PELTON TURBINE

1.0 INTRODUCTION

There are two types of turbines, reaction and the impulse, the difference being the manner of

head conversion. In the reaction turbine, the fluid fills the blade passages, and the head

change or pressure drop occurs within the runner. An impulse turbine first converts the water

head through a nozzle into a high-velocity jet, which then strikes the buckets at one position

as they pass by. The runner passages are not fully filled, and the jet flow past the buckets is

essentially at constant pressure. Impulse turbines are ideally suited for high head and

relatively low power. The Pelton turbine used in this experiment is an impulse turbine.

The Pelton turbine consists of three basic components: a stationary inlet nozzle, a runner and

a casing. The runner consists of multiple buckets mounted on a rotating wheel. The jet

strikes the buckets and imparts momentum. The buckets are shaped in a manner to divide the

flow in half and turn its relative velocity vector nearly 180°.

2.0 OBJECTIVE

To determine the characteristics of Pelton Turbine operation by using several speed.

3.0 LEARNING OUTCOMES

At the end of the course, students should be able to apply the knowledge and skills they have

learned to:

a. Understand the basic operating system of the Pelton turbine.

b. Understand the factors which influence the efficiency of turbine.

4.0 THEORY

A Pelton Turbine characteristic operation curve can be derived by using the same method as a

pump. It is because the velocity is usually assumed as an independent parameter when the

plotting of power, efficiency, torque and discharge are carried out. Mechanical Power, Pm

(watt) = Rotation (τ , Nm) ¿ Circular velocity (ω , rad/sec). Where, T=Force(N) ¿

Radius(m)(Nm) and ω=2 π radius /min

60 sec/min (rad/s) where, 1 revolution is equal to 2π radius.

Meanwhile, Water Power, Pw= ρ gHQ where,ρ is water density (100kg/m3), g is gravity

constant (9.81m/s2), H is head at inlet point (m) and Q is flowrate (m3/s). Wheel efficiency,

η %=Pm

Pw

×100. To convert the unit of ‘rpm’ to ‘radians per minute’ is given by, x rpm=(x

revolution/min)=(x x2π radian)/min.

5.0 EQUIPMENTS

i. Pelton Turbine

ii. Tachometer

iii. Stopwatch

6.0 PROCEDURES

a) The Pelton turbine equipment was put on the hydraulic bench and by using the provided

connecter, connected it to the water supply.

b) By using clip, the optic tachometer was tighten.

c) Then, make sure that the turbine drum is free from any load (0.0N).

d) The valve controller was fully open. Then, the tachometer was level until the rotation

reaches the maximum value of 2000 rotation/minute or rpm.

e) The reading of tachometer, flow rate, pressure at inlet point (H) and load, W2 (N) were

recorded. The brake equipment were put on the turbine drum. Then, the brake was level

on the right spring at W1. Start with the W1 = 1.0N.

f) All the readings were recorded in the Table 6.1.

g) Step 3-6 were repeated with W varies in the range of 1.5N to 6.0N.

7.0 RESULT

RPM 10229 9863.5 8417.8 7613.7 6750.0 5653.1 4640.6 3189.8 2418.2 1053.5

ω

(rad/s)1071.1

781032.9

0881.51 797.30 706.85

8591.99 485.96

2334.03

5253.23 110.32

W1 (N)0 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

W2 (N) 0 1.60 2.60 3.50 4.40 5.60 6.80 7.40 8.60 9.30

W2 –W1

(N)0 0.60 1.10 1.50 1.90 2.60 3.30 3.40 4.10 4.30

Drum Radiusx10-3m

30 30 30 30 30 30 30 30 30 30

Rotationτ (Nm)

0 0.018 0.033 0.045 0.072 0.078 0.099 0.114 0.138 0.150

Pm(W) 0 18.59 29.09 35.88 50.89 46.18 48.11 38.08 34.90 16.55

Volume (l)

5 5 5 5 5 5 5 5 5 5

Volume (m³)

0.005 0.005 0.005 0.005 0.005 0.005 0.005 0.005 0.005 0.005

Time(s) 29.13 25.30 23.31 21.10 19.22 18.40 16.97 16.22 16.00 15.85

Flowrate (m3/s) X10-4

1.716 1.976 2.145 2.370 2.601 2.717 2.946 3.083 3.125 3.155

Pressure (mH2O)

24.0 24.0 24.0 24.0 24.0 24.0 24.0 24.0 24.0 24.0

Pw (W) 40.360 46.476 50.450 55.742 61.176 63.904 69.291 72.512 73.500 74.206

Efficienc

y η (%)

0 39.999 57.661 64.368 83.186 72.265 62.432 50.429 47.483 22.303

Table 1: Recorded Data for Pelton Turbine

8.0 DATA ANALYSIS

Calculated Angular Velocity (w)

Due to the recorded angular velocity’s unit is based on RPM (Revolution per minutes), hence

all the recorded value must be converted to SI unit, rad/s.

Given formula: rad/s = (2Π / 60) x RPM (1 revolution = 2Π radius)

By converting this, the unit of w (Angular velocity) will be corrected to rad/s.

W1 = RPM1 X 2Π /60 = (10229 X 2Π)/60 = 10 71 . 1 78 rad/s

W2 = RPM2 X 2Π /60 = (9863.5X 2Π)/60 = 1032.90 rad/s

W3 = RPM3 X 2Π /60 = (8417.8 X 2Π)/60 = 881.51 rad/s

W4 = RPM4 X 2Π /60 = (7613.7 X 2Π)/60 = 797.30 rad/s

W5 = RPM5 X 2Π /60 = (6750.0 X 2Π)/60 = 706.858rad/s

W6 = RPM6 X 2Π /60 = (5653.1X 2Π)/60 = 591.99rad/s

W7 = RPM7 X 2Π /60 = (4640.6X 2Π)/60 = 485.962 rad/s

W8 = RPM8 X 2Π /60 = (3189.8X 2Π)/60 = 334.035 rad/s

W9 = RPM9 X 2Π /60 = (2418.2X 2Π)/60 = 253.23 rad/s

Calculated Rotation τ (Nm)

Given formula τ (Nm) = Force (N) x Radius (m)

Given drum Radius, r = 30 x 10-3m

The force for rotation is the difference of w2 – w1which tabulated on Table 1.

Τ1 = (w2 – w1) x r = 0 x (30x 10-3) = 0 .000Nm

Τ2 = (w2 – w1) x r = 0.60 x (30x 10-3) = 0.018Nm

Τ3 = (w2 – w1) x r = 1.10 x (30x 10-3) = 0.0 33 Nm

Τ4 = (w2 – w1) x r = 1.50 x (30x 10-3) = 0.0 45 Nm

Τ5 = (w2 – w1) x r = 1.90 x (30x 10-3) = 0.0 72 Nm

Τ6 = (w2 – w1) x r = 2.60 x (30x 10-3) = 0.078Nm

Τ7 = (w2 – w1) x r = 3.30 x (30x 10-3) = 0. 099 Nm

Τ8 = (w2 – w1) x r = 3.40 x (30x 10-3) = 0.1 14 Nm

Τ9 = (w2 – w1) x r = 4.10 x (30x 10-3) = 0.138Nm

Calculated Mechanical Power (Pm)

Given formula: Pm = Torque (τ) x Circular Velocity (w)

The values of τ and w had been calculated and tabulated on the calculation (1) and (2)

and Table 1 respectively.

Pm1 = τ1 x w1 = 0 x 1071.178 = 0 .000 w

Pm2 = τ2 x w2 = 0.018 x 892.757 = 18.59 w

Pm3 = τ3 x w3 = 0.033 x 819.432 = 29.09 w

Pm4 = τ4 x w4 = 0.045 x 706.858 = 35.88 w

Pm5 = τ5 x w5 = 0.072 x 608.537 = 50.89 w

Pm6 = τ6 x w6 = 0.078 x 526.500 = 46.18 w

Pm7 = τ7 x w7 = 0.099 x 485.462 = 48 . 11 w

Pm8 = τ8 x w8 = 0.114x 334.035 = 38 . 08 w

Pm9 = τ9 x w9 = 0.138 x 214.403 = 34.90 w

Calculated Flow Rate (Q)

Given formula Q = volume/time =m3/s

The volume are initially fixed as 5 little = 0.005m3

Q1 = 0.005/ 29.13 = 1.716 x10 -4 m 3 s

Q2 = 0.005/ 25.30 = 1.976 x10 -4 m 3 s

Q3 = 0.005/ 23.31 = 2.145 x10 -4 m 3 s

Q4 = 0.005/ 21.10 = 2.370 x10 -4 m 3 s

Q5 = 0.005/19.22 = 2.601 x10 -4 m 3 s

Q6 = 0.005/ 18.40 = 2.717 x10 -4 m 3 s

Q7 = 0.005/16.97 = 2.946 x10 -4 m 3 s

Q8 = 0.005/16.22 = 3.083 x10 -4 m 3 s

Q9 = 0.005/16.00 = 3.125 x10 -4 m 3 s

Calculated Water Power (Pw)

Given formula: Pw = flow rate x pressure

Pw1 = 1.716 x24.0 = 40.360w

Pw2 = 1.976 x 24 = 46.476w

Pw3 = 2.145 x 24 = 50.450 w

Pw4 = 2.370 x 24 = 55.742w

Pw5 = 2.601x 24 = 61.176 w

Pw6 = 2.717 x 24 = 63.904 w

Pw7 = 2.946 x 24 = 69.291 w

Pw8 = 3.083 x 24 = 72.512 w

Pw9 = 3.125 x 24 = 73.500w

Calculated Efficiency Ƞ (%)

Given formula: Wheel efficiency Ƞ% = Pm/PW X100

Ƞ1 = 0/ 40.360 x100 = 0%

Ƞ2 = 18.59/46.476 x100 = 39.999%

Ƞ3 = 29.09/50.450 x100 = 57.661%

Ƞ4 = 35.88/55.742 x100 = 64.368%

Ƞ5 = 50.89/ 61.176 x100 = 83.186%

Ƞ6 = 46.18/ 63.904 x100 = 72.265%

Ƞ7 = 48.11/ 69.291x100 = 69.432%

Ƞ8 = 38.08/ 72.512x100 = 50.429%

Ƞ9 = 34.90/ 73.500x100 = 47.483%

9.0 DISCUSSION AND SUGGESTION

1. Graphs are plotted as below to show the relationship between rotational power curve,

efficiency curve and discharge versus motor speed.

0 200 400 600 800 1000 12000

0.02

0.04

0.06

0.08

0.1

0.12

0.14

ROTATIONAL POWER CURVE

Angular Velocity, ω (rad/s)

Rota

tiona

l Tor

que,

τ (N

m)

Graph A1: Rotational power curve shows the relationship of angular velocity and rotational

torque produced by Pelton turbine.

0 200 400 600 800 1000 12000

20

40

60

80

100

120

140

EFFICIENCY CURVE

Angular Velocity, ω (rad/s)

Efficie

ncy,

η (%

)

Graph A2: Efficiency curve shows the relationship of angular velocity and efficiency of

Pelton Turbine.

0 200 400 600 800 1000 12000

0.5

1

1.5

2

2.5

3

3.5

DISCHARGE VERSUS MOTOR SPEED

Angular Velocity, ω (rad/s)

Flow

Rat

e, Q

(m3/

s)

Graph A3: Discharge versus motor speed graph shows the relationship of angular velocity

and flow rate produced by Pelton Turbine.

2. For the graph of rotation power curve, we can see the rotational torque has an

inversely proportional relationship to angular velocity. This also can be proven by the

equation of mechanical power which is Pm = τ x ω. When angular velocity increases,

eventually the rotational torque is decreasing.

For the graph of efficiency curve, we noticed that the shape of the graph is similar as

a bell-shaped which means the efficiency of Pelton turbine is low when the angular

velocity is small at the beginning. The efficiency increases as the angular velocity

increase until reaching the angular velocity of 700.962rad/s with the maximum

percentage of efficiency that accounts 85.62%. After all, the efficiency of Pelton

turbine starts to drop as the angular velocity still keep on arising.

For the graph of discharge versus motor speed, we observed that the motor speed

increased which indicated by angular velocity, the discharge of this Pelton turbine

decreased. Thus, it is obvious to prove that the performance of rotation of wheel are

not necessary increased the water flow rate.

From all the graphs regarding to the Pelton turbine that has been plotted, it has obeyed

the standard Pelton wheel turbine performance on motor speed, power and efficiency.

3. To calculate the velocity where the maximum power is reached, we applied the

conversion formulae of velocity and angular velocity.

Velocity, v = Radius, r X Angular velocity, ω

Maximum output power for experimental result of Pelton Turbine, Pmax = 50.89 W

Angular velocity of the total experiment, ω = 110.32 rad/s

v = r ω

= 30 x 10-3 x 110.32

= 3.310 m/s

As we keep the water power constant throughout the experiment without any losses of

it when the turbine is turned on, we will obtained high efficiency if the mechanical

power of the turbine is high. The varies of the water power as well as the mechanical

power of this Pelton Turbine happened due to the rotational velocity of the turbines

has kept multiplied in order to get a set of readings.

4. In order to achieve the most accuracy sets of readings for Pelton turbine experiment,

some safety factors should be taken throughout the processes.

a) Make sure the valve controller does not exceed the maximum pressure which

reads 25mH2O. It may causes failure of water nozzle and Pelton wheel cups.

b) For every various load testing, reading for rotation speed of drum brake should

been taken only after the highest speed obtained. It can be determined by the

highest frequency sound arise or keep checking the tachometer reading while

adjusting the valve controller.

c) Tachometer which used to record the rotation or revolution per minutes of

stainless steel drum brake should always locate on the same marking tape

which means it must be static all the time until getting the reading.

d) After a set of testing is done, the tension of load belt should be released and

run the Pelton turbine at zero load condition for at least five minutes to ensure

that the turbine are free from loading.

e) The hydraulic bench valve which connected to water supply should always

open or unchanged so that the initial water flow or flow rate can be constant

from time to time before affected by Pelton wheel.

5. Furthermore, an experiment normally exist a minor error which is done by human or

accidentally mistakes. All the gross and accidental error should be taken in

consideration to ensure that the result obtained has reaches an acceptable accuracy.

We are advised to do the experiment indoor so that there is no disturbance from air

movement such as wind to the load belt while the experiment is carry on.

10.0 CONCLUSION

From this Pelton wheel turbine experiment, we understand the affect on the

characteristic of Pelton turbine operation by using several speeds such as the changes

in rotational torque, flow rate of the water as well as the efficiency of the turbine. It is

clearly shown in the graphs that the increases of load on Pelton wheel turbine will

influence the rotation of wheel by function of dynamometer which absorbed it

mechanical power. Hence, the relationship between turbine rotational speed (RPM)

with power output (W), wheel torque (τ) and turbine efficiency as plotted on graph

A1, A2 and A3. Based on the result, the power output and turbine efficiency obtained

are not directly proportional to the turbine rotational speed, the highest power output

only happened on specific rotational speed. Hence to ensure the highest efficiency on

power generation, control the rotational speeds are significantly.