Question 6 of 8: Two-branch duct system balanced against the fan characteristic

Question 6: Two-branch duct system balanced against the fan characteristic (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. The total flow and the two outlet flows with the damper fully open, and then the total flow and the damper loss factor required to make the two outlets deliver equally.

Approach. Work in total pressure. The system is one series run (fan exit expander plus 15 m of 500 mm duct) feeding two parallel legs from B, so three conditions close the problem: the two legs must suffer the same total-pressure loss, their flows must add to the fan flow, and the fan characteristic must supply the sum of the common loss and the leg loss at that flow.

1. Set up velocity pressures. Every fitting loss is a factor times a velocity pressure, $p_v = \tfrac12\rho(Q/A)^2$, evaluated at the area the data specifies. With $\rho = 1.2$ kg/m$^3$ that is $p_v = 0.6(Q/A)^2$ Pa. The expander at the fan takes the maximum velocity, which is at the smaller 400 mm × 300 mm fan exit; the expanders at the outlets likewise take the 400 mm duct velocity rather than the larger grille velocity.
2. Account for the discharge. The fan draws from still atmosphere and both outlets discharge to still atmosphere, so over the whole path the fan total pressure is entirely consumed: $$P_t(Q_{\text{tot}})=\Delta p_{\text{common}}(Q_{\text{tot}})+\Delta p_{\text{leg}}(Q_{\text{leg}})$$ where each leg loss includes the velocity pressure thrown away at its grille, $p_v$ at the grille area, in addition to the grille's own 0.4 factor. Omitting that exit term is the single most common way to get this problem wrong.
3. Write the duct friction. The friction chart for galvanised sheet steel ($\varepsilon = 0.15$ mm) is reproduced analytically by the Colebrook–White relation $$\frac{1}{\sqrt f}=-2\log_{10}\!\left(\frac{\varepsilon}{3.7D}+\frac{2.51}{Re\sqrt f}\right), \qquad \Delta p_f = f\,\frac{L}{D}\,\frac{\rho v^2}{2}$$ which for these ducts agrees with the published chart to within its reading accuracy.
4. Assemble the two legs. Leg D leaves the tee as an off-take, so it takes the 0.5 factor on its own velocity pressure, then 15 m of friction, then the damper, the expander, the grille and the exit. Leg C continues as the main, so it takes only 0.2 on the downstream velocity pressure, but pays 45 m of friction and a bend: $$\Delta p_D = (0.5+\zeta_{\text{damper}}+0.25)p_{v,400}+\Delta p_f(15\ \text{m})+(0.4+1.0)p_{v,\text{gr}}$$ $$\Delta p_C = (0.2+0.3+0.25)p_{v,400}+\Delta p_f(45\ \text{m})+(0.4+1.0)p_{v,\text{gr}}$$ Leg D is fitting-dominated and short; leg C is friction-dominated and three times as long.
5. Part (a) — solve with the damper fully open. Setting $\zeta_{\text{damper}} = 0.2$ and iterating on $Q_{\text{tot}}$ until $\Delta p_D(Q_D) = \Delta p_C(Q_C)$ with $Q_D+Q_C = Q_{\text{tot}}$ and the fan supplying the total, the balance converges on $$\boxed{\;Q_{\text{tot}} = 1.873\ \text{m}^3\text{s}^{-1},\quad Q_D = 1.034,\quad Q_C = 0.839\ \text{m}^3\text{s}^{-1}\;}$$ at a fan total pressure of $P_t = 200-12(1.873)^2 = 157.9$ Pa. The common run absorbs 64.2 Pa and each leg the remaining 93.7 Pa. The shorter leg takes 23% more air, as it must. Velocities are 15.6 m/s at the fan exit, 9.5 m/s in the 500 mm main and 8.2 and 6.7 m/s in the two branches — all within normal low-velocity practice.
6. Check the pressure budget term by term. Both legs must total 93.7 Pa, and they do so by quite different routes:

   Loss	Leg D ($Q = 1.034$), Pa	Leg C ($Q = 0.839$), Pa
   Branch at B	20.31 (0.5 on off-take)	5.35 (0.2 on main)
   Duct friction	27.40 (15 m)	55.37 (45 m)
   Bend	—	8.03
   Damper, fully open	8.12	—
   Expander	10.15	6.69
   Grille	7.92	5.22
   Exit velocity pressure	19.79	13.04
   Leg total	93.69	93.70
   Common run (expander 36.54 + friction 27.67)	64.21 Pa at $Q_{\text{tot}} = 1.873$
   Fan total pressure	157.90 Pa = 64.21 + 93.69

7. Part (b) — throttle the damper for equal outlets. The damper sits in leg D, which is the leg that naturally takes more air, so closing it is the right adjustment. Impose $Q_D = Q_C = Q_{\text{tot}}/2$. Leg C now has no adjustable element, so it alone fixes the operating point: solving $\Delta p_C(Q_{\text{tot}}/2) = P_t(Q_{\text{tot}})-\Delta p_{\text{common}}(Q_{\text{tot}})$ gives $$\boxed{\;Q_{\text{tot}} = 1.775\ \text{m}^3\text{s}^{-1},\quad Q_C = Q_D = 0.887\ \text{m}^3\text{s}^{-1}\;}$$ at $P_t = 200-12(1.775)^2 = 162.2$ Pa, with 57.8 Pa in the common run and 104.4 Pa available to each leg.
8. Find the damper setting required. At 0.887 m$^3$/s leg D loses only 63.4 Pa with the damper wide open, so the damper must absorb the shortfall, $104.41-63.38 = 41.03$ Pa. Dividing by the branch velocity pressure at that flow, $p_{v,400} = 0.6(0.887/0.12566)^2 = 29.93$ Pa, $$\boxed{\;\zeta_{\text{damper}} = \frac{41.03}{29.93}=1.37\;}$$ The damper therefore moves from $\zeta = 0.2$ to $\zeta = 1.37$, roughly a $30\,{}^{\circ}$ blade closure on a single-blade damper. Balancing costs 5.2% of the total flow and pushes the fan 4.3 Pa further up its curve — both in the expected direction, which is a useful check that the solution is not spurious.