PELTON & FRANCIS TURBINES

FACULTY OF CIVIL AND ENVIRONMENTAL ENGINEERING DEPARTMENT OF WATER & ENVIROMENTAL ENGINEERING WATER ENGINEERING LABORATORY

LAB REPORT Subject Code Code & Experiment Title Course Code Date Section / Group Name Members of Group Lecturer/Instructor/Tutor Received Date

Comment by examiner

BFC 21201 MKA – 03 ; PELTON & FRANCIS TURBINES 2 BFF/1 21 NOVEMBER 2011 2/5 AFANDI BIN ABD WAHID 1.MUHAMMAD IKHWAN BIN ZAINUDDIN 2.MOHD HASIF BIN AZMAN 3.MUHAMMAD HUZAIR BIN ZULKIFLI CIK AMNANI BIN ABU BAKAR EN JAMILULLAIL BIN AHMAD TAIB 29 NOVEMBER 2011

Received

(DF100122) (DF100018) (DF100079) (DF100040)

STUDENTS’ ETHICAL CODE (SEC) DEPARTMENT OF WATER & ENVIRONMENTAL ENGINEERING FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITI TUN HUSSEIN ONN MALAYSIA BATU PAHAT, JOHOR

“I declare that I have prepared this report with my own efforts. I also declare not receive or give any assistance in preparing this report and make this affirmation in the belief that nothing is in, it is true”

………………………………………. (STUDENT SIGNATURE) NAME : AFANDI BIN ABD WAHID MATRIC NO. : DF100122 DATE : 29 NOVEMBER 2011

STUDENTS’ ETHICAL CODE (SEC) DEPARTMENT OF WATER & ENVIRONMENTAL ENGINEERING FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITI TUN HUSSEIN ONN MALAYSIA BATU PAHAT, JOHOR

“I declare that I have prepared this report with my own efforts. I also declare not receive or give any assistance in preparing this report and make this affirmation in the belief that nothing is in, it is true”

………………………………………. (STUDENT SIGNATURE) NAME : MUHAMMAD IKHWAN BIN ZAINUDDIN MATRIC NO. : DF100018 DATE : 29 NOVEMBER 2011

STUDENTS’ ETHICAL CODE (SEC) DEPARTMENT OF WATER & ENVIRONMENTAL ENGINEERING FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITI TUN HUSSEIN ONN MALAYSIA BATU PAHAT, JOHOR

“I declare that I have prepared this report with my own efforts. I also declare not receive or give any assistance in preparing this report and make this affirmation in the belief that nothing is in, it is true”

………………………………………. (STUDENT SIGNATURE) NAME : MOHD HASIF BIN AZMAN MATRIC NO. : DF100079 DATE : 29 NOVEMBER 2011

STUDENTS’ ETHICAL CODE (SEC) DEPARTMENT OF WATER & ENVIRONMENTAL ENGINEERING FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITI TUN HUSSEIN ONN MALAYSIA BATU PAHAT, JOHOR

“I declare that I have prepared this report with my own efforts. I also declare not receive or give any assistance in preparing this report and make this affirmation in the belief that nothing is in, it is true”

……………………………………. (STUDENT SIGNATURE) NAME : MUHAMMAD HUZAIR BIN ZULKIFLI MATRIC NO. : DF100040 DATE : 29 NOVEMBER 2011

PART A: PELTON TURBINE 1.0

INTRODUCTION A turbine converts energy in the form of falling water into rotating shaft power. The selection of the best turbine for any particular hydro site depends on the site characteristics, the dominant ones being the head and flow available. Selection also depends on the desired running speed of the generator or other device loading the turbine. Other considerations such as whether the turbine is expected to produce power under partflow conditions also play an important role in the selection. All turbines have a powerspeed characteristic. They will tend to run most efficiently at a particular speed, head and flow combination. A turbine design speed is largely determined by the head under which it operates. Turbines can be classified as high head, medium head or low head machines. Turbines are also divided by their principle way of operating and can be either impulse or reaction turbines. HIGH HEAD

MEDIUM HEAD

IMPULSE

Pelton

Cross-flow multi-jet

TURBINES

Turgo

Pelton Turgo

REACTION

LOW HEAD Cross-flow

Francis

Propeller Kaplan

TURBINES Table 1 - Groups of impulse and reaction turbines

The rotating element (called `runner') of a reaction turbine is fully immersed in water and is enclosed in a pressure casing. The runner blades are profiled so that pressure differences across them impose lift forces, like those on aircraft wings, which cause the runner to rotate. In contrast, an impulse turbine runner operates in air, driven by a jet (or jets) of water. Here the water remains at atmospheric pressure before and after making contact with the runner blades. In this case a nozzle converts the pressurized low velocity water into a high speed jet. The runner blades deflect the jet so as to maximize the change of momentum of the water and thus maximizing the force on the blades.

Impulse turbines are usually cheaper than reaction turbines because there is no need for a specialist pressure casing, or for carefully engineered clearances. However, they are only suitable for relatively high heads. 2.0

OBJECTIVE 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 skill they have learned to:

4.0

i.

Understand the basic operating system of the Pelton turbine.

ii.

Understand the factors which influence the efficiency of turbine

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 60 sec/ min

is equal to 2  radius. 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 Pw

‘radians per minute’ is given by, x rpm=( x revolution/min)=( x x2  radian)/min.

5.0

EQUIPMENT

Figure 1: Stopwatch

Figure 2: Tachometer

Figure 3: Pelton Turbine 6.0

PROCEDURES 1. Put the Pelton turbine equipment on the hydraulic bench and connected it to the water supply by using the provided connector. 2. Tighten the optic tachometer by using clip. 3. Make sure that the turbine drum is free from any load. 4. Open the valve controller fully. Then level the tachometer until the rotation reaches the maximum valve of 2000 rotation/minute or rpm. 5. Record the reading of tachometer, flow rate, pressure at inlet point and load. Put the brake equipment on the turbine drum. Then level the brake on the right spring at W1. Start with W1= 1.0N. 6. Record all at table 6.1 7. Repeat 3-6 with W varies in the range of 1.5N to 6.0N.

7.0

RESULT RPM

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

104.37

9863.5

8417.8

7613.7

6870.6

6201.8

5653.1

4612.5

2418.2

1053.5

1092.9 6

1032.90

881.51

797.30

719.49

649.45

591.99

483.02

253.23

110.32

0

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

0

1.6

2.6

3.5

4.4

5.2

6.1

7.5

9.1

10.0

0

0.6

1.1

1.5

1.9

2.2

2.6

3.5

4.6

5.0

30

30

30

30

30

30

30

30

30

30

0

0.018

0.033

0.045

0.057

0.066

0.078

0.105

0.138

0.150

0

18.59

29.09

35.88

41.01

42.86

46.18

50.71

34.95

16.55

Volume (l)

6 liter 0.006m3

Volume (m3) Time(s) Flowrate (m3/s) ( x 10-4 ) Pressure (mH2O) Pw (W)

33.44

25.31

24.31

22.10

20.41

18.94

18.40

17.97

17.79

17.50

1.794

2.371

2.468

2.715

2.940

3.167

3.261

3.339

3.371

3.429

23.0

18.0

17.0

15.0

13.0

12.0

11.5

11.0

10.5

10.0

40.48

41.87

41.16

39.95

37.49

37.28

36.79

36.03

34.72

33.64

Efficiency  (%)

0

44.40

70.67

89.81

109.39

114.97

125.52

140.79

100.66

49.19

Table 2 – Result of Pelton Turbine experiment

8.0

DATA ANALYSIS

 (rad/s) RPM = 10437

RPM = 9863.5

RPM = 8417.8

=

=

=

RPM x 2π 60

RPM x 2π 60

RPM x 2π 60

= 10437 x 2π 60

= 9863.5x 2π 60

= 8417.8x 2π 60

= 1092.96 rad/s

= 1032.90 rad/s

= 881.51 rad/s

RPM = 7613.7

RPM = 6870.6

RPM = 6201.8

=

=

=

RPM x 2π 60

RPM x 2π 60

RPM x 2π 60

= 7613.7x 2π 60

= 6870.6x 2π 60

= 6201.8x 2π 60

= 797.30 rad/s

= 719.49 rad/s

= 649.45 rad/s

RPM = 5653.1

RPM = 4612.5

RPM = 2418.2

=

=

=

RPM x 2π 60

RPM x 2π 60

RPM x 2π 60

= 5653.1x 2π 60

= 4612.5x 2π 60

= 2418.2x 2π 60

= 561.99 rad/s

= 483.02 rad/s

= 253.23 rad/s

RPM = 1053.5

=

RPM x 2π 60

= 1053.5x 2π 60 = 110.32 rad/s

W2 – W1 (N) W2 = 0

W2 = 1.6

W2 = 2.6

W1 = 0

W1 = 1.0

W1 = 1.5

W2 – W1 = 0 – 0

W2 – W1 = 1.6 – 1.0

W2 – W1 = 2.6 – 1.5

=0N

= 0.6 N

= 1.1 N

W2 = 3.5

W2 = 4.4

W2 = 5.2

W1 = 2.0

W1 = 2.5

W1 = 3.0

W2 – W1 = 3.5 – 2.0

W2 – W1 = 4.4 – 2.5

W2 – W1 = 5.2 – 3.0

= 1.5 N

= 1.9 N

= 2.2 N

W2 = 6.1

W2 = 7.5

W2 = 9.1

W1 = 3.5

W1 = 4.0

W1 = 4.5

W2 – W1 = 6.1 – 3.5

W2 – W1 = 7.5 – 4.0

W2 – W1 = 9.1 – 4.5

= 2.6 N W2 = 10.0 W1 = 5.0 W2 – W1 = 10.0 – 5.0 = 5.0 N

= 3.5 N

= 4.6 N

Rotation,  (Nm) Drum Radius = 30 x10-3 W2 – W1 (N) = 0 N  (Nm) = W2 – W1 (N) x Drum Radius = 0 x 30 x10 = 0 Nm

-3

= 0.6 x 30 x10 = 0.018 Nm

W2 – W1 (N) = 1.5 N  (Nm) = W2 – W1 (N) x Drum Radius = 1.5 x 30 x10 = 0.045 Nm

-3

W2 – W1 (N) = 2.6 N  (Nm) = W2 – W1 (N) x Drum Radius = 2.6 x 30 x10 = 0.078 Nm

W2 – W1 (N) = 0.6 N  (Nm) = W2 – W1 (N) x Drum Radius

-3

-3

W2 – W1 (N) = 1.9 N  (Nm) = W2 – W1 (N) x Drum Radius -3

= 1.9 x10 = 0.057 Nm

= 1.1 x 30 x10 = 0.033 Nm

-3

-3

W2 – W1 (N) = 2.2 N  (Nm) = W2 – W1 (N) x Drum Radius = 2.2 x 30 x10 = 0.066 Nm

W2 – W1 (N) = 3.5 N  (Nm) = W2 – W1 (N) x Drum Radius = 3.5 x 30 x10 = 0.105 Nm

W2 – W1 (N) = 1.1 N  (Nm) = W2 – W1 (N) x Drum Radius

-3

W2 – W1 (N) = 4.6 N  (Nm) = W2 – W1 (N) x Drum Radius = 4.6 x 30 x10 = 0.138 Nm

-3

W2 – W1 (N) = 5.0 N  (Nm) = W2 – W1 (N) x Drum Radius = 5.0 x 30 x10 = 0.150 Nm

-3

Pm (W) Drum Radius = 30 x10-3

rotation,  = 0.0 Nm Pm =  x  =1092.96 x 0.0 =0W rotation,  = 0.045 Nm Pm =  x  =797.30 x 0.045 = 35.88 W rotation,  = 0.078 Nm Pm =  x  = 591.99 x 0.078 = 46.18 W rotation,  = 0.150 Nm Pm =  x  =110.32 x 0.150 = 16.55 W

rotation,  = 0.018 Nm Pm =  x  =1032.90 x 0.018 = 18.59 W rotation,  = 0.057 Nm Pm =  x  =719.49 x 0.057 = 41.01 W rotation,  = 0.105 Nm Pm =  x  = 483.02x 0.105 = 50.71 W

rotation,  = 0.033 Nm Pm =  x  =881.51 x 0.033 = 29.09 W rotation,  = 0.066 Nm Pm =  x  =649.45 x 0.066 = 42.86 W rotation,  = 0.138 Nm Pm =  x  =253.23 x 0.138 = 34.95 W

Volume (m3) 1 m3 = 1000 liter 6 liter = 6 / 1000 = 0.006 m3

Flowrate (m3/s) (x10-4) volume (m3) = 0.006 m3 time (s) = 33.44 s 3 Q = volume (m ) / time (s) = 0.006 / 33.44 -4 3 = 1.794 x10 m /s

time (s) = 25.31 s 3 Q = volume (m ) / time (s) = 0.006 / 25.31 -4 3 = 2.371 x10 m /s

time (s) = 24.31 s 3 Q = volume (m ) / time (s) = 0.006 / 22.4 -4 3 = 2.468 x10 m /s

time (s) = 22.10 s 3 Q = volume (m ) / time (s) = 0.006 / 22.10 -4 3 = 2.715 x10 m /s

time (s) = 20.41 s 3 Q = volume (m ) / time (s) = 0.006 / 20.41 -4 3 = 2.940 x10 m /s

time (s) = 18.94 s 3 Q = volume (m ) / time (s) = 0.006 / 18.94 -4 3 = 3.167 x10 m /s

time (s) = 18.40 s 3 Q = volume (m ) / time (s) = 0.006 / 18.40 -4 3 = 3.261 x10 m /s

time (s) = 17.97 s 3 Q = volume (m ) / time (s) = 0.006 / 17.97 -4 3 = 3.339 x10 m /s

time (s) = 17.79 s 3 Q = volume (m ) / time (s) = 0.006 / 17.79 -4 3 = 3.371 x10 m /s

time (s) = 17.50 s 3 Q = volume (m ) / time (s) = 0.006 / 17.50 -4 3 = 3.429 x10 m /s

Pw (W)

H = 23 -4 Q = 1.87 x10 Pw = gHQ = 1000 x 9.81 x 23 x -4 1.794 x10 = 40.48 W

H = 18 -4 Q = 2.34x10 Pw = gHQ = 1000 x 9.81 x 18 x -4 2.371 x 10 = 41.87 W

H = 17 -4 Q = 2.68 x 10 Pw = gHQ = 1000 x 9.81 x 17 x -4 2.468 x10 = 41.16 W

H = 15 -4 Q = 2.76 x x10 Pw = gHQ = 1000 x 9.81 x 15 x -4 2.715 x 10 = 39.95 W

H = 13 -4 Q = 1.87 x x10 Pw = gHQ = 1000 x 9.81 x 13 x -4 2.940 x 10 = 37.49 W

H = 12 -4 Q = 3.16 x x10 Pw = gHQ = 1000 x 9.81 x 12 x -4 3.167 x 10 = 37.28 W

H = 11.5 -4 Q = 3.55x10 Pw = gHQ = 1000 x 9.81 x 11.5 x -4 3.261 x 10 = 36.79 W

H = 11 -4 Q = 3.75 x 10 Pw = gHQ = 1000 x 9.81 x 11 x -4 3.339 x 10 = 36.03 W

H = 10.5 -4 Q = 3.77 x10 Pw = gHQ = 1000 x 9.81 x 10.5 x -4 3.371 x 10 = 34.72 W

H = 10 -4 Q = 3.55x10 Pw = gHQ = 1000 x 9.81 x 10 x -4 3.429 x 10 = 33.64 W

Efficiency  (%)

Pm = 0 Pw = 40.48  = (Pm / Pw) x 100 = (0 / 44.03) x 100 =0

Pm = 18.59 Pw = 41.87  = (Pm / Pw) x 100 = (18.59/ 41.87) x 100 = 44.40

Pm = 29.09 Pw = 41.16  = (Pm / Pw) x 100 = (29.09/ 41.16) x 100 = 70.67

Pm = 35.88 Pw = 39.95  = (Pm / Pw) x 100 = (35.88/ 39.95) x 100 = 89.81

Pm = 41.01 Pw = 37.49  = (Pm / Pw) x 100 = (41.01/ 37.49) x 100 = 109.39

Pm = 42.86 Pw = 37.28  = (Pm / Pw) x 100 = (42.86/ 37.28) x 100 = 114.97

Pm = 46.18 Pw = 36.79  = (Pm / Pw) x 100 = (46.18/ 36.79) x 100 = 125.52

Pm = 50.71 Pw = 36.03  = (Pm / Pw) x 100 = (50.71/ 36.03) x 100 = 140.74

Pm = 34.95 Pw = 34.72  = (Pm / Pw) x 100 = (34.95/ 34.72) x 100 = 100.66

Pm = 16.55 Pw = 33.64  = (Pm / Pw) x 100 = (16.55/ 33.64) x 100 = 49.19

9.0

QUESTIONS 1. Plot graph of: a. The rotation power curve (Refer Graph Rotation Power Curve Versus Motor Speed) b. Efficiency curve (Refer Graph Efficiency Versus Motor Speed) c. Discharge versus motor speed (Refer Graph Discharge Versus Motor Speed) 2. Give a comment on the graph obtained i. Torque rotations were dependent to motor speed. When the motor speed increase the torque rotation will decrease and so on reversely. ii. The efficiency increase with the increase of motor speed until an optimum condition after that the increasing of motor speed will only decrease the efficiency. iii. For the graph of motor power versus the motor speed. The graph does not resemble any trait, it is inconstant. This may be due to some error during the experiment. It is suspected the inconstant flow rate of water is the cause. When the velocity of motor is increased, the water power will decrease. As

Pw  gHQ , so water is depend on the

value of flow rate and time. 3. Calculated the velocity where the maximum power is reached. Give your comment based on the level of maximum efficiency. From the graph, maximum power is reached at 1520 RPM. During 4612.5 RPM the motor efficiency is also at its optimum level, which is 140.74%.

4. State five (5) safety factors that have been taken in the experiment? Five safety factors have been taken in the experiment: a) Avoid taking parallax reading b) Make sure all the equipments (e.g. stopwatch, Pelton machine, and tachometer) are in good condition before starting the experiment. c) Make sure there is enough water in reserve tank so that the flow rate is constant. d) When using tachometer to take read of the rotation, the tachometer is hold until a stabled range of reading could be seen and the average value is taken. e) It is the best if the reading is taken by the same person. f) Replace the brake if it is ineffective. 10.0

DISCUSSION The 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, (see conservation of mass), thus allowing the water flow to continue uninterrupted. Often two buckets are mounted side-by-side, thus splitting the water jet in half (see photo). This balances the side-load forces on the wheel, and helps to ensure smooth, efficient momentum transfer of the fluid jet to the turbine wheel. Because water and most liquids are nearly incompressible, almost all of the available energy is extracted in the first stage of the hydraulic turbine. Therefore, Pelton wheels have only one turbine stage, unlike gas turbines that operate with compressible fluid.

Figure 4: Runner of a Pelton turbine

Figure 5: bucket shape In large scale hydro installation Pelton turbines are normally only considered for heads above 150 m, but for micro-hydro applications Pelton turbines can be used effectively at heads down to about 20 m. Pelton turbines are not used at lower heads because their rotational speeds become very slow and the runner required is very large and unwieldy. If runner size and low speed do not pose a problem for a particular installation, then a Pelton turbine can be used efficiently with fairly low heads. If a higher running speed and smaller runner are required then there are two further options: - Increasing The Number Of Jets. Having two or more jets enables a smaller runner to be used for a given flow and increases the rotational speed. The required power can still be attained and the part-flow efficiency is especially good because the wheel can be run on a reduced number of jets with each jet in use still receiving the optimum flow. - Twin Runners. Two runners can be placed on the same shaft either side by side or on opposite sides of the generator. This configuration is unusual and would only be used if the number of jets per runner had already been maximized, but it allows the use of smaller diameter and hence faster rotating runners.

11.0

CONCLUSION As a conclusion, the characteristics of Pelton Turbine operation by using several speed. There are several speeds that influence the characteristics of Pelton Turbine besides the power of turbine, rotation and the flow rate. It is also influence the efficiency of turbine. From the graph, value of the maximum power is reached at 1520 RPM. During 4612.5 RPM the motor efficiency is also at its optimum level, which is 140.74%.

12.0

REFFERENCE

i. www.jfccivilengineer.com/turbines.htm. Accessed on 27 November 2011 ii. Mifflin, Boston, MA.White, F.M. (1994). Fluid Mechanics, 3rd edition, McGraw-Hill, Inc., New York, NY. iii. R. E. Featherstone, C. Naluri. (1995.) Civil Engineering Hydraulics. Bodmin, Cornwall: Blackwell Science

PART B: FRANCIS TURBINE 1.0

INTRODUCTION In Francis Turbine water flow is radial into the turbine and exits the Turbine axially. Water pressure decreases as it passes through the turbine imparting reaction on the turbine blades making the turbine rotate. Read more about design and working principle of Francis Turbine in this article. Francis Turbine is the first hydraulic turbine with radial inflow. It was designed by American scientist James Francis. Francis Turbine is a reaction turbine. Reaction Turbines have some primary features which differentiate them from Impulse Turbines. The major part of pressure drop occurs in the turbine itself, unlike the impulse turbine where complete pressure drop takes place up to the entry point and the turbine passage is completely filled by the water flow during the operation. For power generation using Francis Turbine the turbine is supplied with high pressure water which enters the turbine with radial inflow and leaves the turbine axially through the draft tube. The energy from water flow is transferred to the shaft of the turbine in form of torque and rotation. The turbine shaft is coupled with dynamos or alternators for power generation. For quality power generation speed of turbine should be maintained constant despite the changing loads. To maintain the runner speed constant even in reduced load condition the water flow rate is reduced by changing the guide vanes angle.

2.0

OBJECTIVE To determine the relationship between the head, flow rate, velocity, power and efficiency of Francis Turbine

3.0

LEARNING OUTCOMES i.

Understand the basic operating system of the Francis Turbine.

ii.

Understand on the factors which influence the efficiency of turbine.

4.0

THEORY Hydraulic power can be obtain at the turbine inlet (usually known in watt unit) and can be calculated as, Phyd = Htot Q, and Q as the volume discharge that can be read from the measuring equipment (m3/s) and  is a water per unit of volume (9820 N/m3). Htot is a total head (m) which can be calculated (using theorem Bernoulli method) in a circuit section flow just before the turbine in a location of pressure as total of three parameter, Htot = Hman + Hkin + Hpres which is Hman is a the differences of pressure head position which measured the pressure and the turbine shaft bar. As using the experiment table, Hman ~ 0.35 m (if the pressure decreasing at the turbine point out which cause by the mixer which count as an available head, Hman can be calculated as the height differences between the pressure gauging point position and the point of water level at the outflow of reservoir) is a kinetic parameter which cause by water velocity, Vm (in m/s), at the pressure gauging location, where the pipe section is Sm (this value is not S and v value at the inlet section). Hkin = Hpres =

vm 2 Q , where as g is the gravity constant (9.81 m/s2) and vm = . Sm 2g

m is a resultant parameter from water pressure, m (in unit N/m2) as measured by 

pressure gauging. In calculation table, the readable gauging value which has been read by four gauging, all the readable value from the calculation can be added by the related values, vm, Hman, Hkin, Hpres, Htot, Phd, which has been described first. This following relationship, has been used to calculated the value of mechanical power out Pmec (usually in watt), Pmec = c = 2nC / 60, where as , is the turbine angular velocity (in rad/s), n is a turbine average velocity obtain by reader equipment (RPM) and c is a braking torque velocity at the turbine shaft as obtain by reader equipment (Nm).

 P  Lastly, the turbine overall efficiency,    mec  is calculated as the ratio of power at Phyd   outlet point compared to the generated power. Last table arrangement for the experimental value and the calculation value (for each z) can be explained in a provided table.

5.0

EQUIPMENTS

Figure 1, 2 and 3: Francis Turbnine Equipments 6.0

PROCEDURE i.

The delivery control shutter closed.

ii.

The distributor leverage at the chosen value (50% and 100%) was fixed.

iii.

The delivery control shutter was opened until desired flow rate.

iv.

The reading of pressure (m), speed (n), flow rate (Q), and voltage (v) for each distributor leverage value in Table 6.1 was recorded.

v.

Step 3-4 steps were repeated with different distributor leverage value.

PRECAUTION:

Must ensure the delivery control shutter should be closed before switch on the equipment. Carelessness may cause equipment damage.

7.0

719

9.4

1.00

60

46000

4.19

587

7.7

1.32

80

35000

4.33

433

4.2

1.60

100

51200

4.08

737.4

9.48

0.972

98.20

Efficiency, (Pm / P hyd) x 100 (%)

4.08

Hydraulic power, Phyd (watt)

52000

Mechanical power, Pm (watt)

40

Total head, Htot (m)

0.69

Pressure, Hpres (m)

10.4

Kinetic, Hkin (m)

902

1.97

0.198

6.31

6.868

109.54

259.95

42.14

2.01

0.206

5.91

6.466

94.46

250.17

37.76

2.08

0.221

5.30

5.871

75.29

235.23

32.01

2.14

0.233

4.68

5.263

61.47

216.55

28.39

2.21

0.249

3.56

4.159

45.34

176.84

25.64

1.93

0.190

8.15

8.690

143.57

322.57

44.51

1.97

0.198

7.54

8.088

120.11

306.58

39.18

0.206

7.13

7.686

91.94

297.30

30.92

6.72

7.288

73.93

290.57

25.44

5.91

6.485

55.19

261.74

21.09

0.00196

0.35

80000

3.78

1371

19.5

0.41

20

74000

3.86

1147

15.5

1.03

40

70000

3.94

878

11.4

1.30

60

2.01

66000

4.06

706

9.0

1.67

80

2.07

58000

4.11

527

6.9

1.80

100

2.10

100 %

69600

3.95

925.8

12.46

Distance between turbine & shaft pressure tap, Hman (m)

3.94

Water speed, Vm (m/s)

58000

Pipe cross section, Sm, (m2)

20

Electrical power, Pel, (W)

0.25

Average of Current reading, I (ampere)

15.2

Average of Voltage, V (volt)

1046

Average of Speed, n (RPM)

3.86

Average of Flowrate, Q (m3/s) (x 10-3)

62000

Average of pressure, Pm, (N/m2)

Current reading, I (%)

Voltage, V (volt)

Current reading, I (ampere)

Speed, n (RPM)

Flowrate, Q (m3/s) (x 10-3)

Pressure, Pm (N/m2)

Z (%) 50 %

RESULT

1.241

Table 1: Result of Francis Turbine.

0.218

0.225

8.0

DATA ANALYSIS

A) Data at Z=50% Pressure, Pm (N/m2) 1 bar = 100kPa

100000 Pa

100000 N/m2

= 0.62 bar x 100000 N/m2 = 62000 N/m2 1 bar = 100kPa

100000 Pa

100000 N/m2

= 0.58 bar x 100000 N/m2 = 58000 N/m2 1 bar = 100kPa

100000 Pa

100000 N/m2

= 0.52 bar x 100000 N/m2 = 52000 N/m2 1 bar = 100kPa

100000 Pa

100000 N/m2

= 0.46 bar x 100000 N/m2 = 46000 N/m2 1 bar = 100kPa

100000 Pa

= 0.35 bar x 100000 N/m2 = 35000 N/m2

100000 N/m2

Flowrate, Q (m3/s) 13.9 x103 l x 1m3 h 1000l

x

1h = 3.86 x 10-3 m3/s 3600s

14.2 x103 l x 1m3 h 1000l

x

1h = 3.94 x 10-3 m3/s 3600s

14.7 x103 l x 1m3 h 1000l

x

1h = 4.08 x 10-3 m3/s 3600s

15.1 x103 l x 1m3 h 1000l

x

1h = 4.19 x 10-3 m3/s 3600s

15.6 x103 l x 1m3 h 1000l

x

1h = 4.33 x 10-3 m3/s 3600s

Water Speed, Vm (m/s)

Vm = Q / Sm = 3.86 x10 3 / 0.00196 = 1.97 m/s Vm = Q / Sm = 3.94 x10 3 / 0.00196 = 2.01 m/s Vm = Q / Sm = 4.08x10 3 / 0.00196 = 2.08 m/s Vm = Q / Sm = 4.19 x10 3 / 0.00196 = 2.14 m/s Vm = Q / Sm = 4.33x10 3 / 0.00196 = 2.21 m/s

Kinetic, Hkin (m)

H kin = Vm / 2g = 1.97 2 /( 2  9.81) = 0.198 m 2

2





2





2





2





H kin = Vm / 2g = 2.012 /( 2  9.81) = 0.206 m

H kin = Vm / 2g = 2.082 /( 2  9.81) = 0.221 m H kin = Vm / 2g = 2.14 2 /( 2  9.81) = 0.233 m

H kin = Vm / 2g = 2.212 /( 2  9.81) = 0.249 m Pressure, Hpres (m)

H pres = Pm /  =

62000 = 6.31 m 9820

H pres = Pm /  =

58000 = 5.91 m 9820

H pres = Pm /  =

52000 = 5.30 m 9820

H pres = Pm /  =

46000 = 4.68 m 9820

H pres = Pm /  =

35000 = 3.56 m 9820

Total Head, Htot (m)

H tot = H man  H kin  H pres = 0.35 + 0.198 + 6.31 = 6.858 m H tot = H man  H kin  H pres = 0.35 + 0.206 + 5.91 = 6.466 m H tot = H man  H kin  H pres = 0.35 + 0.221 + 5.30 = 5.871 m H tot = H man  H kin  H pres = 0.35 + 0.233 + 4.68 = 5.263 m H tot = H man  H kin  H pres = 0.35 + 0.249 + 3.56 = 4.159 m

Mechanical Power, Pm (watt)

Pm = 2n / 60  2 (1046) / 60 = 109.54 watt Pm = 2n / 60  2 (902) / 60 = 94.46 watt

Pm = 2n / 60  2 (719) / 60 = 75.29 watt Pm = 2n / 60  2 (587) / 60 = 61.47 watt Pm = 2n / 60  2 (433) / 60 = 45.34 watt

Hydraulic Power, Phyd (watt)

Phyd = H totQ = (9820)(6.858)(3.86 x 10-3) = 259.95 watt Phyd = H totQ = (9820)(6.466)(3.94 x 10-3) = 250.17 watt Phyd = H totQ = (9820)(5.871)(4.08 x 10-3) = 235.23 watt

Phyd = H totQ = (9820)(5.263)(4.19 x 10-3) = 216.55 watt Phyd = H totQ = (9820)(4.159)(4.33 x 10-3) = 176.84 watt

Efficiency, Pm / Phyd (watt) ( Pm / Phyd ) x 100 = (109.54 / 259.95) x 100 = 42.14 %

( Pm / Phyd ) x 100 = (94.46 / 250.17) x 100 = 37.76 %

( Pm / Phyd ) x 100 = (75.29 / 235.23) x 100 = 32.01 %

( Pm / Phyd ) x 100 = (61.47 / 216.55) x 100 = 28.39 %

( Pm / Phyd ) x 100 = (45.34 / 176.84) x 100 = 25.64 %

Average Data at Z = 50% i.

Average of Pressure, Pm (N/m2) = (62000 + 58000 + 52000 + 46000 + 35000) / 5 = 51200 N/m2

ii.

Average of Flowrate, Q (m3/s) = (3.86 + 3.94 + 4.08 + 4.19 + 4.33) x 10-3 / 5 = 4.08 x 10-3 m3/s

iii.

Average of Speed, n (RPM) = (1046 + 902 + 719 + 587 + 433) / 5 = 737.4 RPM

iv.

Average of Voltage, V (volt) = (15.2 + 10.4 + 9.4 + 7.7 + 4.2) / 5 = 9.48 Volt

v.

Average of Current reading, I (ampere) = (0.25 + 0.69 + 1.00 + 1.32 + 1.60) / 5 = 0.972 Ampere

Output Torque at Z = 50% n = average RPM Pm = Mechanical Power = (Pm x 60) / 2πn = (109.54 x 60) / 2π(737.4) = 1.419 Nm = (Pm x 60) / 2πn = (94.46 x 60) / 2π(737.4) = 1.223Nm = (Pm x 60) / 2πn = (75.29 x 60) / 2π(737.4) = 0.975 Nm = (Pm x 60) / 2πn = (61.47 x 60) / 2π(737.4) = 0.796 Nm = (Pm x 60) / 2πn = (45.34 x 60) / 2π(737.4) = 0.587 Nm

B) Data at Z=100% Pressure, Pm (N/m2) 1 bar = 100kPa

100000 Pa

100000 N/m2

= 0.80 bar x 100000 N/m2 = 80000 N/m2 1 bar = 100kPa

100000 Pa

100000 N/m2

= 0.74 bar x 100000 N/m2 = 74000 N/m2 1 bar = 100kPa

100000 Pa

100000 N/m2

= 0.70 bar x 100000 N/m2 = 70000 N/m2 1 bar = 100kPa

100000 Pa

100000 N/m2

= 0.66 bar x 100000 N/m2 = 66000 N/m2 1 bar = 100kPa

100000 Pa

= 0.58 bar x 100000 N/m2 = 58000 N/m2

100000 N/m2

Flowrate, Q (m3/s) 13.6 x103 l x 1m3 h 1000l

x

1h = 3.78 x 10-3 m3/s 3600s

13.9 x103 l x 1m3 h 1000l

x

1h = 3.86 x 10-3 m3/s 3600s

14.2 x103 l x 1m3 h 1000l

x

1h = 3.94 x 10-3 m3/s 3600s

14.6 x103 l x 1m3 h 1000l

x

1h = 4.06 x 10-3 m3/s 3600s

14.8 x103 l x 1m3 h 1000l

x

1h = 4.11 x 10-3 m3/s 3600s

Water Speed, Vm (m/s)

Vm = Q / Sm = 3.78x10 3 / 0.00196 = 1.93 m/s Vm = Q / Sm = 3.86 x10 3 / 0.00196 = 1.97 m/s Vm = Q / Sm = 3.94 x10 3 / 0.00196 = 2.01 m/s Vm = Q / Sm = 4.06 x10 3 / 0.00196 = 2.07 m/s Vm = Q / Sm = 4.11x10 3 / 0.00196 = 2.10 m/s

Kinetic, Hkin (m) 2





2





H kin = Vm / 2g = 1.932 /( 2  9.81) = 0.190 m H kin = Vm / 2g = 1.97 2 /( 2  9.81) = 0.198 m

H kin = Vm / 2g = 2.012 /( 2  9.81) = 0.206 m 2

H kin = Vm / 2g = 2.07 2 /( 2  9.81) = 0.218 m 2

2





H kin = Vm / 2g = 2.10 2 /( 2  9.81) = 0.225 m Pressure, Hpres (m)

H pres = Pm /  =

80000 = 8.15 m 9820

H pres = Pm /  =

74000 = 7.54 m 9820

H pres = Pm /  =

70000 = 7.13 m 9820

H pres = Pm /  =

66000 = 6.72 m 9820

H pres = Pm /  =

58000 = 5.91 m 9820

Total Head, Htot (m)

H tot = H man  H kin  H pres = 0.35 + 0.190 + 8.15 = 8.690 m H tot = H man  H kin  H pres = 0.35 + 0.198 + 7.54 = 8.088 m H tot = H man  H kin  H pres = 0.35 + 0.206 + 7.13 = 7.686 m H tot = H man  H kin  H pres = 0.35 + 0.218 + 6.72 = 7.288 m H tot = H man  H kin  H pres = 0.35 + 0.255 + 5.91 = 6.485 m

Mechanical Power, Pm (watt)

Pm = 2n / 60  2 (1371) / 60 = 143.57 watt Pm = 2n / 60  2 (1147) / 60 = 120.11 watt

Pm = 2n / 60  2 (878) / 60 = 91.94 watt Pm = 2n / 60  2 (706) / 60 = 73.93 watt Pm = 2n / 60  2 (527) / 60 = 55.19 watt

Hydraulic Power, Phyd (watt)

Phyd = H totQ = (9820)(8.690)(3.78 x 10-3) = 322.57 watt Phyd = H totQ = (9820)(8.088)(3.86 x 10-3) = 306.58 watt Phyd = H totQ = (9820)(7.686)(3.94 x 10-3) = 297.30 watt

Phyd = H totQ = (9820)(7.288)(4.06 x 10-3) = 290.57 watt Phyd = H totQ = (9820)(6.485)(4.11 x 10-3) = 261.74 watt

Efficiency, Pm / Phyd (watt) ( Pm / Phyd ) x 100 = (143.57 / 322.57) x 100 = 44.51%

( Pm / Phyd ) x 100 = (120.11 / 306.58) x 100 = 39.18 %

( Pm / Phyd ) x 100 = (91.94 / 297.30) x 100 = 30.92%

( Pm / Phyd ) x 100 = (73.93 / 290.57) x 100 = 25.44%

( Pm / Phyd ) x 100 = (55.19 / 261.74) x 100 = 21.09 %

Average Data at Z = 100% vi.

Average of Pressure, Pm (N/m2) = (80000 + 74000 + 70000 + 66000 + 58000) / 5 = 69600 N/m2

vii.

Average of Flowrate, Q (m3/s) = (3.78 + 3.86 + 3.94 + 4.06 + 4.11) x 10-3 / 5 = 3.95 x 10-3 m3/s

viii.

Average of Speed, n (RPM) = (1371 + 1147 + 878 + 706 + 527) / 5 = 925.8 RPM

ix.

Average of Voltage, V (volt) = (19.5 + 15.5 + 11.4 + 9.0 + 6.9) / 5 = 12.46 Volt

x.

Average of Current reading, I (ampere) = (0.41 + 1.03 + 1.03 + 1.30 + 1.67 + 1.80) / 5 = 1.241 Ampere

Output Torque at Z = 100% n = average RPM Pm = Mechanical Power = (Pm x 60) / 2πn = (143.57 x 60) / 2π(925.8) = 1.481 Nm = (Pm x 60) / 2πn = (120.11 x 60) / 2π(925.8) = 1.239 Nm = (Pm x 60) / 2πn = (91.94 x 60) / 2π(925.8) = 0.948 Nm = (Pm x 60) / 2πn = (73.93 x 60) / 2π(925.8) = 0.763 Nm = (Pm x 60) / 2πn = (55.19 x 60) / 2π(925.8) = 0.569 Nm

9.0

QUESTIONS 1. Plot a graph of: a.

Turbine velocity versus flowrate (Refer Graph Turbine Velocity Versus Flowrate)

b.

Turbine velocity versus output torque (Refer Graph Turbine Velocity Versus Output Torque)

c.

Turbine velocity versus hydraulic power (Refer Graph Turbine Velocity Versus Hydraulic Power)

d.

Turbine velocity versus mechanical power (Refer Graph Turbine velocity versus mechanical power)

e.

Turbine velocity versus efficiency (Refer Graph Turbine velocity versus efficiency)

2. Give your commnet(s) from the graph obtained. From the graph turbine velocity versus flow rate, the flow rate increase with decreasing of the turbine velocity. However the graph for turbine velocity versus output torque and the graph for turbine velocity versus efficiency is linear graph which is the data for Z=50% is parallel to the data for Z=100%. This case is also happened to the graph for hydraulic power, but the graph is curve. The turbine velocity versus mechanical power graph is increasing at a rate equal even though the difference in the value of z is 50% and 100%. 3. State five (5) safety factors that have been taken in the experiment? i. Make sure that the power is turned on properly before any work is carried out. ii. Make sure all of the equipment in good condition. iii. Make sure the distributor leverage at the chosen value. iv. Read the constant value of data to get a best result. v. Readings taken three times to get accurate results by averaging the data.

10.0

DISCUSSION From these experiments, we can see the relationship between the flow rate (Q), velocity (v), speed (n), power and efficiency of a Francis turbine. Generally, the function of turbines is to get electricity converted from natural hydraulic power. Francis turbine may also be used for pumped storage, where a reservoir is filled by the turbine (acting as a pump) during low power demand, and then reversed and used to generate power during peak demand.

11.0

CONCLUSION The experimental objective is achieved. In producing the highest power through turbines, all factors must be taken. The same flow rate produces a different velocity. While the velocity of water will produce a different power and efficiency according to the different capabilities of the turbine. Factors such as the flow rate, velocity, power and efficiency of this interaction and should be taken into account. Results of experiments carried out showed the relationship between the head, flow rate, velocity, power and efficiency were produce the desired energy. Aperture-opening in these experiments showed differences and advantages between an aperture with the aperture so that we can to reflect the actual operating situation of a Francis turbine.

12.0

REFERENCE i. http://www.brighthub.com/engineering/mechanical/articles/27407.aspx (Accessed on 27 November 2011

ii. Mifflin, Boston, MA.White, F.M. (1994). Fluid Mechanics, 3rd edition, McGraw-Hill, Inc., New York, NY. iii. R. E. Featherstone, C. Naluri. (1995.) Civil Engineering Hydraulics. Bodmin, Cornwall: Blackwell Science