250-L tank contains propane at 30°C, 90% quality. The tank is heated to 300°C. Calculate the heat transfer during the process

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Accepted Answer

[latex]\begin{aligned}& V =250 L =0.25 m ^{3} \& T _{1}=30^{\circ} C =303.2 K , x _{1}=0.90 \& ext { Heat to } T _{2}=300{ }^{\circ} C =573.2 K \& M =44.094, T _{ c }=369.8 K , P _{ c }=4.25 MPa \& R =0.18855, \quad C _{ P 0}=1.6794\end{aligned}[/latex]

[latex]T _{ r1 }=0.82 ightarrow[/latex] Fig. D.1:

[latex]Z _{1}=\left(1- x _{1} ight) Z _{ f1 }+ x _{1} Z _{ g1 }=0.1 imes 0.05+0.9 imes 0.785=0.711[/latex]

Fig D.2: [latex]\frac{ h _{1}^{*}- h _{1}}{ RT _{ c }}=0.1 imes 4.43+0.9 imes 0.52=0.911[/latex]

[latex]\begin{aligned}& P _{ r }^{ SAT }=0.30 P _{1}^{ SAT }=1.275 MPa \& m =\frac{1275 imes 0.25}{0.711 imes 0.18855 imes 303.2}=7.842 kg \& P _{ r 2}=\frac{7.842 imes Z _{2} imes 0.18855 imes 573.2}{0.25 imes 4250}=\frac{ Z _{2}}{1.254}\end{aligned}[/latex]

at [latex]T _{ r 2}=1.55[/latex] Trial and error on [latex]P _{ r 2}[/latex]

[latex]P _{ r 2}=0.743 \Rightarrow P _{2}=3.158 MPa , \quad Z _{2}=0.94,\left( h ^{*}- h ight)_{2}=0.35 RT _{ C }[/latex]

[latex]\begin{aligned}&\left( h _{2}^{*}- h _{1}^{*} ight)=1.6794(300-30) \quad=453.4 kJ / kg \&\left( h _{1}^{*}- h _{1} ight)=0.911 imes 0.18855 imes 369.8=63.5 kJ / kg \&\left( h _{2}- h _{2} ight)=0.35 imes 0.18855 imes 369.8=24.4 kJ / kg\end{aligned}[/latex]

[latex]\begin{aligned}Q _{12} &= m \left( h _{2}- h _{1} ight)-\left( P _{2}- P _{1} ight) V =7.842(-24.4+453.4+63.5)-(3158-1275) imes 0.25 \&=+3862-471= 3 3 9 1 kJ\end{aligned}[/latex]

Question 12.99: 250-L tank contains propane at 30°C, 90% quality. The tank i...

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