HVAC · Heat-Transfer · Problem 7PDFSolution in PDF ↓
HVAC · Heat-Transfer · Problem 7
Problem & Solution
Video Synthesis
Problem: 48 degree chilled water flows in a 200-foot long 8-inch steel pipe with two inches of insulation through a room maintained at 73 degrees.
Given: 4 inches; 8 inches; 4 inches, and then the distance to the outside of the insulation, since the insulation has 2 inch...
Approach: Let's start with some dimensions.
Calc: We have the pipe itself, which is an 8-inch pipe.
Calc: So that means it has an inner radius, not an inner radius that would imply it's to the inside of the pipe, but to the outside o...
Result: Once you know that the name of the game is going to be to set these two quantities equal, the rest is just being careful with the units and the sub...
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Student questions asked in live office hours about this problem
OH 19: HVAC: Heat Transfer #7
Q: The problem gives a thermal resistance value in BTU·in/(hr·ft²·°F), but you refer to it as thermal conductivity—why can you treat it either way if the units are technically different?
A: The problem statement actually says 'thermal conductivity,' so the units are the guide—BTU·in/(hr·ft²·°F) are units of conductivity k (or equivalently, k expressed per inch of thickness). The solution uses it correctly by working through the units rather than labeling it 'resistance,' which would imply different units. If units work out in the end, you've used it correctly regardless of what label you attached.
OH 40: HVAC: Heat Transfer #7
Q: Why do you set convection and radiation heat gain from outside the insulation equal to conduction through the insulation—shouldn't they add up separately?
A: At steady state, all the heat flowing in from convection and radiation at the outer surface must equal all the heat flowing out through conduction to the chilled water pipe—energy can't accumulate at the insulation surface. Setting them equal is the steady-state energy balance at the outer surface of the insulation. The three modes aren't added together as separate contributions to a total; they're equated as inflow and outflow at the same boundary.
OH 53: HVAC: Heat Transfer #7
Q: Why didn't you include the thermal resistance of the steel pipe wall itself in the calculation?
A: Steel is an excellent conductor—its thermal resistance is negligibly small compared to insulation materials. For practical purposes, the pipe wall adds essentially zero resistance, so including it would complicate the calculation without meaningfully changing the answer. It's one of those engineering simplifications that's universally accepted.
OH 64: HVAC: Heat Transfer #7
Q: Why do you set convective and radiative heat transfer from outside the insulation equal to conduction through it—is the reason that the room temperature is maintained at 73°F? Also, why use 6 inches for R2 instead of converting to feet?
A: Yes—the steady room temperature means we can treat this as a steady-state problem where heat in equals heat out at every interface. For R2, as long as you're consistent with units throughout the cylinder wall formula (using the log-mean radius correctly), you can work in inches if k has the matching units; what matters is that units cancel properly. If you converted to feet and got 50.5°F, check that all your thermal conductivity units matched the radius units.
OH 72: HVAC: Heat Transfer #7
Q: The reference handbook shows T1 and T2 in the opposite order from the solution—is this because we have chilled water absorbing heat rather than a hot pipe emitting heat?
A: Exactly—for a chilled water pipe, heat flows inward (from the room into the pipe), so the temperature gradient is reversed compared to a hot pipe. Always set up delta T so it yields a positive number consistent with the actual direction of heat transfer. As a general rule for heat transfer problems, arrange your delta T so the result is positive; if it comes out negative, flip the subtraction.
OH 76: HVAC: Heat Transfer #7
OH 119 · April 28, 2026
Q: Why does the outside surface area term in the pipe insulation heat transfer formula reduce from 2πR₂L to just R₂?
A: When deriving the overall heat transfer coefficient for a cylinder, you set two expressions for Q equal: U₀·2πR₂L·ΔT on the left and 2πLK/ln(R₂/R₁) on the right. Since 2πL appears on both sides it cancels algebraically, leaving only R₂. The key is recognizing that the two expressions are being equated — once you see that, the cancellation is straightforward.
Q: The problem statement says 'thermal resistance' but the units given are BTU·in/(hr·ft²·°F)—should I use thermal conductivity instead based on the units?
A: Yes—the units are your guide, not the label. BTU·in/(hr·ft²·°F) are units of thermal conductivity (k), not resistance (R); calling it 'resistance' in the problem statement appears to be a typo or error. Use the units to confirm what physical quantity you're dealing with, and proceed with conductivity.