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 41P

(a)

To determine

The mass of the gas treating it as an ideal gas mixture.

(a)

Expert Solution
Check Mark

Answer to Problem 41P

The mass of the gas treating it as an ideal gas mixture is 4.441×106lbm.

Explanation of Solution

Write the expression to calculate the mole number of the substance (N).

N=mM (I)

Here, the mass of the substance is m and the molar mass of the substance is M.

Write the expression to calculate the mole fraction of the substance (y).

y=NNm (II)

Here, total mole number of the products is Nm.

Write the expression to calculate the individual gas constant (R).

R=RuM (III)

Here, universal gas constant is Ru.

Write the expression to calculate the mass of the gas from ideal gas expression.

m=PVRT (IV)

Here, pressure of the gas is P, volume of the gas is V, and the temperature is T.

Conclusion:

From the Table A-1 of “Molar mass, gas constant, and critical-point properties”, obtain the molar masses as follows:

MCH4=16.0lbm/lbmolMC2H6=30.0lbm/lbmol

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

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

Substitute 25lbm for mC2H6 and 30.0lbm/lbmol for MC2H6 in Equation (I).

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

Calculate the total number of moles (Nm).

Nm=NCH4+NC2H6=4.6875lbmol+0.8333lbmol=5.5208lbmol

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

yCH4=4.6875lbmol5.5208lbmol=0.8491

Substitute 0.8333lbmol for NC2H6 and 5.5208lbmol for Nm in Equation (II).

yC2H6=0.8333lbmol5.5208lbmol=0.1509

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

Mm=100 lbm5.5208lbmol=18.11lbm/lbmol

Substitute 10.73psiaft3/lbmolR for Ru and 18.11lbm/lbmol for Mm in Equation (III).

R=10.73psiaft3/lbmolR18.11lbm/lbmol=0.5925psiaft3/lbmR

Substitute 2000 psia for P, (1×106 ft3) for V, 0.5925psiaft3/lbmR for R, and 760 R for T in Equation (IV).

m=2000 psia×(1×106 ft3)0.5925psiaft3/lbmR×(760 R)=4.441×106lbm

Thus, the mass of the gas treating it as an ideal gas mixture is 4.441×106lbm.

(b)

To determine

The mass of the gas using the compressibility chart.

(b)

Expert Solution
Check Mark

Answer to Problem 41P

The mass of the gas using the compressibility chart is 4.684×106lbm.

Explanation of Solution

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

PR=PPcr (V)

Here, pressure at the critical point is Pcr.

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

TR=TTcr (VI)

Here, temperature at the critical point is Tcr.

Write the expression to calculate the compressibility factor of the mixture (Zm).

Zm=yCH4ZCH4+yC2H6ZC2H6 (VII)

Here, compressibility factor for CH4 is ZCH4 and compressibility factor for C2H6 is ZC2H6.

Write the expression to calculate the mass from using the compressibility factor.

m=PVZmRT (VIII)

Conclusion:

Substitute 760R for Tm and 343.9R for Tcr,CH4 in Equation (VI).

TR,CH4=760R343.9R=2.210

Substitute 2000psia for PCH4 and 673psia for Pcr,CH4 in Equation (V).

PR,CH4=2000psia673psia=2.972

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced pressure of 2.972 and reduced temperature of 2.210 as 0.98

Substitute 760R for Tm and 549.8R for Tcr,C2H6 in Equation (VI).

TR,C2H6=760R549.8R=1.382

Substitute 1500psia for PC2H6 and 708psia for Pcr,C2H6 in Equation (V).

PR,C2H6=1500psia708psia=2.119

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced pressure of 2.119 and reduced temperature of 1.382 as 0.77

Substitute 0.8491 for yCH4, 0.98 for ZCH4, 0.1509 for yC2H6, and 0.77 for ZC2H6 in Equation (VII).

Zm=0.8491(0.98)+0.1509(0.77)=0.9483

Substitute 2000 psia for P, (1×106 ft3) for V, 0.9483 for Zm, 0.5925psiaft3/lbmR for R, and 760 R for T in Equation (VIII).

m=2000 psia×(1×106 ft3)0.9483×0.5925psiaft3/lbmR×(760 R)=4.684×106lbm

Thus, the mass of the gas using the compressibility chart is 4.684×106lbm.

(c)

To determine

The mass of the gas using Dalton’ law.

(c)

Expert Solution
Check Mark

Answer to Problem 41P

The mass of the gas using the Daltonls law is 4.575×106lbm.

Explanation of Solution

Write the expression to calculate the reduced volume specific volume (vR).

vR=V/mRTcr/Pcr (IX)

Conclusion:

From the Table A-2E of “Ideal gas specific heats of various common gases”, obtain the gas constants as follows:

RCH4=0.6688psiaft3/lbmRRC2H6=0.3574psiaft3/lbmR

Substitute (1×106 ft3) for V, 4.441×106lbm for m, 0.6688psiaft3/lbmR for RCH4, 343.9R for Tcr,CH4, and 673psia for Pcr,CH4 in Equation (IX).

vR,CH4=(1×106 ft3)/4.441×106lbm0.6688psiaft3/lbmR×343.9R/673psia=0.8782

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced specific volume of 0.8782 and reduced temperature of 2.210 as 0.98

Substitute (1×106 ft3) for V, 4.441×106lbm for m, 0.3574psiaft3/lbmR for RC2H6, 549.8R for Tcr,C2H6, and 708psia for Pcr,C2H6 in Equation (IX).

vR,C2H6=(1×106 ft3)/4.441×106lbm0.3574psiaft3/lbmR×549.8R/708psia=3.244

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced specific volume of 3.244 and reduced temperature of 1.382 as 0.92

Substitute 0.8491 for yCH4, 0.98 for ZCH4, 0.1509 for yC2H6, and 0.92 for ZC2H6 in Equation (VII).

Zm=0.8491(0.98)+0.1509(0.92)=0.9709

Substitute 2000 psia for P, (1×106 ft3) for V, 0.9709 for Zm, 0.5925psiaft3/lbmR for R, and 760 R for T in Equation (VIII).

m=2000 psia×(1×106 ft3)0.9709×0.5925psiaft3/lbmR×(760 R)=4.575×106lbm

Thus, the mass of the gas using the Daltonls law is 4.575×106lbm.

(d)

To determine

The mass of the gas using Kay’s rule.

(d)

Expert Solution
Check Mark

Answer to Problem 41P

The mass of the gas using Kay’s rule is 4.579×106lbm.

Explanation of Solution

Write the formula to calculate the pseudo critical temperature (Tcr,m).

Tcr,m=yCH4Tcr,CH4+yC2H6Tcr,C2H6 (X)

Write the formula to calculate the pseudo critical pressure (Pcr,m).

Pcr,m=yCH4Pcr,CH4+yC2H6Pcr,C2H6 (XI)

Conclusion:

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

Tcr,m=0.8491(343.9R)+0.1509(549.8R)=375R

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

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

Substitute 760R for Tm and 375R for Tcr,m in Equation (VI).

TR=760R375R=2.027

Substitute 2000psia for PCH4 and 678.3psia for Pcr,m in Equation (V).

PR=2000psia678.3psia=2.949

From the Table A-15 of “Nelson-Obert generalized compressibility chart”, obtain the compressibility factor at reduced pressure of 2.949 and reduced temperature of 2.027 as 0.97

Substitute 2000 psia for P, (1×106 ft3) for V, 0.97 for Zm, 0.5925psiaft3/lbmR for R, and 760 R for T in Equation (VIII).

m=2000 psia×(1×106 ft3)0.97×0.5925psiaft3/lbmR×(760 R)=4.579×106lbm

Thus, the mass of the gas using Kay’s rule is 4.579×106lbm.

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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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