Resolução Termodinâmica Çengel capítulo 3 completo, Exercícios de Termodinâmica

Baixe Resolução Termodinâmica Çengel capítulo 3 completo e outras Exercícios em PDF para Termodinâmica, somente na Docsity! 3-1 Chapter 3 PROPERTIES OF PURE SUBSTANCES Pure Substances, Phase Change Processes, Property Diagrams 3-1C Yes. Because it has the same chemical composition throughout. 3-2C A liquid that is about to vaporize is saturated liquid; otherwise it is compressed liquid. 3-3C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor. 3-4C No. 3-5C No. 3-6C Yes. The saturation temperature of a pure substance depends on pressure. The higher the pressure, the higher the saturation or boiling temperature. 3-7C The temperature will also increase since the boiling or saturation temperature of a pure substance depends on pressure. 3-8C Because one cannot be varied while holding the other constant. In other words, when one changes, so does the other one. 3-9C At critical point the saturated liquid and the saturated vapor states are identical. At triple point the three phases of a pure substance coexist in equilibrium. 3-10C Yes. 3-11C Case (c) when the pan is covered with a heavy lid. Because the heavier the lid, the greater the pressure in the pan, and thus the greater the cooking temperature. 3-12C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into a vapor. At subcritical pressures, there is always a distinct surface between the phases. Property Tables 3-13C A given volume of water will boil at a higher temperature in a tall and narrow pot since the pressure at the bottom (and thus the corresponding saturation pressure) will be higher in that case. 3-14C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperature and thus the corresponding saturation pressure inside the pan drops. An easy way of removing the lid is to reheat the food. When the temperature rises to boiling level, the pressure rises to atmospheric value and thus the lid will come right off. 3-2 3-15C The molar mass of gasoline (C8H18) is 114 kg/kmol, which is much larger than the molar mass of air that is 29 kg/kmol. Therefore, the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than the surrounding air. As a result, the warm mixture of air and gasoline on top of an open gasoline will most likely settle down instead of rising in a cooler environment 3-16C Ice can be made by evacuating the air in a water tank. During evacuation, vapor is also thrown out, and thus the vapor pressure in the tank drops, causing a difference between the vapor pressures at the water surface and in the tank. This pressure difference is the driving force of vaporization, and forces the liquid to evaporate. But the liquid must absorb the heat of vaporization before it can vaporize, and it absorbs it from the liquid and the air in the neighborhood, causing the temperature in the tank to drop. The process continues until water starts freezing. The process can be made more efficient by insulating the tank well so that the entire heat of vaporization comes essentially from the water. 3-17C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance. 3-18C No. Because in the thermodynamic analysis we deal with the changes in properties; and the changes are independent of the selected reference state. 3-19C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified temperature or pressure. It can be determined from hfg = hg - hf . 3-20C Yes; the higher the temperature the lower the hfg value. 3-21C Quality is the fraction of vapor in a saturated liquid-vapor mixture. It has no meaning in the superheated vapor region. 3-22C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the hfg . 3-23C Yes. It decreases with increasing pressure and becomes zero at the critical pressure. 3-24C No. Quality is a mass ratio, and it is not identical to the volume ratio. 3-25C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus . TfPT @, vv ≅ 3-26 [Also solved by EES on enclosed CD] Complete the following table for H2 O: T, °C P, kPa v, m3 / kg Phase description 50 12.352 4.16 Saturated mixture 120.21 200 0.8858 Saturated vapor 250 400 0.5952 Superheated vapor 110 600 0.001051 Compressed liquid 3-5 u=IntEnergy(Fluid$,p=p,x=x) endif Endif If Prop1$='Enthalpy, kJ/kg' Then h=Value1 If Prop2$='Enthalpy, kJ/kg' then Call Error('Both properties cannot be Enthalpy, h=xxxF2',h) if Prop2$='Pressure, kPa' then p=value2 T=Temperature(Fluid$,h=h,P=p) s=entropy(Fluid$,h=h,P=p) v=volume(Fluid$,h=h,P=p) u=intenergy(Fluid$,h=h,P=p) x=quality(Fluid$,h=h,P=p) endif if Prop2$='Temperature, C' then T=value2 p=Pressure(Fluid$,T=T,h=h) s=entropy(Fluid$,T=T,h=h) v=volume(Fluid$,T=T,h=h) u=intenergy(Fluid$,T=T,h=h) x=quality(Fluid$,T=T,h=h) endif if Prop2$='Entropy, kJ/kg-K' then s=value2 p=Pressure(Fluid$,h=h,s=s) T=Temperature(Fluid$,h=h,s=s) v=volume(Fluid$,h=h,s=s) u=intenergy(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 p=Pressure(Fluid$,h=h,v=v) T=Temperature(Fluid$,h=h,v=v) s=entropy(Fluid$,h=h,v=v) u=intenergy(Fluid$,h=h,v=v) x=quality(Fluid$,h=h,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 p=Pressure(Fluid$,h=h,u=u) T=Temperature(Fluid$,h=h,u=u) s=entropy(Fluid$,h=h,u=u) v=volume(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,u=u) endif if Prop2$='Quality' then x=value2 p=Pressure(Fluid$,h=h,x=x) T=Temperature(Fluid$,h=h,x=x) s=entropy(Fluid$,h=h,x=x) v=volume(Fluid$,h=h,x=x) u=IntEnergy(Fluid$,h=h,x=x) endif endif If Prop1$='Entropy, kJ/kg-K' Then 3-6 s=Value1 If Prop2$='Entropy, kJ/kg-K' then Call Error('Both properties cannot be Entrolpy, h=xxxF2',s) if Prop2$='Pressure, kPa' then p=value2 T=Temperature(Fluid$,s=s,P=p) h=enthalpy(Fluid$,s=s,P=p) v=volume(Fluid$,s=s,P=p) u=intenergy(Fluid$,s=s,P=p) x=quality(Fluid$,s=s,P=p) endif if Prop2$='Temperature, C' then T=value2 p=Pressure(Fluid$,T=T,s=s) h=enthalpy(Fluid$,T=T,s=s) v=volume(Fluid$,T=T,s=s) u=intenergy(Fluid$,T=T,s=s) x=quality(Fluid$,T=T,s=s) endif if Prop2$='Enthalpy, kJ/kg' then h=value2 p=Pressure(Fluid$,h=h,s=s) T=Temperature(Fluid$,h=h,s=s) v=volume(Fluid$,h=h,s=s) u=intenergy(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 p=Pressure(Fluid$,s=s,v=v) T=Temperature(Fluid$,s=s,v=v) h=enthalpy(Fluid$,s=s,v=v) u=intenergy(Fluid$,s=s,v=v) x=quality(Fluid$,s=s,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 p=Pressure(Fluid$,s=s,u=u) T=Temperature(Fluid$,s=s,u=u) h=enthalpy(Fluid$,s=s,u=u) v=volume(Fluid$,s=s,s=s) x=quality(Fluid$,s=s,u=u) endif if Prop2$='Quality' then x=value2 p=Pressure(Fluid$,s=s,x=x) T=Temperature(Fluid$,s=s,x=x) h=enthalpy(Fluid$,s=s,x=x) v=volume(Fluid$,s=s,x=x) u=IntEnergy(Fluid$,s=s,x=x) endif Endif if x<0 then State$='in the compressed liquid region.' if x>1 then State$='in the superheated region.' If (x<1) and (X>0) then State$='in the two-phase region.' If (x=1) then State$='a saturated vapor.' if (x=0) then State$='a saturated liquid.' 3-7 end "Input from the diagram window" {Fluid$='Steam' Prop1$='Temperature' Prop2$='Pressure' Value1=50 value2=101.3} Call Find(Fluid$,Prop1$,Prop2$,Value1,Value2:T,p,h,s,v,u,x,State$) T[1]=T ; p[1]=p ; h[1]=h ; s[1]=s ; v[1]=v ; u[1]=u ; x[1]=x "Array variables were used so the states can be plotted on property plots." ARRAYS TABLE h KJ/kg P kPa s kJ/kgK T C u KJ/kg v m3/kg x 2964.5 400 7.3804 250 2726.4 0.5952 100 0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,0 0 100 200 300 400 500 600 700 s [kJ/kg-K] T [C ] 8600 kPa 2600 kPa 500 kPa 45 kPa Steam 10-4 10-3 10-2 10-1 100 101 102 103 0 100 200 300 400 500 600 700 v [m3/kg] T [C ] 8600 kPa 2600 kPa 500 kPa 45 kPa Steam 3-10 3-31 Complete the following table for Refrigerant-134a: T, °C P, kPa v, m3 / kg Phase description -8 320 0.0007569 Compressed liquid 30 770.64 0.015 Saturated mixture -12.73 180 0.11041 Saturated vapor 80 600 0.044710 Superheated vapor 3-32 Complete the following table for Refrigerant-134a: T, °C P, kPa u, kJ / kg Phase description 20 572.07 95 Saturated mixture -12 185.37 35.78 Saturated liquid 86.24 400 300 Superheated vapor 8 600 62.26 Compressed liquid 3-33E Complete the following table for Refrigerant-134a: T, °F P, psia h, Btu / lbm x Phase description 65.89 80 78 0.566 Saturated mixture 15 29.759 69.92 0.6 Saturated mixture 10 70 15.35 - - - Compressed liquid 160 180 129.46 - - - Superheated vapor 110 161.16 117.23 1.0 Saturated vapor 3-34 Complete the following table for H2 O: T, °C P, kPa v, m3 / kg Phase description 140 361.53 0.05 Saturated mixture 155.46 550 0.001097 Saturated liquid 125 750 0.001065 Compressed liquid 500 2500 0.140 Superheated vapor 3-35 Complete the following table for H2 O: T, °C P, kPa u, kJ / kg Phase description 143.61 400 1450 Saturated mixture 220 2319.6 2601.3 Saturated vapor 190 2500 805.15 Compressed liquid 466.21 4000 3040 Superheated vapor 3-11 3-36 A rigid tank contains steam at a specified state. The pressure, quality, and density of steam are to be determined. Properties At 220°C vf = 0.001190 m3/kg and vg = 0.08609 m3/kg (Table A-4). Analysis (a) Two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. The pressure of the steam is the saturation pressure at the given temperature. Then the pressure in the tank must be the saturation pressure at the specified temperature, kPa 2320== °C@220satTP Steam 1.8 m3 220°C (b) The total mass and the quality are determined as 0.0269=== =+=+= = × == = × == 518.1 13.94 kg 1.51894.132.504 kg 3.941 /kgm 0.08609 )m (1.82/3 kg .2504 /kgm 0.001190 )m (1.81/3 3 3 3 3 t g gft g g g f f f m m x mmm m m v V v V (c) The density is determined from 3mkg/ 287.8=== =+=−+= 003474.0 11 /kgm 003474.0)08609.0)(0269.0(001190.0)( 3 v vvvv ρ fgf x 3-37 A piston-cylinder device contains R-134a at a specified state. Heat is transferred to R-134a. The final pressure, the volume change of the cylinder, and the enthalpy change are to be determined. Analysis (a) The final pressure is equal to the initial pressure, which is determined from kPa 90.4=      +=+== 22 2 2atm12 kg.m/s 1000 kN 1 /4m) (0.25 )m/s kg)(9.81 (12kPa 88 /4 ππD gm PPP p (b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10°C and at the final state of 90.4 kPa and 15°C are (from EES) v1 = 0.2302 m3/kg h1 = 247.76 kJ/kg QR-134a 0.85 kg -10°C v 2 = 0.2544 m3/kg h2 = 268.16 kJ/kg The initial and the final volumes and the volume change are 3m 0.0205=−=∆ === === 1957.02162.0 m 2162.0/kg)m kg)(0.2544 85.0( m 1957.0/kg)m kg)(0.2302 85.0( 33 22 33 11 V vV vV m m (c) The total enthalpy change is determined from kJ/kg 17.4=−=−=∆ kJ/kg 247.76)6kg)(268.1 85.0()( 12 hhmH 3-12 3-38E The temperature in a pressure cooker during cooking at sea level is measured to be 250°F. The absolute pressure inside the cooker and the effect of elevation on the answer are to be determined. Assumptions Properties of pure water can be used to approximate the properties of juicy water in the cooker. Properties The saturation pressure of water at 250°F is 29.84 psia (Table A-4E). The standard atmospheric pressure at sea level is 1 atm = 14.7 psia. Analysis The absolute pressure in the cooker is simply the saturation pressure at the cooking temperature, psia 29.84== °Fsat@250abs PP H2O 250°F It is equivalent to atm 2.03=      = psia 14.7 atm 1psia 84.29absP The elevation has no effect on the absolute pressure inside when the temperature is maintained constant at 250°F. 3-39E The local atmospheric pressure, and thus the boiling temperature, changes with the weather conditions. The change in the boiling temperature corresponding to a change of 0.3 in of mercury in atmospheric pressure is to be determined. Properties The saturation pressures of water at 200 and 212°F are 11.538 and 14.709 psia, respectively (Table A-4E). One in. of mercury is equivalent to 1 inHg = 3.387 kPa = 0.491 psia (inner cover page). Analysis A change of 0.3 in of mercury in atmospheric pressure corresponds to psia 0.147 inHg 1 psia 0.491inHg) 3.0( =      =∆P P±0.3 inHg At about boiling temperature, the change in boiling temperature per 1 psia change in pressure is determined using data at 200 and 212°F to be F/psia 783.3 psia )538.11709.14( F)200212( °= − °− = P T ∆ ∆ Then the change in saturation (boiling) temperature corresponding to a change of 0.147 psia becomes F0.56°°=∆°=∆ =psia) 147F/psia)(0. 783.3(F/psia) 783.3(boiling PT which is very small. Therefore, the effect of variation of atmospheric pressure on the boiling temperature is negligible. 3-15 3-45 Saturated steam at Tsat = 30°C condenses on the outer surface of a cooling tube at a rate of 45 kg/h. The rate of heat transfer from the steam to the cooling water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The condensate leaves the condenser as a saturated liquid at 30°C. Properties The properties of water at the saturation temperature of 30°C are hfg = 2429.8 kJ/kg (Table A- 4). Analysis Noting that 2429.8 kJ of heat is released as 1 kg of saturated vapor at 30°C condenses, the rate of heat transfer from the steam to the cooling water in the tube is determined directly from 30°C kW 30.4=kJ/h 341,109kJ/kg) .8kg/h)(2429 45(evap === fghmQ && 3-46 The average atmospheric pressure in Denver is 83.4 kPa. The boiling temperature of water in Denver is to be determined. Analysis The boiling temperature of water in Denver is the saturation temperature corresponding to the atmospheric pressure in Denver, which is 83.4 kPa: (Table A-5) C94.6°== kPa [email protected] 3-47 The boiling temperature of water in a 5-cm deep pan is given. The boiling temperature in a 40-cm deep pan is to be determined. Assumptions Both pans are full of water. Properties The density of liquid water is approximately ρ = 1000 kg/m3. Analysis The pressure at the bottom of the 5-cm pan is the saturation pressure corresponding to the boiling temperature of 98°C: (Table A-4) kPa 94.39Csat@98 == oPP The pressure difference between the bottoms of two pans is kPa 3.43 skg/m 1000 kPa 1 m) )(0.35m/s )(9.807kg/m (1000 2 23 =        ⋅ == hgP ρ∆ L = 35 m D = 3 cm 40 cm 5 cm Then the pressure at the bottom of the 40-cm deep pan is P = 94.39 + 3.43 = 97.82 kPa Then the boiling temperature becomes (Table A-5) C99.0°== kPa [email protected] TT 3-48 A cooking pan is filled with water and covered with a 4-kg lid. The boiling temperature of water is to be determined. Analysis The pressure in the pan is determined from a force balance on the lid, PA = PatmA + W or, kPa 102.25 skg/m 1000 kPa 1 m) (0.1 )m/s kg)(9.81 (4kPa) (101 22 2 =       ⋅ += += π A mgPP atm The boiling temperature is the saturation temperature corresponding to this pressure, P Patm W = mg (Table A-5) C100.2kPa 102.25@sat °== TT 3-16 3-49 EES Problem 3-48 is reconsidered. Using EES (or other) software, the effect of the mass of the lid on the boiling temperature of water in the pan is to be investigated. The mass is to vary from 1 kg to 10 kg, and the boiling temperature is to be plotted against the mass of the lid. Analysis The problem is solved using EES, and the solution is given below. "Given data" {P_atm=101[kPa]} D_lid=20 [cm] {m_lid=4 [kg]} "Solution" "The atmospheric pressure in kPa varies with altitude in km by the approximate function:" P_atm=101.325*(1-0.02256*z)^5.256 "The local acceleration of gravity at 45 degrees latitude as a function of altitude in m is given by:" g=9.807+3.32*10^(-6)*z*convert(km,m) "At sea level:" z=0 "[km]" A_lid=pi*D_lid^2/4*convert(cm^2,m^2) W_lid=m_lid*g*convert(kg*m/s^2,N) P_lid=W_lid/A_lid*convert(N/m^2,kPa) P_water=P_lid+P_atm T_water=temperature(steam_iapws,P=P_water,x=0) mlid [kg] Twater [C] 1 100.1 2 100.1 3 100.2 4 100.3 5 100.4 6 100.5 7 100.6 8 100.7 9 100.7 10 100.8 1 2 3 4 5 6 7 8 9 10 100 100.1 100.2 100.3 100.4 100.5 100.6 100.7 100.8 100.9 mlid [kg] T w at er [ C ] r 0 1 2 3 4 5 6 7 8 9 30 40 50 60 70 80 90 100 110 z [km] P w at e [ kP a] Effect of altitude on boiling pressure of water in pan with lid mass of lid = 4 kg P w at er [ kP a] 0 1 2 3 4 5 6 7 8 9 70 75 80 85 90 95 100 105 z [km] T w at er [ C ] mass of lid = 4 kg Effect of altitude on boiling temperature of water in pan with lid T w at er [ C ] 3-17 3-50 A 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 = PatmA + W P Patm W = mg or, kPa 119.61 skg/m 1000 kPa 1 m 0.01 )m/s kg)(9.81 (20kPa) (100 22 2 atm =       ⋅ += += A mgPP The boiling temperature is the saturation temperature corresponding to this pressure, (Table A-5) C104.7°== kPa 119.61@satTT 3-51 A rigid tank that is filled with saturated liquid-vapor mixture is heated. The temperature at which the liquid in the tank is completely vaporized is to be determined, and the T-v diagram is to be drawn. Analysis This is a constant volume process (v = V /m = constant), H2O 75°C and the specific volume is determined to be /kgm 0.1667 kg 15 m 2.5 3 3 === m V v When the liquid is completely vaporized the tank will contain saturated vapor only. Thus, T 1 2 /kgm 0.1667 32 == gvv The temperature at this point is the temperature that corresponds to this vg value, (Table A-4) C187.0°== = /kgm 0.1667@sat 3gvTT v 3-52 A rigid vessel is filled with refrigerant-134a. The total volume and the total internal energy are to be determined. Properties The properties of R-134a at the given state are (Table A-13). R-134a 2 kg 800 kPa 120°C /kgm 0.037625 kJ/kg .87327 C012 kPa 008 3= =    = = v u T P o Analysis The total volume and internal energy are determined from kJ 655.7 m 0.0753 3 === === kJ/kg) kg)(327.87 (2 /kg)m 25kg)(0.0376 (2 3 muU mvV 3-20 3-56 EES Problem 3-55 is reconsidered. The effect of pressure on the total mass of water in the tank as the pressure varies from 0.1 MPa to 1 MPa is to be investigated. The total mass of water is to be plotted against pressure, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. P[1]=800 [kPa] P[2]=P[1] T[2]=350 [C] V_f1 = 0.1 [m^3] V_g1=0.9 [m^3] spvsat_f1=volume(Steam_iapws, P=P[1],x=0) "sat. liq. specific volume, m^3/kg" spvsat_g1=volume(Steam_iapws,P=P[1],x=1) "sat. vap. specific volume, m^3/kg" m_f1=V_f1/spvsat_f1 "sat. liq. mass, kg" m_g1=V_g1/spvsat_g1 "sat. vap. mass, kg" m_tot=m_f1+m_g1 V[1]=V_f1+V_g1 spvol[1]=V[1]/m_tot "specific volume1, m^3" T[1]=temperature(Steam_iapws, P=P[1],v=spvol[1])"C" "The final volume is calculated from the specific volume at the final T and P" spvol[2]=volume(Steam_iapws, P=P[2], T=T[2]) "specific volume2, m^3/kg" V[2]=m_tot*spvol[2] mtot [kg] P1 [kPa] 96.39 100 95.31 200 94.67 300 94.24 400 93.93 500 93.71 600 93.56 700 93.45 800 93.38 900 93.34 1000 10-3 10-2 10-1 100 101 102 100 101 102 103 104 105 v [m3/kg] P [k P a] 350 C Steam 1 2 P=800 kPa 100 200 300 400 500 600 700 800 900 1000 93 93.5 94 94.5 95 95.5 96 96.5 P[1] [kPa] m to t [k g] 3-21 3-57E Superheated water vapor cools at constant volume until the temperature drops to 250°F. At the final state, the pressure, the quality, and the enthalpy are to be determined. Analysis This is a constant volume process (v = V/m = constant), and the initial specific volume is determined to be (Table A-6E) /lbmft 3.0433 F500 psia 180 3 1 1 1 =    = = voT P H2O 180 psia 500°F At 250°F, vf = 0.01700 ft3/lbm and vg = 13.816 ft3/lbm. Thus at the final state, the tank will contain saturated liquid-vapor mixture since vf < v < vg , and the final pressure must be the saturation pressure at the final temperature, psia 29.84== F250@sat oPP T 1 2 (b) The quality at the final state is determined from 0.219= − − = − = 01700.0816.13 01700.00433.32 2 fg fx v vv (c) The enthalpy at the final state is determined from v Btu/lbm 426.0=×+=+= 41.945219.063.218fgf xhhh 3-22 3-58E EES Problem 3-57E is reconsidered. The effect of initial pressure on the quality of water at the final state as the pressure varies from 100 psi to 300 psi is to be investigated. The quality is to be plotted against initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. T[1]=500 [F] P[1]=180 [psia] T[2]=250 [F] v[ 1]=volume(steam_iapws,T=T[1],P=P[1]) v[2]=v[1] P[2]=pressure(steam_iapws,T=T[2],v=v[2]) h[2]=enthalpy(steam_iapws,T=T[2],v=v[2]) x[2]=quality(steam_iapws,T=T[2],v=v[2]) P1 [psia] x2 100 0.4037 122.2 0.3283 144.4 0.2761 166.7 0.2378 188.9 0.2084 211.1 0.1853 233.3 0.1665 255.6 0.1510 277.8 0.1379 300 0.1268 10-2 10-1 100 101 102 103 104 0 200 400 600 800 1000 1200 1400 v [ft3/lbm] T [° F] 1600 psia 780 psia 180 psia 29.82 psia 0.05 0.1 0.2 0.5 1.2 1.3 1.4 1.5 Btu/lbm-R Steam 1 2 100 140 180 220 260 300 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 P[1] [psia] x[ 2] 3-25 3-63E A rigid tank contains saturated liquid-vapor mixture of R-134a. The quality and total mass of the refrigerant are to be determined. Analysis At 50 psia, vf = 0.01252 ft3/lbm and vg = 0.94791 ft3/lbm (Table A-12E). The volume occupied by the liquid and the vapor phases are 33 ft 12andft 3 == gf VV R-134a 15 ft3 50 psia Thus the mass of each phase is lbm .6612 /lbmft 0.94791 ft 12 lbm .63239 /lbmft 0.01252 ft 3 3 3 3 3 === === g g g f f f m m v V v V Then the total mass and the quality of the refrigerant are mt = mf + mg = 239.63 + 12.66 = 252.29 lbm 0.05018=== lbm 252.29 lbm 12.66 t g m m x 3-64 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final temperature and the volume change are to be determined, and the process should be shown on a T-v diagram. Analysis (b) At the final state the cylinder contains saturated liquid-vapor mixture, and thus the final temperature must be the saturation temperature at the final pressure, H2O 300°C 1 MPa (Table A-5) C179.88°== MPa sat@1TT (c) The quality at the final state is specified to be x2 = 0.5. The specific volumes at the initial and the final states are (Table A-6) /kgm 0.25799 C300 MPa 1.0 3 1 1 1 =    = = voT P T 1 2 /kgm .097750 )001127.019436.0(5.0001127.0 5.0 MPa 1.0 3 22 2 2 = −×+= +=    = = fgf xx P vvv Thus, v 3m 20.128−=−=−= /kgm0.25799)5kg)(0.0977 (0.8)(∆ 312 vvV m 3-26 3-65 The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be determined. Analysis This is a constant volume process (v = V /m = constant), and the initial specific volume is equal to the final specific volume that is (Table A-4) /kgm 48392.0 3C150@21 === °gvvv °C 150 250 1 2 T since the vapor starts condensing at 150°C. Then from Table A-6, H2O T1= 250°C P1 = ? T MPa 0.60=    = °= 13 1 1 /kgm 0.39248 C025 P v v 3-66 Water is boiled in a pan by supplying electrical heat. The local atmospheric pressure is to be estimated. Assumptions 75 percent of electricity consumed by the heater is transferred to the water. Analysis The amount of heat transfer to the water during this period is kJ 2700s) 6030(kJ/s) 2)(75.0(timeelect =×== fEQ The enthalpy of vaporization is determined from kJ/kg 2269 kg 1.19 kJ 2700 boil === m Qh fg Using the data by a trial-error approach in saturation table of water (Table A-5) or using EES as we did, the saturation pressure that corresponds to an enthalpy of vaporization value of 2269 kJ/kg is Psat = 85.4 kPa which is the local atmospheric pressure. 3-27 3-67 Heat is supplied to a rigid tank that contains water at a specified state. The volume of the tank, the final temperature and pressure, and the internal energy change of water are to be determined. Properties The saturated liquid properties of water at 200°C are: vf = 0.001157 m3/kg and uf = 850.46 kJ/kg (Table A-4). Analysis (a) The tank initially contains saturated liquid water and air. The volume occupied by water is 33 11 m 001619.0/kg)m 001157.0kg)( 4.1( === vV m which is the 25 percent of total volume. Then, the total volume is determined from 3m 0.006476== )001619.0( 25.0 1 V (b) Properties after the heat addition process are kg/m 004626.0 kg 1.4 m 0.006476 33 2 === m V v kJ/kg 5.22011 kg/m 004626.0 2 2 2 2 3 2 = = °=     = = u P T x kPa 21,367 C371.3 v (Table A-4 or A-5 or EES) (c) The total internal energy change is determined from kJ 1892==−=∆ kJ/kg 850.46)-kg)(2201.5 4.1()( 12 uumU 3-68 Heat is lost from a piston-cylinder device that contains steam at a specified state. The initial temperature, the enthalpy change, and the final pressure and quality are to be determined. Analysis (a) The saturation temperature of steam at 3.5 MPa is [email protected] MPa = 242.6°C (Table A-5) Q Steam 3.5 MPa Then, the initial temperature becomes T1 = 242.6+5 = 247.6°C Also, (Table A-6) kJ/kg 1.2821 C6.247 MPa 5.3 1 1 1 =    °= = h T P (b) The properties of steam when the piston first hits the stops are (Table A-5) /kgm 001235.0 kJ/kg 7.1049 0 MPa 5.3 3 2 2 2 12 = =    = == v h x PP Then, the enthalpy change of steam becomes kJ/kg -1771=−=−=∆ 2821.17.104912 hhh (c) At the final state (Table A-4 or EES) 0.0006 kPa 1555 = =     °= == 3 3 3 3 23 C200 /kgm 001235.0 x P T vv The cylinder contains saturated liquid-vapor mixture with a small mass of vapor at the final state. 3-30 3-75 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is 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.m3/kg.K (Table A-1). Analysis Initially, the absolute pressure in the tire is Tire 25°C kPa310100210atm1 =+=+= PPP g Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from kPa 336kPa) (310 K 298 K 323 1 1 2 2 2 22 1 11 ===→= P T TP T P T P VV Thus the pressure rise is ∆P P P= − = − =2 1 336 310 26 kPa The amount of air that needs to be bled off to restore pressure to its original value is kg 0.0070=−=−=∆ = ⋅⋅ == = ⋅⋅ == 0.08360.0906 kg 0.0836 K) K)(323/kgmkPa (0.287 )m kPa)(0.025 (310 kg 0.0906 K) K)(298/kgmkPa (0.287 )m kPa)(0.025 (310 21 3 3 2 1 2 3 3 1 1 1 mmm RT Pm RT Pm V V 3-76E An automobile tire is under inflated with air. The amount of air that needs to be added to the tire to raise its pressure to the recommended value is 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 = 0.3704 psia.ft3/lbm.R (Table A-1E). Analysis The initial and final absolute pressures in the tire are Tire 0.53 ft3 90°F 20 psig P1 = Pg1 + Patm = 20 + 14.6 = 34.6 psia P2 = Pg2 + Patm = 30 + 14.6 = 44.6 psia Treating air as an ideal gas, the initial mass in the tire is lbm 0.0900 R) R)(550/lbmftpsia (0.3704 )ft psia)(0.53 (34.6 3 3 1 1 1 = ⋅⋅ == RT Pm V Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes lbm 0.1160 R) R)(550/lbmftpsia (0.3704 )ft psia)(0.53 (44.6 3 3 2 2 2 = ⋅⋅ == RT Pm V Thus the amount of air that needs to be added is ∆m m m= − = − =2 1 0.1160 0.0900 0.0260 lbm 3-31 3-77 The pressure and temperature of oxygen gas in a storage tank are given. The mass of oxygen in the tank is to be determined. Assumptions At specified conditions, oxygen behaves as an ideal gas Properties The gas constant of oxygen is R = 0.2598 kPa.m3/kg.K (Table A-1). Pg = 500 kPaAnalysis The absolute pressure of O2 is O2 V = 2.5 m3 T = 28°C P = Pg + Patm = 500 + 97 = 597 kPa Treating O2 as an ideal gas, the mass of O2 in tank is determined to be kg 19.08= +⋅⋅ == K273)K)(28/kgmkPa (0.2598 )m kPa)(2.5 (597 3 3 RT Pm V 3-78E A rigid tank contains slightly pressurized air. The amount of air that needs to be added to the tank to raise its pressure and temperature to the recommended values is to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tank remains constant. Properties The gas constant of air is R = 0.3704 psia.ft3/lbm.R (Table A-1E). Analysis Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be lbm 33.73 R) R)(550/lbmftpsia (0.3704 )ft 3psia)(196. (35 ft 196.3 psia 20 R) R)(530/lbmftpsia 4lbm)(0.370 (20 3 3 2 2 2 3 3 1 11 = ⋅⋅ == = ⋅⋅ == RT P m P RTm V V Air, 20 lbm 20 psia 70°F Thus the amount of air added is ∆m m m= − = − =2 1 33.73 20.0 13.73 lbm 3-79 A rigid tank contains air at a specified state. The gage pressure of the gas in the tank is to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Treating air as an ideal gas, the absolute pressure in the tank is determined from Pg kPa 1069.1 m 0.4 K) K)(298/kgmkPa kg)(0.287 (5 3 3 = ⋅⋅ == V mRTP Air 400 L 25°C Thus the gage pressure is kPa 972.1=−=−= 971069.1atmPPPg 3-32 3-80 Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank and the final equilibrium pressure when the valve is opened are to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and the mass of air in the first tank are determined to be kg 5.846 K) K)(298/kgmkPa (0.287 )m kPa)(1.0 (500 kPa 200 K) K)(308/kgmkPa kg)(0.287 (5 3 3 1 1 3 1 11 = ⋅⋅ =      = = ⋅⋅ =      = A A B B RT Pm P RTm V V 3m2.21 A B Thus, × Air m=5 kg T=35°C P=200 kPa Air V=1 m3 T=25°C P=500 kPa kg 10.8465.05.846 m 3.212.211.0 3 =+=+= =+=+= BA BA mmm VVV Then the final equilibrium pressure becomes kPa284.1 m 3.21 K) K)(293/kgmkPa kg)(0.287 (10.846 3 3 2 2 = ⋅⋅ == V mRTP 3-35 3-86 The specific volume of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1, R = 0.08149 kPa·m3/kg·K, Tcr = 374.2 K, Pcr = 4.059 MPa Analysis (a) From the ideal gas equation of state, )( kPa 900 K) K)(343/kgmkPa (0.08149 3 error 13.3%/kgm 0.03105 3=⋅⋅== P RT v (b) From the compressibility chart (Fig. A-15), R-134a 0.9 MPa 70°C 894.0 .9170 K 374.2 K 343 0.222 MPa 4.059 MPa 0.9 =        === === Z T TT P PP cr R cr R Thus, error)(1.3%/kgm 0.02776 3=== /kg)m 03105(0.894)(0. 3idealvv Z (c) From the superheated refrigerant table (Table A-13), } /kgm 0.027413 3=°== vC07 MPa .90TP 3-87 The specific volume of nitrogen gas is to be determined using the ideal gas relation and the compressibility chart. The errors involved in these two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of nitrogen are, from Table A-1, R = 0.2968 kPa·m3/kg·K, Tcr = 126.2 K, Pcr = 3.39 MPa Analysis (a) From the ideal gas equation of state, )error %4.86( kPa 10,000 K) K)(150/kgmkPa (0.2968 3 /kgm0.004452 3= ⋅⋅ == P RT v (b) From the compressibility chart (Fig. A-15), N2 10 MPa 150 K 54.0 1.19 K 126.2 K 150 2.95 MPa 3.39 MPa 10 =        === === Z T TT P PP cr R cr R Thus, error) (0.7% /kg)m 04452(0.54)(0.0 3ideal /kgm0.002404 3=== vv Z 3-36 3-88 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1, R = 0.4615 kPa·m3/kg·K, Tcr = 647.1 K, Pcr = 22.06 MPa Analysis (a) From the ideal gas equation of state, error) (3.7%/kgm 0.09533 3=⋅⋅== kPa 3500 K) K)(723/kgmkPa (0.4615 3 P RT v (b) From the compressibility chart (Fig. A-15), H2O 3.5 MPa 450°C 961.0 .121 K 647.1 K 723 0.159 MPa 22.06 MPa 3.5 =       === === Z T TT P PP cr R cr R Thus, error) (0.4%/kgm 0.09161 3=== /kg)m 09533(0.961)(0. 3idealvv Z (c) From the superheated steam table (Table A-6), } /kgm 0.09196 3=°== v C450MPa .53TP 3-89E The temperature of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a tables. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1E, R = 0.10517 psia·ft3/lbm·R, Tcr = 673.6 R, Pcr = 588.7 psia Analysis (a) From the ideal gas equation of state, R527.2 R)/lbmftpsia (0.10517 /lbm)ft 86psia)(0.13 (400 3 3 = ⋅⋅ == R PT v (b) From the compressibility chart (Fig. A-15a), 03.1 1.15 R) R)(673.65/lbmftpsia (0.10517 psia) 7/lbm)(588.ft (0.1386 / 0.678 psia 588.7 psia 400 3 3 actual =        = ⋅⋅ == === R crcr R cr R T PRT P PP v v Thus, R 693.8=×== 6.67303.1crRTTT (c) From the superheated refrigerant table (Table A-13E), R) (700F240°=    = = TP /lbmft 0.13853 psia 400 3v 3-37 3-90 The pressure of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a tables. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1, R = 0.08149 kPa·m3/kg·K, Tcr = 374.2 K, Pcr = 4.059 MPa Analysis The specific volume of the refrigerant is /kgm 0.016773 kg 1 m 0.016773 3 3 === m V v R-134a 0.016773 m3/kg 110°C (a) From the ideal gas equation of state, kPa 1861=⋅⋅== /kgm 0.016773 K) K)(383/kgmkPa (0.08149 3 3 v RTP (b) From the compressibility chart (Fig. A-15), 39.0 2.24 kPa) K)/(4059 K)(374.2/kgmkPa (0.08149 /kgm 0.016773 1.023 K 374.2 K 383 3 3 crcr actual cr =        = ⋅⋅ == === R R R P /PRT T TT v v Thus, kPa 1583=== kPa) 4059(0.39)(crPPP R (c) From the superheated refrigerant table (Table A-13), kPa 1600=    = = PT /kgm 0.016773 C110 3v o 3-91 Somebody claims that oxygen gas at a specified state can be treated as an ideal gas with an error less than 10%. The validity of this claim is to be determined. Properties The critical pressure, and the critical temperature of oxygen are, from Table A-1, T Pcr cr= =154.8 K and 5.08 MPa Analysis From the compressibility chart (Fig. A-15), 79.0 1.034 K 154.8 K 160 0.591 MPa 5.08 MPa 3 =        === === Z T TT P PP cr R cr R O2 3 MPa 160 K Then the error involved can be determined from %. .Z 626 790 1111Error ideal −=−=−= − = v vv Thus the claim is false. 3-40 Other Equations of State 3-95C The constant a represents the increase in pressure as a result of intermolecular forces; the constant b represents the volume occupied by the molecules. They are determined from the requirement that the critical isotherm has an inflection point at the critical point. 3-96 The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas, van der Waals, and Beattie-Bridgeman equations. The error involved in each case is to be determined. Properties The gas constant, molar mass, critical pressure, and critical temperature of nitrogen are (Table A-1) R = 0.2968 kPa·m3/kg·K, M = 28.013 kg/kmol, Tcr = 126.2 K, Pcr = 3.39 MPa Analysis The specific volume of nitrogen is N2 0.0327 m3/kg 175 K /kgm 0.0327 kg 100 m 3.27 33 === m V v (a) From the ideal gas equation of state, error) (5.5%kPa 1588=⋅⋅== /kgm 0.0327 K) K)(175/kgmkPa (0.2968 3 3 v RTP (b) The van der Waals constants for nitrogen are determined from a R T P b RT P cr cr cr cr = = ⋅ ⋅ = ⋅ = = ⋅ ⋅ × = 27 64 8 2 2 (27)(0.2968 kPa m / kg K) (126.2 K) (64)(3390 kPa) 0.175 m kPa / kg (0.2968 kPa m / kg K)(126.2 K) 8 3390 kPa 0.00138 m / kg 3 2 2 6 2 3 3 Then, error) (0.7%kPa 1495=− − × =− − = 22 (0.0327) 0.175 0.001380.0327 1750.2968 vv a b RTP (c) The constants in the Beattie-Bridgeman equation are /kmolKm104.2 0.05084 0.9160 0.0069110.050461 132.339 0.9160 0.026171136.23151 334 ⋅×= =      −−=      −= =      −=      −= c bBB aAA o o v v since /kmolm 0.9160/kg)m .0327kg/kmol)(0 (28.013 33 === vv M . Substituting, ( ) ( ) ( ) error) (0.07%kPa 1504= −+        × × − × = −+      −= 23 4 2 232 0.9160 132.3390.050840.9160 1750.9160 104.21 (0.9160) 5178.314 1 v v vv AB T cTRP u 3-41 3-97 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, van der Waals equation, and the steam tables. Properties The gas constant, critical pressure, and critical temperature of steam are (Table A-1) R = 0.4615 kPa·m3/kg·K, Tcr = 647.1 K, Pcr = 22.06 MPa Analysis The specific volume of steam is /kgm 0.3520 kg 2.841 m 1 3 3 === m V v H2O 1 m3 2.841 kg 0.6 MPa (a) From the ideal gas equation of state, K457.6 K/kgmkPa 0.4615 /kg)m kPa)(0.352 (600 3 3 = ⋅⋅ == R PT v (b) The van der Waals constants for steam are determined from /kgm 0.00169 kPa 22,0608 K) K)(647.1/kgmkPa (0.4615 8 kPa/kgm 1.705 kPa) 0(64)(22,06 K) (647.1K)/kgmkPa 5(27)(0.461 64 27 3 3 26 22322 = × ⋅⋅ == ⋅= ⋅⋅ == cr cr cr cr P RT b P TR a Then, ( ) ( ) K 465.9=−       +=−      += 0.001690.352 (0.3520) 1.705600 0.4615 11 22 baP R T v v (c) From the superheated steam table (Tables A-6), K) 473( /kgm 00.352 MPa 0.6 3 =°=    = = C200TP v 3-42 3-98 EES Problem 3-97 is reconsidered. The problem is to be solved using EES (or other) software. The temperature of water is to be compared for the three cases at constant specific volume over the pressure range of 0.1 MPa to 1 MPa in 0.1 MPa increments. The %error involved in the ideal gas approximation is to be plotted against pressure. Analysis The problem is solved using EES, and the solution is given below. Function vanderWaals(T,v,M,R_u,T_cr,P_cr) v_bar=v*M "Conversion from m^3/kg to m^3/kmol" "The constants for the van der Waals equation of state are given by equation 3-24" a=27*R_u^2*T_cr^2/(64*P_cr) b=R_u*T_cr/(8*P_cr) "The van der Waals equation of state gives the pressure as" vanderWaals:=R_u*T/(v_bar-b)-a/v_bar**2 End m=2.841[kg] Vol=1 [m^3] {P=6*convert(MPa,kPa)} T_cr=T_CRIT(Steam_iapws) P_cr=P_CRIT(Steam_iapws) v=Vol/m P_table=P; P_vdW=P;P_idealgas=P T_table=temperature(Steam_iapws,P=P_table,v=v) "EES data for steam as a real gas" {P_table=pressure(Steam_iapws, T=T_table,v=v)} {T_sat=temperature(Steam_iapws,P=P_table,v=v)} MM=MOLARMASS(water) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/MM "Particular gas constant" P_idealgas=R*T_idealgas/v "Ideal gas equation" "The value of P_vdW is found from van der Waals equation of state Function" P_vdW=vanderWaals(T_vdW,v,MM,R_u,T_cr,P_cr) Error_idealgas=Abs(T_table-T_idealgas)/T_table*Convert(, %) Error_vdW=Abs(T_table-T_vdW)/T_table*Convert(, %) P [kPa] Tideal gas [K] Ttable [K] TvdW [K] Errorideal gas [K] 100 76.27 372.8 86.35 79.54 200 152.5 393.4 162.3 61.22 300 228.8 406.7 238.2 43.74 400 305.1 416.8 314.1 26.8 500 381.4 425 390 10.27 600 457.6 473 465.9 3.249 700 533.9 545.3 541.8 2.087 800 610.2 619.1 617.7 1.442 900 686.4 693.7 693.6 1.041 1000 762.7 768.6 769.5 0.7725 3-45 3-101 EES Problem 3-100 is reconsidered. Using EES (or other) software, the pressure results of the ideal gas and Beattie-Bridgeman equations with nitrogen data supplied by EES are to be compared. The temperature is to be plotted versus specific volume for a pressure of 1000 kPa with respect to the saturated liquid and saturated vapor lines of nitrogen over the range of 110 K < T < 150 K. Analysis The problem is solved using EES, and the solution is given below. Function BeattBridg(T,v,M,R_u) v_bar=v*M "Conversion from m^3/kg to m^3/kmol" "The constants for the Beattie-Bridgeman equation of state are found in text" Ao=136.2315; aa=0.02617; Bo=0.05046; bb=-0.00691; cc=4.20*1E4 B=Bo*(1-bb/v_bar) A=Ao*(1-aa/v_bar) "The Beattie-Bridgeman equation of state is" BeattBridg:=R_u*T/(v_bar**2)*(1-cc/(v_bar*T**3))*(v_bar+B)-A/v_bar**2 End T=150 [K] v=0.041884 [m^3/kg] P_exper=1000 [kPa] T_table=T; T_BB=T;T_idealgas=T P_table=PRESSURE(Nitrogen,T=T_table,v=v) "EES data for nitrogen as a real gas" {T_table=temperature(Nitrogen, P=P_table,v=v)} M=MOLARMASS(Nitrogen) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/M "Particular gas constant" P_idealgas=R*T_idealgas/v "Ideal gas equation" P_BB=BeattBridg(T_BB,v,M,R_u) "Beattie-Bridgeman equation of state Function" PBB [kPa] Ptable [kPa] Pidealgas [kPa] v [m3/kg] TBB [K] Tideal gas [K] Ttable [K] 1000 1000 1000 0.01 91.23 33.69 103.8 1000 1000 1000 0.02 95.52 67.39 103.8 1000 1000 1000 0.025 105 84.23 106.1 1000 1000 1000 0.03 116.8 101.1 117.2 1000 1000 1000 0.035 130.1 117.9 130.1 1000 1000 1000 0.04 144.4 134.8 144.3 1000 1000 1000 0.05 174.6 168.5 174.5 10-3 10-2 10-1 70 80 90 100 110 120 130 140 150 160 v [m3/kg] T [K ] 1000 kPa Nitrogen, T vs v for P=1000 kPa EES Table Value Beattie-Bridgeman Ideal Gas 3-46 Special Topic: Vapor Pressure and Phase Equilibrium 3-102 A glass of water is left in a room. The vapor pressures at the free surface of the water and in the room far from the glass are to be determined. Assumptions The water in the glass is at a uniform temperature. Properties The saturation pressure of water is 2.339 kPa at 20°C, and 1.706 kPa at 15°C (Table A-4). Analysis The vapor pressure at the water surface is the saturation pressure of water at the water temperature, H2O 15°C kPa 1.706=== °Csat@15@satsurface water , water PPP Tv Noting that the air in the room is not saturated, the vapor pressure in the room far from the glass is kPa 1.404==== ° kPa) 339.2)(6.0(Csat@20@satair , air PPP Tv φφ 3-103 The vapor pressure in the air at the beach when the air temperature is 30°C is claimed to be 5.2 kPa. The validity of this claim is to be evaluated. Properties The saturation pressure of water at 30°C is 4.247 kPa (Table A-4). WATER 30°C Analysis The maximum vapor pressure in the air is the saturation pressure of water at the given temperature, which is kPa 4.247=== °Csat@30@satmax , air PPP Tv which is less than the claimed value of 5.2 kPa. Therefore, the claim is false. 3-104 The temperature and relative humidity of air over a swimming pool are given. The water temperature of the swimming pool when phase equilibrium conditions are established is to be determined. Assumptions The temperature and relative humidity of air over the pool remain constant. Properties The saturation pressure of water at 20°C is 2.339 kPa (Table A-4). Analysis The vapor pressure of air over the swimming pool is Patm, 20°C kPa0.9357kPa) 339.2)(4.0(Csat@20@satair , air ==== °PPP Tv φφ POOL Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore, kPa0.9357air ,surface water , == vv PP and C6.0°=== kPa 9357.0@sat@satwater v TTT P Discussion Note that the water temperature drops to 6.0°C in an environment at 20°C when phase equilibrium is established. 3-47 3-105 Two rooms are identical except that they are maintained at different temperatures and relative humidities. The room that contains more moisture is to be determined. Properties The saturation pressure of water is 2.339 kPa at 20°C, and 4.247 kPa at 30°C (Table A-4). Analysis The vapor pressures in the two rooms are Room 1: kPa 1.699==== ° kPa) 247.4)(4.0(Csat@301@sat11 1 PPP Tv φφ Room 2: kPa 1.637==== ° kPa) 339.2)(7.0(Csat@202@sat22 2 PPP Tv φφ Therefore, room 1 at 30°C and 40% relative humidity contains more moisture. 3-106E A thermos bottle half-filled with water is left open to air in a room at a specified temperature and pressure. The temperature of water when phase equilibrium is established is to be determined. Assumptions The temperature and relative humidity of air over the bottle remain constant. Properties The saturation pressure of water at 70°F is 0.3633 psia (Table A-4E). Analysis The vapor pressure of air in the room is Thermos bottle psia 0.1272psia) 3633.0)(35.0(Fsat@70@satair , air ==== °PPP Tv φφ Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore, 70°F 35% psia0.1272air ,surface water , == vv PP and F41.1°=== psia 1272.0@sat@satwater v TTT P Discussion Note that the water temperature drops to 41°F in an environment at 70°F when phase equilibrium is established. 3-107 A person buys a supposedly cold drink in a hot and humid summer day, yet no condensation occurs on the drink. The claim that the temperature of the drink is below 10°C is to be evaluated. Properties The saturation pressure of water at 35°C is 5.629 kPa (Table A-4). Analysis The vapor pressure of air is 35°C 70% kPa940.3kPa) 629.5)(7.0(Csat@35@satair , air ==== °PPP Tv φφ The saturation temperature corresponding to this pressure (called the dew-point temperature) is C28.7°=== kPa [email protected]@satsat v TTT P That is, the vapor in the air will condense at temperatures below 28.7°C. Noting that no condensation is observed on the can, the claim that the drink is at 10°C is false. 3-50 3-113 (a) On the P-v diagram, the constant temperature process through the state P= 300 kPa, v = 0.525 m3/kg as pressure changes from P1 = 200 kPa to P2 = 400 kPa is to be sketched. The value of the temperature on the process curve on the P-v diagram is to be placed. 10-4 10-3 10-2 10-1 100 101 102 100 101 102 103 104 105 106 v [m3/kg] P [kP a ] 133.5°C SteamIAPWS 300 0.525 200 400 1 2 (b) On the T-v diagram the constant specific vol-ume process through the state T = 120°C, v = 0.7163 m3/kg from P1= 100 kPa to P2 = 300 kPa is to be sketched.. For this data set, the temperature values at states 1 and 2 on its axis is to be placed. The value of the specific volume on its axis is also to be placed. 10-3 10-2 10-1 100 101 102 0 100 200 300 400 500 600 700 v [m3/kg] T [°C ] 198.7 kPa 300 kPa 100 kPa SteamIAPWS 1 2 0.7163 3-51 3-114 The pressure in an automobile tire increases during a trip while its volume remains constant. The percent increase in the absolute temperature of the air in the tire is to be determined. Assumptions 1 The volume of the tire remains constant. 2 Air is an ideal gas. Properties The local atmospheric pressure is 90 kPa. TIRE 200 kPa 0.035 m3 Analysis The absolute pressures in the tire before and after the trip are P P P P P P 1 2 200 90 290 220 90 310 = + = + = = + = + = gage,1 atm gage,2 atm kPa kPa Noting that air is an ideal gas and the volume is constant, the ratio of absolute temperatures after and before the trip are 1.069= kPa 290 kPa 310= 1 2 1 2 2 22 1 11 P P T T T P T P =→= VV Therefore, the absolute temperature of air in the tire will increase by 6.9% during this trip. 3-115 A hot air balloon with 3 people in its cage is hanging still in the air. The average temperature of the air in the balloon for two environment temperatures is to be determined. Assumptions Air is an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis The buoyancy force acting on the balloon is N 44,700 m/skg1 N 1)m )(4189m/s )(9.8kg/m (1.089 kg/m 1.089 K) K)(288/kgmkPa (0.287 kPa 90 m4189/3m)(1043/4 2 323 balloonair cool 3 3air cool 333 balloon =      ⋅ = = = ⋅⋅ == === V V gF RT P r B ρ ρ ππ Hot air balloon D = 20 m The vertical force balance on the balloon gives gmmm WWWFB )( peoplecageairhot peoplecageairhot ++= ++= mcage = 80 kg Patm = 90 kPa T = 15°C Substituting,       ⋅ ++= 2 2 airhot m/skg 1 N 1)m/s kg)(9.8 195kg 80(N 44,700 m which gives kg 4287airhot =m Therefore, the average temperature of the air in the balloon is K 306.5= ⋅⋅ == K)/kgmkPa kg)(0.287 (4287 )m kPa)(4189 (90 3 3 mR PT V Repeating the solution above for an atmospheric air temperature of 30°C gives 323.6 K for the average air temperature in the balloon. 3-52 3-116 EES Problem 3-115 is to be reconsidered. The effect of the environment temperature on the average air temperature in the balloon when the balloon is suspended in the air is to be investigated as the environment temperature varies from -10°C to 30°C. The average air temperature in the balloon is to be plotted versus the environment temperature. Analysis The problem is solved using EES, and the solution is given below. "Given Data:" "atm---atmosphere about balloon" "gas---heated air inside balloon" g=9.807 [m/s^2] d_balloon=20 [m] m_cage = 80 [kg] m_1person=65 [kg] NoPeople = 6 {T_atm_Celsius = 15 [C]} T_atm =T_atm_Celsius+273 "[K]" P_atm = 90 [kPa] R=0.287 [kJ/kg-K] P_gas = P_atm T_gas_Celsius=T_gas - 273 "[C]" "Calculated values:" P_atm= rho_atm*R*T_atm "rho_atm = density of air outside balloon" P_gas= rho_gas*R*T_gas "rho_gas = density of gas inside balloon" r_balloon=d_balloon/2 V_balloon=4*pi*r_balloon^3/3 m_people=NoPeople*m_1person m_gas=rho_gas*V_balloon m_total=m_gas+m_people+m_cage "The total weight of balloon, people, and cage is:" W_total=m_total*g "The buoyancy force acting on the balloon, F_b, is equal to the weight of the air displaced by the balloon." F_b=rho_atm*V_balloon*g "From the free body diagram of the balloon, the balancing vertical forces must equal the product of the total mass and the vertical acceleration:" F_b- W_total=m_total*a_up a_up = 0 "The balloon is hanging still in the air" -10 -5 0 5 10 15 20 25 30 0 10 20 30 40 50 60 70 80 90 100 Tatm,Celsius [C] T ga s ,C el si us [C ] 3 people 6 people 9 people Tatm,Celcius [C] Tgas,Celcius [C] -10 17.32 -5 23.42 0 29.55 5 35.71 10 41.89 15 48.09 20 54.31 25 60.57 30 66.84 3-55 3-122 Two rigid tanks that contain hydrogen at two different states are connected to each other. Now a valve is opened, and the two gases are allowed to mix while achieving thermal equilibrium with the surroundings. The final pressure in the tanks is to be determined. Properties The gas constant for hydrogen is 4.124 kPa·m3/kg·K (Table A-1). Analysis Let's call the first and the second tanks A and B. Treating H2 as an ideal gas, the total volume and the total mass of H2 are A B kg0.3080.0600.248 kg 0.060 K) K)(303/kgmkPa (4.124 )m kPa)(0.5 (150 kg 0.248 K) K)(293/kgmkPa (4.124 )m kPa)(0.5 (600 m 1.00.50.5 3 3 1 1 3 3 1 1 3 =+=+= = ⋅⋅ =      = = ⋅⋅ =      = =+=+= BA B B A A BA mmm RT P m RT P m V V VVV × H2 V = 0.5 m3 T=30°C P=150 kPa H2 V = 0.5 m3 T=20°C P=600 kPa Then the final pressure can be determined from kPa365.8 m 1.0 K) K)(288/kgmkPa kg)(4.124 (0.308 3 3 2 = ⋅⋅ == V mRT P 3-123 EES Problem 3-122 is reconsidered. The effect of the surroundings temperature on the final equilibrium pressure in the tanks is to be investigated. The final pressure in the tanks is to be plotted versus the surroundings temperature, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data" V_A=0.5 [m^3] T_A=20 [C] P_A=600 [kPa] V_B=0.5 [m^3] T_B=30 [C] P_B=150 [kPa] {T_2=15 [C]} "Solution" R=R_u/MOLARMASS(H2) R_u=8.314 [kJ/kmol-K] -10 -5 0 5 10 15 20 25 30 330 340 350 360 370 380 390 T2 [C] P 2 [k P a] V_total=V_A+V_B m_total=m_A+m_B P_A*V_A=m_A*R*(T_A+273) P_B*V_B=m_B*R*(T_B+273) P_2*V_total=m_total*R*(T_2+273) P2 [kPa] T2 [C] 334.4 -10 340.7 -5 347.1 0 353.5 5 359.8 10 366.2 15 372.5 20 378.9 25 385.2 30 3-56 3-124 A large tank contains nitrogen at a specified temperature and pressure. Now some nitrogen is allowed to escape, and the temperature and pressure of nitrogen drop to new values. The amount of nitrogen that has escaped is to be determined. Properties The gas constant for nitrogen is 0.2968 kPa·m3/kg·K (Table A-1). Analysis Treating N2 as an ideal gas, the initial and the final masses in the tank are determined to be kg 0.92 K) K)(293/kgmkPa (0.2968 )m kPa)(20 (400 kg 136.6 K) K)(296/kgmkPa(0.2968 )m kPa)(20 (600 3 3 2 2 2 3 3 1 1 1 = ⋅⋅ == = ⋅⋅ == RT P m RT P m V V N2 600 kPa 23°C 20 m3 Thus the amount of N2 that escaped is kg 44.6=−=−= 0.92136.621 mmm∆ 3-125 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, the generalized chart, and the steam tables. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1, MPa 22.06 K, 647.1 ,K/kgmkPa 0.4615 crcr 3 ==⋅⋅= PTR Analysis (a) From the ideal gas equation of state, H2O 0.02 m3/kg 400°C kPa 15,529=⋅⋅== /kgm 0.02 K) K)(673/kgmkPa (0.4615 3 3 v RTP (b) From the compressibility chart (Fig. A-15a), 57.0 1.48 K) K)(647.1/kgmkPa (0.4615 kPa) 0/kg)(22,06m (0.02 / 040.1 K 647.1 K 673 3 3 crcr actual cr =        = ⋅⋅ == === R R R P PRT T TT v v Thus, kPa 12,574=×== 060,2257.0crPPP R (c) From the superheated steam table, (from EES) kPa 12,576=    = °= PT /kgm 0.02 C400 3v 3-126 One section of a tank is filled with saturated liquid R-134a while the other side is evacuated. The partition is removed, and the temperature and pressure in the tank are measured. The volume of the tank is to be determined. Analysis The mass of the refrigerant contained in the tank is kg 11.82 /kgm 0.0008458 m 0.01 3 3 1 1 === v V m Evacuated R-134a P=0.8 MPa V =0.01 m3 since /kgm 0.0008458 3MPa0.8@1 == fvv At the final state (Table A-13), } /kgm 0.05421 C20 kPa 004 3222 =°== vTP Thus, 3m 0.641==== /kg)m 1kg)(0.0542 (11.82 322tank vVV m 3-57 3-127 EES Problem 3-126 is reconsidered. The effect of the initial pressure of refrigerant-134 on the volume of the tank is to be investigated as the initial pressure varies from 0.5 MPa to 1.5 MPa. The volume of the tank is to be plotted versus the initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data" x_1=0.0 Vol_1=0.01[m^3] P_1=800 [kPa] T_2=20 [C] P_2=400 [kPa] "Solution" v_1=volume(R134a,P=P_1,x=x_1) Vol_1=m*v_1 v_2=volume(R134a,P=P_2,T=T_2) Vol_2=m*v_2 P1 [kPa] Vol2 [m 3] m [kg] 500 0.6727 12.41 600 0.6612 12.2 700 0.6507 12 800 0.641 11.82 900 0.6318 11.65 1000 0.6231 11.49 1100 0.6148 11.34 1200 0.6068 11.19 1300 0.599 11.05 1400 0.5914 10.91 1500 0.584 10.77 500 700 900 1100 1300 1500 0.58 0.6 0.62 0.64 0.66 0.68 P1 [kPa] Vo l2 [m 3] 3-60 3-134 (a) On the P-v diagram the constant temperature process through the state P = 280 kPa, v = 0.06 m3/kg as pressure changes from P1 = 400 kPa to P2 = 200 kPa is to be sketched. The value of the temperature on the process curve on the P-v diagram is to be placed. 10-4 10-3 10-2 10-1 100 101 101 102 103 104 105 v [m3/kg] P [kP a ] -1.25°C R134a 280 400 1 200 2 0.06 (b) On the T-v diagram the constant specific volume process through the state T = 20°C, v = 0.02 m3/kg from P1 = 1200 kPa to P2 = 300 kPa is to be sketched. For this data set the temperature values at states 1 and 2 on its axis is to be placed. The value of the specific volume on its axis is also to be placed. 10-4 10-3 10-2 10-1 100 -100 -50 0 50 100 150 200 250 v [m3/kg] T [°C ] 1200 kPa 572 kPa 300 kPa R134a 0.02 1 2 75 0.6 20 3-61 Fundamentals of Engineering (FE) Exam Problems 3-135 A rigid tank contains 6 kg of an ideal gas at 3 atm and 40°C. Now a valve is opened, and half of mass of the gas is allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the tank is (a) 186°C (b) 59°C (c) -43°C (d) 20°C (e) 230°C Answer (a) 186°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). "When R=constant and V= constant, P1/P2=m1*T1/m2*T2" m1=6 "kg" P1=3 "atm" P2=2.2 "atm" T1=40+273 "K" m2=0.5*m1 "kg" P1/P2=m1*T1/(m2*T2) T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=m1*(T1-273)/(m2*W1_T2) "Using C instead of K" P1/P2=m1*T1/(m1*(W2_T2+273)) "Disregarding the decrease in mass" P1/P2=m1*T1/(m1*W3_T2) "Disregarding the decrease in mass, and not converting to deg. C" W4_T2=(T1-273)/2 "Taking T2 to be half of T1 since half of the mass is discharged" 3-136 The pressure of an automobile tire is measured to be 190 kPa (gage) before a trip and 215 kPa (gage) after the trip at a location where the atmospheric pressure is 95 kPa. If the temperature of air in the tire before the trip is 25°C, the air temperature after the trip is (a) 51.1°C (b) 64.2°C (c) 27.2°C (d) 28.3°C (e) 25.0°C Answer (a) 51.1°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). "When R, V, and m are constant, P1/P2=T1/T2" Patm=95 P1=190+Patm "kPa" P2=215+Patm "kPa" T1=25+273 "K" P1/P2=T1/T2 T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=(T1-273)/W1_T2 "Using C instead of K" (P1-Patm)/(P2-Patm)=T1/(W2_T2+273) "Using gage pressure instead of absolute pressure" (P1-Patm)/(P2-Patm)=(T1-273)/W3_T2 "Making both of the mistakes above" W4_T2=T1-273 "Assuming the temperature to remain constant" 3-62 3-137 A 300-m3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 25% of the mass is liquid and the 75% of the mass is vapor, the total mass in the tank is (a) 451 kg (b) 556 kg (c) 300 kg (d) 331 kg (e) 195 kg Answer (a) 451 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_tank=300 "m3" P1=200 "kPa" x=0.75 v_f=VOLUME(Steam_IAPWS, x=0,P=P1) v_g=VOLUME(Steam_IAPWS, x=1,P=P1) v=v_f+x*(v_g-v_f) m=V_tank/v "kg" "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" T=TEMPERATURE(Steam_IAPWS,x=0,P=P1) P1*V_tank=W1_m*R*(T+273) "Treating steam as ideal gas" P1*V_tank=W2_m*R*T "Treating steam as ideal gas and using deg.C" W3_m=V_tank "Taking the density to be 1 kg/m^3" 3-138 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion-type electric heating element. The coffee maker initially contains 1 kg of water. Once boiling started, it is observed that half of the water in the coffee maker evaporated in 18 minutes. If the heat loss from the coffee maker is negligible, the power rating of the heating element is (a) 0.90 kW (b) 1.52 kW (c) 2.09 kW (d) 1.05 kW (e) 1.24 kW Answer (d) 1.05 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_1=1 "kg" P=101.325 "kPa" time=18*60 "s" m_evap=0.5*m_1 Power*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Power*time=m_evap*h_g "Using h_g" W2_Power*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Power=2*Power "Assuming all the water evaporates"