A mass of 5kg of saturated water liquid vapor mixture of wat

A mass of 5kg of saturated water liquid vapor mixture of water is containe din a piston-cylinder device at 125 kPa. Initially, 2 kg of water is in th eliquid phase and the rest in the vapor phase. Heat is now transferred to the water and the piston which is resting on a set of stops. The Piston starts to move when the pressure inside reaches 300 kPa. Heat transfer continues until the total volume increases 20 Percent. Determine, the initial an dfinal temperatures, the mass of the water when piston moves and the work during this process.

Solution

Initial quality x1 = (5-2)/5 = 0.6
P1 = 125 kPa
From steam properties at P1 = 125 kPa and x1 = 0.6, we get T1 = 106 deg C, v1 = 0.825 m^3/kg, u1 = 1690 kJ/kg, h1 = 1790 kJ/kg

Volume V1 = m*v1
= 5*0.825
= 4.125 m^3

Let the state when piston starts to move be denoted by suffix 2.

P2 = 300 kPa
v2 = v1 = 0.825 m^3/kg

From steam properties at P2 = 300 kPa and v2 = 0.825 m^3/kg, we get T2 = 268 deg C, x2 = superheated vapor, u2 = 2760 kJ/kg, h2 = 3000 kJ/kg

Since x2 is superheated state, hence mass of water when piston starts to move = 0.

State 1 to state2 is a constant volume process and hence no work is done.
State2 to state3 is a constant pressure process.
V3 = 1.2*V2.....but V2 = V1
= 1.2*4.125
= 4.95 m^3

Work done = P2*(V3 - V2)
= 300*(4.95 - 4.125)
= 247.5 kJ