Problem: The supply fan and duct distribution for a restaurant are shown.
Given: 800 feet per minute; 500 CFM outlets requires a minimum terminal pressure of 0; 2 inches of water
Approach: The velocity is not to exceed 1,800 feet per minute.
Calc: Well we've got 5 outlets and they're all 400 CFM each.
Calc: So that's 2000 CFM total, right?
Result: So 1630 over 45 quantity squared is 0.166 inches of water and now we can substitute that in to get static pressure by itself 0.4 minus 0.166 and th...
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Student questions asked in live office hours about this problem
OH 65: HVAC: HVAC-15
Q: In HVAC-15, why calculate velocity pressure at the initial section (2000 CFM) rather than another section, and what's the difference between total pressure and external static pressure for fans?
A: Velocity pressure at the fan discharge (highest flow section) captures the kinetic energy contribution at the point where you're specifying the fan. Fan total pressure = static pressure + velocity pressure at discharge; external static pressure is only the duct losses and doesn't include that velocity head.
OH 97: HVAC: HVAC-15
Q: Is it correct to find the fan's total pressure by adding the velocity pressure at 1800 FPM to the 0.2-inch terminal pressure, getting 0.42 inches total?
A: Yes — you basically narrated the solution in your question, so those are the right steps. That said, understand why each step works rather than memorizing the sequence, so you can adapt when the problem setup changes.
OH 105: HVAC: HVAC-15
Q: For HVAC-15, I intersected 2000 CFM and 1800 FPM on the friction chart, then used V/4005² for velocity pressure and solved for static pressure of 0.24 inches WG — is this accurate?
A: You got away with it this time, but the approach isn't quite optimal — this problem requires reading the provided friction chart at the correct operating point, not an arbitrarily chosen point. Make sure you're reading the chart at the exact conditions for the duct section you're analyzing.