Question 7 of 8: Round duct sizing by the equal-friction method

Question 7: Round duct sizing by the equal-friction method (20 marks)

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 small low-velocity supply system of three outlets totalling 320 cfm, with the run lengths scaled from the figure and the terminal pressure requirement marked at each outlet.

Find. A round duct size for every section, the resulting pressure balance on each of the three runs, and the damper settings needed to balance the system.

Assumptions stated (the question asks for them explicitly):

• Galvanised steel round duct, absolute roughness $\varepsilon = 0.0003$ ft (ASHRAE "medium smooth"), carrying standard air at $0.075\ \text{lb}/\text{ft}^{3}$.
• The lengths marked on the figure are taken as total equivalent lengths, i.e. the equivalent lengths of the elbows, take-offs and transitions are already included. If instead fittings were added at the customary 50 % of measured length, the design friction rate would fall from 0.128 to 0.085 in wg per 100 ft and every duct would come out one nominal size larger.
• The riser at A is not dimensioned on the figure; 5 ft is allowed for it. Because that run is not the index run, the assumption affects only the damper setting at outlet 1.
• The pressure quoted at each outlet is the total pressure the diffuser and its neck require; the plenum pressure of 0.15 in wg is total pressure, so velocity-pressure regain is not credited — conservative for a system this small.
• Duct sizes are selected from the standard round series (4, 5, 6, 7, 8, 9, 10, 12 in), rounding up from the calculated diameter.

Approach. Identify the index run — the one that consumes the most pressure — and spend the available pressure on it: subtract its terminal loss, divide what remains by its total equivalent length to get a friction rate, then size every section of the system at that same rate. Finally re-total each run at the selected sizes and damper off the surplus at the two short outlets.

1. Total the flows and find the index run. The three outlets draw $100+100+120=320$ cfm. Three routes leave the plenum, and their duct lengths are 30 ft to outlet 1, $25+18+10+33=86$ ft to outlet 2, and $25+18+18=61$ ft to outlet 3. Outlet 2 is both the longest and carries the second-largest terminal loss, so it is the index run.
2. Set the design friction rate. Of the 0.15 in wg available, the index outlet's diffuser takes 0.040, leaving 0.110 in wg to be spent on 86 ft of duct: $$\text{friction rate}=\frac{0.150-0.040}{86}\times100 =\boxed{0.128\ \text{in wg per 100 ft}}$$ This is a normal low-pressure design rate; the customary band is 0.08 to 0.15 in wg per 100 ft for commercial supply ductwork, chosen to keep velocities and hence noise down.
3. Size each section at that rate. For round duct the friction loss per 100 ft is $$\frac{\Delta p}{100\ \text{ft}}=f\,\frac{100}{D}\left(\frac{V}{4005}\right)^{2}$$ with $f$ from the Colebrook equation, $1/\sqrt{f}=-2\log_{10}\!\left[\varepsilon/3.7D+2.51/(Re\sqrt{f})\right]$, $D$ in feet and $V$ in feet per minute. Solving for the diameter that gives 0.128 in wg per 100 ft at each flow, and rounding up to the standard series:
4. Tabulate the selection. The required and selected diameters, with the velocity and the actual loss at the selected size:

   Duct selection at 0.128 in wg per 100 ft

   Section	cfm	Calculated $D$, in	Selected $D$, in	Velocity, fpm	Loss per 100 ft	Section loss, in wg
   1 plenum–A	320	8.46	9	724	0.0945	0.0236
   2 A–B	220	7.36	8	630	0.0848	0.0153
   3 B–C	100	5.48	6	509	0.0825	0.0083
   4 C–outlet 2	100	5.48	6	509	0.0825	0.0272
   5 B–outlet 3	120	5.87	6	611	0.1147	0.0207
   6 riser at A	100	5.48	6	509	0.0825	0.0041

   Every velocity is between 500 and 730 fpm, well inside the 600–1000 fpm band that keeps a small office system quiet, so no size needs to be revisited for noise.
5. Total each run and check the index. Adding the section losses along each route and adding the terminal loss: $$\text{outlet 2 (index)}:\ 0.0236+0.0153+0.0083+0.0272+0.040=\boxed{0.114\ \text{in wg}}$$ $$\text{outlet 3}:\ 0.0236+0.0153+0.0207+0.035=0.095\ \text{in wg}$$ $$\text{outlet 1}:\ 0.0236+0.0041+0.050=0.078\ \text{in wg}$$ The index run needs 0.114 in wg against the 0.15 available, so the system works, with 0.036 in wg in hand — the margin arises because every duct was rounded up to the next standard size, which is exactly why one rounds that way.
6. Balance the system. The two short runs must be throttled to the same total, or all the air will short-circuit to them: $$\text{damper at outlet 1}:\ 0.150-0.078=0.072\ \text{in wg}$$ $$\text{damper at outlet 3}:\ 0.150-0.095=0.055\ \text{in wg}$$ Volume dampers are therefore required at both take-offs, and are needed even in a system as small as this because the pressure to be absorbed at outlet 1 is nearly half the total available. Proportional balancing against measured flows is the commissioning step that finally sets them.
7. Sanity-check the result. A useful cross-check on any equal-friction design is that the main should be roughly one nominal size above the sum of the branches it serves: 9 in for 320 cfm, 8 in for 220 cfm and 6 in for 100–120 cfm follow the expected progression, and the fan needs only $0.15\ \text{in wg}\times320\ \text{cfm}/(6356\times0.55)\approx0.014$ hp of air power — a fractional-horsepower fan, consistent with a small terminal system.