1/27/13Chapter 2 : Pelton Turbine | Fluid Machinery Fluid Machinery Recommend 6k CHAPTERS Home Topics Chapter 1 : General Concepts Chapter 2 : Pelton Turbine Chapter 3 : Francis and Kaplan Turbine Chapter 4 : Centrifugal Pumps Chapter 5 : Similarity Relations and Performance Characteristics Chapter 6 : Reciprocating Pumps Chapter 7 : Hydraulic devices and Systems Home Chapter 2 : Pelton Turbine Q. 1. Classify Hydraulic turbine. Ans. According to the type of energy at inlet (a) Impulse turbine (b) Reaction turbine. According to the direction of flow through runner (a) Tangential flow turbine (b) Radial flow turbine (c) Axial flow turbine (d) Mixed flow turbine. According to the head at inlet of turbine (a) High head turbine (b) Medium head turbine (c) Low head turbine. According to the specific speed of turbine (a) Low specific speed turbine (b) Medium specific speed turbine (c) High specific speed turbine. According to the name of the inventor (a) Pelton turbine (b) Francis turbine (c) Kaplan turbine. Q. 2. What are the factors to be considered in deciding for a particular hydro electric project. Ans. (1) Water availability (2)Water storage (3)Head of the water (4)Distance from load centre ptumech.loremate.com/fluid-machinery/node/12 1/38 1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery (5)Access to site (6)Ground water data (7)Environment aspects of site selection (8)Consideration of water pollution effects. Q. 3. Show that the maximum hydraulic efficiency of a pelton bucket is 100%. Ans. V = Absolute velocity, Hydraulic =V-v efficiency= Q. 4. Distinguish between impulse turbines and reaction turbines. Impulse turbine turbine 1. All the available fluid Only a portion of fluid energy is converted in kinetic energy. 2. Blades are in action only when they are in the front of the nozzle. 3. Water may be allowed to enter a part or whole of the wheel circumference. 4. The wheel does not run full and air has free access to the buckets. 5. Unit is installed energy is converted into kinetic energy. Blades are in action all the time. Water is admitted over the circumference of the wheel. Water completely fills the vane passages throughout the operation of the Reaction turbine. Unit is kept entirely above the tail race. submerged in water 6. There is no loss when below the tail race. the flow is regulated. There is always a loss 2/38 ptumech.loremate.com/fluid-machinery/node/12 1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery when the regulated. flow is Q. 5. Draw the performance characteristics curves for both impulse and reaction turbines and discuss their nature. Ans. Head, speed and output are the important factors for designing a turbine. So it required to know the operating conditions of the turbine under these variable factors. Information can be obtained practically by running the turbine system. The results are drawn in the form of curves are known as the characteristic curves. (i) Main or Constant Head Characteristics: When the head is maintained constant the speed is varied by quantity of water flow through the inlet the brake power is measured. The main characteristics of Francis turbine are identical to those of Kaplan turbine the discharge characteristics, however, differ the following information is obtained: →For pelton turbine discharge curves are the horizontal lines. →For Kaplan turbine discharge curve rises as the speed increases. →Power and efficiency curves are parabolic in nature. →For pelton (impulse) turbine the maximum efficiency for different gate openings occurs at the same speed. →For Francis (reaction) turbines the maximum efficiency for different gate openings usually occurs at different speeds. ptumech.loremate.com/fluid-machinery/node/12 3/38 Results are graphically represented as shown in the figure: ptumech.loremate. discharge and head H may vary the brake power P is measured.com/fluid-machinery/node/12 4/38 . Overall efficiency is then calculated. Constant head characteristics for Pelton and Kaplan turbines (ii) Operating or constant speed characteristics: The speed is kept constant.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Fig. com/fluid-machinery/node/12 5/38 . →The performance of the Kaplan and pelton wheel is much superior at the low heads and at part load. →Different turbines have the same maximum overall efficiency of about 85% at full load.loremate. ptumech. The curves are draws after obtaining the data from various other curves like versus and versus . (iii) Constant Efficiency curves. →Propeller turbine gives the poorest performance at part load.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Fig. Constant speed curves for a hydraulic turbine The following information is collected: →Kaplan turbine is most efficient at all ranges of the output. A curve for the best performance is obtained by joining the peak points of various iso of various iso efficiency curves as shown in the figure. These curves are also called as iso-efficiency curves. 1. 6.com/fluid-machinery/node/12 6/38 . Sketch layout of a typical hydroelectric power plant and label it. Head of the water 4. Constant efficiency curves for a reaction turbine Q. What are the factors to be considered in deciding for a particular hydro electric project.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Fig. Distance from load centre 5.loremate. Water availability 2. Ans. Q. Access to site ptumech. Water storage 3. 7. Ans. Therefore. =V–v Hydraulic efficiency V = Absolute For maximum efficiency 2V -4v =0 Or It means that velocity of the wheel. Q.D/kN of water (Substituting v = ) (1 + cos ) KN ptumech. maximum W.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery 6.com/fluid-machinery/node/12 7/38 . Consideration of water pollution effects. velocity. for maximum hydraulic efficiency. should be half of the velocity. Ans. Environment aspects of site selection 8.loremate. Show that the maximum hydraulic efficiency of a pelton bucket is 100%. 8. Ground water data 7. loremate. ptumech.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Maximum hydraulic efficiency: taking cos =1. Ans. Q.e.com/fluid-machinery/node/12 8/38 .9. i. Sketch a pelton turbine bucket and show its working proportions. = = 1 orl00%. 11.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Q. Q. 10. Bucket of Pelton turbine The striking Jet of water is divided into two parts by the splitter and each part of the jet flows side ways round the smooth inner surface of the bucket and leaves it with relative velocity almost opposite in direction to the original jet.loremate. What the function of notch in pelton turbine? ptumech. In pelton wheel each bucket is divided vertically into two parts by a splitter that has a sharp edge at the centre and the buckets look like a double hemispherical cup.com/fluid-machinery/node/12 9/38 . Why the buckets of pelton wheel are provided with an under-cut? What role does the splitter play in the pelton turbine? Ans. 3.1. What are the materials used for the buckets of pelton turbine? Ans. These are known as hydrographs and flow duration curves when the river flow data is calculated on daily. 12. 2. The curves or graphs can be plotted between the river flow and time. they are made of special bronze or steel alloys with nickel. Thus. Water-storage-The output of hydropower plant is not uniform due to wide variations of rainfall. Explain the various factors which decide the choice for a particular hydraulic turbine for a hydraulic power project.com/fluid-machinery/node/12 10/38 . weekly. A notch made near the edge of the outler rim of each bucket is carefully sharpened to ensure a lossfree entry of the Jet into the buckets i. Distance from load centre-To be economical on ptumech. monthly and yearly basis.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Ans. To have uniform power output water storage is needed so that excess flow at certain times may be stored to make it available at the times of low flow. Ans. The buckets are the most important part of the pelton turbine they have to be designed to withstand the full force of the jet. the path of the jet is not obstructed by the incoming buckets.e. 13. Water Availability—The estimates of the average quantity of water available should be prepared on the basis of actual measurement.loremate. Head of water—The level of water in the reservoir for a proposed plant should always be within limits throughout the year. Q. 4. chromium or stainless steel. Q. (iii) To preserve important historic. Q.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery transmission of electric power. cultural and natural aspects of the site. (ii) To avoid health hazards. Ground water data-The underground movement of water has important effects on the stability of ground slopes and also on the amount and type of grounding required to prevent the leakage. Hydraulic turbines are directly coupled to alternators which must run continuously at constant speed. Environment aspects of site selection—The project should be designed on the basis so that it fulfils the following requirement related with environment (i) To assure safe. Access to site—It is always desirable factor to have a good access to the site of the plant.loremate. the routes and distances should be carefully considered because cost depends upon the route selected for the transmission line. 8. productive and culturally pleasing surroundings. The power produced by water turbine is ptumech. healthful. 7. 14. governing Ans.com/fluid-machinery/node/12 11/38 . so that electricity is produced at constant frequency. The effects effect the economy and reliability of the power plant. 5. Consideration of water pollution-The effects of polluted water on the power plant is one of the major considerations in selecting the site of hydraulic power plant. Write short note on mechanism for hydraulic turbines. The transport facilities must taken into the considerations 6. ptumech. In pelton turbine. Ans. Q.p. 15. the flow area is changed by moving the spear inside the nozzle and in reaction turbine. Triangle of the velocities V = Absolute velocity of entering water = Relative velocity of water = Velocity of flow at inlet.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery directly proportional to the available head and discharge through the turbine. Fig. The quantity of water flowing can be controlled by varying the area of flow at the turbine inlet.m. the area of flow is varied by rotating the guide vanes with the help of governor in a controlling unit. = Corresponding values at outlet D= Diameter of wheel d= Diameter of the nozzle N= Revolution 9f the wheel in r.loremate.com/fluid-machinery/node/12 12/38 . Obtain Hydraulic efficiency and work done by pelton turbine (Impulse turbine). =0°. Work done per kN of water = is negative as ptumech. Consider case in which the value of shown in figure. So. =v and =v-v = (V-v) The relation between two velocity triangles is =v and Force.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery =Angle of blade tip at outlet H= Total head of water In case.loremate. KN of water in the direction of motion of jet Work done = Force x Distance= = Hydraulic efficiency.com/fluid-machinery/node/12 13/38 . a=0°. 1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Q. i. Actual velocity. — It represents the ratio of the peripheral velocity to the theoretical velocity of the jet. The diameter is calculated from the formula U D u the pitch or mean diameter.45 . m=D/d 5. Value of = 0. Velocity of Jet – Theoretical velocity.0.99 (Friction loss) 2.e.47 3. Number of jets—Pelton wheel has one nozzle or ptumech.loremate. Explain the various design aspects of pelton wheel. 1. 4. 0. Jet ratio — m represents the ratio of the pitch circle diameter of the jet diameter. . Mean diameter of the wheel — D refers to the diameter of the wheel measured upto the centers of the buckets. 16. Speed Ratio.com/fluid-machinery/node/12 14/38 . 97 .0 . Ans. V Value. Define the term Net or effective head. (ii) The jet of water must be fully utilized so that no water from the jet goes waste. The head available at the entrance to the turbine is called Net or effective head.com/fluid-machinery/node/12 Overall efficiency. Gross Head.2d 7.8d to 1. H = Power supplied the jet = WQH = delivered by the bucket wheel= (7) (8) ptumech. of buckets are decided on the following principles: (i) The number of buckets should be as few as possible so that there is little loss due to friction. Volumetric efficiency. Number of buckets — No. L = 2d to 3d Depth. A number of nozzles may be employed when more power is required. (1) Work done by pelton wheel. Working proportions-The working proportion of the turbine bucket are generally specified in terms of jet diameter d. H= Where is the difference of Head race and tail race. Q. (3) (4) (5) (6) Power Maximum. W = (2) Efficiency of Pelton wheel. Ans.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery one jet.loremate. is the loss in head due to friction in penstock. T = 0. 17. B = 3d to 4d Radial length. and usually adopted values are Axial width. 6. 15/38 . Problem 1. A double jet Pelton wheel has a specified speed of 16 and is required to deliver 1000 kW The supply of water to the turbine is through a pipeline from a reservoir whose level is 350 m above the nozzles Allowing 5% for friction loss in pipe make calculations for speed in rev/mm.5 m d (Jet dia) =? N (Speed)? D (mean dia of bucket circle) =? = 0.com/fluid-machinery/node/12 16/38 . of Jets= 2 =16 P=l000kW=l000x W H = 350 — (0.loremate.05 x 350) = 332. (10) Mechanical efficiency. Take velocity co-efficient = 0.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery (9) Hydraulic efficiency. Ans. speed ratio = 0.98 Ku = 0.98. No.85 Power for single jet = = 500 kW 16= ptumech.46 and overall efficiency = 85%. diameter of jets and mean diameter of bucket circle.46 = 0. P. jet diameter is not exceed one-sixth of the wheel diameter.com/fluid-machinery/node/12 17/38 .pm.p.36 For single jet q= = 0.772 kW ptumech. speed = 750 r.772 kW.loremate. S. overall efficiency = 86%. head = 380 meters. = 11. Given Shaft power. Shaft power = 11. Determine: (1) Wheel diameter (2) No. of jets required (3) Diameter of the jet Solution.85= Q= = 0.m 0. A pelton wheel is to be designed for the following specifications.0538m Problem 2.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery N= = 1016 r. u = = Speed ratio x = 0. Velocity of jet. m/s The velocity of wheel. of jet.85 = Or 0.loremate.985 But Dia.45 x 38.p. N = 750 r. Overall efficiency. H =380m Speed. d = = 0.86 Ratio of jet dia to wheel dia. or 0. Speed ratio. q = Area of jet x velocity of jet (0165) x85.05 Now ptumech. D= = = 0.85 m/s But U= 38.05 = 0. = 86 %. = Coefficient of velocity.165 m Discharge of one jet.m.45 = 85.com/fluid-machinery/node/12 18/38 .989 M Ans.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Head. 1 m Area of the jet.P = 206 kW Power absorbed in mechanical resistance = 4.loremate.007854 Discharge of the nozzle. = 80 m Diameter of jet. Q = Number of jets 2jets.30 = 0.com/fluid-machinery/node/12 19/38 .007854 x ptumech.5 kW Now. S. d = 100 mm = 0. Given Head at the base of nozzle. Solution. Q = area of jet x velocity of jet 0. The following data is related to a pelton wheel: Head at the base of the nozzle 80 m Diameter of the jet 100 mm Discharge of the nozzle = Power at the shaft = 206 Kw Power absorbed in mechanical resistance = 4. Q = 0.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery 0.5 kW Determine (i) Power lost in nozzle (ii) Power lost due to hydraulic resistance in the runner.30 Shaft power. Problem 3.86 Total discharge. a = = 0. discharge. 44 — 218.59 + Power lost in runner + 45 Power lost in runner = 235.59 + 4.85 = 16.com/fluid-machinery/node/12 20/38 .44 — (206 + 16.85 kW (1) Power at the base of the nozzle = Power of the jet + Power lost in Nozzle 235.44= 218.09 = 8.5) = 235.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Power at the base of the nozzle in kW = =235.35 kW. Problem 4.44=206 + 16. A pelton wheel is supplied with ptumech.loremate.44 Power corresponding to kinetic energy of the jet in kW = 218.44 — 227.85 + Power lost in nozzle Power lost in nozzle = 235.59 kW (2) Also power at base of nozzle = Power at shaft + Power lost m nozzle + Power lost in runner + Power lost due to mechanical resistance 235. 985 m/s u = 14 m/s K =1 (Neglecting friction in buckets) =180°—165° =150° =cos15°0.267 + 1.267— 14) + (1 + 0.267 [(29.9659 = 29.9659)] 14 = 800 (15.95. Power developed is given by w = 9810 W =9810x0.9659) 14 =193. = ptumech. Calculate the power delivered to the shaft and overall efficiency of the machine.985 and mechanical efficiency 0. The buckets deflect the jet through 165° and the mean bucket speed 14m/s.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery water under a head of 45m and at a rate of 48 /min. Solution.com/fluid-machinery/node/12 21/38 .8=7848N/s v = 0.008 kW Hydraulic efficiency.loremate. Assume coefficient of viscosity 0. com/fluid-machinery/node/12 22/38 .38 Area of each jet.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Overall efficiency. v = kv = Equipment 0.85 Total discharge through the wheel. Solution. when the water is discharged from the wheel in a direction parallel to the axis of rotation (ii) coefficient of velocity of nozzle ku 0.37 = 2A x 80.97 and blade speed ratio ku 0.9320 = 93.98 x 0. Q= Loremate.46 (iii) relative velocity of water at exit from the bucket is 0. the cross-sectional area of each jet and the hydraulic efficiency of the turbine. Make the following assumptions (i) Overall efficiency is 85%. A = Velocity of bucket.loremate. 1. operating under an available head of 350 m. Power available from the turbine shaft 4000x (9810xQx350)x0.37 Velocity of jet. A pelton wheel is required to develop 4000 kW at 400 revs/min. Calculate the bucket pitch circle diameter. u= = 38. There are two equal jets and the bucket deflection angle is 165°.com Like 6.86 times the relative velocity at inlet.95 = 0.20% Problem 5.12 m/s | Heat And Mass Transfer Related: Heat Capacity | Heat Transfer P ptumech.97 x | Heat Transfer Search Total discharge.099 = = 1. = 0. 46.4742) (1+ 0. overall efficiency is 0.8% Problem 6.com/fluid-machinery/node/12 (180 — 165) = 250 — 20 23/38 . H = = 230 m ptumech.12 = =1.82m Since the jet gets deflected 165°.474 — 0. 15° Bucket to jet speed ratio.9128 or 91. The project is required to produce a total shaft output of 13. Through three pipes the friction loss is estimated to be 20 m.858 = 85. water available under a head of 250 m is delivered to the power house through three pipes each 250 m long. At hydroelectric power plant.97 and 0. Invoking the relation for the hydraulic efficiency of a pelton wheel.loremate.02 respectively. (relative to K. Determine (i) the number of pelton wheels to be used (ii) Jet diameter (in) diameter of supply pipe Take velocity coefficient for the nozzle and Darcy’s friction factor as 0. Net available head.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Also 38. The wheel speed is 650 rpm.86 cos 15°) = 0. of jet) = 2 x (0.85 and speed ratio is 0.E. Solution.5.25 MW by installing a number of single jet pelton wheels whose specific speed is not to exceed 38.28% (relative to power) = 0. 97= = 0.167 m Total discharge for 5 machines = 5 x 1.com/fluid-machinery/node/12 = 24/38 .881 m Loss of head. QP = .38= Hence jet diameter. Q = Also 1.38 = 6. D = Power available from turbine P = (w Q H) = (9810 x Q x 230) x 0.85 Discharge. 29. ptumech.loremate. = = 2829 = = 65.68.46 x Velocity of jet = 0.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery We know Power available.9 Discharge per pipe.16 u = 0. say 5 (ii) Velocity of jet. d = 0. V = m/s Tangential velocity of bucket.16 = 29.97 m/s also Diameter of wheel. P = kW Number of machines =4.46 x 65. overall efficiency 80%. Given =95% = 80% = 0.46. H = 400m.8 Size of jet: Let jet We know that velocity d = Diameter of the of the Jet.25 = 85.com/fluid-machinery/node/12 = 67.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery 20= =0. find the mean diameter of the runner.933m. Solution. v = 85. Find the total flow in liters /second and size of the jet. Assume generator efficiency 95%.9 ptumech. A pelton wheel is required to generate 3750 kW under an effective head of 400 meters.loremate. 0.97. speed ratio = 0. If the jet ratio is 10. speed ratio 0. coefficient of velocity 0.9m/s Total discharge = discharge through the Jet Q 1.5 25/38 . Problem 7.97.46 Total flow of water in liters/second P = 375OkW. Overall efficiency. Given H = 350m.4m/s Peripheral velocity (v) D =37.4m 2. Diameter of the wheel N= And peripheral velocity of the wheel V =0.24 = 1.136m = 136mm. Design a pelton wheel for a head of 350 m at a speed of 300 r p m Take overall efficiency of the wheel as 85% and ratio of jet to the wheel diameter as 1/10 Solution.4/15. 300r.0185 Or d = 0.m. Diameter of jet. Problem 8. =240mm 3.p.46V=0.5 = 0.com/fluid-machinery/node/12 26/38 .25/67.7=2.2 X d = 1.24 = 0.48m 5.2m 4. A pelton wheel has a mean bucket speed of 12 m/s and is supplied with water at a rate of 750 liters per second under a head of 35 ptumech. No.loremate.2=37.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery = 1.2 x 0. Depth of the buckets = 1. of buckets Problem 9.46x81. Width of the buckets = 5 x d = 5 x 0. 1. Neglect friction in the bucket.98. = 735.68 -12) (1+0.68 m/s u =12 m/s k =1 (for neglecting the friction in the buckets) = (180°— = cos ) = (180 — 160°) = 200 = 0.5 N/s =25. the power developed by the turbine in metric h. The power developed by the turbine is given as W=wQ W =9810 W = 9810x 0.9397 Thus by substitution.com/fluid-machinery/node/12 27/38 .p. Take the coefficient of velocity as 0. is = 324. If the bucket deflects the jet through an angle of 160°.816kW Since I metric h. find the power developed by the turbine and its hydraulic efficiency.699 metric h. we get [(25.5 W. Also determine the overall efficiency of the turbine if its mechanical efficiency is 80%.9397) x12W =238816 W= kW=238. Solution.loremate.75 = 7357.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery m.p. The hydraulic efficiency of the turbine is given as ptumech.p. 773 or 77.80 = 0.com/fluid-machinery/node/12 28/38 .80 = 0.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery = 0. Kinetic energy of the jet) = 51840 W = 51 84 Kw Power lost m the nozzle = (5651 -51 84) = 467 kW Power supplied to the number is equal to the ptumech.loremate. Power at the base of the nozzle (9810 x 018 x 32) =56510W=56. (in) in mechanical friction.966 or 96.966 x 0. (ii) in the runner.6% The overall efficiency of the turbine given by equation = 80% or 0.5lkW Velocity of flow through the nozzle v= = Power at the nozzle exit (e g. The following data were obtained from a test on a Pelton wheel: (a) Head at the base of the nozzle = 32 m (b) Discharge of the nozzle = 018 (C) Area of the jet= 7500 sq mm (d) Mechanical available at the shaft = 44 kW (e) Mechanical efficiency = 94% Calculate the power lost (z) in the nozzle.3% Problem 10. Solution. Assume suitable values for the velocity coefficient and the speed ratio. V = = 0.81 kW As a check on computation. the difference of power at the base of the nozzle and the power available at the shaft must be equal to the sum of the power lost in the nozzle.51kW And =12.84kW Power developed by the runner = = 46.51kW Problem 11 How does a single Jet Pelton wheel differ from a multi-jet wheel A Pelton wheel is required to develop 6 MW when working under a head c 300 m it rotates with a speed of 550 rpm assuming jet ratio as 10 and overall efficiency as 85%. (i) Diameter of wheel (ii) quantity of water required and (iii) number of jets.81) = 5.67+5.loremate.81— 44) = 2. Thus.12 x 9.03 kW Power lost in mechanical friction = (46.29 ptumech. Solution. = 35. 75.18 m/s Velocity of Jet.03+281) (Assuming = 0.81 x 300 (4.84 — 46.98 -.com/fluid-machinery/node/12 29/38 .81 kW Power lost in the runner = (51. calculate.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery kmetic energy of the Jet =51. i the runner and in mechanical friction.51—44) =12. we have (56. 9) Tangential (peripheral) velocity of wheel. 398= d = 0. 2.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery m/s (assuming K. Revised jet diameter follows from the relation.com/fluid-machinery/node/12 30/38 . Q = (c) The discharge through the wheel is supplied by the jets. d = (b) Power available from the turbine shaft is. ptumech.70 And hence we employ three jets.8856 = 2. 0. P= Total discharge through the Pelton wheel.398/0. n 2.46) Also Bucket pitch circle diameter. Thus Number of jets.. D= Diameter of jet. A Pelton wheel of 12 m mean bucket diameter works under a head of 650 m.1164 m Problem 12.loremate. 89.u) cos 15° = u. the velocity triangle at outlet will be: From expression (i) and (ii).6 . (c) impulsive force and the power developed by the wheel.98/1. (b) ratio of bucket speed of jet velocity. the blade angle at exit. and (e) efficiency of the wheel with power input to buckets as reference input. Take Solution.41 m/s ptumech.85 (109. Velocity of jet Let bucket speed Relative velocity at inlet Relative velocity at outlet.821 = 49. (d) available power (water power) and the power input to buckets.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery The jet deflection is 165° and its relative velocity is reduced over the buckets by 15% due to friction. 0.com/fluid-machinery/node/12 31/38 .97. = 0.980. If the water is to leave the bucket without any whirl. determine: (a) rotational speed of the wheel.821u=u Blade speed u = 89.loremate. Since the jet gets deflected through 165°. =180—165=15° As the jet leaves the bucket without any whirl. 41 = Rotational speed of wheel. speed of wheel = ptumech.4508 (c) Discharge through the wheel Impulsive force on the buckets. =0.41 = 4657 x W = 4657 kW (d) Available power (water power) = wQH = 9810 x 0.com/fluid-machinery/node/12 32/38 . The following data relate to a Pelton wheel : Head = 72 m.6 = 94256 N Power develop by wheel.loremate. Speed to jet speed. P = impulsive force x distance moved = = 94256 x 46.86 x 109. N = (b) Ratio of bucket.86 x 650 =5455x W=5455kW Power input to buckets = =5165x (e) Efficiency of wheel. W=5165kW Problem 13. = 1000 x 0.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery 49. 37m Diameter of Jet Discharge 0. speed ratio = 0.85 Q= = 9810 Velocity of jet.45. Power available from turbine shaft = (WQH) x 115.98.81 m/s = 17.com/fluid-machinery/node/12 33/38 .1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery 240 rpm. X x Q x72 x 0. overall efficiency = 85%. Design the Pelton wheel. V= = 36.loremate..28 m/s Diameter of wheel 17. shaft power of the wheel = 115 kW.081m81 mm ptumech.28 = D =1. Solution. coefficient of velocity 0.91 = d =0. Given: Total head=300 m Length = 2. diameter of the Pelton wheel and the Jet diameter when the followings are available. of buckets on the wheel Z = Problem 14.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Size of buckets Width of buckets =5xd=5x81=405mm Depth of buckets = 1. wheel speed 650 rpm. rpm).94 respectively and the overall efficiency of the wheel 87%.97 and 0.2 mm No.46 ptumech. Water under a head of 300 m is available for the hydel-plant situated at a distance of 235 km from the surface the frictional losses of energy for transporting water are equivalent to 26 J/N.35 km = 2350 m Frictional losses = 26 (J/N) 26 (Nm/ N) (as J = Nm) 26 m Net head.46. kW. A number of Pelton wheels are to be installed generating a total output of 18 MW Determine the number of units to be installed. Cv and Cd for the nozzle are 0.2 x 81= 97.p.loremate. specific speed not to exceed 30 (m.com/fluid-machinery/node/12 34/38 kW N= 650 .m.2 x d = 1. ratio of bucket to jet speed 0. Ratio of bucket to jet speed =0. Solution. H = 300 -26 = 274 m Total output = 18 MW = 18 x r. Find: (i) Number of units to be installed (ii) Dia. we get P = (ii) Dia.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery = O.715m/s. 35/38 .94 = 87% Squaring both sides.com/fluid-machinery/node/12 = 0.loremate.46 Or = 0.p. = 0. But ratio of bucket to jet speed = 0.12 = 32.87 And where H is in m.28) as =30 in kW and = 0.m.46 x ptumech.97.46 x 71. P N in r. of jet of water (d) (i) Number of units to be installed Let output of each unit in kW Using equation P = Power (18. of Pelton wheel (D) (iii) Dia. of Pelton wheel (D) Velocity of jet is given by. Water power= =9. of Pelton wheel = 0. (iii) Dia.com/fluid-machinery/node/12 36/38 .945 Ans. of jet (d) We know Or Total water power in kW = Water power in kW per unit = =2.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery But Dia.loremate.8lxQx274 But discharge (Q) through one unit is also given by Q= Or ptumech.955x kW But water power in kW per unit is given by equation as. Ans.1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery Or d. P= (where k = 0.1424 m = 142. A pelton wheel is supplied with water under a head of 45m and at a rate of 48 .8 /sec The power developed is given by.9176 =269.4 mm.92kW Overall efficiency.com/fluid-machinery/node/12 37/38 .loremate. Given: H =45 m. Calculate the power delivered to shaft and overall efficiency of the machine.8 x 175. Problem 15. Ans. The buckets deflect the jet through 165° and the mean bucket speed is 14 m/s. Q = 48 min =0.952 x 1. = = 0. ptumech. Assume coefficient of velocity 0985 and mechanical efficiency 095.95) = 0. loremate.79% ptumech.com/fluid-machinery/node/12 38/38 .1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery = 0.9279 = 92.