Question 6 of 8: Equal-friction duct sizing

Problem 6: Equal-friction duct sizing (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 five-section low-velocity system carrying 300 cfm total, with the three terminal losses printed on the figure and 0.14 in. wg of total pressure at the plenum.

Find. A round duct size for each of the five sections, and the balancing each run then needs.

Approach. Identify the run with the least pressure per foot available — the index run — set the design friction rate from it, size every section at that rate, round up to commercially available diameters, and finish by checking each path against the 0.14 in. wg budget.

1. Find the index run. Each run must fit its duct friction plus its terminal loss inside 0.14 in. wg, so its allowable friction rate is $(0.14-\Delta p_{term})/L \times 100$: $$\text{run 1 (A+B1)}:\ \frac{0.14-0.050}{32}(100)=0.281\ \text{in. wg/100 ft}$$ $$\text{run 2 (A+B+D)}:\ \frac{0.14-0.040}{51}(100)=0.196\ \text{in. wg/100 ft}$$ $$\text{run 3 (A+B+C)}:\ \frac{0.14-0.036}{43}(100)=0.242\ \text{in. wg/100 ft}$$ The smallest governs, so run 2 — the 100 cfm grille, longest at 51 ft — is the index run and the design rate is $$\boxed{\Delta p/L = 0.196\ \text{in. wg per 100 ft}}$$
2. Size at that rate. For each section the diameter is the one satisfying $$\frac{\Delta p}{L}=\frac{f}{D}\cdot\frac{\rho V^2}{2g_c},\qquad \frac{1}{\sqrt f}=-2\log_{10}\!\left(\frac{\varepsilon}{3.7D}+\frac{2.51}{Re\sqrt f}\right)$$ with standard air ($\rho = 0.075\ \text{lb/ft}^3$, $\nu = 1.57\times10^{-4}\ \text{ft}^2/\text{s}$) and galvanised roughness $\varepsilon = 0.0003$ ft — which is precisely what the ASHRAE friction chart plots. The calculated diameters are 7.57, 4.62, 6.74, 5.38 and 5.02 in.
3. Round up to available sizes. Round duct is made in whole-inch increments, and rounding up keeps every section at or below the design rate: $$A \to 8\ \text{in},\quad B_1 \to 5\ \text{in},\quad B \to 7\ \text{in},\quad C \to 6\ \text{in},\quad D \to 6\ \text{in}$$ Velocities are then 859, 587, 823, 611 and 509 fpm — all comfortably inside the 1,000–1,200 fpm ceiling that keeps a low-pressure system quiet, so no size has to be increased for noise.
4. Section losses at the selected sizes. Recomputing the friction rate at the rounded diameters: $$A:\ 0.150 \Rightarrow 0.0299,\quad B_1:\ 0.134 \Rightarrow 0.0161,\quad B:\ 0.163 \Rightarrow 0.0130$$ $$C:\ 0.115 \Rightarrow 0.0172,\quad D:\ 0.082 \Rightarrow 0.0190\ \ \text{in. wg}$$ (rate in in. wg/100 ft, then the section loss over its own length).
5. Check each path against the budget. $$\text{run 1}:\ 0.0299+0.0161+0.050=0.096\ \text{in. wg}$$ $$\text{run 2}:\ 0.0299+0.0130+0.0190+0.040=0.102\ \text{in. wg}$$ $$\text{run 3}:\ 0.0299+0.0130+0.0172+0.036=0.096\ \text{in. wg}$$ Every path fits inside 0.14 in. wg, and the index run has the largest demand as it should.
6. Balancing. Because each diameter was rounded up, all three runs finish with surplus pressure: 0.044, 0.038 and 0.044 in. wg respectively. The system is therefore self-balancing to within 0.006 in. wg — the runs are within 6 % of one another — and the practical instruction is to fit a balancing damper in each branch and set them to absorb the surplus, or, better, to reduce the plenum pressure to about 0.102 in. wg and save the fan energy. The near-equality of the three surpluses is the equal-friction method working as intended.