Perimeter duct branch sized by the equal-friction method: 22-Mec-B2 Environmental Control in Buildings, December 2013, Q6

Question 6: Perimeter duct branch sized by the equal-friction method (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.

Quantity	Symbol	Value
Total pressure available at plenum	$p_{t}$	0.15 in. wg
Diffuser A (via sections 1–2)	—	80 cfm, boot loss 0.050 in. wg, 35 + 15 = 50 ft
Diffuser B (via sections 1–3–4)	—	100 cfm, boot loss 0.040 in. wg, 35 + 15 + 35 = 85 ft
Diffuser C (via sections 1–3–5)	—	120 cfm, boot loss 0.036 in. wg, 35 + 15 + 23 = 73 ft
Total system airflow	$\dot V$	80 + 100 + 120 = 300 cfm
Duct material	—	round galvanised steel, $\varepsilon$ = 0.0003 ft

Find. A round-duct size for each of the five sections by the equal-friction method, and the actual total-pressure loss of each of the three runs with those sizes in place.

[Figure not reproduced: Figure 6.1 — Duct layout redrawn from the examination figure, with the five sections and the sizes determined below. Section 1 carries the full 300 cfm; section 3 carries 220 cfm after the 80 cfm branch is taken off. See the official exam paper.]

Approach. Equal friction means every section is sized at one common friction rate. The rate is set by the most demanding run — the one with the least pressure left over per foot after its boot loss is paid. Size each section at that rate, round up to the next commercially available diameter, then recompute the real loss of each run with the rounded sizes.

Assign a flow and a length to each section. Working back from the diffusers: $$\text{sec 1: }300\ \text{cfm},\ 35\ \text{ft};\quad \text{sec 2: }80\ \text{cfm},\ 15\ \text{ft};\quad \text{sec 3: }220\ \text{cfm},\ 15\ \text{ft}$$ $$\text{sec 4: }100\ \text{cfm},\ 20+15 = 35\ \text{ft};\quad \text{sec 5: }120\ \text{cfm},\ 8+15 = 23\ \text{ft}$$
Find the governing run and hence the design friction rate. Each run must fit its duct friction plus its boot loss inside the 0.15 in. wg available, so the friction rate it can afford is $(0.15 - \Delta p_{\text{boot}})/L$: $$\text{Run A: }\frac{0.15-0.050}{50}(100) = 0.200;\qquad \text{Run B: }\frac{0.15-0.040}{85}(100) = 0.129;\qquad \text{Run C: }\frac{0.15-0.036}{73}(100) = 0.156\ \text{in. wg/100 ft}$$ Run B — the longest — is the most demanding, so the whole system is sized at $$\boxed{\text{design friction rate} = 0.129\ \text{in. wg per 100 ft}}$$ Sizing at run A’s 0.200 would leave run B short of pressure; sizing everything at run B’s rate guarantees the index run works and leaves surplus on the short runs, which the balancing dampers absorb.
Size each section at that rate. For round galvanised duct the friction loss is $$\frac{\Delta p}{L} = f\,\frac{1}{D}\left(\frac{V}{4005}\right)^2 ,\qquad \frac{1}{\sqrt f} = -2\log_{10}\!\left(\frac{\varepsilon}{3.7D} + \frac{2.51}{Re\sqrt f}\right)$$ Solving for the diameter that gives 0.129 in. wg/100 ft at each flow, and rounding up to the next available size, gives the table that follows. This is exactly the operation the ASHRAE friction chart performs graphically: enter with cfm, follow the 0.129 friction line, read the diameter.
Check the resulting velocities. All five sections fall between 400 and 700 fpm. For a below-floor perimeter system serving occupied space that is comfortably inside the recommended range (main ducts up to about 1000 fpm, branches 600–900 fpm), so no section needs to be re-sized on noise grounds — a check the equal-friction method does not make for you.
Actual total-pressure loss of each run. With the rounded sizes in place the real losses are lower than the design allowance, because every diameter was rounded up. Summing the section losses along each path and adding the boot: $$\text{Run A} = 0.0296 + 0.0083 + 0.050 = \boxed{0.088\ \text{in. wg}}$$ $$\text{Run B} = 0.0296 + 0.0128 + 0.0291 + 0.040 = \boxed{0.112\ \text{in. wg}}$$ $$\text{Run C} = 0.0296 + 0.0128 + 0.0266 + 0.036 = \boxed{0.105\ \text{in. wg}}$$
Interpret the result. Every run needs less than the 0.15 in. wg available, so the system will deliver at least the design air with the plenum pressure given; run B, the index run, has the smallest margin (0.038 in. wg) and therefore governs. Because the three runs do not lose the same amount, they will not naturally divide the air 80/100/120 — the short run A would take more than its share. The surplus on each run must be dissipated by the balancing dampers: $$\text{damper A} = 0.062,\qquad \text{damper B} = 0.038,\qquad \text{damper C} = 0.045\ \text{in. wg}$$ That is the meaning of the qualifier “assuming that the proper amount of air is flowing” in the question.

Section	Flow (cfm)	Length (ft)	Required D (in.)	Selected D	Velocity (fpm)	Loss (in. wg)
Section 1	300	35	8.25	9 in.	679	0.0296
Section 2	80	15	5.04	6 in.	407	0.0083
Section 3	220	15	7.35	8 in.	630	0.0128
Section 4	100	35	5.48	6 in.	509	0.0291
Section 5	120	23	5.86	6 in.	611	0.0266

Check — fittings and equivalent length. The figure gives no fitting data, so elbow and take-off losses are not itemised; the elbows at the risers and boots are taken as included in the quoted boot pressure drops, and the tees are assumed loss-free on the straight-through path. In a real design each fitting would be converted to an equivalent length (typically 10–20 diameters per elbow) and added to the run length, which would roughly double the friction component and reduce the damper margins — but the index run would still be run B and the sizing procedure would be unchanged.

Final results
Quantity	Result
Design friction rate (equal-friction)	0.129 in. wg per 100 ft — set by run B
Section 1 (300 cfm, 35 ft)	9 in. diameter, 679 fpm, 0.0296 in. wg
Section 2 (80 cfm, 15 ft)	6 in. diameter, 407 fpm, 0.0083 in. wg
Section 3 (220 cfm, 15 ft)	8 in. diameter, 630 fpm, 0.0128 in. wg
Section 4 (100 cfm, 35 ft)	6 in. diameter, 509 fpm, 0.0291 in. wg
Section 5 (120 cfm, 23 ft)	6 in. diameter, 611 fpm, 0.0266 in. wg
Actual loss, run A / B / C	0.088 / 0.112 / 0.105 in. wg (all < 0.15 available)
Balancing damper drop, run A / B / C	0.062 / 0.038 / 0.045 in. wg

Concept & Theory

Topic: Low-pressure duct sizing by the equal-friction method.

Key relations: $\Delta p/L = f(1/D)(V/4005)^2$ with the Colebrook friction factor; design rate $= (p_{\text{avail}} - \Delta p_{\text{boot}})/L_{\text{index}}$; run loss $= \Sigma\,\Delta p_{\text{sections}} + \Delta p_{\text{boot}}$; damper drop $= p_{\text{avail}} - $ run loss.

Why this works: Equal friction sizes every duct so that the pressure gradient along it is the same, which makes the total loss of any run simply proportional to its length. That turns the design into a single screening question — which run has the least pressure per foot to spend — and that run becomes the index run governing the whole system. Rounding to commercial diameters always rounds the loss down, so the finished system is guaranteed to have surplus pressure everywhere except possibly on the index run, and the surplus is deliberately wasted in dampers to force the design flow split.

Common pitfall: Sizing at the rate implied by the shortest run (the largest number), which starves the index run; and forgetting that the section carrying air to two diffusers must carry the sum of their flows — section 3 here carries 220 cfm, not 100 or 120. Also, equal friction says nothing about noise: velocities must be checked separately against the recommended limits.

Source: ASHRAE Handbook — Fundamentals, Ch. 21 “Duct Design” (friction chart, equal-friction and static-regain methods, recommended velocities); McQuiston, Parker & Spitler, Ch. 12.

Question 6 of 8: Perimeter duct branch sized by the equal-friction method

Question 6: Perimeter duct branch sized by the equal-friction method (20 points)