Carbon dioxide collected from a fermentation process at 40 F, 15 lbf / in .^{2} should be brought to 438 R, 590 lbf / in .^{2} in a steady flow process. Find the minimum amount of work required and the heat transfer. What devices are needed to accomplish this change of state?
Question 12.174E: Carbon dioxide collected from a fermentation process at 40 F...
R =\frac{35.1}{778}=0.04512 Btu / lbm R
T _{ ri }=\frac{500}{547.4}=0.913, \quad P _{ ri }=\frac{15}{1070}=0.014
From D.2 and D.3: \left(\frac{ h ^{*}- h }{ RT _{ C }}\right)=0.02, \quad\left(\frac{ s ^{*}- s }{ R }\right)=0.01 R
T _{ re }=\frac{438}{547.4}=0.80, \quad P _{ re }=\frac{590}{1070}=0.551From D.2 and D.3: h _{ e }^{*}- h _{ e }=4.50 RT _{ c }, \quad s _{ e }^{*}- s _{ e }=4.70 R
\begin{aligned}&\left( h _{ i }- h _{ e }\right)=-\left( h _{ i }^{*}- h _{ i }\right)+\left( h _{ i }^{*}- h _{ e }^{*}\right)+\left( h _{ e }^{*} -h _{ e }\right) \\&\begin{aligned}=-& 0.04512 \times 547.4 \times 0.02+0.203(500-438)+0.04512 \times 547.4 \times 4.50 \\&=123.2 Btu / lbm\end{aligned}\end{aligned}\begin{aligned}&\left( s _{ i }- s _{ e }\right)=-\left( s _{ i }^{*}- s _{ i }\right)+\left( s _{ i }^{*}- s _{ e }^{*}\right)+\left( s _{ e }^{*} -s _{ e }\right) \\&=-0.04512 \times 0.01+0.203 \ln \frac{500}{438}-0.04512 \ln \frac{15}{590}+0.04512 \times 4.70 \\&=0.4042 Btu / lbm R\end{aligned}
\begin{aligned}& w ^{ rev }=\left( h _{ i }- h _{ e }\right)- T _{0}\left( s _{ i }- s _{ e }\right)=123.2-500(0.4042)=-78.4 Btu / lbm \\& q ^{ rev }=\left( h _{ e }- h _{ i }\right)+ w ^{ rev }=-123.2-78.9= – 2 0 2 . 1 Btu / lbm\end{aligned}
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