A 0.35-ft3 well-insulated rigid can initially contains refrigerant-134a at 90 psia and 30°F. Now a crack develops in the can, and the refrigerant starts to leak out slowly. Assuming the refrigerant remaining in the can has undergone a reversible, adiabatic process, determine the final mass in the can when the pressure drops to 20 psia. Use the tables for R-1

m = 1.37 lbm

Explanation:

We are given that;

P1 = 90 psia

T1 = 30°F

From the table i attached, at T = 30°F, entropy, s1 = sf = 0.04752 Btu/lbm.R

We are also given;

P2 = 20 psia.

At this, s2 = s1 = 0.04752 Btu/lbm.R

From the table i attached, sf at 20 psia is; sf = 0.02605 Btu/lbm.R and sfg = 0.19962 Btu/lbm.R

Now, formula for quality of steam at Pressure P2 is;

X2 = (s2 - sf)/sfg

Plugging in the relevant values to obtain;

X2 = (0.04752 - 0.02605)/0.19962

X2 = 0.1076

Now, v2 = vf + x2•vfg

From the table i attached, at 20 psia, vf = 0.01182, vg =2.27772 and vfg = vg - vf = 2.27772 - 0.01182 = 2.2659 ft³/lbm

Thus,

v2 = 0.01182 + 0.1076*2.2659 = 0.2556 ft³/lbm

Now, let's find mass of the refrigerant from, m = V/v2

m = 0.35/0.2556 = 1.37 lbm