Discuss the pressure changes due to the condensation of vapor containing non-condensable gases.

1. For unsaturated steam, at equilibrium the partial pressure of water vapor should equal the saturated vapor pressure; therefore, it should change as the condensation temperature changes. The overall pressure of the mixed gas, of course, will change. For example, when opening the lid of a thermos bottle that has contained boiling water for some time, more resistance is felt, and this is due to the decrease in the overall mixing pressure inside. 2. It can be calculated based on the sum of the partial pressures of each gas, such as the total mixing pressure.

At first, the water vapor was superheated steam, with its partial pressure equal to the mole ratio; when it cooled to around 90 degrees, it turned into saturated water vapor and began to condense. After condensation, the MOL value of water vapor decreases, and the pressure changes! But the air can also change; how does it change? And this one of mine is a condenser; the pressure at the inlet remains at 1 Bar. The outlet pressure should be equal to the inlet pressure minus the pressure loss. But the gas is indeed less, yet the space in the condenser remains just as large! Chaotic, a bit chaotic! We need to untangle it strand by strand!

This problem is like that of a condenser; non-condensable gases should be removed, so you won’t have to worry about it.

This problem is indeed a bit complex to calculate. 1. How to determine the temperature at which condensation begins and the variation in the condensation temperature? When a confined space contains air, the pressure at the time of water condensation equals the saturated vapor pressure at that temperature plus the partial pressure of air. Thus, the condensation temperature of water at that moment can be determined from the pressure during condensation; this is a self-reinforcing process. It can be calculated using cyclic interpolation methods. It can be seen that in a confined space, condensation is a process in which the condensation temperature (pressure) changes continuously. 2. The outlet pressure should be equal to the inlet pressure minus the pressure loss plus the condensation pressure drop. From this formula, it is possible to understand why the outlet of some condensation devices experiences a vacuum condition with a pressure lower than atmospheric pressure.

The first one is understandable. But the condenser should not be a closed space. Secondly, the outlet pressure = inlet pressure - pressure drop - condensation pressure drop. How is the condensation pressure drop determined? Water vapor is constantly condensing; the temperature at which it condenses corresponds to its vapor partial pressure, but the partial pressure of air is changing! Since the pressure is 1 Bar and steam is supplied continuously at the inlet, the overall gas pressure remains almost at 1 Bar. The vapor partial pressure decreases, while the air partial pressure increases. I would like to ask whether it is possible to consider having water vapor in the gas; this water vapor must be saturated vapor, and its partial pressure should be the saturation pressure at that temperature. The total heat transfer amount Q is calculated in this way.

This post was last edited by breath on 2009-8-22 17:05. 1. Ordinary heat exchangers are considered to be closed spaces, unless the volume of the container changes, or there are openings connecting it to other spaces, or other substances are introduced during the process. 2. This is the difference between quantitative static equilibrium and continuous dynamic equilibrium. The actual heat exchange process is a state of continuous dynamic equilibrium, which is a point within the variation of static equilibrium. Taking the condensation pressure drop into account is indeed quite complex; it changes continuously during the heat exchange process, and differentiation, interpolation, and integration are required to calculate it. Generally, it involves simplified rough calculations and estimates based on experience; usually, the parameters at the outlet are used for estimation. In the case of condensation, the inlet pressure cannot be simply considered constant; it is also affected by the condensation pressure drop, such as the phenomenon of condensed liquid flowing back into the steam inlet pipeline. This is mainly due to excessive cooling capacity, which results in a large sudden condensation pressure drop; as a result, shock vibrations occur during the condensation heat exchange process. Yes, the actual calculation takes into account a state of equilibrium, where the water vapor is necessarily saturated, and its partial pressure corresponds to the saturation pressure at that temperature.

Yes, in the calculations it is assumed to be a closed space in dynamic equilibrium. I used HTRI to simulate this condition, and the outlet pressure was almost equal to the inlet pressure. The final outlet temperature is 45°C, and the partial pressure of saturated steam at the outlet is 9.6 KPa. I would like to ask, what is the partial pressure of air? Is it 10 Kpa, or 100-9.6-pressure drop? This affects the component that ends up without condensed steam. I think it's the latter. But if it is the latter, the amount of condensation will be very large. (Difference from HTRI simulations)

Yes, of course, it’s the latter. The temperature has dropped to 45°C, so of course the amount of condensation is high! It is estimated that most of the water vapor has been condensed. However, it seems that HTRI’s analysis of pressure changes during the condensation phase transition has significant errors; it appears to consider only the fluid resistance, without giving adequate account to the pressure drop caused by the phase transition. Or, your model data can only calculate these. Does HTRI seem to have a dedicated module for condensation phase transition analysis?

First, you need to confirm the scope. If it is a single condenser isolated from the system, then when the temperature drops to 45 degrees, the vapor pressure becomes equal to the saturated vapor pressure at that temperature, and the pressure throughout the condenser decreases. If it is connected to the system, the pressure will not change, but the water vapor partial pressure will decrease, as the reduced pressure will be compensated for by the system.

I’d like to ask, how is the voltage division calculated when it’s connected to a system? I’m facing this same problem now: a vacuum pump is connected after the condenser, and the overall system pressure is 20 KPa. The media involved are dry air, water, and propane. The temperature at the exit of the condenser is 15 degrees; at this temperature, the saturated vapor pressure of water is 1.7 KPa, while that of propane is 19.7 KPa. So what should be the partial pressure of dry air? I’d appreciate some guidance on this