Report 3a Pelton Turbine_Hydraulic

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, Pm (watt) = Rotation (  , Nm)  Circular velocity (  , rad/sec). Where, T = Force(N)  Radius(m)(Nm) and  

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

Meanwhile, Water Power, Pw  gHQ where,  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,  % 

Pm  100 . To convert the unit of ‘rpm’ to ‘radians per minute’ is given by, Pw

x rpm = ( x revolution/min) = ( x x2  radian)/min. 1

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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, 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. 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 1.5N to 6.0N.

6.0 RESULT

RPM

10231

9355.2

9148.4

8888.7

10472

7127.1

5300.6

6322.8

1488.8

482



1071.38 8

979.67 4

958.01 8

930.82 2

1096.6 25

746.34 8

555.07 8

662.12 2

155.907

50.475

0 0

1.0 1.7

1.5 2.2

2.0 2.7

2.5 3.7

3.0 4.6

3.5 5.3

4.0 6.2

4.5 7.1

5.0 7.7

0

0.7

0.7

0.7

1.2

1.6

1.8

2.2

2.6

2.7

30

30

30

30

30

30

30

30

30

30

0

20.573

20.118

19.547

39.479

35.825

29.974

43.700

12.161

4.088

Rotation  (Nm) Volume (l)

0

0.021

0.021

0.021

0.036

0.048

0.054

0.066

0.078

0.081

5

5

5

5

5

5

5

5

5

5

Volume (m³) Time(s)

0.005

0.005

0.005

0.005

0.005

0.005

0.005

0.005

0.005

0.005

29.72

29.32

28.93

28.44

28.12

27.90

27.69

27.24

26.87

26.54

Flowrate (m3/s) X10-4

1.682

1.705

1.728

1.758

1.592

1.634

1.521

1.739

1.411

1.727

Pressure (mH2O) Pw (W)

24.0

24.0

24.0

24.0

24.0

24.0

24.0

24.0

24.0

24.0

39.601

40.143

40.684

41.390

37.482

38.471

35.810

40.943

33.221

40.660

51.25

49.45

47.23

105.33

93.12

83.70

106.73

36.61

10.05

(rad/s) W1 (N) W2 (N) W2 –W1 (N) Drum Radius x10-3m Pm(W)

Efficiency 0  (%)

Table 1: Recorded Data for Pelton Turbine

7.0 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 =

(10231 x 2Π)/60

=

1071.388 rad/s

W2

= RPM2 x 2Π /60 =

(9355.2 x 2Π)/60

=

979.674 rad/s

W3

= RPM3 x 2Π /60 =

(9148.4 x 2Π)/60

=

958.018 rad/s

W4

= RPM4 x 2Π /60 =

(8888.7 x 2Π)/60

=

930.822 rad/s

W5

= RPM5 x 2Π /60 =

(10472 x 2Π)/60

=

1096.652 rad/s

W6

= RPM6 x 2Π /60 =

(7127.1 x 2Π)/60

=

746.348 rad/s

W7

= RPM7 x 2Π /60 =

(5300.6 x 2Π)/60

=

555.078 rad/s

W8

= RPM8 x 2Π /60 =

(6322.8 x 2Π)/60

=

662.122 rad/s

W9

= RPM9 x 2Π /60 =

(1488.8 x 2Π)/60

=

155.907 rad/s

W10

= RPM10 x 2Π /60 = (482.0 x 2Π)/60

=

50.475 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 (30 x 10-3)

= 0.000Nm

Τ2

= (w2

– w1) x r

= 0.70 x (30 x 10-3)

= 0.021Nm

Τ3

= (w2

– w1) x r

= 0.70 x (30 x 10-3)

= 0.021Nm

Τ4

= (w2

– w1) x r

= 0.70x (30 x 10-3)

= 0.021Nm

Τ5

= (w2

– w1) x r

= 1.20 x (30 x 10-3)

= 0.036Nm

Τ6

= (w2

– w1) x r

= 1.60 x (30 x 10-3)

= 0.048Nm

Τ7

= (w2

– w1) x r

= 1.80 x (30 x 10-3)

= 0.054Nm

Τ8

= (w2

– w1) x r

= 2.20 x (30 x 10-3)

= 0.066Nm

Τ9

= (w2

– w1) x r

= 2.60 x (30 x 10-3)

= 0.078Nm

Τ10

= (w2

– w1) x r

= 2.70 x (30 x 10-3)

= 0.081Nm

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 ω1

=

0 x 1071.388

=

0.000 w

Pm2

=

τ2 x ω 2

=

0.021 x 979.674

=

20.573w

Pm3

=

τ3 x ω 3

=

0.021 x 958.018

=

20.118w

Pm4

=

τ4 x ω 4

=

0.021 x 930.822

=

19.574w

Pm5

=

τ5 x ω 5

=

0.036 x 1096.625

=

39.479w

Pm6

=

τ6 x ω 6

=

0.048 x 746.348

=

35.825w

Pm7

=

τ7 x ω 7

=

0.054 x 555.078

=

29.974w

Pm8

=

τ8 x ω 8

=

0.066 x 662.122

=

43.700w

Pm9

=

τ9 x ω 9

=

0.078 x 155.907

=

12.161w

Pm10

=

τ10 x ω 10

=

0.081 x 50.475

=

4.088w

Calculated Flow Rate (Q) Given formula Q = volume/time =m3/s 

The volume are initially fixed as 5 little = 0.005m3

Q1

=

0.005/29.72

=

1.682 x 10-4 m3s

Q2

=

0.005/29.32

=

1.705 x 10-4 m3s

Q3

=

0.005/28.93

=

1.728 x 10-4 m3s

Q4

=

0.005/28.44

=

1.758 x 10-4 m3s

Q5

=

0.005/31.40

=

1.592 x 10-4 m3s

Q6

=

0.005/30.60

=

1.634 x 10-4 m3s

Q7

=

0.005/32.88

=

1.521 x 10-4 m3s

Q8

=

0.005/28.75

=

1.739 x 10-4 m3s

Q9

=

0.005/35.44

=

1.411 x 10-4 m3s

Q10

=

0.005/28.96

=

1.727 x 10-4 m3s

Calculated Water Power (Pw) Given formula: Pw = ρgHQ Pw1

=

1000 x 9.81 x 24.0 x 1.682(10-4)

=

39.601 w

Pw2

=

1000 x 9.81 x 24.0 x 1.705(10-4)

=

40.143w

Pw3

=

1000 x 9.81 x 24.0 x 1.728(10-4)

=

40.684 w

Pw4

=

1000 x 9.81 x 24.0 x 1.758(10-4)

=

41.390 w

Pw5

=

1000 x 9.81 x 24.0 x 1.592(10-4)

=

37.482 w

Pw6

=

1000 x 9.81 x 24.0 x 1.634(10-4)

=

38.471 w

Pw7

=

1000 x 9.81 x 24.0 x 1.521(10-4)

=

35.810 w

Pw8

=

1000 x 9.81 x 24.0 x 1.739(10-4)

=

40.943 w

Pw9

=

1000 x 9.81 x 24.0 x 1.411(10-4)

=

33.221 w

Pw10

=

1000 x 9.81 x 24.0 x 1.727(10-4)

=

40.660 w

Calculated Efficiency Ƞ (%) Given formula: Wheel efficiency Ƞ% = Pm/PW X100 Ƞ1

=

0/ 39.601

x100

=

0%

Ƞ2

=

20.573/ 40.143 x100

=

51.25%

Ƞ3

=

20.118/ 19.547 x100

=

49.45%

Ƞ4

=

19.547/ 39.479 x100

=

47.23%

Ƞ5

=

39.479/ 37.482 x100

=

105.33%

Ƞ6

=

35.825/ 38.471 x100

=

93.12%

Ƞ7

=

29.974/ 35.810 x100

=

83.70%

Ƞ8

=

43.700/ 40.943 x100

=

106.73%

Ƞ9

=

12.161/ 33.221 x100

=

36.61%

Ƞ10

=

4.088/ 40.660 x100

=

10.05%

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.

ROTATIONAL POWER CURVE Rotational Torque, τ (Nm)

0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 0

200

400

600

800

1000

1200

Angular Velocity, ω (rad/s)

Graph A1: Rotational power curve shows the relationship of angular velocity and rotational torque produced by Pelton turbine.

EFFICIENCY CURVE 120 Efficiency, η (%)

100 80 60 40 20 0 0

200

400

600

800

1000

1200

Angular Velocity, ω (rad/s)

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

DISCHARGE VERSUS MOTOR SPEED 2 Flow Rate, Q (m3/s) (x10-4)

1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

200

400

600

800

1000

1200

Angular Velocity, ω (rad/s)

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 Pm = τ x ω. For the graph of efficiency 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 close to 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

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, ω Maximum output power for experimental result of Pelton Turbine, Pmax = 43.700 W Angular velocity of the total experiment, ω = 50.475 rad/s v

=rω = 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 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 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 tape which 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.

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 optimum working 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

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11.0 ATTACHMENT