Water Loss Calculations in a Cooling Tower

The evaporation of water requires about 1000 Btu/lb, depending on the temperature/pressure condition. From the Btu rejected by your process, it is thus easy to estimate the amount of water evaporated. (Similar in SI units.)

To keep contaminants from building up in the cooling tower water, it is necessary to treat and/or "bleed off" (i.e., drain) some water. Without other treatment, a rule of thumb is that the amount bled off should equal the amount evaporated. This is not precise, and would depend on how "hard" the water is, how dusty the atmosphere is, etc. If these conditions are mild, then less bleed-off; if severe, then more. But anyway, this is a reasonable starting point.

Dear Mr. Tornado,

I want to supplement to your answer.

There is one more - item of Water Loss in the Cooling Tower, it is known as DRIFT LOSS - where minute water particles are carried away by the out going air - which will be hot, saturated or near to saturated condition depending upon the Relative Humidity.

The drift loss will be 0.2% to 0.5% of the total quantity of water circulated. The Drift Eliminators play a vital role in controlling this DRIFT LOSS.

You have referred " Water requires about 1000 BTU/Lb. Actually it is 970 BTU/Lb. or 540 K.Cal/Kg. of water or 2257.9 KJ/Kg.of water at atmospheric pressure of 14.696 PSI or 1.03 Kg/cm^2. which is at Sea-Level. If the Altitude goes-up, the latent heat will INCREASE - as referred in the Steam Tables.

Thanks,

RAJESWARI.

Meter the tower make-up water flow.

PWSlack gave a good answer. In case of no bleeding condition and no other leakages, the simplest way to find rate of water loss is to measure rate of make-up water added.

Calculating theoretically is also very simple. RH of ambient air is the biggest factor. Less the RH, more the water loss. At cooling tower outlet air quality will be 100% RH (fully saturated in contact with water to be cooled). Water is evaporated to make air saturated which is termed as water loss. Calculate specific humidity of air at ambient temperature and RH and specific humidity of out going air at outlet temperature and 100% RH. Difference of these two figures will give water added in per kg of air. Multiplying it with air weight flow rate will give minimum water loss rate. Actual loss may be more because of free water particles travelling along with air which depends on type & quality of packing.

That doesn't sound like a "simplified way".

Measurement ≠ calculation.

Measurement will tell you how much water is actually being consumed, but it won't tell you if that is the "right" amount of water.