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22-Mec-B2 Environmental Control in Buildings · December 2014

Question 3 of 8: Equal-friction sizing of a perimeter duct branch

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Professional Engineers of Ontario / Engineers Canada annual examination, 07-Mec-B2 Environmental Control in Buildings, December 2014. Three hours, open book, any non-communicating calculator. Eight problems of 20 points each; candidates are required to solve five. ASHRAE psychrometric charts (SI and IP) and an HFC-134a pressure-enthalpy diagram are attached to the paper. All eight problems are solved here.

Reference texts.

Reading the numbers. Chart properties here are computed from the ASHRAE psychrometric formulations rather than scaled off the printed chart, so a candidate working graphically should expect agreement to about the width of a pencil line — roughly ±0.3 °F on a dew point and ±1 % on a humidity ratio.

Question 3: Equal-friction sizing of a perimeter duct branch (20 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A below-floor perimeter branch fed from a plenum, with three outlets whose flows, run lengths and diffuser-boot losses are printed on the layout.

Given data — Problem 3 (total pressure available at the plenum, 0.13 in. wg)
SectionLength (ft)Air quantity (cfm)Boot loss (in. wg)
S1 — plenum to the first take-off25420—
B1 — branch to the first outlet151200.050
S2 — first to second take-off10300—
B2 — branch to the second outlet151600.036
S3 — second take-off to the last outlet (8 + 12 + 15)351400.040

Find. A round-duct size for every section by the equal-friction method, and the actual total-pressure loss of each of the three runs.

[Figure not reproduced: Problem 3 — the branch as printed on the paper. Air leaves the plenum and travels left to right; the index run is the longest path, 25 + 10 + 8 + 12 + 15 = 70 ft, ending at the 140 cfm outlet. See the official exam paper.]

Approach. Identify the index (longest) run, spend the available pressure less that run's boot loss uniformly along it to set a design friction rate, size every section at that rate, round to commercial diameters, and then recompute the real loss of each run at the sizes actually selected.

  1. Total the air quantity and identify the index run. The branch carries $120+160+140 = 420\ \text{cfm}$. Three runs leave the plenum; the longest is the one serving the 140 cfm outlet, $L = 25+10+8+12+15 = 70\ \text{ft}$. Equal friction sizes the whole system from this run because it is the one with the least pressure to spare per foot.
  2. Set the design friction rate. Of the 0.130 in. wg available, 0.040 in. wg is consumed by the index run's diffuser boot, leaving 0.090 in. wg for 70 ft of duct: $$\left(\frac{\Delta p}{100\ \text{ft}}\right)_{\mathrm{design}} = \frac{0.130-0.040}{70}\times 100 = \boxed{0.129\ \text{in. wg per 100 ft}}$$
  3. Size every section at that rate. Using the ASHRAE friction correlation for galvanised round duct, $$\frac{\Delta p}{100\ \text{ft}} = \frac{0.109136\ Q^{1.9}}{D^{5.02}}$$ with $Q$ in cfm and $D$ in inches, solving for $D$ at 0.1286 in. wg/100 ft gives the calculated diameters listed in the table below. Reading the same values off the ASHRAE friction chart at the intersection of each flow with the 0.13 in. wg/100 ft line gives the same sizes to within about 0.1 in.
  4. Round to commercial sizes. Round duct is made in whole-inch diameters, so S1 (9.52 in.) becomes 10 in., S2 (8.38 in.) becomes 8 in., B1 (5.93 in.) becomes 6 in. and B2 (6.61 in.) becomes 7 in. Section S3 calculates at 6.28 in. and is rounded up to 7 in. rather than down, because it lies on the index run: at 6 in. its loss would be 0.0567 in. wg and the run total would reach 0.138 in. wg, more than the 0.130 in. wg available. This is the one judgement call in the problem.
  5. Recompute the actual loss of each section. Substituting the selected diameters back into the friction equation gives the rates and losses tabulated below; the velocities are all between 520 and 860 fpm, comfortably inside the 900 fpm normally accepted for a below-floor perimeter system on noise grounds.
  6. Total each run and find the balancing requirement. Summing duct losses along each path and adding the boot, $$\Delta p_1 = 0.093,\quad \Delta p_2 = 0.092,\quad \boxed{\Delta p_3 = 0.108\ \text{in. wg}}$$ Every run is within the 0.130 in. wg available, so the branch will deliver its design air. The two short runs need dampering — 0.037 in. wg at outlet 1 and 0.038 in. wg at outlet 2 — and even the index run finishes with 0.022 in. wg in hand.
Section sizing at a design rate of 0.1286 in. wg per 100 ft
SectioncfmL (ft)$D$ calc. (in.)$D$ selected (in.)Rate (in. wg/100 ft)$\Delta p$ (in. wg)Velocity (fpm)
S1420259.52100.10050.0251770
S2300108.3880.16250.0163859
S3140356.2870.07470.0261524
B1120155.9360.12080.0181611
B2160156.6170.09630.0144599

The result is a system that balances almost by itself, which is what the equal- friction method is meant to achieve on a short branch like this one. The spread between the runs is only 0.016 in. wg, so modest damper settings at the two nearer outlets will hold the design split; if the spread had been large, the proper response would be to resize the short branches down a diameter rather than to throttle them, since a damper generating that much pressure drop is also generating noise directly above the occupied space. The 0.022 in. wg of surplus on the index run is the system's tolerance for fitting losses that the problem does not tabulate — elbows, the take-off tees and the boot entries — and on a real drawing those would be added as equivalent lengths before the sizes were released.

Problem 3 — actual total-pressure loss by run (0.130 in. wg available)
RunSectionsDuct loss (in. wg)Boot (in. wg)Total (in. wg)Damper needed (in. wg)
1 — 120 cfm outletS1 + B10.0430.0500.0930.037
2 — 160 cfm outletS1 + S2 + B20.0560.0360.0920.038
3 — 140 cfm outlet (index)S1 + S2 + S30.0680.0400.1080.022