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The application of Appropriate Technology

# Worked Example 7: Parallel Pipes

Consider the parallel pipes shown in Figure 19.

Figure 19

The two key expressions are from Section 9 i), firstly that the flow in equals the flow out, so we can say:

$Q=Q_1+Q_2$

And for pressure continuity at point B, the frictional head loss between A and B in both pipes must be equal, so:

$f_{h1}=f_{h2}$

Numerical Example

A supply line is divided at a junction (A) into two 100m long pipes one of 1″ dia. and the other of ½” dia. which run parallel and connect at junction (B) further down the gradient. If the flow rate through the supply pipe is 1LPS, what are the flow rates through each parallel pipe and the frictional head loss between points A and B ?

We know the total flow through the system (Q) is 1LPS so we can substitute this into the flow equation above:

$1=Q_1+Q_2$

And we can rewrite this as:

$Q_2=1-Q_1$

We will assume that the flow through the 1″ dia. pipe is designated Q1 and through the ½” dia. pipe, Q2. We will now choose values of Q2 and calculate the frictional head loss (fh2) for pipe 2 from the friction charts. Using the rearranged flow equation we will calculate the corresponding flow rate in pipe 1 (Q1) and from the friction charts the frictional head loss in pipe 1 (fh1). This data is shown below:

Q2 (LPS) Frictional head loss fh2 (m/100m) Q1 (LPS) Frictional head loss fh1 (m/100m)
0.06 1.05 0.94 12.30
0.13 3.77 0.87 10.67
0.19 7.99 0.81 9.32
0.25 13.61 0.75 8.01

As both pipes are 100m long we can plot the two frictional head losses on a graph against the flow rate for Pipe 2 (Q2).

Figure 20

The point where the two lines intersect is the flow rate in Pipe 2 when the two frictional head losses are similar. In this case it is approximately:

$Q_2=\bf{0.20~LPS}$

Which means that the flow in Pipe 1 is:

$Q_1=1-Q_2=1-0.2=0.80~\bf{LPS}$

And from the graph, the frictional head loss at the point of intersection is approximately:

$f_{h1}=f_{h2}=9.0~m$