Heat Transfer in Steam Boiler

Initial Conditions:
- Volume of steam = 5m3
- Volume of water = 5m3
- Pressure = 1 MPa

Final Conditions:
- Volume of steam = 1m3
- Volume of water = 4m3
- Pressure = 1 MPa

Process:
- Steam (dry) is separated at constant pressure from the saturated mixture until 4m3 of water is left.

Calculation of Heat Transfer:
- First, we need to determine the mass of steam and water in the initial and final conditions. We can use the ideal gas law to determine the mass of steam:
- PV = mRT
- m = PV/RT
- For steam at 1 MPa, the density is 0.592 kg/m3
- Therefore, the mass of steam in the initial condition is 5 x 0.592 = 2.96 kg
- The mass of water in the initial condition is also 5 x 1000 = 5000 kg
- In the final condition, the mass of steam is 1 x 0.592 = 0.592 kg and the mass of water is 4 x 1000 = 4000 kg
- Next, we can use the specific enthalpy values for steam and water at 1 MPa to determine the heat transfer:
- The specific enthalpy of steam at 1 MPa is 2783 kJ/kg
- The specific enthalpy of water at 1 MPa is 419 kJ/kg
- Therefore, the initial enthalpy of the system is:
- (2.96 kg x 2783 kJ/kg) + (5000 kg x 419 kJ/kg) = 16,138.32 kJ
- The final enthalpy of the system is:
- (0.592 kg x 2783 kJ/kg) + (4000 kg x 419 kJ/kg) = 3,613.28 kJ
- The heat transfer during the process is the difference between the initial and final enthalpy values:
- 16,138.32 kJ - 3,613.28 kJ = 12,525.04 kJ

Conclusion:
- The heat transfer during the process is 12,525.04 kJ. This calculation assumes that the process is adiabatic (no heat is added or removed from the system) and that there is no work done.