HVAC · Psychrometrics · Problem 5PDFSolution in PDF ↓
HVAC · Psychrometrics · Problem 5
Problem & Solution
Video Synthesis
Problem: Psychometrics Problem 5 A cooling tower cools 150GPM of condenser water, maintaining a 15 degree delta T.
Given: 150GPM of condenser water, maintaining a 15 degree delta T; 500 GPM delta T and we know all those things, right? We k...
Approach: On a design day, air enters the cooling tower at 90 degrees dry bulb and 75 degrees wet bulb and leaves at 105 degrees dry bulb...
Calc: Here's our basin, we got some water and then we've got the tower and we've got air entering on the sides at condition 1 and the...
Calc: Then the water, we have some condenser water return and some condenser water supply and there's a delta T between those 15 degr...
Result: So that's our numerator in B2's power dividing by 4.5 and then dividing by H2 minus H1 which is 69.4 minus 38.4 which works out to 31 and then if y...
Office Hours
1
Student questions asked in live office hours about this problem
OH 98: HVAC: Psychrometrics Module #5
Q: Why do we use 4.5 CFM Delta H (total heat) to equate with 500 GPM Delta T for water heat gain in a cooling tower, rather than 1.1 CFM Delta T?
A: The 500 GPM Delta T represents heat released by water, which transfers to air through evaporation—a phase change process. We use 4.5 CFM Delta H because it captures total enthalpy change (both latent and sensible) in the air; we can't use 1.1 CFM Delta T because sensible heat alone doesn't account for the evaporative cooling mechanism that actually pulls energy from the water. The key is that evaporation may increase moisture without raising air temperature, so you must track total enthalpy change, not just temperature change.