A piston/cylinder arrangement has a load on the piston so it maintains constant pressure. It contains 1 lbm of steam at 80 lbf / in .^{2}, 50% quality. Heat from a reservoir at 1300 F brings the steam to 1000 F. Find the second-law efficiency for this process. Note that no formula is given for this particular case, so determine a reasonable expression for it.
Question 8.185E: A piston/cylinder arrangement has a load on the piston so it...
1: P _{1}, x _{1} \Rightarrow v _{1}=2.7458 ft ^{3} / lbm , h _{1}=732.905 Btu / lbm,
s _{1}=1.0374 Btu / lbm R2: P _{2}= P _{1}, T _{2} \Rightarrow v _{2}=10.831 ft ^{3} / lbm , h _{2}=1532.6 Btu / lbm
s _{2}=1.9453 Btu / lbm Rm \left( u _{2}- u _{1}\right)={ }_{1} Q _{2}-{ }_{1} W _{2}={ }_{1} Q _{2}- P \left( V _{2}- V _{1}\right)
\begin{aligned}&{ }_{1} Q _{2}= m \left( u _{2}- u _{1}\right)+\operatorname{Pm}\left( v _{2}- v _{1}\right)= m \left( h _{2}- h _{1}\right)=799.7 Btu \\&{ }_{1} W _{2}=\operatorname{Pm}\left( v _{2}- v _{1}\right)=119.72 Btu \\&{ }_{1} W _{2 \text { to atm }} = P _{0} m \left( v _{2}- v _{1}\right)=22 Btu\end{aligned}
Useful work out ={ }_{1} W _{2}-{ }_{1} W _{2 \text { to atm }}=119.72-22=97.72 Btu
\Delta \phi_{\text {reservoir }}=\left(1-\frac{ T _{0}}{ T _{\text {res }}}\right) { }_{1} Q _{2}=\left(1-\frac{536.67}{1759.67}\right) 799.7=556 Btun _{ II }= W _{ net } / \Delta \phi=\frac{97.72}{556}= 0 . 1 7 6
Remark: You could argue that the stored exergy (exergy) should be accounted for in the second law efficiency, but it is not available from this device alone.
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