10 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at 100 kPa, with a quality of 50%. It is now heated so the volume triples. The mass of the piston is such that a cylinder pressure of 200 kPa will float it, as in Fig.3.171. Find the final temperature and the heat

Question

10 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at 100 kPa, with a quality of 50%. It is now heated so the volume triples. The mass of the piston is such that a cylinder pressure of 200 kPa will float it, as in Fig.3.171. Find the final temperature and the heat transfer in the process.

Accepted Answer

Take CV as the water.

Continuity Eq.: [latex]m _{2}= m _{1}= m[/latex];

Energy Eq.: [latex]m \left( u _{2}- u _{1} ight)={ }_{1} Q _{2}-{ }_{1} W _{2}[/latex]

Process: v = constant until [latex]P = P _{ ext {lift }}[/latex], then P is constant.

State 1: Two-phase so look in Table B.1.2 at 100 kPa

[latex]\begin{array}{l}u _{1}=417.33+0.5 imes 2088.72=1461.7 kJ / kg \v _{1}=0.001043+0.5 imes 1.69296=0.8475 m ^{3} / kg\end{array}[/latex]

State 2: [latex]v _{2}, P _{2} \leq P _{ ext {lift }} \Rightarrow v_{2}=3 imes 0.8475=2.5425 m ^{3} / kg[/latex]

Interpolate: [latex]T _{ 2 }= 8 2 9 ^{\circ} C , \quad u _{2}=3718.76 kJ / kg[/latex]

[latex]\Rightarrow V _{2}= mv _{2}=25.425 m ^{3}[/latex]

From the process equation (see P-V diagram) we get the work as

[latex]{ }_{1} W _{2}= P _{ lift }\left( V _{2}- V _{1} ight)=200 kPa imes 10 kg (2.5425-0.8475) m ^{3} / kg =3390 kJ[/latex]

From the energy equation we solve for the heat transfer

[latex]\begin{aligned}{ }_{1} Q _{2} &= m \left( u _{2}- u _{1} ight)+{ }_{1} W _{2}=10 kg imes(3718.76-1461.7) kJ / kg +3390 kJ \&= 2 5 9 6 1 ~ k J\end{aligned}[/latex]

Question 3.171: 10 kg of water in a piston cylinder arrangement exists as sa...

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