Question 6.227E: A 1 gallon jug of milk at 75 F is placed in your refrigerato...

A 1 gallon jug of milk at 75 F is placed in your refrigerator where it is cooled down to the refrigerators inside temperature of 40 F. Assume the milk has the properties of liquid water and find the entropy generated in the cooling process.

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C.V. Jug of milk. Control mass at constant pressure.
Continuity Eq.:              m _{2}= m _{1}= m;

Energy Eq.3.5:              m \left( u _{2}- u _{1}\right)={ }_{1} Q _{2}-{ }_{1} W _{2}

Entropy Eq.6.37:          m \left( s _{2}- s _{1}\right)=\int d Q / T +{ }_{1} S _{2  gen}

State 1:   Table F.7.1:        v _{1} \cong v _{ f }=0.01606   ft ^{3} / Ibm ,          h _{1}= h _{ f }=43.085   Btu / lbm ;

s _{ f }=0.08395   Btu / lbm  R

State 2: Table F.7.1:        h _{2}= h _{ f }=8.01   Btu / lbm , s _{2}= s _{ f }=0.0162   Btu / lbm  R

Process:        P = constant = 14.7 psia      \Rightarrow{ }_{1} W _{2}= mP \left( v _{2}- v _{1}\right)

V _{1}=1 Gal =231   in ^{3} \Rightarrow m =231 / 0.01606 \times 12^{3}=8.324   lbm

Substitute the work into the energy equation and solve for the heat transfer

{ }_{1} Q _{2}= m \left( h _{2}- h _{1}\right)=8.324   lbm (8.01-43.085)   Btu / lbm =-292   Btu

The entropy equation gives the generation as

\begin{aligned}{ }_{1} S _{2  gen}  &= m \left( s _{2}- s _{1}\right)-{ }_{1} Q _{2} / T _{\text {refrig }} \\&=8.324(0.0162-0.08395)-(-292 / 500) \\&=-0.564+0.584= 0 . 0 2   Btu / R\end{aligned}

…………………………………..

Eq.6.37 : S_{2}-S_{1}=\int_{1}^{2} d S=\int_{1}^{2} \frac{\delta Q}{T}+{ }_{1} S_{2 gen}

Eq.3.5: E_{2}-E_{1}={ }_{1} Q_{2}-{ }_{1} W_{2}

 

1
F.7.1
F.7.1'

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