A new compound is used in an ideal Rankine cycle where saturated vapor at 400 F enters the turbine and saturated liquid at 70 F exits the condenser. The only properties known for this compound are a molecular mass of 80 lbm/lbmol, an ideal gas heat capacity of Cpo = 0.20 Btu/lbm-R and Tc = 900 R,

Question

A new compound is used in an ideal Rankine cycle where saturated vapor at 400 F enters the turbine and saturated liquid at 70 F exits the condenser. The only properties known for this compound are a molecular mass of 80 lbm/lbmol, an ideal gas heat capacity of Cpo = 0.20 Btu/lbm-R and Tc = 900 R, Pc = 750 psia. Find the specific work input to the pump and the cycle thermal efficiency using the generalized charts.

Accepted Answer

[latex]\begin{aligned}& T _{1}=400 F =860 R \& x _{1}=1.0 \& T _{3}=70 F =530 R \& x _{3}=0.0\end{aligned}[/latex]

Properties known:

[latex]\begin{aligned}& M =80, C _{ PO }=0.2 Btu / lbm R \& T _{ C }=900 R , P _{ C }=750 lbf / in. ^{2} \& T _{ r 1}=\frac{860}{900}=0.956, \quad T _{ r 3}=\frac{530}{900}=0.589\end{aligned}[/latex]

From D.1: [latex]P _{ r1 }=0.76, P _{1}=0.76 imes 750=570 lbf / in .^{2}= P _{4}[/latex]

[latex]\begin{aligned}& P _{ r 3}=0.025, P _{3}=19 lbf / in .^{2}= P _{2}, \quad Z _{ f 3}=0.0045 \& v _{ f 3}= Z _{ f 3} RT _{3} / P _{3}=\frac{0.0045 imes 1545 imes 530}{19 imes 144 imes 80}=0.0168 ft ^{3} / lbm \& w _{ P }=-\int_{3}^{4} vdP \approx v _{ f 3}\left( P _{4}- P _{3} ight)=-0.0168(570-19) imes \frac{144}{778}=-1.71 Btu / lbm \& q _{ H }+ h _{4}= h _{1}, ext { but } h _{3}= h _{4}+ w _{ P }=> q _{ H }=\left( h _{1}- h _{3} ight)+ w _{ p }\end{aligned}[/latex]

From D.2,

[latex]\begin{aligned}&\left( h _{1}^{*}- h _{1} ight)=(1.9859 / 80) imes 900 imes 1.34=30.0 Btu / lbm \&\left( h _{3}^{*}- h _{3} ight)=(1.9859 / 80) imes 900 imes 5.2=116.1 Btu / lbm \&\left( h _{1}^{*}- h _{3}^{*} ight)= C _{ P 0}\left( T _{1}- T _{3} ight)=0.2(400-70)=66.0 Btu / lbm \&\left( h _{1}- h _{3} ight)=-30.0+66.0+116.1=152.1 Btu / lbm \& q _{ H }=152.1+(-1.71)=150.4 Btu / lbm\end{aligned}[/latex]

Turbine, [latex]\left( s _{2}- s _{1} ight)=0=-\left( s _{2}^{*}- s _{2} ight)+\left( s _{2}^{*}- s _{1}^{*} ight)+\left( s _{1}^{*}- s _{1} ight)[/latex]

From D.3,

[latex]\begin{aligned}&\left( s _{1}^{*}- s _{1} ight)=(1.9859 / 80) imes 1.06=0.0263 Btu / lbm R \&\left( s _{2}^{*}- s _{1}^{*} ight)=0.20 \ln \frac{530}{860}-0.02482 \ln \frac{19}{570}=-0.0124 Btu / lbm R\end{aligned}[/latex]

Substituting,

[latex]\begin{aligned}&\left( s _{2}^{*}- s _{2} ight)=-0.0124+0.0263=0.0139=\left( s _{2}^{*}- s _{ f 2} ight)- x _{2} s _{ fg 2} \&0.0139=0.02482 imes 8.77- x _{2} imes 0.02482(8.77-0.075) \&\Rightarrow x _{2}=0.9444 \&\left( h _{2}^{*}- h _{2} ight)=\left( h _{2}^{*}- h _{ f 2} ight)- x _{2} h _{ fg 2}\end{aligned}[/latex]

From D.2,

[latex]\begin{aligned}& h _{ fg 2}=0.02482 imes 900(5.2-0.07)=114.6 Btu / lbm \&\left( h _{2}^{*}- h _{2} ight)=116.1-0.9444 imes 114.6=7.9 Btu / lbm \& w _{ T }=\left( h _{1}- h _{2} ight)=-30.0+66.0+7.9=43.9 Btu / lbm \&\eta_{ TH }=\frac{ w _{ NET }}{ q _{ H }}=\frac{43.9-1.7}{150.4}= 0 . 2 8 1\end{aligned}[/latex]

Question 12.182E: A new compound is used in an ideal Rankine cycle where satur...

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