A distributor of bottled propane, C3H8, needs to bring propane from 350 K, 100 kPa to saturated liquid at 290 K in a steady flow process. If this should be accomplished in a reversible setup given the surroundings at 300 K, find the ratio of the volume flow rates Vin/Vout, the heat transfer and the

Question

A distributor of bottled propane, [latex]C _{3} H _{8}[/latex], needs to bring propane from 350 K, 100 kPa to saturated liquid at 290 K in a steady flow process. If this should be accomplished in a reversible setup given the surroundings at 300 K, find the ratio of the volume flow rates [latex]\dot{ V }_{ in } / \dot{ V }_{ out }[/latex], the heat transfer and the work involved in the process.

Accepted Answer

From Table A.2: [latex]T _{ ri }=\frac{350}{369.8}=0.946, \quad P _{ ri }=\frac{0.1}{4.25}=0.024[/latex]

From D.1, D.2 and D.3,

[latex]\begin{aligned}& Z _{ i }=0.99 \&\left( h _{ i }^{*}- h _{ i } ight)=0.1886 imes 369.8 imes 0.03=2.1 kJ / kg \&\left( s _{ i }^{*}- s _{ i } ight)=0.1886 imes 0.02=0.0038 kJ / kg K \& T _{ re }=\frac{290}{369.8}=0.784, ext { and } x =0\end{aligned}[/latex]

From D.1, D.2 and D.3,

[latex]P _{ re }=0.22, P _{ e }=0.22 imes 4.25=0.935 MPa ext { and } Z _{ e }=0.036[/latex]

[latex]\begin{aligned}&\left( h _{ e }^{*} -h _{ e } ight)=0.1886 imes 369.8 imes 4.57=318.6 kJ / kg \&\left( s _{ e }^{*}- s _{ e } ight)=0.1886 imes 5.66=1.0672 kJ / kg K \&\left( h _{ e }^{*}- h _{ i }^{*} ight)=1.679(290-350)=-100.8 kJ / kg\end{aligned}[/latex]

[latex]\begin{aligned}&\left( s _{ e }^{*}- s _{ i }^{*} ight)=1.679 \ln \frac{290}{350}-0.1886 \ln \frac{0.935}{0.1}=-0.7373 kJ / kg K \&\left( h _{ e }- h _{ i } ight)=-318.6-100.8+2.1=-417.3 kJ / kg \&\left( s _{ e }- s _{ i } ight)=-1.0672-0.7373+0.0038=-1.8007 kJ / kg K\end{aligned}[/latex]

[latex]\frac{\dot{ V }_{ in }}{\dot{ V }_{ out }}=\frac{ Z _{ i } T _{ i } / P _{ i }}{ Z _{ e } T _{ e } / P _{ e }}=\frac{0.99}{0.036} imes \frac{350}{290} imes \frac{0.935}{0.1}= 3 1 0 . 3[/latex]

[latex]\begin{aligned}& w ^{ rev }=\left( h _{ i }- h _{ e } ight)- T _{0}\left( s _{ i }- s _{ e } ight)=417.3-300(1.8007)=- 1 2 2 . 9 kJ / kg \& q ^{ rev }=\left( h _{ e }- h _{ i } ight)+ w ^{ rev }=-417.3-122.9=- 5 4 0 . 2 kJ / kg\end{aligned}[/latex]

Question 12.112: A distributor of bottled propane, C3H8, needs to bring propa...

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