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Question 6.244E: A cylinder/piston contains 5 lbm of water at 80 lbf/in.^2, 1...

A cylinder/piston contains 5 lbm of water at 80 lbf / in. ^{2}, 1000 F. The piston has cross-sectional area of 1 ft ^{2} and is restrained by a linear spring with spring constant 60 lbf/in. The setup is allowed to cool down to room temperature due to heat transfer to the room at 70 F. Calculate the total (water and surroundings) change in entropy for the process.

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State 1: Table F.7.2    v _{1}=10.831   ft ^{3} / lbm , \quad u _{1}=1372.3    btu / lbm,

s _{1}=1.9453   Btu / lbm R

State 2: T _{2} & on line in P-v diagram.

P = P _{1}+\left( k _{ s } / A _{ cyl }^{2}\right)\left( V – V _{1}\right)

Assume state 2 is two-phase,

\Rightarrow P _{2}= P _{ sat }\left( T _{2}\right)=0.3632   lbf / in ^{2}

 

\begin{aligned}v _{2} &= v _{1}+\left( P _{2}- P _{1}\right) A _{ cyl }^{2} / mk _{ s }=10.831+(0.3632-80) 1 \times 12 / 5 \times 60 \\&=7.6455   ft ^{3} / lbm = v _{ f }+ x _{2}  v _{ fg }=0.01605+ x _{2}  867.579\end{aligned}

 

x _{2}=0.008793, u _{2}=38.1+0.008793 \times 995.64=46.85   btu / lbm,

 

s _{2}=0.0746+0.008793 \times 1.9896=0.0921   Btu / lbm  R

 

\begin{aligned}{ }_{1} W _{2} &=\frac{1}{2}\left( P _{1}+ P _{2}\right) m \left( v _{2}- v _{1}\right) \\&=\frac{5}{2}(80+0.3632)(7.6455-10.831) \frac{144}{778}=-118.46   Btu\end{aligned}

 

{ }_{1} Q _{2}= m \left( u _{2}- u _{1}\right)+{ }_{1} W _{2}=5(46.85-1372.3)-118.46=-6746   Btu

 

\begin{aligned}\Delta S _{\text {tot }}=& S _{\text {gen tot }}= m \left( s _{2}- s _{1}\right)-{ }_{1} Q _{2} / T _{\text {room }} \\&=5(0.0921-1.9453)+6746 / 529.67= 3 . 4 7   Btu / R\end{aligned}

 

 

 

1
F.7.2
F.7.2'
F.7.2''
F.7.2'''

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