constant as it fl ows throughout the duct. This reasonable assumption also allows the use of water (in this article) to more readily illustrate the pressures involved in fl uid fl ow. Drawing (a) in Figure 2 shows a round pipe carrying water with a section of clear tube attached at right angles (‘normally’) to the side of the pipe. Looking at drawing (a) with the water


fl owing smoothly through the straight pipe, the height, z, of the column of fl uid gives what is known as the ‘static head’ of the water at that point – the vertical pipe is a simple ‘manometer’. This refl ects the static energy in the water, since the movement of the water is in line with the direction of the pipe and so the velocity of the water will not impose a pressure at the entry to this manometer tube. If, as an example, the height of the water


in the manometer, z, was 0.20m, then this would be the value of the static head and the static pressure (relative to the air around the pipe) at that point can be determined from pressure = ρ g z (Pa), where ρ is the density of the water, (nominally 1000kg/m3


) and g


is the acceleration due to gravity, 9.81 m/s2. So, in this particular case the (relative)


static pressure = ρ g z = 1000 x 9.81 x 0.20 = 1962Pa. This static pressure happens to be


positive relative to the air outside the pipe (the atmospheric pressure), and so any leaks in the pipe would push water out into the air. If, however, this were a length of pipe being used to draw water from a reservoir below into a pump (in suction) the relative static pressure would be negative; and if this simple manometer tube were still attached to the pipe, it would suck air into the system. If a tube is added to the inside of the original pipe facing the direction of


High velocity energy


Figure 2: Measuring the (a) static head and (b) total head of water fl owing through a pipe


(a)


the fl ow, known as a ‘Pitot tube’ (as shown in drawing (b) in Fig 2), then the height of the water in this manometer would be greater as this will now additionally refl ect the velocity energy of the fl owing water (that is always positive) as well as the static head. The manometer column height, z, will give the sum of the static head + velocity head. Used in conjunction with an adjacent static head reading (as in drawing (a) in Fig 2) the velocity head may be determined by subtracting the static head from the combined total head (static head + velocity head); and this assumes that the potential head is the same for both measurements, and so cancels itself out. And these can readily be converted into


static and velocity pressures, as before, using pressure = ρ g z. The manometer cannot refl ect the ‘potential pressure’ – in a level piece of pipework, potential pressure will not alter but as the pipe rises the potential pressure will also rise and there will be an equivalent drop in the static pressure. For the fl ow of air in ductwork in low-rise buildings the changes in potential pressure are almost always neglected, as these are relatively small (due to the low density of the air). So practically, when considering the pressure of ducted air systems, the potential pressure is assumed to be constant; duct


velocity pressure (pv) + duct static pressure (ps) = duct total pressure (pt). If the average velocity, c, of the air (m/s) is known, the value of velocity pressure


Lower velocity


(pv) can be calculated from 0.5 ρ c2 (and frequently this is shortened to 0.6 c2 by applying a ‘standard’ density of air of 1.2 kg/ m3


Low velocity energy


). This value is based on air fl ow, with


the speed c being measured normally to the direction of the air’s travel (ie in line


and the area, A (m2


Resistance to change As the air enters the ductwork system – in Figure 3 through a simple louvred entry – the air will be accelerated from the still air outside the louvre to a velocity determined by the volume fl owrate, qv (m3


pv, of virtually zero and once in the duct pv = 0.6x62


= 21.60Pa. This gain in velocity


Static head Velocity


head


pressure will need to be matched by a drop in static pressure. As the air fl ows through the louvre it


(b) Total head


will have to overcome some resistance and hence suffer an additional pressure loss. This can be calculated using the zeta factor, ζ, for the fi tting (obtainable from section 4.11 of CIBSE Guide C 2007 or from manufacturers’ data). The resulting drop in static pressure is


given by ∆ps = ζ x pv, and the value of pv is normally taken as that downstream of a fi tting (but this may vary and should be clearly indicated in the tables in Guide C and elsewhere). The ζ for this particular


42 CIBSE Journal August 2011 www.cibsejournal.com


In another CPD article to be published in the future the knowledge discussed here will be further applied to consider pressure profi les and fan requirements


with the duct) – in most ‘real’ applications the velocity of the air will alter across a duct due to the friction at the side of the duct, to obstructions and changes in direction.


Pressure (pv) + duct static pressure (ps) = duct total pressure (pt), where the term ‘duct static pressure’ will be taken to


mean ‘duct static pressure relative to the surrounding air’. If the average velocity, c, of the air (m/s)


is known, the value of velocity pressure (pv) can be calculated from 0.5 ρ c2


(frequently this is shortened to 0.6 c2 ‘standard’ density of air of 1.2 kg/m3


by applying a ).


In a theoretical world, if the duct was frictionless, then at two points in a duct (for example, points A and B in Figure 3) Bernoulli’s Equation would mean that the total pressure at A would equal the total


pressure at B, ie, ptA = ptB. But, of course in real ducts there is friction and so ptA = (ptB + frictional losses).


/s) of the air, ) of the duct. In this case


as c = qv/A = 0.6/0.1 = 6m/s. Before the air enters the duct it has a velocity pressure,


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