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Unformatted text preview: Li/di)(V 2/2g);
i
plus the Colebrook formula (6.74) for each friction factor fi. There is no need to use resistance
ideas such as Eq. (6.109c). Specify that fi 0 and Rei 4000. Then, if one enters Q ∑Qi
(99.8/3600) m3/s, EES quickly solves for hf 20.3 m. Conversely, if one enters hf
20.3 m , EES solves for Q 99.8 m3/h.  v v EES  eText Main Menu  Textbook Table of Contents  Study Guide 6.8 MultiplePipe Systems 379 Consider the third example of a threereservoir pipe junction, as in Fig. 6.24c. If all
flows are considered positive toward the junction, then
Q1 Q2 Q3 0 (6.110) which obviously implies that one or two of the flows must be away from the junction.
The pressure must change through each pipe so as to give the same static pressure pJ
at the junction. In other words, let the HGL at the junction have the elevation
hJ zJ pJ
g where pJ is in gage pressure for simplicity. Then the head loss through each, assuming p1 p2 p3 0 (gage) at each reservoir surface, must be such that
h1 2...
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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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