EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220100257056
Author: CENGEL
Publisher: YUZU
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Chapter 13.3, Problem 71P

a)

To determine

The total entropy change and exergy destruction by treating the mixture as an ideal gas.

a)

Expert Solution
Check Mark

Answer to Problem 71P

The entropy generated is 8.496kJ/K.

The energy destroyed is 2,489kJ.

Explanation of Solution

Write the entropy balance equation to obtain the expression of entropy generation in terms of H2andN2.

SinSout+Sgen=ΔSsystemQinTb,surr+Sgen=m(s2s1)Sgen=m(s2s1)+QinTsurrSgen=mH2(s2s1)H2+mN2(s2s1)N2QinTsurr (I)

Here, mass of H2 is mH2, mass of N2 is mN2, boundary temperature is Tb,surr, surrounding temperature is Tsurr, entropy at state 1 and 2 is s1ands2, entropy generation is Sgen, heat input is Qin, outlet entropy is Sout, and inlet entropy is Sin.

Write the expression to obtain the energy destroyed during a process (Xdestroyed).

Xdestroyed=T0Sgen (II)

Here, initial temperature is T0.

Conclusion:

Refer Table A-2b, “Ideal gas specific heats of various common gases”, obtain the specific heat at constant pressure of H2andN2 at 200K and 160K as, 13.6kJ/kgK and 1.039kJ/kgK respectively.

Obtain the value of (s2s1)H2 from Equation (I).

(s2s1)H2=(cplnT2T1RlnP20P1)H2 (III)

Here, constant pressure specific heat is cp, initial temperature and pressure is T1andP1, and apparent gas constant is R,

The partial pressure of H2 gas remains constant in the total mixture pressure, thus calculate (s2s1)H2 from Equation (III).

(s2s1)H2=(cplnT2T1)H2 (IV)

Substitute 13.60kJ/kgK for cp, 200 K for T2, and 160 K for T1 in Equation (IV).

(s2s1)H2=((13.60kJ/kgK)ln200K160K)=3.034kJ/kgK

Obtain the value of (s2s1)N2 from Equation (I).

(s2s1)N2=(cplnT2T1RlnP20P1)N2 (V)

The partial pressure of N2 gas remains constant in the total mixture pressure, thus calculate (s2s1)N2 from Equation (V).

(s2s1)N2=(cplnT2T1)N2 (VI)

Substitute 1.039kJ/kgK for cp, 200 K for T2, 160 K for T1 in Equation (VI).

(s2s1)N2=((1.039kJ/kgK)ln200K160K)=0.231kJ/kgK

Substitute 6 kg for mH2, 21 kg for mN2, 3.034kJ/kgK for (s2s1)H2, 0.231kJ/kgK for (s2s1)N2, 4273 kJ for Qin and 293 K for Tsurr in Equation (I).

Sgen=(6kg)(3.034kJ/kgK)+(21kg)(0.231kJ/kgK)4273kJ293K=8.496kJ/K

Thus, the entropy generated is 8.496kJ/K.

Substitute 293 K for T0 and 8.496kJ/K for Sgen in Equation (II).

Xdestroyed=(293K)(8.496kJ/K)=2,489kJ

Thus, the energy destroyed is 2,489kJ.

b)

To determine

The total entropy change and exergy destruction by treating the mixture as a non ideal gas using Amagat’s law.

b)

Expert Solution
Check Mark

Answer to Problem 71P

The entropy generation is 9.390kJ/K.

The energy destroyed is 2,751kJ.

Explanation of Solution

Write the expression to obtain the initial reduced temperature of H2 (TR1,H2).

TR1,H2=Tm,1Tcr,H2 (VII)

Here, critical temperature of H2 is Tcr,H2.

Write the expression to obtain the initial and final reduced pressure of H2 (PR1,H2).

PR1,H2=PR2,H2=PmPcr,H2 (VIII)

Here, critical temperature of H2 is Pcr,H2.

Write the expression to obtain the final reduced temperature of H2 (TR2,H2).

TR2,H2=Tm,2Tcr,H2 (IX)

Here, critical temperature of H2 is Tcr,H2.

Write the expression to obtain the initial reduced temperature of N2 (TR1,N2).

TR1,N2=Tm,1Tcr,N2 (X)

Here, critical temperature of N2 is Tcr,N2.

Write the expression to obtain the initial and final reduced pressure of N2 (PR1,N2).

PR1,N2=PR2,N2=PmPcr,N2 (XI)

Here, critical temperature of N2 is Pcr,N2.

Write the expression to obtain the final reduced temperature of N2 (TR2,N2).

TR2,N2=Tm,2Tcr,N2 (XII)

Here, critical temperature of N2 is Tcr,N2.

Write the expression to obtain the entropy change for H2 (ΔSH2).

ΔSH2=ΔSH2,ideal (XIII)

Write the expression to obtain the entropy change for N2 (ΔSN2).

ΔSN2=NN2Ru(Zs1Zs2)+ΔSN2,ideal (XIV)

Here, number of moles of N2 is NN2, and universal gas constant is Ru.

Write the expression to obtain the surrounding entropy change (ΔSsurr).

ΔSsurr=QsurrT0 (XV)

Here, surrounding heat is Qsurr.

Write the expression to obtain the entropy generation (Sgen).

Sgen=ΔSH2+ΔSN2+ΔSsurr (XVI)

Write the expression to obtain the energy destroyed during a process (Xdestroyed).

Xdestroyed=T0Sgen (XVII)

Here, initial temperature is T0.

Conclusion:

Substitute 160 K for Tm,1 and 33.3 K for Tcr,H2 in Equation (VII).

TR1,H2=160K33.3K=4.805

Substitute 5 MPa for Pm and 1.30 MPa for Pcr,H2 in Equation (VIII).

PR,H2=5MPa1.30MPa=3.846

Substitute 200 K for Tm,2 and 33.3 K for Tcr,H2 in Equation (IX).

TR2,H2=200K33.3K=6.006

Refer Figure A-30, “Generalized entropy departure chart”, obtain the value of Zs1  and Zs2 for H2 as 1 and 1 by taking TR1,H2 as 4.805, PR1,H2 as 3.846, and TR2,H2 as 6.006.

Substitute 160 K for Tm,1 and 126.2 K for Tcr,N2 in Equation (X).

TR1,N2=160K126.2K=1.268

Substitute 5 MPa for Pm and 3.39 MPa for Pcr,N2 in Equation (XI).

PR,N2=5MPa3.39MPa=1.475

Substitute 200 K for Tm,2 and 126.2 K for Tcr,N2 in Equation (XII).

TR2,N2=200K126.2K=1.585

Refer Figure A-30, “Generalized entropy departure chart”, obtain the value of Zs1 and Zs2 for N2 as 0.8 and 0.4 by taking TR1,N2 as 1.268, PR1,N2 as 1.475 and TR2,N2 as 1.585.

Substitute 18.21kJ/K for ΔSH2,ideal in Equation (XIII).

ΔSH2=18.21kJ/K

Substitute 0.75 kmol for NN2, 8.314kPam3/kmolK for Ru, 0.8 for Zs1, 0.4 for Zs2, and 4.87kJ/K for ΔSN2,ideal in Equation (XIV).

ΔSN2=0.75kmol(8.314kPam3/kmolK)(0.80.4)+4.87kJ/K=7.37kJ/K

Substitute –4,745 kJ for Qsurr and 293 K for T0 in Equation (XV).

ΔSsurr=4,745kJ293K=16.19kJ/K

Substitute 7.37kJ/K for ΔSN2, 18.21kJ/K for ΔSH2, and 16.19kJ/K for ΔSsurr in Equation (XVI).

Sgen=18.21kJ/K+7.37kJ/K16.19kJ/K=9.390kJ/K

Thus, the entropy generation is 9.390kJ/K.

Substitute 293 K for T0 and 9.390kJ/K for Sgen in Equation (XVII).

Xdestroyed=(293K)(9.390kJ/K)=2,751kJ

Thus, the energy destroyed is 2,751kJ.

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Chapter 13 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 13.3 - Prob. 11PCh. 13.3 - Prob. 12PCh. 13.3 - Prob. 13PCh. 13.3 - Is a mixture of ideal gases also an ideal gas?...Ch. 13.3 - Express Daltons law of additive pressures. Does...Ch. 13.3 - Express Amagats law of additive volumes. Does this...Ch. 13.3 - How is the P-v-T behavior of a component in an...Ch. 13.3 - Prob. 18PCh. 13.3 - Prob. 19PCh. 13.3 - Prob. 20PCh. 13.3 - Prob. 21PCh. 13.3 - Consider a rigid tank that contains a mixture of...Ch. 13.3 - Is this statement correct? The volume of an...Ch. 13.3 - Is this statement correct? The temperature of an...Ch. 13.3 - Is this statement correct? The pressure of an...Ch. 13.3 - Prob. 26PCh. 13.3 - Prob. 27PCh. 13.3 - Prob. 28PCh. 13.3 - 13–29 A gas mixture at 350 K and 300 kPa has the...Ch. 13.3 - Prob. 30PCh. 13.3 - Prob. 31PCh. 13.3 - A rigid tank that contains 2 kg of N2 at 25C and...Ch. 13.3 - Prob. 33PCh. 13.3 - Prob. 34PCh. 13.3 - Prob. 35PCh. 13.3 - Prob. 36PCh. 13.3 - A 30 percent (by mass) ethane and 70 percent...Ch. 13.3 - Prob. 38PCh. 13.3 - Prob. 39PCh. 13.3 - Prob. 40PCh. 13.3 - Prob. 41PCh. 13.3 - Prob. 42PCh. 13.3 - Prob. 43PCh. 13.3 - Is the total internal energy of an ideal-gas...Ch. 13.3 - Prob. 45PCh. 13.3 - Prob. 46PCh. 13.3 - 13–47C Is the total internal energy change of an...Ch. 13.3 - Prob. 48PCh. 13.3 - Prob. 49PCh. 13.3 - The volumetric analysis of a mixture of gases is...Ch. 13.3 - Prob. 52PCh. 13.3 - Prob. 53PCh. 13.3 - Prob. 54PCh. 13.3 - Prob. 55PCh. 13.3 - Prob. 56PCh. 13.3 - An insulated tank that contains 1 kg of O2at 15C...Ch. 13.3 - Prob. 59PCh. 13.3 - Prob. 60PCh. 13.3 - Prob. 61PCh. 13.3 - Prob. 62PCh. 13.3 - Prob. 63PCh. 13.3 - Prob. 64PCh. 13.3 - Prob. 66PCh. 13.3 - Prob. 67PCh. 13.3 - Prob. 69PCh. 13.3 - A pistoncylinder device contains 6 kg of H2 and 21...Ch. 13.3 - Prob. 71PCh. 13.3 - Prob. 72PCh. 13.3 - Prob. 73PCh. 13.3 - Prob. 74PCh. 13.3 - Prob. 75PCh. 13.3 - Prob. 76PCh. 13.3 - Prob. 77PCh. 13.3 - Prob. 78PCh. 13.3 - Prob. 80PCh. 13.3 - Prob. 81PCh. 13.3 - Fresh water is obtained from seawater at a rate of...Ch. 13.3 - Prob. 83PCh. 13.3 - Prob. 84RPCh. 13.3 - The products of combustion of a hydrocarbon fuel...Ch. 13.3 - Prob. 89RPCh. 13.3 - Prob. 91RPCh. 13.3 - Prob. 92RPCh. 13.3 - A spring-loaded pistoncylinder device contains a...Ch. 13.3 - Prob. 94RPCh. 13.3 - Reconsider Prob. 1395. Calculate the total work...Ch. 13.3 - A rigid tank contains a mixture of 4 kg of He and...Ch. 13.3 - Prob. 97RPCh. 13.3 - Prob. 100RPCh. 13.3 - An ideal-gas mixture whose apparent molar mass is...Ch. 13.3 - 13–102 An ideal-gas mixture consists of 2 kmol of...Ch. 13.3 - An ideal-gas mixture consists of 2 kmol of N2and 4...Ch. 13.3 - Prob. 104FEPCh. 13.3 - Prob. 105FEPCh. 13.3 - An ideal-gas mixture consists of 3 kg of Ar and 6...Ch. 13.3 - Prob. 107FEPCh. 13.3 - Prob. 108FEPCh. 13.3 - Prob. 109FEPCh. 13.3 - Prob. 110FEP
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