A cylinder contains ethylene, C _{2} H _{4} , at 222.6 lbf / in .^{2} , 8 F. It is now compressed isothermally in a reversible process to 742 lbf / in .^{2} . Find the specific work and heat transfer.
Question 12.175E: A cylinder contains ethylene, C2H4 , at 222.6 lbf/in.^2 , 8 ...
Ethylene C _{2} H _{4}, \quad R =55.07 ft lbf / lbm R =0.07078 Btu / lbm R
State 1: P _{1}=222.6 lbf / in .^{2}, \quad T _{2}= T _{1}=8 F =467.7 R
State 2: P _{2}=742 lbf / in .^{2}
T _{ r 2}= T _{ r1 }=\frac{467.7}{508.3}=0.920 ; \quad P _{ r1 }=\frac{222.6}{731}=0.305From D.1, D.2 and D.3: Z_{1}=0.85, \left(\frac{ h ^{*}- h }{ RT _{ C }}\right)_{1}=0.40
\left( h _{1}^{*}- h _{1}\right)=0.07078 \times 508.3 \times 0.40=14.4 Btu / lbm\left( s _{1}^{*}- s _{1}\right)=0.07078 \times 0.30 \quad \square \square=0.0212 Btu / lbm R
P _{ r 2}=\frac{742}{731}=1.015 \text { (comp. liquid) }
From D.1, D.2 and D.3: Z_{2}=0.17
\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}\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}
\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}
{ }_{1} w _{2}={ }_{1} q _{2}-\left( u _{2}- u _{1}\right)=-149.1+107.0=- 4 2 . 1 Btu / lbm
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