Pipes in Series with multiple demands
11-05-2016, 05:45 PM (This post was last modified: 11-25-2016 09:55 AM by Ángel Martin.)
Post: #1
 Ángel Martin Senior Member Posts: 1,368 Joined: Dec 2013
Pipes in Series with multiple demands
Pipes in Series with multiple demands. [ TUBSER ]
From the author’s Engineering Collection, included in the ETSII4 module (ETI4 on the CL Library)

This program has two modes of utilization. In DESIGN mode, it calculates the optimal diameters Di for a set of pipes in series with multiple demands, when the costs relation is known. In OPERATION mode it calculates the total head required to meet the demand with known pipe diameters.

The equation for optimal diameter di for a pipe with friction loss factor fi and flow qi is shown below:

di = (lambda *[ fi qi^2 ]^(1/a+5); with a cost relation given as Ci = A di^a

where lambda reflects the combined configuration in series,

lambda = [ (k/pi HT) SUM { Li(fi qi^2)^(a/a+5) } ]^(1/5) ; i = 1, 2,.. n

Where HT is the available total head (or piezometric height between ends of the pipes), and k is a constant defined as k = 8/g pi^2

Note that qi is the flow through each of the pipes – not the discharged flow at the points of demands.

Whereas the formula for the total available head when the diameters are known (i.e. not the design case) is given by the expression below:

DH = k SUM [ fi Li qi^2 / Di^5 ]; i = 1,2… n

Example.

Obtain the optimal diameters for the pipeline with 15 m of total available head, formed by 4 individual pipes of fibrocement (friction loss factor = 0.018, and cost relation given by C = 0.228 D^0.567 with D in meters); with the following conditions:

Lengths (m), flows q(l/s)
L1=500 L2=200 L3=315 L4=180
q1=100 q2=4 q3=86 q4=74

The results are provided below: Diameter (m)
D1=0.2589; D2=0.2532; D3=0.2453; D4=0.2324;
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