The effect of opening and closing branch dampers

In any system it is usual to ensure that the correct quantity of airflows along the branches off the index run by partially closing regulating dampers in the various branches. (The alternative to this would be to size down the branch ducts so that they absorbed the correct amounts of energy by virtue of their reduced diameters.) Only the index run has its damper fully open. Thus, in a simple system, as illustrated in Figure 15.29(a), if the index run is A-B-C, then the natural energy loss along BB’ is less than it is along BC, for the flow of the design air quantities. This situation is clearly anomalous; only one total pressure can exist at B, hence the loss along BB’ must be artificially increased by the partial closure of a regulating damper. Then

Aptbc = Aptbb"

The effect of opening and closing branch dampers

Volume

Fig. 15.28 The effect of changing the inclination of guide vanes at the inlet of a centrifugal fan.

Two interesting questions arise. What happens if the damper in BB’ is fully closed? What happens if it is fully opened? The answers are best provided by means of an example.

EXAMPLE 15.15

Suppose the simple system shown in Figure 15.29(a) is designed to deliver 1 m3 s“1 with a total pressure loss of 625 Pa, the air quantity handled being divided equally between the two duct runs, BC and BB’. The total pressure loss is made ujj as follows:

Plant 375 Pa when 1.0 m3 s-1 is flowing Duct A’B 60 Pa when 1.0 m3 s-1 is flowing Duct BC 190 Pa when 0.5 m3 s-1 is flowing

Duct BB’ 60 Pa when 0.5 m3 s-1 is flowing and its regulating damper is wide open.

Assume that there is negligible energy loss across the damper in branch BB’ when it is completely open.

Calculate:

(a) the pressure drop across the damper in BB’ when it is adjusted to give an airflow rate of 0.5 m3 s’1 in the branch,

(.b) the total airflow rate handled when the damper in BB’ is fully closed, and (c) the total airflow rate handled when the damper in BB’ is fully open.

Answer

(a) Pressure to be absorbed by the partially closed damper in BB’

= APtbc — Aptbb’

= 190 — 60 = 130 Pa

(.b) Any assumption can be made about the air quantity flowing along BC in order to plot a new system characteristic for the revised arrangement with the damper in BB’ closed. It is, however, most convenient to assume an airflow rate of 1 m3 s“1 along BC, since the loss of energy from A to B is already known for this rate.

Plant

A’

B

C

N

Regulating

Damper

(a)

B’

The effect of opening and closing branch dampers

Airflow in m3 s~1 (b)

Fig. 15.29 (a) Plant and duct diagrams for example 15.15. (b) System and fan characteristics for

Example 15.15.

Vl

подпись: vl

Hence

подпись: hence

APtab = 375 + 60 = 435 Pa when 1 m3 s_1 is flowing Ap*c = 190 x (1.0/0.5)2 = 760 Pa when 1 m3 s-1 is flowing

APtac = 435 + 760 = 1195 Pa when 1 m s is flowing

0.

Pa

подпись: pa

462 Airflow in ducts and fan performance A new system characteristic can now be established:

подпись: 462 airflow in ducts and fan performance a new system characteristic can now be established:

2 0.4 0.6 0.8 1.0

48 191 430 765 1195

This is plotted in Figure 15.29(b). The intersection of this new system curve with the fan characteristic shows that 0.72 m3 s’1 is handled when the damper in BB’ is fully closed.

(c) If the damper in BB’ is fully open, then the loss of total pressure from B to C must equal that from B to B’. Any value for this loss of pressure may be assumed and a new system curve established. For ease of computation, assume that the pressure drop along BB’ is 190 Pa. Hence the air quantity flowing along BB’ may be calculated:

Qby = 0.5 m3 s“1 with a pressure drop of 60 Pa

The effect of opening and closing branch dampers

= 0.89 m3 s 1 with a pressure drop of 190 Pa

When 0.89 m3 s_1 flows through BB’ the amount passing through AB is 0.89 + 0.50 = 1.39 m3 s"1. Since the design pressure drop from A to B is 375 + 60 = 435 Pa when 1.0 m3 s“1 is handled we have

A/?tab = 435 x (1.39/1.0)2 = 840 Pa

And the total system loss is 840 + 190 = 1030 Pa. This is the starting point for establishing a new system characteristic curve, according to an assumed square law, as shown in Figure 15.29(b). The intersection with the fan curve occurs at 1.07 m3 s“1, and this is the total air quantity handled by the system, when the damper in BB’ is left fully open. Thus, for the particular conditions of this example, an overload of 7 per cent is imposed on the system.

It is clear that a risk exists, when a system is first started, of overloading the fan motor if the branch dampers are open. This could cause the motor to burn out. Accordingly, it is sound practice to include a main damper which can be shut when the system is first started and then opened gradually until the approximate design air quantity is handled. It is also essential to add a margin of 25 to 35 per cent to the fan power when selecting the motor required to drive the fan. Incidentally, this margin provides something in hand if the fan does not at first deliver the required air quantity, and the speed of its rotation must be increased. If the power characteristic is known for a fan running at a certain speed, then the power curve for any other speed can be determined and plotted by means of Fan Law A.3, as given in section 15.16. See the discussion on proportional balancing in section 13.14 and on margins in section 15.18.

Posted in Engineering Fifth Edition


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