HVAC · Psychrometrics · Problem 3PDFSolution in PDF ↓
HVAC · Psychrometrics · Problem 3
Problem & Solution
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Problem: A 12-foot by 15-foot room with a 9-foot ceiling has a relative humidity of 25% and a dry bulb temperature of 70°.
Given: 0.00393 pounds of water per pound of dryer and that's the water that's being added to the air and now let's work towa...
Approach: A humidifier is turned on and run continuously adding 0.04 gallons of water per hour to the room.
Calc: It was down around 25% at the starting point and the goal would be to have it be more comfortable, something like 50% relative ...
Calc: Then the question becomes how long is it going to take to get that relative humidity up to 50% where we'd like it?
Result: Now I want to approach to this problem would be to figure out what the mass of water is in state one and what the mass of water is in state two and...
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Student questions asked in live office hours about this problem
OH 44: HVAC: Psychrometrics-3
Q: In practice problem 3, where does the 0.04 gallons per hour volume flow rate come from at the end to calculate time?
A: That volume flow rate was given in the problem statement itself—it's highlighted in the materials. The student who asked just overlooked it initially, but you should be able to pull that value out and use it to find the time.
OH 66: HVAC: Psychometrics Module #3
Q: Is it valid to use standard density of air and water instead of specific volume at the initial condition to convert humidity ratio into cubic feet of water per cubic feet of air?
A: Your alternate approach works and gives a close answer, but I recommend learning both methods—yours and the recorded solution—so you can solve it two different ways and cross-check your work on exam day for added confidence.
OH 74: HVAC: Psychometrics Module #3
Q: Does it matter whether I use the specific volume from the initial air state or final air state when calculating the mass of dry air in a humidifier problem?
A: For a humidifier in a fixed volume with no condensation, you'll get essentially the same answer either way because you're only adding water vapor to a fixed mass of dry air. However, best practice is to use entering conditions when a coil is involved; in this case, try both approaches and verify you get the same mass of dry air (~120.1 lbs) to build confidence in the method.