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The application of Appropriate Technology

Numerical Example

Figure 22 below shows the topographic survey results for a proposed gravity flow water system.

Figure 22

Figure 22

Assuming that the allowable pipe pressure head is 100m, the safe yield from the spring is 0.25LPS and that the design parameters from Jordan are adhered to (see Appendix 4.), design a system that will supply water to the community for the minimum cost.

Answer

The design of this water system will be approached in the following phases:

  1. Requirement for and placement of break pressure tanks.
  2. Design of pipe work.
  3. Check chosen pipe work for low pressure head parameter.

1. Requirement for and placement of break pressure tanks

Consider the control valve at the reservoir tank being closed to begin with. This is the maximum static head condition (Appendix 4 No. 1). All the static heads can be calculated from Equation (35) below:

H_{max}=h_1-h_2

Consider the section of the system from the spring (point 1) to low point 2. The static head is:

H_{max~1-2}=200-50=150~m

which exceeds the 100m allowable pipe pressure head given above.

Consider the section of the system from the spring (point 1) to high point 3. The static head is:

H_{max~1-3}=200-125=75~m

which is within the above limit.

Consider the section of the system from the high point 3 to the reservoir tank (point 4). The static head is:

H_{max~3-4}=125-0=125~m

which exceeds the 100m allowable pipe pressure head given above.

It is clear that the pipes at points 2 and 4 will blow unless we introduce break pressure tanks to relieve the pressure.

The first break pressure tank (hereafter BP1) must relieve the pressure at point 2 but still allow the water to flow over point 3.

The maximum height (hBP1) we can place BP1 at is given by:

h_{BP1}-h_2<= 100

so:

h_{BP1}-50<=100

and therefore:

h_{BP1}<=150

and:

h_{BP1}-h_3>0

so:

h_{BP1}-125>0

and therefore:

h_{BP1}>125

So the height of BP1 must be less than or equal to 150 and greater than 125, to satisfy the conditions. As there will be frictional losses in the pipes we should situate BP1 at its maximum allowable height, which is in this case 150m. This corresponds to a location approximately 225m from the spring tank.

The second break pressure tank (hereafter BP2) must relieve the pressure at point 4 but not exceed the pressure head limits in the pipe work between it and BP1.

The maximum height (hBP2) we can place BP2 at is given by:

h_{BP1}-h_{BP2}<=100

so:

150-h_{BP2}<=100

and therefore:

h_{BP2}>=50

And

h_{BP2}-h_4<=100

so:

h_{BP2}-0<=100

and therefore:

h_{BP2}<=100

So the height of BP1 must be greater than or equal to 50 and less than or equal to 100, to satisfy the conditions. So it will be placed at a height of 70m. This corresponds to a location approximately 2500m from the reservoir tank.

The two break pressure tank locations are added to the topographic survey, and shown below in Figure 23.

Figure 23

Figure 23

y inspection of Figure 23, the maximum static heads (after the introduction of the two break pressure tanks) are as follows:

At BP1:

H_{max~BP1}=200-150=50~m

At Point 2:

H_{max~2}=150-50=100~m

At Point 3:

H_{max~3}=150-125=25~m

At BP2:

H_{max~BP2}=150-70=80~m

At Point 4:

H_{max~4}=70-0=70~m

All of which are acceptable.

2. Design of pipe work

The safe yield of the water source is 0.25LPS. So in general we want to make sure that at no point in the system is more than 0.25LPS being drawn, as this will either empty a break pressure tank or the spring tank, allowing air into the system. We can do this in two ways.

  1. Design the natural flow situations such that just less than 0.25LPS is being drawn.
  2. Use control valves at the break pressure tanks and reservoir tank to limit the flow to just less than 0.25LPS.

We will consider the second option because the calculations are simpler and the control of the system easier.

Consider the section of pipe between the spring tank (point 1) and BP1:

This section of pipe is 225m long and the maximum static head is 50m (see above). From the friction tables for a controlled flow of 0.25LPS we get the following data.

Pipe Diameter Frictional head loss (m/100m) Frictional head loss for 225m of pipe (m) Velocity (m/s)
½” 13.61 30.62 1.28
¾ “ 3.47 7.81 0.73
1″ 1.07 2.41 0.45
1¼ “ 0.28 0.63 0.26

It is clear that we can use ½” dia. pipe here, as it will not reduce the maximum static head below the 10m low pressure head limit (see Appendix 4. No. 3) and the velocity of the water lies within acceptable parameters (see Appendix 4 No. 4).

The residual head at the BP1 valve (ΔHBP1) is given by Equation (34):

\Delta H_{BP1}=\left (h_1-h_{BP1}\right )-f_{h1-BP1}

Substituting the numerical values into this equation for ½” dia. pipe we get:

\Delta H_{BP1}=\left (200-150\right )-30.62=19.38~m

This is an acceptable residual head based upon the limits set in Appendix 4 No. 6.

Consider the section of pipe between BP1 and BP2:

This section of pipe is 2500 – 225 = 2375m long. The main feature we have to contend with is the topographical peak at point 3. This only lies 25m below BP1, so we can’t afford to “burn off” too much frictional head between BP1 and point 3 or else the residual head will drop below the limit of 10m (see Appendix 4. No. 3). So we need to burn off no more than 25 – 10 = 15m of head between BP1 and point 3.

The distance between BP1 and point 3 is from Figure 23, 1800 – 225 = 1575m. From the friction tables for a controlled flow of 0.25LPS we get the following data.

Pipe diameter Frictional head loss (m/100m) Frictional head loss for 1575m of pipe (m) Velocity (m/s)
½” 13.61 214.36 1.28
¾ “ 3.47 54.65 0.73
1″ 1.07 16.85 0.45
1¼ “ 0.28 4.41 0.26

We clearly cannot use ½” and ¾” diameter pipes here as they “burn off” far more than 15m of head.

We could use 1¼ ” dia. pipe here, but there are two problems. Firstly the water velocity is very low at 0.26 m/s (see Appendix 4 No. 4) and secondly it is a little more costly than 1″ dia. pipe (Appendix 4 No. 8).

So consider using 1″ dia. pipe. The water velocity is still below the limit of 0.7 m/s (see Appendix 4 No. 4), but the head loss is almost acceptable. There seems to be only one solution. Use 1″ pipe between BP1 and point 3, and place a wash out at the lowest point between BP1 and point 3, which happens to be point 2.

The pressure head at point 3 (H3) is given by Equation (34):

H_3=\left (h_{BP1}-h_3\right )-f_{hBP1-3}

Substituting the numerical values into this equation for 1″ dia. pipe we get :

H_3=\left (150-125\right )-16.85=8.15~m

This is a little less than the minimum low pressure head limit of 10m (see Appendix 4 No. 3) but is acceptable in the circumstances.

Now consider the second section of pipe between point 3 and BP2. The remaining pressure head is 8.15m (from above) and so we can add this to the remaining head between point 3 and BP2 to get the overall head.

H_{3-BP2}=\left (125-70\right )+8.15=63.15~m

The most desirable residual head (Appendix 4 No.6) at a valve or tap is around 15m. So we are looking to “burn off” something like :

f_{h3-BP2}=63.15-15.00=48.15~m

The distance between point 3 and BP2 is from Figure 23, 2500 – 1800 = 700m. From the friction tables for a controlled flow of 0.25LPS we get the following data.

Pipe diameter Frictional head loss (m/100m) Frictional head loss for 700m of pipe (m) Velocity (m/s)
½” 13.61 95.27 1.28
¾ “ 3.47 24.29 0.73
1″ 1.07 7.49 0.45
1¼ “ 0.28 1.96 0.26

We clearly cannot use ½” diameter pipe here as it “burns off” far more than 48.15m of head.

¾ ” dia. pipe here looks good, as although it only burns off approximately half of the required head, it produces a water velocity within the parameters required (Appendix 4 No. 4) and is cheaper than the 1″ dia. pipe (Appendix 4 No. 8).

The residual head at the BP2 valve (ΔHBP2) is given by :

\Delta H_{BP2}=H_3+\left (h_3-h_{BP2}\right )-f_{h3-BP2}

Substituting the numerical values into this equation for ¾ ” dia. pipe we get :

\Delta H_{BP2}=8.15+\left (125-70\right )24.29=38.86~m

This is in the high end of the residual head range based upon the limits set in Appendix 4 No. 6.

We can improve on this and reduce costs by using a combination of ¾” dia. pipe and ½” dia. pipe. Consider the combination pipes Equation (41).

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

Where:
X = Length of smaller diameter pipe, to give desired frictional head loss (m).
H = Desired frictional head loss (m).
fhl = 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 48.15m, L = 700m, fhl = 3.47m/100m and fhs = 13.61m/100m (from above), so :

X=\dfrac{(100\times 48.15-(3.47\times 700))}{(13.61-3.47)} = 235.3~m

This close to 200m, and as the pipe usually comes in 100m lengths, we should employ 200m of ½” dia. pipe and 500m of ¾ ” dia. pipe.

This combination of pipes will produce a total frictional head loss (fhcomb) of :

   \begin{array}{lll}  f_{hcomb}&=&f_{h{1/2}''}+f_{h{3/4}''}\\  &=&(13.61\times 200/100)+(3.47\times 500/100)=44.57~m  \end{array}

So the residual head at the BP2 control valve will be :

\Delta H_{BP1}=8.15+\left (125-70\right )-44.57=18.58~m

This very close to the desired residual head based upon the limits set in Appendix 4 No. 6.

Consider the section of pipe between BP2 and the reservoir tank (point 4) :

This section of pipe is 3200 – 2500 = 700m long and the maximum static head is at point 4 and is 70m (see above). From the friction tables for a controlled flow of 0.25LPS we get the following data.

Pipe diameter Frictional head loss (m/100m) Frictional head loss for 700m of pipe (m) Velocity (m/s)
½” 13.61 95.27 1.28
¾ “ 3.47 24.29 0.73
1″ 1.07 7.49 0.45
1¼ “ 0.28 1.96 0.26

If we try to achieve the desired residual head at the reservoir tank valve of 15m (see Appendix 4 No. 6) then we need to “burn off” 70 – 15 = 55m of head through friction. Studying the data above suggests that a combination of ½” and ¾ ” pipe would be the optimum solution. Using the combination pipes Equation (41) as previously:

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

In this case our desired head loss (H) is 55 m, L = 700m, fhl = 3.47m/100m and fhs = 13.61m/100m (from above), so:

X=\dfrac{\left (100\times 55-\left (3.47\times 700\right )\right )}{\left (13.61-3.47\right )}=302.86~m

This close to 300m, and as the pipe usually comes in 100m lengths, we should employ 300m of ½” dia. pipe and 400m of ¾” dia. pipe.

This combination of pipes will produce a total frictional head loss of :

   \begin{array}{lll}  f_{hcomb}&=&f_{h{1/2}''}+f_{h{3/4}''}\\  &=&(13.61\times 300/100)+(3.47\times 400/100)=54.71~m  \end{array}

The residual head at the reservoir tank control valve (ΔH4) is given by equation (34) :

\Delta H_4 = (h_{BP2}-h_4)-f_{hBP2-4}

Substituting the numerical values into this equation for the combination of ½” dia. and ¾ ” dia. pipe we get:

\Delta H_{BP1}=\left (70-0\right )-54.71=15.29~m

This is an acceptable residual head based upon the limits set in Appendix 4 No. 6.

3. Check chosen pipe work for low pressure head parameter

At this stage of the design we need to check that we have not reduced the pressure head in the pipe below the 10m limit set in Appendix 4 No. 3. We can do this most simply by plotting the Hydraulic Grade Line (see Jordan) onto the topographic survey. To do this we need to tabulate all the chosen pipe sections, their lengths and their respective friction head losses based upon a controlled flow of 0.25LPS. This data is shown below:

Section Pipe dia. Length (m) Frictional head loss (m/100m) Total frictional head loss (m/100m)
1–BP1 ½” 225 13.61 30.62
BP1–3 1″ 1575 1.07 16.85
3–BP2 ¾” 500 3.47 17.35
3–BP2 ½” 200 13.61 27.22
BP2-4 ¾” 400 3.47 13.88
BP2-4 ½” 300 13.61 40.83

These lines are plotted onto the topographic survey as shown in Figure 24.

Examination of the HGL shows that it the pressure head never becomes negative (i.e. the HGL never goes below the ground surface) and its minimum is at point 3, which we have already considered. Therefore the chosen pipe work design is acceptable.

Figure 24

Figure 24