Calculation of condensation volume
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I would appreciate some guidance from experts: Semi-water gas contains many non-condensable gases. If some of these gases are condensed, but their boiling point is higher than that of the gas coming out of the condenser, what method can be used to calculate the amount of gas that gets condensed? Can Raoult’s law be applied for this purpose? :), thank you
It is complicated to carry out calculations from the liquid phase; instead, the amount of condensation should be calculated from the gas phase. The specific method is as follows: Determine the temperature at which cooling is to take place, find the saturated vapor pressures of each component at this temperature, divide the saturated vapor pressure by the total pressure to obtain the fraction of that component in the gas phase, and then multiply this fraction by the total volume to get the volume of that component； The amount of condensation of that component is obtained by subtracting the volume of that component at the cooling temperature from its volume in the original state.
The saturated vapor pressure at the condensation temperature divided by the total pressure, and then multiplied by the total pressure – isn’t the result the saturated vapor pressure?
This post was last edited by *aoye613 on 2009-6-23 21:36. Here’s an example to illustrate: Suppose the pressure of semi-water gas is 0.03 MPa and its temperature is 40 degrees, with the water in a saturated state; now it is cooled to 20 degrees. Calculate the amount of water that condenses. At 40 degrees Celsius, the saturated vapor pressure is 7.3766 kPa; therefore, the water vapor content in each m3 of semi-water gas is 7.3766/(0.03×1000) = 0.246 m3 ； At 20 degrees, the saturated vapor pressure is 2.3346 kPa. Let the amount of condensed water vapor be x; then (0.246 – x)/(1 – x) = 2.3346/(0.03 x 1000) ； The calculated value is x = 0.17 m3 ； That is, the amount of condensed water vapor is 0.17 m3.