Questions about vapor pressure

Compare the enthalpy values of the two types of steam; it’s a simple calculation.

It should be taken into consideration; the essence of heating is exactly that.

For an accurate calculation, it is necessary to know the enthalpy values of the two types of steam before entering the heater and their enthalpy values after exiting the heater. When the enthalpy difference is large, the amount of steam required is small. Roughly estimated ； The amounts of steam used in the two cases should be similar, as when using steam for heating, it is primarily the latent heat of the steam that is utilized; since the latent heat of the two types of steam is comparable, the amount of steam used does not change much. At 7 kilograms, the saturation temperature of steam is high, and the temperature difference for heat transfer in the heater is large; as a result, the heat transfer area required by the heater can be smaller.

It’s a very simple calculation – it’s the enthalpy value of the heater

What was said upstairs is all correct; the calculation of enthalpy is very simple!

Regarding the question raised by the poster, in terms of steam consumption alone, the enthalpy difference between steam at 0.5 MPa(G) and 0.7 MPa(G) is indeed not significant; the enthalpy difference between these two types of steam can be considered negligible. But can this issue be understood solely based on enthalpy values and steam usage? This is a heater whose heat exchange area remains constant. According to the heat transfer equation Q = K.S.ΔT, when all other conditions remain unchanged, the overall heat transfer coefficient K stays roughly constant. The heat exchange area does not change, but ΔT increases. Assuming this heater is a reboiler, let’s look at the parameters of the two types of steam: Pressure P = 0.7 MPa(G), Temperature T = 170.41 °C, Specific enthalpy H = 2768.30 kJ/kg; Pressure P = 0.5 MPa(G), Temperature T = 158.83 °C, Specific enthalpy H = 2756.14 kJ/kg. When the heater functions as a reboiler, the temperature difference increases by about 12 °C. If the designed temperature difference for this reboiler is 40 °C, then this increase in temperature difference is quite significant. In other words, with all other conditions unchanged, the heat load on the reboiler increases by 12/40 = 30%! ! The greater the design temperature difference, the smaller this increase is. What does an 30% increase in the heat load of a reboiler mean? Steam consumption will definitely go up! How could it decline? ? Think about it: when the heating steam at the bottom of the distillation column increases the pressure, and the amount of product taken from the top of the column remains unchanged, will the reflux flow at the top of the column increase? It is obvious that the amount of vaporization has increased; even if it’s not a reboiler, the temperature of the medium being heated will definitely rise! Let’s talk about another experience that everyone has: the pressure before the steam control valve is 1.5 MPa(G), and it is reduced to 0.5 MPa(G) by the control valve for use in the heater. If you now increase the pressure behind the valve to 0.7 MPa(G), the opening degree of the valve will definitely increase, and the steam consumption will surely rise – it will never decrease. For the medium being heated, it is obvious that: with a constant flow rate, its temperature upon leaving the heater will increase by about 12°C; while if the temperatures at the inlet and outlet of the heat exchanger remain constant, the flow rate will increase (which means that the equipment’s processing capacity increases, allowing for an expansion of production). As for the exact amount of increase, it can be calculated using the formula Q = W·Cp·(T_out – T_in) = K·S·ΔT – the calculation is quite simple. Once the parameters of a heat exchanger unit are determined, changes in its operating parameters are interdependent; it is incorrect to discuss separately the impact of differences in steam pressure and enthalpy on steam consumption. It is recommended that the original poster do some careful calculations. The heat load of your equipment is quite high; based on your data, the heat load when using 0.5MPa(G) steam is: Q = 10x1000x(2756.14 – 670.5) = 2.09x10^7 kJ/h = 4.98x10^6 kcal/h. That’s 5 million kcal of heat per hour – such a heater must be a very large unit, with an area of several hundred square meters. It seems like the original poster wants to expand production; be thorough in your calculations. . . Since Chase was mentioned, it seems like this problem is just an example from the principles of chemical engineering. In fact, many of the problems people encounter on site are quite basic ones; they are things that are taught in university, and one should know them as long as they take a little time to write them down on paper

There’s logic in the 7th floor; it needs to be calculated carefully.

Just calculate the heat by checking the enthalpy value!

It is necessary to calculate the enthalpy of steam; the enthalpy of steam at different pressures varies. However, steam used as a heat source is generally not superheated steam

Perhaps the original poster is referring to another situation: if the reboiler is of the submerged type, the steam pressure provided is 0. 7MPA is generally achievable by increasing the liquid level in the reboiler, reducing the area of the heat exchange tubes exposed to steam, and increasing the area submerged in the liquid. The area that needs to be exposed to steam must be reduced. There is a very slight reduction in usage, which may only be observed over time. Everyone is welcome to discuss. This post was last edited by Albertlu on 2009-2-23 20:15 ]

I think what was said on the 7th floor makes sense; the problems encountered in reality are quite simple. The key is to consider things from multiple perspectives and try to avoid focusing on just one aspect.