A mixture of 50% carbon dioxide and 50% water by mass is brought from 2800 R, 150 lbf/in.^2 to 900 R, 30 lbf/in.^2 in a polytropic process through a steady flow device. Find the necessary heat transfer and work involved using values from F.4.

Question

A mixture of 50% carbon dioxide and 50% water by mass is brought from 2800 R, 150 [latex]lbf / in .^{2}[/latex] to 900 R, 30 [latex]lbf / in .^{2}[/latex] in a polytropic process through a steady flow device. Find the necessary heat transfer and work involved using values from F.4.

Accepted Answer

Process [latex]Pv ^{ n }[/latex] = constant leading to

[latex]n \ln \left( v _{2} / v _{1} ight)=\ln \left( P _{1} / P _{2} ight) ; \quad v = RT / P[/latex]

n = ln(150/30) / ln(900 × 150/30 × 2800) =3.3922

[latex]\begin{aligned}& R _{\operatorname{mix}}= \Sigma c _{ i } R _{ i }=(0.5 imes 35.1+0.5 imes 85.76) / 778=0.07767 Btu / lbm R \& C _{ P mix }= \Sigma c _{ i } C _{ Pi }=0.5 imes 0.203+0.5 imes 0.445=0.324 Btu / lbm R\end{aligned}[/latex]

Work is from Eq.7.18:

[latex]\begin{aligned}w &=-\int vd P =-\frac{ n }{ n -1}\left( P _{ e } v _{ e }- P _{ i } v _{ i } ight)=-\frac{ nR }{ n -1}\left( T _{ e }- T _{ i } ight) \&=-\frac{3.3922 imes 0.07767}{2.3922}(900-2800)= 2 0 9 . 3 \frac{ B tu }{ lbm }\end{aligned}[/latex]

Heat transfer from the energy equation

[latex]q=h_{e}-h_{i}+w=C_{p}\left(T_{e}-T_{i} ight)+w=-406.3 Btu / lbm[/latex]

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Eq.7.18 :
[latex]\begin{aligned}w &=-\int_{i}^{e} v d P \quad ext { and } \quad P v^{n}= ext { constant }=C^{n} \w &=-\int_{i}^{e} v d P=-C \int_{i}^{e} \frac{d P}{P^{1 / n}} \&=-\frac{n}{n-1}\left(P_{e} v_{e}-P_{i} v_{i} ight)=-\frac{n R}{n-1}\left(T_{e}-T_{i} ight)\end{aligned}[/latex]

Question 11.185E: A mixture of 50% carbon dioxide and 50% water by mass is bro...

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