Introduction

Pelton Wheel Turbine is an impulse or a constant pressure water turbine. In this case water head

is very high. Pelton wheel consists of a wheel called rotor. The rotor of the turbine consists of a

circular disc with a number of double spoon shaped buckets evenly distributed over the

periphery. The water is the supplied from the reservoir. In such type of Turbine available

hydraulic energy of the water is converted into the kinetic energy at atmospheric pressure by

means of the nozzle. Each nozzle directs the jet along a tangent to the circle through the centers

of the buckets. Each bucket consists of a splitter which divides the incoming jet in to two equal

portions and after flowing round the smooth inner surface of the bucket the water leaves with a

relative velocity almost opposite in direction to the original jet. The change in momentum of the

water jet in passing over the buckets exerts tangential force on the wheel causing it to rotate.

Thus converts the hydraulic energy into the mechanical energy by means of the shaft rotation.

Objective

To study the operation of Pelton Wheel and also to determine the efficiency and power output of

Pelton Turbine.

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Procedures

1. First of all, perform the general start-up procedures.

2. Fully open the throttle valve and allow the water to circulate until all air bubbles have

dispersed.

3. Open and adjust the spear valve for a particular nozzle opening.

4. Tighten up the tensioning screw on the pulley wheel until the turbine is almost stalled

( rotor just turning).

5. Decide on suitable increments in force to give adequate sample points and note the value

of the pulley brake.

6. Slacken off the tensioning screw so no force is being applied to the turbine.

7. Tighten the screw to give the first increment in force for the brake. When readings are

steady enough, record all the readings again.

8. Repeats step 7 above for a gradually increasing set of fb values. The final sample point

will correspond to the turbine stalling.

9. The data may now be used for analysis and to plot the pelton turbine characteristics

curve.

10. Now decrease the volume flow rate to a new setting by changing the throttle valve

position and at the same time also change the spear valve position to maintain the

pressure at 1.0 kgf/cm.

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RESULT ANALYSIS

Formula Used :

1. Hi =P 1

ρ× g2. Ph =ρ× g × Hi× Q3. T = Fb × r4. Pb = 2 × π × N × T

5. Et = PbPh

×100%

Experiment 1 : Turbine Characteristic

g = 9.81 m/s2

π = 3.142

r = 0.04 m

ρ = 1000 kg/m3

Flow rate Measurement Data :

V1(L)

T1(MIN)

T2(MIN)

Q(LPM)

Q(m3/s)

10 9.3 12.9 0.90 9.0E-0410 11.0 12.0 0.86 8.70E-0410 12.1 12.1 0.82 8.29E-0410 12.3 12.5 0.80 8.07E-0410 12.5 12.5 0.80 8.01E-0410 11.9 11.7 0.84 8.48E-0410 12.5 12.4 0.80 8.05E-0410 13.2 12.4 0.78 7.82E-0410 12.6 12.4 0.80 8.01E-04

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M1(g)

M2(g)

Fb1(N)

Fb2(N)

Fb(N)

N1(rpm)

N2(rpm)

N(rpm)

N(HZ)

40 30 0.3924 0.2943 0.0981 1765 1770 1767.5 29.46190 40 1.8639 0.3924 1.4715 1600 1596 1598 26.63340 40 3.3354 0.3924 2.943 1367 1354 1360.5 22.68490 40 4.8069 0.3924 4.4145 1199 1186 1192.5 19.88640 40 6.2784 0.3924 5.886 963.9 955.4 959.65 15.99790 50 7.7499 0.4905 7.2594 867.1 853.3 860.2 14.33940 70 9.2214 0.6867 8.5347 612.6 627.3 619.95 10.331090 150 10.6929 1.4715 9.2214 568.8 586.7 577.75 9.621250 150 12.2625 1.4715 10.791 36.7 84.6 60.65 1.01

Q(m3/s)

Fb(N)

N(HZ)

P1(bar)

Hi(m)

Ph(w)

T(Nm)

Pb(W)

Et(%)

9.0E-04 0.0981 29.46 1.0 9.99 88.20 0.00392 7.249 8.2188.70E-04 1.4715 26.63 1.0 9.99 85.26 0.0588 10.89 12.778.29E-04 2.943 22.68 1.0 9.99 81.24 0.1177 21.78 26.808.07E-04 4.4145 19.88 1.0 9.99 79.08 0.1765 32.67 41.318.01E-04 5.886 15.99 1.0 9.99 78.49 0.2354 43.57 55.518.48E-04 7.2594 14.33 1.0 9.99 83.10 0.2903 53.74 64.668.05E-04 8.5347 10.33 1.0 9.99 78.89 0.3413 63.18 80.087.82E-04 9.2214 9.62 1.0 9.99 76.63 0.3688 68.27 89.098.01E-04 10.791 1.01 1.0 9.99 78.49 0.4316 79.90 101.79

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0 5 10 15 20 25 30 350

0.050.1

0.150.2

0.250.3

0.350.4

0.450.5

Torque vs Turbine Rotational Speed

Torque vs Turbine Rota-tional Speed

Turbine Rotational Speed

Torq

ue

0 5 10 15 20 25 30 350

20

40

60

80

100

120

Power output vs Turbine Rotational Speed

Power output vs Turbine Rotational Speed

Turbine Rotational speed (hz)

pow

er o

utpu

t (W

)

5

0 5 10 15 20 25 30 350

102030405060708090

100

Efficiency vs Turbine Rotational Speed

Efficiency vs Turbine Rota-tional Speed

Turbine Rotational speed (hz)

Efficie

ncy(

%)

Sample Calculation

Volumetric Flow Rate, Q

Q = V T

= 10

(0.155+0.215 ) /2

= 54.05 LPM

= 0.90 L

min ×

1 m1000 L

× 1min60 s

= 9.0 E-04m3/s

Brake Force, Fb

Fb =Fb1 – Fb2

= (m1×g) (m2×g)

= (0.04 kg × 9.81m/s2 ) (0.03 × 9.81m/s2)

=0.0981 N

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N = N 1+N 2

2

= 1765+1770

2

= 1767.5 rpm

=1767.5 × 1min60 s ×

1 Hz1/ s

= 29.46 Hz

Input Head to Turbine, Hi

H = (P1) Ρ×g

= 1.0 ×98066.5 pa

1000kgm 3

× 9.81m / s

= 9.99 m

Hydraulic Power at Input, Ph

Ph = p×g×Hi×Q

= 1000 kg/m3 × 9.81 m/s3×9.99 m ×9.0E - 04m3/s

= 88.20 W

Torque, T

T = Fb × r

= 0.0981 × 0.04 m

= 0.00392 Nm

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Brake Power, Pb

Pb = 2 × π × N × T

= 2 × 3.142 × 29.46 × 0.00392 Nm

= 7.249 W

Turbine Efficiency, Et

Et = PbPh

× 100%

= 7.249W88.20 W

× 100%

= 8.218 %

Discussion

The working principle of Pelton wheel turbine is water flows along the tangent to the path of the

runner. Nozzles direct forceful streams of water against a series of spoon-shaped buckets

mounted around the edge of a wheel. As water flows into the bucket, the direction of the water

velocity changes to follow the contour of the bucket. When the water-jet contacts the bucket, the

water exerts pressure on the bucket and the water is decelerated as it does a "u-turn" and flows

out the other side of the bucket at low velocity. In the process, the water's momentum is

transferred to the turbine. This "impulse" does work on the turbine. For maximum power and

efficiency, the turbine system is designed such that the water-jet velocity is twice the velocity of

the bucket. A very small percentage of the water's original kinetic energy will still remain in the

water; however, this allows the bucket to be emptied at the same rate it is filled, thus allowing

the water flow to continue uninterrupted.

From the results obtained, we can see how Pelton Wheel reacts to different kind of input.

Different flow rates give different value of work input. The slower the flow rates, the larger the

work being put into the wheel. The efficiency of the slower flow rates is also better than faster

one. The speed of the wheel also dropped when much weight being dropped until it stopped

suddenly when the weight is too much for it to go against.

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Conclusion

As a conclusion from the experiment that had been performed, we can conclude that different

range of flow rates and rotational speeds influences the performance of Pelton wheel turbine.

The combination of flow rate and jet velocity manipulates the power or work input. The bigger

the diameter nozzle the faster the flow rates but lower in velocity jet. Therefore we need the

perfect combination of both. In general, impulse turbine is high-head, low flow rate device. So

we can assume that our experiment is successful due to the result we obtained.

REFERENCE

Frank M. White. 2008. Fluid Mechanics. Sixth Edition. New York: Mc Graw Hill

International Edition. pp341-446.

Brady, James E. Engineering Thermodynamic. New York: John Wiley & Sons, 1997.

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