HVAC · Thermodynamics · Problem 6PDFSolution in PDF ↓
HVAC · Thermodynamics · Problem 6
Problem & Solution
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Problem: Thermo 6 saturated liquid water at 50 psi a is cooled to 80 degrees Fahrenheit at constant pressure.
Given: 50 psi a is cooled to 80 degrees Fahrenheit at constant pressure; 50 PSIA and we know that it's a saturated liquid, w...
Approach: What is the change in enthalpy during cooling?
Calc: So let's draw that on the saturation curve vapor dome just to remind ourselves where this is where at some point here has a sta...
Calc: And we're cooling from 1 to 2.
Result: So we get 201 BTO per pound mass, which is close enough to answer choice B.
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Student questions asked in live office hours about this problem
OH 115: HVAC: Thermo #6
Q: If the problem had a saturated water mixture (quality between 0 and 1) instead of saturated liquid, could we still use pressure alone to find temperature from the steam tables?
A: Yes, absolutely. Inside the vapor dome, any given pressure corresponds to exactly one temperature—this is the defining feature of saturated steam tables. Whether quality is zero, one, or anywhere in between, that pressure-temperature relationship holds because all those states lie on the same isobar (constant pressure line). That's why the tables are organized by both pressure and temperature for your convenience.
OH 39: HVAC: Thermo-6
Q: Can you find the K value using table extrapolation instead of the analytical method for a problem involving CP and CV at 1,020°F?
A: Yes, that's a neat approach! The key mistake is converting 1,020°F to absolute temperature by adding 460 to get 1,480 Rankine, then using the table (even extrapolating to 1,500R gives K ≈ 1.35, which matches the analytical answer). Know both methods, but always verify you're using the correct temperature units—Fahrenheit vs. Rankine—for your table lookup.
OH 51: HVAC: Thermo-6
Q: Can you just look up the enthalpy on the steam table and subtract to get the change in enthalpy for water cooled from saturated liquid at 50 PSIA to 80°F at constant pressure?
A: Yes, you'll be very close doing it that way, and I'm fine with it as a practical approach, but I want to use this to show you why ΔH = Cp·ΔT isn't always valid. State 2 at 80°F and 50 PSIA is actually in the subcooled region, not on the saturation curve, so you can't look it up directly in the steam table—you'd need to account for the pressure difference, which shows where the approximation breaks down.
OH 71: HVAC: Thermo Module #6
Q: For the water compression problem in thermo #6, I used the table values for saturated liquid (HF at 50 psi and 80°F) and got 202 by taking the difference—is this method correct, or is the solution's answer of 201 from a different approach?
A: Both methods work here. The solution likely used the compressed water table assumption (enthalpy unchanged from saturated state), which introduces minor error, while your CP·ΔT approach is slightly more precise. Know both methods—the difference is tight enough that you'll get away with either one, and this flexibility will help you on the exam.