THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<
9th Edition
ISBN: 9781266657610
Author: CENGEL
Publisher: MCG CUSTOM
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Chapter 9.12, Problem 145P
To determine

The total exergy destruction for each process of an ideal dual cycle

Expert Solution & Answer
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Answer to Problem 145P

The total exergy destruction in an ideal dual cycle is 25.6Btu/lbm.

Explanation of Solution

Draw the ideal dual cycle on Pv diagram.

THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<, Chapter 9.12, Problem 145P

Consider, the pressure is Pi , the specific volume is vi, the temperature is Ti, the entropy is si, the enthalpy is hi corresponding to ith state.

Write the expression of temperature and volume relation for the isentropic compression process 1-2.

T2=T1(v1v2)k1=T1(r)k1 (I)

Here, compression ratio is r and specific heat ratio is k.

Write the expression of pressure, and volume relation for the isentropic expansion process 1-2.

P2=P1(v1v2)k=P1rk (II)

Write the expression of pressure ratio relation.

Px=P3=rpP2 (III)

Here, pressure ratio is rp.

Write the expression of temperature, and pressure relation for the constant volume heat addition process 2x.

Tx=T2(PxP2) (IV)

Write the expression of temperature, and volume relation for the constant pressure heat addition process x3.

T3=Tx(v3vx)=Txrc (V)

Here, cutoff ratio is rc.

Write the expression of temperature, and volume relation for the constant pressure heat addition process 3-4.

T4=T3(v3v4)k1=T3(rcr)k1 (VI)

Write the expression to calculate the heat added to the cycle during process 41.

qout=cv(T4T1) (VII)

Write the expression to calculate the heat added to the cycle during process 2x.

q2x=cv(TxT2) (VIII)

Here, heat input to the process 2x is q2x.

Write the expression to calculate the heat added to the cycle during process x3.

qx3=cp(T3Tx) (IX)

Here, specific heat at constant pressure is cp and the work done by the process x3 is qx3.

Write the expression of net heat addition to the cycle (qin,net).

qin,net=q2x+qx3 (X).

Write the expression for exergy destruction during the process of the cycle.

xdest=T0sgen=T0(ΔsqinTsource+qoutTsink)

Here, temperature of the surroundings is T0, and the temperature of the heat source is Tsource.

Write the expression of entropy change for the process 2x of an ideal dual cycle (sxs2).

sxs2=cvlnTxT1+Rlnvxv1 (XI)

Here, specific heat of air at constant volume is cv.

Write the expression for the exergy loss for the isothermal process 2x(xdest,2x).

xdest,2x=T0(sxs2qin,2xTsource) (XII)

Write the expression of entropy change for the isothermal process x3 of an ideal dual cycle (s3sx).

s3sx=cplnT3Tx+RlnP3Px (XIII)

Write the expression for the exergy loss for the process x3(xdest,x3).

xdest,x3=T0(s3sxqin,x3Tsource) (XIV)

Write the expression of entropy change for the process 41 of an ideal dual cycle (s1s4).

s1s4=cvlnT1T4+Rlnv1v4 (XV)

Write the expression for the exergy loss for the process 41 (xdest,41).

xdest,41=T0(s1s4+qoutTsink) (XVI)

Here, temperature of the sink is Tsink.

Write the expression to calculate the total in an ideal dual cycle.

xdest,total=xdest,1-2+xdest,2x+xdest,x3+xdest,3-4+xdest,41 (XVII)

Conclusion:

From Table A-1E, “Molar mass, gas constant, and critical-point properties”, obtain the following properties of air at room temperature.

k=1.4cv=0.171Btu/lbmRcp=0.240Btu/lbmR

From Table A-2Ea, “Ideal-gas specific heats of various common gases”, obtain the value for gas content (R) as 0.3704psiaft3/lbmR for air at room temperature.

Substitute 530R for T1, 20 for r, and 1.4 for k in Equation (I).

T2=(530R)(20)1.41=1757R

Substitute 14psia for P1, 20 for r and 1.4 for kin Equation (II).

P2=(14psia)(20)1.4=928psia

Substitute 1.2 for rp and 928psia for P2 in Equation (III).

Px=(1.2)(928psia)=1114psia

Substitute 1757R for T2, 1114psia for Px, and 928psia for P2 in Equation (IV).

Tx=(1757R)(1114psia928psia)=2109R

Substitute 2109R for Tx and 1.4 for rc in Equation (V).

T3=(2109R)(1.3)=2742R

Substitute 2742R for T3, 20 for r and 1.3 for rc in Equation (VI).

T4=(2742R)(1.320)1.41=918.8R

Substitute 0.171Btu/lbmR for cv, 1757R for T2 and 2109R for Tx in Equation (VIII).

q2x=(0.171Btu/lbmR)(2109R1757R)=60.19Btu/lbm

Substitute 0.240Btu/lbmR for cv, 2742R for T3 and 2109R for Tx in Equation (IX).

qx3=(0.240Btu/lbmR)(2742R2109R)=157.9Btu/lbm

Substitute 60.19Btu/lbm for q2x and 157.9Btu/lbm for qx3 in Equation (X).

qin,net=60.19Btu/lbm+157.9Btu/lbm=212.1Btu/lbm

Substitute 0.171Btu/lbmR for R , 918.8K for T4 and 530R for T1 in

Equation (VII).

qout=0.171Btu/lbmR(918.8K530R)=66.48Btu/lbm

The exergy loss for the isothermal process 1-2 (xdest,1-2) is equal to zero.

xdest,1-2=0.

Here vx=v2

Substitute vx for v2 , 0.171Btu/lbmR for cv , 1757R for T2 and 2109R for Tx in Equation (XI).

sxs2=(0.171Btu/lbmR)ln2109R1757R+Rlnvxvx=(0.171Btu/lbmR)ln2109R1757R+Rln(1)=0.03123Btu/lbmR

Substitute 530R for T0 , 0.03123Btu/lbmR for (sxs2) , 60.19Btu/lbm for qin,2x and 2742R for Tsource in Equation (XII).

xdest,2x=530R((0.03123Btu/lbmR)60.19Btu/lbm2742R)=4.917Btu/lbm

Substitute P3 for Px , 0.240Btu/lbmR for cp , 2742R for T3 and 2109R for Tx in Equation (XIII).

s3sx=(0.240Btu/lbmR)ln2742R2109R+RlnP3P3=0.06299Btu/lbmR

Substitute 530R for T0 , 0.06299Btu/lbmR for (s3sx) , 151.9Btu/lbm for qin,x3 and 2742R for Tsource in Equation (XIV).

xdest,x3=530R((0.06299Btu/lbmR)151.9Btu/lbm2742R)=4.024Btu/lbm

The exergy loss for the isothermal process 3-4 (xdest,3-4) is equal to zero.

xdest,3-4=0

Here, v1=v4

Substitute v1 for v4 , 0.171Btu/lbmR for cv , 530R for T1 and 918.8R for T4 in Equation (XV).

s1s4=(0.171Btu/lbmR)ln530R918.8R+Rlnv1v1=(0.171Btu/lbmR)ln530R918.8R+Rln(1)=0.09408Btu/lbmR

Substitute 530R for T0 , 0.09408Btu/lbmR for s1s4, 66.48Btu/lbm for qout and 530R for Tsink in Equation (XVI).

xdest,41=530R((0.09408Btu/lbmR)+66.48Btu/lbm530R)=16.62Btu/lbm

During heat rejection process the largest exergy destruction in an ideal dual cycle occur.

Substitute 0 for xdest,1-2 , 0 for xdest,3-4, 4.917Btu/lbm for xdest,2x, 4.024Btu/lbm for xdest,x3 and 16.62Btu/lbm for xdest,41 in Equation (XVII).

xdest,total=0+4.917Btu/lbm+4.024Btu/lbm+0+16.62Btu/lbm=25.6Btu/lbm

Thus, the total exergy destruction in an ideal dual cycle is 25.6Btu/lbm.

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Chapter 9 Solutions

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