HVAC · Thermodynamics · Problem 12PDFSolution in PDF ↓
HVAC · Thermodynamics · Problem 12
Problem & Solution
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Problem: Thermo-12, a 1200 cubic foot room, is filled with atmospheric air at 95 degrees.
Given: 14.7 pound force per inch squared times 144 inches squared per square foot times the volume which is 1200 cubic feet ...
Approach: So I'll write P with a circle to show constant pressure and I found in the reference manual this formula here for a constant sy...
Key formula: formula is going to give us the work per unit mass and then we're going to have to multiply by the mass
Calc: We know the temperature at state 1 and at state 2.
Calc: For example, suppose we were going to solve for the specific volume and find V1 and V2.
Result: It's 95 degrees plus 460 and now let's just check out the units the pound force will go away the inches squared will go away the ranking will go aw...
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Student questions asked in live office hours about this problem
OH 15: Thermo 12
Q: Where does the equation W = p∆V come from, especially since I see W = V∆p for constant volume processes?
A: The formula is in the reference handbook section 3.1.1, but rather than just finding formulas, focus on understanding why they work. If you dig deeper into the derivation and reasoning behind this equation, you'll get much more value than just plugging numbers in.
OH 18: OH18-02-THERMO-12
Q: Can you use mass = density × volume instead of PV = mRT for this problem, even though air temperature isn't standard? I got 90 pounds instead of your 85.8 pounds.
A: Using standard air density (0.075 lb/ft³ at ~70°F) when the actual temperature is higher will overstate density by about 5%, which is why you got 90 instead of 85.8. Whether this matters depends on answer choice spacing—here the choices are far enough apart that both answers lead to the correct selection, but when choices are close, you'd need to account for actual temperature-adjusted density or use PV = mRT.
OH 115: HVAC: Thermo #12
Q: If volume is constant in both states, shouldn't work be zero? Also, why did we use temperature from state 1 (95°F) to find mass instead of state 2 temperature or delta T?
A: The key subtlety is this is an open system—air mass can change even though room volume stays constant, so density and specific volume change between states. We use state 1 conditions to find the initial mass, then calculate specific work using the pressure-specific volume relationship (p·Δv, where v is specific volume), which accounts for the different conditions at each state.
OH 53: HVAC: Thermo #12
Q: In a cooling problem where air temperature changes from 95°F to 70°F, why is mass calculated only at the initial 95°F condition if mass is a function of temperature?
A: It's a closed system—a sealed room with fixed mass and no mass crossing the boundary—so the mass of air doesn't actually change during cooling. We calculate mass at the initial condition (95°F) because we have complete initial state information; the mass is constant throughout the process regardless of temperature changes.
OH 77: HVAC: Thermo Module #12
Q: I used density from tables and interpolation to find mass (85.86 lbs), but the solution used ideal gas law—is my approach wrong or an alternative method? Also, why doesn't Q = mcp·ΔT work to find work done on air during constant pressure cooling?
A: Your density method is completely valid—ideal gas law (ρ = P/RT) gives the same result as table interpolation since both depend on the same physical properties. For the second part, you can't assume work equals heat gain; use the first law of thermodynamics for closed systems: Q - W = ΔU + ΔKE + ΔPE, which accounts for internal energy change, not just sensible heat.
OH 94: HVAC: Thermo Module #12
Q: How did we get a friction factor of 0.19 for 1000 CFM and 12-inch duct diameter, when I'm calculating 0.26?
A: You need to carefully identify the correct line on the friction factor chart—the 1000 CFM line is slightly lower than it appears, and when you follow it across to intersect with the 12-inch duct diameter line, you'll get 0.19. Count from the 500 CFM line upward to confirm you're on the right horizontal line.
OH 111: HVAC: Thermo Module #12
Q: Why can't I use Q = mCP ΔT with density and volume at state 1 to solve this cooling problem?
A: Q = mCP ΔT is for heat transfer, not work. For constant pressure work in a closed system, you need to use W = PΔV. Since pressure is constant and the gas is ideal, you can manipulate the ideal gas law (PV = mRT) to express work in terms of temperature change, but the coefficient is R, not CP.