The total mass of air in the room iskg 284.6)K 288)(K /kg m kPa 0.287()m 240)(kPa 98(m240m 865331133=⋅⋅===⨯⨯=RT P m V VWe first take the entire room as our system,which is a closed system since no mass leaks in or out. The power rating of the electric heater is determined by applying the conservation of energy relation to this constant volume closed system:()()12avg ,out in fan,in e,in fan,in e,energiesetc. potential, kinetic, internal,in Change system massand work,heat,by nsfer energy tra Net 0)=PE =KE (since T T mc Q W W t U Q W W E E E out out in -=-+∆∆∆∆=-+∆=-vSolving for the electrical work input giveskW5.40=⨯-⋅+-=∆--=+s) 60C/(15)1525)(C kJ/kg 0.718)(kg 284.6()kJ/s 0.2()kJ/s 200/60(/)(12in fan,out in e, tT T W Q W mc v (b ) We now take the heating duct as the system, which is a control volume since masscrosses the boundary. There is only one inlet and one exit, and thus mm m 12==. The energy balance for this adiabatic steady-flow system can be expressed in the rate formas)()(0)pe ke (since 01212in fan,in e,21in fan,in e,energiesetc. potential, kinetic, internal,in change of Rate (steady) 0systemmassand work,heat,by nsfer energy tra net of Rate T T c m h h mW W Q h m h mW W E E E E E p outin out in -=-=+≅∆≅∆==++==∆=-Thus, ()()C 6.7 =⋅+=+=-=∆K kJ/kg 1.005kg/s 50/60kJ/s )2.040.5(infan,in e,12p c m W W T T T5-107R-134a is condensed in a condenser. The heat transfer per unit mass is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. Analysis We take the pipe in which R-134a is condensed as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as21o u t 21o u to u t21o u tin energiesetc. potential, kinetic, internal,in change of Rate (steady) 0systemmassand work,heat,by nsfer energy tra net of Rate out in )(0h h q h h m QQ h m h m E E E E E -=-=+===∆=-The enthalpies of R-134a at the inlet and exit of the condenser are (Table A-12, A-13).kJ/kg61.1010kP a 900kJ/kg13.295C 60kP a 900kPa 900@22111==⎭⎬⎫===⎭⎬⎫︒==f h h x P h T PSubstituting,kJ/kg 193.5=-=61.10113.295out q5-112Helium flows from a supply line to an initially evacuated tank. The flow work of the helium in the supply line and the final temperature of the helium in the tank are to be determined.Properties The properties of helium are R = 2.0769 kJ/kg.K, c p = 5.1926 kJ/kg.K, c v = 3.1156 kJ/kg.K (Table A-2a).Analysis The flow work is determined from its definition but we first determine the specific volume/kg m 0811.4kP a)200(K)27320kJ/kg.K)(1 0769.2(3line =+==P RT vkJ/kg816.2===/kg)m 1kPa)(4.081 200(3flow v P wNoting that the flow work in the supply line is converted to sensible internal energy in the tank, the final helium temperature in the tank issat. liq.60︒Cdetermined as followsK655.0=−→−=−→−==+===tank tank tank tank -line line linetank kJ/kg.K) 1156.3(kJ/kg 7.2040kJ/kg7.2040K) 27320kJ/kg.K)(1 1926.5(T T T c u T c h h u p vAlternative Solution : Noting the definition of specific heat ratio, the final temperature in the tank can also be determined fromK 655.1=+==K) 273120(667.1line tank kT T which is practically the same result.5-119A rigid tank initially contains superheated steam. A valve at the top of the tank is opened, and vapor is allowed to escape at constant pressure until the temperature rises to 500︒C. The amount of heat transfer is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process by using constant average properties for the steam leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of water are (Tables A-4 through A-6)kJ/kg3468.3,kJ/kg 3116.9/kgm 0.17568C 500MP a 2kJ/kg3024.2,kJ/kg 2773.2/kgm 0.12551C 300MP a 2223222113111===⎭⎬⎫︒=====⎭⎬⎫︒==h u T P h u T P v v Analysis We take the tank as the system, which is a control volume since masscrosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u , respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance :21system out in m m m m m m e -=→∆=-Energy balance :)0 (since 1122in energiesetc. potential, kinetic,internal,in Change system massand work,heat,by nsfer energy tra Net out in ≅≅≅-=-∆=-pe ke W u m u m h m Q E E E e eThe state and thus the enthalpy of the steam leaving the tank is changing during thisprocess. But for simplicity, we assume constant properties for the exiting steam at the average values. Thus,kJ/kg 3246.22kJ/kg3468.33024.2221=+=+≅h h h e The initial and the final masses in the tank arekg 1.138/kgm 0.17568m 0.2kg1.594/kg m 0.12551m 0.23322233111======v V v V m mThen from the mass and energy balance relations,kg 0.456138.1594.121=-=-=m m m e()()()()()()kJ606.8=-+=-+=kJ/kg 2773.2kg 1.594kJ/kg 3116.9kg 1.138kJ/kg 3246.2kg 0.4561122u m u m h m Q e e in5-131An insulated piston-cylinder device with a linear spring is applying force to the piston. A valve at the bottom of the cylinder is opened, and refrigerant is allowed to escape. The amount of refrigerant that escapes and the final temperature of the refrigerant are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process assuming that the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible.Properties The initial properties of R-134a are (Tables A-11 through A-13)kJ/kg11.354kJ/kg 03.325/kg m 02423.0C 120MPa 2.1113111===⎭⎬⎫︒==h u T P v Analysis We take the tank as the system, which is a control volume since masscrosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u , respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance : 21system out in m m m m m m e -=→∆=- Energy balance :)0 (since 1122in b,energiesetc. potential, kinetic,internal,in Change system massand work,heat,by nsfer energy tra Net out in ≅≅≅-=-∆=-pe ke Q u m u m h m W E E E e e23212322233111m 0.502.33m 0.5kg02.33/kgm 0.02423m 0.8v v V v V -=-======m m m v m m eNoting that the spring is linear, the boundary work can be determined fromkJ 270m 0.5)-0.8(2kPa600)(1200)(232121in b,=+=-+=V V P P W Substituting the energy balance,kJ/kg) kg)(325.03 02.33(m 5.0m 5.002.3327022323-⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛--u h e v v (Eq. 1) where the enthalpy of exiting fluid is assumed to be the average of initial and finalenthalpies of the refrigerant in the cylinder. That is,2kJ/kg) 11.354(2221h h h h e +=+=Final state properties of the refrigerant (h 2, u 2, and v 2) are all functions of finalpressure (known) and temperature (unknown). The solution may be obtained by a trial-error approach by trying different final state temperatures until Eq. (1) is satisfied. Or solving the above equations simultaneously using an equation solver with built-in thermodynamic functions such as EES, we obtainT 2 = 96.8︒C , m e = 22.47 kg, h 2 = 336.20 kJ/kg, u 2 = 307.77 kJ/kg, v 2 = 0.04739 m 3/kg, m 2 = 10.55 kg。
工程热力学第三版答案【英文】第9章
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 -=-=-=。
工程热力学第三版答案【英文】第3章
3-23T, ︒C P, kPa h , kJ / kg x Phase description 120.21 200 2045.8 0.7 Saturated mixture 140 361.53 1800 0.565 Saturated mixture 177.66 950 752.74 0.0 Saturated liquid 80 500 335.37 - - - Compressed liquid 350.08003162.2- - -Superheated vapor3-26Complete the following table for Refrigerant-134a :T, ︒C P, kPa v , m 3 / kg Phase description -12 320 0.000750 Compressed liquid 30 770.64 0.0065 Saturated mixture 18.73 550 0.03741 Saturated vapor 606000.04139Superheated vapor3-29A rigid container that is filled with water is cooled. The initial temperature and the final pressure are to be determined.Analysis This is a constant volume process. The specific volume is/kgm 150.0kg1m 150.03321====m Vv vThe initial state is superheated vapor. The temperature is determined to be6)-A (Table /kg m 150.0MPa21311 C 395︒=⎭⎬⎫==T P vThis is a constant volume cooling process (v = V /m = constant). The final state is saturated mixture and thus the pressure is the saturation pressure at the final temperature:4)-A (Table /kg m 150.0C40C 40 @sat 23122 kPa 7.385==⎭⎬⎫==︒=︒P P T v vvQ3-33A spring-loaded piston-cylinder device is filled with R-134a. The water now undergoes a process until its volume increases by 40%. The final temperature and the enthalpy are to be determined.Analysis From Table A-11E, the initial specific volume is/lbmft 5463.3)01143.04300.4)(80.0(01143.0311=-+=+=fgfx vvvand the initial volume will be3311ft0.7093/lbm)ft 3lbm)(3.546 2.0(===v V mWith a 40% increase in the volume, the final volume will be3312ft0.9930)ft 7093.0(4.14.1===V VThe distance that the piston moves between the initial and final conditions isft3612.0ft)1(ft)7093.09930.0(44/232=-=∆=∆=∆ππDA x pVVAs a result of the compression of the spring, the pressure difference between the initial and final states ispsia1.42lbf/in42.1in)12(in)123612.0(lbf/in) 37(44/222==⨯=∆=∆=∆=∆ππDxk A x k A F P ppThe initial pressure is11E)-A (Table psia 87.9F 30- @sat 1 ==︒P PThe final pressure is then psia 29.1142.187.912=+=∆+=P P P and the final specific volume is/lbmft 965.4lbm2.0ft 0.99303322===mV vAt this final state, the temperature and enthalpy areEES) (from /lbm ft 965.4psia29.1111322 Btu/lbm119.9F81.5=︒=⎭⎬⎫==h T P vNote that it is very difficult to get the temperature and enthalpy from Table A-13E accurately.3-39vA piston-cylinder device fitted with stops contains water at a specified state. Now the water is cooled until a final pressure. The process is to be indicated on the T -v diagram and the change in internal energy is to be determined. Analysis The process is shown on T-v diagram. The internal energy at the initial state is6)-A (Table kJ/kg 9.2728 C 250kPa 300111 =⎭⎬⎫︒==u T PState 2 is saturated vapor at the initial pressure. Then,5)-A (Table /kg m 6058.0 vapor)(sat. 1kPa3003222 =⎭⎬⎫==v x PProcess 2-3 is a constant-volume process. Thus,5)-A (Table kJ/kg 3.1163 /kg m 6058.0kPa1003333 =⎭⎬⎫==u P vThe overall change in internal energy iskJ/kg1566=-=-=∆3.11639.272831u u u3-46A vertical piston-cylinder device is filled with water and covered with a 20-kg piston that serves as the lid. The boiling temperature of water is to be determined.Analysis The pressure in the cylinder is determined from a force balance on the piston, PA = P atm A + Wor,kPa119.61s kg/m 1000kPa1m 0.01)m/s kg)(9.81 (20kPa) (100222atm =⎪⎪⎭⎫ ⎝⎛⋅+=+=Amg P PThe boiling temperature is the saturation temperature corresponding to this pressure,C104.7︒==k Pa119.61@sat T T (Table A-5)3-58A piston-cylinder device that is filled with R-134a is heated. The volume change is to be determined.v250︒QAnalysis The initial specific volume is13)-A (Table /kg m 33608.0 C 20kPa 603111 =⎭⎬⎫︒-==v T Pand the initial volume is3311m0.033608/kg)m 8kg)(0.3360 100.0(===v V mAt the final state, we have13)-A (Table /kg m 0.50410 C 100kPa 603222 =⎭⎬⎫︒==v T P3322m0.050410/kg)m 0kg)(0.5041 100.0(===v V mThe volume change is then3m0.0168=-=-=∆0336080050410012..V V V3-73An elastic tank contains air at a specified state. The volume is doubled at the same pressure. The initial volume and the final temperature are to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Analysis According to the ideal gas equation of state,F590R 1050ft 404.93︒==−→−+=−→−==+⋅⋅==2212123R460)(652R460)R)(65/lbmol ft psia 73lbmol)(10. 3.2(psia) (32T T T T TnR P u V V V V V3-77An automobile tire is inflated with air. The pressure rise of air in the tire when the tireis heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined.Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant.Properties The gas constant of air is R = 0.287 kPa.m 3/kg.K (Table A-1). Analysis Initially, the absolute pressure in the tire is2TirekPa310100210atm 1=+=+=P P P gTreating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined fromkPa336kPa) (310K298K 3231122222111===−→−=P T T P T P T P V VThus the pressure rise is∆P P P =-=-=2133631026kPaThe amount of air that needs to be bled off to restore pressure to its original value iskg0.0070=-=-=∆=⋅⋅===⋅⋅==0.08360.0906kg0.0836K)K)(323/kg m kPa (0.287)m kPa)(0.025 (310kg0.0906K)K)(298/kg m kPa (0.287)m kPa)(0.025 (310213321233111m m m RT P m RT P m V V3-90CO 2 gas flows through a pipe. The volume flow rate and the density at the inlet and the volume flow rate at the exit of the pipe are to be determined.Properties The gas constant, the critical pressure, and the critical temperature of CO 2 are (Table A-1) R = 0.1889 kPa·m 3/kg·K,T cr = 304.2 K, P cr = 7.39 MPaAnalysis(a ) From the ideal gas equation of state, error)(2.1%/kg m 0.062973=⋅⋅==kPa)(3000K) K)(500/kg m kPa 89kg/s)(0.18 (23111P RT m Verror)(2.1%mkg/ 31.763=⋅⋅==K)K)(500/kg m kPa (0.1889kPa)(30003111RT P ρerror)(3.6%/kg m 0.056673=⋅⋅==kPa)(3000K) K)(450/kg m kPa 89kg/s)(0.18 (23222P RT m V(b ) From the compressibility chart (EES function for compressibility factor is used)9791.01.64K 304.2K 5000.407MPa 7.39MPa3111,1=⎪⎪⎭⎪⎪⎬⎫======Z T T T P P P cr R cr R9656.01.48K 304.2K 4500.407MPa 7.39MPa3222,2=⎪⎪⎭⎪⎪⎬⎫======Z T T T P P P crR cr RThus, /kgm 0.061653=⋅⋅==kPa) (3000K) K)(500/kg m kPa 89kg/s)(0.18 (0.9791)(231111P RT m Z V3mkg/ 32.44=⋅⋅==K)K)(500/kg m kPa .1889(0.9791)(0kPa)(300031111RT Z P ρ/kg m 0.054723=⋅⋅==kPa)(3000K) K)(450/kg m kPa 89kg/s)(0.18 (0.9656)(232222P RT m Z V。
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4. Consider a device with one inlet and one exit.
If the volume flow rates at the inlet and at the
exit are the same, is the flow through the device necessarily steady？ Why?
heat rejection. The highest possible efficiency of
this heat engine is
[a] 6.2％
[b] 15.6 ％
[c] 50.0 ％
[d] 93.8 ％
[e] 100 ％
5.3 A heat engine receives heat from a source at 100 ℃ and rejects the waste heat to a sink at 50 ℃ .If heat is supplied to this engine at a rate of
e) What is the change in entropy of the surroundings?
f) Is this process reversible or irreversible? How do you know? (Note: you do not have to prove your answer mathematically to get full credit. Explain in a sentence how you know whether the process is reversible or irreversible)
2. On a P-v diagram, what does the area under the process curve represent?
3. Determine the energy required to accelerate a 2000-kg car from 20 to 70 km/h on an uphill road with a vertical rise of 40m.
CHAPTER 5 THE SECOND LAW OF THERMODYNAMICS
5.1 The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 22℃ to 8 ℃ at an average rate of 8kg/h .If the COP of this refrigerator is 3.1,the required power input to this refrigerator is
c) What is the final temperature of the system Tfinal expressed in terms of TA and TB ?
b) What is the change in entropy of compartment A? What is the change in entropy of compartment B?
100kj/s,the maximum power this heat engine can
produce is
[a] 25.4kW [c] 174.6 kW [e] 100.0kW
[b] 55.4kW [d] 95.0kW
5.4 Steam is compressed from 8 MPa and 300 ℃ to 10 MPa isentropically. The final temperature of the steam is
c) What is the thermodynamic efficiency of this cycle?
2. The sketch below shows a perfectly insulated container with two compartments separated by a non-adiabatic, frictionless piston. Both compartments are at the same pressure (PA= PB) and contain an equal amount of the same gas ( mA =mB ). Initially compartment A is at a higher temperature than compartment B(TA>TB).
[a] -38 ℃ [e]144 ℃
[b]30 ℃
[c]40 ℃
[d] 53 ℃
2.2 A 5 m3 rigid vessel contains steam at 20 MPa and 400
℃.The mass of the steam is
[a] 5.0kg [b] 0.322kg [c] 322kg [e]680kg
b) Is the final internal energy Ufinal of the system (both compartments together) lower, equal or higher than initial internal energy Uinitial of the system? Why? (A sentence or two is expected, perhaps bolstered by some equations.)
[a] 290 ℃ [b] 300 ℃ [c] 320 ℃ [d] 330 ℃ [e] 340 ℃
5. A 2 m3 rigid tank contains nitrogen gas at 500kPa and 300K. Now heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 800kPa. The work done during this process is
[a] 705kg [b] 500kg
[c]258kg [d] 635kg
[e]2809
Chapter 3 The first Law of Thermodynamics
A reversible cycle plots as a perfect circle on a T-S diagram with maximum and minimum temperatures 600 K and 300K and a maximum and minimum entropy of 600 kJ/K and 300kJ/K.
A system is capable of executing a cyclic process as indicated in the P V sketch; it may be executed either clockwise abca or counterclockwise adca. (a) When going clockwise to state c, 80KJ of heat flow from the system. Find the work along the path ca. (b) When going counterclockwise to state c, 70KJ of heat flow from the system. Find the work during the process adc.
10 kg of air contained in a piston cylinder
apparatus at 5 bar,300K, expands isothermally
until the pressure is 1 bar. Determine:
(i) the work done.
a) What is the net work of this cycle?
b) Indicate the path in the cycle along which heat is rejected. How do you know?(A sentence or two is required, perhaps bolstered by an equation).
(ii) the internal energy change of the air.
(iii) the heat transfer.
Assume that air behaves as an ideal gas and pressure variation is gradual and continuous.
[a] 28W [c] 88W [e] 403W
[b] 42W [d] 130W
5.2 A heat engine cycle is executed with steam
in the saturation dome. The pressure of steam
1MPa during heat addition, and 0.5MPa during
(A) 600 kJ
(B) 1000 kJ
(C) 0 kJ(D) 5源自0 kJ(E) 1600 kJ
6. A 2 m3 rigid tank contains nitrogen gas at 500kPa and 300K. Now heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 800kPa. The work done during this process is