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However, there are two, 90 degree turns in the branch that will cause a large pressure drop, not accounted for in our calculations. To compensate for this drop, we should increase the effective length of the branch. For a 90 degree, mitered elbow, the equivalent straight duct length is given by the following formula,
L90 elbow ( ft.) = 65 * D(in.) . 12
For a
L90 elbow = 65 * 126 = 32.5 ft.
So, the total equivalent straight duct length is,
Le = 71 ft. + 2 * 32.5 ft. = 136 ft. .
Now, the required duct diameter should be recalculated,
⎡ | 1.82 | ⎤ | 0.2058 | |
DA = ⎢ | 0.00123174 *136 *120 | ⎥ | = 6.67 in. ≅ 7 in. | |
0.1 | ||||
⎣ | ⎦ |
|
Thus, for section ”A”, the required duct diameter is 7 inches.
Now, it is necessary to determine the allowable pressure drops for all other segments in the branch. The equation for duct diameter can be rearranged to give the pressure drop for any flowrate, providing the duct length, the flowrate, and the duct diameter are known. The equation looks like this,
ΔP(in. W.G.) = | 0.00123174 * Le( ft.)*V (CFM )1.82 | . |
| ||
| D(in.)4.86 |
For section “B”, if the flowrate is 120 CFM, the length is 11 ft., and the duct diameter
ΔPB = 0.00123174 *11*1201.82 = 0.00644 in. W.G. .
74.86
For section “C”, including a
ΔPC = 0.00123174 * (6 + 37.9)*1201.82 = 0.0257 in. W.G. .
74.86
66