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# Worked Example 2: Natural Flow With Pipes of Different Diameters and Lenghts

Consider the same situation as in Worked Example 1, but in this case there are two pipes of different diameters (d1, d2) and lengths (L1, L2) connected together as shown in Figure 15 below.

Figure 15

Write the Bernoulli equation between points A and B and B and C, assuming the kinetic energy term to be negligible.

$A\to B~~~P_1 + \rho.g.h_1 = P_2 + \rho.g.h_2 + f_{h1}.\rho.g$ $B\to C~~~P_2+\rho.g.h_2 = P_3 + \rho.g.h3 + f_{h2}.\rho.g$

Rearrange the first equation to get P2:

$P_2=P_1+\rho.g.h_1-\rho.g.h_2-f_{h1}.\rho.g$

And substitute it into the second:

$P_1+\rho.g.h_1-\rho.g.h_2-f_{h1}.\rho.g+\rho.g.h_2=P_3+\rho.g.h_3+f_{h2}.\rho.g$

Simplify, noting that P1= P3 (both are at atmospheric pressure), to give:

$\rho.g.h_1-f_{h1}.\rho.g=\rho.g.h_3+f_{h2}.\rho.g$

Rearrange to put actual head on one side and frictional head on the other:

$\rho.g.h_1-\rho.g.h_3=f_{h1}.\rho.g + f_{h2}.\rho.g$

Now divide by ρ.g to convert to head:

$h_1-h_3=f_{h1}+f_{h2}$

(32)

This is the general equation for two pipes of different diameters and lengths joined together in a natural flow situation.

It should be noted that for any number of pipes n, with a total difference in head between the top and bottom tanks of Δh, the general equation can be written:

$\Delta h=f_{h1}+f_{h2}+f_{h3}+\cdots +f_{hn}$

(33)

So we can say that for any number of pipes n, the sum of the frictional head losses for each pipe, at some given flow rate Q will be equal to the total difference in head between the top and bottom tanks.

Numerical Example

A spring tank is 200m from a reservoir tank, the difference in height between the two is 10m. It is proposed to use 100m of ½” dia. pipe and 100m of 1″ dia. pipe to link the two, what will be the volumetric flow rate of the water assuming a natural flow situation ?

Knowing h1 – h3 is 10m we can substitute this into Equation 32 to give us:

$10=f_{h1}+f_{h2}$

As both the ½” and 1″ dia. pipes are 100m long we can compare the friction chart directly with the actual head of 10m.

The friction chart gives us the following range of data:

Flow Rate (LPS) Frictional Head Loss (m/100m) ½” dia Frictional Head Loss (m/100m) 1″ dia. Total (m)
0.06 1.05 0.09 1.14
0.13 3.77 0.30 4.07
0.19 7.99 0.62 8.61
0.25 13.61 1.07 14.68

Studying this data we can see that the natural flow rate lies somewhere between 0.19 and 0.25 as the total frictional head “burned off” is 10 m and the limits either side of this are 8.61m and 14.68m. We need to interpolate to find the correct answer.

Using the interpolation method (see Appendix 5) we can find the flow rate to give us exactly 10m of frictional head loss for the combination of pipes

The variables for the general interpolation Equation (30) are given below:

X1 = 8.61, X2 = 14.68, Y1 = 0.19, Y2 = 0.25 and X3 = 10.00.

Substitute these into the general interpolation Equation 30 to give:

$Y_3=0.19+\dfrac{\left (0.25-0.19\right )\times \left (10.00-8.61\right )}{\left (14.68-8.61\right )}$

And so, the approximation is Y3 = 0.2037 which can be rounded down to 0.20 LPS.

This is the flow that will produce a frictional head loss of 10m over a combination of 100m of ½” dia. pipe and 100m of 1″ dia. pipe and is therefore the natural flow rate of the system.