Advanced Air Duct Design Part 1

8/13/2019 Advanced Air Duct Design Part 1

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Method of Duct Design Lecture No.(1) By Badran M. Salem

1

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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3

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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4

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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5

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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6

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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7

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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8

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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9

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


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



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10

0146.0

00014.0071.1

00015.0

1035.510413.18

071.1768.7184.1Re 56

f d

Vd


change.

smV

md

/768.7

071.1


smV d AV Qcal /998.6768.7071.144322


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


0163.0

000197.07632.0

00015.0

1022.310413.18

7632.0557.6184.1Re 56

f

d

Vd


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11

smV

md

/275.6

780.0


0164.0

000192.0780.0

00015.0

1015.310413.18

780.0275.6184.1Re 56

f d

Vd


change.

smV

md

/275.6

780.0


smV d AV Qcal /9984.2275.678.044322


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



12/35


12

0157.0

000171.08764.0

00015.0

1074.310413.18

8764.0630.6184.1Re 56

f d

Vd

smV

md

/747.6

869.0


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.


smV d AV Qcal /002.4747.6869.044322


The results of calculation must be illustrated in a table to use it in the

calculation of air fan power.


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13

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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14

Duct friction chart-low flow rate.


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15

Duct friction chart-high flow rate.


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16

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.


smQ

QQQQ

t

t

/8134 3221


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


A-B 30 8 8.00 1.135 8.094


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17

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.


smQ

QQQQ

t

t

/8134 3221


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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18

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


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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19

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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20

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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21

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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22

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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23

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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24

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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25

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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26

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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27

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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28

Reshaping the duct system with air handling unit.

The Air Duct System can be simplified as follows:


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29

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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30

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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31

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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32

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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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


34/35


34

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.


35/35


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

Documents

Advanced Air Duct Design Part 1