30 CHAPTER 5 TURBINES 5. 1 Introduction Hydraulic turbines are the machines which use the energy of water and convert it to mechanical energy. The mechanical energy developed by a turbine is used in running an electric generator which is directly coupled to the shaft of the turbine. The electric generator thus develops electric power, which is known as hydro-electric power. 5. 2 Elements of Hydraulic Power Plants Fig. 5. 1 General Layout of a Hydraulic Power Plant 31 Fig. 5. 1 shows a general layout of hydraulic power plant, in which an artificial storage reservoir formed by constructing a dam has been shown. 5. 3

General Classification of Turbines Turbines are hydraulic machines that convert energy into rotating mechanical energy which in turn generators to produce electrical energy. Originally developed from the water wheels, hydraulic turbines are the prime mortars of importance in modern water power development. According to their hydraulic action, turbines are broadly divided into two classes. (1) Impulse Turbine: Impulse turbines are more efficient for high heads. At the inlet to the turbine runner, pressure head can be completely converted into kinetic head in the form of a jet of water issuing from one or more nozzles.

The free jet will be at atmospheric pressure before as well as after striking the vanes. The turbines are regulated by nozzles which may be a simple straight flow type or a deflector type. The impulse turbines are commonly represented by Pelton Wheels. Turgo turbine is also an impulse turbine but with different buckets, when compared with pelton. Turgo and cross flow turbines are relatively new developments in this class. The main advantages of these turbines are: â€¢ They can be easily adopted to power variation with almost constant efficiency. â€¢

The penstock overpressure and the runner overspeed control are easier. â€¢ The turbine enables an easier maintenance. â€¢ Due to the jet the manufacturer of these turbines impose a better solid particle control, conducting, consequently, to a lower abrasion effect. (2) Reaction Turbine: A turbine can be made to rotate under the action of water flowing under pressure through the runner. In such turbines the penstocks, the inlet passage to the runner, passage between the runner vanes, all form a continuous passage for the flow under a pressure which continuously decreases from inlet to outlet.

The turbine runner directly converts both kinetic energy as well as the pressure energy into mechanical energy. Reaction turbines are represented in modern practice by two principal types: the Francis turbine where the flow is directed radial to the runner axis and the Propeller type 32 where the flow is axial to the runner axis. Propeller turbines may be fixed blade or adjustable blade types. Kaplan turbine has adjustable blades. The main advantages of these turbines are: â€¢ It needs lesser installation space. â€¢ It provides a greater net head and a better protection against downstream high flood levels. It can have greater runner speed. â€¢ It can attain greater efficiencies for high power values. In order to distinguish different turbines, the hydraulically salient features like pressure, head, flow direction and magnitude, speed and power etc. The general classification of hydraulic turbines is illustrated in Fig. 5. 2. Hydraulic Turbines Impulse Turbines Pelton Turgo Reaction Turbines Cross-Flow Fixed -Blade Propeller Kaplan Francis Tubular Deriaz Bulb Fig. 5. 2 General Classification of Turbines 5. 4 Number of Units It is normally cost effective to have a minimum number of units at a given small ydropower installation. Multiple units may, however, be necessary from the operational point of view so that even one unit breaks down or is in the routine maintenance, the power generation can be achieved to a certain extent. The efficiency curves of turbines show that the 33 efficiency of power generation from hydraulic turbines considerably decreases at low flow ratios or power ratios. In multiple units, it is possible to maintain the higher efficiency even in low flows and the low loads by running a certain number of the units at a time depending upon the available discharge and the load demand.

Multiple units thus, make the most effective use of water where the flow as well as the load variations are significant. 5. 5 Limits of Use of Turbine Types The selection of best turbines for any particular small hydropower site depends on the site characteristics, the dominant beings the head and the available flow. There are some limits on the range of these parameters in the selection of turbines. Each turbine type is best suited to a certain range of pressure head and the flow rate. For instance, Pelton wheels operate with low flows discharged under great pressures where as Propeller turbines are effective in high flows under low heads.

Francis turbines fall in the medium category covering a wide range of different heads and discharges. The common practice of SHP systems is to develop standard unit sizes of equipment that will operate over a range of heads and flows. Either charts or nomographs are used to select appropriate units for site specific application. One such chart showing the head-flow range of normal SHP schemes applicable to each type of turbine is given in Fig5. 3. The graph also indicates the approximate power generation for each combination of the head and the discharge applicable to SHP schemes. Fig. . 3 Head-Flow Ranges for Different Turbines 34 5. 6 Pelton Wheel Pelton wheel is well suited for operating under high heads. A pelton turbine has one or more nozzles discharging jets of water which strike a series of buckets mounted on the periphery of a circular disc. The runner consists of a circular disc with a number of buckets evenly spaced round its periphery. The buckets have a shape of a double semi-ellipsoidal cups. The pelton bucket is designed to deflect the jet back through 165Â° which is the maximum angle possible without the return jet interfering with the next bucket.

Fig. 5. 4 Pelton bucket General arrangement of a pelton wheel is shown in the Fig. 5. 5. For SHP schemes, Pelton wheels are easier to fabricate and are relatively cheaper. The turbines are in general, not subjected to the cavitation effect. The turbines have access to working parts so that the maintenance or repairs can be effected in a shorter time. Fig. 5. 5 General Arrangement of a Pelton Wheel 35 Traditionally, micro hydro pelton wheels were always single jet because of the complexity and the cost of flow control governing of more than one jet.

Advantages of multi-jet: -Higher rotational speed -Smaller runner -Less chance of blockage Disadvantages of multi-jet: -Possibility of jet interference on incorrectly designed systems -Complexity of manifolds 5. 7 Francis Turbine Francis turbine is a mixed flow type, in which water enters the runner radially at its outer periphery and leaves axially at its center. Fig. 5. 6 illustrates the Francis turbine. The runner blades are profiled in a complex manner and the casing is scrolled to distribute water around the entire perimeter of the runner.

The water from the penstock enters a scroll case which completely surrounds the runner. The purpose of the scroll case is to provide an even distribution of water around the circumference of the turbine runner, maintaining an approximately constant velocity for the water so distributed. The function of guide vane is to regulate the quantity of water supplied to the runner and to direct water on to the runner at an angle appropriate design. A draft tube is a pipe or passage of gradually increasing cross sectional area which connects the runner exit to the tail race. Fig. 5. 6 Francis Turbine 36 . 8 Kaplan Turbine It is an axial flow turbine which is suitable for relatively low heads. From Fig. 5. 7, it will be seen that the main components of Kaplan turbine such as scroll casing, guide vanes, and the draft tube are similar to those of a Francis turbine. Fig. 5. 7 5. 9 Kaplan Turbine Specific Speed The specific speed of any turbine is the speed in r. p. m of a turbine geometrically similar to the actual turbine but of such a size that under corresponding conditions it will develop 1 metric horsepower when working under unit head. Ns = NP H5/ 4 â€¦â€¦â€¦â€¦â€¦(5. 1) where Ns = specific speed

P = power in HP 5. 10 Characteristic Curves The turbines are generally designed to work at particular values of H,Q,P,N and ? o which are known as the designed conditions. It is essential to determine exact behaviour of the turbines under the varying conditions by carrying out tests either on the actual turbines or on their small scale models. The results of these tests are usually graphically represented and the resulting curves are known as characteristic curves. 37 -constant head characteristic curves -constant speed characteristic curves -constant efficiency characteristic curves

In order to obtain constant head characteristics curves the tests are performed on the turbine by maintaining a constant head and a constant gate opening and the speed is varied by changing the load on the turbine. A series of values of N are thus obtained and corresponding to each value of N, discharge Q and the output power P are measured. A series of such tests are performed by varying the gate opening, the head being maintained constant at the previous value. From the data of the tests the values of Qu, Pu, nu and ? o are computed for each gate opening. Then with Nu as abscissa the values of

Qu, Pu and ? o for each gate opening are plotted. The curves thus obtained for pelton wheel and the reaction turbines for four different gate openings are shown in Fig. 5. 8. Fig. 5. 8 Constant head characteristics for Pelton wheel and reaction turbines 38 5. 11 Cavitation in turbines When the pressure in any part of the turbine reaches the vapour pressure of the flowing water, it boils and small bubbles of vapour form in large numbers. These bubbles are carried along by the flow, and on reaching the high pressure zones these bubbles suddenly collapse as the vapour condenses to liquid again.

The alternate formation and collapse of vapour bubbles may cause severe damage to the surface which ultimately fails to fatigue and the surface becomes badly scored and pitted. This phenomenon is known as cavitation. In order to determine whether cavitation will occur in any portion of the turbine, D. Thomas has developed a dimensionless parameter called Thomas’cavitation factor ? which is expressed as ?= Ha ? Hv ? Hs H â€¦â€¦â€¦â€¦â€¦(5. 2) where Ha = atmospheric pressure head Hv = vapour pressure head Hs = suction pressure head For Francis turbines: Critical cavitation factor ? c = 0. 625 (Ns/444)2 â€¦â€¦.. (5. 3) For Propeller turbines: ? c = 0. 28 + [ Example 5. 1 1 Ns 3 ( )] 7. 5 444 â€¦â€¦â€¦.. (5. 4) Estimate the maximum height of straight conical draft tube of a 18000 h. p. Francis turbine running at 150 r. p. m under a net head of 27 m. The turbine is installed at a station where the effective atmospheric pressure is 10. 6 m of water. The draft tube must sink at least 0. 77 m below the tail race. Ns = NP H5/ 4 = 327 39 ?c = 0. 625 (Ns/444)2 = 0. 339 Cavitation factor ? = Ha -Hv -Hs H Ha -Hv = 10. 6 m, H =27 m 0. 339 = 10. 6 -Hs 27 Hs = 1. 45 m Max length of the draft tube = 1. 45 + 0. 7 = 2. 22 m 5. 12 Governing of Turbines All the modern turbines are directly coupled to the electric generators. The generators are always required to run at constant speed irrespective of the variations in the load. This constant speed N of the generator is given by the expression N= 60 f p â€¦â€¦â€¦â€¦.. (5. 5) where f = frequency (usually 50) p = numbers of pairs of poles 5. 13 Water Hammer A gate or valve at the end of the penstock pipe controls the discharge to the turbine. As soon as this governor regulated gate opening is altered, the pipe flow has to be adjusted to the new magnitude of flow.

In doing so, there are rapid pressure oscillations in the pipe, often accompanied by a hammering like sound. Hence this phenomenon is called as water hammer. 5. 14 Jet Speed The velocity of flow of the jet depends upon the total net head H at the base of the nozzle and is given by the nozzle equation: v = C v ? 2gH where the discharge coefficient velocity of the nozzle is taken as 0. 95. â€¦â€¦â€¦â€¦(5. 6) 40 5. 15 Bucket speed V= ? DN 60 â€¦â€¦â€¦â€¦(5. 7) The bucket speed should be half of the jet speed. In practice, losses in the turbine cause the maximum efficiency to occur at slightly less than a half, typically 0. 46. V =0. 6 v 5. 16 Design of Pelton Wheel Runner diameter: Runner diameter can be found out from the rpm equation. D= 38 ? H N â€¦â€¦â€¦. (5. 8) where N = runner speed(rpm) H = net head Nozzle diameter: The nozzle diameter is given by the nozzle equation: d = 0. 54 ? Q 0. 5 1 ? 0. 25 H n jet â€¦â€¦â€¦. (5. 9) Jet ratio: Jet ratio D/d is a size parameter for the turbine. It has a value in a range of 10 to 24. For the high efficiency Pelton wheel design, the ratio of the runner diameter to the nozzle should be more than 9. Number of buckets: The number of buckets required for the efficient operation of the Pelton turbine is calculated as:

N buc = 0. 5 ? D + 15 d â€¦â€¦â€¦â€¦.. (5. 10) In practice, the selection and the detail design of the turbine units are carried out by the manufactures based on the model performances. 41 Example 5. 2 Powerhouse is equipped with a vertical shaft pelton turbine. The generator is provided with 6 pairs of poles. Design discharge is 1. 4 m3/s and net head is 425 m. The turbine will provide 6500 hp. Take coefficient of nozzle as 0. 95. Determine (a) the specific speed (b) velocity of jet (c) jet diameter (d) pitch circle diameter of the wheel (e) number of buckets (a) N= 60 f p = 60 x 50/6 = 500 rpm

Ns = = NP H5/ 4 500 6500 4255 / 4 = 20. 9 Use single jet pelton turbine (b) velocity of jet v = C v ? 2gH = 0. 95 ? 2×9. 81×425 = 86. 75 m/s (c) jet diameter d = 0. 54 ? = 0. 54 ? Q 0. 5 1 ? 0. 25 H n jet 1. 4 0. 5 425 0. 25 ? = 0. 14 m =14 cm 1 1 42 (d) pitch circle diameter D= 38 ? H N = 1. 57 m (e) Number of buckets N buc = 0. 5 ? N buc = 0. 5 ? D + 15 d 1. 57 + 15 0. 14 = 20. 6 = 21 5. 17 Work done of Pelton Wheel In turbines, the water flows on to the runner, which itself is rotating with a certain speed. The water flows over the runner and leaves the runner at its outlet point.

We can speak of absolute velocity of water before it flows in the runner, the relative velocity of water w. r. t the runner and again the absolute velocity of water after it has left the runner. In order to ascertain the relationship between these velocities, the velocity vector diagram prove to be very useful. Fig. 5. 9 shows the velocity triangles at the tips of the bucket of a pelton wheel. At the outlet tip velocity triangles are different depending upon the magnitude of u corresponding to which it is slow, medium or fast runner. Inlet velocity diagram V

Vri Vai =Vwi 43 Outlet velocity diagram V Vwo ? ? Vfo Vro Vao Fig. 5. 9 Velocity triangles V = bucket velocity Vai = absolute velocity of jet at inlet tip Vai = Cv 2 gH1 Vao = absolute velocity of jet at outlet tip Vri = relative velocity of jet at inlet = Vai-V Vro = relative velocity of jet at outlet = k. Vri Vwi = velocity of whirl at inlet =Vai Vwo = velocity of whirl at outlet = V-VroCos ? Vfo = velocity of flow at outlet Mass/sec m =? Q =? a Vai=? ?/4 d2 Vai â€¦â€¦â€¦(5. 11) Workdone on the bucket/sec (power developed by turbine) P = m (Vwi -Vwo) V â€¦â€¦â€¦(5. 12)

Maximum hydraulic efficiency ? h max = 1 (1 + kCos? ) 2 â€¦â€¦â€¦(5. 13) The hydraulic efficiency is maximum when the bucket speed is equal to half of the velocity of jet. 44 Example 5. 3 The head available at entrance to the nozzle supplying a pelton wheel is 300 m and the coefficient of velocity for the nozzle is 0. 98. The wheel diameter is 1. 8 m and the nozzle diameter is 125 mm. The buckets deflect the jet through 165Â°. Assuming the relative velocity of the jet is reduced by 16%, calculate the theoretical speed in rev per min for the maximum hydraulic efficiency.

What is the hydraulic efficiency when running at this speed, and what is the power developed? Deflection angle =165Â° = (180- ? ) ? = 15Â° k = 0. 84 hf hn H’=Vai2/2g H For max hydraulic efficiency V/Vai =0. 5 Vai = Cvv2gH1 = 75 m/s V =Vai/2 = 37. 5 m/s V =? DN/60 N =60V/? D = 398 rpm ?hmax = 1/2 (1+ k Cos ? ) = 90. 55 % mass/sec = m = ? Q = ? ?/4 d2 Vai = 920 kg/sec H1 45 Inlet diagram: V Vri From velocity diagram Vwi = Vai =75 m/s Vri = Vai- V =37. 5 m/s Vwi = Vai Outlet diagram: Vwo = V -Vro Cos ? V = V – k Vri Cos ? Vwo Vro Power = m V ( Vwi -Vwo) = 234600 Watts Vao =7 m/s 46 CHAPTER 6

CENTRIFUGAL PUMP 6. 1 Introduction Centrifugal pumps are classified as rotodynamic type of pumps in which dynamic pressure is developed which enables the lifting of liquids from a lower to a higher level. The basic principle on which a centrifugal works is that when a certain mass of liquid is made to rotate by an external force, it is thrown away from the central axis of rotation and a centrifugal head is impressed which enable it to rise to a higher level. Now, if more liquid is constantly made available at the centre of rotation, a continuous supply of liquid at a higher level may be ensured.

Since in these pumps the lifting of the liquid is due to centrifugal action, these pumps are called ‘centrifugal pumps’. 6. 2 Advantages of centrifugal pumps over reciprocating pumps The main advantage of a centrifugal pump is that its discharging capacity is very much greater than a reciprocating pump which can handle relatively small quantity of liquid only. A centrifugal pump can be operated at very high speeds without any danger of separation and cavitation . The maintenance cost of a centrifugal pump is low and only periodical check up is sufficient .

But for a reciprocating pump the maintenance cost is high because the parts such as valves etc. , may need frequent replacement. 6. 3 Component Parts of a Centrifugal Pump The main component parts of a centrifugal pump are: -impeller -casing -suction pipe -delivery pipe 47 Fig. 6. 1 Component part of a centrifugal pump 6. 4 Workdone by the Impeller The expression of the workdone by the impeller of a centrifugal pump on the liquid flowing through it may be derived in the same way as for a turbine. The liquid enters the impeller at its centre and leaves at its periphery. Fig. 6. shows a portion of the impeller of a centrifugal pump with one vane and the velocity triangles at the inlet and outlet tips of the vane. V is absolute velocity of liquid, u is tangential velocity of the impeller, Vr is relative velocity of liquid, Vf is velocity of flow of liquid, and Vw is velocity of whirl of the liquid at the entrance to the impeller. Similarly V1,u1,Vr1,Vf1 and Vw1 represent their counterparts at the exit point of the impeller. 48 Fig. 6. 2 Velocity triangles for an impeller vane ? = the impeller vane angle at the entrance ? = the impeller vane angle at the outlet = the angle between the directions of the absolute velocity of entering liquid and the peripheral velocity of the impeller at the entrance ? = the angle between the absolute velocity of leaving liquid and the peripheral velocity of the impeller at the exit point Work done per second by the impeller on the liquid may be written as Work done = W ( Vw1 u1 – Vw u) ————-(6. 1) g where W kg of liquid per second passes through the impeller. Since the liquid enters the impeller radially ? = 90 and hence Vw = 0. Thus equation (6. 1) becomes Work done = W (Vw1u1) ————-(6. ) g 6. 5 Head of a Pump The head of a centrifugal pump may be expressed in the following two ways: (a) Static head (b) Manometric head (or total head or gross head) 49 Fig. 6. 3 Head on a centrifugal pump (a) Static Head Static head is the vertical distance between the liquid surfaces in the pump and the tank to which the liquid is delivered by the pump. Static head (or lift) Hs = hs + hd where hs = static suction lift hd = static delivery lift â€¦â€¦â€¦.. (6. 3) 50 (b) Manometric Head Manometric head is the total head that must be produced by the pump to satisfy external requirements.

If there are no energy losses in the impeller and the casing of the pump, then the manometric head Hm will be equal to the energy given to the liquid by the V ? 1u1 g impeller, i. e Hm = . But if losses occur in the pump then V ? 1u1 ? losses of head in the pump g â€¦â€¦â€¦.. (6. 4) Applying Bernoulliâ€™s equation between the points, O at the liquid surface in the pump and 1 in the suction pipe just at the inlet to the pump (i. e. , at the centre line of the pump), the following expression is obtained if the liquid surface in the sump is taken as datum. 0= p s Vs2 + + hs + h f s ? 2g ps Vs2 ? [ + hs + h fs ] 2g ? â€¦â€¦â€¦. (6. 5) where ps is the pressure at point l ; Vs is the velocity of flow in the suction pipe ; hs is the suction lift and hfs is the head loss in the suction pipe which includes the head loss due to friction and the other minor losses. It may however be pointed out that if the pump is situated below the level of the liquid surface in the sump, hs will be negative. Equation (6. 5) indicates that at the inlet to the pump there is always a suction or vacuum pressure developed which will be recorded by the vacuum gauge provided at this point as shown in Fig. . 3. The head expressed by equation (6. 5) is called the suction head of the pump. Also, applying Bernoulliâ€™s equation between points 1 and 2, which is just at the outlet of the impeller and is assumed to be at the same level as point 1, then since the impeller imparts a head equal to (Vw1u1/g) to the liquid the following expression is obtained: p s V s2 V? 1u1 p 2 V12 + + = + + hLi ? 2g g ? 2g â€¦â€¦â€¦.. (6. 6) where p2 is the pressure and V1 is the absolute velocity of the liquid leaving the impeller and hLi is the loss of head in the impeller. 51 6. Specific Speed of Centrifugal Pumps In order to compare the performance of different pumps, it is necessary to have some term which will be common to all centrifugal pumps. The term used for this purpose is the specific speed. The specific speed of a centrifugal pump is the speed at which the specific pump must run to deliver unit quantity against unit head, the efficiency being the same as the actual pump. Ns = NQ H 3/ 4 â€¦â€¦â€¦â€¦(6. 7) where Ns= specific speed N = rotational speed(rpm) H = total head 6. 7 Performance of Pumps- Characteristic Curves

A pump is usually designed for one speed, flow rate and head in actual practice, the operation may be at some other condition of head on flow rate, and for the changed conditions, the behaviour of the pump may be quite different. Therefore, in order to predict the behaviour and performance of a pump under varying conditions, tests are performed and the results of the tests are plotted. The curves thus obtained are known as the characteristic curves of the pump. The following three types of characteristic curves are usually prepared for the centrifugal pumps : a) Main and operating characteristics. (b) Constant efficiency or Muschel curves . (c) Constant head and constant discharge curves. Main and Operating Characteristics In order to obtain the main characteristic curves of a pump it is operated at different speeds. For each speed the rate of flow Q is varied by means of a delivery valve and for the different values of Q the corresponding values of manometric head Hm, shaft H. P. , P , and overall efficiency ? are measured or calculated. The same operation is repeated for different speeds of the pump. Then Q v/s Hm ; Q v/s P and Q v/s ? urves for different speeds are plotted, so that three sets of curves, as shown in Fig. 6. 4 are obtained, which represent the 52 main characteristics of a pump. The main characteristics are useful in indicating the performance of a pump at different speeds. During operation a pump is normally required to run at a constant speed, which is its designed speed, (same as the speed of the driving motor). As such that particular set of main characteristics which corresponds to the designed speed is mostly used in the operations of a pump and is, therefore, known as the operating characteristics.

A typical set of such characteristics of a pump is shown in Fig. 6. 5 Fig. 6. 4 Main characteristics of a centrifugal pump Fig. 6. 5 Operating characteristic curves of a centrifugal pump 53 6. 8 Parallel or Series Operation of Pumps Pumps in series Centrifugal pumps generate a relatively low head delivering a fairly high rate of discharge. Normally a pump with a single impeller can be used to deliver the required discharge against a maximum head of about 100 m. But if the liquid is required to be delivered against a still larger head then it can be done by using two or more pumps in series. Fig. 6. 6 Three stage centrifugal pump

If the required head is more than that can be provided by one pump, the pumps are connected in series. The same discharge passes through both pumps but the head developed by one pump add the other. The total head developed is obtained by adding together the value of the head of each pump corresponding to the relevant discharge. Pumps in Parallel The multi-stage pumps or the pumps in series as described earlier are employed for delivering a relatively small quantity of liquid against very high heads. However, when a large quantity of liquid is required to be pumped against a relatively small head, then it may 4 not be possible for a single pump to deliver the required discharge. In such cases two or more pumps are used which are so arranged that each of these pumps working separately lift the liquid from a common sump and deliver it to a common collecting pipe through which it is carried to the required height Fig. 6. 7. Since in this case each of the pumps deliver the liquid against the same head, the arrangement is known as pumps in parallel. If Q1, Q2, Q3â€¦.. , Qn are the discharging capacities of n pumps arranged in parallel then the total discharge delivered by these pumps will be

Qt = (Q1+Q2+Q3+â€¦â€¦+Qn) â€¦â€¦â€¦â€¦. (6. 8) If the discharging capacity of all the n pumps is same, equal to Q , then the total discharge delivered by these pumps will be Qt = nQ Fig. 6. 7 Two centrifugal pumps arranged in parallel 55 A centrifugal pump, having four stages in parallel, delivers 11m3/min of Example 6. 1 liquid against a head of 24. 7m, the diameter of the impeller being 225mm and the speed 1700 rpm. A pump is to be made up with a number of identical stages in series, of similar constriction to those in the first pump, to run at 1250 rpm, and to deliver 14. 5 m3/min, against a head of 248m.

Find the number of stages required for the second pump. 1st Pump 2nd Pump Q = 11m3/min Q = 14. 5m3/min H = 24. 7 m H = 248 m N = 1700 rpm N =1250 rpm D = 225 mm Specific speed N s = NQ H 3/ 4 Q for one pump = 11/4 = 2. 75 m3/min Ns = 1700 2. 75 24. 7 3 / 4 = 254 For 2nd pump, with identical stages in series i. e multi-stage pump, if each stage is similar to those of each stage is similar to those of the first pump. The specific speed of each stage Ns = 254 Ns = 254 = NQ H 3/ 4 1250 14. 5 H 3/ 4 H = 49. 64 m Total head required = 248 m No of stages required = 248/49. 64 = 5 stages 56 CHAPTER 7

DIMENSIONAL ANALYSIS, HYDRAULIC SIMILITUDE AND MODEL INVESTIGATION 7. 1 Dimensional Analysis Dimensional analysis is a mathematical method of obtaining the equations, changing units, determining a convenient arrangement of variable of a physical relation. In an equation expressing a physical relationship between quantities, absolute numerical and dimensional equality must exit. In general, all such physical relationships can be reduced to the fundamental quantities of mass M, length L and time T. It is a based on the assumption that the phenomenon can be expressed by a dimensionally homogeneous equation, with certain variable.

The dimensional analysis is widely used in research work for developing design criteria and also for conducting model tests. 7. 2 Dimensions and Units All physical quantities are measured by comparison. This comparison is always made with respect to some arbitrarily fixed value for each independent quantity, called dimension(e. g. , length, mass, time, etc. ). Since there is no direct relationship between these dimensions, they are called fundamental dimensions. Some other quantities such as area, volume, velocity, force etc. can not be expressed in terms of fundamental dimensions and thus may be alled derived dimensions. There are two systems for fundamental dimensions namely FLT (i. e force, length, time) and MLT (i. e. , mass, length, time). One common system employed in dimensional analysis is the M,L,T system. Table is a listing of some of the quantities used in fluid flow, together with their symbols and dimensions. 57 Quantity Symbol Dimensional Form Length l L Time t T Mass m M Velocity v L T-1 Acceleration a L T-2 Force F M L T-2 Pressure P M L-1 T-2 Discharge Q L3 T-1 Power P M L2 T-3 W,E M L2 T-2 Density ? M L-3 Dynamic viscosity Âµ M L-1 T-1 Kinematic viscosity ? L2 T-1 Surface tension M T-2 Work,energy 7. 3 Methods of Dimensional Analysis The methods of dimensional analysis are: -Buckingham’s ? theorem -Ralyeigh’s method Buckingham’s ? Theorem If there are ‘n’ variables in a dimensionally homogeneous equation, and if these variables contain ‘m’ fundamental dimensions such as (M,L,T) , they may be grouped into (nm) non-dimensional independent ? terms. Mathematically, if a variable x1 depends upon independent variables x2, x3,x4, â€¦. ,xn, the functional equation may be written as x1 = f (x2, x3, x4, â€¦. , xn) The equation may be written in its general form as 58 f1 ( x1, x2, x3,â€¦â€¦. xn) = C In this equation there are ‘n’ variables. If there are ‘m’ fundimental dimensions, the according to ? theorem f2 ( ? 1,? 2,? 3,â€¦.. , ? n-m ) = C1 e. g Q = f (d,H,Âµ,? ,g) f1 (Q, d, H, Âµ, ? , g) = C n = 6; m = 3; (n-m) = 3 f2 ( ? 1, ? 2, ? 3) = C1 Procedure 1. First of all, write the functional relationship with the given data. 2. Then write the equation in its general form. 3. Choose ‘m’ repeating variables and write separate expressions for each term. Every ? term will contain the repeating variables and one of the remaining variables. The repeating variables are written in exponential form. 4.

With the help of the principle of dimensional homogeneity, find out the values of the exponents by obtaining simultaneous equations. 5. Substitute the value of these exponents in the ? term. 6. After the ? terms are obtained, write the functional relation in the required form. e. g Q = f ( d, H, Âµ, ? , g) f1 (Q, d, H, Âµ, ? , g) =C n = 6 , m = 3 , (n-m) = 3 f2 (? 1,? 2,? 3)=C1 Choose ? , g, d as repeating variable, ?1 = ? a1 gb1 dc1 Q ? 2 = ? a2 gb2 d c2 H ? 3 = ? a3 gb3 dc3 Âµ 59 Selection of Repeating Variable 1. The variables should be such that none of them is dimensionless. 2. No two variables should have the same dimensions. . Independent variables should be as far as possible, be selected as repeating variable. ? > fluid property ? > flow characteristics l > geometric characteristics Example 7. 1 A V-notch weir is a vertical plate with a notch angle ? cut into the top of it and placed across an open channel. The liquid in the channel is backed up and forced to flow through the notch. The discharge Q is some function of the elevation H of upstream liquid surface above the bottom of the notch. In addition it depends upon gravity and upon the velocity of approach Vo to the weir. Determine the form of discharge equation: ?V ? Q = gH 5/2 f ? ,? ? ? gH ? ? ? Q = f (H, g , Vo , ? ) f1 ( Q, H ,g ,Vo, ? ) = C Choose g and H as repeating variables n = 5; n-m =3 ; m=2 ?1 = Ha1 gb1 Q = (L)a1 (LT-2)b1 L3 T-1 ?2 =Ha2 gb2 Vo = (L)a2 (LT-2)b2 LT-1 ?3 = ? (M)o (L)o (T)o = (L)a1 (LT-2)b1 L3 T-1 a1+b1+3 =0 ? a1 = -5/2 -2b1-1 =0 ? b1 = -1/2 ?1 = H-5/2 g-1/2 Q = Q vg H 5/2 (M)o (L)o (T)o = (L)a2 (LT-2)b2 L T-1 a2 = -1/2 60 b2 = -1/2 ?2 =H-1/2 g-1/2 Vo= Vo vgH Q f2 ( gH 5/ 2 Q gH 5/ 2 , Vo ,? gH ) = C1 Vo ,? gH ) Vo ,? gH ) =f( Q = gH 5/2 f ( Example 7. 2 Q = VD 2 f [ Prove that the discharge over a spillway is given by the relation gD H ,] V D where V= velocity of flow

D = depth of throat H = Head of water g = Acceleration due to gravity Q= f (V,D,H,G) f1 (Q,V,D,H,G) = C Choose V and D as repeating variables n = 5, m =2, n-m =3 ?1= Va1 Db1 Q = (LT-1)a1 (L)b1 (L3T-1) ?2 = Va2 Db2 H = (LT-1)a2 (L)b2 (L) ?3 = Va3 Db3 g = (LT-1)a3 (L)b3 (LT-2) M0L0T 0 = (LT-1)a1 (L)b1 (L3T-1) 0 = -a1-1 ; a1 = -1 0 = a1+b1+3 ; b1= -2 ?1 = V -1 D

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