Answer:
W = 31.393 Btu
Explanation:
given data
piston-cylinder assembly contains water = 0.5 lb
p1 = 40 lbf/in² = 40 × 144 lbf/ft²
T1 = 300° F
p2 = 14.7 lbf/in² = 14.7 × 144 lbf/ft²
pv 1.2 = constant.
solution
we use here superheated table A 4E to get value for p1 = 40 and t = 300
v1 = 11.04 ft³/lbm
so we know
[tex]P_2 \times V_2^{1.2} = P_1 \times V_1^{1.2}[/tex] ,...............1
[tex]14.7 \times V_2^{1.2} = 40 \times (11.04)^{1.2}[/tex]
solve it we get
v2 = 25.425 ft³/lbm
and here
W = [tex](\frac{P_1V_1 - P_2V_2}{n-1}) m[/tex] ...............2
put here value we get
W = [tex](\frac{40\times 144\times (11.04) -14.7 \times 144\times (25.425) }{1.2-1})0.5[/tex]
solve it we get
W = 31.393 Btu
