Question 8.184E: Air in a piston/cylinder arrangement, shown in Fig. P8.135, ...

Air in a piston/cylinder arrangement, shown in Fig. P8.135, is at 30 lbf / in .^{2}, 540 R with a volume of 20 ft ^{3}. If the piston is at the stops the volume is 40 ft ^{3} and a pressure of 60 lbf / in .^{2} is required. The air is then heated from the initial state to 2700 R by a 3400 R reservoir. Find the total irreversibility in the process assuming surroundings are at 70 F.

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

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

Process:                    P = P _{0}+\alpha\left( V – V _{0}\right) \quad \text { if } \quad V \leq V _{\text {stop }}

Information:          P _{\text {stop }}= P _{0}+\alpha\left( V _{\text {stop }}- V _{0}\right)

Eq. of state ⇒        T _{\text {stop }}= T _{1} P _{\text {stop }} V _{\text {stop }} / P _{1} V _{1}=2160< T _{2}

So the piston will hit the stops      \Rightarrow V _{2}= V _{ stop }

P _{2}=\left( T _{2} / T _{\text {stop }}\right) P _{\text {stop }}=(2700 / 2160) 60=75   psia =2.5  P _{1}

State 1:

\begin{aligned}m_{2} &=m_{1}=\frac{P_{1} V_{1}}{R T_{1}} \\&=\frac{30 \times 20 \times 144}{53.34 \times 540} \\&=3.0   lbm\end{aligned}

 

{ }_{1} W _{2}=\frac{1}{2}\left( P _{1}+ P _{\text {stop }}\right)\left( V _{\text {stop }}- V _{1}\right)=\frac{1}{2}(30+60)(40-20)=166.6   Btu

 

{ }_{1} Q _{2}= m \left( u _{2}- u _{1}\right)+{ }_{1} W _{2}=3(518.165-92.16)+166.6=1444.6   Btu

 

\begin{aligned}& s _{2}- s _{1}= s _{ T 2}^{ o }- s _{ T1 }^{ o }- R \ln \left( P _{2} / P _{1}\right) \\&\quad=2.0561-1.6398-(53.34 / 778) \ln (2.5)=0.3535   Btu / lbm  R\end{aligned}

 

Take control volume as total out to reservoir at  T _{ RES }

{ }_{1} S _{2  \text { gen tot }} = m \left( s _{2}- s _{2}\right)-{ }_{1} Q _{2} / T _{ RES }= 0 . 6 3 5 6 ~ B t u / R

 

{ }_{1} I _{2}= T _{0}\left({}_{1} S _{2  gen}  \right)=530 \times 0.6356= 3 3 7   \text { Btu }

 

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