A twenty pound-mass of water in a piston/cylinder with constant pressure is at 1100 F and a volume of 22.6 ft ^{3}. It is now cooled to 100 F. Show the P–v diagram and find the work and heat transfer for the process.
Question 3.302E: A twenty pound-mass of water in a piston/cylinder with const...
C.V. Water
Energy Eq.3.5: { }_{1} Q _{2}= m \left( u _{2}- u _{1}\right)+{ }_{1} W _{2}= m \left( h _{2}- h _{1}\right)
Process Eq.: Constant pressure \Rightarrow \quad { }_{1} W _{2}= mP \left( v _{2}- v _{1}\right)
Properties from Table F.7.2 and F.7.3
State 1: T _{1}, \quad v _{1}=22.6 / 20=1.13 ft ^{3} / lbm , P _{1}=800 lbf / in ^{2}, \quad h _{1}=1567.8 Btu / lbm
State 2: 800 lbf / in ^{2}, 100 F
\Rightarrow \quad v _{2}=0.016092 ft ^{3} / lbm , h _{2}=70.15 Btu / lbm
The work from the process equation is found as
\begin{aligned}{ }_{1} W _{2} &=20 lbm \times 800 psi \times(0.016092-1.13) ft ^{3} / lbm \times 144 in ^{2} / ft ^{2} \\&=-2566444 lbf – ft =-3299 Btu\end{aligned}
The heat transfer from the energy equation is
{ }_{1} Q _{2}=20 lbm \times(70.15-1567.8) Btu / lbm = – 2 9 9 5 3 \text { Btu }
………………………….
Eq.3.5 : E_{2}-E_{1}={ }_{1} Q_{2}-{ }_{1} W_{2}
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