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Method of Duct Design Lecture No.(1) By Badran M. Salem
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Advanced Air Duct Design
Introduction
Correct air diffusion, as well as the proper quantity of conditional air
conditioned, is essential for comfortable conditions in forced systems. A
well designed air duct system, either commercial or industrial, must
consider most of the following system design factors: (1) space
availability; (2) space air diffusion; (3) noise levels; (4) duct leakage; (5)
duct heat gain and losses; (6) balancing; (7) fire and smoke control; (8)
initial investment cost; (9) system operating cost.
Any deficiency in duct design may result in a system that does not
operate properly. These deficiencies include system, which are
excessively expensive to own and / or operate. Poor air distribution can
cause discomfort, and lack sound attenuators may result in objectionable
noise levels. Poor duct construction or lack of duct sealing can cause
inadequate airflow rates at the terminal units. Inadequate airflow is also
caused by excessive heat gains/losses, which can be avoided with proper
duct insulation. Poor design of the branches concerning main ducts may
result in unbalanced systems. As a part of our course, we redesigned the
duct system in the building. Moreover, we resized all the ducts depending
on the actual data that we have calculated.
Duct Design MethodsThe most common methods of air duct system design are: (1) equal
friction, velocity reduction, static regain and variations such as total
pressure, and constant velocity. The choice of design method is the
designer s and the system design with the minimum owning and
operating cost depends on both the application and ingenuity of the
designer. No single duct design method will automatically produce the
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Method of Duct Design Lecture No.(1) By Badran M. Salem
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progressively in the direction of the return intakes. With the ducts sized
and the fittings known, the total pressure losses can be calculated, the
pressure gradients plotted, and the minimum pressure loss or critical path
of the system established.
A refinement of this method involves sizing the branch ducts to
dissipate the pressure available at the entrance to each. The pressure loss
of the ductwork between the fan and first branch take-off is subtracted
from the total fan pressure to obtain the available pressure at the first
junction. Through trial, a branch velocity is found that results in the
branch pressure loss being equal to or less than the pressure available.The procedure is repeated for each branch.
Static Regain Method
The static regain method is design procedure in which the ducts are
sized so that the increase in static pressure (static regain) at each take-off
offsets the pressure loss of the succeeding section of ductwork. Thismethod is especially suited for high velocity installations having long
runs with many take-offs or terminal units. With this design procedure,
approximately the same static pressure exists at the entrance to each
branch, which simplifies outlet or terminal unit selection and system
balancing. With the ductwork sized by this method, the systems total
pressure losses can be calculated. The major disadvantage of this method,however, is that excessively large ducts (low velocities) result at the end
of long duct runs.
The total pressure design method is adaptation of the static regain
method. This method is advantageous since a designer has knowledge of
the intermediate system pressures and control of duct sizes and velocities.
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Constant Velocity Method
Since the constant velocity design method is generally applied to
exhaust systems conveying particulates, and since these systems are
usually round, this method is applied for round ducts only.
Duct Design Procedures
The general procedure for duct design is as follows:
1. Study the plans of the building and arrange the supply and return
outlets to provide proper distribution of air within each space. Adjustcalculated actual air quantities for duct heat gains or losses and duct
leakage. Also, adjust the supply, return, and/or exhaust air quantities
to meet space pressurization requirements.
2. Select outlet sizes from manufacturers data.
3. Sketch the duct system, connecting the supply outlets and return
intakes with the central station apparatus, taking recognizance of the building construction, and avoiding all structural obstructions and
equipment. The space allocated for the supply and return ducts often
dictates the system layout and the shape of the ductwork.
4. Determine the size of main and all branches by the selected design
method.
5.
Calculate the total pressure requirements of all duct sections, bothsupply and return, and then plot the total pressure grade line.
6. To design a system with the minimum owning and operating costs,
repeat steps 4 and 5 with different duct sizes. It is necessary to
estimate the cost of the initial design and the incremental cost
variations due to the redesigns.
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7. Layout the system in detail. If significant duct routing and fitting
variations have occurred from the original design, recalculate the
pressure losses.
8. Redesign the duct branches to minimize the balancing necessary by
dampers.
9. Analyze the design for objectionable noise levels and specify sound
attenuators as necessary.
10. Select the fan.
Air Duct Design
Equal Friction Method
This method is used to determine the diameter of air duct, so that
the duct friction loss per unit length at various duct section always remain
constant. The final dimensions of sized ducts should be rounded to
standard size. The procedure of this method is as follows:
1. Select a suitable velocity in the main duct from sound level
considerations as given in the table.
2. Knowing the air flow rate from the load estimation.
3. The sized of the main duct and friction loss are determined.
4. The remaining ducts are then sized respectively.
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Example:
The system shown below, find the size of the main duct and branches.
First we will use the Equal friction method.
The main duct A-B.
The total flow rate is,
smQ
QQQQ
t
t
/8134 3221
We assume the velocity in the main duct A B and size the duct
md
V d Q
smV
t
128.1884
4
/82
Friction losses in the main duct A-B
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2
2V d fL
P
Assume the roughness of the surface from the table and calculate the
value,
000133.0128.1
00015.0d
We can calculate the Reynolds number for air in the main duct after
finding the viscosity and density of the air from the table.
The properties of the air is at 25 oC.
56 1080.510413.18
128.18184.1Re
Vd
From Moody chart, we can find the fraction factor, f
m Pa L P
Pa P
f
/487.0
61.142
8128.1
300145.0184.1
0145.02
The branch B-E.
42
22
2
82
,4
d
QV
d
QV
AV Q
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smd Q
V
m L P
fQ
d
d Q
d f
L P
/281.54
491.0487.0
10145.0184.188
8
2
525 2
2
42
2
By using the value of d and V, we can calculate the Reynolds number
again and modify the value of d and V if possible as follows,
The first modification of d and V gives,
000305.0491.0
00015.0
1066.110413.18
491.0281.5184.1Re 56
d
Vd
016.0 f
smV
md
/073.5
501.0
The second modification of d and V gives,
016.0
000299.0513.0
00015.0
1063.110413.18501.0073.5184.1
Re 56
f d
Vd
The same fraction factor, so the value of d and V does not change.
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smV
md
/073.5
501.0
We can check the volume flow rate is less or higher the design value,
smV d AV Qcal /000069.1073.5501.044322
This is acceptable value.
The branch B-C.
smd Q
V
m
L P
fQd
/489.7091.1744
091.1487.0
7016.0184.188
22
52
2
5 2
2
The first modification of d and V gives,
000137.0091.1
00015.0
1025.510413.18
0696.179.7184.1Re 56
d
Vd
0146.0 f
The same fraction factor we used, so the value of d and V does not
change.
smV
md
/768.7
071.1
The second modification of d and V gives,
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0146.0
00014.0071.1
00015.0
1035.510413.18
071.1768.7184.1Re 56
f d
Vd
The same fraction factor we used, so the value of d and V does not
change.
smV
md
/768.7
071.1
We can check the volume flow rate is less or higher the design value,
smV d AV Qcal /998.6768.7071.144322
This is acceptable value.
The branch C-F.
smd Q
V
m
L P
fQd
/557.67621.0
344
7632.0487.0
30146.0184.188
22
52
2
5 2
2
The first modification of d and V gives,
0163.0
000197.07632.0
00015.0
1022.310413.18
7632.0557.6184.1Re 56
f
d
Vd
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smV
md
/275.6
780.0
The second modification of d and V gives,
0164.0
000192.0780.0
00015.0
1015.310413.18
780.0275.6184.1Re 56
f d
Vd
The same fraction factor we used, so the value of d and V does not
change.
smV
md
/275.6
780.0
We can check the volume flow rate is less or higher the design value,
smV d AV Qcal /9984.2275.678.044322
This is acceptable value.
The branch C-D.
smd Q
V
m
L P
fQd
/630.68551.0
444
8764.0487.0
40164.0184.188
22
52
2
5 2
2
The first modification of d and V gives,
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0157.0
000171.08764.0
00015.0
1074.310413.18
8764.0630.6184.1Re 56
f d
Vd
smV
md
/747.6
869.0
The second modification of d and V gives,
0159.0
000173.0869.0
00015.0
1077.310413.18
869.0747.6184.1Re 56
f d
Vd
smV
md
/747.6
869.0
The difference between second and first modification is not so much, so
the value of d and V does not change.
We can check the volume flow rate is less or higher the design value,
smV d AV Qcal /002.4747.6869.044322
This is acceptable value.
The results of calculation must be illustrated in a table to use it in the
calculation of air fan power.
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Results
Section L (m) Q (m3/s) V (m/s d (m) Qcal
A-B 30 8 8.00 1.128 8.000
B-C 15 7 7.77 1.071 6.998
C-D 75 4 6.75 0.869 4.002
B-E 30 1 5.07 0.501 1.000
C-F 15 3 6.275 0.780 2.998
For galvanized steel air duct and air properties at 20 oC, the following
charts could be used for estimate the duct size and velocity at a given
pressure drop per meter and volume flow rate.
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Duct friction chart-low flow rate.
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Duct friction chart-high flow rate.
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The above two charts recommended only for air and the duct which
made from galvanized steel with = 0.00015 m and rounded section. But
if another duct material it should be use correction factor.
We now recalculate the given example by using the duct chart.
The main duct A-B.
The total flow rate is,
smQ
QQQQ
t
t
/8134 3221
We assume the velocity in the main duct A B and size the duct
smV /8
From chart at Q = 8 m3/s and V = 8 m/s, the equivalent diameter and pressure loss are,
m Pa P
md
/5.0
135.1
We use the value of P=0.5 Pa/m is constant through all branches and
determine the velocity and diameter from chart and tabulated the resultsas follows,
Resluts
Section L (m) Q (m3/s) V (m/s d (m) Qcal
A-B 30 8 8.00 1.135 8.094
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Velocity Reduction Method
In this method the main duct is designed in the same manner as in
the equal friction method. Thereafter, arbitrary reductions are made in the
air velocity as we go down the duct run. Equivalent diameters are found,
as before from the friction chart. We now recalculate the given example
by using the velocity reduction method and using the chart.The main duct A-B.
The total flow rate is,
smQ
QQQQ
t
t
/8134 3221
We assume the velocity in the main duct A B and size the duct
smV /8
From chart at Q = 8 m3/s and V = 8 m/s, the equivalent diameter and
pressure loss are,
m Pa P
md
/5.0
135.1
B-C 15 7 7.8 1.080 7.145
C-D 75 4 6.9 0.875 4.149
B-E 30 1 4.80 0.515 0.999
C-F 15 3 6.30 0.780 3.010
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We assume the velocity as follows,
B-C, V = 7 m/s
B-E, V = 7 m/s, C-D, C-F, V = 6 m/s.
After that, we determine the duct diameter and friction loss from the chart
and put the results in table as follows,
Results
Section L (m) Q (m3/s) V (m/s d (m) Qcal
A-B 30 8 8.00 1.135 8.094
B-C 15 7 7.00 1.130 7.020
C-D 75 4 6.00 0.925 4.032
B-E 30 1 7.00 0.430 1.017
C-F 15 3 6.00 0.800 3.016
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Air Fan Power
Total, Static and dynamic Pressure.
Bernoullis equation
V V
S T S T V
T
V S
T V S
T Z V S
P V
V P
P P P P P
P
P P
P P P dZ
P P P P
C gZ V
P
C gZ V P
2or
2
Pitotube. bymeasuringare,where,
Pressure.Total:
PressureDynamicorPressureVelocity:Pressure,Static:Where,
,,0If
2
2
2
2
2
For frictionless flow between two sections as follows,
T V S V S P P P P P 2211
But due to friction loss
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2.and1 betweenlosseshydraulicordrop pressuretotal:where,2211
L
LV S V S
P
P P P P P
If the Fan or Blower is introduced between two section .
duct.in thedrop pressurelossestherepresents:where,222
pressure.headFan total the
calledisandFan workthetoduerise pressuretheiswhere,
222211
2211
L
LS S
LV S V S
h
hV P W V P
FTP
P P P FTP P P
As flow continues in a duct, the static pressure of air continuously drops.
This drop in pressure takes place due to two factors,
Duct to friction (friction loss)
diameter meanhydraulic: where,,2
2
d V
d fL
P F
Change of direction or velocity (dynamic loss)
constant:where,,2
2
K V
K P M
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Method of Duct Design Lecture No.(1) By Badran M. Salem
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For Enlargement.
21122211
V V S S
V S V S
P P P P
P P P P
Due to friction loss,
221121 V S V S T T L P P P P P P P
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Air flow through a simple duct system.
Suction-side
Friction of inlet grill, P i :
)(
)()(
Similarly,)(
)(
)(
0
552215
52215225
2212
2112
11
111
1
V iS
iT T
V iS
T T
V iS
V S iT
iT
P P P P P
P P P P P P
P P P P
P P P
P P P
P P C P
P P
Discharge-side.
The pressure loss at outlet in discharge grill, P e :
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Method of Duct Design Lecture No.(1) By Badran M. Salem
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8865221
5221886
56
6886666
8868686
888
88
)(
FTPPressure,TotalFan
V ei
iV e
T T
V V eV T S
V eT T
eV T S
V eT
P P P P P P
P P P P P P
P P FTP
P P P P P P P
P P P P P P
P P P P
P P P
By applying the modified Bernoullis equation between 5 -6,
Constant, , Z, ,0
adiabatic,is processfantheIf
)(2
656565
56
25
2656
.
6565
Z V V Q
Z Z g V V P P
mW Q S S
886655221.
552216886
.
65
Power
Power
V eV V i
V iV V e
P P P P P P P P m
P P P P P P P P m
W
If P V5 =P V6 , V 5 =V 6
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Method of Duct Design Lecture No.(1) By Badran M. Salem
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T act
e
ed F
V V K V d L f m
V P P
m
Power Power
is, powerFanactualThe
,efficiencytotalhasFanairtheIf
222
2Power
T
222.
2.
Air Fan System Characteristics.
Consider a straight air duct system in which the total pressure drop
is calculated by adding the losses of different sections, which are in turn
proportional to their respective velocity pressure as follows:
222
2.
2.2.2.
222
1211
211
21
21
21
21
21
21
21
A A K
Ad L
f V P
AV
AV
K AV
d L
f P
V V K V d L f P
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Method of Duct Design Lecture No.(1) By Badran M. Salem
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222
.2
2.
.
1211
211
21
where,
,
follows,aswirtten becanlossdynamicandfrictionThe
rate.flowair volumetheistheduct,airof sectiononeFor
A A K
Ad L
f R
V
P R
V R P
V
Where, R is the air duct flow resistance. So, by analogy with electricity,
we can derive the concept of resistance R of the duct system.
Air Duct System in Series.
The friction loss and dynamic loss be calculated in each section
and we use the equation power directly as follow:
321 P P P P
Air Duct System in Parallel.
The friction loss and dynamic loss be calculated in each branch and we
use the equation power directly as follow:
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Method of Duct Design Lecture No.(1) By Badran M. Salem
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321
321
.
3
.
2
.
1
.
1111
R R R R
R P
R P
R P
R P
V V V V
t
t
After could calculate the friction and dynamic loss in air duct system in
series or in parallel and calculate the equivalent resistance, we will tray to
calculate the power required in the last example.
Power of Air FanTo calculate the Fan power we reshape the duct system again
according to the diameter which we calculated as follows.
The pressure drop in air handling unit is as follows:
Damper : 050 Pa
Filter : 100 Pa
Cooling coil : 150 Pa
Eliminators : 050 Pa
Heating Coil : 150 pa
Mixing and suction to fan : 050 Pa
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The dynamic loss coefficients K is as follows:
Fan discharge to main duct : 0.30Standard 90 oC elbow : 0.75Reduction : 0.05Exit grille : 0.50
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Reshaping the duct system with air handling unit.
The Air Duct System can be simplified as follows:
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Air Handling unit, R 1 :
4
22.1
2
/12581.78
527.464
/527.464
184.1501505015010050
mV
P R
sm
P P
AH
AH
Section A-B, R 2 :
42
222
/14285.0)128.130
0209.03.0(9986.01
21
1
2
11
2
1
m R Ad
L f
A
K R
Section B-C, R 3 :
43
223
/12112.0)07.1
150208.005.0(
8086.01
21
1211
21
m R
Ad L
f A
K R
Section C-D, R 4 :
44
4
224
/10596.6
)1868.075
0225.050.075.005.0(3502.01
21
1211
21
m R
R
Ad L
f A
K R
Section B-E, R 5 :
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Method of Duct Design Lecture No.(1) By Badran M. Salem
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45
5
225
/1151.44
)1513.0
30026.050.075.0(
0427.0
1
2
1
1211
21
m R
R
Ad L
f A
K R
Section C-F, R 6 :
46
6
226
/1970.5
)1778.015
0232.050.075.0(2259.01
21
1211
21
m R
R
Ad L
f A
K R
R 4 , R 6 in parallel shape, R t1 :
41
641
/15149.1
970.5
1
0596.6
1111
m R
R R R
t
t
R 3 , R t1 in series shape, R t2 : 4
132 /17261.15149.12112.0 m R R R t t
R 5 , R t2 in parallel shape, R t3 :
4
3
253
/120324.1
7261.1
1
151.44
1111
m R
R R R
t
t t
R 1 , R 2 , R t3 in series shape, R t :
4321 /18895.820324.14282.02581.7 m R R R R t t
222.
/930.56888895.8 smV R P P
t d F
Fan Power.
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Method of Duct Design Lecture No.(1) By Badran M. Salem
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KW W V R
V RV P P
m
t
t d F
4.5Power 886.538888895.8184.1Power
Power
33.
2...
A First Approximate Method.
We can apply the modified Bernoullis equation on the longest line only,from A to D as follows,
222Power 222.
eV V d L
f V
K mW
KW
W DC
C B
B AC A
7.5Power
948.5679332.97411.10387.27527.4648184.1Power
2
78.6)15.075.005.0(
184.1
702.075
279.7
05.0184.1
702.015
28
3.0184.1
702.030184.1
501505015010050
8184.1Power
2
2
2
/
A Second Approximate Method.
We apply modify Bernoullis equation in all section of the duct system asfollows,
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E B
DC
C B
B AC A
284.4
)15.075.0(184.1
702.0301184.1
278.6
15.075.005.02
702.0754184.1
2
79.705.0
184.1
702.0157184.1
28
3.0184.1
702.030184.1
5015050150100508184.1Power
2
2
2
2
/
KW
W F C
4.5Power
750.5350806.177263.52037.375234.8641.4659Power
231.6
)15.075.0(2702.015
3184.1
2
Problems in Air Duct Design
1- For air conditioning system shown below, calculate the duct size and
air velocity in each branch by using equal friction and velocity
reduction method. For the longest branch A-E and by using modified
Bernoull equation, estimate the power required for the electric air fan.
Assume that = 0.13.
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Method of Duct Design Lecture No.(1) By Badran M. Salem
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Modified Bernoull Equation.
2- Calculate the duct size and velocity of each branch in the below
network and also suggest a suitable design structure to calculate the
power required. The air distributor gives equal flow rate of 15 m 3/min.
3- In the duct layout shown below, the outlets are deliver 25 m 3/min at 1,
15 m 3/min at 2 and 30 m 3/min at 3. Also, select air velocity of 8 m/s in
the section A. Determine the size of duct system using equal friction
method and determine the static pressure requirement for the air fan.
22)(222
12
21
2212 V
K
V
d
fL
z z g
V V P P
mW Qt
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4- Size the duct in the problem 3 by using Velocity reduction method and
find the power required for the air fan.
5- For a system shown below, size the ducts on a rate of friction pressure
drop of 0.7536 Pa/m and the air flow rate from the fan is 4 m 3/s. The
two outlets delivers equal masses of air.
a) Modify the diameter of branch duct to outlet 1 so that no
damper is required at the outlet.
b) Calculate the fan total and static pressure, also the power
required.
The pressure drops in each equipment are as follows:
Filter 100 Pa. Damper 50 Pa.
Cooling coil 150 Pa. Mixing section 50 Pa.
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The dynamic loss coefficient K for all expander are to be taken as
applying to the difference the upstream and downstream velocity
pressure, and for all reducers as applying to the downstream velocity
pressure only. The values are given in the following table.
SectionK
Condition
Inlet 1.4 Mean face velocity = 4 m/s
Expander AB 0.35M ean face veloci ty at fi l ter = 1.5 m/s
Reducer BC 0.02 Mean face velocity at damper = 3 m/s
Reducer EF to fan section 0.02
Reducer GH at fan
discharge
0.3
Straight through duct
suction.
0.25
Elbow 0.23
Grille 0.5