PELTON & FRANCIS TURBINES

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FACULTY OF CIVIL AND ENVIRONMENTAL

ENGINEERING

DEPARTMENT OF WATER & ENVIROMENTAL ENGINEERING

WATER ENGINEERING LABORATORY

LAB REPORTSubject Code BFC 21201

Code & Experiment Title MKA – 03 ; PELTON & FRANCIS TURBINES

Course Code 2 BFF/1

Date 21 NOVEMBER 2011

Section / Group 2/5

Name AFANDI BIN ABD WAHID (DF100122)

Members of Group 1.MUHAMMAD IKHWAN BIN ZAINUDDIN (DF100018)

2.MOHD HASIF BIN AZMAN (DF100079)

3.MUHAMMAD HUZAIR BIN ZULKIFLI (DF100040)

Lecturer/Instructor/Tutor CIK AMNANI BIN ABU BAKAR

EN JAMILULLAIL BIN AHMAD TAIB

Received Date 29 NOVEMBER 2011

Comment by examiner Received



STUDENTS’ ETHICAL CODE

(SEC)

DEPARTMENT OF WATER & ENVIRONMENTAL

ENGINEERINGFACULTY 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




(SEC)


ENGINEERING

FACULTY OF CIVIL & ENVIRONMENTAL

ENGINEERING


BATU PAHAT, JOHOR




………………………………………. (STUDENT SIGNATURE)

NAME : MUHAMMAD IKHWAN BIN ZAINUDDIN






(SEC)


ENGINEERING


ENGINEERING


BATU PAHAT, JOHOR




………………………………………. (STUDENT SIGNATURE)

NAME : MOHD HASIF BIN AZMAN






(SEC)


ENGINEERING


ENGINEERING


BATU PAHAT, JOHOR




……………………………………. (STUDENT SIGNATURE)

NAME : MUHAMMAD HUZAIR BIN ZULKIFLI





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 part-

flow conditions also play an important role in the selection. All turbines have a power-

speed 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 LOW HEAD

IMPULSE

TURBINES

Pelton

Turgo

Cross-flow multi-jet

Pelton Turgo

Cross-flow

REACTION

TURBINES

Francis Propeller

Kaplan

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 therunner 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.

http://www.microhydropower.net/#Pelton

http://www.microhydropower.net/#Turgo

http://www.microhydropower.net/#Crossflow





http://www.microhydropower.net/#Francis

http://www.microhydropower.net/#Propeller

http://www.microhydropower.net/#Kaplan

http://www.microhydropower.net/#Kaplan

http://www.microhydropower.net/#Propeller

http://www.microhydropower.net/#Francis










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:

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

ii. 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, P m (watt) = Rotation ( , Nm) Circular velocity ( , rad/sec).

Where, T =Force( N ) Radius(m)( Nm) and minsec/60

min/2 radius (rad/s) where, 1 revolution

is equal to 2 radius. Meanwhile, Water Power, gHQ P w where, is water density

(100kg/m3), g is gravity constant (9.81m/s

2), 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.



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

RPM104.37 9863.5 8417.8 7613.7 6870.6 6201.8 5653.1 4612.5 2418.2 1053.5

(rad/s)

1092.96

1032.90 881.51 797.30 719.49 649.45 591.99 483.02 253.23 110.32

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

W2 (N)0 1.6 2.6 3.5 4.4 5.2 6.1 7.5 9.1 10.0

W1 – W2 (N)0 0.6 1.1 1.5 1.9 2.2 2.6 3.5 4.6 5.0

Drum Radius

x10-3

m 30 30 30 30 30 30 30 30 30 30

Rotation

(Nm)0 0.018 0.033 0.045 0.057 0.066 0.078 0.105 0.138 0.150

Pm(W)0 18.59 29.09 35.88 41.01 42.86 46.18 50.71 34.95 16.55

Volume (l)6 liter

Volume (m3) 0.006m

3

Time(s) 33.44 25.31 24.31 22.10 20.41 18.94 18.40 17.97 17.79 17.50Flowrate (m /s)

( x 10-4

)1.794 2.371 2.468 2.715 2.940 3.167 3.261 3.339 3.371 3.429

Pressure (mH O) 23.0 18.0 17.0 15.0 13.0 12.0 11.5 11.0 10.5 10.0

Pw (W)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 x 2π

60

= 10437 x 2π 60

= 1092.96 rad/s

RPM = 9863.5

= RPM x 2π

60

= 9863.5x 2π 60

= 1032.90 rad/s

RPM = 8417.8

= RPM x 2π

60

= 8417.8x 2π 60

= 881.51 rad/s

RPM = 7613.7

= RPM x 2π

60

= 7613.7x 2π 60

= 797.30 rad/s

RPM = 6870.6

= RPM x 2π

60

= 6870.6x 2π 60

= 719.49 rad/s

RPM = 6201.8

= RPM x 2π

60

= 6201.8x 2π 60

= 649.45 rad/s

RPM = 5653.1

= RPM x 2π

60

= 5653.1x 2π 60

= 561.99 rad/s

RPM = 4612.5

= RPM x 2π

60

= 4612.5x 2π 60

= 483.02 rad/s

RPM = 2418.2

= RPM x 2π

60

= 2418.2x 2π 60

= 253.23 rad/s

RPM = 1053.5

= RPM x 2π

60

= 1053.5x 2π

60

= 110.32 rad/s



W2 – W1 (N)

W2 = 0

W1 = 0

W2 – W1 = 0 – 0

= 0 N

W2 = 1.6

W1 = 1.0

W2 – W1 = 1.6 – 1.0

= 0.6 N

W2 = 2.6

W1 = 1.5

W2 – W1 = 2.6 – 1.5

= 1.1 N

W2 = 3.5

W1 = 2.0

W2 – W1 = 3.5 – 2.0

= 1.5 N

W2 = 4.4

W1 = 2.5

W2 – W1 = 4.4 – 2.5

= 1.9 N

W2 = 5.2

W1 = 3.0

W2 – W1 = 5.2 – 3.0

= 2.2 N

W2 = 6.1

W1 = 3.5

W2 – W1 = 6.1 – 3.5

= 2.6 N

W2 = 7.5

W1 = 4.0

W2 – W1 = 7.5 – 4.0

= 3.5 N

W2 = 9.1

W1 = 4.5

W2 – W1 = 9.1 – 4.5

= 4.6 N

W2 = 10.0

W1 = 5.0

W2 – W1 = 10.0 – 5.0

= 5.0 N



Rotation, (Nm)

Drum Radius = 30 x10-3

W2 – W1 (N) = 0 N (Nm) = W2 – W1 (N) x Drum

Radius

= 0 x 30 x10-3

= 0 Nm

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

Radius

= 0.6 x 30 x10-3

= 0.018 Nm

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

Radius

= 1.1 x 30 x10-3

= 0.033 Nm

W2 – W1 (N) = 1.5 N (Nm) = W2 – W1 (N) x Drum

Radius

= 1.5 x 30 x10-3

= 0.045 Nm

W2 – W1 (N) = 1.9 N (Nm) = W2 – W1 (N) x Drum

Radius

= 1.9 x10-3

= 0.057 Nm

W2 – W1 (N) = 2.2 N (Nm) = W2 – W1 (N) x Drum

Radius

= 2.2 x 30 x10-3

= 0.066 Nm

W2 – W1 (N) = 2.6 N (Nm) = W2 – W1 (N) x Drum

Radius

= 2.6 x 30 x10-3

= 0.078 Nm

W2 – W1 (N) = 3.5 N (Nm) = W2 – W1 (N) x Drum

Radius

= 3.5 x 30 x10-3

= 0.105 Nm

W2 – W1 (N) = 4.6 N (Nm) = W2 – W1 (N) x Drum

Radius

= 4.6 x 30 x10-3

= 0.138 Nm

W2 – W1 (N) = 5.0 N (Nm) = W2 – W1 (N) x Drum

Radius

= 5.0 x 30 x10-3

= 0.150 Nm

Pm (W)

Drum Radius = 30 x10-3

rotation, = 0.0 Nm

Pm = x

=1092.96 x 0.0= 0 W

rotation, = 0.018 Nm

Pm = x

=1032.90 x 0.018= 18.59 W


Pm = x

=881.51 x 0.033= 29.09 W


Pm = x

=797.30 x 0.045= 35.88 W


Pm = x

=719.49 x 0.057= 41.01 W


Pm = x

=649.45 x 0.066= 42.86 W


Pm = x

= 591.99 x 0.078= 46.18 W


Pm = x

= 483.02x 0.105= 50.71 W


Pm = x

=253.23 x 0.138= 34.95 W


Pm = x

=110.32 x 0.150= 16.55 W



Volume (m3)

1 m = 1000 liter

6 liter = 6 / 1000= 0.006 m

3

Flowrate (m3/s) (x10

-4)

volume (m3) = 0.006 m

3

time (s) = 33.44 sQ = volume (m

3) / time (s)

= 0.006 / 33.44= 1.794 x10

-4m

3/s


3) / time (s)

= 0.006 / 25.31= 2.371 x10

-4m

3/s


3) / time (s)

= 0.006 / 22.4= 2.468 x10

-4m

3/s


3) / time (s)

= 0.006 / 22.10= 2.715 x10

-4m

3/s


3) / time (s)

= 0.006 / 20.41= 2.940 x10

-4m

3/s


3) / time (s)

= 0.006 / 18.94= 3.167 x10

-4m

3/s


3) / time (s)

= 0.006 / 18.40= 3.261 x10

-4m

3/s


3) / time (s)

= 0.006 / 17.97= 3.339 x10

-4m

3/s


3) / time (s)

= 0.006 / 17.79= 3.371 x10

-4m

3/s

time (s) = 17.50 s

Q = volume (m3) / time (s)= 0.006 / 17.50= 3.429 x10

-4m

3/s



Pw (W)

H = 23Q = 1.87 x10

-4

Pw = gHQ = 1000 x 9.81 x 23 x

1.794 x10-4

= 40.48 W

H = 18Q = 2.34x10

-4

Pw = gHQ = 1000 x 9.81 x 18 x

2.371 x 10-4

= 41.87 W

H = 17Q = 2.68 x 10

-4

Pw = gHQ = 1000 x 9.81 x 17 x

2.468 x10-4

= 41.16 W

H = 15Q = 2.76 x x10

-4

Pw = gHQ

= 1000 x 9.81 x 15 x2.715 x 10

-4

= 39.95 W

H = 13Q = 1.87 x x10

-4

Pw = gHQ

= 1000 x 9.81 x 13 x2.940 x 10

-4

= 37.49 W

H = 12Q = 3.16 x x10

-4

Pw = gHQ

= 1000 x 9.81 x 12 x3.167 x 10

-4

= 37.28 W

H = 11.5Q = 3.55x10

-4

Pw = gHQ

= 1000 x 9.81 x 11.5 x3.261 x 10

-4

= 36.79 W

H = 11Q = 3.75 x 10

-4

Pw = gHQ

= 1000 x 9.81 x 11 x3.339 x 10

-4

= 36.03 W

H = 10.5Q = 3.77 x10

-4

Pw = gHQ

= 1000 x 9.81 x 10.5 x3.371 x 10

-4

= 34.72 W

H = 10Q = 3.55x10

-4

Pw = gHQ

= 1000 x 9.81 x 10 x3.429 x 10-4

= 33.64 W



Efficiency (%)

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

= (0 / 44.03) x 100= 0

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

= (18.59/ 41.87) x 100= 44.40

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

= (29.09/ 41.16) x 100= 70.67

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

= (35.88/ 39.95) x 100= 89.81

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

= (41.01/ 37.49) x 100= 109.39

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

= (42.86/ 37.28) x 100= 114.97

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

= (46.18/ 36.79) x 100= 125.52

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

= (50.71/ 36.03) x 100= 140.74

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

= (34.95/ 34.72) x 100= 100.66

Pm = 16.55Pw = 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 gHQ P

w

, 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

http://www.jfccivilengineer.com/turbines.htm





PART B: FRANCIS TURBINE

1.0 INTRODUCTION

In Francis Turbine water flow is radial into the turbine and exits the Turbineaxially. 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/m

3). 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 + H pres 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 = g

vm

2

2

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

mS

Q.

H pres =

mis a resultant parameter from water pressure, m (in unit N/m

2) 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 relatedvalues, vm, Hman, Hkin, H pres, Htot, Phd, 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 = 2nC / 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).

Lastly, the turbine overall efficiency,

hyd

mec

P P is calculated as the ratio of power at

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 RESULT

Z(%)

Pressure,P m

N/m2

Flowrate,Q

(m3/s)(x10-3)

Speed,n

(RPM)

Voltage,V

volt

Currentreading,I

(ampere)

Currentreading,I

(%)

Averageofpressure,P

m ,

N/m2

AverageofFlowrate,Q

(m3/s)(x10-3)

AverageofSpeed,n

(RPM)

AverageofVoltage,V

volt

AverageofCurrentreading,I

(ampere)

Electricalpower,Pel ,

(W)

Pipecrosssection,Sm,

(m2)

Waterspeed,Vm

m/s

Distancebetweenturbine&

shaftpressuretap,Hman

m Kinetic,Hkin

m Pressure,Hpres

m Totalhead,Htot

(m)

Mechanicalpower,Pm

(watt)

Hydraulicpower,Phyd

(watt)

Efficiency,(Pm/Phyd )

x100

(%)

50 %

62000 3.86 1046 15.2 0.25 20

51200 4.08 737.4 9.48 0.972

98.20 0.00196

1.97

0.35

0.198 6.31 6.868 109.54 259.95 42.14

58000 3.94 902 10.4 0.69 402.01 0.206 5.91 6.466 94.46 250.17 37.76

52000 4.08 719 9.4 1.00 602.08 0.221 5.30 5.871 75.29 235.23 32.01

46000 4.19 587 7.7 1.32 802.14 0.233 4.68 5.263 61.47 216.55 28.39

35000 4.33 433 4.2 1.60 1002.21 0.249 3.56 4.159 45.34 176.84 25.64

100 %

80000 3.78 1371 19.5 0.41 20

69600 3.95 925.8 12.46 1.241

1.93 0.190 8.15 8.690 143.57 322.57 44.51

74000 3.86 1147 15.5 1.03 40 1.97 0.198 7.54 8.088 120.11 306.58 39.18

70000 3.94 878 11.4 1.30 60 2.01 0.206 7.13 7.686 91.94 297.30 30.92

66000 4.06 706 9.0 1.67 80 2.070.218

6.72 7.288 73.93 290.57 25.44

58000 4.11 527 6.9 1.80 100 2.10 0.225 5.91 6.485 55.19 261.74 21.09

Table 1: Result of Francis Turbine.



8.0 DATA ANALYSIS

A) Data at Z=50%

Pressure, P m (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/m

2

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 100000 N/m2

= 0.35 bar x 100000 N/m2

= 35000 N/m2



Flowrate, Q (m3/s)

13.9 x103 l x 1m

3x 1h = 3.86 x 10

-3m

3/s

h 1000l 3600s

14.2 x103 l x 1m

3x 1h = 3.94 x 10

-3m

3/s

h 1000l 3600s

14.7 x103 l x 1m

3x 1h = 4.08 x 10

-3m

3/s

h 1000l 3600s

15.1 x103 l x 1m

3x 1h = 4.19 x 10

-3m

3/s

h 1000l 3600s

15.6 x103 l x 1m3 x 1h = 4.33 x 10-3

m3/s

h 1000l 3600s

Water Speed, Vm (m/s)

mV = mS/Q = 00196.0/1086.3 3 x = 1.97 m/s

mV = mS/Q = 00196.0/1094.3 3 x = 2.01 m/s

mV = mS/Q = 00196.0/1008.4 3 x = 2.08 m/s

mV = mS/Q = 00196.0/1019.4 3 x = 2.14 m/s

mV = mS/Q = 00196.0/1033.4 3 x = 2.21 m/s



Kinetic, Hkin (m)

kinH = g2/V2

m = )81.92/(97.1 2 = 0.198 m

kinH = g2/V2

m = )81.92/(01.2 2 = 0.206 m

kinH = g2/V2

m = )81.92/(08.2 2 = 0.221 m

kinH = g2/V2

m = )81.92/(14.2 2 = 0.233 m

kinH = g2/V 2m = )81.92/(21.2 2 = 0.249 m

Pressure, Hpres (m)

presH = /m P =9820

62000= 6.31 m

presH = /m P =982058000 = 5.91 m

presH = /m P =9820

52000= 5.30 m

presH = /m P =9820

46000= 4.68 m

presH = /m P =9820

35000= 3.56 m



Total Head, Htot (m)

totH = preskinman HHH = 0.35 + 0.198 + 6.31 = 6.858 m





Mechanical Power, Pm (watt)

mP = 60/)1046(260/2 n = 109.54 watt

mP

= 60/)902(260/2 n = 94.46 watt

mP = 60/)719(260/2 n = 75.29 watt

mP = 60/)587(260/2 n = 61.47 watt

mP = 60/)433(260/2 n = 45.34 watt



Hydraulic Power, Phyd (watt)

hydP = QH tot = (9820)(6.858)(3.86 x 10-3

) = 259.95 watt

hydP = QH tot = (9820)(6.466)(3.94 x 10-3

) = 250.17 watt

hydP = QH tot = (9820)(5.871)(4.08 x 10-3

) = 235.23 watt

hydP = QH tot = (9820)(5.263)(4.19 x 10-3

) = 216.55 watt

hydP = QH tot = (9820)(4.159)(4.33 x 10-3) = 176.84 watt

Efficiency, Pm / Phyd (watt)

( hydm P/P ) x 100 = (109.54 / 259.95) x 100 = 42.14 %

( hydm P/P ) x 100 = (94.46 / 250.17) x 100 = 37.76 %

( hydm P/P ) x 100 = (75.29 / 235.23) x 100 = 32.01 %

( hydm P/P ) x 100 = (61.47 / 216.55) x 100 = 28.39 %

( hydm P/P ) x 100 = (45.34 / 176.84) x 100 = 25.64 %



Average Data at Z = 50%

i. Average of Pressure, P m (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) / 2π n = (109.54 x 60) / 2π (737.4) = 1.419 Nm

= (Pm x 60) / 2π n = (94.46 x 60) / 2π (737.4) = 1.223Nm

= (Pm x 60) / 2π n = (75.29 x 60) / 2π (737.4) = 0.975 Nm

= (Pm x 60) / 2π n = (61.47 x 60) / 2π (737.4) = 0.796 Nm

= (Pm x 60) / 2π n = (45.34 x 60) / 2π (737.4) = 0.587 Nm



B) Data at Z=100%

Pressure, P m (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 100000 N/m2

= 0.58 bar x 100000 N/m2

= 58000 N/m2



Flowrate, Q (m3/s)

13.6 x103 l x 1m

3x 1h = 3.78 x 10

-3m

3/s

h 1000l 3600s

13.9 x103 l x 1m

3x 1h = 3.86 x 10

-3m

3/s

h 1000l 3600s

14.2 x103 l x 1m

3x 1h = 3.94 x 10

-3m

3/s

h 1000l 3600s

14.6 x103 l x 1m

3x 1h = 4.06 x 10

-3m

3/s

h 1000l 3600s

14.8 x103 l x 1m3 x 1h = 4.11 x 10-3

m3/s

h 1000l 3600s

Water Speed, Vm (m/s)

mV = mS/Q = 00196.0/1078.3 3 x = 1.93 m/s

mV = mS/Q = 00196.0/1086.3 3 x = 1.97 m/s

mV = mS/Q = 00196.0/1094.3 3 x = 2.01 m/s

mV = mS/Q = 00196.0/1006.4 3 x = 2.07 m/s

mV = mS/Q = 00196.0/1011.4 3 x = 2.10 m/s



Kinetic, Hkin (m)

kinH = g2/V2

m = )81.92/(93.1 2 = 0.190 m

kinH = g2/V2

m = )81.92/(97.1 2 = 0.198 m

kinH = g2/V2

m = )81.92/(01.2 2 = 0.206 m

kinH = g2/V2

m = )81.92/(07.2 2 = 0.218 m

kinH = g2/V 2m = )81.92/(10.2 2 = 0.225 m

Pressure, Hpres (m)

presH = /m P =9820

80000= 8.15 m

presH = /m P =982074000 = 7.54 m

presH = /m P =9820

70000= 7.13 m

presH = /m P =9820

66000= 6.72 m

presH = /m P =9820

58000= 5.91 m



Total Head, Htot (m)






Mechanical Power, Pm (watt)

mP = 60/)1371(260/2 n = 143.57 watt

mP

= 60/)1147(260/2 n = 120.11 watt

mP = 60/)878(260/2 n = 91.94 watt

mP = 60/)706(260/2 n = 73.93 watt

mP = 60/)527(260/2 n = 55.19 watt



Hydraulic Power, Phyd (watt)

hydP = QH tot = (9820)(8.690)(3.78 x 10-3

) = 322.57 watt

hydP = QH tot = (9820)(8.088)(3.86 x 10-3

) = 306.58 watt

hydP = QH tot = (9820)(7.686)(3.94 x 10-3

) = 297.30 watt

hydP = QH tot = (9820)(7.288)(4.06 x 10-3

) = 290.57 watt

hydP = QH tot = (9820)(6.485)(4.11 x 10-3) = 261.74 watt

Efficiency, Pm / Phyd (watt)

( hydm P/P ) x 100 = (143.57 / 322.57) x 100 = 44.51%

( hydm P/P ) x 100 = (120.11 / 306.58) x 100 = 39.18 %

( hydm P/P ) x 100 = (91.94 / 297.30) x 100 = 30.92%

( hydm P/P ) x 100 = (73.93 / 290.57) x 100 = 25.44%

( hydm P/P ) x 100 = (55.19 / 261.74) x 100 = 21.09 %



Average Data at Z = 100%

vi. Average of Pressure, P m (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) / 2π n = (143.57 x 60) / 2π (925.8) = 1.481 Nm

= (Pm x 60) / 2π n = (120.11 x 60) / 2π (925.8) = 1.239 Nm

= (Pm x 60) / 2π n = (91.94 x 60) / 2π (925.8) = 0.948 Nm

= (Pm x 60) / 2π n = (73.93 x 60) / 2π (925.8) = 0.763 Nm

= (Pm x 60) / 2π n = (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

http://www.brighthub.com/engineering/mechanical/articles/27407.aspx

http://www.brighthub.com/engineering/mechanical/articles/27407.aspx

Documents

PELTON & FRANCIS TURBINES