Question 4.206E: A tank contains 10 ft^3 air at 15 psia, 540 R. A pipe flowin...

A tank contains 10 ft ^{3} air at 15 psia, 540 R. A pipe flowing air at 150 psia, 540 R is connected to the tank and it is filled slowly to 150 psia. Find the heat transfer to reach a final temperature of 540 R.

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C.V. The tank volume and the compressor.
This is a transient problem (filling of tank).

Continuity Eq.4.20:    m _{2}- m _{1}= m _{ in }

Energy Eq.4.21:    m _{2} u _{2}- m _{1} u _{1}={ }_{1} Q _{2}-{ }_{1} W _{2}+ m _{ in } h _{ in }

Process:     Constant volume    { }_{1} W _{2}=0,

States:    u _{1}= u _{2}= u _{ in }= u _{540} ; \quad h _{ in }= u _{ in }+ RT _{ in }

\begin{array}{l}m _{1}= P _{1} V _{1} / RT _{1}=15 \times 144 \times 10 /(53.34 \times 540)=0.75   lbm \\m _{2}= P _{2} V _{2} / RT _{2}=150 \times 144 \times 10 /(53.34 \times 540)=7.499   lbm\end{array}

 

Heat transfer from the energy equation

\begin{aligned}{ }_{1} Q _{2} &= m _{2} u _{2}- m _{1} u _{1}- m _{ in } h _{ in }=\left( m _{1}+ m _{ in }\right) u _{1}- m _{1} u _{1}- m _{ in } u _{ in }- m _{ in } RT _{ in } \\&= m _{1} u _{1}- m _{1} u _{1}+ m _{ in } u _{1}- m _{ in } u _{ in }- m _{ in } RT _{ in }=- m _{ in } R T _{ in } \\&=-(7.499-0.75)  lbm \times 53.34   ft – lbf / lbm – R \times 540   R \\&=-194395   ft – lbf =- 2 5 0   Btu\end{aligned}

………………………………

Eq.4.20: \left(m_{2}-m_{1}\right)_{ C.V. }=\sum m_{i}-\sum m_{e}

Eq.4.21:

\begin{aligned}E_{2}-E_{1}=Q_{ C.V. }-W_{ C.V. } &+\sum m_{i}\left(h_{i}+\frac{1}{2} V _{i}^{2}+g Z_{i}\right) \\&-\sum m_{e}\left(h_{e}+\frac{1}{2} V _{e}^{2}+g Z_{e}\right)\end{aligned}
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