Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 16.6, Problem 94RP

a)

To determine

The amount of heat required for the process.

a)

Expert Solution
Check Mark

Answer to Problem 94RP

The amount of heat required for the process is 324,033kJ.

Explanation of Solution

Write the energy balance equation for the reported process.

EinEout=ΔEsystem (I)

Here, input energy is Ein, output energy is Eout, and the change in system energy is ΔEsystem.

Write the expression to obtain the amount of heat required for the process (Qin).

Qin=N(ˉu2ˉu1) (II)

Here, number of moles is N, internal energy of the system at state 1 is ˉu1, and internal energy of the system at state 2 is ˉu2.

Write the expression to obtain the internal energy of the system at state 1 (ˉu1).

ˉu1=h1RuT1 (III)

Here, enthalpy of the system at state 1 is h1, universal gas constant is Ru, and temperature of the system at state 1 is T1.

Write the expression to obtain the internal energy of the system at state 2 (ˉu2).

ˉu2=h2RuT2 (IV)

Here, enthalpy of the system at state 2 is h2, and temperature of the system at state 2 is T2.

Write the expression to obtain the change in enthalpy of the system (h2h1).

h2h1=21cpdT (V)

Conclusion:

Substitute (h1RuT1) for ˉu1, and (h2RuT2) for ˉu2 in Equation (II).

Qin=N(ˉu2ˉu1)=N(h2RuT2(h1RuT1))=N(h2h1Ru(T2T1)) (VI)

Refer Table A-2c, “Ideal-gas specific heats of various common gases”, obtain the specific heat relation as a+bT+cT2+dT3.

Substitute a+bT+cT2+dT3 for cp in Equation (V) and then integrate.

h2h1=21cpdT=21(a+bT+cT2+dT3)dT=a[T]21+b[T22]21+c[T33]21+d[T44]21=a[T2T1]+b2[T22T21]+c3[T32T31]+d4[T42T41] (VII)

Here, constants are a, b, c and d.

Refer Table A-2c, “Ideal-gas specific heats of various common gases”, obtain the values of constants a, b, c and d for methane as 19.89, 5.024×102, 1.269×105, and 11.01×109 respectively.

Substitute 19.89 for a, 5.024×102 for b, 1.269×105 for c, 11.01×109 for d, 1,000 K for T2, and 298 K for T1 in Equation (VII).

h2h1=a[T2T1]+b2[T22T21]+c3[T32T31]+d4[T42T41]={19.89(1,000K298K)+(5.024×1022)((1,000K)2(298K)2)+(1.269×1053)((1,000K)3(298K)3)+(11.01×1094)((1,000K)4(298K)4)}=13,962.78+22,889+4,1182,730=38,239kJ/kmol

Substitute 38,239kJ/kmol for (h2h1), 8.314kJ/kmolK for Ru, 1,000 K for T2, and 298 K for T1 in Equation (VI).

Qin=N(h2h1Ru(T2T1))=(10mol)(38,239kJ/kmol(8.314kJ/kmolK)(1,000K298K))=324,033kJ

Thus, the amount of heat required for the process is 324,033kJ.

b)

To determine

The amount of heat required for the process.

b)

Expert Solution
Check Mark

Answer to Problem 94RP

The amount of heat required for the process is 245,200kJ.

Explanation of Solution

Write the stoichiometric reaction for the dissociation process.

CH4C+2H2

From the stoichiometric reaction, infer that the stoichiometric coefficient for methane (vCH4) is 1, for hydrogen (vH2) is 2, and for carbon (vC) is 1.

Write the expression to obtain the actual reaction for the dissociation process.

CH4xCH4+yC+zH2 (VIII)

From the actual reaction, infer that the equilibrium composition contains x amount of methane (NCH4), y amount of carbon (NC), and z amount of hydrogen (NH2).

Write the expression to obtain the total number of moles (Ntotal).

Ntotal=NCH4+NC+NH2 (IX)

Here, number of moles of CH4 is NCH4, number of moles of C is NC, and number of moles of H2 is NH2.

Write the expression to obtain the equilibrium constant (Kp) for the dissociation process.

Kp=NvCCNvH2H2NvCH4CH4(PNtotal)(vC+vH2vCH4) (X)

Here, pressure is P.

Write the expression to obtain the mole fraction of Methane (yCH4).

yCH4=NCH4Ntotal (XI)

Write the expression to obtain the mole fraction of carbon (yC).

yC=NCNtotal (XII)

Write the expression to obtain the mole fraction of hydrogen (yH2).

yH2=NH2Ntotal (XIII)

Write the expression to obtain the amount of heat required for the process (Qin).

Qin=N(yCH4cV, CH4T2+yH2cV, H2T2+yCcV, CT2)NcV, CH4T1 (XIV)

Here, specific heat of methane is cV, CH4, specific heat of hydrogen is cV, H2, and specific heat of carbon is cV, C.

Conclusion:

Write the carbon balance equation from Equation (VIII).

1=x+yy=1x (XV)

Write the hydrogen balance equation from Equation (VIII).

4=4x+2zz=22x (XVI)

Substitute x for NCH4, y for NC, z for NH2, 1x for y, and 22x for z in Equation (IX).

Ntotal=x+y+z=x+1x+22x=32x (XVII)

Substitute e2.328 for Kp, x for NCH4, y for NC, z for NH2, 1x for y, 22x for z.

32x for Ntotal,1 atm for P, 2 for vH2, and 1 for both vC and vCH4 in Equation (X).

e2.328=(1x)(22x)2x(1atm32x)(1+21)0.0975x=(1x)(22x)2(1(32x)2)0.0975x[9+4x212x]=(1x)(4+4x28x)0.8775x+0.39x31.17x2=(4+4x28x4x4x3+8x2)

3.61x313.17x2+12.8775x4=0x=0.641

Substitute 0.641 for x in equation (XV).

y=1x=10.641=0.359

Substitute 0.641 for x in equation (XVI).

z=22x=22(0.641)=0.718

Substitute 0.641 for x in equation (XVII).

Ntotal=32x=32(0.641)=1.718

Substitute 0.641 for x, 0.359 for y, and 0.718 for z in Equation (VIII).

CH4xCH4+yC+zH2CH40.641CH4+0.359C+0.718H2

Substitute 0.641 for x, and 1.718 for Ntotal in Equation (XI).

yCH4=0.6411.718=0.37

Substitute 0.359 for x, and 1.718 for Ntotal in Equation (XII).

yC=0.3591.718=0.2

Substitute 0.718 for x, and 1.718 for Ntotal in Equation (XIII).

yH2=0.7181.718=0.41

Substitute 10 kmol for N, 0.37 for yCH4, 63.3kJ/kmolK for cV, CH4, 0.41 for yH2, 21.7kJ/kmolK for cV, H2, 0.2 for yC, 0.711kJ/kmolK for cV, C, 1,000 K for T2, and 298 K for T1 in Equation (XIV).

Qin={(10kmol)((0.37)(63.3kJ/kmolK)(1,000K)+(0.41)(21.7kJ/kmolK)(1,000K)+(0.2)(0.711kJ/kmolK)(1,000K))(10kmol)(27.8kJ/kmolK)(298K)}=245,200kJ

Thus, the amount of heat required for the process is 245,200kJ.

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Chapter 16 Solutions

Thermodynamics: An Engineering Approach

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