Question 6.237E: A cylinder/piston contains carbon dioxide at 150 lbf/in.^2, ...

A cylinder/piston contains carbon dioxide at 150 lbf / in .^{2}, 600 F with a volume of 7 ft ^{3}. The total external force acting on the piston is proportional to V ^{3}. This system is allowed to cool to room temperature, 70 F. What is the total entropy generation for the process?

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C.V. Carbon dioxide gas of constant mass m _{2}= m _{1}= m out to ambient.

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 \text { gen }}={ }_{1} Q _{2} / T _{ amb }+{ }_{1} S _{2  gen}

Process:      P = CV ^{3} \text { or } PV ^{-3}=\text { constant } ,  which is polytropic with n = -3

State 1:      P _{1}=150   lbf / in ^{2}, T _{1}=600   F =1060   R , V _{1}=7   ft ^{3}  Ideal gas

m =\frac{ P _{1} V _{1}}{ RT _{1}}=\frac{150 \times 144 \times 7}{35.10 \times 1060}=4.064   lbm

Process:    P = C V ^{3} \text { or } PV ^{-3}= const.    polytropic with n = -3.

P _{2}= P _{1}\left( T _{2} / T _{1}\right)^{\frac{ n }{ n -1}}=150\left(\frac{530}{1060}\right)^{0.75}=89.2   lbf / in ^{2}

 

\&   V_{2}=V_{1}\left(T_{1} / T_{2}\right)^{\frac{1}{n-1}}=V_{1} \times \frac{P_{1}}{P_{2}} \times \frac{T_{2}}{T_{1}}=7 \times \frac{150}{89.2} \times \frac{530}{1060}=5.886

 

{ }_{1} W _{2}=\int PdV =\frac{ P _{2} V _{2}- P _{1} V _{1}}{1- n }=\frac{(89.2 \times 5.886-150 \times 7)}{1+3} \times \frac{144}{778}=-24.3   Btu

 

{ }_{1} Q _{2}=4.064 \times 0.158 \times(530-1060)-24.3=-346.6   Btu

 

m \left( s _{2}- s _{1}\right)=4.064 \times\left[0.203 \times \ln \left(\frac{530}{1060}\right)-\frac{35.10}{778} \ln \left(\frac{89.2}{150}\right)\right]=-0.4765   Btu / R

 

\Delta S _{ SURR }=-{ }_{1} Q _{2} / T _{ amb }=364.6 / 530=+0.6879   Btu / R

 

From Eq.6.37 or 6.39

\begin{aligned}{ }_{1} S _{2  gen}  &= m \left( s _{2}- s _{1}\right)-{ }_{1} Q _{2} / T _{ amb }=\Delta S _{ NET }=\Delta S _{ CO 2}+\Delta S _{ SURR } \\&=-0.4765   Btu / R +0.6879   Btu / R =+ 0 . 2 1 1 4   Btu / R\end{aligned}

 

Notice:
n = -3, k = 1.3
n < k

……………………………………

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

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

Eq.6.39:
\begin{aligned}\left(S_{2}-S_{1}\right)_{\text {total }}=&\left(S_{2}-S_{1}\right)_{A}+\left(S_{2}-S_{1}\right)_{B}+\left(S_{2}-S_{1}\right)_{C} \\=& \int \frac{\delta Q_{a}}{T_{a}}-\int \frac{\delta Q_{b}}{T_{b}}+S_{\operatorname{gen} A}+\int \frac{\delta Q_{b}}{T_{b}}-\int \frac{\delta Q_{c}}{T_{c}}+S_{\operatorname{gen} B} \\&+\int \frac{\delta Q_{c}}{T_{c}}-\int \frac{\delta Q_{a}}{T_{a}}+S_{ gen C} \\=& S_{\operatorname{gen} A}+S_{\operatorname{gen} B}+S_{\operatorname{gen} C} \geq 0\end{aligned}

 

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