ashrea duct work
TRANSCRIPT
ASHRAE Calculations
ASHRAE Fundamentals Handbook has a methodology for calculating the friction loss in ducts and fittings. Based on these formulas, Dryer-Ells can be measured the same as straight pipe. This permits use of the maximum allowed run length.
• A ten inch radius smooth interior ninety degree bend will only have an equivalent vent length of 1-1/2 feet (the same length as model LT90). • A ten inch radius smooth interior forty-five degree bend will only have an equivalent vent length of 3/4 feet (the same length as model LT45). ASHRAE Formula as Confirmed JB Engineering and Code Consulting, P.C.:
Calculation Detail (Using Model LT90 As an Example) The pressure loss due to friction in a round duct is calculated using the following
equation: Where:PL = Pressure Loss in Duct in Inches of Water Column L = Length of Duct in Feet FL = Friction Loss per 100 feet of Duct at Specified Velocity of Flow The pressure loss through a duct fitting is calculated using the following equation:
Where:TP = Total Pressure Loss in Inches of WC C = Fitting Coefficient Vp = Velocity Pressure at Upstream Connection in Inches of WC The pressure loss through a fitting is often calculated by establishing the equivalent length of duct for each fitting. By using this method of calculating duct pressure loss, the equivalent length of each fitting is added to the total duct length to establish the pressure loss through the duct system. To establish the equivalent length for a given fitting, the fitting equation is set as being equal to the duct length equation. Solving for “L” establishes the equivalent duct length for a specific fitting with a given velocity of flow through the duct. The equation
becomes:
A standard three section, 4 inch dryer exhaust duct 90° elbow with a 4 inch radius has been established as having an equivalent length of 5 feet in Section 504.6.1 of the 2003 ICC International Mechanical Code and Section M1501.3 of the ICC International Residential Code. The coefficient for a three section elbow in SMACNA Duct Design Handbook is 0.42. The coefficient for a 4 inch smooth 90° elbow with a 10 inch radius is 0.12. To establish the equivalent length of dryer exhaust duct for the 4 inch smooth 90° elbow with a 10 inch radius, you cancel out the equivalent factors and are left with the following equation:
Where:L10 = Equivalent length for the 10 inch radius smooth elbow L4 = Equivalent length for the 4 inch radius, three section elbow C10 = Coefficient for the 10 inch radius smooth elbow C4 = Coefficient for the 4 inch radius, three section elbow
Solving the equation results in: Rounding up, the 4 inch smooth 90° elbow with a 10 inch radius would have an equivalent length of dryer vent of 1-1/2 feet.
Equal Friction Method
The equal friction method of sizing ducts is easy and straightforward to use
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The equal friction method of sizing ducts is often preferred because it is quite easy to use. The method can be summarized to
1. Compute the necessary air flow volume (m3/h, cfm) in every room and branch of the 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 the duct
sizes throughout the distribution system 6. Determine the total resistance in the duct system by multiplying the static
resistance with the equivalent length of the longest run 7. Compute balancing dampers
1. Compute the air volume in every room and branch
Use the actual heat, cooling or air 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. Avoid over-sizing the main system by multiplying the accumulated
volume with a factor less than one (This is probably the hard part - and for larger systems sophisticated computer-assisted indoor climate calculations are often required).
3. Determine the maximum acceptable airflow velocity in the main ducts
Select the maximum velocity in the main duct on basis of the application environment. To avoid disturbing noise 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 the actual 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. Use the equivalent length which is
• the actual length + additional lengths for bends, T's, inlets and outlets
7. Calculate balancing dampers
Use the total resistance in 6) and the volume flow throughout the system to calculate necessary dampers and the 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 flow velocities throughout the system. The reduced velocities are in general within the noise limits of the application environment.
The method can increase the numbers of reductions compared to other methods, and often a poorer pressure balance in the system require more adjusting dampers. This may increase the system cost compared to other methods.
Example - Equal Friction Method
The equal friction method can be done manual or more or less semi automatic with a spreadsheet as shown in the table below.
The table is based on the diagram above. Air flow and friction loss from a diagram is added. Minor pressure loss 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 dampers to balance the system.
The excel template can be downloaded here!
Rooms and Rates of Outdoor Make Up Air
Recommended rates of outdoor make up air in some common types of rooms - banks, assembly halls, hotels and many more
The rates of outdoor make up air in the table below can be used as a guideline in design of ventilation systems.
Type of Building and Room
Air Shifts per hour
(1/h)
Airflow (l/s per m2 floor area)
Outdoor Airflow
(l/s) Sheep 3 per animal Horses 40 per animal Hens 3 per animal Cows 100 per animal Chickens 2 per animal Pigs, sow 60 per animal
Animals
Piglet 15 per animal Apartments 15 per person
Assembly halls 5 - 10 10 - 15 per person
Auditoriums 10 - 15 per person
Production 8 Bakeries Pastry room 6 Service 2 - 3 Archive 1 Banks Staff 2 - 3
Baths 5 - 8 15
Barber Shops 10 - 15 per person
Bars 15 - 20 per person
Beauty Shops 10 - 15 per person
Bowling Alleys 10 - 15 per person
Cafeterias 4 10 - 15 per person
Cinemas/Theatres 5 - 10 7 - 10 per person
Type of Building and Room
Air Shifts per hour
(1/h)
Airflow (l/s per m2 floor area)
Outdoor Airflow
(l/s) Sterile room 15 Laboratory 15 Packing room 4
Chemist's shops
Store 1 Churches 2 10 per person
Clubhouses 15 - 20 per person
Cocktail Lounges 15 - 20 per person
Computer Rooms 10 - 15 per person
Court Houses 10 - 15 per person
Dance halls 15 - 20 per person
Dental Clinics, Offices 10 - 15 per
person Department Stores (cfm/Sq.Ft)
0.3
Dining Halls 10 - 15 per person
Drug Shop 10 - 15 per person
Engine rooms 4 Exhibition halls 10 - 20
Fire Stations 10 - 15 per person
Garages 5 - 6 > 1 General 15 per person Therapy 9 Operating theatre 14
Hospitals
Autopsy 9
Reception 15 - 20 per person Hotels
Room 3 Industries Factory 10 - 20 per
Type of Building and Room
Air Shifts per hour
(1/h)
Airflow (l/s per m2 floor area)
Outdoor Airflow
(l/s) person
Carpenter work shop 2 - 5
Car repair shop 4 Car repair shop, exhaust outlet 60 - 80 per car
Mechanical work shop 3 - 4
Welding 12 - 15 Assembly 4 - 5
Jails 10 - 15 per person
Jewelry Shop 10 - 15 per person
Libraries 10 - 15 per person
Lunch Rooms, Luncheonettes 10 - 15 per
person Municipal Buildings 10 - 15 per
person
Museums 10 - 15 per person
Smithies 6 - 7 Small 20 - 40 Large 10 - 20 < 7 m2 15 > 7 m2 20 Industrial in general 25 - 30
Gas ovens 550/m2 outside wall
Electric ovens 300/m2 outside wall
Grill table 400/m2 outside wall
Frying pan 300
Kitchens
Boiling pan 100 l 100
Type of Building and Room
Air Shifts per hour
(1/h)
Airflow (l/s per m2 floor area)
Outdoor Airflow
(l/s) Boiling pan 200 l 200
Coffee machine 60 Refrigerated cabinet 7 - 10
Cooling and freezer room 0.3 - 0.5
Store 2 - 4 Laundries 10 - 15 Lavatories 5 - 10 10
Night Clubs 15 - 25 per person
Malls 10 - 15 per person
Motels 15 - 20 per person
Module 3 - 8 10 - 20 per person
Landscape 10 - 20 per person
Conference room 15 - 30 per
person
Lecture room 10 - 20 per person
Dining hall 15 - 25 per person
Staff, changing clothes 8 - 12
Dining room 8 - 10
Offices
Resting room 15 per person
Police Stations 10 - 15 per person
Post Offices 10 - 15 per person
Precision Manufacturing 10 - 15 per
person
Restaurants 5 - 10 10 10 - 15 per person
Type of Building and Room
Air Shifts per hour
(1/h)
Airflow (l/s per m2 floor area)
Outdoor Airflow
(l/s) Retail Stores (cfm/Sq.Ft) 0.3
Schools Classroom 4 - 5 10 - 15 per person
Shops 10 - 15 per person
Shopping Centers 10 - 15 per person
Sports halls 3 - 4 Supermarkets 10 per person Swimming pools 5 - 10
Taverns 10 - 15 per person
Town Halls 10 - 15 per person
• 1 dm3(litre)/s = 10-3 m3/s = 3.6 m3/h = 0.03532 ft3/s = 2.1189 ft3/min (cfm) = 13.200 Imp.gal (UK)/min = 15.852 gal (US)/min = 792 Imp. gal (UK)/h
Sizing Ducts
The ductwork of ventilation systems are often sized with either the Velocity, the Constant Pressure Loss (or Equal Friction Loss) - or the Static Pressure Recovery Methods
The Velocity Method
Proper air flow velocities for the application considering the environment are selected. Sizes of ducts are then given by the continuity equation like:
A = q / v (1)
where
A = duct cross sectional area (m2)
q = air flow rate (m3/s)
v= air speed (m/s)
Alternatively in Imperial units
A = 144 q / vi i i (1b)
where
A = duct cross sectional area (sq.in.)
q = air flow rate (cfm)
v= air speed (fpm)
A proper velocity will depend on the application and the environment. The table below indicate commonly used velocity limits:
Comfort Systems Industrial Systems High Speed SystemsType of Duct m/s ftm m/s ftm m/s ftm
Comfort Systems Industrial Systems High Speed SystemsType of Duct m/s ftm m/s ftm m/s ftm
Main ducts 4 - 7 780 - 1380 8 - 12 1575 - 2360 10 - 18 1670 - 3540
Main branch ducts
3 - 5 590 - 985 5 - 8 985 - 1575 6 - 12 1180 - 2360
Branch ducts 1 - 3 200 - 590 3 - 5 590 - 985 5 - 8 985 - 1575
Be aware that high velocities close to outlets and inlets may generate unacceptable noise.
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 used as a template for the rest of the system. The pressure (or friction) loss is kept at a constant level throughout the system. 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 other methods.
The Static Pressure Recovery Method
With the static pressure recovery method the secondary and branch ducts are selected to achieve more or less the same static pressure in front of all outlets or inlets. The major advantage of the method are more common conditions for outlets and inlets. Unfortunate the method is complicated to use and therefore seldom used.
Rectangular and Circular HVAC Ducts - Equivalent Diameter
Equivalent diameters for rectangular and circular ducts - air flows between 100 - 50,000 cfm
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The table below can be used to compare equivalent diameters for rectangular and round circular ducts. The table is based on the ducts friction loss formula.
The rectangular dimensions and the air flow volume are adapted to the equal friction loss method of sizing ventilation duct systems. An approximate friction loss of 0.8 inches water gauge per 100 ft duct (6.6 Pa/m) is used.
Air flow - q - (Cubic Feet per Minute, cfm) (m3/s)
Rectangular Duct Sizes (Inches)
Equivalent Diameter Round Duct Sizes - de - (Inches)
Velocity - v - (ft/min) (m/s)
Friction Loss (Inches water gauge per 100 ft duct)
200 (0.09)
3 x 7 4 x 5
4.9 4.9
1527 (7.8) 0.88
300 (0.14)
4 x 7 5 x 6
5.7 6.0
1635 (8.3) 0.82
400 (0.19)
4 x 9 5 x 7 6 x 6
6.4 6.4 6.6
1736 (8.8) 0.80
500 (0.24) 6 x 7 7.1 1819
(9.2) 0.78
750 (0.35)
5 x 12 6 x 10 7 x 8
8.3 8.4 8.2
1996 (10.1) 0.77
1000 (0.47)
7 x 10 8 x 9
9.1 9.3
2166 (11) 0.79
Air flow - q - (Cubic Feet per Minute, cfm) (m3/s)
Rectangular Duct Sizes (Inches)
Equivalent Diameter Round Duct Sizes - de - (Inches)
Velocity - v - (ft/min) (m/s)
Friction Loss (Inches water gauge per 100 ft duct)
1250 (0.59)
8 x 10 9 x 9
9.8 9.8
2386 (12.1) 0.88
1500 (0.71)
8 x 12 10 x 10
10.7 10.9
2358 (11.9) 0.77
1750 (0.83)
8 x 14 9 x 12 10 x 11
11.5 11.3 11.5
2469 (12.5) 0.78
2000 (0.94)
8 x 15 10 x 12
11.8 12.0
2589 (13.2) 0.81
2500 (1.2)
10 x 14 12 x 12
12.9 13.1
2712 (13.8) 0.8
3000 (1.4) 12 x 14 14.1 2767
(14.1) 0.75
3500 (1.7) 12 x 15 14.6 3010
(15.3) 0.84
4000 (1.9)
10 x 22 14 x 15
15.9 15.8
2938 (14.9) 0.73
4500 (2.1)
12 x 19 14 x 16
16.4 16.4
3068 (15.6) 0.76
5000 (2.4)
10 x 25 12 x 20 15 x 16
16.9 16.8 16.9
3248 (16.5) 0.82
6000 (2.8)
14 x 20 15 x 18
18.2 17.9
3358 (17.1) 0.8
7000 (3.3)
12 x 26 16 x 20
19.0 19.5
3482 (17.7) 0.8
8000 (3.8)
12 x 30 14 x 25
20.2 20.2
3595 (18.3) 0.8
9000 (4.3)
12 x 34 15 x 25
21.4 21.0
3671 (18.6) 0.78
10000 (4.7)
12 x 36 16 x 25 20 x 20
21.9 21.7 21.9
3858 (19.6) 0.83
12500 (5.9)
12 x 45 16 x 30 20 x 24
24.1 23.7 23.9
4012 (20.4) 0.8
Air flow - q - (Cubic Feet per Minute, cfm) (m3/s)
Rectangular Duct Sizes (Inches)
Equivalent Diameter Round Duct Sizes - de - (Inches)
Velocity - v - (ft/min) (m/s)
Friction Loss (Inches water gauge per 100 ft duct)
15000 (7.1)
16 x 36 18 x 30 23 x 25
24.7 25.2 26.2
4331 (22) 0.87
17500 (8.3)
16 x 40 20 x 32 25 x 25
27.0 27.5 27.3
4337 (22) 0.79
20000 (9.4)
20 x 35 25 x 28
28.6 28.9
4483 (22.8) 0.79
25000 (11.8)
16 x 55 20 x 43 25 x 38
31.0 31.5 33.5
4709 (23.9) 0.78
30000 (14.2)
20 x 50 30 x 32
33.7 33.9
4815 (24.5) 0.74
35000 (16.5)
20 x 55 30 x 35
35.2 35.4
5179 (26.3) 0.81
40000 (18.9)
25 x 48 30 x 40
37.4 37.8
5243 (26.6) 0.77
45000 (21.2) 32 x 40 39.1 5397
(27.4) 0.77
50000 (23.6)
32 x 45 35 x 40
41.3 40.9
5222 (26.5) 0.66
Duct Velocity
Calculate velocities in circular and rectangular ducts - imperial and SI-units - online calculator
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Imperial Units
The velocity of air in a ventilation duct can be expressed in imperial units like
v = q / A = 576 q / (π d ) = 144 q / (a b ) (1)i i i i i2
i i i
where
v = air velocity (ft/min)i
q = air flow (cfm)i
A = area of duct (square feet)i
d = diameter of duct (inches)i
a = width of duct (inches)i
b = width of duct (inches)i
Imperial Units Air Flow Velocity Calculator
Air velocity can be calculated with the calculator below. Add air volume - q - and diameter - d - (or length a and b).
1000 Air volume - qi - (cfm)
12 Diameter - di - (inches)
or alternatively
Length side - ai - (inches)
Length side - bi - (inches)
SI - Units
Air velocity in a duct can alternatively be expressed in SI units like
v = q / A = 4 q / (π d ) = q / (a b ) (2)m m m m m2
m m m
where
v = air velocity (m/s)m
q = air flow (m /s)m3
A = area of duct (m )m2
d = diameter of duct (m)m
a = width of duct (m)m
b = width of duct (m)m
SI Units Air Flow Velocity Calculator
Air velocity can be calculated with the calculator below. Add air volume - q - and diameter - d - (or length a and b).
1 Air volume - qm - (m3/s)
Diameter - dm - (m)
or alternatively
0.5 Length side - am - (m)
0.5 Length side - bm - (m)
Friction Loss in Ducts
Friction loss or major loss in ducts - equations and online calculator for rectangular and circular ducts - imperial units
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The major loss, or friction loss, in a circular duct in galvanized steel with turbulent flow can for imperial units be expressed
Δp = (0.109136 q1.9) / de5.02 (1)
where
Δp = friction (head or pressure loss) (inches water gauge/100 ft of duct)
de = equivalent duct diameter (inches)
q = air volume flow - (cfm - cubic feet per minute)
For rectangular ducts the equivalent diameter must be calculated.
Air Ducts Friction Loss Calculator
The friction loss calculator below is based on formula (1):
800 Air Volume Flow - q - (cfm)
10 Equivalent duct diameter - de - (inches)
• velocities in rectangular ducts
Pressure loss and air flow velocity for some common duct sizes and air flow volumes can be taken from the table below:
Pressure Loss (inches water gauge per 100 feet duct) Air velocity (ft/min)
Duct Size (inches) Air Volume (cfm) 4 5 6 8 10 12 16 0.65 0.21 0.09 0.02 0.01 100 1146 733 509 286 183 0.8 0.32 0.08 0.02 0.01 200 1467 1019 573 367 255
Pressure Loss (inches water gauge per 100 feet duct) Air velocity (ft/min)
Duct Size (inches) Air Volume (cfm) 4 5 6 8 10 12 16 1.19 0.28 0.09 0.04 0.01 400 2037 1146 733 509 286 0.34 0.14 0.03 800 1467 1019 573 0.12 1600 1146
The air velocity should not exceed certain limits to avoid unacceptable noise generation.
Minor Loss Coefficients for Air Duct Components
Minor loss - pressure or head loss - coefficients for air duct distribution systems components
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Minor Loss - Head or Pressure Loss in Air Duct Components - can be expressed as
hminor_loss = ξ v2/ 2 g (1)
where
hminor_loss = minor head loss (m, ft)
ξ = minor loss coefficient
v = flow velocity (m/s, ft/s)
g = acceleration of gravity (m/s2, ft/s2)
Minor loss can also be expressed as pressure loss instead of head loss.
Minor loss coefficients for different components common in air duct distribution systems
Component or Fitting Minor Loss Coefficient- ξ -
90o bend, sharp 1.3 90o bend, with vanes 0.7 90o bend, rounded radius/diameter duct <1 0.5
90o bend, rounded radius/diameter duct >1 0.25
45o bend, sharp 0.5 45o bend, rounded radius/diameter duct <1 0.2
45o bend, rounded radius/diameter duct >1 0.05
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
Component or Fitting Minor Loss Coefficient- ξ -
Enlargement, abrupt (due to speed before reduction) (v1= velocity before enlargement and v2 = velocity after enlargement)
(1 - v2 / v1)2
Enlargement, tapered angle < 8o (due to speed before reduction) (v1= velocity before enlargement and v2 = velocity after enlargement)
0.15 (1 - v2 / v1)2
Enlargement, tapered angle > 8o (due to speed before reduction) (v1= velocity before enlargement and v2 = velocity after enlargement)
(1 - v2 / v1)2
Grilles, 0.7 ratio free area to total surface 3 Grilles, 0.6 ratio free area to total surface 4 Grilles, 0.5 ratio free area to total surface 6 Grilles, 0.4 ratio free area to total surface 10 Grilles, 0.3 ratio free area to total surface 20 Grilles, 0.2 ratio free area to total surface 50
Air Ducts Minor Loss Coefficient Diagrams
Minor loss coefficient diagrams for air ductwork, bends, expansions, inlets and outlets - SI units
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Minor loss in air ducts based on a summarized minor loss coefficient and air flow velocity can be estimated with the diagram below:
• 1 Pa = 1 N/m2 = 1.4504x10-4 lb/in2 = 1x10-5 bar = 4.03x10-3 in water = 0.336x10-3 ft water = 0.1024 mm water = 0.295x10-3 in mercury = 7.55x10-3 mm mercury = 0.1024 kg/m2 = 0.993x10-5 atm
Minor loss for some common ductwork components like bends, expansions, inlets and outlets can be estimated with the diagrams below.
Minor Loss Coefficients - Bends
The minor loss coefficients in 90o bends can be estimated with the diagram below.
The minor loss in bends 0 - 180o can be estimated with the equation
ξ0-180 = ξ90 α / 90 (1)
where
ξ0-180 = minor loss coefficient for the bend with the actual angle
ξ90 = minor loss coefficient for the 90o bend according the diagram above
α = angle of the actual bend
The minor loss coefficient in a 90o bend like the one below is 1.0.
Minor Loss Coefficients - Expansions
Minor Loss Coefficients - Inlets
Minor Loss Coefficients - Outlets
Air Ducts Friction Loss Diagram
Major loss diagram air ducts - SI units
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Friction loss in standard air ducts are indicated in the diagram below.
• 1 m/s = 196.85 ft/min • 1 m3/s = 3600 m3/h = 1000 dm3(liter)/s = 35.32 ft3/s = 2118.9 ft3/min = 13200
Imp.gal (UK)/min = 15852 gal (US)/min • 1 mm H2O = 9.81 Pa = 9.807x10-6 N/mm2 = 0.0987 10-3 bar = 1 kp/m2 = 0.09678
10-3 atm = 1.422 10-3 psi (lbf/in2)
Example - Air Duct and Friction Loss
The friction loss in a 500 mm main duct in comfort system with air flow 1 m3/s can be estimated as indicated below to 0.05 mm H2O/m (~ 0.5 Pa/m).
The air velocity in the duct is 5 m/s.
Friction Loss - Temperature and Pressure
Friction loss in air ducts depends on actual temperature and air pressure
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Most of the friction loss diagrams and tables available for air ducts are estimated for air at Normal Temperature and Pressure - NTP - conditions (20oC and 1 bar).
Temperature
If the actual temperature condition is otherwise the friction loss can be compensated like
Δp = k Δp (1)t NTP
where
Δp = actual friction loss (pressure or head)
k = temperature compensating coefficient t
Δp = friction loss at NTP conditions (pressure or head)NTP
The temperature compensating coefficient can be estimated with the diagram below:
Pressure
If the actual pressure is otherwise the friction loss can be compensated like
Δp = k Δp (2)p NTP
where
Δp = actual friction loss (pressure or head)
k = pressure compensating coefficient p
Δp = friction loss at NTP conditions (pressure or head)NTP
The pressure compensating coefficient can be estimated with the diagram below:
• 1000 mbar = 1 bar = 105 Pa (N/m2) = 0.1 N/mm2 = 10,197 kp/m2 = 10.20 m H2O = 0.9869 atm = 14.50 psi (lbf/in2) = 106 dyn/cm2 = 750 mmHg
Rectangular Duct Sizes
Dimensions of common rectangular air ducts used in ventilation systems
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Common rectangular duct sizes used in air handling ventilation systems:
Height (mm) Width (mm) 100 150 200 250 300 400 500 600 800 1000 1200200 1) 1) 2) 3) 3) 3) 3) 3) 3) 3) 3)
250 2) 2) 2) 2) 3) 3) 3) 3) 3) 3) 3)
300 1) 1) 1) 2) 2) 3) 3) 3) 3) 3) 3)
400 1) 1) 1) 2) 1) 2) 3) 3) 3) 3) 3)
500 3) 1) 1) 2) 1) 1) 2) 3) 3) 3) 3)
600 3) 1) 1) 2) 1) 1) 1) 2) 3) 3) 3)
800 3) 3) 1) 2) 1) 1) 1) 1) 2) 3) 3)
1000 3) 3) 3) 2) 1) 1) 1) 1) 1) 2) 3)
1200 3) 3) 3) 3) 1) 1) 1) 1) 1) 1) 2)
1400 3) 3) 3) 3) 3) 2) 2) 2) 2) 2) 2)
1600 3) 3) 3) 3) 3) 1) 1) 1) 1) 1) 1)
1800 3) 3) 3) 3) 3) 3) 2) 2) 2) 2) 2)
2000 3) 3) 3) 3) 3) 3) 3) 3) 1) 1) 1)
1) Preferred, 2) Acceptable, 3) Not common
Rectangular Air Ducts Velocity Diagram
Velocity diagram rectangular air ducts
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The diagram below can be used to estimate air velocity in rectangular air ducts.
Air Velocities in Ducts
Recommended maximum air velocities in ventilation ducts
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Air velocities in ducts should not exceed certain limits to avoid high pressure losses and unacceptable noise generation. The values below are common guidelines for some typical applications.
Air Velocity Air Ducts (m/s) (ft/s)
Combustion air ducts 12 - 20 40 - 66 Air inlet to boiler room 1 - 3 3.3 - 9.8 Warm air for house heating 0.8 - 1.0 2.6 - 3.3 Vacuum cleaning pipe 8 - 15 26 - 49 Compressed air pipe 20 - 30 66 - 98 Ventilation ducts (hospitals) 1.8 - 4 5.9 - 13 Ventilation ducts (office buildings) 2.0 - 4.5 6.5 - 15
Air Ducts Friction Loss Diagram
Major loss diagram air ducts - Imperial units ranging 10 - 100 000 cfm
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Friction loss (head loss) in standard air ducts are indicated in the diagram below:
The diagram is based on standard air 0.075 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
The friction loss in a 20 inches duct with air flow 4000 cfm can be estimated to approximately 0.23 inches water per 100 feet duct as shown in the diagram below. The air velocity can be estimated to approximately 1850 feet per minute.
Sizing Circular Ducts
A rough guide to maximum air volume capacity of circular ducts in comfort, industrial and high speed ventilation systems
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Maximum air velocity in the ducts should be kept below certain limits to avoid unacceptable generation of noise.
The values from the table below can be used to rough sizing of ducts in comfort, industrial and high speed ventilation systems.
Maximum Air Volume Flow (m3/h) Comfort systems Industrial systems High speed systems
Main ducts
Secondary ducts
Connecting ducts
Main ducts
Secondary ducts
Connecting ducts
Main ducts
Secondary ducts
Connecting ducts
Speed (m/s)
Diameter
(mm)
Area
(m2)
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
Maximum Air Volume Flow (m3/h) Comfort systems Industrial systems High speed systems
Main ducts
Secondary ducts
Connecting ducts
Main ducts
Secondary ducts
Connecting ducts
Main ducts
Secondary ducts
Connecting ducts
Speed (m/s)
Diameter
(mm)
Area
(m2)
5.5 4 2 10 6.5 4 14 9 6.5
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 1121
6 7291 4487 15703 10095 7291
800 0.502 9948 7235 3617 1808
6 11756 7235 25321 16278 11756
1000 0.785
15543 11304 5652 2826
0 18369 11304 39564 25434 18369
1250 1.227
24286 17663 8831 4415
6 28702 17663 61819 39741 28702
Converting Round Duct Diameter to Area
Converting round duct areas to square feet and square meters
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Duct Diameter Area (in) (mm) (ft2) (m2) 8 203 0.3491 0.032 10 254 0.5454 0.051 12 305 0.7854 0.073 14 356 1.069 0.099 16 406 1.396 0.130 18 457 1.767 0.290 20 508 2.182 0.203 22 559 2.640 0.245 24 609 3.142 0.292 26 660 3.687 0.342 28 711 4.276 0.397 30 762 4.900 0.455 32 813 5.585 0.519 34 864 6.305 0.586 36 914 7.069 0.657
Velocities in Ventilation Ducts
Recommended velocities in ventilation ducts
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The duct velocity in air condition and ventilation systems should not exceed certain limits to avoid unnecessary noise generation and pressure drop in the duct work.
The limits of velocities depends on the actual application. The background noise in an industrial building is significant higher than the noise in a public building and more duct generated noise can be accepted.
Commonly accepted duct velocities can be found in the table below.
Velocity - v Public buildings Industrial plant Service
(m/s) (ft/min) (m/s) (ft/min) Air intake from outside 2.5 - 4.5 500 - 900 5 - 6 1000 - 1200 Heater connection to fan 3.5 - 4.5 700 - 900 5 - 7 1000 - 1400 Main supply ducts 5.0 - 8.0 1000 - 1500 6 - 12 1200 - 2400 Branch supply ducts 2.5 - 3.0 500 - 600 4.5 - 9 900 - 1800 Supply registers and grilles 1.2 - 2.3 250 - 450 1.5 - 2.5 350 - 500 Low level supply registers 0.8 - 1.2 150 - 250 Main extract ducts 4.5 - 8.0 900 - 1500 6 - 12 1200 - 2400 Branch extract ducts 2.5 - 3.0 500 - 600 4.5 - 9 900 - 1800
Recommended Air Velocities in Ducts
Ducts and recommended air velocities
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Air flow velocity in ducts should be kept within certain limits to avoid noise and unacceptable friction loss and energy consumption.
Low and Medium Pressure Ducts
• Maximum friction rate 0.1 - 0.2 inches W.G./100 ft • Velocity 1,500 - 2,000 ft/min (8 - 10 m/s)
Air Flow Rate Maximum Velocity(m3/h) (CFM) (m/s) (ft/min) < 300 < 175 2.5 490 < 1,000 < 590 3 590 < 2,000 < 1,200 4 785 < 4,000 < 2,350 5 980 < 10,000 < 5,900 6 1,180 > 10,000 > 5,900 7 1,380
High Pressure Ducts
• Maximum friction rate less than 0.4 inches W.G./100 ft • Velocity 2,000 - 3,500 ft/min (10 - 18 m/s)
Shafts Air Flow Rate Maximum Velocity (m3/h) (CFM) (m/s) (ft/min) < 5,000 < 2,950 12 2,350 < 10,000 < 5,900 15 2,950 < 17,000 < 10,000 17 3,350 < 25,000 < 14,700 20 3,940 < 40,000 < 23,500 22 4,300 < 70,000 < 41,000 25 4,900 < 100,000 < 59,000 30 5,800
It is common to keep main duct velocity above 20 m/s (3940 ft/min).
Corridors Air Flow Rate Maximum Velocity(m3/h) (CFM) (m/s) (ft/min) < 5,000 < 2,950 10 2,000 < 10,000 < 5,900 12 2,350 < 17,000 < 10,000 15 2,950 < 25,000 < 14,700 17 3,350 < 40,000 < 23,500 20 3,940
User Areas
• Offices, receptions, lounges and similar
Air Flow Rate Maximum Velocity(m3/h) (CFM) (m/s) (ft/min) < 5,000 < 2,950 10 2,000 < 10,000 < 5,900 12 2,350 < 17,000 < 10,000 14 2,750 < 25,000 < 14,700 16 3,150
Air Flow and Velocities due to Natural Draft
Air flow - volume and velocity - due to stack or flue effect caused by indoor hot and outdoor cold temperature difference
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A temperature difference between the outside and inside air will create a "natural draft" forcing air to flow through the building.
The direction of the flow depends on the temperatures. If inside temperature is higher than outside temperature, inside air density is less than outside air density, and inside air will flow up and out of the upper parts of the building. Cold outside air will flow into the lower parts of the building.
If outside temperature is higher than inside air temperature - the air flow will be in the opposite direction.
Natural Draft Head
The natural draft is caused by the difference in outside and inside air density. The natural draft head can therefore be expressed as
dpmmH2O = (ρo - ρr) h (1)
where
dpmmH2O = head in millimeter water column (mm H2O)
ρo = density outside air (kg/m3)
ρr = density inside air (kg/m3)
h = height between outlet and inlet air (m)
Natural Draft Pressure
Equation (1) can be modified to SI pressure units like
dp = g (ρo - ρr) h (1b)
where
dp = pressure (Pa, N/m2)
g = acceleration of gravity - 9.81 (m/s2)
Density and Temperature
With air density of 1.293 kg/m at 0 C, the air density at any temperature can be expressed as
3 o
ρ = (1.293 kg/m3) (273 K) / (273 K + t) (2)
or
ρ = 353 / (273 + t) (2b)
where
ρ = density of air (kg/m3)
t = the actual temperature (oC)
Equation (1) above can easily be modified by replacing the densities with equation (2).
Natural Draft Pressure Calculator
The calculator below can be used to calculate the natural draft pressure generated by the inside and outside temperature difference.
-20 outside temperature (oC)
20 inside temperature (oC)
10 height (m)
Major and Minor System Loss
The natural draft force will be balanced to the major and minor loss in ducts, inlets and outlets. The major and minor loss in the system can be expressed as
dp = λ (l / d ) (ρ v / 2)h r2 + Σξ 1/2 ρ vr
2 (3)
where
dp = pressure loss (Pa, N/m , lb /ft )2f
2
λ = D'Arcy-Weisbach friction coefficient
l = length of duct or pipe (m, ft)
dh = hydraulic diameter (m, ft)
Σ ξ = minor loss coefficient (summarized)
Air Flow and Air Velocity
Equation (1) and (3) can be combined to express the air velocity through the duct
v = [ (2 g (ρo - ρr) h ) / ( λ l ρ / dr h + Σξ ρ ) ] (4)r 1/2
Equation (4) can also be modified to express the air flow volume through the duct
q = π dh2 /4 [ (2 g (ρo - ρr) h ) / ( λ l ρ / dr h + Σξ ρr ) ] (5)1/2
where
q = air volume (m /s)3
Natural Draft Air Flow and Velocity Calculator
The calculator below can be used to calculate the air flow volume and velocity in a duct similar to the drawing above. The friction coefficient used is 0.019 which is appropriate for normal galvanized steel ducts.
-10 outside temperature (oC)
20 inside temperature (oC)
8 height (m)
0.2 duct hydraulic diameter (m)
3.5 duct length (m)
1 Σξ minor loss coefficient (summarized)
Example - Natural Draft
Calculate the air flow caused by natural draft in a normal family house with two floors. The height of the hot air column from ground floor to outlet air duct above roof is approximately 8 m. The outside temperature is -10 oC, the inside temperature is 20 oC.
A duct of diameter 0.2 m goes from 1. floor to the outlet above the roof. The length of the duct is 3.5 m. Air leakages through the building are neglected. The minor coefficients are summarized to 1.
The density of the outside air can be calculated like
ρo = (1.293 kg/m3) (273 K) / ((273 K) + (-10 oC))
= 1.342 kg/m3
The density of the inside air can be calculated like
ρr = (1.293 kg/m3) ( 273 K) / ((273 K) + (20 oC))
= 1.205 kg/m3
The velocity through the duct can be calculated like
v = [ (2 (9.81m/s2) ((1.342 kg/m3) - (1.205 kg/m3)) (8 m) ) / ( 0.019 (3.5m)(1.205 kg/m3)/(0.2 m) + 1(1.205 kg/m3) ) ] 1/2
= 3.7 m/s
The air flow can be calculated like
q = (3.7 m/s) 3.14 (0.2 m) / 4 2
= 0.12 m /s3
Note!
that these equations can be used for dry air, not for mass flow and energy loss calculations where air humidity may have vast effects.
Natural Draft Chart - SI and Imperial Units
Air Return Intakes - Sizes and Capacities
Sizes and capacities of air return intakes
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The diagrams below indicates aproximate capacities of air return intakes:
• Air Return Intakes - Sizes and Capacities (pdf)
Nominal Size
(inches x inches)
Aproximate Capacity
(cfm) 10 x 4 100 10 x 5 125 10 x 6 155 10 x 8 215 12 x 4 120 12 x 5 155 12 x 6 190 12 x 8 260 12 x 10 325 14 x 4 145 14 x 5 180 14 x 6 220 14 x 8 305 24 x 4 260 24 x 5 320 24 x 6 390 24 x 8 490 24 x 10 665 24 x 12 795 30 x 4 300 30 x 5 400 30 x 6 485 30 x 8 660 30 x 10 830 30 x 12 1015 30 x 14 1160 30 x 16 1355
Screen Efficiency
The free intake area will be reduced by an insect or bird screen.
• 1/2" mesh screen -> 90% free area • 1/4" mesh screen -> 80% free area • insect screen -> 50% free area
Air Intakes and Outlets
Ventilation systems - air intakes and outlets and rules of thumbs
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Intakes
• Intakes should be at least 0.15 m (0.5 ft) above the terrain. In areas with traffic the intake should be at least 5 m (16 ft) above the terrain.
• The distance between intake, firewalls and surrounding buildings must be according the rules of the local authorities.
• Short cutting intake air with used outlet air must be avoided. • An air intake should be located in a position where the wind influence on the
pressure conditions within the system is limited. • Air velocities in the intake openings should not exceed 2.5 m/s (~ 500 ft/min)
Outlets
• Outlets should go direct out in unrestricted area. • Short cuts with intake air, window openings and residence areas must be avoided. • The distance between outlet, firewalls and surrounding buildings must be
according the rules of the local authorities. • An air outlet should be located in a position where the wind influence on the
pressure conditions within the system is limited. • The air velocity through the outlet should not exceed 3 - 15 m/s (~ 500 - 3000
ft/min).
