THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
8th Edition
ISBN: 9781307434316
Author: CENGEL
Publisher: INTER MCG
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Chapter 13.3, Problem 97RP
To determine

The second law efficiency and the exergy destruction during the expansion process.

Expert Solution & Answer
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Answer to Problem 97RP

The second law efficiency is 83.7%.

The exergy destruction during the expansion process is 33.7 Btu/lbm.

Explanation of Solution

Write the expression to obtain the mole number of N2 (NN2).

NN2=mN2MN2 (I)

Here, molar mass of N2 is MN2 and mass of N2 is mN2.

Write the expression to obtain the mole number of He (NCH4).

NHe=mHeMHe (II)

Here, molar mass of He is MHe and mass of He is mHe.

Write the expression to obtain the mole number of CH4 (NCH4).

NCH4=mCH4MCH4 (III)

Here, molar mass of CH4 is MCH4 and mass of CH4 is mCH4.

Write the expression to obtain the mole number of C2H6 (NC2H6).

NC2H6=mC2H6MC2H6 (IV)

Here, molar mass of C2H6 is MC2H6 and mass of C2H6 is mC2H6.

Write the expression to obtain the equation to calculate the mole number of the mixture (Nm).

Nm=NN2+NHe+NCH4+NC2H6 (V)

Write the expression to obtain the molar mass of the gas mixture (Mm).

Mm=mmNm (VI)

Write the expression to obtain the equation to calculate the constant-pressure specific heat of the mixture (cp).

cp=mfN2cp,N2+mfHecp,He+mfCH4cp,CH4+mfC2H6cp,C2H6 (VII)

Here, mass fraction of N2, He, CH4, andC2H6 are mfN2, mfHe, mfCH4, and mfC2H6 respectively and constant pressure specific heat of N2, He, CH4, andC2H6 are cp,N2, cp,He, cp,CH4, and cp,C2H6 respectively.

Write the expression to obtain the gas constant of the mixture (R).

R=RuMm (VIII)

Here, the universal gas constant is Ru.

Write the expression to obtain the constant volume specific heat (cv).

cv=cpR (IX)

Write the expression to obtain the specific heat ratio (k).

k=cpcv (X)

Write the expression to obtain the temperature at the end of the expansion for the isentropic process (T2s).

T2s=T1(P2P1)(k1)/k (XI)

Write the expression to obtain the actual outlet temperature (T2).

T2=T1ηturb(T1T2s) (XII)

Here, efficiency of the turbine is ηturb.

Write the expression to obtain the entropy change of the gas mixture (s2s1).

s2s1=cplnT2T1RlnP2P1 (XIII)

Write the expression to obtain the actual work output (wout).

wout=cp(T1T2) (XIV)

Write the expression to obtain the reversible work output (wrev,out).

wrev,out=woutT0(s1s2) (XV)

Write the expression to obtain the second law efficiency (ηII).

ηII=woutwrev,out (XVI)

Write the expression to obtain the exergy destruction (xdest).

xdest=wrev,outwout (XVII)

Conclusion:

Refer Table A-1E, “Molar mass, gas constant, and critical point properties”, obtain the molar masses of N2, He, CH4 and C2H6 as 28.0lbm/lbmol, 4.0lbm/lbmol, 16.0lbm/lbmol and 30.0lbm/lbmol.

Substitute 0.15 lbm for mN2 and 28lbm/lbmol for MN2 in Equation (I).

NN2=0.15lbm28lbm/lbmol=0.005357lbmol

Substitute 0.5 lbm for mHe and 4lbm/lbmol for MHe in Equation (II).

NHe=0.5lbm(4lbm/lbmol)=0.0125lbmol

Substitute 0.6 lbm for mCH4 and 16lbm/lbmol for MCH4 in Equation (III).

NCH4=0.6lbm(16lbm/lbmol)=0.0375lbmol

Substitute 0.2 lbm for mC2H6 and 30lbm/lbmol for MC2H6 in Equation (IV).

NC2H6=0.2lbm(30lbm/lbmol)=0.006667lbmol

Substitute 0.005357lbmol for NN2, 0.0125lbmol for NHe, 0.0375lbmol for NCH4, and 0.006667lbmol for NC2H6 in Equation (V).

Nm=0.005357lbmol+0.00125lbmol+0.00375lbmol+0.006667lbmol=0.06202lbmol

Substitute 1 lbm for mm and 0.06202lbmol for Nm in Equation (VI).

Mm=1lbm0.06202lbmol=16.12lbm/lbmol

Refer Table A-2Ea, “Ideal gas specific heats of various common gases”, obtain the constant pressure specific heats of N2, He, CH4, and C2H6 as 0.248 Btu/lbmR, 1.25 Btu/lbmR, 0.532 Btu/lbmR and 0.427 Btu/lbmR.

Substitute 0.15 for mfN2, 0.05 for mfHe, 0.60 for mfCH4, 0.20 for mfC2H6, 0.248 Btu/lbmR for cp,N2, 1.25 Btu/lbmR for cp,He, 0.532 Btu/lbmR for cp,CH4, and 0.427 Btu/lbmR for cp,C2H6 in Equation (VII).

cp=[(0.15)(0.248 Btu/lbmR)+(0.05)(1.25 Btu/lbmR)+(0.60)(0.532 Btu/lbmR)+(0.20)(0.427 Btu/lbmR)]=0.5043 Btu/lbmR

Substitute 1.9858 lbm/lbmolR for Ru and 16.12lbm/lbmol for Mm in Equation (VIII).

R=1.9858 lbm/lbmolR16.12lbm/lbmol=0.1232 Btu/lbmR

Substitute 0.1232 Btu/lbmR for R and 0.5043 Btu/lbmR for cp in Equation (IX).

cv=0.5043 Btu/lbmR0.1232 Btu/lbmR=0.3811 Btu/lbmR

Substitute 0.5043 Btu/lbmR for cp and 0.3811 Btu/lbmR for cv in Equation (X).

k=0.5043 Btu/lbmR0.3811 Btu/lbmR=1.323

Substitute 1.323 for k, 400°F for T1, 15 psia for P2, and 200 psia for P1 in Equation (XI).

T2s=(400°F+460)R(15psia200 psia)(1.3231)/1.323=456.8 R

Substitute 400°F for T1, 456.8 for T2s and 0.85 for ηturb in Equation (XII).

T2=(400°F+460)R0.85((400°F+460)R456.8 R)=517.3 R

Substitute 0.5043 Btu/lbmR for cp, 517.3 R for T2, 400°F for T1, 0.1232 Btu/lbmR for R, 15 psia for P2, and 200 psia for P1 in Equation (XIII).

s2s1={(0.5043 Btu/lbmR)ln(517.3 R(400°F+460)R)(0.1232 Btu/lbmR)ln(15 psia200 psia)}=0.06270Btu/lbmR

Substitute 0.5043 Btu/lbmR for cp, 517.3 R for T2 and 400°F for T1 in Equation (XIV).

wout=0.5043 Btu/lbmR((400°F+460)R517.3 R)=172.8 Btu/lbm

Substitute 172.8 Btu/lbm for wout, 77°F for T0 and (0.06270Btu/lbmR) for (s1s2) in Equation (XV).

wrev,out=172.8 Btu/lbm(77°F+460)R(0.06270Btu/lbmR)=206.5 Btu/lbm

Substitute 172.8 Btu/lbm for wout and 206.5 Btu/lbm for wrev,out in Equation (XVI).

ηII=172.8 Btu/lbm206.5 Btu/lbm=0.837=83.7%

Thus, the second law efficiency is 83.7%.

Substitute 172.8 Btu/lbm for wout and 206.5 Btu/lbm for wrev,out in Equation (XVII).

xdest=206.5 Btu/lbm172.8 Btu/lbm=33.7 Btu/lbm

Thus, the exergy destruction during the expansion process is 33.7 Btu/lbm.

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Chapter 13 Solutions

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

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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