CEN370 DUCT SIZING Dr. N.

Zakhia
Purpose
Duct system delivers conditioned air to each diffuser in the desired
space at a specified total pressure, Pt. This can be done by selecting the
proper fan to overcome the total pressure losses in the system.
Pressures in ducts:
1
From Bernoulli’s equation: 𝑃 + 2 𝜌𝑉 2 + 𝜌𝑔𝑍 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
P = static pressure, Pst, pa
1
𝜌𝑉 2
≡ Pv ≡ theoretical dynamic or velocity pressure, Pa
2
𝜌𝑔𝑍 ≡ Elevation pressure, Pa ≈ 0
1
𝑃 + 2 𝜌𝑉 2 ≡ Pstag = stagnation or total pressure (Pt).
1
∴ 𝑃𝑠𝑡𝑎𝑔 − 𝑃𝑠𝑡 = 2 𝜌𝑉 2 ≡ 𝑃v ← velocity or dynamic pressure
For air at std, 𝜌 = 1.204 kg/m3 = 0.075 lbm/ft3. ⇒ thus, theoretically,
̅
𝑉 2 ̅
𝑉 2
𝜌 (1097) = (4005) ← 𝐸𝑛𝑔𝑙𝑖𝑠ℎ 𝑢𝑛𝑖𝑡𝑠, (𝑖𝑛. 𝑤𝑔)
𝑃v = { 2 2
̅
𝑉 ̅
𝑉
𝜌 (1.414) = (1.29) → 𝑆𝐼 𝑢𝑛𝑖𝑡𝑠 (𝑃𝑎)

• 𝑃𝑡 always decreases in the direction of the flow.

• The fan total pressure, FTP, must overcome the total head loss of the
system ⇒ FTP = Fan inlet pressure – fan discharge pressure
Return & ventilation Ducts
(Fan suction side) Fan Duct system (Fan supply side)

Residential buildings: 125 Pa < Pt < 250 Pa

Commercial buildings:
– Low-pressure duct system ⇒ Pt ≤ 500 Pa
– Medium-pressure duct system ⇒ 500 Pa < Pt < 1500 Pa.
Steps in duct sizing
1. Determine the CFM for each room:
𝑠𝑒𝑛𝑠𝑖𝑏𝑙𝑒 𝑠𝑝𝑎𝑐𝑒 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑜𝑟 ℎ𝑒𝑎𝑡𝑖𝑛𝑔 𝑙𝑜𝑎𝑑𝑠,𝐵𝑡𝑢ℎ
𝐶𝐹𝑀 = 1.08 x (𝐷𝐵𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 −𝐷𝐵𝑙𝑒𝑎𝑣𝑖𝑛𝑔 ),𝑖𝑛 °𝐹

2. Draw a layout on the plan showing AHU, ducts, diffusers, and grilles
3. Size the main duct and all branches
Calculation:
The fan specs are known before duct sizing (low, Medium or high static
fan unit) ⇒ thus, Pstatic,fan and 𝑚̇𝑓𝑎𝑛 are available from the catalogue.
Furthermore, ∆Plosses of all other elements of the system are known
except for the supply and return ducts.
Therefore, the Pnet,static available for the ducts is:
𝑷𝒏𝒆𝒕 𝒔𝒕𝒂𝒕𝒊𝒄,𝒅𝒖𝒄𝒕 = 𝑷𝒔𝒕𝒂𝒕𝒊𝒄 𝒇𝒂𝒏 − 𝑷𝒍𝒐𝒔𝒔𝒆𝒔
where: ∆𝑃𝑠𝑡𝑎𝑡𝑖𝑐,𝐹𝑎𝑛 = 𝑃𝑠𝑡,𝑑𝑢𝑐𝑡 + ∆𝑃𝑅 + 𝑃v
∆𝑃𝑓
• 𝑃𝑠𝑡,𝑑𝑢𝑐𝑡 = ∑ 𝐿𝑒𝑓𝑓 x ( ) + ∑ 𝑃𝑑 where:
𝑚
Leff = effective length = Lduct + Le
Le = total equivalent length of fittings …
∆Pf/meter = ∆P due to friction loss per meter of duct length,
from the flow chart
ΣPd = total dynamic losses for all fittings and equipment such
as fan coils, dampers, grills, …
• ∆𝑃𝑅 = Pressure drop in the Return duct, Pa.
̅2 2
𝑉
𝐶𝑜 (4005) → (𝑖𝑛𝑐ℎ. 𝑤𝑔) V = fan face velocity
• 𝑃v ≡ { 2 Co ≡ loss factors due to fittings
̅2
𝑉
𝐶𝑜 (1.29) → (𝑃𝑎) in the duct (tabulated)
Duct shape:
Round, oval, or rectangular shape can be used. However, rectangular
𝑙𝑒𝑛𝑔𝑡ℎ 𝑊
shape must have aspect ratio: 𝐴𝑅 = = < 4 to avoid vibration
𝑤𝑖𝑑𝑡ℎ 𝐻
and noises.
If AR > 4 ⇒ Expensive to fabricate, hard to
H
install due to large perimeters, high friction
W
losses and may produce vibration and noises
The formula that converts round pipes to rectangular shape:
(𝑊 x 𝐻)0.625
𝐷𝑒 = 1.3 [ ( ] where W = 4 H (see table 2 below)
𝑊+ 𝐻)0.25

Classification of duct systems:

Ducts are classified based on the cooling load, application types, air
pressure and turbulence in ducts as:
Low pressure systems Vair ≤ 10 m/s , Pst ≤ 5 cm H2O (g) (0.49 kPa or 0.071 psi)
Medium pressure systems Vair ≤ 10 m/s, Pst ≤ 15 cm H2O (g) (1.47 kPa, 0.213 psi)
High pressure systems Vair > 10 m/s, 15 < Pst ≤ 25 cm H2O (g) (2.45 kPa or 0.355 psi)

Note: high velocities in the ducts results in:

a) Smaller ducts ⇒ low initial cost and low space requirement
b) Higher pressure drops ⇒ larger fan power consumption
c) Increased noise ⇒ need for noise attenuation
ASHRAE recommendation: Residences: 3 m/s to 5 m/s
Theatres: 4 to 6.5 m/s
Restaurants: 7.5 m/s to 10 m/s
If nothing is specified, then a velocity of 5 to 8 m/s is used for main
ducts and 4 to 6 m/s for the branches.
Recommended values of 𝑷𝒇 ⁄𝒎 for duct sizing:
For the supply side of the Fan:
a) Low Velocity Duct: 1000 to 1500 ft/min (5.0 to 7.5 m/s)
𝑃𝑓
= 0.1 - 0.15 inch/100 ft (1 - 1.23 Pa/m)
𝑚

b) High Velocity Duct: 4000 - 5000 ft/min (20.0 to 25.0 m/sec)

𝑃𝑓
(𝑚) = 0.7 inch/100 ft (5.7 Pa/m)
𝑚𝑎𝑥
𝑃𝑓
For the suction side of the Fan: = 0.65 Pa/m
𝑚

Table 2. Maximum duct velocity - TRANE

Low-velocity systems
Schools, theatres,
Designation Private residences Public buildings Industrial buildings
Maximum velocities (fpm)
Main ducts 800 - 1200 1100 - 1600 1300 - 2200
Branch ducts 700 - 1000 800 - 1300 1000 - 1800
Branch risers 650 - 800 800 - 1200 1000 - 1600

Figures and charts for duct sizing (see next pages)

 Figure 1. Friction Pressure Loss per 100ft of duct (English units)
CFM
 Figure 2. Friction Pressure Loss per meter length of duct – SI units

Pressure Friction Loss, Pa/m

Air Quantity, L/s

 (𝑊 x 𝐻)0.625
𝐷𝑒 = 1.3 [ ( ] where W = 4 H
𝑊+ 𝐻)0.25
Common methods used for duct sizing:
The figure below shows the schematic of a typical supply air duct
layout. As shown, the supply air from the fan is distributed to five
outlets (1 to 5), located in five different conditioned zones. The duct
runs A to I denote the portions of the duct to different outlets. The
duct run with the highest ∆Pf is called “the index run”. The supply
airflow rates to each space can be found from the load and
psychrometric calculations. From the building layout, the length of
each duct run is known.
4
Figure A.
5

cfm
Fa
n
1
2 3
The design of an air conditioning duct system in large buildings may
require the use of a computer software. However, the following
methods are most commonly used for simpler layouts such as the one
shown in the above figure.
Applications Methods
1. Equal Friction loss method.
Low and medium pressure 2. Velocity-reduction method
systems 3. Equal pressure drop method
(Balanced Capacity method)
High/large pressure system 4. Static regain method
1. Equal friction loss method
• This method is straight forward and easy to use.
• It gives an automatic reduction of the Vair throughout the system.
The reduced velocities are, in general, within the noise limits.
Advantages and disadvantages of the “equal friction loss method”:
• This method usually yields a better design than the “velocity-
reduction” method since ∆P is dissipated as friction in the duct
runs, rather than in the balancing dampers.
• This method is generally suitable when the ducts are not too long.
(i.e. residential applications). It can be used for both supply and
return ducts. However, similar to velocity method, the equal
friction method also requires partial closure of dampers in all but
the index run, which may generate noise.
• If the ducts are too long ⇒ the total pressure drop will be high
and due to dampening, ducts near the fan get over-pressurized.
• This method is not recommended for VAV systems and is better
suited for CAV (constant air volume) systems.
• It is usually applied for normal capacity (CFM < 2250 cfm) or
𝑄̇𝑐𝑜𝑜𝑙𝑖𝑛𝑔 < 60,000 Btu/h.
• The short runs will have to be damped, which may cause
considerable noise. The dampering pressure for supply and return
duct is found by deducting the total pressure from pressure losses
of coils, filter, grills, and accessory (all found in the manufacturer’s
catalogue). The remaining pressure is divided between the supply
and the return.
Procedure steps of “Equal Friction Pressure loss” method:
∆𝑃𝑓
This method assumes ( ) in the main and branch ducts is kept equal.
𝐿
∆𝑃𝑓 ∆𝑃𝑓 ∆𝑃𝑓 ∆𝑃𝑓
Meaning: ( 𝐿
) =( 𝐿
) =( 𝐿
) =( 𝐿
) =⋯
𝐴 𝐵 𝐶 𝐷
Steps are:
1. Compute the air flow rate in the main duct by summing the flow
rates of individual branches:
𝑄𝐴 = 𝑄1 + 𝑄2 + 𝑄3 + 𝑄4 + 𝑄5
∆𝑃𝑓
2. Select a suitable ( ) from the friction chart
𝐿
3. Calculate the diameter in the main duct, 𝑫𝒆𝒒,𝑨 ⇒ can be found from:
𝑄
• 𝑄𝐴 = 𝑉𝐴 𝐷𝐴 ⇒ 𝐷𝐴 = 𝑉𝐴 or
𝐴
∆𝑃𝑓
• Since ( ) is constant for all the duct runs, 𝐷𝐴 and all other 𝐷’𝑠
𝐿
can be found from the friction chart knowing Q for each duct.
4. Calculate the pressure drops for the longest run as:
∆𝑃𝑓 ∆𝑃𝑓
∆𝑃𝑓,𝐴 = ( ) . 𝐿𝐴 ; ∆𝑃𝑓,𝐵 = ( ) . 𝐿𝐵 ; …
𝐿 𝐿
5. Compute the dynamic pressure losses, Pv, due to fittings: 2 methods:
• Based on fitting types loss factor, Co (will be provided) or
• Using effective length method.
6. Check the pressure drops for the other branches and determine the
static head of the fan based on the highest ∆P.
7. Next, the fan is selected to suit the index run with the highest ∆P.
8. Finally, dampers will be selected and installed in all duct runs to
balance the total pressure loss.
Example:
consider a typical duct layout shown below. Use “Equal friction
method” to size this duct system. The velocity of air in the main duct is
8 m/s and ρair = 1.2 kg/m3. The fittings loss coefficients are:
𝐶𝑜 = 0.88 for the outlets
45° convergent Tee (WYE) Main duct
𝐶𝑏 = 0.515 for the branches
𝐶𝑚 = 0.18 for the main. VC Vm
45°
𝐶𝑒𝑙𝑏𝑜𝑤 = 0.53 for the 90° elbow.
Deduce the required FTP and the
1.5 Db
amount of dampering in each duct. Db

Vb Branch

2 m3/s 2 1 m3/s 3
Return
D F
15 m 12 m 18 m 66m
m
Fresh air
Outside

AHU A C E
+
FAN
66m
m
B 45° convergent Tee 90° Elbow
45° convergent Tee

1 m3/s 1

Solution:
1. Compute the total air flow rate of the main duct:
𝑄𝑡𝑜𝑡 = 𝑄𝐵 + 𝑄𝐷 + 𝑄𝐹 = 1 + 2 + 1 = 4 𝑚3 /𝑠 and V = 8 m/sec
∆𝑷𝒇
2. =? in the main duct (A): it can be found from the friction chart:
𝑳
∆𝑃𝑓,𝐴
At Q = 4 m3/sec and V = 8 m/s ⇒ ≈ 0.78 Pa/m. This method
𝐿
requires that ∆Pf/L is the same for all ducts.
3. Compute D’s =?
• DA =? can be read from the same figure 2 or calculated as:
𝑄 4
QA = VA. AA ⇒ 𝐴 = 𝑉 = 8 = 0,5 𝑚2 ⇒ 𝐷𝐴 = 0.798 m

All other D’s are found from the chart (figure 2), as:
∆𝑃𝑓,𝐴
• DB =?at QB = 1 m3/sec and = 0.78 ⇒ Deq,B =0.475 m
𝐿
∆𝑃𝑓,𝐴
• DC =? at QC = 3 m3/sec and = 0.78 ⇒ DB =0.72 m
𝐿
∆𝑃𝑓,𝐴
• Deq,D =? at QD = 2 m3/sec and = 0.78 ⇒ Deq,D =0.62 m
𝐿
∆𝑃𝑓,𝐴
• Deq,E =? at QE = 3 m3/sec and = 0.78 ⇒ Deq,E =0.475 m
𝐿

HWK: convert the above diameters into rectangular ducts.

Q = 4 m3/s
Air Quantity, m3/s Figure 2. Friction loss ∆Pf/m – SI units

Pressure Friction Loss, Pa/m

∆Pf, A /m ≈ 0.78 Pa
Calculation of total pressure drop to balance the system:
1. From fan to outlet 1: ∆𝑃𝐴→𝐵 = ∆𝑃𝑓,𝐴 + ∆𝑃𝑓,𝐵 + ∆𝑃𝑒𝑙𝑏−𝑏 + ∆𝑃𝑜𝑢𝑡𝑙𝑒𝑡 1
∆𝑃𝑓,𝐴 and ∆𝑃𝑓,𝐵 ≡ frictional ∆P in sections A and B. They can be found
from the charts.
∆𝑃𝑒𝑙𝑏−𝑏 ≡ dynamic ∆P of the elbow branch:

2 m3/s 2 1 m3/s 3

D F
15 m 12 m 18 m 66m
m

A C E

FAN 66m
m
B

1 m3/s 1

∆𝑃𝑜𝑢𝑡𝑙𝑒𝑡 1 ≡ ΔPGrill = dynamic ∆P of the outlet 1.

Therefore:
∆𝑃𝑓
• ∆𝑃𝑓,𝐴 = ( 𝐿
) . 𝐿𝐴 = (0.78)(15𝑚) = 11.7 𝑃𝑎
∆𝑃𝑓
• ∆𝑃𝑓,𝐵 = ( 𝐿
) . 𝐿𝐵 = (0.78)(6𝑚) = 5.1 𝑃𝑎

𝑉𝐵2
• ∆𝑃𝑒𝑙𝑏−𝑏 = 𝐶𝑐,𝑏 (𝜌 ) ⇒ VB =?
2
∆𝑃𝑓,𝐴
From the chart at = 0.78 and 𝑄̇𝐵 = 1 𝑚3 /𝑠 ⇒ VB ≈ 5.8 m/sec
𝐿𝐴

𝜌𝑉𝐵2 1.2 𝑥 5.82

Thus, ∆𝑃𝑒𝑙𝑏−𝑏 = 𝐶𝑐,𝑏 ( ) = 0.515 ( ) = 10.4 𝑃𝑎
2 2

𝜌𝑉𝐵2 1.2 𝑥 5.82

• ∆𝑃𝑜𝑢𝑡𝑙𝑒𝑡 1 = 𝐶𝑜 ( ) = 0.88 ( ) = 17.76 𝑃𝑎
2 2

∴∆𝑷𝑨−𝑩 = 11.7 + 5.1 + 10.4 + 17.76 = 44.96 𝑃𝑎

2. From fan to outlet 2:
∆𝑷𝑨−𝑪−𝑫 = ∆𝑃𝑓,𝐴 + ∆𝑃𝑓,𝐶 + ∆𝑃𝑓,𝐷 + ∆𝑃𝑒𝑙𝑏−𝑚,𝑐 + ∆𝑃𝑒𝑙𝑏−𝑏,𝐷 + ∆𝑃𝑜𝑢𝑡𝑙𝑒𝑡 2
≈ 70.50 𝑃𝑎 1 m3/s 3
2 m3/s 2

D F
15 m 12 m 18 m 66m
m

A C E

FAN 66m
m
B

1 m3/s 1

3. From fan to exit outlet 3:

∆𝑷𝑨−𝑪−𝑬−𝑭 = ∆𝑃𝑓,𝐴 + ∆𝑃𝑓,𝐶 + ∆𝑃𝑓,𝐸 + ∆𝑃𝑓,𝐹 + 𝑃𝑒𝑙𝑏−𝑚,𝐶 + ∆𝑃𝑒𝑙𝑏−𝑚,𝐸
+∆𝑃𝑒𝑙𝑏𝑜𝑤 + ∆𝑃𝑜𝑢𝑡𝑙𝑒𝑡 3 ≈ 93.50 𝑃𝑎
Since the duct from the fan to outlet 3 has the highest ∆P, thus, it is
called “the index run”. ⇒ the Fan Total Pressure FTP = 93.50 Pa

Installing dampers for each outlet excluding the index run:

Amount of dampering required for branch B = FTP - ΔPA-B = 48.54 Pa
Amount of dampering required for branch D = FTP - ΔPA-C-D = 23.0 Pa

2 m3/s 2 1 m3/s 3
Damper (23.0 Pa)
D F
15 m 12 m 18 m 66m
m

A C E

FAN 66m
m
B

1 m3/s 1 Damper (48.54 Pa)

2. Velocity Reduction Method:
This method is recommended for circular duct. The pressure drop in
circular duct is lower than that of rectangular. In this system, control
dampers are needed to balance the duct.
• ∆Pf α Pv along the duct. The dampers are even more important for
the velocity-reduction method than equal friction loss. This is due
to less symmetry of the branch ducts so dampers will have to do
more adjusting.
• Once Vair in the main duct & branches are known ⇒ CFM is found
⇒ diameter and ∆Pf/L can be found.
This method is best suited for sizing the supply and return ducts. In the
case of supply systems, an initial guess is needed for the velocity (from
Table shown below), after the fan outlet. This starting velocity should
be fairly high (as long as it is in the range shown in following Table),
because it will become progressively smaller at the various branch
takeoffs. This is the reason it is called “velocity-reduction method”.

Velocity (m/s)
Residences Hotels Schools
Main supply 5 6 7.5
Main return 4 5 6
Branch supply 3 4 6
Branch return 3 4 5
Steps of calculation:
1. Select suitable velocities in the main and branch ducts from
ASHRAE Table.
2. Find the diameters of the main and branch ducts from airflow
rates and velocities for circular ducts.
3. From the velocities and duct dimensions obtained in the
previous step, find the ∆Pf for the main and branches using
friction chart or an equation.
4. From the duct layout, dimensions and airflow rates, find the Pv for
all the bends and fittings.
5. Select a fan that can provide sufficient Pst for the index run.
6. Balancing dampers have to be installed in each run. The damper in
the index run is left completely open, while the other dampers are
throttled to reduce the flow rate to the required design values.
Advantages/disadvantages of the velocity-reduction method:
It is a simple method to design both supply and return air-ducts.
However, selection of suitable velocities in different duct runs requires
experience. Wrong selection of velocities can lead to very large ducts,
which leads to occupying large space in the building and increases the
cost. Furthermore, if your selection of velocities led to very small ducts,
it will result in large pressure drop and hence necessitates the selection
of a large fan leading to higher fan cost and running cost. In addition,
the method is not very efficient as it requires partial closing of all the
dampers except the one in the longest or index run, so that the total
pressure drop in each run will be same.
Example: redo the previous example using the “velocity-reduction
method”
2 m3/s 2 1 m3/s 3
Return
D F
15 m 12 m 18 m 66m
m
Fresh air
Outside

AHU A C E
+
FAN
66m
m
B 45° convergent Tee 90° Elbow
45° convergent Tee

1 m3/s 1

Solution:
1. V =? Select a velocity of 5 m/s for the downstream and branches
while Vmain remains 8 m/sec.
2. D=?
Segment A: Flow rate, Q = 4 m3/s and VA = 8 m/s ⇒
AA = QA/VA = 4/8 = 0.5 m2 ⇒ DA = 0.798 m
Segment B: Flow rate, Q = 1 m3/s and VB = 5 m/s ⇒
AB = QB/VB = 1/5 = 0.2 m2 ⇒ DB = 0.505 m
Segment C: Flow rate, QC = 3 m3/s and VC = 5 m/s ⇒
AC = QC/VC = 3/5 = 0.6 m2 ⇒ DC = 0.874 m
Segment D: Flow rate, QD = 2 m3/s and VD = 5 m/s ⇒
AD = QD/VD = 2/5 = 0.4 m2 ⇒ DD = 0.714 m
Segments E&F: Flow rate, QE,F = 1 m3/s and VE,F = 5 m/s ⇒
AE = AF = QE/VE = 1/5 = 0.2 m2 ⇒ DE = DF = 0.505 m
3. Calculation of pressure drop (balancing the system):
Section A-B: ∆𝑃𝐴→𝐵 = ∆𝑃𝑓,𝐴 + ∆𝑃𝑓,𝐵 + ∆𝑃𝑒𝑙𝑏−𝑏 + ∆𝑃𝑜𝑢𝑡𝑙𝑒𝑡 1
∆𝑃𝑓,𝐴
∆𝑃𝑓,𝐴 =? at QA = 4 m3/s and VA = 8 m/s ⇒ = 0.78 Pa/m
𝐿
∆𝑃𝑓 𝐴
∆𝑃𝑓 𝐴 = ( ) 𝐿𝐴 = 0,78 𝑥 15 𝑚 = 11,7 Pa
𝐿
∆𝑃𝑓,𝐵
∆𝑃𝑓,𝐵 =? at Q = 1 m3/s and VB = 5 m/s ⇒ = 0.52 Pa/m
𝐿
∆𝑃𝑓 𝐵
∆𝑃𝑓 𝐵 = ( ) 𝐿𝐵 = 0,52 𝑥 6 𝑚 = 3,12 Pa
𝐿

𝜌𝑉𝐷2 1,2 x 52
∆𝑃𝑒𝑙𝑏−𝑏 = 𝐶𝑏 ( ) = 0,515 ( ) = 16,5 𝑃𝑎
2 2

𝜌𝑉𝐷2 1,2 x 52
∆𝑃𝑜𝑢𝑡𝑙𝑒𝑡(1) = 𝐶𝑜 ( ) = 0,88 ( ) = 13,2 𝑃𝑎
2 2

∴∆𝑷𝑨−𝑩 = 11.7 + 3.12 + 16.5 + 13.2 = 44.52 𝑃𝑎

Section A-C-D:
∆𝑷𝑨−𝑪−𝑫 = ∆𝑃𝑓,𝐴 + ∆𝑃𝑓,𝐶 + ∆𝑃𝑓,𝐷 + ∆𝑃𝑒𝑙𝑏−𝑚,𝑐 + ∆𝑃𝑒𝑙𝑏−𝑏,𝐷 + ∆𝑃𝑜𝑢𝑡𝑙𝑒𝑡 2
≈ 51.41 𝑃𝑎 ← redo the calculation then justify the answer
Section A-C-E-F:
∆𝑷𝑨−𝑪−𝑬−𝑭 = ∆𝑃𝑓,𝐴 + ∆𝑃𝑓,𝐶 + ∆𝑃𝑓,𝐸 + ∆𝑃𝑓,𝐹 + 𝑃𝑒𝑙𝑏−𝑚,𝐶 + ∆𝑃𝑒𝑙𝑏−𝑚,𝐸

+∆𝑃𝑒𝑙𝑏𝑜𝑤 + ∆𝑃𝑜𝑢𝑡𝑙𝑒𝑡 3 ≈ 69.3 𝑃𝑎 ← justify the answer

Thus, the run with maximum ∆P is A-C-E-F (the index run). Thus, the
Fan Total Pressure FTP = 69.3 Pa
Dampering:
Amount of dampering required at 1 = FTP – ΔPA-B = 24.78 Pa
Amount of dampering required at 2 = FTP – ΔPA-C-D = 17.89 Pa
Conclusion: It is seen that:
a) The Velocity method results in larger “D” due to the velocities
selected in branch and downstream. This led to lower FTP.
b) The Equal Friction method results in smaller “D” but larger FTP.
c) For dampering, the velocity reduction method requires less
dampering at outlet 1 but more at outlet 2.

Damper (25.0 Pa)

Damper (18.0 Pa)

3. Equal pressure drop method (Balanced Capacity)

2
4

AHU 1 3
+
FAN

The main idea: ΔP12 = ΔP135 = ΔP134.

The basic idea is to make ∆Ptot EQUAL ∆P for all parallel ducts from the
fan to each outlet.
For instance, the duct 1-3-4 is the longest run (index) and duct 3-5 is a
branch. Then we have to modify the duct size in the branch 3-5 so that
∆Ptot in this branch is equal to ∆Ptot in duct 3-4. This condition forces
the designer to locate the diffuser at a specific distance from the fan
to ensure equal ∆P.
Steps: similar to equal friction method
1. The pressure loss per unit length of the index is determined the
same way as “equal friction method”.
2. Adjust (by iterations) the duct sizes in the branches such that the
total pressure loss in each branch is equal to its parallel section of
the longest run.
Example: Size the ducting system shown below using the “balanced
capacity - equal pressure drop” method. Use the equivalent length
method to compute the losses. This layout is considered low velocity
system. Thus, assume max velocity 5 m/s. Identify the dampers
required in order to balance the system.
0.60 m3/s
E

6m

AHU A 5m B 6m C 4m D
+ 1.10 m3/s
FAN
4m

Branch Fittings: F
Wye, 45 degree 0.20 m3/s
Solution:
1. Find the flow rates:
𝑄𝐴𝐵 = 0.6 + 0.2 + 1.1 = 1.9 𝑚3 /𝑠𝑒𝑐
𝑄𝐵𝐶 = 0.2 + 1.1 = 1.3 𝑚3 /𝑠𝑒𝑐
2. Calculate ∆P and the diameter in the main duct:
• Given Vmax = VAB = 5 m/sec in duct AB (next to the fan) ⇒ Thus:

4𝑄
𝑄𝐴𝐵 = 𝑉𝐴𝐵 x 𝐴𝐴𝐵 ⇒ 𝐷𝐴𝐵 = √𝜋 𝑉𝐴𝐵 = 0.70 𝑚 (or from the chart)
𝐴𝐵

• From the chart at 𝑄𝐴𝐵 = 1.9 m3/sec and VAB = 5 m/sec ⇒ read:
∆𝑃𝐴𝐵 ∆𝑃𝐴𝐵
= 0.36 Pa/m. This method requires that = 0.36 Pa/m
𝐿 𝐿
remains the same for all ducts to their exits.
 Figure 2. Friction loss ∆Pf/m – SI units
Q = 4 m3/s
Air Quantity, m3/s

Pressure Friction Loss, Pa/m

∆Pf, A /m ≈ 0.36 Pa
3. Calculate the diameters and velocities of the remaining ducts:
∆𝑃𝐵𝐶
Duct BC: @ Q = 1.3 m3/s and = 0.36 Pa/m
𝐿
From the chart ⇒ DBC = 0.60 m and VBC = 4.6 m/s
∆𝑃𝐶𝐷
Duct CD: @ Q = 1.1 m3/s and = 0.36 Pa/m
𝐿
From the chart ⇒ DCD = 0.56 m and VCD = 4.4 m/s
∆𝑃𝐶𝐹
Duct CF: @ Q = 0.2 m3/s and = 0.36 Pa/m
𝐿
From the chart ⇒ DCF = 0.29 m and VCF = 2.9 m/s
∆𝑃𝐵𝐸
Duct BE: @ Q = 0.6 m3/s and = 0.36 Pa/m
𝐿
From the chart ⇒ DBE = 0.45 m and VBE = 3.8 m/s
Since ∆P = 0.36 Pa/m is equal over each duct, this creates an
unbalance duct. Therefore, the system must be balanced.
4. Balancing the system:
This means accounting for the losses in the system as well as using
dampers to balance cfm. This is carried out by calculating ∆P for
each duct (to its exit) in the system. Meaning:
𝑃𝐴 = ∆𝑃𝐴→𝐸 = ∆𝑃𝐴→𝐹 = ∆𝑃𝐴→𝐷
0.60 m3/s
E

6m

AHU A 5m B 6m C 4m D
+ 1.10 m3/s
FAN
4m
Branch Fittings: F
Wye, 45 degree 0.20 m3/s
➢ Let’s first make ∆𝑃𝐴→𝐹 = ∆𝑃𝐴→𝐷 where:
∆𝑃𝐴→𝐹 = ∆𝑃𝐴→𝐵 + ∆𝑃𝐵→𝐶 + ∆𝑃𝐶→𝐹 This requires
∆𝑃𝐴→𝐷 = ∆𝑃𝐴→𝐵 + ∆𝑃𝐵→𝐶 + ∆𝑃𝐶→𝐷 ∆𝑃𝐶→𝐷 = ∆𝑃𝐶→𝐹
∆𝑃
• ∆𝑃𝐶→𝐷 =? Using the effective length method, ∆𝑃𝐶→𝐷 = ( 𝐿 ) . 𝐿𝑒𝑓𝑓
𝑒

𝐿𝑒𝑓𝑓 = 𝐿𝐶𝐷 + 𝐿𝑒,𝑡ℎ𝑟𝑜𝑢𝑔ℎ

Where: 𝐿𝐶𝐷 = actual length of the duct CD = 4 m
𝐿𝑒
| = 10 (converging in the direction of CD) ⇒ table 3
𝐷 𝑡ℎ𝑟𝑜𝑢𝑔ℎ

𝐿
𝐿𝑒,𝑡ℎ𝑟𝑜𝑢𝑔ℎ = ( 𝐷𝑒| ) 𝐷𝐶𝐷 = 10 x 0.56 = 5.6 m
𝑡ℎ𝑟𝑜𝑢𝑔ℎ

∴𝐿𝑒𝑓𝑓 = 4 + 5.6 = 9.6 m

∆𝑃
Thus, ∆𝑃𝐶→𝐷 = ( 𝐿 ) . 𝐿𝑒𝑓𝑓 = 0.36 x 9.6 = 3.46 𝑃𝑎
𝑒

• ∆𝑃𝐶→𝐹 =? 𝐿𝑒𝑓𝑓 = 𝐿𝐶𝐹 + 𝐿𝑒,𝑏𝑟𝑎𝑛𝑐ℎ where:

𝐿𝑒 𝐿
𝐷
| = 20 ⇒ 𝐿𝑒,𝑏𝑟𝑎𝑛𝑐ℎ = ( 𝐷𝑒 | ) 𝐷𝐶𝐹 = 20 x 0.29 = 5.8 m
𝑏𝑟𝑎𝑛𝑐ℎ 𝑏𝑟𝑎𝑛𝑐ℎ

∆𝑃
∴ 𝐿𝑒𝑓𝑓 = 4 + 5.8 = 9.8 m ⇒ ∆𝑃𝐶𝐹 = ( 𝐿 ) 𝐿𝑒𝑓𝑓 = 0.36 x 9.8 = 3.53 Pa.
𝑒

Since ∆𝑃𝐶𝐷 ≠ ∆𝑃𝐶𝐹 ⇒ they must be balanced by introducing a

damper to duct CD (∆𝑃𝐶𝐷 < ∆𝑃𝐶𝐹 ). Thus:
(∆𝑃𝐶𝐷 )𝐷𝑎𝑚𝑝𝑒𝑟 = ∆𝑃𝐶𝐹 − ∆𝑃𝐶𝐷 = 3.53 − 3.46 = 0.047 𝑃𝑎
➢ Now, make ∆𝑃𝐴→𝐸 = ∆𝑃𝐴→𝐹 or ∆𝑃𝐴→𝐸 = ∆𝑃𝐴→𝐷 where:
∆𝑃𝐴→𝐸 = ∆𝑃𝐴→𝐵 + ∆𝑃𝐵→𝐸 Branch CF is already balanced.
∆𝑃𝐵→𝐸 𝑚𝑢𝑠𝑡 = ∆𝑃𝐵→𝐶 + ∆𝑃𝐶→𝐹
∆𝑃𝐴→𝐹 = ∆𝑃𝐴→𝐵 + ∆𝑃𝐵→𝐶 + ∆𝑃𝐶→𝐹 Where ∆𝑃𝐶→𝐹 = 3.53 Pa.
 Do same above calculation: ∆𝑃𝐵→𝐶 = 3.89 Pa and ∆𝑃𝐵→𝐸 = 5.4 Pa
Since ∆𝑃𝐵→𝐸 < (∆𝑃𝐵→𝐶 + ∆𝑃𝐶→𝐹 = 3.89 + 3.53 = 7.42 Pa) ⇒ a
damper is needed:
(∆𝑃𝐵𝐸 )𝐷𝑎𝑚𝑝𝑒𝑟 = ∆𝑃𝐵𝐸 − ∆𝑃𝐵𝐶 = 7.42 − 5.4 = 2.0 𝑃𝑎
Therefore, this system requires 2 dampers for stability, one in duct
CD and another in duct BE. Notice that duct ACF has the highest
losses before the addition of dampers (this is the index run). Thus,
the FTP may be found as:
∆𝑃𝐴 = ∆𝑃𝐴𝐵 + ∆𝑃𝐵𝐸
Or ∆𝑃𝐴 = ∆𝑃𝐴𝐵 + ∆𝑃𝐵𝐶 + ∆𝑃𝐶𝐷
Or ∆𝑃𝐴 = ∆𝑃𝐴𝐵 + ∆𝑃𝐵𝐶 + ∆𝑃𝐶𝐹
∆𝑃
Where: ∆𝑃𝐴𝐵 = ( 𝐿 ) . 𝐿𝑒𝑓𝑓 = 0.36 x 5 = 1.8 𝑃𝑎
𝑒

Thus: ∆𝑃𝐴 = ∆𝑃𝐴𝐵 + ∆𝑃𝐵𝐸 = 1.8 + (5.4 + 2.0) = 9.22 𝑃𝑎

∆𝑃𝐴 = ∆𝑃𝐴𝐵 + ∆𝑃𝐵𝐶 + ∆𝑃𝐶𝐹 = 1.8 + 3.89 + 3.53 = 9.22 𝑃𝑎
∆𝑃𝐴 = ∆𝑃𝐴𝐵 + ∆𝑃𝐵𝐶 + ∆𝑃𝐶𝐷 = 1.8 + 3.89 + 3.53 = 9.22 𝑃𝑎
Now the system is BALANCED.

0.60 m3/s
E
Damper BE
6m

AHU A 5m 6m C 4m D
+ 1.10 m3/s
FAN B
4m

F Damper CD
0.20 m3/s
Table - 3
 HVAC engineering Practice rules of thumb
1. Dampers.
a) Volume Control Dampers: used to control air
flow. They are of two types:
• Parallel blade dampers: The blades rotate in
one direction. These dampers are best suited to
fully-open or fully-close the air flow (cfm) or for
fine control between 80% to 100% full flow.
• Opposed blade dampers: The blades rotate in
opposite directions. These dampers are best
suited to regulate cfm over a wide range
AMCA (Air Movement and Control Association)
recommends using an opposed-blade damper when
volume control is needed.
Balancing/volume adjusting dampers should be installed close to the
main supply, as far away as possible from the outlets. Terminal
dampers such as those used in registers and diffusers should not be
considered in branch balancing as they are meant to be used for fine
adjustment only and would normally be in an almost fully open
position to prevent unnecessary noise.
b) Fire and Smoke Dampers
A fire damper is installed adjacent to duct outlets to interrupt the
passage of flame. Fire dampers are equipped with a fusible link {rated
for 165°F up to 286°F (74 °C to 141°C)} that holds the blades open until
the link melts.
Should the ductwork fall away, the damper needs to stay in the wall or
floor to maintain the integrity of the wall or floor. One should actually
think of the fire damper as part of the wall system itself.
Smoke dampers are defined as a device designed to resist the passage
of smoke through the HVAC system.
In today’s design, fire and smoke dampers come in one piece.
2. Diffusers, Grilles & Registers, plenum ceiling see EG and kbe files for
proper use and installation (posted on blackboard).
• Diffusers: They supply air in various directions through deflecting
vanes. 2 types: (see EG and kbe files on blackboard)
– Ceiling diffusers
– Slot diffusers

Linear bar diffuser Linear slot diffuser

• Grilles: typically used for return ducts. In general, Grilles are not used
in supply ducts due to their inability to control the air.

• Registers: look like grilles but are comprised of one-way or two-way

adjustable air stream deflectors and dampers to restrict the amount

of air flow required to be returned, supplied or exhausted.
Steps for Selecting Air outlets:
a. Determine the cfm and size for each room.
b. Determine V, Throw, NC and ∆P across the diffuser.
c. Outlet air velocity: the normal air velocity used for comfortable air
distribution is 50 fpm at the end of the throw (acceptable range is
from 25 to 75 fpm)
d. Pressure drop: It is the ∆P across the diffuser. From catalogue
e. Select the appropriate diffuser.
Throw: it is the horizontal distance from a diffuser at a specified
velocity. For example, T50 = 15 ft, indicates that at a distance of 15 ft
from the diffuser, the velocity of the air is 50 fpm. Usually, the throw is
shown in the following format: [T150 -T100 - T50]. For example, in the
table below, air flow of 60 cfm results in a V = 150 fpm at 7 ft from the
diffuser, a V = 100 fpm at 9 ft from the diffuser, and a V = 50 fpm at 12
ft from the diffuser.
The typical characteristics and performance of diffuser are shown in
the table below
Air flow (cfm) 50 60 70 85 95 110 120
Velocity (fpm) 400 500 600 700 800 900 1000
∆P (in. H2O) 0.056 0.090 0.131 0.175 0.225 0.290 0.355
Noise (NC) 14 20 24 28 32 35 38
Throw (ft) 5-8-13 7-9-12 8-12-19 9-13-18 10-15-21 12-17-24 13-19-31

supply
air Diffuser

Length Airflow

T50 T100 T150 T150 T100 T50

V=150 fpm V=100 fpm v=50 fpm

Location of Air supply diffusers and Outlets (see table entitled “supply
air outlets performance” listed below)
The following key point should be noted:
a. Locate diffusers so that the T50 length is nearly equivalent to the
characteristic length.
b. When cooling is the dominant factor, install ceiling diffusers or
high wall outlets that discharge air parallel to the ceiling.
 c. Place the returns high when cooling is the dominant factor, and
low when heating is the dominant factor.
d. Special applications such as bars, kitchens, lavatories, dining
rooms, club rooms, etc, the return and exhaust inlets must be
located near or at the ceiling level to collect the warm air "build-
up," odors, smoke, and fumes.
Supply Air Outlet Performance-Location

Recommended flow velocity for supply grilles

- side wall supply diffuser: 400 - 600 ft/min. (2- 3 m/s)
- supply Ceiling diffuser: 800 ft/min. (4 m/s)
- Return grille: Less than 400 ft/min. (2 m/s)
Return Air Devices
All devices used for air supply are suitable for return air. Grilles are
commonly used due to their lower cost. The location of return grilles is
not as critical as supply devices. However, it is a wise rule to locate
return air inlets far from outlets.
3. Sound/noise. The SPL limitation:

Recommendation design criteria for noise level

SPACE NC level
Board room, teleconferencing rooms 20 to 25
Conference rooms 30
Private offices, hotel rooms, apartments 35
Open-plan offices, lobbies, toilets, corridors, computer
40
terminal rooms, retail spaces
Storage, locker rooms, laboratories without fume hoods 45
Kitchens, laundries, computer rooms 50
Garage, laboratories with fume hoods 55

Sound Control: most noise is generated from fans. Therefore:

• Isolate fans from their supports by using vibration isolators and from
the ductwork by using flexible connections.
• Make duct connection transitions as gradual as possible.
• Use duct velocities recommended for quietness.
• Avoid abrupt changes in direction in ducts. Use wide radius elbows
or turning vanes.
• Avoid obstructions in the ductwork. Install dampers only when
required.
• Balance the system so that throttling of dampers is minimized.
• Select air outlets as recommended by the manufacturer.
THE SUPPLY DUCT SYSTEM
The two most common supply duct systems are the ‘extended plenum’
system and the ‘radial’ system. The other options are spider and
perimeter loop systems.
1. Extended Plenum Systems

Diffuser Trunk
Branches

24 ft ≈ 7.3 m

(Single plenum)

24 ft 24 ft
maximum maximum

18 in.
minimum

Supply
plenum

18 in.
minimum

Perimeter outlets

Extended plenum duct system (double plenum)

2. Reducing Plenum System (50% rule): used for longer distance

18 in.
minimum 600 cfm
450 ft/min

Supply
plenum

4 ft
minimum
1200 cfm
900 ft/min

15 to 20 ft 24 ft
maximum

3. Reducing Trunk System

Reducing trunk duct system

 Example: Reducing Extended Plenum System

A Tee performs
AHU Supply the same function
as a reducer
Return

Return

Primary-secondary trunk system.

 4. Radial System: there is no trunk duct, or branch ducts; instead,
the short, direct duct runs maximize air flow. The radial system is
most economical and easiest to install, but is not practical if the
air handling unit cannot be centrally located.

Radial duct system

5. Perimeter Loop System
A perimeter loop system uses a perimeter duct fed from a central
supply plenum using several feeder ducts. This system is typically built
on slab in cold climates.

Perimetric loop system

Practical Guidelines for Designing Ductwork
Here are some key guidelines for designing ductwork that should be
followed to get the most out of your system:
1. Configuration: Ducts should be designed so that the length of each
run (each section of ductwork) is short enough to provide proper
control of air flow and stability of construction. Radial or trunk-and-
branch configurations have shorter runs and generally work best.
Ducts should be located, if possible, within the conditioned space.
2. Go straight: this is the most important rule of all. From an energy
perspective, air wants to go straight and will lose energy if you make
it bend. From a cost perspective, straight duct costs less than fittings.
Fittings are expensive because they must be hand assembled even if
the pieces are automatically cut by plasma cutters. So, when laying
out a system, try to reduce the number of bends and turns.
3. Make sure ducts are the correct size: Ductwork that’s too small
won’t be able to carry enough air to heat or cool your building. Ducts
that are too large can lose both air and energy, cutting system
efficiency, and wasting money. Use trusted industry standards and
procedures, such as those published by ASHRAE, to size your ducts.
4. Make sure there are enough return ducts: duct system requires
enough return ducts to bring warm air back to the HVAC unit to be
conditioned again. Each room that receives heating or cooling should
have at least one return duct. As a rule of thumb, use 2 cfm for each
in2 of return air opening. for example, 20” x 20” grille equals 400 in2
gross area of grille, which means 800 cfm of recommended airflow.
5. Be careful where you install ducts: Ducts placed in conditioned
spaces are more efficient than those placed in unconditioned spaces.
If located within conditioned space, conductive and radiative losses,
leakage losses, and equipment cabinet losses are reduced or
regained into the building space. If it is not feasible to locate
ductwork within conditioned spaces, the ducts should be properly
sealed and insulated. The trunk ducts are usually located above
corridors in the “false ceiling” to minimize noise transmission to
allow easy access without disturbing the building occupants.
6. Seal and insulate: Make sure all ductwork sections fit together
tightly. Connections can be mechanically sealed with sheet metal
screws or other fasteners to improve connection strength. Seal
connections with mastic or metal tape. Cover the ductwork with
insulation such as rigid fiber board insulation.
Rules of Thumb:
• 10% Rule: For supply ducts longer than 10 feet, the air is reduced
in that run by 10% for every 5 feet. For example, a 30-foot run
yields a reduction of 40% (30-10=20, 20÷5=4, 4×10 = 40%).
Keep the supply duct length as close to 10 feet as possible but
never less than 6 feet. Use the fewest number of bends if possible.
• 24-inch Rule: Use at least 24 inches of straight plenum before any
fitting, such as an elbow, tee, or takeoff. Electric duct heaters
require 48 inches. Avoid elbows directly off units. The maximum
total plenum length should be restricted to 150 ft. For the plenum,
maximize length and minimize restrictions.
• 60/40 Rule: When using a tee, split the flow as close to 50/50 as
possible, no more than 60/40. Always use a turning vane.
 • 70/30 Rule: Turn the tee 90° to make a side branch with no more
than 30 percent of the air. Do not use a turning vane.
• Takeoffs: Maintain distance between takeoffs as evenly as
possible. Space the takeoffs at least 6 inches apart and 12 inches
from the end cap.
• Fittings: Use long and radii duct fittings instead of short or
mitered fittings wherever possible.
• Add acoustic silencers on the fan discharge.
• Use flexible ducts only at the terminal ends of diffusers. The
flexible ducts should be completely stretched and restricted to a
6-foot length.
• Duct Hanger Spacing: (according to SMACNA requirements). The
Maximum ductwork hanger spacing:
– Horizontal: 8 feet maximum
– Vertical: 16 feet and at each floor
however, Engineering Practices suggest:
– Horizontal Ducts with A < 4 ft2 ⇒ 8 feet maximum
– Horizontal Ducts with 4 ft2<A<10 ft2 ⇒ 6 feet maximum
– Horizontal Ducts A> 10 ft2⇒ 4 feet maximum
– Vertical Round Ducts: 12 feet maximum
– Vertical Rectangular Ducts: 10 feet maximum
• Duct insulation and sealing: in general, One-inch-thick fiberglass
blanket (Rth = 16) is sufficient for thermal protection.
Steps for balancing the System
1. Inspect the complete system, meaning locate all ducts, openings,
and dampers.
2. Open all dampers in the ducts and at the grilles.
3. Check the velocities at each outlet.
 4. Measure the "free" grille area.
5. Calculate the air flow volume at each outlet.
6. Total the cfm.
7. Determine the floor areas of each room. Then get Atot
𝐴
8. Get the cfm for each room: 𝑐𝑓𝑚𝑟𝑜𝑜𝑚 = (𝐴𝑟𝑜𝑜𝑚) 𝑐𝑓𝑚𝑡𝑜𝑡𝑎𝑙
𝑓𝑙𝑜𝑜𝑟
𝑎𝑟𝑒𝑎
9. Adjust duct dampers and grille dampers to obtain these values.
10. Recheck all outlet grilles. In some cases, it may be necessary to
overcome excess duct resistance by installing an air duct booster.
These are fans used to increase air flow when a duct is too small,
too long, or has too many elbows.
Duct supports and joints- see SMACNA shared file
The support distance for duct work is typical based upon deflection,
stress and cylinder buckling analysis.

spacing

For sheet metal ductwork, the space between the hangers should not
exceed the distances listed below:

Horizontal Duct Vertical duct

Ducts Size Ducts Type Max. Spacing, ft (m)
Max. Spacing, round 12 (3.6)
Area, Diameter
ft (m)
ft2 (m2) inch (mm) rectangular 10 (3)
A < 4 (0.4) D < 5 (125) 8 (2.5)
4 (0.4) < A < 10 (1) 5 (125) < D < 38 (1000) 6 (2)
A > 10 (1) D > 38 (1000) 4 (1.2)
 Typical Duct Fabrication and Supporting

Typical 12mm diam Galvanized 24 Aluminum

duct hanger sheet metal cladding for
exposed duct only

25mm thick Fire

retardant Fiberglass duct
insulation with Aluminum
foil vapor barrier

End tabs or flaps for

Air duct joints formed by stripping
from the vapor barrier
seal overlaps with a quick
tacking adhesive
Angle undercut with
Flame resistant quick neoprene packer
Tacking adhesive to be
Galvanized 18 Tie wire
coated 100% of duct 50mm x 50mm x 102mm
spaced @ 305mm
surface area Gauge #20 Galvanized
sheet metal corner beading
Typical Upper Hanger Attachment
Typical Lower Hanger Attachment
DUCT CLEANING
There is also the question of when duct cleaning should be done, and
how the job could be validated. Duct cleaning may be warranted in any
of the following situations:
• There is substantial evidence of visible mold growth inside the
hard surface of ducts or on other components of the heating and
cooling system. Get laboratory analysis carried out for the sample
before taking the decision.
• If you have insulated air ducts and the insulation gets wet or
moldy, it cannot be effectively cleaned and should be removed
and replaced.
• Ducts are infested with vermin, e.g. (rodents or insects).
• Ducts are clogged with excessive amounts of dust or debris, that
me be released into the home from the supply registers.
Duct Cleaning Methods
Methods of duct cleaning vary, although standards have been
established by industry associations concerned with air duct cleaning.
Typically, a service provider will use specialized tools to dislodge dirt
and other debris in ducts, and then vacuum them out with a high-
powered vacuum cleaner. Common duct cleaning methods include: (1)
Contact vacuuming, (2) Air sweeping and (3) Power brushing.
1. Vacuum Method: vacuuming involves cleaning the interior duct
surfaces by way of existing openings and outlets or, when
necessary, through openings cut into the ducts. The vacuum unit
should only use HEPA (High Efficiency Particle Air) collection
equipment; conventional equipment may release extremely fine
particles into the atmosphere, instead of gathering them.
 Starting at the return side of the system, the vacuum cleaner head
is inserted into the section of the duct to be cleaned at the
opening furthest upstream, and then the vacuum cleaner is
turned on. Vacuuming proceeds downstream slowly enough to
allow the vacuum to pick up all dirt and dust particles.

Contact vacuuming usually requires larger access windows than

other methods, in order to allow the cleaning equipment to reach
into the last corner of the duct. The head of the vacuum cleaner is
introduced into the duct using the nearest opening at the beginning
of the duct network. Hovering then starts, following the direction of
air flow, and slowly enough to capture and gather up all the dirt.
2. Air Sweep Method: In the air sweep or air washing method, a
vacuum collection unit is connected to the downstream end of the
duct section. The vacuum unit should use HEPA filtering, if it is
exhausting into an occupied space. The isolated section of duct
being cleaned should be subjected to a minimum of 1" negative air
pressure to draw loosened materials into the vacuum collection
system. Take care not to collapse the duct.
3. Power (Mechanical) Brushing Method:
In this method, a vacuum collection unit is connected to the duct in
the same way as with the air sweep method. Pneumatic or electric
rotary brushes are used to dislodge dirt and dust particles, which
become airborne and are then drawn into the vacuum unit. Brushing
operations will usually require larger access openings than the
previous method. Nevertheless, fewer openings are needed. Certain
types of mechanical brushes can reach up to 24 ft. in both directions.
ROUND DUCT FITTINGS & TRANSITIONS
RECTANGULAR DUCT FITTINGS & TRANSITIONS

End of this module