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# Worked Example 6: DistributionSystem – The General Equation

The general Equation (22) for the residual head for any tap in the distribution system is derived in Section 8 v: $\dfrac{P_n}{p.g}=h_t-h_n-f_{t-A}-f_{A-n}-\frac{1}{2}.\dfrac{v_A^2}{g}$

If we assume the kinetic energy term is negligible and write the residual head at any tap as Hn, then we get: $H_n=h_t-h_n-f_{t-A}-f_{A-n}$

(37)

This is the simplified general equation for the residual head of any tap in a distribution system of the form shown in Figure 18.

Numerical Example

A reservoir tank is at an elevation of 40m above a community that requires 3 public tap stands, each giving a flow of 0.25LPS. The tank supply line has been laid and is 100m of 1″ dia. pipe. The preferred residual head at each tap is 10m, the minimum residual head is 7m and the topographic survey results are as follows :

Tap site no. Height above datum (m) Distance from supply line junction (m)
1 2 50
2 20 100
3 7 100

What are the optimum diameter pipes that should be used for each tap and what are the residual heads at each tap ?

First of all we need to calculate the overall flow rate through the supply line, this is given by: $q_A=q_1+q_2+q_3=n.q$

Where n is the number of taps and q is the maximum flow through each tap. So: $q_A=3\times 0.25=0.75~LPS$

Calculate the frictional head loss in the 100m of 1″ dia. supply pipe due to a flow rate of 0.75LPS.

The friction tables give us the following data:

Flow Rate (LPS) Frictional Head Loss (m/100m)
0.69 6.96
0.76 8.19

Using the interpolation method (see Appendix 5) we can find frictional head loss for a flow rate of 0.75LPS.

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

X1 = 0.69, X2 = 0.76, Y1 = 6.96, Y2 = 8.19 and X3 = 0.75.

Substitute these into the general interpolation Equation (30) to give: $Y_3=6.96+\dfrac{\left (8.19-6.96\right )\times \left (0.75-0.69 \right)}{(0.76-0.69)}$

And so, the approximation is Y3 = 8.014m/100m which can be rounded up to 8.01 m/100m.

As we have 100m of pipe, we can say: $f_{t-A}=8.01m$

We can now substitute this value and the height of the tank into the general Equation (37) above to gives: $H_n=40-h_n-8.01-f_{A-n}=31.99-h_n-f_{A-n}$

If we assume we are trying to achieve an optimum residual head at each tap of 10m, then we can substitute this value as Hn in the above equation and rearrange it to give the required frictional head loss fA-n in each distribution pipe. $f_{A-n}=31.99-h_n-10=21.99-h_n$

So for each pipe in tabular form:

Tap site no. Height above datum (m) Required frictional head loss (m) to give 10m of residual head
1 2 19.99
2 20 1.99
3 7 14.99

Consider tap 1.

Examination of the friction chart shows us that the highest frictional loss for a flow of 0.25LPS is for a ½” dia. pipe and is 13.61m/100m. In this case we have 50m of pipe, so the actual head loss is: $f_{A-1}=13.61\times 50/100=6.8~m$

This is the best we are going to do so we will select 50m of ½” dia. pipe giving us a residual head at tap 1 of: $H_n=31.99-2.00-6.81=\bf{23.18~m}$

Consider tap 2.

Examination of the friction chart gives us the following frictional head losses for a flow rate of 0.25LPS:

½” 13.61
¾ “ 3.47
1″ 1.07
1¼ “ 0.28

We require a head loss of 1.99m and we have 100m of pipe, so it is clear that the 1″ dia. pipe is closest.

We will therefore select 100m of 1″ dia. pipe giving us a residual head at tap 2 of: $H_n=31.99-20.00-1.07=\bf{10.92~m}$

Consider tap 3.

Examination of the friction chart shows us that the highest frictional loss for a flow of 0.25LPS is for a ½” dia. pipe and is 13.61m/100m. In this case we have 100m of pipe, so the actual head loss is: $f_{A-3}=13.61~m$

This is the best we are going to do so we will select 100m of ½” dia. pipe giving us a residual head at tap 1 of: $H_n=31.99-7.00-13.61=\bf{11.38~m}$