Question

Consider 1 kilogram of saturated water at 70 degrees celsius in a closed and rigid container which is heated to steam (superheated water) at 300 degrees celsius and 5 MPa by adding 1500 kJ of heat.

a) What is the fraction of vapour and liquid at the initial state?

b) What is the work done as a result of the heating process?

Answer 1

At initial state we have saturated steam at 70 oC

as the question itself says saturated water we have saturated water

1.) Hence, fraction of water = 1 and fraction of vapor = 0 at initial state

H at initial state = 293 Kj/Kg

amount of water = 1 Kg

H1 = 293 KJ ( from steam table)

Final state 300 oC and P = 5 MPa

H2 =2925.64 KJ ( from steam table)

2.) Work done = H2-H1 = 2925.64 - 293 = 2632.64 KJ is the amount of heat required or the work that has to be done to the system.