Question

A rigid tank (volume constant) contains 1.2 lbm of R-134a, initially at 90 psia and 160 F. The fluid is then cooled to a final temperature of 50 °F a) Calculate the final pressure b) Calculate the amount of thermal energy (heat) removed during the cooling process

Answer 1

Answer:

(a) 74.023 psi (b) 133.4 KJ

Explanation:

We have given m=1.2 lbm

p₁=90 psi

T₁=160°F=459.67+160=619.67

T₂=50°F=459.67+50=509.67

(a) We know that for constant volume pressure is directly proportional to temperature

$\frac{P1}{T1}$=$\frac{P2}{T2}$

$\frac{90}{(160+459.67)}$=$\frac{P2}{(50+459.67}$

P₂=74.023 psi

(b) for R-134a

At P₁=90 psi ,T₁=160°F

h₁=458.658 KJ/KG (from the table)

At P₂=74.023 psi, T₂=50°F

h₂=219.593b (from the table)

m=1.2 lbm=.5443 kg

Q=.5443 ×(458.658-213.533)

Q=133.4 KJ