Question 8 of 8: Duct sizing for equal branch friction, and the effect of hot low-pressure air

Question 8: Duct sizing for equal branch friction, and the effect of hot low-pressure air (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.

Find. A round-duct size for each branch that produces the same 0.15 in. wg of friction without exceeding 2 000 fpm; the total friction loss of the same volumetric flow of hot, low-pressure air through the same ducts; and the fan speed needed to deliver it.

Approach. Convert each branch's 0.15 in. wg total loss into a friction rate per 100 ft, solve the Darcy–Weisbach equation with Colebrook–White friction for the diameter at that rate, then check each result against the velocity limit and enlarge where it governs. For part (b) recompute the same ducts at the hot-air density and viscosity; for part (c) apply the fan laws.

1. Part (a) — turn the equal-total-loss requirement into a friction rate for each branch. Equal total loss over unequal lengths means unequal loss per unit length: $$\text{A:}\ \frac{0.15}{100}\times 100 = 0.150,\qquad \text{B:}\ \frac{0.15}{150}\times 100 = 0.100,\qquad \text{C:}\ \frac{0.15}{35}\times 100 = 0.429\ \text{in.\ wg per 100 ft}$$ This is a balanced-pressure-loss design rather than the constant-friction-rate method: because the branches all discharge from one junction, making their losses equal is what makes the system self-balancing without dampers.
2. Solve for each diameter from Darcy–Weisbach. With $$\Delta p_f = f\,\frac{L}{D}\,\frac{\rho V^2}{2}, \qquad \frac{1}{\sqrt f} = -2\log_{10}\!\left(\frac{\varepsilon}{3.7D} + \frac{2.51}{Re\sqrt f}\right)$$ at ρ = 0.0749 lb/ft³ and the ASHRAE friction-chart roughness for galvanised steel ε = 0.0003 ft (0.09 mm), the required diameters are $$D_A = 24.7\ \text{in} \;(V = 1\,799\ \text{fpm}), \qquad D_B = 23.0\ \text{in} \;(V = 1\,384\ \text{fpm}), \qquad D_C = 13.2\ \text{in} \;(V = 2\,100\ \text{fpm})$$ Branch A and branch B satisfy the velocity limit, but branch C does not — and that is the reason the limit is stated in the question.
3. Enlarge branch C until the velocity limit is met. The smallest duct that keeps 2 000 cfm at or below 2 000 fpm needs A = 1.00 ft², so $$D_C = \sqrt{\frac{4 \times 1.00}{\pi}} \times 12 = \boxed{13.5\ \text{in}} \qquad (V = 2\,000\ \text{fpm exactly})$$ At that size branch C loses only 0.133 in. wg rather than 0.150, so it is now the least resistant path and will take more than its share of air unless a balancing damper adds back the missing 0.017 in. wg. Reporting that damper requirement is part of the answer: the velocity constraint has broken the self-balancing property the equal-friction design was chosen for.
4. Give the practical selections. Ducts are made in standard sizes, so the design sizes round up to the next available diameter, and the actual losses follow:

   Branch	cfm	Length	Required D	Selected	Actual V	Actual loss
   A	6 000	100 ft	24.7 in	26 in ø	1 627 fpm	0.117 in. wg
   B	4 000	150 ft	23.0 in	24 in ø	1 273 fpm	0.122 in. wg
   C	2 000	35 ft	13.5 in	14 in ø	1 871 fpm	0.113 in. wg

   Rectangular equivalents follow from D_e = 1.30(ab)^0.625/(a+b)^0.25 if the installation needs flat ducts. The selected sizes happen to land within 8 % of one another in loss, which is as close to balanced as standard sizes allow.
5. Add the main run to get the fan's duty. The 40 in × 24 in main has a face area of 6.667 ft², so V = 12 000/6.667 = 1 800 fpm (within the limit), and its circular equivalent is $$D_e = 1.30\,\frac{(40 \times 24)^{0.625}}{(40 + 24)^{0.25}} = 33.6\ \text{in}$$ Over the 25 ft + 6 ft = 31 ft to the junction it loses 0.037 in. wg, so the index (longest-resistance) run totals $$\Delta p_{total,(a)} = 0.037 + 0.150 = \boxed{0.188\ \text{in.\ wg}}$$ This is friction only: no loss coefficients are given for the fan-outlet transition, the tee or the grilles, and a real selection would add 0.2 to 0.4 in. wg of fitting and terminal losses.
6. Part (b) — fix the hot-air properties. At 150 °F and 14.0 psia, $$\rho_h = \frac{14.0 \times 144}{53.35 \times 609.67} = 0.0620\ \text{lb/ft}^3 \quad\left(\frac{\rho_h}{\rho_s} = 0.828\right), \qquad \frac{\mu_h}{\mu_s} = 1.113$$ Because the volumetric flow is unchanged, every velocity in the system is unchanged; only the density and viscosity move.
7. Recompute the losses. Friction loss is proportional to density at a given velocity, but the Reynolds number falls by the ratio (ρ_h/ρ_s)(μ_s/μ_h) = 0.744, so the friction factor rises a little and the loss does not fall quite in proportion to density: $$\Delta p_{total,(b)} = 0.032\ (\text{main}) + 0.129\ (\text{branch B, now the index run}) = \boxed{0.161\ \text{in.\ wg}}$$ That is 0.861 times the design value, against the 0.828 that a pure density scaling would predict. The 4.0 % difference is the viscosity effect, and it is smaller than the accuracy with which the friction chart can be read — so the practical answer is 0.188 × 0.828 = 0.155 in. wg, and the exact figure is 0.161.
8. Part (c) — the fan speed. The fan laws at constant size give Ṁ ∝ N, Δp ∝ ρN² and P ∝ ρN³. The requirement is the same 12 000 cfm, and flow depends on speed alone, so $$\frac{\dot V_2}{\dot V_1} = \frac{N_2}{N_1} = 1 \quad\Longrightarrow\quad \boxed{N_2 = 100\ \%\ \text{of the part-(a) speed}}$$ and the consistency of that answer is the point of the question: at unchanged speed the fan's pressure falls to 0.188 × 0.828 = 0.155 in. wg and the system's requirement falls to almost exactly the same figure, because both scale with density. A fan is a constant-volume machine; hot air does not change what it can move, only the pressure it develops and the power it draws. Its shaft power falls by the same density ratio, to 82.8 % of the design value — which is why a fan sized on hot air can overload its motor if the system is ever started cold.
9. State the second-order correction honestly. Carrying the friction-factor change from step 7, the system needs 0.161 rather than 0.155 in. wg, a 3.7 % shortfall that would take √1.037 = 1.018, or about 102 % speed, to make up. The first-order answer of 100 % is the one the fan laws give and the one the question is testing; the 2 % is well inside the tolerance of a fan curve and of the friction chart, and is reported rather than carried.