The work done that the process is 1180.4kJ/kg.
Temperature = 10°C
Quality fo steam = 84% = 0.84
Final pressure = 3MPa.
The internal energy at state 1 will be:
= 42.020 + 0.814(2346.6)
= 1952.2 kJ/kg
The internal energy at state 2 will be:
= 0.1511 + 0.814(8.7488)
= 7.2726 kJ/kg
From the superheated water, at P2 = 3MPa, the internal energy is 3132.6 kJ/kg. The work done will be:
= ∆U
= m(u2 - u1)jJ1180.4kJ/kg
= 3132.6 - 1952.2
= 1180.4kJ/kg
Learn more about work done on:
brainly.com/question/25573309
#SPJ1
