Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Question
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Chapter 8.8, Problem 75P

a)

To determine

The final mass (m2) in the tank is 3.90kg_.

a)

Expert Solution
Check Mark

Answer to Problem 75P

The final temperature of the helium is 321.7 K_.

Explanation of Solution

Write the expression for the final mass (m2) stored in the tank.

m2=Vv2 (I)

Here, volume of the rigid tank is V.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800kPa.

vf=0.0008457m3/kgandvg=0.025645m3/kguf=94.80kJ/kgandug=246.82kJ/kgsf=0.35408kJ/kgKandsg=0.91853kJ/kgK

Here, internal energy of saturated liquid and saturated vapor is ufandug respectively, specific volume of saturated liquid and saturated vapor is vfandvg respectively, and specific entropy of saturated liquid and saturated vapor is sfandsg respectively.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800kPa.

he,hf@800kPa=95.48kJ/kgse,sf@800kPa=0.35408kJ/kgK

Here, enthalpy of saturated liquid is hf.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800kPa.

v2,vg@800kPa=0.025645m3/kgu2,ug@800kPa=246.82kJ/kgs2,sg@800kPa=0.91853kJ/kgK

Substitute 0.1m3 for V and 0.025645m3/kg for v2 in equation (1).

m2=0.1m30.025645m3/kg=3.899kg3.90kg

Thus, the final mass (m2) in the tank is 3.90kg_.

b)

To determine

The reversible work (Wrev,out) associated with the process

b)

Expert Solution
Check Mark

Answer to Problem 75P

The reversible work (Wrev,out) associated with the process is 16.6kJ_.

Explanation of Solution

Write the expression for the mass balance for the control volume system.

minmout=Δmsystemme=m1m2 (II)

Here, mass of the refrigerant at the inlet is min, mass of the refrigerant at the exit is mout, change in mass within the system is Δmsystem, initial and final mass of the refrigerant is m1andm2 respectively, and mass at the exit is me.

Write the expression for the energy balance for the system.

EinEout=ΔEsystemQin=mehe+m2u2m1u1 (III)

Here, net energy transfer in to the control volume is Ein, net energy transfer exit from the control volume is Eout, change in internal energy of system is ΔEsystem, amount of heat transfer to the tank from the source is Qin, mass of refrigerant is m, initial internal energy is u1 and final internal energy is u2.

Write the expression for the initial mass (m1) stored in tank.

m1=mf+mg=0.3Vfvf+(10.3)Vgvg=0.3Vfvf+0.7Vgvg (IV)

Here, mass of the liquid refrigerant is mf, mass of the vapor refrigerant is mg, volume of the liquid refrigerant is Vf and volume of the vapor refrigerant is Vg.

Write the expression for the initial internal energy (U1) stored in tank.

U1=m1u1=mfuf+mgug=0.3Vfvf(uf)+0.7Vgvg(ug) (V)

Write the expression for the initial entropy (S1) stored in tank.

S1=m1s1=mfsf+mgsg=0.3Vfvf(sf)+0.7Vgvg(sg) (VI)

Write the expression for the entropy balance for refrigerant.

SinSout+Sgen=ΔSsystemQinTsourcemese+Sgen=m2s2m1s1Sgen=m2s2m1s1+meseQinTsource (VI)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout, and source temperature is Tsource.

Write the expression for the reversible work (Wrev,out) associated with the process.

Wrev,out=Xdestroyed+Wact,out

Here, exergy destroyed is Xdestroyed and actual work associated with the process is Wact,out.

Since, the process does not involve any actual work, substitute 0 for Wact,out.

Wrev,out=Xdestroyed+0Wrev,out=Xdestroyed=T0Sgen (VII)

Here, dead state temperature is T0.

Conclusion:

Substitute 0.1m3 for Vf, 0.0008457m3/kg for vf, 0.025645m3/kg for vg and 0.1m3 for Vg in Equation (IV).

m1=0.3(0.1m3)0.0008457m3/kg+0.7(0.1m3)0.025645m3/kg=0.03m30.0008457m3/kg+0.07m30.025645m3/kg=38.202kg

Substitute 0.1m3 for Vf, 0.0008457m3/kg for vf, 0.025645m3/kg for vg, 94.80kJ/kg for uf, 246.82kJ/kg for ug and 0.1m3 for Vg in Equation (V)

U1=m1u1=0.3(0.1m3)0.0008457m3/kg(94.80kJ/kg)+0.7(0.1m3)0.025645m3/kg(246.82kJ/kg)=3,362.73kJ+673.81kJ=4,036.4kJ

Substitute 0.1m3 for Vf, 0.0008457m3/kg for vf, 0.025645m3/kg for vg, 0.35408kJ/kgK for sf, 0.91853kJ/kgK for sg, 0.1m3 for Vg in Equation (VI)

S1=m1s1=0.3(0.1m3)0.0008457m3/kg(0.35408kJ/kgK)+0.7(0.1m3)0.025645m3/kg(0.91853kJ/kgK)=0.03m30.0008457m3/kg(0.35408kJ/kgK)+0.07m30.025645m3/kg(0.91853kJ/kgK)=12.55kJ/K+2.507kJ/K

=15.067kJ/K

Substitute 38.202kg for m1 and 3.90kg for m2 in Equation (II).

me=m1m2=38.202kg3.90kg=34.30kg

Substitute 34.30kg for me, 95.48kJ/kg for he, 38.202kg for m1, 3.90kg for m2, 246.82kJ/kg for u2 and 4036.4kJ for m1u1 in Equation (III) .

Qin=mehe+m2u2m1u1=[(34.30kg)(95.48kJ/kg)]+[(3.90kg)(246.82kJ/kg)]4,036.4kJ=3,274.96kJ+962.598kJ4,036.4kJ=201kJ

Substitute 3.899kg for m2, 15.067kJ/K for m1s1, 34.30kg for me, 0.35408kJ/kgK for se, 0.91853kJ/kgK for s2, 201kJ for Qin and 60°C for Tsource in Equation (VI).

Sgen={[(3.899kg)(0.91853kJ/kgK)]15.067kJ/K+[(34.30kg)(0.35408kJ/kgK)]201kJ60°C}=3.5813kJ/K15.067kJ/K+12.1449kJ/K201kJ(60+273)K=0.6592kJ/K201kJ333K=0.6592kJ/K0.6036kJ/K

=0.0556kJ/K

Substitute 25°C for T0 and 0.0556kJ/K for Sgen in Equation (VII).

Wrev,out=(25°C)(0.0556kJ/K)=[(25+273)K](0.0556kJ/K)=(298K)(0.0556kJ/K)=16.6kJ

Thus, the reversible work (Wrev,out) associated with the process is 16.6kJ_.

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Chapter 8 Solutions

Thermodynamics: An Engineering Approach

Ch. 8.8 - 8–11C Consider a process during which no entropy...Ch. 8.8 - Prob. 12PCh. 8.8 - 8–13E Saturated stem is generated in a boiler by...Ch. 8.8 - One method of meeting the extra electric power...Ch. 8.8 - Prob. 15PCh. 8.8 - A heat engine that receives heat from a furnace at...Ch. 8.8 - Consider a thermal energy reservoir at 1500 K that...Ch. 8.8 - A heat engine receives heat from a source at 1100...Ch. 8.8 - A heat engine that rejects waste heat to a sink at...Ch. 8.8 - Prob. 21PCh. 8.8 - A freezer is maintained at 20F by removing heat...Ch. 8.8 - Prob. 23PCh. 8.8 - Can a system have a higher second-law efficiency...Ch. 8.8 - A mass of 8 kg of helium undergoes a process from...Ch. 8.8 - Prob. 26PCh. 8.8 - Which is a more valuable resource for work...Ch. 8.8 - Which has the capability to produce the most work...Ch. 8.8 - A pistoncylinder device contains 8 kg of...Ch. 8.8 - The radiator of a steam heating system has a...Ch. 8.8 - A well-insulated rigid tank contains 6 lbm of a...Ch. 8.8 - Prob. 33PCh. 8.8 - Prob. 35PCh. 8.8 - Prob. 36PCh. 8.8 - A pistoncylinder device initially contains 2 L of...Ch. 8.8 - A 0.8-m3 insulated rigid tank contains 1.54 kg of...Ch. 8.8 - An insulated pistoncylinder device initially...Ch. 8.8 - An insulated rigid tank is divided into two equal...Ch. 8.8 - Prob. 41PCh. 8.8 - Prob. 42PCh. 8.8 - Prob. 43PCh. 8.8 - Prob. 44PCh. 8.8 - Prob. 45PCh. 8.8 - Prob. 46PCh. 8.8 - A pistoncylinder device initially contains 1.4 kg...Ch. 8.8 - Prob. 48PCh. 8.8 - Prob. 50PCh. 8.8 - Prob. 51PCh. 8.8 - Air enters a nozzle steadily at 200 kPa and 65C...Ch. 8.8 - Prob. 55PCh. 8.8 - Prob. 56PCh. 8.8 - Argon gas enters an adiabatic compressor at 120...Ch. 8.8 - Prob. 58PCh. 8.8 - Prob. 59PCh. 8.8 - Prob. 60PCh. 8.8 - Combustion gases enter a gas turbine at 900C, 800...Ch. 8.8 - Prob. 62PCh. 8.8 - Refrigerant-134a is condensed in a refrigeration...Ch. 8.8 - Prob. 64PCh. 8.8 - Refrigerant-22 absorbs heat from a cooled space at...Ch. 8.8 - Prob. 66PCh. 8.8 - Prob. 67PCh. 8.8 - Prob. 68PCh. 8.8 - Prob. 69PCh. 8.8 - Air enters a compressor at ambient conditions of...Ch. 8.8 - Hot combustion gases enter the nozzle of a...Ch. 8.8 - Prob. 72PCh. 8.8 - Prob. 73PCh. 8.8 - Prob. 74PCh. 8.8 - Prob. 75PCh. 8.8 - Prob. 76PCh. 8.8 - Prob. 77PCh. 8.8 - An insulated vertical pistoncylinder device...Ch. 8.8 - Prob. 79PCh. 8.8 - Prob. 80PCh. 8.8 - Prob. 81PCh. 8.8 - Steam is to be condensed on the shell side of a...Ch. 8.8 - 8–83 Air enters a compressor at ambient conditions...Ch. 8.8 - Prob. 84PCh. 8.8 - Prob. 85PCh. 8.8 - Prob. 86RPCh. 8.8 - Prob. 87RPCh. 8.8 - Steam enters an adiabatic nozzle at 3.5 MPa and...Ch. 8.8 - Prob. 89RPCh. 8.8 - Prob. 91RPCh. 8.8 - A well-insulated, thin-walled, counterflow heat...Ch. 8.8 - Prob. 93RPCh. 8.8 - Prob. 94RPCh. 8.8 - Prob. 95RPCh. 8.8 - Prob. 96RPCh. 8.8 - Prob. 97RPCh. 8.8 - Prob. 98RPCh. 8.8 - Prob. 99RPCh. 8.8 - Prob. 100RPCh. 8.8 - Prob. 101RPCh. 8.8 - A pistoncylinder device initially contains 8 ft3...Ch. 8.8 - Steam at 7 MPa and 400C enters a two-stage...Ch. 8.8 - Steam enters a two-stage adiabatic turbine at 8...Ch. 8.8 - Prob. 105RPCh. 8.8 - Prob. 106RPCh. 8.8 - Prob. 107RPCh. 8.8 - Prob. 108RPCh. 8.8 - Prob. 109RPCh. 8.8 - Prob. 111RPCh. 8.8 - Prob. 112RPCh. 8.8 - A passive solar house that was losing heat to the...Ch. 8.8 - Prob. 114RPCh. 8.8 - Prob. 115RPCh. 8.8 - Prob. 116RPCh. 8.8 - Prob. 117RPCh. 8.8 - Prob. 118RPCh. 8.8 - A 4-L pressure cooker has an operating pressure of...Ch. 8.8 - Repeat Prob. 8114 if heat were supplied to the...Ch. 8.8 - Prob. 121RPCh. 8.8 - Prob. 122RPCh. 8.8 - Reconsider Prob. 8-120. The air stored in the tank...Ch. 8.8 - Prob. 124RPCh. 8.8 - Prob. 125RPCh. 8.8 - Prob. 126RPCh. 8.8 - Prob. 127RPCh. 8.8 - Prob. 128RPCh. 8.8 - Water enters a pump at 100 kPa and 30C at a rate...Ch. 8.8 - Prob. 130RPCh. 8.8 - Nitrogen gas enters a diffuser at 100 kPa and 110C...Ch. 8.8 - Obtain a relation for the second-law efficiency of...Ch. 8.8 - Writing the first- and second-law relations and...Ch. 8.8 - Prob. 134RPCh. 8.8 - Prob. 136FEPCh. 8.8 - Prob. 137FEPCh. 8.8 - A heat engine receives heat from a source at 1500...Ch. 8.8 - Prob. 139FEPCh. 8.8 - Prob. 140FEPCh. 8.8 - A 12-kg solid whose specific heat is 2.8 kJ/kgC is...Ch. 8.8 - Keeping the limitations imposed by the second law...Ch. 8.8 - A furnace can supply heat steadily at 1300 K at a...Ch. 8.8 - Air is throttled from 50C and 800 kPa to a...Ch. 8.8 - Prob. 145FEP
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