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.
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.
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 RState 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 lbmSubstitute 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 BtuThe 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}