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CHAPTER 21
1959 Guide
Table 2.. Grariar Equivalents of Rectangular Ducts for Equal Friction and Capacity
Air Duct Design
Table 2___ Circular Equivalents of Rectangular Ducts for Equal Friction and Capacity {Concluded) Diatcnaons to Incbet
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und other duct elements have been determined experimentally
and are given in Tables 3 and 4. It should be noted, however,
that absolutely reliable dynamic loss coefficients are not yet
available for all- duct elements and that the information
available for pressure losses due to area changes is generally
restricted to symmetrical area changes.
Jig. 7, which shows the relation of velocity pressure to
velocity for standard air "(V = 4005\/IQ, can be conven
iently
to find the total dynamic pressure loss for any
duct element with known dynamic lota coefficient C.
PRESSURE LOSSES IN ELBOWS .
Dynamic-loss coefficients are nearly independent of the
air-velocity and are affected by the roughnes of the duct
walls only in the
of bends, for which the dynamic losses
are often grouped with the friction losses to facilitate design
calculations. An ASHAF survey* of available data has indi
cated that the method of expressing the combined dynamic
and friction losses due to an elbow as equivalent to the loss
in a length L of similar straight duct is justified for design
purposes, owing to the relation of the loss to the correspond
ing friction factor f.
Fig. 8 gives the additional equivalent length of duct in
terms of widths W for elbows in rectangular ducts; Fig. 9
. gives the equivalent length of duct in terms of diameters D
for round ducts. When these curves for additional equivalent length are used, the straight lengths of duct between elbows should be measured to the intersection of their center lines.
Example 6: Use of Fig. 8 for the calculation of elbow losses.
Given the portion of a duct system drown in Fig. 10, it is required to determine the pressure loss between points A and D. Air at standard conditions is being supplied at the rate of KfOO rfm in a 6 by 24 in.' galvanised duct of average construction. Elbows No. 1 and 2 have center fine radii of 18 and 19 in, re spectively.
Solution: For elbow No. 1 the radius ratio is Ri/Wi = 18/24 -- 0.75 and the aspect ratio is Hi/Wt = 6/24 = 025. The additional equivalent length for elbow No. 1 in terms of W is obtained from Fig. 8; (L/Wh = 11.5. Thus U. - 1L5 X 24/12 = 23 additional equivalent feet. Similarly for elbow No. 2, the ratio radius is Rt/Wt -- 9/6 -- 1-5 and the aspect ratio is Ht/Wt -- 24/6 = 4.0; Fig. 8 gives (.L/W)> = 6, so L* = 6 X 6/12 = 3 additional equivalent feet.
The total length of the straight runs from A to D is l -- + + lo- = 7 + 20 + 5 = 32ft and the additional equivalent
length due to the elbows isL = Zn + Ia = 23 + 3 = 26 ft. Thus the equivalent length of the system from A to D is 1 + L = 32 + 28 = 58 ft of 6 by 24 in. duct.
The diameter of a circular duct, equivalent in friction and capacity to this rectangular duct, is 12.4 in. as given by the table of circular equivalents, Table 2. At a delivery rate of