# The Mathematics of Pumping Water

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1 The Matheatics of Puping Water AECOM Design Build Civil, Mechanical Engineering INTRODUCTION Please observe the conversion of units in calculations throughout this exeplar. In any puping syste, the role of the pup is to provide sufficient pressure to overcoe the operating pressure of the syste to ove fluid at a required flow rate. The operating pressure of the syste is a function of the flow through the syste and the arrangeent of the syste in ters of the pipe length, fittings, pipe size, the change in liquid elevation, pressure on the liquid surface, etc. To achieve a required flow through a puping syste, we need to calculate what the operating pressure of the syste will be to select a suitable pup. Figure : Typical Vertical Turbine Water Pups MATEMATICAL MODEL AND CALCULATIONS Consider the puping arrangeent shown in Figure below: Figure : Puping Arrangeent Water is puped fro the reservoir into a receiving tank. This kind of arrangeent is used to lift water fro a reservoir, or river, into a water treatent works for treatent before the water goes into the supply network. The water level in the reservoir varies but the discharge level in the receiving tanks reains constant as the water is discharged fro a point above the water level. The pup is required to pass forward a flow of /hr to the receiving tank. The operating pressure of a puped syste is calculated in the SI unit of eters (). To aintain diensional consistency, any pressure values used within the calculations are therefore converted fro kpa into using the following conversion; kpa 0.0 (as easured by a water filed U tube anoeter) For the above syste, the operating pressure or the total syste head,, is defined as:, s D P RT P RES s D ( P P ) + + () Static head () RT Dynaic head () RES Pressure on the surface of the water in the receiving tank () Pressure on the surface of the water in the reservoir () Although the atospheric pressure changes with height, the change in pressure that occurs over the puping height is often so sall that it can be considered negligible. In this exeplar, the change in pressure over the elevation fro the reservoir to the receiving tank is not that significant and hence is negligible, i.e., P P 0. RT RES Therefore, equation () becoes: + () s The static head s is the physical change in elevation between the surface of the reservoir and the point of discharge into the receiving tank. As the water level in the reservoir can vary, the static head for the syste will vary between a axiu and a iniu value: D

2 and S S in ax TWL BWL discharge level reservoir T WL discharge level reservoir B WL Top Water Level (reservoir) Botto Water Level (reservoir) If the discharge point is at a level of 0.5 above the ean sea level (also known as Above Ordnance Datu (AOD) in technical language) and the reservoir level varies between 05. AOD and 0.6 AOD, then: S S in ax As a result of the variation in the static head, the total syste head,, will also have a axiu and iniu value which we need to calculate here. The dynaic head is generated as a result of friction within the syste. The dynaic head is calculated using the basic Darcy Weisbach equation given by: K Kv D (3) g loss coefficient v velocity in the pipe (/sec) g acceleration due to gravity (/sec ) We can calculate the velocity in pipe using the following forula: Q Q v (4) A flow rate through the pipe ( 3 /sec) A pipe cross sectional area (CSA) ( ) If Q is /hr and the flow is puped through a 0.8 diaeter pipe then: πd π 0. 8 A ence, using equation (4), we get: v The loss coefficient K eleents: fittings. 39 /sec K K + K (5) is ade up of two pipe is associated with the fittings used in the pipeworks of the syste to pup the water fro reservoir to the receiving tank. Values can be obtained fro standard tables and a total K value can be calculated by adding all the fittings values for each individual fitting within the syste. The following table shows the calculation of K for the syste under consideration: fittings Fitting Ites Pipe Entrance (bellouth) 90 o Bend (short radius) 45 o Bend (short radius) Butterfly Valve (Fully Open) Non Return Valve No. of Ites Value Ite Bellouth Outlet Value Table : Calculating ence, the total consideration is K pipe 9.95 K for the syste under fittings consideration for the syste under is associated with the straight lengths of pipe used within the syste and is defined as: f L D fl K pipe (6) D friction coefficient pipe length () pipe diaeter () The friction coefficient f can be found using a odified version of the Colebrook White equation: 0. 5 f (7) k log Re D k Re Roughness factor () Reynol nuber The pipe roughness factor k is a standard value obtained fro standard tables and is based upon the aterial of the pipe, including any internal coatings, and the internal condition of the pipeline i.e. good, noral or poor.

4 ρ Density (Kg/ 3 ) 000 kg/ 3 for water For this pup, at the axiu head of 0.39 and a flow of /hr ( /s) the pup efficiency is 84%. Therefore, using equation (), the power requireent is: P, or P 840 W 84. k W ence, we can say that to overcoe the required head of 0.39, we need a variable speed pup with 84. W. CONCLUSION The accurate calculation of the axiu and iniu total head is critical for the selection of a suitable pup. Selection of an unsuitable pup can result in too uch or too little water being puped. Too little water ight, for exaple, result in custoers not receiving clean drinking water when they turn on the tap. Too uch water ight result in water being wasted or even lead to flooding. The operating pressure of a puping syste can vary due to various factors, e.g. changes in reservoir level, so all the relevant operating conditions need to be assessed to ensure the selected pup is capable of achieving the entire operating range. Using variable speed pups is one way of coping with the variations in syste operating pressure. EXTENSION ACTIVITIES. Calculate ax and for the in syste if the flow is reduced to /hr.. What happens to the pup power if the pup efficiency reduces? 3. Calculate the power requireent of the pup for the following efficiencies: 95% 75% 50% WERE TO FIND MORE. Basic Engineering Matheatics, John Bird, 007, published by Elsevier Ltd.. Engineering Matheatics, Fifth Edition, John Bird, 007, published by Elsevier Ltd. 3. Pressure and ead Losses in Pipes and Ducts, D.S. Miller, 984

5 Figure 3: Graph of Puping Syste Pressure Curves and Pup Operating Speed Curves Mathew Milnes Project Engineer, AECOM Design Build Mathew has worked in the Water Industry designing clean and dirty water treatent plants for the last 0 years. As a Chartered Mechanical Engineer he uses atheatics on a daily basis to calculate the size and perforance of process equipent to provide people with clean drinking water and to ensure their wastewater is treated and disposed of in an environentally acceptable way.

6 INFORMATION FOR TEACERS The teachers should have soe knowledge of terinology used in puping water and the physical eaning behind the handling forulae with the ethod of back substitution plotting graphs using excel sheets anipulating calculations and converting units for unifority Equation can be derived fro an extension to the Euler equation. Please refer to the last page for ore inforation and observe the use of partial derivatives and liit theory. TOPICS COVERED FROM MATEMATICS FOR ENGINEERING Topic : Matheatical Models in Engineering Topic 4: Functions LEARNING OUTCOMES LO 0: Understand the idea of atheatical odelling LO 04: Understand the atheatical structure of a range of functions and be failiar with their graphs LO 09: Construct rigorous atheatical arguents and proofs in engineering context LO 0: Coprehend translations of coon realistic engineering contexts into atheatics ASSESSMENT CRITERIA AC.: State assuptions ade in establishing a specific atheatical odel AC.: Describe and use the odelling cycle AC 4.: Identify and describe functions and their graphs AC 4.: Analyse functions represented by polynoial equations AC 9.: Use precise stateents, logical deduction and inference AC 9.: Manipulate atheatical expressions AC 9.3: Construct extended arguents to handle substantial probles AC 0.: Read critically and coprehend longer atheatical arguents or exaples of applications LINKS TO OTER UNITS OF TE ADVANCED DIPLOMA IN ENGINEERING Unit : Investigating Engineering Business and the Environent Unit 3: Selection and Application of Engineering Materials Unit 4: Instruentation and Control Engineering Unit 5: Maintaining Engineering Plant, Equipent and Systes Unit 6: Investigating Modern Manufacturing Techniques used in Engineering Unit 7: Innovative Design and Enterprise Unit 8: Matheatical Techniques and Applications for Engineers Unit 9: Principles and Application of Engineering Science ANSWERS TO EXTENSION ACTIVITIES. ax 9.85, in 6.5 ( D 0.95). The power requireent goes up as the efficiency reduces % 74.5 kw, 75% 94.3kW, 50% 4.5kW

7 ANNEXE: EXTENSION OF EULER EQUATION In this section, we investigate incopressible flow along a strealine under the action of pressure gradients and gravitational body forces but not friction. ence density is constant and there are no shear forces. During the derivation it will also be necessary to assue that the flow is steady. Consider a sall cylindrical eleent of fluid aligned along a strealine. It has a cross sectional area da, pressure is assued unifor across its en da, and the local velocity is defined q. Applying Newton's laws of otion to the flow through the cylindrical eleent along the strealine, the force (in the direction of otion along the strealine) ass x acceleration. The ass of the eleent and the forces acting on it will be considered later, but first we look at the acceleration of the fluid eleent. Ignoring the possibility that the flow ight be steady, q can change with tie t, and also with position s along the strealine. In other wor, q is a function of t and s, or q f (t, s). ence, if the eleent oves a distance δs in tie δt, then the total change in velocity δq is given by: q δ q δ s s + q δ t t and in the liit as δt ten to zero, the "substantive" derivative is given as: dq dt Li δ t 0 δ q q Li δ t s δ t 0 δ s q + δ t t q q q + s t In other wor, fluid can accelerate because it is oving (at velocity q) through a region with changing velocity, or because the flow is changing with tie. owever, for a steady flow the local velocity at a point does not vary with tie, so the last ter under such circustances will be zero. Looking now at the forces acting on the eleent and applying Newton's laws: p p δ A (p + δ s) δ A ρ δ A δ s g cos θ ρ δ s δ A q s dividing through by δa. δs and defining δz δs cosθ, we have that: p q δ z + ρ q + ρ g 0 s s δ s dividing through by δa. δs and defining δz δs cosθ, we have that: and in the liit as δs ten to zero, p q δ z + ρ q + ρ g 0 s s δ s dp + ρ q dq + ρ g dz 0 dq

8 or dp dq dz + q + g 0 ρ This is a for of Euler's equation, and relates p, q, and z along a strealine. Assuing ρ is constant, and reebering that: dq q d( q ) if the ter above is substituted into Euler's equation, it then becoes possible to integrate it giving: + q ρ p p + p ρ g ρ q + g z constant along a strealine q + + z g + ρ g z constant a strealine constant along a strealine The three equations above are valid for incopressible, frictionless steady flow, and what they state is that total energy is conserved along a strealine. The first of these fors of the Bernoulli equation is a easure of energy per unit ass, the second of energy per unit volue, and the third of "head", equivalent to energy per unit weight. In the second equation, the ter p is the static pressure, {½ ρ q } is the dynaic pressure, ρ gz is the elevational ter, and the SUM of all three is known as the stagnation (or total) pressure, p 0 In the third equation, p/ ρ g is known as the pressure head, q /g as the dynaic head, and the su of the three ters as the ead. The Bernoulli equation is used widely in fluid echanics and hydraulics. Assuing that the flow is actually frictionless and incopressible, what it shows is that if the velocity falls in a flow, then the pressure ust rise and vice versa. For a gas, the elevational ters can be assued negligible. The su {p + ρ gz} is often written as p * the piezoetric pressure. We can then say: p * + ½ ρ q constant along a strealine To easure the static pressure in a fluid flow, it is noral to ake a sall hole in the boundary wall of the flow and to connect the hole to a pressure easuring device a anoeter being the traditional instruent used. To easure the total pressure, it is noral to eploy a device known as a Pitot tube. This is a thin tube that can be pointed directly into the flow such that it is aligned exactly with the local strealines. The other end of the tube is connected to a anoeter (or other pressure easuring device). The strealine that eets the end of the tube within the flow is brought to rest because there is no actual flow through the tube/anoeter syste and therefore all the dynaic pressure is converted to static pressure. The su of these two fors of static pressure is known as the stagnation pressure or total pressure. To easure the dynaic pressure, the ost coon device (and the siplest and cheapest) used is a Pitot static tube. This is a cobination of the two techniques described above within one instruent. It consists of two thin concentric tubes bent into an L shape; the inner tube has an open end which is pointed into the flow (as described above when easuring total pressure), while the outer tube is sealed and strealined at its end but has a nuber of sall holes around its circuference soe way back fro the end. The two tubes are connected across a differential pressure easuring device (again, coonly a U tube anoeter), and the difference in pressure is the dynaic pressure.

9

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