Airflow through a simple duct system

Figure 15.7 shows a simple straight duct on the suction side of a fan, followed by a similar duct of smaller cross-section on its discharge side. There is a grille over both the inlet and outlet openings of the duct.

Airflow through a simple duct system



Friction loss in outlet duct

Fig. 15.7 Airflow through a simple, notional duct system.

Air entering the system is accelerated from rest outside the duct to a velocity V prevailing

Within the duct, once the vena-contracta has been passed and the flow has settled down to

Normal. If, for convenience of illustration, the physical presence of the vena-contracta is ignored but the loss which it causes is included with the frictional loss past the inlet grille, the entry loss can be shown as occurring in the plane of entry. We can then write

Entry loss = 0 — px = — pti static suction = — ptl — pwl = —ps

The static depression immediately within the inlet grille is numerically greater than the velocity pressure by an amount equal to the energy loss past the inlet grille.

The loss due to friction along the duct between plane 1 and plane 2 equals ptl-pt2. The static depression is clearly pt2 — pv2. It is also clear that the total pressure at fan inlet, pt2, represents the loss of energy through the system up to that point, and that it comprises the frictional loss past the inlet grille, the loss past the vena-contracta and the loss due to friction in the duct itself.

On the discharge side of the system it is easier, for purposes of illustration, to start at the discharge grille. The change of static pressure past this grille equals ps4 — 0 but the change of total pressure across the grille equals pt4 — 0, or the friction plus the loss of kinetic energy. Between the plane 3 and 4, the loss of energy due to friction equals pt3 — pt4, in the direction of airflow. The total energy upstream must exceed that downstream if airflow is to take place. Hence, pl3 must be greater than pl4 and so also must ptl be greater than pt2, in a similar way. But pt3 is the total energy at fan outlet and pt2 is the total energy at fan inlet. Consequently, the difference between these two quantities, /?t3 and pl2, must be the energy supplied to the system by the fan. This statement fits in with equation (15.21) and we can write

PtҐ=Pii-Pt2 = I A3 1 + I Pt2 I

Numerically speaking, since pl2 has a negative sign. Thus, the fan total pressure can never be negative.

To sum up:

/7tP = [friction past the inlet grille + energy loss at the vena-contracta]

+ [friction loss in the inlet duct]

+ [friction loss past the outlet grille + kinetic energy lost from the system]

+ [friction loss in the outlet duct]

= (0 — Ai) + (Pti — Pm) + (Pt4 — 0) + 03t3 — pt4)

= (Pt3-Pa)

Posted in Engineering Fifth Edition