CPD PROGRAMME 5 metres Outdoor air O Louvre Expansion Fan 0.5m G 3 metres
Final bend and away The following bend is treated in just the same way as with any other duct fi tting. Determine the value of ζ from tables and, if there is a change in duct area, determine the change in velocity pressure. To determine the correct data in Guide
H Diffuser
Figure 3: An example of a simplifi ed ducted air system√
Air to room R
C needs the Reynolds number, Re, for the air fl ow in the fi tting. This is fairly straightforward as Re = ρ c d / η where d is the diameter of the duct (m) and η is the dynamic viscosity of the air (kg/ms). At around 20C the dynamic viscosity of air is approximately 18x10-6
A B C 0.5m 40º 5 metres 5 metres DE F 0.5m Smooth bend
(to be considered in a future article) so between C and F will be the straight 0.50 m diameter duct of 10 metres length. Again looking at duct sizing diagrams (in Guide C or elsewhere) this gives a pressure drop of 0.21Pa/m duct. So the pressure drop for this straight section is 10 x 0.21 = 2.10Pa.
kg/ms. So in
this case, with the air velocity remaining at 3m/s, Re = 1.2 x 3 x 0.5/(18 x 10-6
) = 1 x 105
louvre (taken from the table in Guide C) is 3.0, so the static pressure loss as the air fl ows through the louvre due to friction, ∆ps = 3.0 x 21.60 = 64.80Pa. Although expressed as a drop in static pressure, this loss will be a direct reduction in total pressure and, increasingly, many people relate the pressure drop in ducted air systems directly in terms of total pressure. This is useful for clarity when selecting fans, but the use of the static pressure can provide a clearer interpretation of the pressure inside the duct that is available to drive air out through supply terminals (such as diffusers and VAV boxes). And so combining the louvre pressure
loss with that required to match the gain in velocity pressure, the static pressure at a point just after point A would be -21.60- 64.80 = -85.40Pa
Straight but still resistant As the air fl ows through a straight duct (between points B and C) the friction of the air against the side walls of the duct (as well as between the air molecules themselves) will cause the air to lose energy (to heat and a little in producing sound). The drop in energy manifests itself as
a pressure drop. Between B and C the air will be travelling at constant velocity, so its velocity pressure will remain constant. The drop in pressure will be quite small (in a low velocity ductwork system) – typically around 1 Pa per metre run of straight ductwork.
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In this particular circular steel ductwork system the value is 1.3Pa/m (and this may be checked by using fi gure 4.3 in Guide C). The length of that duct is 5m, and so the pressure drop is 5 x 1.3 = 6.50Pa.
Static regain The next section is an expansion – in a real application the duct may change shape to pass around an obstacle, to join a device (such as a fan) or, indeed, the expansion may be there to alter the pressure characteristics in the duct. Again, looking at Guide C (Table 4.55) the ζ factor for this expansion can be
determined as ζ = 0.20 (where AreaC/AreaB = 0.20/0.10 = 3 and the angle of the ‘cone’ is 40°). In this case (for the expansion) the pressure loss is calculated using the entering air velocity pressure, ie 21.60Pa. So the pressure loss is 0.20 x 21.60 = 4.32Pa. However, an interesting change takes
place at the same time. As the air passes through the expansion, the velocity will drop
to 0.6/0.2 = 3.00m/s giving pv = 0.6x3.002 =5.40Pa. Hence the velocity pressure has dropped by 21.60-5.40 = 16.20Pa. The loss in velocity pressure will be balanced by a gain in static pressure. So remembering that the pressure drop due to friction/turbulence will be 4.32Pa, the overall change in static pressure will be -4.32 + 16.20 = +11.88Pa, ie, an increase in static pressure – this is known as ‘static regain’. For the time being the fan will be ignored
. Using this value the ζ value can be read
from the Guide C tables for a smooth 90° bend as ζ = 0.290 and hence the pressure loss calculated as 0.290 x 5.4Pa = 1.57Pa, and since there is no change in average velocity the velocity pressure remains at 5.40Pa. The pressure drop for the following straight section (G to H) is simply 3 x 0.21 = 0.63Pa (as per previous straight section). Finally the air is supplied into the room.
If this were just a plain opening at the end of the duct the static pressure at this point immediately before the air leaves the duct would be practically zero (ie, the same as the room or atmospheric pressure) and the duct total pressure would simply be due to the duct velocity pressure. This would effectively mean that there would be one velocity pressure loss as the air left the duct and passed into the room. However, there would more normally
be a diffuser or grille at the end of the duct (and frequently some fl exible duct and/ or a reducer) that will additionally incur a pressure drop related to its ζ factor (or a pressure drop taken from manufacturers’ data). It is important not to forget the fi nal velocity pressure, as it can be signifi cant (although in this case it would only be 5.4Pa). In a CPD article in the near future,
this knowledge will be further applied to consider the system pressure profi les and fan requirements.
© Tim Dwyer August 2011 CIBSE Journal 43
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