Problem: Fluids number 6, a residential gutter with a rectangular cross section transports 2GPM during peak rainfall.
Given: 2GPM during peak rainfall; 6 inches wide and is pitched such that the height of the flowing water is 2 inches; 6 inch...
Approach: So some of the things that we already know about how to find the friction factor are still applicable, but one of the typical f...
Calc: It's this 6 inch wide channel.
Calc: We have this gutter which is an open channel and it's 6 inches wide and at peak rainfall, it is quite a bit of water in there.
Result: And that equals 0.036, which is closest to 0.04.
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Student questions asked in live office hours about this problem
OH 95: HVAC: Fluids Module #6
Q: Why didn't we use the hydraulic diameter to calculate the cross-sectional area (π D_H²/4) instead of using the actual channel dimensions (height × width)?
A: Because we can calculate the actual cross-sectional area exactly from the known dimensions (height × width), we should use that with Q = V × A to find velocity precisely. The hydraulic diameter is only introduced as a substitute when we lack a direct way to define a diameter—like in Reynolds number for non-circular channels—but we never replace a calculable actual area with an approximate one.
OH 103: HVAC: Fluids Module #6
Q: Why does HVAC6 ignore the heat load from occupants when calculating total cooling load, even though people represent a significant load?
A: The occupant count was given to determine CFM (ventilation requirements), not to calculate internal heat load directly. Since both indoor and outdoor conditions were fully defined, you can use the shortcut rule: Q = 4.5 × CFM × ΔH, which requires only the enthalpy difference between outside and room air—no need to build up loads from individual sources like people, lights, and equipment.