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Unformatted text preview: ee pipes in Example 6.17 are now in parallel with the same total head
loss of 20.3 m. Compute the total flow rate Q, neglecting minor losses. Solution
From Eq. (6.109a) we can solve for each V separately
2g 20.3 m 2
2g (1) 273,000.
Guess fully rough flow in pipe 1: f1 0.0262, V1 3.49 m/s; hence Re1 V1d1/
From the Moody chart read f1 0.0267; recompute V1 3.46 m/s, Q1 62.5 m3/h. [This problem can also be solved from Eq. (6.66).]
Next guess for pipe 2: f2 0.0234, V2 2.61 m/s; then Re2 153,000, and hence f2
0.0246, V2 2.55 m/s, Q2 25.9 m3/h.
Finally guess for pipe 3: f3 0.0304, V3 2.56 m/s; then Re3 100,000, and hence f3
0.0313, V3 2.52 m/s, Q3 11.4 m3/h.
This is satisfactory convergence. The total flow rate is
Q Q1 Q2 Q3 62.5 25.9 99.8 m3/h 11.4 Ans. These three pipes carry 10 times more flow in parallel than they do in series.
This example is ideal for EES. One enters the pipe data (Li, di, i); the fluid properties
( , ); the definitions Qi ( /4)d iVi , Rei
Vidi/ , and hf fi(...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
- Spring '08