Question 8.182E: Calculate the irreversibility for the process described in P...

Calculate the irreversibility for the process described in Problem 4.211E, assuming that the heat transfer is with the surroundings at 61 F.

 

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C.V. Cylinder volume out to T _{ o } = 61 F.

Continuity Eq.4.15:        m _{2}- m _{1}= m _{ in }

Energy Eq.4.16:              m _{2} u _{2}- m _{1} u _{1}= m _{ in } h _{\text {line }}+{ }_{1} Q _{2}-{ }_{1} W _{2}

Entropy Eq.7.12:            m _{2} s _{2}- m _{1} s _{1}= m _{ i } s _{ i }+{ }_{1} Q _{2} / T _{ o }+{ }_{1} S _{2  gen}

Process:        P _{1}  is constant to stops, then constant V to state 2 at  P _{2}

State 1:      P _{1}, T _{1} \quad m _{1}=\frac{ P _{1} V }{ RT _{1}}=\frac{45 \times 9 \times 144}{53.34 \times 519.7}=2.104   lbm

State 2:
Open to 60   lbf / in ^{2}, T _{2}=630   R

Table F.5:

\begin{aligned}& h _{ i }=266.13   btu / lbm \\& u _{1}=88.68   Btu / lbm \\& u _{2}=107.62   Btu / lbm\end{aligned}

Only work as piston moves (V changes) while  P = P _{1} \text { until } V = V _{\text {stop }}

 

{ }_{1} W _{2}=\int Pd V = P _{1}\left( V _{\text {stop }}- V _{1}\right)=45 \times(36-9) \frac{144}{778}=224.9   Btu

 

m _{2}= P _{2} V _{2} / RT _{2}=\frac{60 \times 36 \times 144}{53.34 \times 630}=9.256   lbm , \quad m _{ i }=7.152   lbm

 

\begin{aligned}&{ }_{1} Q _{2}= m _{2} u _{2}- m _{1} u _{1}- m _{ i } h _{ i }+{ }_{1} W _{2} \\&\begin{aligned}=& 9.256 \times 107.62-2.104 \times 88.68-7.152 \times 266.13+224.9 \\&=-868.9   Btu\end{aligned}\end{aligned}

Use from table F.4:      C _{ p }=0.24,     R =53.34 / 778=0.06856   Btu / lbm  R,

\begin{aligned}& I = T _{ o } {}_{1}S _{2  gen }= T _{ o }\left[ m _{1}\left( s _{2}- s _{1}\right)+ m _{ i }\left( s _{2}- s _{ i }\right)\right]-{ }_{1} Q _{2} \\&=520.7\left[2.104\left( C _{ p } \ln \frac{630}{519.7}- R \ln \frac{60}{45}\right)+7.152\left( C _{ p } \ln \frac{630}{1100}- R \ln \frac{60}{75}\right)\right] \\&-(-868.9) \\&=520.7(0.05569-0.8473)+868.9 \\&= 4 5 6 . 7   Btu\end{aligned}

 

……………………………………..

Eq.4.15 : 1=\frac{\dot{m}_{1}}{\dot{m}_{3}}+\frac{\dot{m}_{2}}{\dot{m}_{3}}

Eq.4.16 : 0=\frac{\dot{m}_{1}}{\dot{m}_{3}} h_{1}+\frac{\dot{m}_{2}}{\dot{m}_{3}} h_{2}-h_{3}+\dot{Q} / \dot{m}_{3}

Eq.7.12 : \left(m_{2} s_{2}-m_{1} s_{1}\right)_{ c . v .}=\sum m_{i} s_{i}-\sum m_{e} s_{e}+\int_{0}^{t} \sum_{ c.s. } \frac{\dot{Q}_{ c.v. }}{T} d t+{ }_{1} S_{2 gen }

 

1
F.5.1
F.5.2
F.4

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