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Concept explainers
(a)
Draw the P−v and T−s diagrams for the given cycle.
(a)
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Answer to Problem 173RP
The P−v and T−s diagrams for the given cycle are shown as in Figure (1).
Explanation of Solution
Draw the P−v and T−s diagram for the given cycle.
Thus, the P−v and T−s diagrams for the given cycle are shown as in Figure (1)
(b)
The expression for the back work ratio as a function of k and r.
(b)
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Answer to Problem 173RP
The expression for the back work ratio as a function of k and r is 1k−11rk−1rk−1−1r−1_.
Explanation of Solution
Find the work of compression using the first law for process 1-2.
q1−2−w1−2=Δu1−2w1−2=−Δu1−2w1−2=−cv(T2−T1)wcomp=−w1−2
wcomp=−cv(T2−T1) (I)
Here, heat interaction during the process 1-2 is q1−2, work interaction for process 1-2 is w1−2, change in specific internal energy for process 1-2 is Δu1−2, constant volume specific heat is cv, temperature at state 1 and 2 is T1, and T2 respectively.
Write the expression of expansion work.
wexp=w2−3=3∫2Pdv=P(v3−v2)=R(T3−T2) (II)
Here, gas constant is R, specific volume at state 2 and 3 is v2 and v3, and work interaction for process 2-3 is w2−3.
Write the expression of back work ratio using the equations (I) and (II).
wcompwexp=cv(T3−T1)R(T3−T2)=cvRT1T2(T3/T1)−1(T3/T2)−1 (III)
Here, temperature at state 1, 2, and 3 are T1,T2, and T3 respectively.
Conclusion:
Process 1-2: Isentropic
Calculate the ratio of T1/T2 and P2/P1.
T1T2=(v2v1)k−1=1rk−1
P2P1=(v1v2)k=rk
Here, pressure at state 1 and 2 is P1,P2, volume at states 1 and 2 is v1,v2, compression ratio is r, and specific heat ratio is k.
Process 2-3: Constant pressure
Calculate the expression for T3/T2.
P3v3T3=P2v2T2T3T2=v3v2=v1v2=rT3T2=v1v2T3T2=r
Here, volume at state 1 and 2 is v1 and v2.
Process 3-1: Constant volume
Calculate the expression for T3/T1.
P3v3T3=P1v1T1T3T1=P3P1T3T1=P2P1T3T1=rk
Substitute rk for T3T1 and r for T3T2, 1rk−1 for T1T2, and Rk−1 for cv in Equation (III).
wcompwexp=Rk−1R(1rk−1)(rk)−1(r)−1=1k−1(1rk−1)(rk)−1(r)−1
Thus, the expression for the back work ratio as a function of k and r is 1k−11rk−1rk−1r−1_.
(c)
The expression for the cycle thermal efficiency as a function of k and r.
(c)
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Answer to Problem 173RP
The expression for the cycle thermal efficiency as a function of k and r is 1−1k1rk−1rk−1r−1.
Explanation of Solution
Express out the heat addition and heat rejection in the process using first law to the closed system for processes 2-3 and 3-1.
qin=cp(T3−T2)
qout=cv(T3−T1)
Here, constant pressure specific heat is cp.
Express the cycle thermal efficiency.
ηth=1−qoutqin (IV)
Conclusion:
Substitute cp(T3−T2) for qin and cv(T3−T1) for qout in Equation (IV).
ηth=1−cv(T3−T1)cp(T3−T2)=1−1kT1(T3/T1−1)T2(T3/T2−1)
Substitute rk−1 for T3T1, r for T3T2, and 1rk−1 for T1T2.
ηth=1−cv(T3−T1)cp(T3−T2)=1−1k1rk−1(rk−1)(r−1)
Thus, the expression for the cycle thermal efficiency as a function of k and r is 1−1k1rk−1rk−1r−1.
(d)
The value of the back work ratio and thermal efficiency as r goes to unity.
(d)
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Answer to Problem 173RP
The value of the back work ratio and thermal efficiency as r goes to unity is 1_ and 0_.
Explanation of Solution
Recall the expression of back work ratio and apply the limits as r goes to infinity.
wcompwexp=1k−1(1rk−1)(rk−1)−1(r)−1limr→1wcompwexp=1k−1{limr→11rk−1(rk−1)−1(r)−1}limr→1wcompwexp=1k−1{limr→1(rk−1)−1(rk)−rk−1}limr→1wcompwexp=1k−1{limr→1(k−1)rk−2krk−1−(k−1)rk−2}
limr→1wcompwexp=1k−1{k−1k−k+1}=1k−1{k−11}=1
Thus, the value of the back work ratio as r goes to unity is 1_.
Recall the expression of cycle thermal efficiency and apply the limits as r goes to infinity.
ηth=1−1k1rk−1rk−1r−1limr→1ηth=1−1k{limr→11rk−1rk−1r−1}limr→1ηth=1−1k{limr→1rk−1rk−rk−1}limr→1ηth=1−1k{limr→1krk−1krk−1−(k−1)rk−2}
limr→1ηth=1−1k{kk−k+1}=1−1k{k1}=0
Thus, the value of the thermal efficiency as r goes to unity is 0_.
From the results of cycle thermal efficiency and back work ratio values of 0 and 1, it shows that no expansion and net work can be done whether you add heat to the system when there is no compression (r=1).
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Chapter 9 Solutions
Thermodynamics: An Engineering Approach
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