HVAC · Heat-Transfer · Problem 19PDFSolution in PDF ↓
HVAC · Heat-Transfer · Problem 19
Problem & Solution
Video Synthesis
Problem: In 18 inch by 24 inch, uninsulated duct runs through 100 foot auditory.
Given: 19. In 18 inch by 24 inch, uninsulated duct runs through 100 foot auditory; 800 feet per minute; 1.4 BTU per hour foo...
Approach: First let's draw the cross section.
Key formula: formula for turbulent flow and circular tubes
Calc: The outside film coefficient for the duct is 1.4 BTU per hour foot squared degrees F.
Calc: Not to do anything too interesting about an 18 by 24 inch cross section, but just to get our head wrapped around this problem.
Result: And I think if you made a slightly different assumption, you would still come out with answer choice C.
Office Hours
10
Student questions asked in live office hours about this problem
OH 18: HVAC: Heat Transfer-19
Q: For Heat Transfer 19, why can't I use the uninsulated duct heat gain/loss formula from section 9.3.6.7 of the reference handbook?
A: That formula requires Q (heat flux per unit area) as an input — and that's exactly what you're trying to find — so it creates a circular dependency. You need to build up the overall heat transfer coefficient from first principles and solve for Q that way.
OH 19: HVAC: Heat Transfer-19
Q: For Heat Transfer 19, how do you know when to use equation 5.3.7 to find the Reynolds number?
A: Before that specific answer: you need the Reynolds number only when the problem doesn't give you the film coefficient (h) and you have to calculate it through the Nusselt number correlation. When h is provided directly, skip the Reynolds number entirely.
OH 65: HVAC: Heat Transfer-19
Q: For Heat Transfer 19, my first approach was to use LMTD since the air temperature changes along the duct — but you used an average bulk temperature instead; are you assuming some averaged heat loss?
A: Yes, using an average bulk temperature is an approximation that acknowledges the temperature changes along the duct length — LMTD is more rigorous, but the average temperature approach is an acceptable simplification for PE exam-level problems. Both approaches should land you on the same answer choice.
OH 70: HVAC: Heat Transfer-19
Q: For Heat Transfer 19, since so many simplifying assumptions are being made, wouldn't it be reasonable to just use the average of 75°F and 55°F as the film temperature and use only the interior film coefficient?
A: This is one of the hardest problems in the bundle — top two or three — and all the complexity is there to expose you to forced convection in a duct with both inside and outside thermal resistances. Understand why the full approach is set up the way it is before taking shortcuts, otherwise the shortcut may fail you on a slightly different problem.
OH 95: HVAC: Heat Transfer-19
Q: I understand the problem setup but I'm confused about why the film temperature isn't used in Heat Transfer 19 — can you clarify?
A: Heat Transfer 19 and 22 are the two hardest problems in this module, and the film temperature question is a common one. In this problem I don't give you the duct surface temperature directly, so you're working from inlet temperature and other given data rather than applying a film temperature lookup.
OH 97: HVAC: Heat Transfer-19
Q: Is this a forced convection problem because there's a fan implied in the duct — and I got 65°F exit temperature using a differential film temperature approach instead of 60°F; is that a coincidence?
A: Yes, the presence of airflow in the duct means forced convection — that's the correct reasoning. The 65°F vs 60°F discrepancy from your film temperature approach is a close approximation but not exact; 60°F is the correct answer, and the closeness isn't a coincidence — your approach is in the right direction.
OH 105: HVAC: Heat Transfer-19
Q: For Heat Transfer 19, the turbulent flow equation says it applies for 'uniform surface temperature or uniform heat flux' — but the surface temperature varies along the duct, so is using an average temperature valid?
A: Yes — using the average or mean temperature along the duct length is a practical simplification that works well for PE exam problems. The handbook equation assumptions are idealized, but applying them with a representative average temperature gives a valid result here.
OH 106: HVAC: Heat Transfer-19
Q: For Heat Transfer 19, I tried equations 9.3.6.2 and 9.3.6.3 and both give answer B or C — how do I choose the correct hydraulic diameter equation for a rectangular duct?
A: The choice depends on the application: for HVAC duct friction/pressure loss sizing, use the equivalent diameter formula that matches friction to a round duct. For heat transfer in a rectangular duct, use the hydraulic diameter D_h = 4A/P, which characterizes the cross-section geometry.
OH 112: HVAC: Heat Transfer-19
OH 119 · April 28, 2026
Q: Why is K = 0.0145 used in this convection problem instead of K = 1.4, which is what the student expected for air?
A: The variable K in the Nusselt number equation (Nu = hD/k) is thermal conductivity, not the ratio of specific heats — two completely different quantities that happen to share the same symbol. Thermal conductivity for air is about 100× smaller than the specific heat ratio and is obtained from the Properties of Air table, not from the thermodynamics section. Slowing down to confirm what each variable represents is critical when the same letter is reused across topics.
Q: In Heat Transfer 19 you used the cross-flow cylinder equation (5.3.5) rather than the horizontal cylinder in stationary fluid equation (5.4.2) — what's the indicator, and how does problem 21 differ?
A: The key indicator is whether a fluid velocity is specified — if the problem gives you FPM or a flow velocity, it's forced convection and you use the cross-flow correlation. In problem 21, no velocity is given, which tells you the air is stationary (natural convection), so you use the large stationary fluid equation.