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How is the heat released when saturated steam at 4.0 MPA is converted to saturated steam at 1.2 MPA? It seems that using only enthalpy for the calculation won’t work, as the enthalpies in these two states are very similar. The enthalpy of saturated steam increases with temperature at first and then decreases; it reaches its highest value around 240–250, after which it drops as the temperature continues to rise. Can anyone explain this? ? ?
It is not suitable for heat transfer; it can be used to generate power by reducing pressure. Isn’t this better? The difference in heat is indeed that much; that’s why high-pressure steam isn’t used as a heat source. I don’t know the answer to your next question – I’ve been trying to learn it*; looking forward to answers from below.
We utilize the thermal energy of steam primarily by taking advantage of its latent heat – that is, the energy released when steam changes from a gaseous state to a liquid state (during phase change). As for the issue of enthalpy decreasing as temperature rises, since the unit of enthalpy is KJ/KG, we do not consider the specific volume of steam, as specific volume is related to pressure. In other words, enthalpy is related to both temperature and pressure.
Reducing 40 kilograms to 12 kilograms is such a waste; there’s a saying that goes, \"Sell fish that are still alive, even if they die after being dropped.\" If it’s necessary to lower the pressure, consider using a saturated hot water tower.
Since the specific enthalpy of the steam remains essentially the same before and after transformation, a throttle valve can be used to allow the steam to expand isentropically, thereby achieving pressure reduction. I have previously worked on a system that converts high-pressure steam into medium-pressure steam
Just choose a temperature-reducing pressure regulator
Choose a pressure reducer based on pressure and flow rate. If the flow rate is not high, using multiple perforated plates can also be used to reduce pressure. Please confirm the units