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Question 12.175E: A cylinder contains ethylene, C2H4 , at 222.6 lbf/in.^2 , 8 ...

A cylinder contains ethylene, C2H4C _{2} H _{4} , at 222.6 lbf/in.2lbf / in .^{2} , 8 F. It is now compressed isothermally in a reversible process to 742 lbf/in.2lbf / in .^{2} . Find the specific work and heat transfer.

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Ethylene  C2H4,R=55.07  ft lbf/lbm R=0.07078  Btu/lbm RC _{2} H _{4}, \quad R =55.07   ft  lbf / lbm  R =0.07078   Btu / lbm  R

State 1:  P1=222.6  lbf/in.2,T2=T1=8 F=467.7  RP _{1}=222.6   lbf / in .^{2}, \quad T _{2}= T _{1}=8  F =467.7   R

State 2:  P2=742  lbf/in.2P _{2}=742   lbf / in .^{2}

Tr2=Tr1=467.7508.3=0.920;Pr1=222.6731=0.305T _{ r 2}= T _{ r1 }=\frac{467.7}{508.3}=0.920 ; \quad P _{ r1 }=\frac{222.6}{731}=0.305

From D.1, D.2 and D.3:    Z1=0.85, (h∗−hRTC)1=0.40Z_{1}=0.85,  \left(\frac{ h ^{*}- h }{ RT _{ C }}\right)_{1}=0.40

(h1∗−h1)=0.07078×508.3×0.40=14.4  Btu/lbm\left( h _{1}^{*}- h _{1}\right)=0.07078 \times 508.3 \times 0.40=14.4   Btu / lbm

 

(s1∗−s1)=0.07078×0.30    □□=0.0212  Btu/lbm R\left( s _{1}^{*}- s _{1}\right)=0.07078 \times 0.30        \quad \square \square=0.0212   Btu / lbm  R

 

Pr2=742731=1.015 (comp. liquid) P _{ r 2}=\frac{742}{731}=1.015 \text { (comp. liquid) }

 

From D.1, D.2 and D.3:      Z2=0.17Z_{2}=0.17

(h2∗−h2)=0.07078×508.3×4.0=143.9(s2∗−s2)=0.07078×3.6=0.2548(h2∗−h1∗)=0(s2∗−s1∗)=0−0.07078ln⁡742222.6=−0.0852\begin{aligned}&\left( h _{2}^{*}- h _{2}\right)=0.07078 \times 508.3 \times 4.0=143.9 \\&\left( s _{2}^{*}- s _{2}\right)=0.07078 \times 3.6=0.2548 \\&\left( h _{2}^{*}- h _{1}^{*}\right)=0 \\&\left( s _{2}^{*}- s _{1}^{*}\right)=0-0.07078 \ln \frac{742}{222.6}=-0.0852\end{aligned}

 

1q2=T(s2−s1)=467.7(−0.2548−0.0852+0.0212)=–149.1  Btu/lbm(h2−h1)=−143.9+0+14.4=−129.5\begin{aligned}&_{1} q _{2}= T \left( s _{2}- s _{1}\right)=467.7(-0.2548-0.0852+0.0212)= – 1 4 9 . 1   Btu / lbm \\&\left( h _{2}- h _{1}\right)=-143.9+0+14.4=-129.5\end{aligned}

 

(u2−u1)=(h2−h1)−RT(Z2−Z1)=−129.5−0.07078×467.7(0.17−0.85)=−107.0\begin{aligned}\left( u _{2}- u _{1}\right) &=\left( h _{2}- h _{1}\right)- RT \left( Z _{2}- Z _{1}\right) \\&=-129.5-0.07078 \times 467.7(0.17-0.85)=-107.0\end{aligned}

 

1w2=1q2−(u2−u1)=−149.1+107.0=−42.1  Btu/lbm{ }_{1} w _{2}={ }_{1} q _{2}-\left( u _{2}- u _{1}\right)=-149.1+107.0=- 4 2 . 1   Btu / lbm

 

 

 

D.1
D.2
D.3

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