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# EPANET & System Modeling

EPANET is a Windows 95/98/NT/XP program that models water distribution piping systems, allowing the user to model and design efficient and effective networks without needing to fully understand the intricacies of hydraulic theory and fluid mechanics. Epanet performs extended-period simulation of the hydraulic and water quality behaviour within pressurized pipe networks, both for gravity-flow and pump systems.

EPANET is public domain software that may be freely copied and distributed.

## Preface

i. Glossary Term Description Units a Acceleration Metres/Second2 A Area Metres2 At Reservoir Supply Pipe Area Metres2 D,d Pipe Diameter Metres/inches E Energy Joules Ek Kinetic Energy Joules Ep Potential Energy Joules F Force Newtons f Friction Factor – fA-n Frictional Head Loss in Tap Supply Pipe Metres fh Frictional Head Loss Metres fhl Frictional […]

## Part 1: Introduction

There are basically two equations needed to design gravity flow water systems: The Continuity Equation The Bernoulli Equation With these two relations and an understanding of frictional effects most systems can be designed and/or analysed.

## Part 2: General Concepts

The following are the basic equations which can be used to derive the Continuity and Bernoulli Equations: (a) Velocity = Distance Moved/Time Metres/Second (m/s)(1) Example: If I run 100 metres in 10 seconds (in a straight line), I have a velocity of 100/10 = 10 metres per second (m/s). (b) Acceleration = Velocity Change/Time Metres/Second2 […]

## Part 3: Derivation of the Continuity Equation

Consider an incompressible fluid (water is almost incompressible) flowing along a pipe, as in Figure 1. Its volume (V) is given by: Therefore the volume passing per second (the volumetric flow rate Q) is given by: But we can write velocity as distance moved/time (see Equation (1)), so we can replace L/t by v: (9) […]

## Part 4: The Continuity Equation for Multiple Pipe System

The rule for multiple flow paths for incompressible fluids is: This is written mathematically as: (10) Consider the pipe system shown below (in section) in Figure 3: In this case the flow in is given by: And the flow out is given by: So from Equation X we can write: (11) This is true for […]

## Part 5: Energy in a Perfect System – The Bernoulli Equation

Anywhere in a perfect system (i.e. there are no frictional effects), for an incompressible fluid there are three types of energy existing: Pressure Energy. Example: If you blow up the tyre of a car with a pump you are turning your physical energy of working the pump into pressure energy in the tyre. Kinetic Energy. […]

## Part 6. Imperfect Systems – Friction and the Bernoulli Equation

Of course in the real world, systems are not perfect. Energy is lost from a real system as friction. This is the sound and heat generated by the fluid as it flows through the pipes. These forms of energy are lost by the fluid rubbing against the walls of the pipe, rubbing against itself and […]

## Part 7: Pumps and Turbines – The Bernoulli Equation

Essentially a pump adds energy to a system and a turbine takes it away. Therefore typically in the Bernoulli Equation the pump pressure (Pp) is added to the left-hand side of the equation and the turbine pressure (Pt) is added to the right. So for a system containing a pump and a turbine the Bernoulli […]

## Part 8. Typical Scenarios

i) Natural Flow When the flow between say two tanks at different heights is unrestricted (i.e. no flow control valves), then a situation known as “natural flow” occurs. This means that the flow of water will increase until the frictional energy losses combined with the kinetic energy exactly equal the potential energy of the water […]

## Part 9: Special Scenarios

i) Parallel Pipes Figure 8 shows a pipe network where the feeder pipe (1) divides into two pipes (2,3) of different diameters at point A and rejoins at point B to become another pipe section (4). There are two important points here : a) The flow into the junctions, must equal the flow out (see […]

## Appendix 1: Kinetic Energy of a Fluid

Consider a cylinder of a fluid that is travelling a velocity (v) as shown in Figure 10. This body contains kinetic energy (energy of movement). If we imagine bringing this body to rest then this kinetic energy will be turned into another form. What is the force that is required to bring this body to […]

## Appendix 2: Potential Energy of a Fluid

Consider a column of a fluid of height h and area A, as shown in Figure 11. To raise this column of water to a total height of h, means doing work against gravity. The force required to do this is given by Equation 3, except we replace the “a” term with the acceleration due […]

## Appendix 3. Friction Losses and the Reynolds Number

The frictional head loss for fluids flowing in pipes is calculated by the following equation: (27) Where: f is the friction factor (see below for calculation) L is the pipe length (m) v is the average fluid velocity (m/s) D is the pipe diameter (m) g is the acceleration due to gravity (9.81 m/s/s) The […]

## Appendix 4. Water System Design Parameters

The following are a list of design parameters that are important in the design of gravity-flow water systems: Maximum Pressure Limits: The taps and valves closed state, should be the maximum pressure condition for the system. Maximum head limits for the pipe work will be used to carry out the calculations. This scenario is used […]

## Appendix 5: Interpolation

Introduction Interpolation is a method for finding the value of a variable when non-linear data is provided at discrete points (i.e. we do not know the mathematical function of the variable). It involves drawing straight lines between these discrete points and then using the geometric rule for similar triangles to derive the value of a […]

## Appendix 6: Frictional Head Loss Chart

Polyethylene (PE) SDR-Pressure Rated Tube [(2306, 3206, 3306) SDR 7, 9, 11.5, 15 C=150]

## Worked Example 1: Natural Flow

Consider a collector or spring tank at an elevation h1, supplying a reservoir tank at an elevation h2. Taking the equation from Section 8 i: And assuming the kinetic energy term is negligible gives us: Simplifying gives us: (31) This means that for the natural flow condition the actual head (h1 – h2) must equal […]

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

## Worked Example 3: Simple Tap Syatem (Tap Open)

Consider a similar situation to Worked Example 1, except that a control valve has been inserted into the line, immediately before the discharge point (a tap). Taking the equation from Section 8 ii: Assuming the kinetic energy term to be negligible and rewriting P2 – P1 as the residual pressure at the valve or ΔP […]

## Worked Example 4: Simple Tap System (Tap Closed)

Consider Worked Example 3 and Figure 16, except that in this case the control valve has been shut. Taking the equation from Section 8 iii, we have: We can rewrite P2 – P1 as the maximum static pressure at the valve (Pmax) and divide by ρ.g to convert this equation from pressure (Pa) to head […]

## Worked Example 5: Pump Requirement

Consider the arrangement of a collector tank, pump and reservoir tank shown in Figure 17. Taking the pump pressure requirement equation from Section 8 iv: And assuming the kinetic energy term is negligible gives us:   (36) This is the general equation for calculating the pressure requirement for a pump. Combined with two other Equation […]

## 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: If we assume the kinetic energy term is negligible and write the residual head at any tap as Hn, then we get: (37) This is the simplified general equation for the residual head of […]

## 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 […]

## 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 […]

## 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 […]