Question

A rigid tank of volume 50 m³ contains a pure substance as a saturated liquid vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are: 𝑻𝒔𝒂𝒕 = 143.61 °C, 𝒗𝒇 = 0.001084 m³ /kg, 𝒗𝒈 = 0.46242 m³/kg. The total mass of liquid vapour mixture in the tank is ___________kg (round off to the nearest integer).

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Answer :

Correct answer is : 135.08

mf = 0.2m, mg = 0.8m, vf = 0.001084 m3/kg, vg = 0.46242 m3/kg, V = 50 m3

V = vf × mf + vg × mg

50 = 0.001084 × 0.2m + 0.46242 × 0.8m

m = 135.08 kg

∴ mass of liquid vapour mixture in the tank is 135.08 kg.

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