Calculation of the steam mixture with two parameters
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There are now two types of steam: one with a pressure of 1.4126 MPa(a), a temperature of 365.4°C, a flow rate of 6.58 t/h, and an enthalpy of 3183.6 kJ/kg; the other has a pressure of 2.018 MPa(a), a temperature of 503.6°C, a flow rate of 0.35 t/h, and an enthalpy of 3475 kJ/kg. If these two types of steam are mixed, how can the temperature and pressure of the resulting steam be calculated? Experts, please give me some advice. Thank you!
No, it’s just a heat exchanger; the two types of steam are mixed together before entering the heater.
This is how I calculated it; it’s for reference only. First, the average enthalpy and entropy of the steam are calculated to be 3198.31 kJ/kg and 7.199 kJ/(kg·K), respectively. Ignoring the heat loss during the mixing process, according to the law of conservation of energy, the enthalpy value after mixing should also be 3198.31 kJ/kg. However, knowing only the enthalpy value is not sufficient to determine the steam state; temperature or pressure is also needed to do so. And the pressure is related to the pressure of the backend devices, which is also hard to determine. So, my thinking is that the gas mixing process is a process of entropy increase, and the entropy after mixing should be no less than 7.199 kJ/(kg.K). Assuming the pressure after mixing is 1.42 MPa, the temperature can be determined to be 372.5 degrees Celsius using the pressure and enthalpy values.
Thank you; it’s indeed difficult to determine the pressure after mixing, while the temperature is easier to measure.
It’s not that it’s difficult to calculate; rather, you must specify the pressure (or temperature). Otherwise, this problem would have infinitely many solutions.
If it’s not a injector, the high pressure must pass through a pressure relief valve before it can mix with the low pressure. If it is a injector, the flow ratio needs to be considered. I think that’s the case
Having studied it, it’s indeed worth thinking about in depth.