Airflow through supply branches

Loss occurs along two paths when air flows through a brach-piece: there is a loss of energy for flow through the main and also for flow through the branch. The exact magnitude of the loss depends on the way in which the branch-piece is constructed but it is, nevertheless, possible to generalise. ASHRAE (1997a) gives extensive information on loss coefficients. The CIBSE (1986a) quotes multiplying factors, referred to the velocity pressure in the downstream main or the branch, as appropriate. Both of these are dependent on the ratio of the downstream to upstream velocities. Figure 15.16 illustrates this: there is a loss of total pressure between points 1 and 2, when airflow through the main is considered, and also a loss of total pressure (not necessarily the same) when air flows from point 1 to point

3, by way of the branch.

EXAMPLE 15.6

Q>

O, 2 ——

A,——- ►I V2

подпись: q>
o, 2 
a, ►i v2

Fig. 15.16 Airflow through a supply branch piece.

подпись: 
fig. 15.16 airflow through a supply branch piece.
Given that, in Figure 15.16, the velocities Vu V2 and V3 are 10, 8 and 6 m s-1, respectively, determine the total pressure loss and the static pressure change through the main (1-2) and the branch (1-3). Assume the section of the branch duct is square.

Answer

The CIBSE (1986a) quotes the following loss coefficients:

For the main,

V2/V{ =0.8, Ci-2 = 0.18

For the branch,

Vj/Vi = 0.6, ^i_3 = 3.5 x the factor for the equivalent bend = 3.5 x 0.23. See CIBSE (1986a)

= 0.805

Hence we have

For the main:

Apt = 0.18 x 0.6 x 82 = 6.9 Pa Aps = 0.6 x 102 — 0.6 x 82 — 6.9 = 14.7 Pa

That is, a static regain has occurred in the main because the fall in velocity pressure exceeded the loss so there is a net increase in the potential energy or the static pressure.

For the branch:

Apt = 0.805 x 0.6 x 62 = 17.4 Pa Aps = 0.6 x 102 — 0.6 x 62 — 17.4 = 21.0 Pa

And a static regain has taken place here also.

Posted in Air Conditioning Engineering


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