Answer
$\Delta u=78.41\ Btu/lbm$
$\Delta h=110.67\ Btu/lbm$
Work Step by Step
The circular profile of the 10 in diameter piston has an area of: $A=\frac{\pi}{4}D^2 = 0.545\ ft²$
Hence the initial height of 1 ft³ of air is: $V=A.x \rightarrow x_0= 1.833\ ft$
Since the final volume is half of the original: $x_1 = 0.9167\ ft$, and $\Delta x = -0.9167\ ft = -11\ in$
With a spring constant of $k = 5\ lbf/in$, the elastic force is $F=-55\ lbf$ and $\Delta P = F/A= -0.70\ psia$
Thus, the final pressure is: $P_2=249.3\ psia$
From the ideal gas relation ($T_1=460°F=919.67°R, R=0.3704\ psia.ft³/lbm.°R$):
$PV=mRT$
The mass is $m=0.7339\ lbm$
Therefore the final temperature is:
$T_2=458.55°R$
From table A-2a $c_{p,\ avg}\approx c_{p,\ 80°F} = 0.240\ Btu/lbm.°R$:
So:
$\Delta h=c_p\Delta T$
$\Delta h=110.67\ Btu/lbm$
$c_{v,\ avg}\approx c_{v,\ 80°F} = 0.171\ Btu/lbm.°R$:
$\Delta u=c_v\Delta T$
$\Delta u=78.41\ Btu/lbm$
