English .   Español  .

The application of Appropriate Technology

# Worked Example 8: Sources at Different Elevations

Consider two water sources at different elevations connected together at point A, which then supplies a reservoir tank at point B.

Figure 21

The Bernoulli equations for source 1 to point A and source 2 to point A are given in Section 9 ii and are stated below:

For source 1:

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

For source 2:

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

If we assume the kinetic energy term is negligible and rearrange the equations for PA we get:

For source 1:

$P_A=\rho.g.(h_1-h_3-f_{h1-A})$

For source 2:

$P_A=\rho.g.(h_2-h_3-f_{h2-A})$

If we convert to head rather than pressure we get:

For source 1:

$H_A=h_1-h_3-f_{h1-A}$

(38)

For source 2:

$H_A=h_2-h_3-f_{h2-A}$

(39)

These are the general equations for two sources joined at a junction, and assume pressure continuity at the junction. If this does not occur the system will experience back pressures and other such effects making it difficult to predict the overall flow rate into the reservoir.

The Bernoulli equation from the junction A to the reservoir B is also given in Section 9 ii and is:

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

$H_3=H_A+h_3-f_{hA-3}$

(40)

Finally the continuity equation must be stated:

$q_1+q+_2=q_3$

Numerical Example

Consider a water system identical to that shown in Figure 21. The pipeline from spring tank 1 to the reservoir tank 3 has been laid and consists of 1100m of 1″ dia. pipe. It has been decided to “Tee” in a second water source, spring tank 2, at a point in the mainline 100 m from spring tank 1, and at an altitude of 80m. This new pipeline is 100m long. Spring tank 1 at an altitude of 100m, has a safe yield of 0.2LPS and spring tank 2, at an altitude of 110m, a safe yield of 0.3LPS. The reservoir tank has a control valve at the pipeline exit, an altitude of 0m and represents the altitude datum line. What diameter (or combination of diameters) of pipe would be the optimum in order to supply the required flow rate to the reservoir tank and what is the residual pressure at the control valve ?

First of all we must calculate the pressure head at point A (HA) by taking Equation (38) for Source 1:

$H_A=h_1-h-3-f_{h1-A}$

The required flow rate through this 100m section of 1″ dia. pipe, is 0.2LPS, so from the friction charts we can find the frictional head loss (fh1-A) by interpolation (see Appendix 5). This is approximately 0.7m. Substitute this into the above equation along with the relevant altitudes to get the pressure head at point A:

$H_A=100-80-0.7=19.3~m$

Now substituting this value into Equation (39) for Source 2 we get:

For source 2 :

$19.3=110-80-f_{h2-A}$

Which means that the required frictional head loss in the 100m of pipe between point A and the spring tank 2 (fh2-A) is:

$f_{h2-A}=110-80-19.3=10.7~m$

The required flow rate is 0.3LPS, so what will each diameter of pipe burn off, the data is shown below:

Pipe Dia. (in) Frictional head loss (m) for 100m of pipe at a flow rate of 0.3LPS
½ “ 18.58
¾” 4.73
1″ 1.47

It is clear that we will have to use a combination of pipe diameters to achieve a frictional head loss close to 10.7m. In this case the combination of ½” and ¾” diameters would be optimum.

The combination pipes equation is given in Jordan [P.203 Technical Appendix C] and is stated below:

$X=\dfrac {\left (100.H-(f_{hl}.L)\right )}{\left( f_{hs}-f_{hl}\right )}$

(41)

Where:
X = Length of smaller diameter pipe, to give desired frictional head loss (m).
H = Desired frictional head loss (m).
hl = Frictional head loss due to larger diameter pipe (m/100m).

fhs = Frictional head loss due to smaller diameter pipe (m/100m).
L = Total length of pipe (m).

In this case our desired head loss (H) is 10.7m, L = 100m, fhl = 4.73m/100m and fhs = 18.58m/100m (from above), so:

$X=\dfrac{\left (100\times 10.7\right )-\left (4.73 \times 100\right )}{\left (18.58-4.73\right )} = 43.10m$

So we should employ 43.10m of ½” dia. pipe and 56.90m of 1″ dia. pipe in order to get pressure continuity at point A.

Finally we need to find the residual head at the reservoir tank valve at point B.

By the continuity equation for flow rates:

$q_1+q_2=q_3$

We can say that the flow rate along the pipe from A to B is:

$q_3=0.2+0.3=0.5~LPS$

The frictional head loss for this flow rate through 1″ dia. pipe is from the charts, 3.86m/100m, so for 1000m of pipe the head loss is:

$f_{hA-3}=3.86\times 1000/100 = 38.6~m$

Substitute this value, the pressure head at point A (HA above) and the height of point A (h3) into Equation (40) to get the residual head H3 at the reservoir valve:

$H_3=19.3+80-38.6=60.7~m$