High-pressure steam condensate conversion
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Hello, friends! Our system has medium-pressure steam (1.6 MP) and condensate (225) that enter a flash tank, resulting in low-pressure steam (0.4 MP) and low-pressure steam condensate (153). Is there any way to calculate the amount of low-pressure steam and low-pressure steam condensate based on the volume of this medium-pressure steam condensate?
You need to perform the calculation based on the enthalpy value of saturated water at medium pressure, the enthalpy value of saturated water at low pressure, and the enthalpy value of saturated water vapor at low pressure. Thus, we have the following system of equations: The enthalpy of saturated water at medium pressure = The enthalpy of saturated water at low pressure + The enthalpy of saturated steam at low pressure. The amount of saturated water at medium pressure = The amount of saturated water at low pressure + The amount of saturated steam at low pressure. By solving these equations, we can determine the amounts of saturated water at low pressure and saturated steam at low pressure.
The condensate of saturated steam at 1.6 MPa should be 201 degrees. I wonder where the 225 degrees for LZ came from? Isn’t it water vapor? The condensate of saturated water vapor at 0.4 MPa should also be 143.6 degrees, rather than 153 degrees. Anyway, I don’t know how it came about through flashing. The only explanation is that this stuff isn’t water. Assuming it is water, with the feed being saturated steam at 1.6 MPa and flash evaporation occurring at 0.4 MPa, we need to determine the amount of vapor produced and the amount of liquid formed. The enthalpy of saturated liquid at 1.6 MPa is 858.6 kJ/kg, while the enthalpy of saturated vapor at that pressure is 2737.6 kJ/kg. The enthalpy of saturated liquid at 0.4 MPa is 604.7 kJ/kg. Assuming the feed volume is 1 unit, the amount of vapor produced is calculated as follows: (858.6 – 604.7) / (2737.6 – 604.7) = 11.9%. Therefore, 88.1% of the feed becomes saturated liquid at 0.4 MPa. In other words, 11.9% of the feed converts into vapor at 0.4 MPa, while 88.1% becomes saturated liquid at the same pressure
The solution upstairs is great. Hehe, I’ve learned it too.
Secondly: I would like to ask the person on the third floor to tell me where I can find information on the calorific values of steam and saturated water
Search on the forum; there is information available on this topic. I’m using the Changsha Youyi version, and I’m afraid it might be considered duplicate content, so I haven’t uploaded it； If you can’t find anyone to tag, my friend will do
Hello, friend upstairs! I couldn’t find it. . . .
A friend on the third floor asked when responding: What does 225 at 1.6 MPa mean? As for 0.4 MPa, a value of 153 seems reasonable, as steam pressures are usually specified in engineering terms as gauge pressure; for example, steam at 1.6 MPa has an absolute pressure of 1.7 MPa, and its temperature should be 204 degrees. For steam at 0.4 MPa, the absolute pressure is 0.5 MPa, so its temperature should be 152 degrees. The enthalpy values used in calculations of enthalpy balance should correspond to these two temperatures.
It suddenly occurred to me that 225 might be a typo; it should be 205……
The solution on floor 3 is good, thank you. I’ve learned something from it