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Question 8.183E: Calculate the exergy of the system (aluminum plus gas) at th...

Calculate the exergy of the system (aluminum plus gas) at the initial and final states of Problem 6.245E, and also the irreversibility.

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State 1:      T _{1}=400   F \quad v _{1}=2 / 2.862=0.6988 \quad P _{1}=300   psi

Ideal gas        v _{2}= v _{1}(300 / 220)(537 / 860)=0.595 ; \quad v _{o}=8.904= RT _{o} / P _{o}

The metal does not change volume so the terms as Eq.8.22 are added

\begin{aligned}&\phi_{1}= m _{ gas } \phi_{ gas }+ m _{ Al } \phi_{ Al  1} \\&= m _{ gas } C _{ v }\left( T _{1}- T _{ o }\right)- m _{ gas } T _{o}\left[ C _{ p } \ln \frac{ T _{1}}{ T _{ o }}- R \ln \frac{ P _{1}}{ P _{ o }}\right]+ m _{ gas } P _{ o }\left( v _{1}- v _{ o }\right) \\&+ m _{ Al }\left[ C \left( T _{1}- T _{o}\right)- T _{o} C \ln \left( T _{1} / T _{o}\right)\right]_{ Al }\end{aligned}

 

\begin{aligned}\phi_{1}=2.862 &\left[0.156(400-77)-537\left(0.201 \ln \frac{860}{537}-\frac{35.1}{778} \ln \frac{300}{14.7}\right)\right.\\&\left.+14.7(0.6988-8.904)\left(\frac{144}{778}\right)\right]+8 \times 0.21\left[400-77-537 \ln \frac{860}{537}\right] \\=& 143.96+117.78= 2 6 1 . 7 4 ~ B t u\end{aligned}

 

\begin{aligned}&\phi_{2}=2.862\left[0.156(77-77)-537\left(0.201 \ln \frac{537}{537}-\frac{35.1}{778} \ln \frac{220}{14.7}\right)\right. \\&\left.+14.7(0.595-8.904)\left(\frac{144}{778}\right)\right]+8 \times 0.21\left[77-77-537 \ln \frac{537}{537}\right] \\&=122.91+0= 1 2 2 . 9 1   Btu\end{aligned}

 

\begin{aligned}{ }_{1} W _{2  CO _{2}}=& \int PdV =0.5\left( P _{1}+ P _{2}\right)\left( V _{2}- V _{1}\right)=[(300+220) / 2](1.703-2) \frac{144}{778} \\&=-14.29   Btu\end{aligned}

 

\begin{aligned}{ }_{1} I_{2} &=\phi_{1}-\phi_{2}+\left(1-T_{o} / T_{H}\right) {}_{1} Q_{2}-{{ }_{1} W_{2}} ^{A C}+P_{o} m\left(V_{2}-V_{1}\right) \\&=261.74-122.91+0-(-14.29)+14.7 \times 2.862 \times \frac{144}{778}(0.595-0.6988) \\&= 1 5 2 . 3   \text { Btu }\end{aligned}

 

 

……………………………………..

Eq.8.22 :
\begin{aligned}\psi &=\left(h_{ tot }-T_{0} s\right)-\left(h_{ tot  0}-T_{0} s_{0}\right) \\&=\left(h-T_{0} s+\frac{1}{2} V ^{2}+g Z\right)-\left(h_{0}-T_{0} s_{0}+g Z_{0}\right)\end{aligned}

 

1

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