EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220102809444
Author: CENGEL
Publisher: YUZU
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Chapter 13.3, Problem 63P
To determine

The work required to pump the gas.

Expert Solution & Answer
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Answer to Problem 63P

The work input is 1.86×108Btu.

Explanation of Solution

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

NCH4=mCH4MCH4 (I)

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 (II)

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

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

Nm=NCH4+NC2H6 (III)

Write the expression to obtain the mole fraction of CH4(yCH4).

yCH4=NCH4Nm (IV)

Write the expression to obtain the mole fraction of C2H6(yC2H6).

yC2H6=NC2H6Nm (V)

Write the expression to obtain the apparent molecular weight of the mixture (Mm).

Mm=mmNm (VI)

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

cp=mfCH4cp,CH4+mfC2H6cp,C2H6 (VII)

Here, mole fraction of CH4,andC2H6 is mfCH4,andmfC2H6, specific heat of CH4 is cp,CH4, and specific heat of C2H6 is cp,C2H6.

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

R=RuMm (VIII)

Here, universal gas constant is Ru.

Write the expression to obtain the pseudo-critical temperature of the mixture (Tcr,m).

Tcr,m=yiTcr,i=yCH4Tcr,CH4+yC2H6Tcr,C2H6 (IX)

Here, critical temperature of CH4,andC2H6 are Tcr,CH4,andTcr,C2H6.

Write the expression to obtain the pseudo-critical pressure of the mixture (Pcr,m).

Pcr,m=yiPcr,i=yCH4Pcr,CH4+yC2H6Pcr,C2H6 (X)

Here, critical pressure of CH4,andC2H6 are Pcr,CH4,andPcr,C2H6.

Write the expression to obtain the reduced temperature (TR).

TR=TmTcr,m (XI)

Write the expression to obtain the reduced pressure (PR).

PR=PmPcr,m (XII)

Write the expression to obtain the mass of the mixture (m).

m=VZmRTP (XIII)

Write the expression to obtain the initial reduced temperature (TR,1).

TR,1=TmTcr,m (XIV)

Here, mixing critical temperature is Tcr,m.

Write the expression to obtain the initial reduced pressure (PR,1).

PR,1=PmPcr,m (XV)

Here, mixing critical pressure is Pcr,m.

Write the expression to obtain the final reduced temperature (TR,2).

TR,2=TmTcr,m (XVI)

Write the expression to obtain the final reduced pressure (PR,2).

PR,2=PmPcr,m (XVII)

Write the expression to obtain the enthalpy change for the ideal gas mixture.

(h1h2)ideal=cp(T1T2) (XVIII)

Here, initial state enthalpy is h1 and final state enthalpy is h2.

Write the expression to obtain the enthalpy change with departure factors.

h1h2=(h1h2)idealRTcr,m(Zh1Zh2) (XIX)

Write the expression to obtain the work input (Win).

Win=m(h1h2) (XX)

Conclusion:

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

Substitute 75lbm for mCH4 and 16.0lbm/mol for MCH4 in Equation (I).

NCH4=75lbm16.0lbm/mol=4.6875lbmol

Substitute 25lbm for mC2H6 and 30lbm/mol for MC2H6 in Equation (II).

NC2H6=25lbm30.0lbm/mol=0.8333lbmol

Substitute 4.6875lbmol for NCH4, and 0.8333lbmol for NC2H6 in Equation (III).

Nm=4.6875lbmol+0.8333lbmol=5.5208lbmol

Substitute 4.6875lbmol for Nm and 5.5208lbmol for NCH4 in Equation (IV).

yCH4=4.6875lbmol5.5208lbmol=0.8941

Substitute 4.6875lbmol for Nm and 5.5208lbmol for NC2H6 in Equation (V).

yC2H6=0.8333lbmol5.5208lbmol=0.1509

Substitute 5.5208 lbmol for Nm and 100 lbm for mm in Equation (VI).

Mm=100kg5.5208lbmol=18.11lb/kmol

From the Table of ideal gas specific heats of various common gases, write the constant pressure specific heats of CH4, and C2H6 at 300°F as 0.532Btu/lbmR and 0.427Btu/lbmR.

Substitute 0.532Btu/lbmR for cp,CH4, 0.427Btu/lbmR for cp,C2H6, 0.75 for mfCH4, and 0.25 for mfC2H6 in Equation (VII).

cp=(0.75)(0.532Btu/lbmR)+(0.25)(0.427Btu/lbmR)=0.506Btu/lbmR

Substitute 1.9858Btu/lbmolR for Ru and 18.11lbm/lbmol for Mm in Equation (VIII).

R=1.9858Btu/lbmolR18.11lbm/lbmol=0.1097Btu/lbmR

From the Table of critical properties, write the critical temperature and pressure for CH4 as 343.9 R and 673 psia and critical temperature and pressure for C2H6 as 549.8 R and 708 psia.

Substitute 0.8491 for yCH4, 343.9 R for Tcr,CH4, 0.1509 for yC2H6, and 549.8 R for Tcr,C2H6 in Equation (IX).

Tcr,m=0.8491(343.9R)+0.1509(549.8R)=375.0R

Substitute 0.8491 for yCH4, 673 psia for Pcr,CH4, 0.1509 for yC2H6, and 708 psia for Pcr,C2H6 in Equation (X).

Pcr,m=0.8491(673psia)+0.1509(708psia)=678.3psia

Substitute 760 R for Tm and 375.0 R for Tcr,m in Equation (XI).

TR=760R375.0R=2.027

Substitute 1,300 psia for Pm and 678.3 psia for Pcr,m in Equation (XII).

PR=1,300psia678.3psia=1.917

From the Table of Nelson-Obert generalized compressibility chart, write the compressibility factor, Zm as 0.963.

Substitute 0.963 for Zm, 2×106ft3 for V, 0.5925psiaft3/lbmR for R, 760 R for T, and 1,300 psia for P in Equation (XIII).

m=2×106ft30.963(0.5925psiaft3/lbmR)760R1,300psia=5.996×106lbm

Substitute 760 R for Tm and 375.0 R for Tcr,m in Equation (XIV).

TR,1=760R375.0R=2.027

Substitute 1,300 psia for Pm and 678.3 psia for Pcr,m in Equation (XV).

PR,1=1,300psia678.3psia=1.917

From the Table of generalized enthalpy departure chart, write the initial enthalpy departure, Zh1 as 0.487.

Substitute 660 R for Tm and 375.0 R for Tcr,m in Equation (XVI).

TR,2=660R375.0R=1.76

Substitute 20 psia for Pm and 678.3 psia for Pcr,m in Equation (XVII).

PR,2=20psia678.3psia=0.0295

From the Table of generalized enthalpy departure chart, write the final enthalpy departure, Zh2 as 0.0112.

Substitute 0.506Btu/lbmR for cp, 760 R for T1, and 660 R for T2 in Equation (XVIII).

(h1h2)ideal=0.506Btu/lbmR(760R660R)=50.6Btu/lbm

Substitute 50.6Btu/lbm for (h1h2)ideal, 375 R for Tcr,m 0.1097Btu/lbmR for R, 0.487 for Zh1, 0.0112 for Zh2 in Equation (XIX).

h1h2=50.6Btu/lbm(0.1097Btu/lbmR)(375R)(0.4870.0112)=31.02Btu/lbm

Substitute 31.02Btu/lbm for h1h2 and 5.996×106lbm for m in Equation (XX).

Win=5.996×106lbm(31.02Btu/lbm)=1.86×108Btu

Thus, the work input is 1.86×108Btu.

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Chapter 13 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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