1 duct asizing methods

7/30/2019 1 Duct ASizing Methods

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1 DUCT SIZING METHODS

The design of the ductworks in ventilation systems are often done by using the

Velocity Method

Constant Pressure Loss Method (or Equal Friction Method)

Static Pressure Recovery Method

The Velocity Method

Proper air flow velocities for the application considering the environment are selected. Sizes of ducts are thengiven by the continuity equation like:

A = q / v (1)

where

A = duct cross sectional area (m

2

)

q = air flow rate (m3/s)

v= air speed (m/s)

A proper velocity will depend on the application and the environment. The table below indicate commonly usedvelocity limits:

Type of Duct Comfort Systems Industrial Systems High Speed Systems

Main ducts 4 - 7 m/s 8 - 12 m/s 10 - 18 m/s

Main branch ducts 3 - 5 m/s 5 - 8 m/s 6 - 12 m/s

Branch ducts 1 - 3 m/s 3 - 5 m/s 5 - 8 m/s

Be aware that high velocities close to outlets and inlets may generate unacceptable noise.
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The Constant Pressure Loss Method (or Equal Friction Loss Method)

A proper speed is selected in the main duct close to the fan. The pressure loss in the main duct are then usedas a template for the rest of the system. The pressure (or friction) loss is kept at a constant level throughout thesystem. The method gives an automatic velocity reduction through the system. The method may add more duct

cross sectional changes and can increase the number of components in the system compared to othermethods.

The Static Pressure Recovery Method

With the static pressure recovery method the secondary and branch ducts are selected to achieve more or lessthe same static pressure in front of all outlets or inlets. The major advantage of the method are more commonconditions for outlets and inlets. Unfortunate the method is complicated to use and therefore seldom used.

2 Equal friction method

The equal friction method of sizing ducts is often preferred because it is quite easy to use. The method can besummarized to

1. Compute the necessary air flow volume (m3/h, cfm) in every room and branch ofthe system

2. Use 1) to compute the total air volume (m3/h, cfm) in the main system

3. Determine the maximum acceptable airflow velocity in the main duct

4. Determine the major pressure drop in the main duct

5. Use the major pressure drop for the main duct as a constant to determine theduct sizes throughout the distribution system

6. Determine the total resistance in the duct system by multiplying the staticresistance with the equivalent length of the longest run

7. Compute balancing dampers


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1. Compute the air volume in every room and branch

Use the actual heat, cooling orair quality requirements for the rooms and calculate the required air volume - q.

2. Compute the total volume in the system

Make a simplified diagram of the system like the one above.

Use 1) to summarize and accumulate the total volume - qtotal- in the system.

Note! Be aware that maximum load conditions almost never occurs in all of the rooms at the same time. Avoidover-sizing the main system by multiplying the accumulated volume with a factor less than one (This isprobably the hard part - and for larger systems sophisticated computer-assisted indoor climate calculations areoften required).

3. Determine the maximum acceptable airflow velocity in the main ducts

Select the maximum velocityin the main duct on basis of the application environment. To avoid disturbingnoise levels - keep maximum velocities within experienced limits:

comfort systems - air velocity 4 to 7 m/s (13 to 23 ft/s)

industrial systems - air velocity 8 to 12 m/s (26 to 40 ft/s)

high speed systems - air velocity 10 to 18 m/s (33 to 60 ft/s)

Use the maximum velocity limits when selecting the size of the main duct.

4. Determine the static pressure drop in main duct

Use a pressure drop table or similar to determine the static pressure drop in the main duct.

5. Determine the duct sizes throughout the system

Use the static pressure drop determined in 4) as a constant to determine the ducts sizes throughout the system.Use the air volumes calculated in 1) for the calculation. Select the duct sizes with the pressure drop for theactual ducts as close to the main duct pressure drop as possible.

6. Determine the total resistance in the system

Use the static pressure from 4) to calculate the pressure drop through the longest part of the duct system. Usethe equivalent length which is

the actual length + additional lengths for bends, T's, inlets and outlets
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7. Calculate balancing dampers

Use the total resistance in 6) and the volume flow throughout the system to calculate necessary dampers andthe theoretical pressure loss through the dampers.

Note about the Equal Friction Method

The equal friction method is straightforward and easy to use and gives an automatic reduction of the air flowvelocities throughout the system. The reduced velocities are in general within the noise limits of the applicationenvironment.

The method can increase the numbers of reductions compared to other methods, and often a poorer pressurebalance in the system require more adjusting dampers. This may increase the system cost compared to othermethods.

Example - Equal Friction Method

The equal friction method can be done manual or more or less semi automatic with a spreadsheet as shown inthe table below.

The table is based on the diagram above. Air flow and friction loss from a diagram is added. Minor pressureloss coefficients must be summarized for for the actual applications.

The pressure loss in each path is summarized on the right and pressure loss is added manually in the dampersto balance the system.

3 MINOR LOSS COEFFICIENTS FOR AIR DUCTS

COMPONENTS

Minor Loss - Head or Pressure Loss in Air Duct Components - can be expressed as

hminor_loss = v2/ 2 g (1)

where

hminor_loss = minor head loss (m, ft)

= minor loss coefficient

v = flow velocity (m/s, ft/s)
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g = acceleration of gravity(m/s2, ft/s2)

Minor loss can also be expressed aspressure loss instead of head loss.

Minor loss coefficients for different components common in air duct distribution systems

Component or Fitting

Minor Loss

Coefficient

- -

90o bend, sharp 1.3

90o bend, with vanes 0.7

90o bend, rounded

radius/diameter duct 10.25

45o bend, sharp 0.5

45o bend, rounded

radius/diameter duct 10.05

T, flow to branch

(applied to velocity in branch)0.3

Flow from duct to room 1.0

Flow from room to duct 0.35

Reduction, tapered 0

Enlargement, abrupt

(due to speed before reduction)

(1 - v2 / v1)2
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(v1= velocity before enlargement and v2

= velocity after enlargement)

Enlargement, tapered angle < 8o




0.15 (1 - v2 / v1)2

Enlargement, tapered angle > 8o




(1 - v2 / v1)2

Grilles, 0.7 ratio free area to total

surface 3


surface4


surface6


surface 10


surface20


surface50

4 AIR DUCTS FRICTION LOSS DIAGRAM


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Friction loss (head loss) in standard air ducts are indicated in the diagram below:

The diagram is based on standard air0.075 lb/ft3 in clean round galvanized metal ducts.

1 inch water = 248.8 N/m2(Pa)= 0.0361 lb/in2 (psi) = 25.4 kg/m2= 0.0739 in

mercury

1 ft3/min (cfm) = 1.7 m3/h = 0.47 l/s

1 ft/min = 5.08x10-3 m/s

1 inch = 25.4 mm = 2.54 cm = 0.0254 m = 0.08333 ft

Example - Friction Loss in Air Duct


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The friction loss in a 20 inches duct with air flow 4000 cfm can be estimated to approximately 0.23 inches waterper 100 feetduct as shown in the diagram below. The air velocity can be estimated to approximately 1850 feetper minute.

5 DESIGN OF VENTILATION SYSTEMS

A ventilation system may be designed more or less according the following procedure:

1. Calculate heat or cooling load, including sensible and latent heat2. Calculate necessary air shifts according the number of occupants and their

activity or any other special process in the rooms

3. Calculate air supply temperature

4. Calculate circulated mass of air

5. Calculate temperature loss in ducts

6. Calculate the outputs of components - heaters, coolers, washers, humidifiers

7. Calculate boiler or heater size

8. Design and calculate the duct system

1. Calculate Heat and Cooling Loads

Calculate heat and cooling loads by

Calculating indoor heat or cooling loads

Calculating surrounding heat or cooling loads

2. Calculate Air Shifts according the Occupants or any Processes


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Calculate the pollution created by persons and their activity and processes.

3. Calculate Air Supply Temperature

Calculate air supply temperature. Common guidelines:

For heating, 38 - 50oC (100-120oF) may be suitable

For cooling where the inlets are near occupied zones - 6 - 8oC (10-15oF) belowroom temperature

For cooling where high velocity diffusing jets are used - 17oC (30oF) below room

temperature

4. Calculate Air Quantity

Air Heating

If air is used for heating, the needed air flow rate may be expressed as

qh = Hh / cp (ts - tr) (1)

where

qh = volume of air for heating (m3/s)

Hh =heat load (W)

cp = specific heat capacity of air (J/kg K)

ts = supply temperature (oC)

tr= room temperature (oC)

= density of air (kg/m3)

Air Cooling

If air is used for cooling, the needed air flow rate may be expressed as

qc= Hc / cp (to - tr) (2)

where

qc= volume of airfor cooling (m3/s)

Hc=cooling load (W)

to = outlet temperature (oC) where to = tr if the air in the room is mixed

Example - heating load:

If the heat load is Hh = 400 W, supply temperature ts = 30oCand the room temperature tr= 22

oC, the air flowrate can be calculated as:


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qh = (400 W) / (1.2 kg/m3) (1005 J/kg K) ((30oC) - (22oC))

= 0.041 m3/s

= 149 m3/h

Moisture

If it is necessary to humidify the indoor air, the amount of supply air needed may be calculated as:

qmh = Qh / (x2 - x1) (3)

where

qm = volume of air for humidifying (m3/s)

Qh = moisture to be supplied (kg/s)

= density of air (kg/m3)

x2= humidity of room air (kg/kg)

x1 = humidity of supply air (kg/kg)

Dehumidifying

If it is necessary to dehumidify the indoor air, the amount of supply air needed may be calculated as:

qmd= Qd / (x1 - x2) (4)

where

qmd= volume of air for dehumidifying (m3/s)

Qd= moisture to be dehumified (kg/s)

Example - humidifying

If added moisture Qh = 0.003 kg/s, room humidityx1 = 0.001 kg/kgand supply air humidityx2= 0.008 kg/kg, theamount of air can expressed as:

qmh = (0.003 kg/s) / (1.2 kg/m3) ((0.008 kg/kg)- (0.001 kg/kg))

= 0.36 m3/s

Alternatively the air quantity is determined by the requirements of occupants or processes.

5. Temperature loss in ducts

The heat loss from a duct can be expressed as:

H = A k ( (t1 + t2) / (2 - tr) ) (5)


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where

H = heat loss (W)

A = area of duct walls(m2)

t1 = initial temperature in duct (oC)

t2= final temperature in duct (oC)

k = heat loss coefficient of duct walls (W/m2K) (5.68 W/m2K for sheet metal ducts, 2.3 W/m2K for insulatedducts)

tr= surrounding room temperature (oC)

The heat loss in the air flow can be expressed as:

H = q cp (t1 - t2) (5b)

where

q = mass of air flowing (kg/s)

cp = specific heat capacity of air (kJ/kg K)

(5) and (5b) can be combined to

H = A k ((t1 + t2) / 2 - tr)) = q cp (t1 - t2) (5c)

For large temperature drops should logarithmic mean temperatures be used.

6. Selecting Heaters, Washers, Humidifiers and Coolers

Units as heaters, filters etc. must on basis of of air quantity and capacity be selected from manufacturescatalogues.

7. Boiler

The boiler rating can be expressed as:

B = H (1 + x) (6)

where

B = boiler rating (kW)

H = total heat load of all heater units in system (kW)

x = margin for heating up the system, it is common to use values 0.1 to 0.2

Boiler with correct rating must be selected from manufacturer catalogues.

8. Sizing Ducts


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Air speed in a duct can be expressed as:

v = Q / A (7)

where

v = air velocity (m/s)

Q = air volume (m3/s)

A = cross section of duct (m2)

Overall pressure loss in ducts can be expressed as:

dpt= dpf+ dps + dpc (8)

where

dpt= total pressure loss in system (Pa, N/m

2

)

dpf= major pressure loss in ducts due to friction (Pa, N/m2)

dps = minor pressure loss in fittings, bends etc. (Pa, N/m2)

dpc= minor pressure loss in components as filters, heaters etc. (Pa, N/m2)

Major pressure loss in ducts due to friction can be expressed as

dpf= R l (9)

where

R = duct friction resistance per unit length (Pa, N/m2per m duct)

l = length of duct (m)

Duct friction resistance per unit length can be expressed as

R = / dh ( v2 / 2) (10)

where

R = pressure loss (Pa, N/m2)

= friction coefficient

dh =hydraulic diameter(m)
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6 A rough guide to maximum air volume capacity ofcircular ducts in comfort, industrial and high speed

ventilation systems

Maximum air velocity in the ducts should be kept below certain limits to avoid unacceptable generation ofnoise.

The values from the table below can be used to rough sizing of ducts in comfort, industrial and high speedventilation systems.

Maximum Air Volume Flow (m3/h)

Diameter(mm)

Area(m2)

Comfort systems Industrial systems High speed systems

Mainducts

Secondaryducts

Connectingducts

Mainducts

Secondaryducts

Connectingducts

Mainducts

Secondaryducts

Connectingducts

Speed (m/s)

5.5 4 2 10 6.5 4 14 9 6.5

63 0.003 62 45 22 112 73 45 157 101 73

80 0.005 99 72 36 181 118 72 253 163 118


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100 0.008 155 113 57 283 184 113 396 254 184

125 0.012 243 177 88 442 287 177 618 397 287

160 0.020 398 289 145 723 470 289 1013 651 470

200 0.031 622 452 226 1130 735 452 1583 1017 735

250 0.049 971 707 353 1766 1148 707 2473 1590 1148

315 0.078 1542 1122 561 2804 1823 1122 3926 2524 1823

400 0.126 2487 1809 904 4522 2939 1809 6330 4069 2939

500 0.196 3886 2826 1413 7065 4592 2826 9891 6359 4592

630 0.312 6169 4487 2243 11216 7291 4487 15703 10095 7291

800 0.502 9948 7235 3617 18086 11756 7235 25321 16278 11756

1000 0.785 15543 11304 5652 28260 18369 11304 39564 25434 18369

1250 1.227 24286 17663 8831 44156 28702 17663 61819 39741 28702

7 Equivalent diameters for rectangular and circular

ducts - air flows between 100 - 50,000 cfm

The table below can be used to compare equivalent diameters for rectangular and round circular ducts. Thetable is based on the ducts friction loss formula.
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The rectangular dimensions and the air flow volume are adapted to the equal friction loss method of sizingventilation duct systems. An approximate friction loss of0.8 inches water gauge per 100 ft duct (6.6 Pa/m) isused.

Air flow- q -

(Cubic Feet perMinute, cfm)

(m3/s)

Rectangular DuctSizes

(Inches)

EquivalentDiameter

Round DuctSizes- de -

(Inches)

Velocity- v-

(ft/min)(m/s)

Friction Loss(Inches water

gauge per 100 ftduct)

200(0.09)

3 x 74 x 5

4.94.9

1527(7.8)

0.88

300(0.14)

4 x 75 x 6

5.76.0

1635(8.3)

0.82

400(0.19)

4 x 95 x 76 x 6

6.46.46.6

1736(8.8)

0.80

500(0.24)

6 x 7 7.11819(9.2)

0.78

750(0.35)

5 x 126 x 107 x 8

8.38.48.2

1996(10.1)

0.77

1000(0.47)

7 x 108 x 9

9.19.3

2166(11)

0.79

1250(0.59)

8 x 109 x 9

9.89.8

2386(12.1)

0.88

1500(0.71)

8 x 1210 x 10

10.710.9

2358(11.9)

0.77

1750(0.83)

8 x 149 x 12

10 x 11

11.511.311.5

2469(12.5)

0.78

2000(0.94)

8 x 1510 x 12

11.812.0

2589(13.2)

0.81
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2500(1.2)

10 x 1412 x 12

12.913.1

2712(13.8)

0.8

3000(1.4)

12 x 14 14.12767(14.1)

0.75

3500(1.7)

12 x 15 14.63010(15.3)

0.84

4000(1.9)

10 x 2214 x 15

15.915.8

2938(14.9)

0.73

4500(2.1)

12 x 1914 x 16

16.416.4

3068(15.6)

0.76

5000(2.4)

10 x 2512 x 2015 x 16

16.916.816.9

3248(16.5)

0.82

6000(2.8)

14 x 2015 x 18

18.217.9

3358(17.1)

0.8

7000(3.3)

12 x 2616 x 20

19.019.5

3482(17.7)

0.8

8000(3.8)

12 x 3014 x 25

20.220.2

3595(18.3)

0.8

9000(4.3)

12 x 3415 x 25

21.421.0

3671(18.6)

0.78

10000(4.7)

12 x 3616 x 2520 x 20

21.921.721.9

3858(19.6)

0.83

12500(5.9)

12 x 45

16 x 3020 x 24

24.1

23.723.9

4012(20.4) 0.8

15000(7.1)

16 x 3618 x 3023 x 25

24.725.226.2

4331(22)

0.87

17500 16 x 40 27.0 4337 0.79


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(8.3)20 x 3225 x 25

27.527.3

(22)

20000(9.4)

20 x 3525 x 28

28.628.9

4483(22.8)

0.79

25000(11.8)

16 x 5520 x 4325 x 38

31.031.533.5

4709(23.9)

0.78

30000(14.2)

20 x 5030 x 32

33.733.9

4815(24.5)

0.74

35000(16.5)

20 x 5530 x 35

35.235.4

5179(26.3)

0.81

40000(18.9)

25 x 4830 x 40

37.437.8

5243(26.6)

0.77

45000(21.2)

32 x 40 39.15397(27.4)

0.77

50000(23.6)

32 x 4535 x 40

41.340.9

5222(26.5)

0.66

8 Friction loss or major loss in ducts - equations

and online calculator for rectangular and circular

ducts - imperial units

The major loss, or friction loss, in a circular duct in galvanized steel with turbulent flow can for imperial units beexpressed

p= (0.109136 q1.9) / de

5.02 (1)

where

p= friction (head or pressure loss) (inches water gauge/100 ft of duct)
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de=equivalent duct diameter(inches)

q= air volume flow - (cfm - cubic feet per minute)

For rectangular ducts the equivalent diametermust be calculated.

Air Ducts Friction Loss Calculator

The friction loss calculator below is based on formula (1):

80Air Volume Flow - q - (cfm)

10Equivalent duct diameter- de - (inches)

Pressure loss and air flow velocity for some common duct sizes and air flow volumes can be taken from thetable below:

Pressure Loss (inches water gauge per 100 feet duct)

Air velocity (ft/min)

Air Volume (cfm)

Duct Size (inches)

4 5 6 8 10 12 16

100

0.65 0.21 0.09 0.02 0.01

1146 733 509 286 183

200

0.8 0.32 0.08 0.02 0.01

1467 1019 573 367 255

400

1.19 0.28 0.09 0.04 0.01

2037 1146 733 509 286

800

0.34 0.14 0.03

1467 1019 573
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1600

0.12

1146

The air velocity should not exceed certain limits to avoid unacceptable noise generation.

9 Hydraulic diameter of ducts and tubes

The hydraulic diameter - dh - is used to calculate the dimensionless Reynolds Numberto determine if a flow isturbulent or laminar. A flow is

laminar if Re < 2300

transient for 2300 < Re < 4000

turbulent if Re > 4000

The hydraulic diameter is also used to calculate the pressure lossin a ducts or pipe.

The hydraulic diameter is not the same as the geometrical diameter in a non-circular duct or pipe and can becalculated with the generic equation

dh = 4 A / p (1)

where

dh = hydraulic diameter (m, ft)

A = area section of the duct (m2, ft2)

p = wetted perimeter of the duct (m, ft)

Note! Inches are commonly used in the Imperial unit system.

Hydraulic Diameter of a Circular Tube or Duct

Based on equation (1) the hydraulic diameter of a circular duct can be expressed as:
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dh = 4 r2 / 2 r

= 2 r (2)

where

r = pipe or duct radius (m, ft)

As we could expect the hydraulic diameter of a standard circular tube or duct is two times the radius.

Hydraulic Diameter of a Circular Tube with an inside Circular Tube

Based on equation (1) the hydraulic diameter of a circular duct or tube with an inside duct or tube can beexpressed as

dh = 4 ( ro2 - ri

2) / (2 ro + 2 ri)

= 2 (ro - ri) (3)

where

ro = inside radius of the outside tube (m, ft)

ri = outside radius of the inside tube (m, ft)

Hydraulic Diameter of Rectangular Tubes or Ducts

Based on equation (1) the hydraulic diameter of a rectangular duct or pipe can be calculated as

dh = 4 a b / (2 (a + b))


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= 2 a b / (a + b) (4)

where

a = width/height of the duct (m, ft)

b = height/width of the duct (m, ft)

Rectangular to Circulate Duct/Tube Hydraulic Diameter Calculator

The calculator below is based on formula (4) an can be used to calculate the hydraulic diameter of rectangularduct or tube. The formula is generic and any unit can be used.

1 duct asizing methods

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