APPENDICES. Consider a cylinder of a fluid that is travelling a velocity (v) as shown in Figure 10. Distance S

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1 1 PPENDICES 1. Kinetic Energy of a Fluid Consider a cylinder of a fluid that is travelling a velocity (v) as shown in Figure 10. v Distance S 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 rest? From Equation (3) : nd from Equation (2) : F = M.a (Where a is the deceleration of the body as it is brought to rest) a = (v 2 - v 1 )/t (In this case the final velocity (v 2 ) of the fluid is zero, so the change in velocity is v, negative, therefore a deceleration) So we can write : F = M.v/t (The body is brought to rest in time t) Now from Equation (6) : E = F.S (The body is brought to rest through distance S) So the kinetic energy E k is given by combining the last two equations : E k = M.v.S/t However, the velocity is changing from v to 0 over the distance S, so the average value of velocity must be taken to calculate the overall energy. This is given by : v av = (v-0)/2 = v/2 So the equation is rewritten as : Now from Equation (1) : E k = M.v/2.S/t v = S/t So we can rewrite the energy equation as : E k = ½.M.v 2 Now the mass of the fluid is given by Equation (5) : M = ρ.v In this case, the Volume of the cylinder is given by : V =.S So the Mass can be expressed as : M = ρ..s This is substituted into the kinetic energy equation to give :

2 2 (23) E k = ½.ρ..S.v 2 In the Bernoulli equation, the energies are represented as pressures. From Equation (4) we get : P = F/ nd from Equation (6) : Combining these two equations we get : E = F.S P = E/(.S) So if we divide the kinetic energy Equation (23) by (.S) we get it in terms of pressure : P = ½.ρ..S.v 2 /(.S) The.S terms cancel out to give the kinetic energy in terms of a pressure (P k ) : (24) P k = ½.ρ.v 2 This equation represents the kinetic energy of a fluid in terms of a pressure. 2. Potential Energy of a Fluid Consider a column of a fluid of height h and area, 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 III, except we replace the a term with the acceleration due to gravity g : F = M.g The mass of the column is given by Equation (5) : M = ρ..h h Figure 11. So we can rewrite the equation for the force required to raise this column of a fluid against gravity as :

3 3 F = ρ..h.g Now from Equation (6), the amount of energy required to carry out this action is : E = F.S In this case not all the fluid is being raised to a height h, instead the average distance moved by the fluid is h/2 : So we can write : (25) E p = ρ..h.g.s where S = h/2 This is the potential energy (E p )of the column. In the Bernoulli equation, the energies are represented as pressures. From Equation (4) we get : P = F/ nd from Equation (6) : Combining these two equations we get : E = F.S P = E/(.S) So if we divide the potential energy Equation (25) by (.S) we get it in terms of pressure : (26) P p = ρ.g.h This equation represents the potential energy of a fluid in terms of a pressure. 3. Friction Losses and the Reynolds Number The frictional head loss for fluids flowing in pipes is calculated by the following equation : (27) f h = f.l.v 2 /2.D.g 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 only variable not available to us immediately is the frictional factor (f). This is dependent on the type of flow occurring in the pipe. There are basically two types of flow (although there is a transition state between them ), these are : a) Laminar flow. Example : This is similar to the stream of smoke from a cigarette in still air. Close to the cigarette the stream of smoke is very uniform and flowing evenly. This is laminar flow. b) Turbulent flow. Example : This is when the stream of smoke from the cigarette becomes unstable, with whorls and eddies. This state occurs in the stream of smoke after the laminar flow. These two states of flow can be described by a dimensionless quantity (just a number) known as the Reynolds Number (N RE ). This number is calculated by the following formula : (28) N RE = v.d/µ k Where v is the average fluid velocity (m/s) D is the pipe diameter (m) µ k is the kinematic viscosity of the fluid (m 2 /s), which is a measure of how thick the fluid is.

4 4 If the Reynolds Number is less than 2000 then the fluid flow is laminar. In this case the friction factor (f) can be calculated by the following equation : (29) f = 64/ N RE If the Reynolds Number is greater than 2000 then the fluid flow is turbulent. In this case the friction factor (f) can be calculated by use of a Moody Chart. 4. Water System Design Parameters The following are a list of design parameters that are important in the design of gravity-flow water systems : 1. 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 at the start of the design to be able to place any break-pressure tanks that may be required. 2. Safe Yield : The safe yield is the minimum flow from the water source. It is important to not draw more than this supply from the system at any point. If this happens then spring boxes and/or break pressure tanks will run dry and air will enter the system. 3. Negative or Low Pressure Head : If the pressure head (P in the Bernoulli Equation) becomes negative at any point in the system then two things may happen. Firstly a siphon effect is occurring which is trying to suck water into the system. This is undesirable as polluted groundwater may be introduced into the system. Secondly, large negative pressures can cause air to come out of solution in the water and cause air-blocks. Jordan [P.52] suggests that the pressure head should, if possible, not fall below 10m (98100 Pa pressure) anywhere in the system and never go negative. 4. Velocity Limits : The flow velocity in the pipelines should not be to great as particles suspended in the water will cause excessive erosion. lso if the velocity is too low then these particles will settle out of the flow and may clog the pipes at low points. This then requires washouts at low points in the system. Jordan [P.53] suggests that the minimum velocity should be 0.7m/s and the maximum 3.0m/s. 5. Natural Flow : Natural Flow (see Section 8 i)) may be allowed to occur in the system at some sections of pipe. Natural flow can be problematic in that the water velocity may exceed the limits set in parameter 4 above and/or increase the flow rate above the safe yield parameter 2. Close attention should be made to these situations. 6. Residual Head : The residual head at a tap stand or valve is important. If it s too high it will cause erosion of the valve and if it is too low then the flow will be minimal. Jordan [P.141] suggests the following limits : bsolute minimum : Low end of desired range : Most desirable : High end of desired range : bsolute maximum : 7m 10m 15m 30m 56m 7. ir-blocks : These occur when there are topographic features between the source and the collecting tank that are lower than the collecting tank. Energy is lost from the system as these air-blocks are compressed and can result in no flow. Jordan [P.55] gives the following design practices to avoid air-blocks : rrange pipe sizes to minimise the frictional head loss between the source and the first air-block. Use larger-sized pipe at the top and smaller sized pipe at the bottom of the critical sections where air is going to be trapped. The higher air blocks are the more critical ones and should be eliminated or minimised first. ir valves can be designed into the system to allow trapped air to escape.

5 8. Cost : Wherever possible smaller pipe diameters should be used, as they are cheaper. Combinations of pipes can often produce cheaper solutions than using just one pipe size. However pipe lengths should be rounded to the nearest 100m length. lso the number of concrete structures such as break-pressure tanks should be minimised 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 variable at a location between the discrete points. Theory Consider the set of data below : X Y This data can be represented graphically as below : B Figure 12. Imagine we want to find out what the value of Y is for a value of X = 2.4. If we know the function of X (in this case it is X 2 ) then it is easy Y = = But if we don t know the function of X (which is more generally the case especially with non-linear data) then how do we find the value for Y? This is where we can use the approximate method known as interpolation.

6 6 Interpolation assumes that we have drawn a straight line between points and B above, this will give us the following general situation (Figure 13.) : Y Interpolated line B Y 2 Y 3 C E ctual line Y 1 D X X 1 X 3 X 2 Figure 13. So how do we find Y 3 knowing, X 1, X 2, X 3, Y 1 and Y 2? Well, we note that the triangles BE and CD are similar, that is they share the same angles. This means that the ratio of their sides will be similar. So we can say : Lengths CD/D = BE/E, this can be rewritten as : (Y 3 Y 1 )/(X 3 X 1 ) = (Y 2 Y 1 )/(X 2 X 1 ) We are trying to find Y 3, so we can rearrange the above equation to isolate Y 3. (Y 3 Y 1 ) = (Y 2 Y 1 )*(X 3 X 1 ) /(X 2 X 1 ) and so : Y 3 = Y 1 + (Y 2 Y 1 )*(X 3 X 1 ) /(X 2 X 1 ) (30) This is the general interpolation equation. We can use this equation to approximate the value of any function if we know the coordinates of the points either side (X 1, X 2 ) and (Y 1, Y 2 ) and the X or Y coordinate of the point we are trying to find. Example Take the example in the theory section above for Y=X 2. What is the value of Y when X=2.4? nswer : We know the following values : X 1 = 2, X 2 = 3, Y 1 = 4, Y 2 = 9 and X 3 = 2.4. Substitute these into the general interpolation equation (30) given above to give :

7 7 Y 3 = 4 + (9 4)*(2.4 2) /(3 2) nd so, the approximation is Y 3 = 6. The actual answer is 5.76 (see above), but this is not a bad approximation if we did not know the function. Notes 1. The above method should be used in the water pipe friction tables to estimate head losses. 2. The more data points the more accurate this method becomes. 3. The coordinates X 1 and Y 1 are always associated together, likewise for coordinates X 2 and Y 2. They define the two discrete points on the curve.

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