## EPANET Links

http://www.epanet.com – One-stop source for software technology, news, and collaboration related to Epanet and water distribution modelling. http://www.zonums.com – Programs and complimentary tools for use with Epanet. http://google-maps-images-downloader.en.softonic.com – Program that allows the user to download Google Map images for use with...

Read More## EPANET Program Download And Manuals

From the links below you can download the EPANET program for free and also download the EPANET manual. Download EPANET 2.0.1.1 (1.5 mb zip file) Download the EPANET Manual (1 mb pdf file) Download EPANET website can be found at ...

Read More## Worked Example 9: Sample Water System Design

Numerical Example Figure 22 below shows the topographic survey results for a proposed gravity flow water system. 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: Requirement for and placement of break pressure tanks. Design of pipe work. 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: Consider the section of the system from the spring (point 1) to low point 2. The static head is: 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: 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: 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: so: and therefore: and: so: and therefore: 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: so: and therefore: And so: and therefore: 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. y inspection of Figure 23, the maximum static heads (after the introduction of the two break pressure tanks) are as follows: At BP1: At Point 2: At Point 3: At BP2: At Point 4: All of which are acceptable. 2. Design...

Read More## 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. 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: For source 2: If we assume the kinetic energy term is negligible and rearrange the equations for PA we get: For source 1: For source 2: If we convert to head rather than pressure we get: For source 1: (38) For source 2: (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: (40) Finally the continuity equation must be stated: 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 ? Answer First of all we must calculate the pressure head at point A (HA) by taking Equation (38) for Source 1: 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: Now substituting this value into Equation (39) for Source 2 we get: For source 2 : Which means that the required frictional head loss in the 100m of pipe between point A and the spring tank 2 (fh2-A) is: 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...

Read More## Worked Example 7: Parallel Pipes

Consider the parallel pipes shown in Figure 19. The two key expressions are from Section 9 i), firstly that the flow in equals the flow out, so we can say: And for pressure continuity at point B, the frictional head loss between A and B in both pipes must be equal, so: Numerical Example A supply line is divided at a junction (A) into two 100m long pipes one of 1″ dia. and the other of ½” dia. which run parallel and connect at junction (B) further down the gradient. If the flow rate through the supply pipe is 1LPS, what are the flow rates through each parallel pipe and the frictional head loss between points A and B ? Answer We know the total flow through the system (Q) is 1LPS so we can substitute this into the flow equation above: And we can rewrite this as: We will assume that the flow through the 1″ dia. pipe is designated Q1 and through the ½” dia. pipe, Q2. We will now choose values of Q2 and calculate the frictional head loss (fh2) for pipe 2 from the friction charts. Using the rearranged flow equation we will calculate the corresponding flow rate in pipe 1 (Q1) and from the friction charts the frictional head loss in pipe 1 (fh1). This data is shown below: Q2 (LPS) Frictional head loss fh2 (m/100m) Q1 (LPS) Frictional head loss fh1 (m/100m) 0.06 1.05 0.94 12.30 0.13 3.77 0.87 10.67 0.19 7.99 0.81 9.32 0.25 13.61 0.75 8.01 As both pipes are 100m long we can plot the two frictional head losses on a graph against the flow rate for Pipe 2 (Q2). The point where the two lines intersect is the flow rate in Pipe 2 when the two frictional head losses are similar. In this case it is approximately: Which means that the flow in Pipe 1 is: And from the graph, the frictional head loss at the point of intersection is...

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