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.

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Ans. (1) Water availability

(2)Water storage

(3)Head of the water

(4)Distance from load centre

(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,   = V - v    

               Hydraulic efficiency=        

                                      

Q. 4. Distinguish between impulse turbines and reaction turbines.        Impulse turbine               Reaction turbine1. All the available 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.

Only a portion of fluid energy is converted into kinetic energy.

Blades are in action all the time.

Water is admitted over the circumference of the wheel.

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4. The wheel does not run full and air has free access to the buckets.

5. Unit is installed above the tail race.

6. There is no loss when the flow is regulated.

Water completely fills the vane passages throughout the operation of the turbine.

Unit is kept entirely submerged in water below the tail race.

There is always a loss when the flow is regulated.

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.

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Fig. Constant head characteristics for Pelton and Kaplan turbines

(ii) Operating or constant speed characteristics: The speed is kept constant, discharge   and head H may vary the brake power P is measured. Overall efficiency is then calculated. Results are graphically represented as shown in the figure:

           

Fig. Constant speed curves for a hydraulic turbine

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The following information is collected:

→Kaplan turbine is most efficient at all ranges of the output.

→Different turbines have the same maximum overall efficiency of about 85% at full load.

→Propeller turbine gives the poorest performance at part load.

→The performance of the Kaplan and pelton wheel is much superior at the low heads and at part load.

(iii) Constant Efficiency curves. These curves are also called as iso-efficiency curves.

The curves are draws after obtaining the data from various other curves

like   versus  and   versus .

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.

                 

Fig. Constant efficiency curves for a reaction turbine

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Q. 6. Sketch layout of a typical hydroelectric power plant and label it.

Ans.

Q. 7. 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

        5. Access to site

        6. Ground water data

        7. Environment aspects of site selection

       8. Consideration of water pollution effects.

Q. 8. Show that the maximum hydraulic efficiency of a pelton bucket is 100%.

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   Ans.                                     V = Absolute velocity,   = V – v

              Hydraulic efficiency

                                            

For maximum efficiency

 

                      2V -4v =0

Or               

It means that velocity of the wheel, for maximum hydraulic efficiency, should be half of the velocity.

Therefore, maximum W.D/kN of water

                              

                                                  (Substituting v = )

                                (1 + cos ) KN

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Maximum hydraulic efficiency:

                                                      

                                                          

taking cos  =1, i.e.                         = 

                                                             

                                                          =    1 orl00%.

Q.9. Sketch a pelton turbine bucket and show its working proportions.

Ans.

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Q. 10. Why the buckets of pelton wheel are provided with an under-cut? What role does the splitter play in the pelton turbine?

Ans. 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.

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                          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.

Q. 11. What the function of notch in pelton turbine?

Ans. A notch made near the edge of the outler rim of each bucket is carefully sharpened to ensure a loss-free entry of the Jet into the buckets i.e. the path of the jet is not obstructed by the incoming buckets.

Q. 12. What are the materials used for the buckets of pelton turbine?

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. Thus, they are made of special bronze or steel alloys with nickel, chromium or stainless steel.

Q. 13. Explain the various factors which decide the choice for a particular hydraulic turbine for a hydraulic power project.

Ans.1. Water Availability—The estimates of the average quantity of water available should be prepared on the basis of actual measurement. The curves or graphs can be plotted between the river flow and time. These are known as hydrographs and flow duration curves when the river flow data is calculated on daily, weekly, monthly and yearly basis.

2. Water-storage-The output of hydropower plant is not uniform due to wide variations of rainfall. 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.

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3. Head of water—The level of water in the reservoir for a proposed plant should always be within limits throughout the year.

4. Distance from load centre-To be economical on transmission of electric power, the routes and distances should be carefully considered because cost depends upon the route selected for the transmission line.

5. Access to site—It is always desirable factor to have a good access to the site of the plant. The transport facilities must taken into the considerations

6. 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.

7. 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, healthful, productive and culturally pleasing surroundings.

(ii) To avoid health hazards.

(iii) To preserve important historic, cultural and natural aspects of the site.

8. 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. The effects effect the economy and reliability of the power plant.

Q. 14. Write short note on governing mechanism for hydraulic turbines.

Ans. Hydraulic turbines are directly coupled to alternators which must run continuously at constant speed, so that electricity is produced at constant frequency. The power produced by water turbine is directly proportional to the available head and discharge through the turbine. The quantity of water flowing can be controlled by varying the area of flow at the turbine inlet.

                   In pelton turbine, the flow area is changed by moving the spear inside the nozzle and in reaction turbine, the area of flow is varied by rotating the guide vanes with the help of governor in a controlling unit.

Q. 15. Obtain Hydraulic efficiency and work done by pelton turbine (Impulse turbine).

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Ans.

             

                            Fig. Triangle of the velocities

              V = Absolute velocity of entering water

                   = Relative velocity of water

                  = Velocity of flow at inlet.

     = Corresponding values at outlet

             D= Diameter of wheel

            d= Diameter of the nozzle

         N= Revolution 9f the wheel in r.p.m.

=Angle of blade tip at outlet

      H= Total head of water

In case, a=0°, =0°, =v and  =v-v

The relation between two velocity triangles is

                                   =v and = (V-v)

Force, KN of water in the direction of motion of jet

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Work done = Force x Distance=

                                                     =

Hydraulic efficiency, 

                                                       

                                           

Consider case in which the value of   is negative as shown in figure.

So, Work done per kN of water =

                                             

                                         

                                                      

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Q. 16. Explain the various design aspects of pelton wheel.

Ans.

1. Velocity of Jet –

 Theoretical velocity, 

Actual velocity, V

Value, , 0. 97 - 0 .99 (Friction loss)

2. Speed Ratio,  — It represents the ratio of the peripheral velocity to the theoretical velocity of the jet.

                     

Value of   = 0.45 - 0.47

3. Mean diameter of the wheel — D refers to the diameter of the wheel measured upto the centers of the buckets. The diameter is calculated from the formula

                                U     14

D u the pitch or mean diameter.

4. Jet ratio — m represents the ratio of the pitch circle diameter of the jet diameter. i.e. m=D/d

5. Number of jets—Pelton wheel has one nozzle or one jet. A number of nozzles may be employed when more power is required.

6. Working proportions-The working proportion of the turbine bucket are generally specified in terms of jet diameter d, and usually adopted values are

                     Axial width, B = 3d to 4d

                    Radial length, L = 2d to 3d        

                    Depth, T = 0.8d to 1.2d

7. Number of buckets — No. 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.

(ii) The jet of water must be fully utilized so that no water from the jet goes waste.

Q. 17. Define the term Net or effective head.

Ans. The head available at the entrance to the turbine is called Net or effective head.

                                  H=

Where  is the difference of Head race and tail race.

is the loss in head due to friction in penstock.

(1) Work done by pelton wheel, W =

(2) Efficiency of Pelton wheel, 

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 (3)                               Maximum, 

(4)                         Gross Head, H =

(5)              Power supplied the jet = WQH =

(6) Power delivered by the bucket wheel=

(7)                               Overall efficiency,   

(8)                       Volumetric efficiency, 

(9)                             Hydraulic efficiency, 

(10)                           Mechanical efficiency, 

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, diameter of jets and mean diameter of bucket circle. Take velocity co-efficient = 0.98, speed ratio = 0.46 and overall efficiency = 85%.

Ans.                               No. of Jets= 2

                                                 =16

                                                   P=l000kW=l000x W

                                                 H = 350 — (0.05 x 350) = 332.5 m

                                       d (Jet dia) =?

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                                      N (Speed)?

                          D (mean dia of bucket circle) =?

                                          = 0.98

                                           Ku = 0.46

                                         = 0.85

            Power for single jet = = 500 kW

                                   

                                      16=

                                N= = 1016 r.p.m

                                       

                             0.85=

                            Q=       = 0.36 

                   For single jet    q=

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                       = 0.0538m

                       

                 

Problem 2. A pelton wheel is to be designed for the following specifications. Shaft power = 11,772 kW, head = 380 meters, speed = 750 r.pm, overall efficiency = 86%, jet diameter is not exceed one-sixth of the wheel diameter. Determine:

(1) Wheel diameter                                  

(2) No. of jets required

(3) Diameter of the jet

Solution. Given

           Shaft power, S.P. = 11,772 kW

                           Head, H =380m

                   Speed, N = 750 r.p.m.

Overall efficiency,  = 86 %, or 0.86

Ratio of jet dia to wheel dia. =

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Coefficient of velocity,  = 0.985

Speed ratio,  = 0.45

Velocity of jet,  = 85.05 m/s

The velocity of wheel, u =

                                            = Speed ratio x

                                           = 0.45 x = 38.85 m/s

But                                 U =

                               38.85 =

Or                                        D = 0.989 M Ans.

But                                    

                            Dia. of jet, d = = 0.165 m              

Discharge of one jet, q = Area of jet x velocity of jet

                                   (0165) x85.05

                                   

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Now                              

                                      0.86

                         Total discharge, Q =

                            Number of jets

                                                    2jets.

Problem 3. 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 = 4.5 kW

                  Determine (i) Power lost in nozzle

                 (ii) Power lost due to hydraulic resistance in the runner.

Solution. Given

               Head at the base of nozzle,  = 80 m

              Diameter of jet, d = 100 mm = 0.1 m

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 Area of the jet, a = = 0.007854

              Discharge of the nozzle, Q = 0.30

             Shaft power, S.P = 206 kW

             Power absorbed in mechanical resistance = 4.5 kW

               Now, discharge, Q = area of jet x velocity of jet

                                                

                        0.30 = 0.007854 x 

                        

Power at the base of the nozzle in kW =

                                                                      =235.44

Power corresponding to kinetic energy of the jet in kW

                                                                

                                                                  

                                                                   = 218.85 kW

 (1) Power at the base of the nozzle =

                                                          Power of the jet + Power lost in Nozzle

                                                        235.44= 218.85 + Power lost in nozzle

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                                     Power lost in nozzle = 235.44 — 218.85 = 16.59 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

              235.44=206 + 16.59 + Power lost in runner + 45

              Power lost in runner = 235.44 — (206 + 16.59 + 4.5)

                                                     = 235.44 — 227.09 = 8.35 kW.

Problem 4. A pelton wheel is supplied with water under a head of 45m and at a rate of 48    /min. The buckets deflect the jet through 165° and the mean bucket speed 14m/s. Calculate the power delivered to the shaft and overall efficiency of the machine. Assume coefficient of viscosity 0.985 and mechanical efficiency 0.95.

Solution. Power developed is given by

                                      

                                        

                                          w = 9810   

                                            

                                                 

                             W =9810x0.8=7848N/s

                               v = 0.985   = 29.267 m/s

                               u = 14 m/s

                              K =1 (Neglecting friction in buckets)

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                                =180°—165°  =150°

                         =cos15°0.9659

                         

                         [(29.267— 14) + (1 + 0.9659)] 14

                          = 800 (15.267 + 1.9659) 14

                          =193.008 kW

       Hydraulic efficiency,  =

                                      

                                       

       Overall efficiency, 

                                           = 0.98 x 0.95

                                            = 0.9320

                                           = 93.20%

Problem 5. A pelton wheel is required to develop 4000 kW at 400 revs/min, operating under an available head of 350 m. There are two equal jets and the bucket deflection angle is 165°. Calculate the bucket pitch circle diameter, the cross-sectional area of each jet and the hydraulic efficiency of the turbine. Make the following assumptions (i) Overall efficiency is 85%, when the water is discharged from the wheel in a direction parallel to the axis of

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rotation (ii) coefficient of velocity of nozzle ku 0.97 and blade speed ratio ku 0.46 (iii) relative velocity of water at exit from the bucket is 0.86 times the relative velocity at inlet.

Solution. Power available from the turbine shaft

                                               

                                     4000x  = (9810xQx350)x0.85

 Total discharge through the wheel,

                        Q= = 1.37

Velocity of jet, v = kv

              = 0.97 x

Total discharge, 

                     1.37 = 2A x 80.38

Area of each jet, A =

Velocity of bucket, u=

                                             = 38.12 m/s

Also                                        

                                               38.12 = =1.82m

Since the jet gets deflected 165°, (180 — 165) = 15°

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Bucket to jet speed ratio, 

Invoking the relation for the hydraulic efficiency of a pelton wheel,

                      (relative to K.E. of jet)

                    = 2 x (0.474 — 0.4742) (1+ 0.86 cos 15°)

                         = 0.9128 or 91.28%

                       (relative to power)

                          

                         = 0.858 = 85.8%

Problem 6. At hydroelectric power plant, water available under a head of 250 m is delivered to the power house through three pipes each 250 m long. Through three pipes the friction loss is estimated to be 20 m. The project is required to produce a total shaft output of 13.25 MW by installing a number of single jet pelton wheels whose specific speed is not to exceed 38.5. The wheel speed is 650 rpm, overall efficiency is 0.85 and speed ratio is 0.46. 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 0.97 and 0.02 respectively.

Solution. Net available head, H = 250 — 20 = 230 m

We know                              

                                              

                 Power available, P = = 2829 kW

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Number of machines = = =4.68; say 5

(ii) Velocity of jet, V = = 65.16 m/s

Tangential velocity of bucket,

                               u = 0.46 x Velocity of jet = 0.46 x 65.16 = 29.97 m/s

also                      ; 29.97=

Diameter of wheel, D = = 0.881 m

Power available from turbine

                           P = (w Q H) 

                = (9810 x Q x 230) x 0.85

Discharge, Q =

Also                           

1.38=

Hence jet diameter, d = 0.167 m

Total discharge for 5 machines = 5 x 1.38 = 6.9

Discharge per pipe, QP =

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Loss of head,  =

                        20=

       =0.933m.

Problem 7. A pelton wheel is required to generate 3750 kW under an effective head of 400 meters. Find the total flow in liters /second and size of the jet. Assume generator efficiency 95%, overall efficiency 80%, coefficient of velocity 0.97, speed ratio 0.46, If the jet ratio is 10, find the mean diameter of the runner.

Solution. Given          P = 375OkW, H = 400m,  =95% = 80%

                                                  = 0.97, speed ratio = 0.46

Total flow of water in liters/second

                                               

Overall efficiency, 0.8

                                        

                                    

Size of jet:

Let                                            d = Diameter of the jet

We know that velocity of the Jet, v  = 85.9m/s

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                   Total discharge = discharge through the Jet

                                          Q

                                      1.25 = 85.9 = 67.5

                                    = 1.25/67.5 = 0.0185

Or                                d = 0.136m = 136mm.

Problem 8. 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.  Given H = 350m,                  N = 300r.p.m.                  

1. Diameter of the wheel

                            

 And peripheral velocity of the wheel

                                      V =0.46V=0.46x81.2=37.4m/s

Peripheral velocity (v)

                    

                 D =37.4/15.7=2.4m

2. Diameter of jet,

                       =240mm

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3. Width of the buckets

                          = 5 x d = 5 x 0.24 =  1.2m

4. Depth of the buckets

                                  = 1.2 X d = 1.2 x 0.24 = 0.48m

5. No. of buckets

                                

Problem 9. 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 m. If the bucket deflects the jet through an angle of 160°, find the power developed by the turbine and its hydraulic efficiency. Take the coefficient of velocity as 0.98. Neglect friction in the bucket. Also determine the overall efficiency of the turbine if its mechanical efficiency is 80%.

Solution. The power developed by the turbine is given as

                                                                  

                          W=wQ

                           W =9810

                           

                         W = 9810x 0.75 = 7357.5 N/s

                             =25.68 m/s

                            u =12 m/s

                k =1 (for neglecting the friction in the buckets)

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= (180°—  ) = (180 — 160°) = 200

  = cos   = 0.9397

                  Thus by substitution, we get

                 [(25.68 -12) (1+0.9397) x12W

                =238816 W= kW=238.816kW

        Since I metric h.p. = 735.5 W, the power developed by the              turbine in metric h.p. is

                                           = 324.699 metric h.p.

The hydraulic efficiency of the turbine is given as

                  

                  

                            = 0.966 or 96.6%

The overall efficiency of the turbine given by equation

                                  

                                   = 80% or 0.80

                                     = 0.966 x 0.80

                                         = 0.773 or 77.3%

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Problem 10. 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, (ii) in the runner, (in) in mechanical friction.

Solution. Power at the base of the nozzle = (9810 x 018 x 32)

                                                                         =56510W=56.5lkW

            Velocity of flow through the nozzle

                                                        v=

             Power at the nozzle exit (e g, 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 kmetic energy of the Jet

                                                                          =51.84kW

    Power developed by the runner = = 46.81 kW        

   Power lost in the runner = (51.84 — 46.81) = 5.03 kW

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   Power lost in mechanical friction = (46.81— 44) = 2.81 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, i the runner and in mechanical friction.

Thus, we have

                                         (56.51—44) =12.51kW

                 And                 (4.67+5.03+281) =12.51kW

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%, calculate. (i) Diameter of wheel (ii) quantity of water required and (iii) number of jets. Assume suitable values for the velocity coefficient and the speed ratio.

Solution.     Velocity of Jet, V =

                                        = 0.98 -.12 x 9.81 x 300 75.18 m/s

                                                                      (Assuming  = 0. 9)

Tangential (peripheral) velocity of wheel,

                                                           

                                                               = 35.29 m/s   (assuming K,, 0.46)

Also                                                     

                                                               

               Bucket pitch circle diameter, D=

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                      Diameter of jet, d =

(b) Power available from the turbine shaft is,

                                     P =

                                

Total discharge through the Pelton wheel, Q =

(c) The discharge through the wheel is supplied by the jets. Thus

                                          

                             Number of jets, n 2.398/0.8856 = 2.70

  And hence we employ three jets.

Revised jet diameter follows from the relation,

                                         2.398=

                                                  d = 0.1164 m

                                                      

Problem 12. A Pelton wheel of 12 m mean bucket diameter works under a head of 650 m. The jet deflection is 165° and its relative velocity is reduced over the buckets by 15% due to friction. If the water is to leave the bucket without any whirl, determine: (a) rotational speed of the wheel, (b) ratio of bucket speed of jet velocity, (c) impulsive force and the power developed by the wheel, (d) available power (water power) and the power input to buckets,

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and (e) efficiency of the wheel with power input to buckets as reference input. Take   = 0.97.

Solution. Velocity of jet 

                   Let bucket speed

            Relative velocity at inlet                              

        Relative velocity at outlet, 

Since the jet gets deflected through 165°, the blade angle at exit,

                                 =180—165=15°

As the jet leaves the bucket without any whirl, the velocity triangle at outlet will be:

                                                                       

   From expression (i) and (ii),

                                                 0.85 (109.6 - u) cos 15° = u;

                                                                    89.98-0.821u=u

                                          Blade speed u = 89.98/1.821 = 49.41 m/s

                                       

                                            49.41 =

Rotational speed of wheel, N =

   (b) Ratio of bucket. Speed to jet speed,

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                                                                 =0.4508

(c) Discharge through the wheel

                                      

Impulsive force on the buckets,

                                           

                                  = 1000 x 0.86 x 109.6 = 94256 N    

Power develop by wheel, P = impulsive force x distance moved =    

                                        = 94256 x 46.41 = 4657 x   W = 4657 kW    

(d) Available power (water power)

                                                       = wQH = 9810 x 0.86 x 650  

                                                            =5455x  W=5455kW      

     Power input to buckets =

                                             =5165x  W=5165kW

(e) Efficiency of wheel, 

                                           

Problem 13. The following data relate to a Pelton wheel : Head = 72 m, speed of wheel = 240 rpm., shaft power of the wheel = 115 kW,

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speed ratio = 0.45, coefficient of velocity 0.98, overall efficiency = 85%. Design the Pelton wheel.

Solution. Power available from turbine shaft

                                                          = (WQH) x

                                                      115. X  = 9810 x Q x72 x 0.85

                                                      Q =     

                                           Velocity of jet, V=

                                                                       

                                                                            = 36.81 m/s

                                                                            

                                                                             = 17.28 m/s

Diameter of wheel

                                     

                            17.28 =

                                  D =1.37m

Diameter of Jet Discharge

                                   

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                                   0.91 =

                                        

                                 d =0.081m81 mm

Size of buckets

                        Width of buckets =5xd=5x81=405mm

                       Depth of buckets = 1.2 x d = 1.2 x 81= 97.2 mm

No. of buckets on the wheel

                                                          Z =

Problem 14. 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. A number of Pelton wheels are to be installed generating a total output of 18 MW Determine the number of units to be installed, diameter of the Pelton wheel and the Jet diameter when the followings are available, wheel speed 650 rpm, ratio of bucket to jet speed 0.46, specific speed not to exceed 30 (m, kW, rpm), Cv and Cd for the nozzle are 0.97 and 0.94 respectively and the overall efficiency of the wheel 87%.

Solution. Given:

                                        Total head=300 m

                                            Length = 2.35 km = 2350 m

                           Frictional losses = 26 (J/N) 26 (Nm/ N) (as J = Nm) 26 m

                                         Net head, H = 300 -26 = 274 m

                                                     Total output = 18 MW = 18 x   kW

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                                                                      N= 650 r.p.m.

                         Ratio of bucket to jet speed =0.46

                                                                  = O.97,   = 0.94

                                                                  = 87% = 0.87

And                                                               =30                where H is in m, P                           in kW and N in r.p.m.

Find: (i) Number of units to be installed

 (ii) Dia. of Pelton wheel (D)

(iii) Dia. of jet of water (d)

(i) Number of units to be installed

Let                                                   P = Power output of each unit in kW

Using equation (18.28) as 

Squaring both sides, we get P = 

                                               

                                                                       

(ii) Dia. of Pelton wheel (D)

          Velocity of jet is given by, 

                                                           

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But ratio of bucket to jet speed = 0.46

Or                                 

 = 0.46 x   = 0.46 x 71.12 = 32.715m/s.

But                                          

             Dia. of Pelton wheel = 0.945 Ans.                          

  (iii) Dia. of jet (d)

We know                                        

Or                                                  

 Total water power in kW = 

   Water power in kW per unit = 

                                                               =2.955x  kW

But water power in kW per unit is given by equation as,

       Water power= 

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                                          =9.8lxQx274

                                    

But discharge (Q) through one unit is also given by

                                     Q= 

Or                                      

Or                                        

d. =   = 0.1424 m = 142.4 mm. Ans.

Problem 15. A pelton wheel is supplied with water under a head of 45m and at a rate of 48 . The buckets deflect the jet through 165° and the mean bucket speed is 14 m/s. Calculate the power delivered to shaft and overall efficiency of the machine. Assume coefficient of velocity 0985 and mechanical efficiency 095.

Ans. Given:

                     H =45 m, Q = 48  min =0.8  /sec

                              

The power developed is given by.

                                    P= 

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                                        (where k = 0.95)

                                        

                                          

                                               = 0.8 x 175.952 x 1.9176

                                                     =269.92kW

    Overall efficiency,                   

                                                 

                                                   = 0.9279 = 92.79%

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