Design of ventilation systemsResistances in ductwork

Ventilation systems consist of various different components like fan, bends, grilles, filters etc. All these components have a resi- stance which needs to be conside- red to select a suitable fan. The pressure loss Pressure losses Figure 9 Pressure losses in a ventilation system – Ducting – Shaped sections (bends, T-pieces) – Other components (filters, heaters, grilles)
A B DuctingDuctingRain repellentgrille A B C C GrilleHeater batteryAttenuatorFanFilter unit90°-bend C A C C ∆ p
stat. (static pressu-re) of the total system is calculated by adding all individual resistances.
C Figure 10 Pressure losses ∆ p [Pa/m] (roughness

ε

= o)V.[m Resistances of shaped sections Pressure drop in circular orrectangular ductingEquivalent diameter d
3 /h], c [m/s], d [mm]
Figure 12Pressed bend 90°

A

Σ ∆ p = ∆ p 1 /L · L 1 + ∆ p 2 /L · L 2 +... [Pa] ∆ P/ L1,2 ...: from table in figure 10[Pa/m]L: Length of ducting [m]Equivalent diameter d h h 2 · w · hb + h d
h = [mm]
Air flow speed in m/s w: Width of ducting [mm]h: Height of ducting [mm] Equivalent diameter d Figure 13 Pressed bend 45° h d
h for rectangular fans w x h [cm]d
h [mm] 30 x 15200 40 x 20260 50 x 25330 60 x 30375
Duct diameter in mm 60 x 35400 70 x 40500 80 x 50600 100 x 50650
Figure 14 Symmetrical reducer Pressure loss per m ducting Correction factor for

ε

∆ p
R = ∆ p
ε =0 · Correction factor Correction factor for roughness ε of different materials Metal ducting1.5Wooden ducting1.5 Flexible ducting7.0Concrete ducting2.0 Pressure drop in shapedsections like bends, T-pieces,change in cross section area Fibrated concrete ducting1.5Bricked ducting3.0

B

Σ ∆ p Figure 11 Resistances of components
F = ∆ p
F1 + ∆ p
F2 +... [Pa] (as a calculation guide)
Figure 15Symmetrical reducer ∆ p Ï� 2 F = ζ · –– c 2 [Pa] ComponentResistance ∆ p F1,2 ...: from tables in figure 12-15[Pa]Equivalent diameter c: air flow velocity [m/s] ∆ p component [PA] Grilles, gravity shutters, louvres*20 –40 ζ : pressure loss coefficient Helios VK-shutters*10 – 20 Heater batteries, heat exchangers*100 –150 Filters clean*40 –60dirty250 –300 Resistances of components
Attenuators*40 –80

C

Σ A Com = ∆
p Com1 + ∆
p Com2 +... [Pa] Valves*10 –200 ∆ p Cyclone500 –750 Com1,2 ...: from table 11 or diagramm *more accurate figures see product page Velocity pressure at exhaust Total resistancecalcuation Air flow speed Ï�

D

∆ p
∆ p =+ + + [Pa] A B C D V.A · 3600
d = · c
2 [Pa]
total c = [m/s] 2
Ï� :Density [kg/m 3 ](air at 20 ºC, 1013 mbar = 1.2 kg/m A:Air flow cross section area [m 2 ]V.:Air flow volume[m 3 3 /h] c:Air flow velocity [m/s] 14