Three kg of water in a piston-cylinder apparatus is initially in the saturated l

Question

Three kg of water in a piston-cylinder apparatus is initially in the saturated liquid state at 0.6 MPa. Energy is added slowly to the water as heat and the piston moves in such a way that the pressure remains constant. How much energy must be added as heat to bring the water to the saturated vapor state? Note: Under this constant pressure condition, the work done by the expanding water on the environment is given by
W =mP(V2 - V1 )
where m is the mass of water, P is pressure and V is the specific volume of water.

Explanation / Answer

At P1 = 0.6 MPa for saturated liquid , from steam table,

specific volume = v1 = 0.0011 m3/Kg

and specific internal energy = u1 = 670.14 Kj /Kg

Now after expansion P2= P1 =0.6 Mpa as pressure is constant.

At P2 = 0.6 MPa and for saturated water vapour, from steam table:

specific volume = v2 = 0.315 m3/Kg

and specific internal energy = u2 = 2566.8 Kj/Kg

Now, Work done by the system = W = m x P x (v2 - v1)

W = 3 x 6x105 x (0.315 - 0.0011) = 565.02 KJ

And change in internal energy = U = m x (u2 - u1)

= 3 x (2566.8 - 670.14)

= 5689 KJ

By first Law :

Q =  U + W

Q = 5689 + 565.02

= 6254.02 KJ = energy added to the system

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