A 1 Gallon tank initially is empty and we want to have 0.03 lbm of R-410A in it. The R-410A comes from a line with saturated vapor at 20 F. To end up with the desired amount we cool the can while we fill it in a slow process keeping the can and content at 20 F. Find the final pressure to reach

Question

A 1 Gallon tank initially is empty and we want to have 0.03 lbm of R-410A in it. The R-410A comes from a line with saturated vapor at 20 F. To end up with the desired amount we cool the can while we fill it in a slow process keeping the can and content at 20 F. Find the final pressure to reach before closing the valve and the heat transfer.

Accepted Answer

C.V. Tank:
Continuity Eq.4.20: [latex]m _{2}-0= m _{ i }[/latex]

Energy Eq.4.21: [latex]m _{2} u _{2}-0= m _{ i } h _{ i }+{ }_{1} Q _{2}[/latex]

State 2: 20 F, [latex]v _{2}= V / m _{2}=1\left(231 / 12^{3} ight) / 0.030=4.456 ft ^{3} / lbm[/latex]

From Table F.9.2 we locate the state between 15 and 20 psia.

[latex]P _{2}=15+(20-15) \frac{4.456-4.6305}{3.4479-4.6305}=15.74 ext { psia }[/latex]

[latex]u _{2}=112.68 Btu / lbm[/latex],

State i Table F.9.1: [latex]h _{ i }=119.07 Btu / lbm[/latex]

Now use the energy equation to solve for the heat transfer

[latex]\begin{aligned}{ }_{1} Q _{2} &= m _{2} u _{2}- m _{ i } h _{ i }= m _{2}\left( u _{2}- h _{ i } ight) \&=0.03 imes(112.68-119.07)=-0.192 Btu\end{aligned}[/latex]

....................................

Eq.4.20: [latex]\left(m_{2}-m_{1} ight)_{ C.V. }=\sum m_{i}-\sum m_{e}[/latex]

Eq.4.21:

[latex]\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} ight) \&-\sum m_{e}\left(h_{e}+\frac{1}{2} V _{e}^{2}+g Z_{e} ight)\end{aligned}[/latex]

Question 4.204E: A 1 Gallon tank initially is empty and we want to have 0.03 ...

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