The condensate from 2 9T flows into the condensate tank; now it is necessary to calculate how much of this can be flashed at 0.3 Mpa...

To calculate the amount of saturated steam flashed under given conditions, it is necessary to consider the principles of energy conservation and mass conservation. It is necessary to know the initial temperature of the condensate (or its thermodynamic properties such as entropy and enthalpy), as well as the enthalpy values of the condensate and the 0.3 Mpa saturated steam. The formula is roughly as follows: \( \frac{(m_{in} \cdot h_{in} - m_{out} \cdot h_{f})}{(h_{fg})} = m_{flash} \) Where: – \( m_{in} \) is the mass of condensate water entering the tank, which is 2.9 T/h in this case. - \( h_{in} is the specific enthalpy of the condensate water entering the tank.) - \( \(m_{out}\) is the mass of water discharged from the tank; if all the water is used for flashing, then \(m_{out} = m_{in} - m_{flash}\). ) - \( h_{f} is the specific enthalpy of liquid water at 0.3 Mpa.) - \( h_{fg} is the heat of vaporization of water at 0.3 Mpa. ) - \( m_{flash} ) is the mass of the flash vapor. You need to consult a steam table or use thermodynamic software to obtain the corresponding enthalpy values, and then determine whether there is heat loss in the condensate water based on the actual process, in order to accurately calculate the amount of flash vapor. In practical applications, other factors such as equipment efficiency may also need to be considered. .

This post was last edited by wwyyxxz on 2024-1-5 00:06. The calculation shows over 200 kilograms, but it should actually be more, as your steam trap generally leaks steam

It’s 29T, not 2.9T. The amount of flash steam should be 2.26 T