At condition I a piston cylinder device contain a liquidvap

At condition I a piston - cylinder device contain a liquid-vapor mixture with quality x_1 = 10.85, at a saturation temperature of T_1 = 105.O degree c. Volume V_1 = 0.0010 m^3 and piston diameter d_p = 0.15m. The System is heated to Condition 5 whereupon the water expand S to V_2 =.0.0015 m^3 this point the piston is a about to engage a linear spring Determine The mass m of the water the pressure \"P_2\" the state of two water The work done in moving from condition I to Condition2 \"_1W_e\" Heating continues and the spring of stiffness R = 100RN/m is compressed until the system pressure P_3 = 200.0 R Pa. Determine Final system temperature T_3 The work done in moving from condition epsilon to condition 3\"_2W_2\" Total work done. At the end of your solution list your an swords a) thru\' g) in a neat clear table.

Solution

From steam mixture properties,

At T1 = 105 deg C and x1 = 0.85, we get v1 = 1.21 m^3/kg, P1 = 121 kPa, u1 = 2200 kJ/kg, h1 = 2350 kJ/kg

a)

m = V1/v1 = 0.001 / 1.21 = 8.264*10^-4 kg

b)

Since heating is happening at constant pressure, we get

P2 = P1 = 121 kPa

c)

v2 = V2 / m = 0.0015 / 8.264*10^-4 = 1.815 m^3/kg

From steam properties at P2 = 121 kPa, we get for sat. vapor v = 1.42 m^3/kg.

Since v2 = 1.815 m^3/kg > 1.42 m^3/kg, we get the quality of steam at 2, x2 = superheated

d)

Work done = P*(V2 - V1)

= 121*(0.0015 - 0.001)

= 0.0605 kJ