22-Mec-B2 Environmental Control in Buildings · May 2013

Question 5 of 8: Duct design methods and static-regain sizing (20 marks)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Professional Engineers of Ontario / EGBC annual examination, 07-Mec-B2 (now 22-Mec-B2) Environmental Control in Buildings, May 2013 sitting. Three hours, open book. Eight problems of 20 points each; the candidate is instructed to solve five and to nominate which five are to be graded. Psychrometric charts and a pressure–enthalpy diagram for ammonia (R-717) are appended to the paper, and candidates are expected to bring an environmental-control text and steam tables. Instruction 1 invites the candidate to state any interpretation assumptions with the answer — that latitude is used explicitly below where the printed data are redundant.

All eight problems are worked here. Every psychrometric state has been recomputed from the ASHRAE formulation for saturation vapour pressure rather than scaled off a chart, so the numbers are tighter than a graphical solution would be; chart-quality agreement (about ±0.2 K and ±0.0002 kg/kg) is all that an examiner expects.

Reference texts for this subject.

W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann — the standard reference for this examination code; Ch. 2–3 (psychrometry), Ch. 6 (cooling loads), Ch. 10 (cooling towers), Ch. 15 (duct design).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 8 (energy estimating and degree-day methods), Ch. 12–13 (fluid flow and duct design).
ASHRAE Handbook – Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 21 (duct design), Ch. 25–27 (heat, air and moisture transfer in the envelope).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression and multistage refrigeration).
ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy, and ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality.
Canadian frame: National Building Code of Canada 2020, National Energy Code of Canada for Buildings 2020, and Environment and Climate Change Canada Canadian Climate Normals for degree-day data.

Psychrometric relations used throughout. At barometric pressure $p$, with saturation vapour pressure $p_{ws}(t)$ from the ASHRAE correlation,

$$W = 0.6220\,\frac{\phi\,p_{ws}(t)}{p - \phi\,p_{ws}(t)}, \qquad h = 1.006\,t + W\,(2501 + 1.86\,t)$$

in SI (kJ per kg of dry air), and in the inch-pound system $h = 0.240\,t + W\,(1061 + 0.444\,t)$ Btu per lb of dry air. The thermodynamic wet-bulb temperature is obtained from the adiabatic-saturation equation, which is what a chart's constant-wet-bulb lines represent.

Question 5: Duct design methods and static-regain sizing (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 straight circular main duct with four branch take-offs. The arrowhead printed at the right-hand end of the main duct fixes the direction of flow as right to left, so air enters upstream of D and the last of it leaves down the branch at A.

Given data
Quantity	Value
Branch flows at A, B, C, D	1.5, 1.5, 1.2, 1.2 m³/s (total 5.4 m³/s)
Straight lengths AB, BC, CD	5 m, 6 m, 5 m
Velocity in section AB	6 m/s
Junction loss factor in the main	$\zeta = 0.1$ on the downstream velocity pressure
Duct shape	circular throughout
Assumed air density and kinematic viscosity	$\rho = 1.2$ kg/m³, $\nu = 1.5\times10^{-5}$ m²/s
Assumed absolute roughness (galvanised steel)	$\varepsilon = 0.15$ mm

Find. (a) a comparison of the equal-friction, balanced-capacity and static-regain sizing methods, and (b) the diameters of AB, BC and CD.

The duct system with the section duties that follow from the printed flow arrow: 4.2 m³/s in CD, 3.0 m³/s in BC and 1.5 m³/s in AB.

(a) The three sizing methods.

The equal-friction method selects one pressure gradient — typically 0.8 to 1.0 Pa per metre for commercial low-velocity work — and sizes every section so that it loses that same amount per unit length. It is quick, it needs only a friction chart or duct calculator, and it produces ducts that reduce in size sensibly along a run. Its weakness is that it takes no account of how long each branch is: short branches near the fan end up with far more static pressure available than they need, long branches with too little, so the system must be balanced with dampers after installation. Throttling at dampers wastes fan power and is a classic source of regenerated noise.

The balanced-capacity method (also called balanced-pressure-loss or the “total-pressure” method) sizes each run so that the accumulated pressure loss from the fan to every terminal is the same. It is essentially the equal-friction idea applied run-by-run with a different gradient for each branch: long runs get a low gradient and generous ducts, short runs a high gradient and tight ducts. The advantage is that the system is self-balancing, needs no throttling and therefore runs at the lowest fan pressure consistent with the layout. The disadvantages are the extra design labour, a proliferation of duct sizes and fittings, and the fact that short branches may end up at velocities high enough to generate noise.

The static-regain method sizes each successive downstream section so that the static pressure recovered from the reduction in velocity exactly pays for the friction and fitting losses in that section. The static pressure at every branch take-off is then substantially the same, so branch dampers are unnecessary and the flow split is stable. It is the standard method for medium- and high-velocity systems and for long runs with many take-offs. The penalties are that the duct sizes reduce slowly, so the downstream ducts are physically large and use more sheet metal and more space than the equal-friction alternative; the calculation is iterative; and the regain actually achieved depends on the transition angle, which is why a regain factor or a junction loss coefficient (here 0.1) must be assumed.

(b) Approach. Size AB from the specified 6 m/s, then apply the static-regain condition at B and at C in turn: the velocity-pressure difference across each transition must equal the friction loss in the downstream section plus the junction loss, which gives the upstream velocity pressure directly and hence the upstream diameter.

Section duties. Working back along the flow from the inlet, the main carries 5.4 m³/s upstream of D, then $5.4 - 1.2 = 4.2$ m³/s in CD, $4.2 - 1.2 = 3.0$ m³/s in BC, and $3.0 - 1.5 = 1.5$ m³/s in AB, all of which is taken by the branch at A. The arithmetic closes exactly, which confirms the reading of the flow arrow.
Section AB from the stated velocity. $$A_{AB} = \frac{\dot{V}}{V} = \frac{1.5}{6.0} = 0.2500\ \text{m}^{2}, \qquad d_{AB} = \sqrt{\frac{4A}{\pi}} = \boxed{0.564\ \text{m}}$$ and its velocity pressure is $p_{v,AB} = \tfrac{1}{2}\rho V^{2} = 0.5(1.2)(6.0)^{2} = 21.6$ Pa.
Friction loss in AB. With $Re = Vd/\nu = 6.0(0.5642)/1.5\times10^{-5} = 2.26\times10^{5}$ and relative roughness $\varepsilon/d = 2.66\times10^{-4}$, the Colebrook equation gives $f = 0.0173$, so $$\Delta p_{f,AB} = f\,\frac{L}{d}\,p_v = 0.01726 \times \frac{5}{0.5642} \times 21.6 = 3.31\ \text{Pa}$$
Static-regain condition at junction B. Equal static pressure upstream and downstream of the take-off requires that the regain pay for the downstream losses: $$p_{v,BC} - p_{v,AB} = \zeta\,p_{v,AB} + \Delta p_{f,AB}$$ so $$p_{v,BC} = 21.6 + 0.1(21.6) + 3.31 = 27.07\ \text{Pa} \;\Rightarrow\; V_{BC} = \sqrt{\frac{2(27.07)}{1.2}} = 6.72\ \text{m/s}$$
Diameter of BC. $$A_{BC} = \frac{3.0}{6.716} = 0.4467\ \text{m}^{2}, \qquad d_{BC} = \boxed{0.754\ \text{m}}$$
Friction loss in BC and the regain condition at C. Now $Re = 6.716(0.7541)/1.5\times10^{-5} = 3.38\times10^{5}$, $\varepsilon/d = 1.99\times10^{-4}$ and $f = 0.0160$, so $$\Delta p_{f,BC} = 0.01603 \times \frac{6}{0.7541} \times 27.07 = 3.45\ \text{Pa}$$ Applying the same balance at C, $$p_{v,CD} = 27.07 + 0.1(27.07) + 3.45 = 33.22\ \text{Pa} \;\Rightarrow\; V_{CD} = 7.44\ \text{m/s}$$
Diameter of CD. $$A_{CD} = \frac{4.2}{7.441} = 0.5644\ \text{m}^{2}, \qquad d_{CD} = \boxed{0.848\ \text{m}}$$
Select manufactured sizes. Round spiral duct is made in 50 mm steps, so specify 560 mm, 750 mm and 850 mm diameters. The velocities become 6.09, 6.79 and 7.40 m/s respectively — all comfortably inside the 5 to 9 m/s band normal for commercial main ducts, and the small departures from the calculated sizes shift the regain balance by well under 1 Pa, which any terminal balancing damper absorbs.

Final results
Section	Flow (m³/s)	Velocity (m/s)	Calculated $d$ (m)	Selected duct (mm)
AB	1.5	6.00	0.564	560
BC	3.0	6.72	0.754	750
CD	4.2	7.44	0.848	850

Check: two readings of the figure are arithmetically self-consistent, and the arrowhead is what decides between them. Taken as printed (flow right to left) the sections carry 1.5, 3.0 and 4.2 m³/s as above. Had the air entered at A instead, the same nodes would carry 3.9, 2.4 and 1.2 m³/s and the identical regain procedure would return roughly 0.910 m, 0.752 m and 0.563 m — the same three diameters in the opposite order, because static regain always makes the high-flow end the large end. The method and the marks are unaffected; only the labelling changes. The roughness, density and viscosity are also assumed, as the paper gives none; using the ASHRAE chart value $\varepsilon = 0.09$ mm instead lowers each friction loss by about 5% and each diameter by under 2 mm.

Topic. Static-regain duct sizing.

Key relations. $p_v = \tfrac{1}{2}\rho V^{2}$; Darcy–Weisbach $\Delta p_f = f (L/d) p_v$ with $f$ from Colebrook; the regain condition $p_{v,1} - p_{v,2} = \zeta p_{v,2} + \Delta p_{f,2}$ between an upstream section 1 and the downstream section 2; $d = \sqrt{4\dot{V}/(\pi V)}$.

Why this works. Bernoulli in the form $p_t = p_s + p_v$ says that if total pressure falls only by the real losses, then any reduction in velocity pressure appears as a rise in static pressure. Sizing so that this rise exactly cancels the section's losses leaves the static pressure the same at every take-off, which is precisely the condition for the branches to draw their design flows without dampers. The junction factor of 0.1 acknowledges that a real tee-with-transition does not convert all of the available velocity pressure: it is the irrecoverable part.

Common pitfall. Applying the regain across the wrong pair of sections — the regain at a junction pays for the losses of the section downstream of that junction, not upstream of it. A second trap is forgetting that the velocity must fall in the direction of flow for a regain to exist at all; if a designer specifies too low a velocity in the first section, the method produces impossible (increasing) velocities and the run has to be restarted from a higher initial velocity.

Source. W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann, Ch. 15 §15.6–15.9 (duct sizing methods and worked static-regain examples); ASHRAE Handbook – Fundamentals (2021), Ch. 21 (duct design), Table 21-1 for roughness categories; McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley, Ch. 13.

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