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vladimir1956 [14]

2 years ago

15

A piston-cylinder assembly contains 0.5 lb of water. The water expands from an initial state where p1 = 40 lbf/in.2 and T1 = 300

o F to a final state where p2 = 14.7 lbf/in.2 During the process, the pressure and specific volume are related by the polytropic process pv 1.2 = constant. Determine the energy transfer by work, in Btu.

1 answer:

VARVARA [1.3K]2 years ago

5 0

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

$P_2 \times V_2^{1.2} = P_1 \times V_1^{1.2}$ ,...............1

$14.7 \times V_2^{1.2} = 40 \times (11.04)^{1.2}$

solve it we get

v2 = 25.425 ft³/lbm

and here

W = $(\frac{P_1V_1 - P_2V_2}{n-1}) m$ ...............2

put here value we get

W = $(\frac{40\times 144\times (11.04) -14.7 \times 144\times (25.425) }{1.2-1})0.5$

solve it we get

W = 31.393 Btu

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