Calculation of moisture precipitation in air-cooled towers
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The air processing volume is 20,000 m3/h. The pressure of the air after compression leaving the air compressor is 0.59 MPaA, and the temperature is 50°C. After passing through the air cooling tower, the exit temperature is 30°C. What is the amount of moisture released by the air cooling tower? Please provide the calculation process
It would be great to have some ideas; thank you
The approach is as follows: 1. First, calculate the water content in the atmospheric air at the compressor inlet. Note that this refers to the atmospheric air at the intake point. For example, at the compressor’s intake, the atmospheric pressure P0 is 100 kPa, the temperature is 20°C, and the relative humidity is 75%. The saturated vapor pressure of water at 20°C is Ps = 2.3392 kPa. With a relative humidity of φ = 75%, the water content in the air is given by v% = Ps × φ / P0 = 2.3392 × 75% / 100 = 1.754%. Therefore, the water content at the compressor inlet is V = 20,000 × 1.754% = 350.88 Nm3/h, which is equivalent to 282 kg/h. The value of 20,000 m3/h is based on standard conditions; it doesn’t matter, and you can use your actual values for the calculation. 2. Calculate the water content in the high-pressure air at the compressor outlet after it has been cooled by the cooling tower; generally, this air is saturated, that is, the relative humidity is 100%. The calculation method remains the same as before. For example, if the temperature here is 30°C, then at 30°C the saturated vapor pressure of water is Ps=4.2467 kPa, and the relative humidity is φ=100%. In this case, the proportion of water in the compressed air is given by v%=Ps·φ/P=4.2467×100%/590=0.72%. . . . . 3. The amount of water to be removed corresponds to the amount of water that precipitates out. It is best to perform the calculations on a dry basis, that is, using absolute dry air without water, as the mass of this absolute dry air remains constant. Calculating on a wet basis can be somewhat complicated, but the principle is the same in either case.
Both states have been determined: State 1 (P1=0.59MPaA, t1=50deg.C, saturated)； State 2 (P1=0.59MPaA, t1=30deg.C, saturated); refer to the air property table ； Subtracting the two states gives the amount of precipitated water.
Agree with the algorithm mentioned above. I have a question, OP: why is the outlet temperature of your air compressor so high? Is the aftercooler clogged? And why does it only reach 30 degrees after cooling? Not taking the usual path.
"So the water content at the inlet of the air compressor is V=20000x1.754%=350.88 Nm3/h=282 kg/h. How is 282 kg/h calculated from 350.88 Nm3/h in this case? I would appreciate some advice from those who are more experienced. . . . . . . . . .