Calculations regarding the phase change of water to produce water vapor
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Steam condensate with a flow rate of about 6 tons/h and a pressure of 0.4 MPa is recovered and then depressurized to atmospheric pressure; as a result, some of the water undergoes phase change to produce steam at atmospheric pressure. How can the mass of this steam be calculated? I would appreciate it if someone could provide guidance, preferably with detailed calculation formulas. Thank you
This estimation is for reference only: at a pressure of 0.1 MPa and a temperature of 100°C, the latent heat of vaporization is 539.1 Kcal/kg, while the enthalpy is 639.2 Kcal/kg. At a pressure of 0.4 MPa, the temperature is 140°C; the latent heat of vaporization is 512.1 Kcal/kg, and the enthalpy is 652.8 Kcal/kg. The amount of heat released per kilogram of water when it goes from superheated water at 0.4 MPa and 140°C to water at 0.1 MPa and 100°C is Q = 652.8 – 639.2 = 13.6 Kcal/kg. Taking the average value of the latent heats of vaporization (539.1 + 512.1)/2 = 525.6 Kcal/kg), the amount of water vapor produced per hour from 6 tons of condensed water after depressurization and cooling is 6*1000*13.6/525.6 = 155.25 kg
I think the conditions are insufficient; one equation of mass conservation and one equation of heat conservation, combined, can be used to find a solution.
By combining the two equations of material balance and energy balance, the flash rate can be determined; without considering gas phase entrainment, the typical flash rate is 10%, and it is lower in systems with lower pressures