22-Mec-B2 Environmental Control in Buildings · May 2016
Question 4 of 8: Equal-friction duct sizing and system pressure loss
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Professional
Engineers of Ontario / Engineers Canada annual examination
07-Mec-B2 Environmental Control in Buildings, May 2016,
three hours, open book. Eight problems of 20 points each;
candidates are required to solve five, and all questions carry the same
value. Psychrometric charts and an R-134a p-h diagram are appended to the
paper. All eight problems are solved here, because the set is intended
as a study resource rather than an examination script.
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. 5–6 (heating
and cooling loads), Ch. 10 (cooling towers), Ch. 15 (fans and duct
design).
McQuiston, Parker & Spitler, Heating, Ventilating and Air
Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3
(moist air), Ch. 6 (heating loads and infiltration), Ch. 8 (energy
estimating and the degree-day method), Ch. 12–13 (fluid flow, fans
and duct design).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering
Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression
refrigeration, multistage systems and heat pumps).
ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air
Quality; ANSI/ASHRAE Standard 55, Thermal Environmental
Conditions for Human Occupancy.
Canadian frame: National Building Code of Canada 2020
and its Appendix C design temperatures; National Energy Code of Canada
for Buildings 2020; Environment and Climate Change Canada heating
degree-day normals; CSA B52 Mechanical Refrigeration Code;
Canadian federal halocarbon regulations (SOR/2003-289) for the
refrigerant discussion in Problem 7.
Check: assumptions carried through this
paper. Cover-page instruction 1 invites a clear statement of any
assumption. Standard barometric pressure of 101.325 kPa is used throughout;
moist-air properties follow the ASHRAE Handbook — Fundamentals Ch. 1
formulation (Hyland–Wexler saturation pressure, so results agree with
the appended chart to chart-reading accuracy rather than being read off it);
R-134a properties are on the IIR datum and agree with the appended p-h
diagram. Problem-specific assumptions — climate data, fuel prices,
emission factors, air-change rates, occupant density, duct roughness and the
coil bypass factor — are stated where they are first used.
Question 4: Equal-friction duct sizing and system pressure loss
(20 marks)
Given. A five-outlet round supply system fed from a
rooftop unit through a 4 ft horizontal run and a 10 ft drop, with
trunk runs of 10, 10, 10, 10 and 8 ft between take-offs and diffusers of
50, 200, 120, 125 and 150 cfm; each diffuser has its own quoted total
pressure drop and a boot worth 20 ft of equivalent length. The unit
develops 0.30 in. wg external total pressure of which the return
system takes 0.12 in. wg, and velocity is capped at 850 ft/min
in the main and 650 ft/min in the branches.
Given data read from the layout
Element
Flow (cfm)
Measured length (ft)
Diffuser loss (in. wg)
Unit discharge to first take-off
645
4 + 10 + 10
—
Trunk M2 / M3 / M4 / M5
595 / 395 / 275 / 150
10 / 10 / 10 / 8
—
Branch B1 (up)
50
25
0.05
Branch B2 (down)
200
8 + 14
0.04
Branch B3 (up)
120
20
0.036
Branch B4 (down)
125
13 + 14
0.03
Branch B5 (up)
150
24
0.04
Find. (a) a round diameter for every section by the equal
friction method, with the balancing dampers marked, and (b) the total
pressure the supply system requires, compared with what the unit can
deliver.
Sized layout. Trunk flows fall from 645 cfm at the unit to 150 cfm in the last section; the circled D marks a balancing damper, required in every branch except the index run.
Approach. Add the branch flows to get the trunk flows,
size the first main section to its velocity limit, read the friction rate
that section produces and hold it constant for every other section, round to
standard diameters while respecting the velocity caps, then add the losses
along each of the five paths and compare the worst with the pressure
available.
Check: friction data and fitting
allowances. Friction is computed from the Darcy–Weisbach
relation with the Colebrook friction factor for galvanised steel
($\varepsilon = 0.0003$ ft) and standard air
($\nu = 1.613\times 10^{-4}\ \text{ft}^2/\text{s}$), which is
the relation the published friction chart is drawn from; reading the chart by
eye gives the same answers to about ±5 %. The only fitting
allowance the paper supplies is the 20 ft equivalent length of the
diffuser boot, and that is the only one used here; a tendered design would
add equivalent lengths for the two discharge elbows, the branch take-offs and
the balancing dampers themselves, which typically raises the index-run loss by
25 –50 %. The margin computed below is comfortably larger than
that.
(a) Sizing by the equal-friction method
Trunk flows. Working back from the far end, each trunk
section carries everything downstream of it:
$$Q_{M1}=50+200+120+125+150=645\ \text{cfm}$$
and successively 595, 395, 275 and 150 cfm after each take-off.
Size the first section to the velocity limit. The main
is capped at 850 ft/min, so
$$A=\frac{Q}{V}=\frac{645}{850}=0.759\ \text{ft}^2
\qquad
D=\sqrt{\frac{4A}{\pi}}=0.983\ \text{ft}=11.80\ \text{in}$$
Rounding up to the next standard size gives
12 in, at which 645 cfm travels at 821 ft/min —
under the 850 ft/min cap, as rounding up guarantees.
Read the design friction rate off that section. With
$D=12$ in and $Q=645$ cfm the velocity is 821 ft/min, the velocity
pressure is $(821/4005)^2=0.0420$ in. wg, the Reynolds number is
$8.5\times 10^{4}$ and Colebrook gives $f=0.0200$. Hence
$$J=f\,\frac{100}{D}\,p_v
=0.0200\times\frac{100}{1.0}\times 0.0420
=0.0840\ \text{in. wg per 100 ft}$$
$$\boxed{\;J=0.0840\ \text{in. wg}/100\ \text{ft, held constant
throughout}\;}$$
Size every other section at that friction rate. For each
flow, solve $J(Q,D)=0.0840$ for $D$ and round to the nearest standard
diameter, then check the velocity against its cap and step up one size if it
is exceeded. No section needed that correction here: the equal-friction rule
delivers falling velocities as the flow falls, which is exactly its intended
behaviour.
Duct schedule
Section
Flow (cfm)
Equivalent length (ft)
$D$ at $J$ (in)
Selected $D$ (in)
Velocity (ft/min)
Loss (in. wg)
M1 — unit to the first take-off (4 + 10 + 10 ft)
645
24
12.00
12
821
0.0202
M2 — trunk
595
10
11.64
12
758
0.0072
M3 — trunk
395
10
9.99
10
724
0.0083
M4 — trunk
275
10
8.72
9
622
0.0072
M5 — trunk
150
8
6.96
7
561
0.0065
B1 — branch, 50 cfm diffuser
50
45
4.63
5
367
0.0261
B2 — branch, 200 cfm diffuser
200
42
7.75
8
573
0.0302
B3 — branch, 120 cfm diffuser
120
40
6.40
6
611
0.0462
B4 — branch, 125 cfm diffuser
125
47
6.50
7
468
0.0276
B5 — branch, 150 cfm diffuser
150
44
6.96
7
561
0.0359
Branch equivalent lengths in the schedule include the 20 ft diffuser
boot; for example branch B4 is 13 + 14 ft of duct plus
20 ft of boot, 47 ft in all. Every main velocity is below
850 ft/min and every branch velocity below 650 ft/min, so the
selection is admissible.
(b) Total pressure loss and the balancing dampers
Pressure available to the supply side. The unit develops
0.30 in. wg external to itself and the return system claims 0.12,
so
$$\Delta p_{avail}=0.30-0.12=0.18\ \text{in. wg}$$
Loss along each path. Every path runs from the unit
along the trunk to its take-off, out the branch and through the diffuser.
Adding the section losses from the schedule and the quoted diffuser drop:
Path totals and damper settings
Path to
Duct loss (in. wg)
Diffuser (in. wg)
Path total (in. wg)
Damper must absorb (in. wg)
B1
0.0462
0.05
0.0962
0.0292
B2
0.0576
0.04
0.0976
0.0279
B3
0.0819
0.036
0.1179
0.0075
B4
0.0706
0.03
0.1006
0.0248
B5
0.0854
0.04
0.1254
0.0000
Identify the index run and state the system loss. The
largest path total is the one that fixes the fan duty:
$$\boxed{\;\Delta p_{system}=0.1254\ \text{in. wg along path
B5 (the index run)}\;}$$
Adding the return system, the unit must develop
$0.1254+0.12=0.2454$ in. wg externally, against the
0.30 in. wg it can produce.
Check the margin.
$$\Delta p_{avail}-\Delta p_{system}=0.18-0.1254
=0.0546\ \text{in. wg}$$
a surplus of 43.5 % over the index run — ample to
absorb the fitting equivalent lengths the paper does not supply.
Locate the dampers. Equal friction sizes for a constant
pressure gradient, not for equal path resistance, so the four shorter paths
would each pass more than their design flow if left alone. Every branch except
the index run therefore needs an opposed-blade balancing damper, set to
dissipate the difference shown in the last column: B5 needs none, and
the others need between 0.0075 and 0.0292 in. wg. Dampers belong in
the branch take-off, well upstream of the diffuser — at least four to
five duct diameters — so that the throttling noise they generate is not
radiated straight into the room. Diffuser face dampers are for trim only.
Two features of the result are worth noting. The imbalance to be absorbed
is small in absolute terms, a few thousandths of an inch of water, because the
paths differ mainly in branch length rather than in trunk length; a layout
with one very short and one very long run would demand much more of its
dampers and would be a candidate for the static-regain method instead. And
branch B3, at 120 cfm in a 6 in duct, is the section closest to its
velocity limit at 611 ft/min — if the room served by it proves
noisy in commissioning, that branch is the one to enlarge.
