ac duct design

7/28/2019 AC Duct Design

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Website: http://guider55.4t.com Refrigeration and Air Conditioning

Air Duct Design

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..

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o20.

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http://guider55.4t.com/http://guider55.4t.com/


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.

.

)Methods of Air Duct Design(

.

.

.

.

)(5.0 222

121 VVP = .

sm /10.

)Equal friction method(

)-(

)-(

)-(.


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Table (10-1) Recommended maximum main duct velocity in m/s.

Main Duct BranchesApplication

Supply Return Supply Return

Residences, Homes 5 4 3 3

Hotels, Hospitals 8 7 6 5Offices 10 8 8 6

Theatres 7 6 5 4

Banks, Restaurants 10 8 8 6

Table (10-2) Air dynamic viscosity and density

Temperature, Co sPa. kg/m3

10 17.708 x 10-6 1.247

20 18.178 x 10-6 1.204

25 18.413x 10-6

1.18430 18.648 x 10

-61.164

40 19.118 x 10-6

1.127

50 19.588 x 10-6 1.092

Table (10-3) Surface roughness of some duct materials

Material Roughness, , mm

Cost Iron 0.26

Steel commercial 0.046

Concrete 0.3 3.0Wood 0.18 0.9

Galvanized Iron 0.15

Aluminum 0.04

Fiber glass 0.10

Smooth Glass, Plastic 0.00

Example 10-1

An air duct system shown below, determine the size of main duct and branches

using equal friction method.


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Main DuctAB

.

smQ

QQQQ

t

t

/8134 3

221

=++=

++=

sm /10.

Assume the velocity in main duct AB

smV /8=

VdQt

2

4

=

md 128.18

84=

=

2

2V

d

fLP =

000133.0128.1

00015.0 ==d

.

5

61080.5

10413.18

128.18184.1Re =

==

Vd

Moody chartRed/f

.

mPaL

P

PaV

d

fLP

f

/487.030

61.14

61.142

8

128.1

300145.0184.1

2

0145.0

22

==

=

==

=

mPa /


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.

Branch BE

fdV

.

42

22

42

22

2

2

8

2,

16,

4

4

d

QV

d

QV

d

QV

VdAVQ

===

==

mLP

fQd

d

Qf

d

Q

d

fV

d

f

L

P

491.0487.0

10145.0184.18

/

8

88

2

52

52

2

52

2

42

22

=

=

=

===

smd

QV /281.5

42

==

dVReMoody chartf

.

First correction of d and V

000305.0491.0

00015.0

1066.110413.18

491.0281.5184.1Re 5

6

==

=

==

d

Vd

016.0=f

smV

md

/073.5

501.0

==

Second correction of dand V

016.0

000299.0501.0

00015.0

1063.110413.18

501.0073.5184.1Re 5

6

=

==

=

==

f

d

Vd

dV.

smV

md

/073.5

501.0

==


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dV

.

smVdAVQcal

/000069.1073.5501.0

44

322 ====

Acceptable value of dand Vfor smQ /1 3= .

Branch BC

fdV

.

smd

QV

mLP

fQd

/489.7091.1

744

091.1487.0

7016.0184.18

/

8

22

52

2

52

2

=

==

=

=

=

dVReMoody chartf

.


000137.0091.1

00015.0

1025.510413.18

091.1489.7184.1Re 5

6

==

=

==

d

Vd

0146.0=f

smV

md

/768.7

071.1

==


0146.0

00014.0071.1

00015.0

1035.510413.18

071.1768.7184.1Re

5

6

=

==

=

==

f

d

Vd

dV.

smV

md

/768.7

071.1

==

.


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smVdAVQcal /998.6768.7071.144

322 ====


Branch CFfdV

.

smd

QV

mLP

fQd

/557.67621.0

344

7632.0487.0

30146.0184.18

/

8

22

52

2

52

2

=

==

=

=

=

dVReMoody chartf

.


0163.0

000197.07632.0

00015.0

1022.310413.18

7632.0557.6184.1Re 5

6

=

==

=

==

f

d

Vd

smV

md

/275.6

780.0

==


0164.0

000192.0780.0

00015.0

1015.310413.18

780.0275.6184.1Re 5

6

=

==

=

==

f

d

Vd

.

smV

md

/275.6

780.0

==

.smVdAVQcal /9984.2275.678.0

44

322 ====



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Branch CD

fdV

.

smd

QV

mLP

fQd

/630.68551.0

444

8764.0487.0

40164.0184.18

/

8

22

52

2

52

2

=

==

=

=

=

dVReMoody chartf

.


0157.0

000171.08764.0

00015.0

1074.310413.18

8764.0630.6184.1Re 5

6

=

==

=

==

f

d

Vd

smV

md

/747.6

869.0

==


0159.0

000173.0869.0

00015.0

1077.310413.18

869.0747.6184.1Re 5

6

=

==

=

==

f

d

Vd

smV

md

/747.6

869.0

==

V, d.

.smVdAVQcal /002.4747.6869.0

44

322 ====


.


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Results

(Pa/m)LP 487.0/ =

Section )( mL )/( 3 smQ )/( smV )( md )/( 3 smQcal f

A-B 30 8 8.00 1.128 8.000 0.0145

B-C 15 7 7.77 1.071 6.998 0.0146

C-D 75 4 6.75 0.869 4.002 0.0159

B-E 30 1 5.07 0.501 1.000 0.016

C-F 15 3 6.275 0.780 2.998 0.0164

CT o20=

barP 01325.1=.

smQt /83=

.

smV /8=

smQ /8 3=smV /8=

)-(mPaP /5.0=

.

Fig. 10-1 Duct friction chart


10/18


.

Results

(Pa/m)LP 5.0/ =

Section )( mL )/( 3 smQ )/( smV )( md )/( 3 smQcal

A-B 30 8 8.00 1.135 8.094

B-C 15 7 7.80 1.080 7.145

C-D 75 4 6.90 0.875 4.149

B-E 30 1 4.80 0.515 0.999

C-F 15 3 6.30 0.780 3.010

)city reduction methodVelo(

.

.

smQt /83=

.

smV /8=

.

mPaP

md

/5.0

135.1

==

.

Assume the velocity in each branch with gradually reduction.

smVFC

smVDC

smVEB

smVCB

/6:

/6:

/7:

/7:

====

.


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Results

Section )( mL )/( 3 smQ )/( smV )( md )/( 3 smQcal (Pa/m)P

A-B 30 8 8.00 1.135 8.094 0.50

B-C 15 7 7.00 1.130 7.020 0.38

C-D 75 4 6.00 0.925 4.032 0.38

B-E 30 1 7.00 0.430 1.017 1.35

C-F 15 3 6.00 0.800 3.016 0.45

BE

)/(5.4 sm

.

Results after decreasing the velocity in branch BE

Section )( mL )/( 3 smQ )/( smV )( md )/( 3 smQcal (Pa/m)P

A-B 30 8 8.00 1.135 8.094 0.50

B-C 15 7 7.00 1.130 7.020 0.38

C-D 75 4 6.00 0.925 4.032 0.38

B-E 30 1 4.50 0.54 1.031 0.40

C-F 15 3 6.00 0.800 3.016 0.45

.

( )

( ) 25.0

625.0

3.1ba

abd

+= (10-1)

)Air outlets(


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m8.1Co1

sm /15.0sm /3.0

.

)Types of Air Outlet(

)Grille outlets(

.

)Ceiling diffuser(

.

)Slot diffuser(

)-(

.

Fig. 10-2 Types of air outlet

sm /5.3~5.2

sm /25.0

XBlow or Throw,

)-(X

L.


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Table (10-4) Blow or throw length of some outlets

Outlet )/( 2mWQ

LX/ Outlet )/( 2mWQ LX/

Grille wall

250

190

125

65

1.8

1.8

1.6

1.5

Ceiling diffuser

Round or Square

250

190

125

65

0.8

0.8

0.8

0.8

)Air Fan(

.

:

2

2

221

2

11

22gZ

VPgZ

VP++=++

(10-2)

Or, CgZVP =++ 2

2

TZVS PPPP =++ (10-3)

If no difference between two sections, 0=dZ

TVS PPP =+ (10-4)

Where, SP static pressure, VP , velocity pressure and TP total pressure.

:

LVSVS PPPPP ++=+ 2211 (10-5)

Where,L

P total pressure drop of friction and minor losses between 1 and 2.


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:

LVSVS PPPFTPPP ++=++ 2211 (10-6)

Where, FTP is the pressure rise due to fan work and is called Fan Total Pressure.

LP

.

MFL PPP += (10-7)

Friction losses,2

2V

d

fLPF =

Minor losses,2

2VKPM

=

:

++=

+

+

=

222

2Power

222

2

e

eMF

VVK

V

d

Lfm

VPPm

&

&

(10-8)

)-

(

.


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Table (10-5) pressure drop in air handling unit and fitting coefficient

Component (Pa)P Fitting KDamper 50 Round inlet 0.4

Filter 100 Fan discharge to main duct 0.3

Cooling coil 150 Branch from main duct 0.4

Eliminator 50 Standard 90o

elbow 0.75

Heating coil 150 Expander 0.75

Air washer 100 Reducer 0.05

Mixing to air fan 50 Damper 0.2

Exit grille 0.5

)Fan Power(

:

Fan Power by applying Modified Bernoullis Equation as

+

+

+

=

+++

=

Exit

e

FittingFrictionAir

e

Air

VVKL

L

PPm

VVK

V

d

Lf

Pm

22

1

222Power

22

Handling

222

Handling

&

&

(10-9)

Example 10-2In example 10-1, reshape the duct system and estimate the required power of the

air fan.

Solution

After reshaping the air duct system and making a clear layout, with applying the

energy equation, the power required for air fan can be obtained.


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Reshaping the air duct system with air handling unit

+++

==

222Power

222

Handling

e

Air

VVK

V

d

Lf

PmW

&

+

+

+

=

Exit

e

FittingFrictionAir

VVKL

L

PPm

22

1

22

Handling

&


17/18


+

+

+

=

Exit

e

FittingFrictionAir

VVKL

L

PPV

22

1Power

22

Handling

&

CB

BA

HandlingAir

+

+

+

+

+++++

=

2

77.705.0

184.1

487.0157184.1

2

83.0

184.1

487.030

184.1

501505015010050

8184.1Power

2

2

( )

kW

W

FC

EB

DC

303.5Power

856.5302005.169566.4683.415644.63811.4607Power

2

28.6)15.02.04.0(

184.1

487.0153184.1

2

07.5)15.02.04.0(

184.1

487.0301184.1

2

75.615.075.02.005.0

184.1

487.0754184.1

2

2

2

==++++=

++++

+

+++++

++++++

+

Problems

1- For air duct system shown below, calculate the size and air velocity in each

branch with using equal friction method. Estimate the power required for air

fan. Assume air flow rate per exit is sm /1 3 .

2- Calculate the size and air velocity with velocity reduction method of each

branch in the below air duct system and air fan power required.


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3- For system shown below, size the main duct and branches with friction pressure

drop of mPa /75.0 and the air fan power.

ac duct design

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