Question

A cylindrical vessel of 5m3 capacity contains wet steam at 100KPa. The volumes of vapour and liquid in the vessel are 4.95m3 and 0.05m3 respectively. Heat is transferred to the vessel until the vessel is filled with saturated vapour. Determine the heat transfer during the process.

Answer 1

Given that:

Volume of vessel V = 5m3

Pressure of steam P = 100KPa

Volume of vapour Vg = 4.95m3

Volume of liquid Vf = 0.05m3

Since, the vessel is a closed container, so applying first law analysis, we have:

The final condition of the steam is dry and saturated but its mass *** the same. The specific volume at the end of heat transfer = vg2

The pressure corresponding to vg= 0.0983 from saturated steam table is 2030kPa or 2.03 bar.

At 2.03 bar U2 = ug2 . m = (47.94 + 2922) × 2600.5 = 132.26 MJ

1Q2 = U2 – U1 = 132.26 – 27.33 = 104.93 MJ.