Example 4
Determining The Flow Rate of Air in a Duct
Reconsider Example 8-4. Now the duct length is doubled while its diameter is maintained constant. If the total head loss is to remain constant, determine the drop in the flow rate through the duct.
Solution
The diameter and the head loss in an air duct are given. The drop in the flow rate is to be determined.
The average velocity, Reynolds number, friction factor, and the head loss relation are expressed as ( D is in m, V is in m/s, and Re and f are dimensionless ).
V = = → V =
Re = → Re =
hL = → 20 m =
This is set of four equations in four unknowns and solving them with an equation solver gives
V̇ = 0.24 m3/s , f = 0.0195 , V = 4.23 m/s , and Re = 68,300
Then, drop in the flow rate becomes
V̇drop = V̇old ̶ V̇new = 0.35 ̶ 0.24 = 0.11 m3/s ( a drop of 31 percent )
Therefore, for a specified head loss ( or available head or fan pumping power ), the flow rate drops by about 31percent from 0.35 to 0.24 m3/s when the duct length doubles.
Alternative Solution
The equation for V as a function of f is
Average velocity through the pipe : V =
Once V is calculated, the Reynolds number can be calculated, from which a corrected friction factor is obtained from Moody Chart. We guess f = 0.04 for illustration :
Iteration
|
f ( guess )
|
V, m/s
|
Re
|
Corrected f
|
1
|
0.04
|
2.955
|
4.724 × 104
|
0.0212
|
2
|
0.0212
|
4.059
|
6.489 ×104
|
0.01973
|
3
|
0.01973
|
4.207
|
6.727 ×104
|
0.01957
|
4
|
0.01957
|
4.224
|
6.754 ×104
|
0.01956
|
5
|
0.01956
|
4.225
|
6.756 ×104
|
0.01956
|
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