A distributor of bottled propane, C3H8, needs to bring propane from 630 R, 14.7 lbf/in.^2 to saturated liquid at 520 R in a steady flow process. If this should be accomplished in a reversible setup given the surroundings at 540 R, find the ratio of the volume flow rates Vin/Vout, the heat transfer

Question

A distributor of bottled propane, [latex]C _{3} H _{8}[/latex], needs to bring propane from 630 R, 14.7 [latex]lbf / in .^{2}[/latex] to saturated liquid at 520 R in a steady flow process. If this should be accomplished in a reversible setup given the surroundings at 540 R, find the ratio of the volume flow rates [latex]V _{ in } / V _{ out }[/latex], the heat transfer and the work involved in the process.

Accepted Answer

R = 35.04/778 = 0.045 04 Btu/lbm R

[latex]T _{ ri }=\frac{630}{665.6}=0.947 P _{ ri }=\frac{14.7}{616}=0.024[/latex]

From D.1, D.2 and D.3 : [latex]Z_{i}=0.99[/latex]

[latex]\begin{aligned}&\left( h _{ i }^{*}- h _{ i } ight)=0.04504 imes 665.6 imes 0.03=0.9 Btu / lbm \&\left( s _{ i }^{*}- s _{ i } ight)=0.04504 imes 0.02=0.0009 Btu / lbm R \& T _{ re }=520 / 665.6=0.781\end{aligned}[/latex]

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

[latex]\begin{aligned}&P_{r e}==0.21, P_{e}=0.21 imes 616=129 lbf / in .^{2}, \quad Z_{ e }=0.035 \&\left(h_{ e }^{*}-h_{ e } ight)=0.04504 imes 665.6 imes 4.58=137.3 Btu / lbm \&\left( s _{ e }^{*}- s _{ e } ight)=0.04504 imes 5.72=0.2576 Btu / lbm R \&\left( h _{ e }^{*}- h _{ i }^{*} ight)=0.407(520-630)=-44.8 Btu / lbm \&\left( s _{ e }^{*}- s _{ i }^{*} ight)=0.407 \ln \frac{520}{630}-0.04504 \ln \frac{132}{14.7}=-0.1770 Btu / lbm R \&\left( h _{ e }- h _{ i } ight)=-137.3-44.8+0.9=-181.2 Btu / lbm \&\left( s _{ e }- s _{ i } ight)=-0.2576-0.1759+0.0009=-0.4326 Btu / lbm R\end{aligned}[/latex]

[latex]\frac{\dot{ V }_{ ext {in }}}{\dot{ V }_{ ext {out }}}=\frac{ Z _{ i } T _{ i } / P _{ i }}{ Z _{ e } T _{ e } / P _{ e }}=\frac{0.99}{0.035} imes \frac{630}{520} imes \frac{129}{14.7}= 3 0 0 . 7[/latex]

[latex]\begin{aligned}& w ^{ rev }=\left( h _{ i }- h _{ e } ight)- T _{0}\left( s _{ i }- s _{ e } ight)=181.2-540(0.4326)=- 5 2 . 4 Btu / lbm \& q ^{ rev }=\left( h _{ e }- h _{ i } ight)+ w ^{ rev }=-181.2-52.4=- 2 3 3 . 6 Btu / lbm\end{aligned}[/latex]

Question 12.178E: A distributor of bottled propane, C3H8, needs to bring propa...

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