1 ENG . KHALED AL- OSSAILY e-mail: [email protected]DUCT DESIGN Equal friction method 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 (m 3 /h, cfm) in every room and branch of the system 2. Use 1) to compute the total air volume (m 3 /h, cfm) in the main system 3. Determine the maximum acceptable airflow velocity in the main duct 4. Determine the static pressure drop in the main duct 5. Use the static 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 necessary air volume - q. Summarize for each branch and accumulate. 2. Compute the total volume in the system Use 1) to summarize and accumulate the total volume - q total - in the system. Note! Be aware that maximum load conditions almost never occurs in all 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 required).
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 static pressure drop in the main duct 5. Use the static 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 necessary air volume - q. Summarize for each branch and accumulate.
2. Compute the total volume in the system
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 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 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.
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.
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 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) A proper velocity will depend on the application and the environment. The table below indicate commonly used velocity 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.
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
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 Ducts Air Velocity
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
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) / de
5.02 (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)
The equivalent diameter of a oval duct or tube can be calculated as (Heyt & Diaz) de = 1.55 A0.625/P0.2 (2) where A = cross-sectional area oval duct (m2, in2) P = perimeter oval duct (m, inches)
The cross-sectional area of an oval duct can be expressed as
A = (π b2/4) + b(a - b) (2a)
where a = major dimension of the flat oval duct (m, in) b = minor dimension of the flat oval duct (m, in) The perimeter of an oval duct can be expressed as P = π b + 2(a - b) (2b)
Note! Equivalent diameter is not the same as hydraulic diameter. The hydraulic diameter express the rate between the area section of the duct or tube, and the wetted perimeter of the duct or tube. Hydraulic diameter is used to determine if the flow is laminar or turbulent and to calculate the pressure loss.
Ventilation system may be designed more or less according the following procedure:
1. Calculate heat or cooling load, including sensible and latent heat
2. 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
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) below room temperature For cooling where high velocity diffusing jets are used -
If air is used for heating, the needed air flow rate may be expressed as
qh = Hh / ρ cp (ts - tr) (1)
where qh = volume of air for heating (m3/s) Hh =heat load (W) cp = specific heat capacity of air (J/kgK) ts = supply temperature (oC) tr = room temperature (oC) ρ = density of air (kg/m3) Air Cooling
If air is used for cooling, the needed air flow rate may be expressed as
qc = Hc / ρ 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 = 0.400 kW, supply temperature ts = 30 oC and the room temperature tr = 22 oC, the air flow rate can be calculated as:
If it is necessary to humidify the indoor air, the amount of supply air needed may be calculated as:
qmh = Qh / ρ (x2 - x1) (3)
where qm = volume of air for humidifying (m3/s) Qh = moisture to be supplied (kg/s) ρ = 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 / ρ (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 humidity x1 = 0.001 kg/kg and supply air humidity x2 = 0.008 kg/kg, the amount of air can expressed as:
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 (kW/m2 K) (5.68 10-3 for sheet metal ducts, 2.3 10-3 for insulated ducts) 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 manufactures catalogues.
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.
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/m2) 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/m2 per m duct) l = lengthof duct (m) Duct friction resistance per unit length can be expressed as R = λ / dh (ρ v2 / 2) (10) where R = pressure loss (Pa, N/m2) λ = friction coefficient dh = hydraulic diameter (m) 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.
The velocity of air in a ventilation duct can be expressed in imperial units like
vi = qi / Ai = 576 qi / (π di2) = 144 qi / (ai bi) (1)
where vi = air velocity (ft/min) qi = air flow (cfm) Ai = area of duct (square feet) di = diameter of duct (inches) ai = width of duct (inches) bi = width of duct (inches) 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
vm = qm / Am = 4 qm / (π dm2) = qm / (am bm) (2)
where vm = air velocity (m/s) qm = air flow (m3/s) Am = area of duct (m2) dm = diameter of duct (m) am = width of duct (m)
bm = width of duct (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).
