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The application of Appropriate Technology

# Part 9: Special Scenarios

### i) Parallel Pipes

Figure 8 shows a pipe network where the feeder pipe (1) divides into two pipes (2,3) of different diameters at point A and rejoins at point B to become another pipe section (4).

Figure 8

There are two important points here :

a) The flow into the junctions, must equal the flow out (see section 4. Equation (11)), so we can write:

$A_1v_1 = A_2v_2 + A_3v_3 = A_4v_4$

b) The pressure drop between points A and B in pipe 2 must equal the pressure drop between points A and B in pipe 3. This because there must be continuity of pressure at point B. So if we wrote out the Bernoulli equations between points A and B for each pipe (2,3)we would get:

For pipe 2:

$P_A + \frac{1}{2}.\rho.v_2^2 + \rho.g.h_A = P_B + \frac{1}{2}.\rho.v_2^2 + \rho.g.h_B + f_{h2}.\rho.g$

For pipe 3:

$P_A +\frac{1}{2}.\rho.v_2^2 + \rho.g.h_A = P_B +\frac{1}{2}.\rho.v_3^2+\rho.g.h_B+f_{h3}.\rho.g$

Examination of these two equations and simplification shows that the frictional head losses must be equal:

$f_{h2}=f_{h3}$

Knowing the two relations in a) and b) it is possible to calculate the velocities in pipes 2 and 3 by trial and error (using the friction charts) and thus the pressure drop between A and B.

Worked Example 7 gives a typical calculation for parallel pipes.

### ii) Water Sources At Different Elevations

Consider the scenario shown in Figure 9, where there are two water sources at different elevations joined together at point A, which then supplies a reservoir tank at B.

Figure 9

The important points to note are the following :

a. The pressures (P1, P2, P3) at all three tanks are equal (atmospheric).
b. The pressure at point A (PA) will be the same for all three pipes.
c. The velocities at the two top tanks will be zero (v1 = v2 = 0), so both of these kinetic energy terms disappear.

Typically the problem here is to design one of the source lines such that the pressure at the junction A is equalised. If the pressure is not equalised in the design then the flows will interfere with each other.

Assume, therefore, that the required flows out of each source are known, which means v1A and v2A are fixed.

The Bernoulli Equations for the two sources to point A, are given below:

For source 1:

$\rho.g.h_1 = P_A + \frac{1}{2}.\rho.v_{1A}^2 + \rho.g.h_3 + f_{h1-A}.\rho.g$

For source 2:

$\rho.g.h_2 = P_A + \frac{1}{2}.\rho.v_{2A}^2 + \rho.g.h_3 + f_{h2-A}.\rho.g$

For source 1 the pressure at the junction can be calculated by rearranging the equation as below:

For source 1:

$P_A = \rho.g.h_1 + \frac{1}{2}.\rho.v_{1A}^2 + \rho.g.h_3 + f_{h1-A}.\rho.g$

Knowing PA means we can work out the pipe size required to burn off the required amount of frictional head to equalise the pressures at the junction.

From the Continuity Equation (11) we can calculate the combined flow of the two source pipes.

$A_3v_{3A} = A_1v_{1A} + A_2v_{2A}$

Now we can write the Bernoulli Equation from the junction to the reservoir tank:

$P_A + \frac{1}{2}.\rho.v_{3A}^2 + \rho.g.h_3 = P_3 + \frac{1}{2}.\rho.v_3^2 + f_{hA-3}.\rho.g$

Assuming that v3A = v3 (this means there is a control valve at the reservoir tank), then we can either calculate the residual pressure at the control valve or calculate the pipe size required to reduce this residual pressure to close to zero (the minimum possible pipe size).

For natural flow conditions the calculation becomes too complicated for a treatment in this document.

Worked Example 8 gives a typical calculation for sources at different elevations.