report 3a pelton turbine_hydraulic

8/11/2019 Report 3a Pelton Turbine_Hydraulic

1/13

PART A: PELTON TURBINE

1.0 INTRODUCTION

The Pelton turbine is an impulse turbine consisting of three basic components: a stationary

inlet nozzle, a runner which consists of multiple buckets on a rotating wheel and a casing.

Functioning of Pelton turbine is simple. When high speed water jet injected through the

nozzle hits buckets of Pelton wheel; it induces an impulsive force. This force rotates the

turbine and the rotating shaft in turn runs a generator and produces electricity . 1

Pelton turbine design is always aimed at extracting maximum power from water jet, or

maximizing efficiency. Power extracted by the bucket, P is the product of jet impulse force

and bucket velocity.

2.0 OBJECTIVE

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

3.0 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, P m

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

Radius( m)( Nm) andminsec/60

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

Meanwhile, Water Power, gHQ P w where, is water density (100 kg/m3), g is

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

efficiency, 100% w

m

P P

. To convert the unit of rpm to radians per minute is given by,

x rpm = ( x revolution/min) = ( x x2 radian)/min.

1

Retrieved from http://www.learnengineering.org/2013/08/pelton-turbine-wheel-hydraulic-turbine.html(Achieved by 19 March 2014)
http://www.learnengineering.org/2013/08/pelton-turbine-wheel-hydraulic-turbine.htmlhttp://www.learnengineering.org/2013/08/pelton-turbine-wheel-hydraulic-turbine.htmlhttp://www.learnengineering.org/2013/08/pelton-turbine-wheel-hydraulic-turbine.html


2/13

4.0 EQUIPMENTS

i. Pelton Turbine

ii. Tachometer

iii. Stopwatch

5.0 PROCEDURES

1. The Pelton turbine equipment was put on the hydraulic bench and connected to the water

supply using the provided connector.

2. The optic tachometer was tighten by using clips.

3. The turbine drum is free from any load (0.0N).

4. The valve controller was fully open. Then, the tachometer was levelled until the rotation

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

5. The reading of tachometer, flow rate, pressure at inlet point ( H ) and load, W 2 ( N ) were

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

on the right spring at W 1. Start with the W 1 = 1.0 N .

6. After the reading of tachometer is stabilised, the bottom of the hydraulic bench was

closed to collect water of desired volume (5 litre) and the time needed for the collection

was also recorded.

7. All the readings were recorded in the Table 6.1.

8. Step 3-7 were repeated with W varies in the range of 1.5 N to 6.0 N .


3/13

6.0 RESULT

Table 1: Recorded Data for Pelton Turbine

RPM 10231 9355.2 9148.4 8888.7 10472 7127.1 5300.6 6322.8 1488.8 482

(rad/s)

1071.388

979.674

958.018

930.822

1096.625

746.348

555.078

662.122

155.907 50.475

W 1 (N)0 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

W 2 (N) 0 1.7 2.2 2.7 3.7 4.6 5.3 6.2 7.1 7.7

W 2 W 1(N)

0 0.7 0.7 0.7 1.2 1.6 1.8 2.2 2.6 2.7

DrumRadiusx10 -3m

30 30 30 30 30 30 30 30 30 30

P m(W) 0 20.573 20.118 19.547 39.479 35.825 29.974 43.700 12.161 4.088

Rotation (Nm)

0 0.021 0.021 0.021 0.036 0.048 0.054 0.066 0.078 0.081

Volume(l)

5 5 5 5 5 5 5 5 5 5

Volume(m)

0.005 0.005 0.005 0.005 0.005 0.005 0.005 0.005 0.005 0.005

Time(s) 29.72 29.32 28.93 28.44 28.12 27.90 27.69 27.24 26.87 26.54

Flowrate(m 3/s)X10 -4

1.682 1.705 1.728 1.758 1.592 1.634 1.521 1.739 1.411 1.727

Pressure(mH 2O) 24.0 24.0 24.0 24.0 24.0 24.0 24.0 24.0 24.0 24.0

Pw (W) 39.601 40.143 40.684 41.390 37.482 38.471 35.810 40.943 33.221 40.660

Efficiency (%)

0 51.25 49.45 47.23 105.33 93.12 83.70 106.73 36.61 10.05


4/13

7.0 DATA ANALYSIS

Calculated Angul ar Velocity (w)

Due to the recorded angular velocitys 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 = (2 / 60) x RPM (1 revolution = 2 radius)

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

W 1 = RPM 1 x 2 /60 = (10231 x 2)/60 = 1071.388 rad/s

W 2 = RPM 2 x 2 /6 0 = (9355.2 x 2)/60 = 979.674 rad/s

W 3 = RPM 3 x 2 /60 = (9148.4 x 2)/60 = 958.018 rad/s

W 4 = RPM 4 x 2 /60 = (8888.7 x 2)/60 = 930.822 rad/s

W 5 = RPM 5 x 2 /60 = (10472 x 2)/60 = 1096.652 rad/s

W 6 = RPM 6 x 2 /60 = (7127.1 x 2)/60 = 746.348 rad/s

W 7 = RPM 7 x 2 /60 = (5300.6 x 2)/60 = 555.078 rad/s

W 8 = RPM 8 x 2 /60 = (6322.8 x 2)/60 = 662.122 rad/s

W 9 = RPM 9 x 2 /60 = (1488.8 x 2)/60 = 155.907 rad/s

W 10 = RPM 10 x 2 /60 = (482.0 x 2)/60 = 50.475 rad/s

Calculated Rotation (Nm)

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

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

The force for rotation is the difference of w 2 w1which tabulated on Table 1. 1 = (w2 w1) x r = 0 x (30 x 10 -3) = 0.000Nm

2 = (w2 w1) x r = 0.70 x (30 x 10-3) = 0.021Nm

3 = (w2 w1) x r = 0.70 x (30 x 10-3) = 0.021Nm

4 = (w2 w1) x r = 0.70x (30 x 10-3) = 0.021Nm

5 = (w2 w1) x r = 1.20 x (30 x 10

-3

) = 0.036Nm


5/13

6 = (w2 w1) x r = 1.60 x (30 x 10 -3) = 0.048Nm

7 = (w2 w1) x r = 1.80 x (30 x 10-3) = 0.054Nm

8 = (w2 w1) x r = 2.20 x (30 x 10-3) = 0.066Nm

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

10 = (w2 w1) x r = 2.70 x (30 x 10 -3) = 0.081Nm

Calcul ated Mechani cal Power (Pm)

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

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

Pm 1 = 1 x 1 = 0 x 1071.388 = 0.000 w

Pm 2 = 2 x 2 = 0.021 x 979.674 = 20.573w

Pm 3 = 3 x 3 = 0.021 x 958.018 = 20.118w

Pm 4 = 4 x 4 = 0.021 x 930.822 = 19.574w

Pm 5 = 5 x 5 = 0.036 x 1096.625 = 39.479w

Pm 6 = 6 x 6 = 0.048 x 746.348 = 35.825w

Pm 7 = 7 x 7 = 0.054 x 555.078 = 29.974w

Pm 8 = 8 x 8 = 0.066 x 662.122 = 43.700w

Pm 9 = 9 x 9 = 0.078 x 155.907 = 12.161w

Pm 10 = 10 x 10 = 0.081 x 50.475 = 4.088w

Calcul ated Flow Rate (Q)

Given formula Q = volume/time =m 3/s

The volume are initially fixed as 5 little = 0.005m 3

Q 1 = 0.005/29.72 = 1.682 x 10-4 m 3s

Q 2 = 0.005/29.32 = 1.705 x 10 -4 m 3s


6/13

Q 3 = 0.005/28.93 = 1.728 x 10 -4 m 3s

Q 4 = 0.005/28.44 = 1.758 x 10 -4 m 3s

Q 5 = 0.005/31.40 = 1.592 x 10 -4 m 3s

Q 6 = 0.005/30.60 = 1.634 x 10 -4 m 3s

Q 7 = 0.005/32.88 = 1.521 x 10 -4 m 3s

Q 8 = 0.005/28.75 = 1.739 x 10 -4 m 3s

Q 9 = 0.005/35.44 = 1.411 x 10 -4 m 3s

Q 10 = 0.005/28.96 = 1.727 x 10-4

m3

s

Calcul ated Water Power (Pw)

Given formula: P w = gHQ

P w1 = 1000 x 9.81 x 24.0 x 1.682(10 -4) = 39.601 w

P w2 = 1000 x 9.81 x 24.0 x 1.705(10 -4) = 40.143w

P w3 = 1000 x 9.81 x 24.0 x 1.728(10 -4) = 40.684 w

P w4 = 1000 x 9.81 x 24.0 x 1.758(10 -4) = 41.390 w

P w5 = 1000 x 9.81 x 24.0 x 1.592(10 -4) = 37.482 w

P w6 = 1000 x 9.81 x 24.0 x 1.634(10 -4) = 38.471 w

P w7 = 1000 x 9.81 x 24.0 x 1.521(10 -4) = 35.810 w

P w8 = 1000 x 9.81 x 24.0 x 1.739(10 -4) = 40.943 w

P w9 = 1000 x 9.81 x 24.0 x 1.411(10 -4) = 33.221 w

P w10 = 1000 x 9.81 x 24.0 x 1.727(10 -4) = 40.660 w


7/13

Calculated Efficiency (%)

Given formula: Wheel efficiency % = P m/PW X100

1 = 0/ 39.601 x100 = 0%

2 = 20.573/ 40.143 x100 = 51.25%

3 = 20.118/ 19.547 x100 = 49.45%

4 = 19.547/ 39.479 x100 = 47.23%

5 = 39.479/ 37.482 x100 = 105.33%

6 = 35.825/ 38.471 x100 = 93.12%

7 = 29.974/ 35.810 x100 = 83.70%

8 = 43.700/ 40.943 x100 = 106.73%

9 = 12.161/ 33.221 x100 = 36.61%

10 = 4.088/ 40.660 x100 = 10.05%


8/13

8.0 DISCUSSION AND SUGGESTION

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

efficiency curve and discharge versus motor speed.

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

torque produced by Pelton turbine.

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

Pelton Turbine.

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0 200 400 600 800 1000 1200

R o t a t i o n a

l T o r q u e

, (

N m

)

Angular Velocity, (rad/s)

ROTATIONAL POWER CURVE

0

20

40

60

80

100

120

0 200 400 600 800 1000 1200

E f f i c i e n c y

,

( % )


EFFICIENCY CURVE


9/13

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

and flow rate produced by Pelton Turbine.

1. For the graph of rotation power curve , we can see when angular velocity increases,

eventually the rotational torque has a decreases. Although irregularity makes it

seemed not exactly inversely proportional, but theoretically it should show an

inversely proportional relationship between angular velocity and rotational torque.

This also can be proven by the equation of mechanical power which is P m = x .

For the graph of eff iciency curve , we noticed that the shape of the graph would be

similar to that of a bell if the two points at the sharp heads were omitted.

This means the efficiency of Pelton turbine is low when the angular velocity is closeto zero as depicted at start. The efficiency increases with the angular velocity until

reaching the angular velocity of about 662.122rad/s with the maximum percentage of

efficiency that accounts 106.73%. After that, the efficiency of Pelton turbine

decreases as the angular velocity rises.

For the graph of discharge versus motor speed , we observed an irregular trend

between the angular velocity and flow rate. Thus, it perhaps depicts inaccurate result

0

0.2

0.4

0.6

0.8

1

1.2

1.41.6

1.8

2

0 200 400 600 800 1000 1200

F l o w R a t e

, Q

( m 3

/ s ) ( x 1 0 - 4

)


DISCHARGE VERSUS MOTOR SPEED


10/13

was obtained since discharge depicted by flow rate versus motor speed depicted by

angular velocity should portray regular trend, decreasing.

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

the standard Pelton wheel turbine performance on power and efficiency with slight

errors. Motor speed on the other hand differs too much as depicted by the fluctuations.

2. 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,

Maximum output power for experimental result of Pelton Turbine, P max = 43.700 W

Angular velocity of the total experiment, = 50.475 rad/s

v = r

= 30 x 10 -3 x 50.475

= 1.514 m/s

9.0 PRECAUTION

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

reads 25mH 2O. 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 reading stables.

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

stainless steel drum brake should always locate on the same marking tapewhich means it must be static all the time until getting the reading.

d) 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.


11/13

10.0 CONCLUSION

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

characteristic of Pelton turbine operation by using different loading to alter the result.

Noticeable changes occurred in the rotational torque, flow rate of the water as well as the

efficiency of the turbine.

However, the graphs obtained signify errors in the experiment. Supposedly, the

readings should show a steady trend in whichever graph. However, graphs obtained fluctuate

hence denote irregularity. It is clearly shown in the graphs that the increase of load on Pelton

wheel turbine will influence the rotation of wheel by function of dynamometer which

absorbed its mechanical power. Factors that may have contributed to the irregularity include

man error such as parallax error and exchanging of observer as well as timeworn equipment.

Furthermore, the efficiency of the turbine can be affected by hydraulic losses or power

losses that occurred due to flow irregularity within the bucket. Hence worn equipment may

have caused irregular water jet and flow within the bucket causing fluctuating data. 2 It can be

shown using Euler's turbo machinery equation that maximum power extraction happens when

bucket speed is half the jet velocity. So it is always desirable to operate Pelton wheel at this

condition. Well managed Pelton turbines can give efficiency as high as 90 %, at optimumworking conditions.

From the experiment, we can conclude that higher velocity leads to lower torque and

hydraulic power but it also causes improvement in efficiency of turbine and mechanical

power.

2

Retrieved from http://www.slideshare.net/iaeme/design-of-high-efficiency-pelton-turbine-for-micro-hydropower (Achieved by 19 March 2014)
http://www.slideshare.net/iaeme/design-of-high-efficiency-pelton-turbine-for-micro-hydropowerhttp://www.slideshare.net/iaeme/design-of-high-efficiency-pelton-turbine-for-micro-hydropowerhttp://www.slideshare.net/iaeme/design-of-high-efficiency-pelton-turbine-for-micro-hydropowerhttp://www.slideshare.net/iaeme/design-of-high-efficiency-pelton-turbine-for-micro-hydropowerhttp://www.slideshare.net/iaeme/design-of-high-efficiency-pelton-turbine-for-micro-hydropowerhttp://www.slideshare.net/iaeme/design-of-high-efficiency-pelton-turbine-for-micro-hydropower


12/13

11.0 ATTACHMENT


13/13

report 3a pelton turbine_hydraulic

Documents