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Unformatted text preview: 3 (6.109a) Q3 (6.109b) If the total head loss is known, it is straightforward to solve for Qi in each pipe and
sum them, as will be seen in Example 6.18. The reverse problem, of determining Qi
when hf is known, requires iteration. Each pipe is related to hf by the Moody relation
2
hf f(L/d)(V2/2g) fQ2/C, where C
gd5/8L. Thus each pipe has nearly quadratic
nonlinear parallel resistance, and head loss is related to total flow rate by
Q2
2
gdi5
2
hf
where Ci
(6.109c)
Ci/fi
8Li
Since the fi vary with Reynolds number and roughness ratio, one begins Eq. (6.109c)
by guessing values of fi (fully rough values are recommended) and calculating a first
estimate of hf. Then each pipe yields a flowrate estimate Qi (Cihf /fi)1/2 and hence
a new Reynolds number and a better estimate of fi. Then repeat Eq. (6.109c) to convergence.
It should be noted that both of these parallelpipe cases — finding either Q or hf —
are easily solved by EES if reasonable initial guesses are given.
EXAMPLE 6.18
Assume that the same thr...
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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
 Sakar

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