Problem 2 An evaporating cooling tower I used to cool water

Problem 2: An evaporating cooling tower I used to cool water from 110 to 80 degrees Fahrenheit. Water enters the tower at a rate of 300,000 lbm/hr. Dry air (no water vapor) flows into the tower at a rate 150,000 lbm/hr. If the rate of wet air flow out of the tower is 160,000 lbm/hr, determine the rate of water evaporation in lbm/hr and the rate of cooled water flow in lbm/hr.

Wet ait 160,000 1pm Warm water () Cz) Dry air ) water ni - 150 000 hr

Solution

In this problem we have to consider not only the continuity equation for conservation of total mass, but also the coservation of each species (air and water) in the absence of any chemical raction. There are only two locations with air flow so we know that the mass flow rate of dry air leaving the tower at section 2 must be the same as the mass flow entering at section 3: 150,000 lbm/hr. Since the total flow rate of dry air plus water at section 2 is 160,000 lbm/hr,

the difference between this total flow and the flow of dry air is

160,000 lbm/hr – 150,000 lbm/hr = 10,000 lbm/hr must be the flow of water that evaporated and leaves the tower.

The flow of cooled water at section 4 is simply the original flow of 300,000 lbm/hr minus the evaporated water flow rate of 10,000 lbm/hr. This gives a cooled water flow rate of 290,000 lbm/hr.