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Question 4.208E: Helium in a steel tank is at 40 psia, 540 R with a volume of...

Helium in a steel tank is at 40 psia, 540 R with a volume of 4 ft ^{3}. It is used to fill a balloon. When the tank pressure drops to 24 psia the flow of helium stops by itself. If all the helium still is at 540 R how big a balloon did I get? Assume the pressure in the balloon varies linearly with volume from 14.7 psia (V = 0) to the final 24 psia. How much heat transfer did take place?

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Take a C.V. of all the helium.
This is a control mass, the tank mass changes density and pressure.

Energy Eq.:    U _{2}- U _{1}={ }_{1} Q _{2}-{ }_{1} W _{2}

Process Eq.:    P = 14.7 + CV

State 1:    P _{1}, T _{1}, V _{1}

State 2:    P _{2}, T _{2}, V _{2}=?

Ideal gas:

P _{2} V _{2}= mRT _{2}= mRT _{1}= P _{1} V _{1}

 

V _{2}= V _{1}\left( P _{1} / P _{2}\right)=4 \times(40 / 24)=6.6667   ft ^{3}

 

V _{ bal }= V _{2}- V _{1}=6.6667-4=2.6667   ft ^{3}

 

\begin{aligned}{ }_{1} W _{2} &=\int P d V = AREA =1 / 2\left( P _{1}+ P _{2}\right)\left( V _{2}- V _{1}\right) \\&=1 / 2(40+24) \times 2.6667 \times 144=12288   lbf – ft =15.791   Btu\end{aligned}

 

U _{2}- U _{1}={ }_{1} Q _{2}-{ }_{1} W _{2}= m \left( u _{2}- u _{1}\right)= m C _{ v }\left( T _{2}- T _{1}\right)=0

 

so    { }_{1} Q _{2}={ }_{1} W _{2}= 1 5 . 7 9 ~ B t u

 

Remark: The process is transient, but you only see the flow mass if you select the tank or the balloon as a control volume. That analysis leads to more terms that must be elliminated between the tank control volume and the balloon control volume.

 

 

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