A 200‑m rigid vessel contains a saturated liquid‑vapor mixture with a vapor quality of 75%. The temperature of the vessel is maintained at 393°K. What is the total mass of the liquid‑vapor mixture inside the vessel?

Given,

Temperature;

T = 393;;K

Convert to Celcius;

T = (393-273) degrees

T = 120°C

Using Table A-4 (Saturated water - Temperature table), at T = 120 C;

vf = 0.001060 m³/kg

vg = 0.89133 m³/kg

Quality is given as;

75% = 0.75

Specific volume is given as;

v = vf + x (vg - vf) = 0.001060 + 0.75(0.89133 _ 0.001060)

v= 0.66876 m³/kg

We know;

v = V/m

0.66876  = 100/m

m = 149.53 kg

answer:

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