04-fluid report-group 3 - copy

F1-00 Hydraulics Bench & AccessoriesBy Group 3

1.0 EXPERIMENT :

Experiment H (Hydraulics Impact Of Jet - Momentum Equation)

1.1 INTRODUCTION :

The momentum equation relates the sum of the forces acting on a fluid element to its

acceleration or rate of change of momentum in the direction of the resultant force.

Whenever the momentum of a stream of fluid is increased in a given direction in passing

from one section to another, there must be a net force acting on the fluid in that direction,

and, by Newton’s third law, there will be an equal and opposite force exerted by the fluid on

the system which is producing the change of momentum. A typical example is the reaction

force exerted when a fluid is discharged in the form of a high-velocity jet, and which is

applied to the propulsion of ships and aircraft through the use of propellers, pure jet engines

and rocket motors. The propulsive force can be determined from the application of the linear

momentum equation to flow through a suitable control volume:

F = m (vout – vin )

F = Total force exerted on the fluid in a control volume in a given direction.

m = Mass flow per unit time.

= Av

= Q

vin = Incoming velocity.

vout = Outgoing velocity.

1.2 OBJECTIVE :

To investigate the validity of theoretical expressions for the force exerted by jet on targets

of various shapes for the momentum equation when the jet strikes a flat plate and a

hemispherical cup. Measuring the force on different targets and comparison with the forces

predicted by momentum theory.

1.3 APPARATUS :

Hydraulics Bench F1-10, Impact of Jet Apparatus F1-16, Stop Watch, Vernier Caliper

KUALA LUMPUR INFRASTRUCTURE UNIVERSITY COLLEGE, KLIUC 1


1.4 LABORATORY EQUIPMENT:

FIGURE 1.1 : IMPACT JET APPARATUS F1-16

1.5 PROCEDURE :

1. The top plate and transparent casing is removed, and the nozzle diameter is measured.

Water is fed through a nozzle and discharged vertically to strike a target carried on a stem

which extends through the cover.

2. A weight carrier is mounted on the upper end of the stem. The dead weight of the moving

parts is counter-balanced by a compression spring. The flat (30o) target is placed on the rod

attached to the weight pan.

KUALA LUMPUR INFRASTRUCTURE UNIVERSITY COLLEGE, KLIUC

Weight Carrier

Knurled Screws

Weight

Spring

Flat Target Plate (30o)Nozzle

Tank

Air Vent

Outlet Pipe

Level Gauge

Top PlateSpirit Level

1/2” bore inlet pipe

Adjustable Feet

Stem

2


3. The apparatus is then reassembled and the inlet pipe is connected to the bench, with the

apparatus in the open channel.

4. The base of the apparatus is leveled with the top plate loosely assembled.

5. The top plate is screwed down to datum on the spirit level.

6. The level gauge is adjusted to suit the datum on the weight pan.

7. A nominal mass is placed on the weight pan, water is allowed to flow by operating the

control valve on the bench. The vertical force exerted on the target plate is measured by

adding the weights supplied to the weight pan until the mark on the weight pan corresponds

with the level gauge.

8. When testing for level, the weight pan should be oscillated to minimize the effect of friction.

9. Take reading of volume and time to find the flow rate, Q. Note the mass on the weight pan.

10. Repeat with additional masses on the weight pan 3 times.

1.6 TECHNICAL DETAILS :

TABLE 1.1 : TECHNICAL DETAILS FOR IMPACT JET APPARATUS F1-16

NO. ITEM VALUE

1 Nozzle diameter 8mm

2 Distance between nozzle & target plate 20mm

3 Diameter of target plate 36mm

4 Target plates 30° target

1.7 SUMMARY OF THEORY :

1. External forces acting on control volume, FY = ρQ ( v – v cos θ) , v = Q A

2. For flat target (30º),

3. Total force exerted on the fluid in a control volume in a given direction, FY

FY = ρQ Q – 0 = ρQ2

A A

FY = gM

4. Error = Weight – FY x 100% Weight

5. Area of target, A = Πd2

4

6. For flat plat, the slopes of the graph, i = ρgA



1.8 RESULTS AND CALCULATIONS (30o TARGET) :

The example of calculation is shown as below (for mass of weight pan of 20g):

1. Nozzle diameter used, d = 8mm

Area,

A=πd2

4

¿ π×82

4¿5 .027×10−5m2

2. Weight, W = mass on weight pan x g

= 20_ x 9.81 1000

= 0.196 N

3. Flow rate, Q = Volume of Water, V Time, t

= __5_ 18

= 0.278 ls-1

4.FY= ρQ( v−vcos θ) , v=Q

A

For flat target (90o),

FY = ρQ Q – 0 = ρQ2

A A

= (1000)(0.278)2

5.027x10-5

= 1.543 N

5. Error = Weight - Fy x 100% Weight

= 0.196 – 1.543 x 100% 0.196

= 6.865%

The result for Experiment H (Hydraulics Impact Of Jet) is summarized as shown in Table 1.1:

Result For 30o Target below:

TABLE 1.2 : RESULT FOR 30O TARGET



The data above can be expressed in the plotted graph shown in Figure 1: Mass On Weight

Pan, M Against Flow rate, Q2 below:

FIGURE 1.2 : MASS OF WEIGHT PAN, M AGAINST FLOWRATE, Q2

CALCULATION OF FLOW RATE FOR FLAT TARGET (30O)

0.077

0.1110.128

0.174 0.174 0.174

0.207

0.250

0.309

0.391

0.000

0.050

0.100

0.150

0.200

0.250

0.300

0.350

0.400

0.450

20 30 40 50 60 70 80 90 100 110

Mass On Weight Pan, M (g)

Flow

rate

, Q2 (ls

-1)

LEGEND:

Slope line for 30o target plate obtained from the measured

values

Theoretical slope line for 180o target plate


MASS ON

WEIGHT PAN, M

WEIGHT, W

VOLUME OF

WATER, V

TIME, t FLOWRATE, Q

FLOWRATE, Q

Q2 VELOCITY, vn

FY ERROR

g N l s ls-1 m3s-1 ls-1 ms-1 N %20 0.196 5 18 0.278 0.0003 0.077 5.556 1.543 6.86530 0.294 5 15 0.333 0.0003 0.111 6.667 2.222 6.55140 0.392 5 14 0.357 0.0004 0.128 7.143 2.551 5.50150 0.491 5 12 0.417 0.0004 0.174 8.333 3.472 6.07960 0.589 5 12 0.417 0.0004 0.174 8.333 3.472 4.89970 0.687 5 12 0.417 0.0004 0.174 8.333 3.472 4.05680 0.785 5 11 0.455 0.0005 0.207 9.091 4.132 4.26590 0.883 5 10 0.500 0.0005 0.250 10.000 5.000 4.663100 0.981 5 9 0.556 0.0006 0.309 11.111 6.173 5.292110 1.079 5 8 0.625 0.0006 0.391 12.500 7.813 6.240

5


1.9 DISCUSSION :

In this experiment, the force, Fy exerted of the water flow to the target is equal to the weight

in pan. From the result, the error between Fy and weight in pan for 30º target we tested is

between 4.056% and 6.858%.

Volume applied is different for every degree of target plate. For 30o plate, volume of water is

decreasing when mass decreasing. So, the flow rate of water also decreasing and only

slightly minor of error is occurred.

From the observation and discussion, for target plate of 30o, 90o, 120o and 180o, volume of

water increasing correspond to the increasing of mass in according time, and flow rate is

varies. In the 180o plate, the error is negative, which mean the plate is nearly to the nozzle

and the water flow efficiently (please refer to Figure 1.2 : Mass On Weight Pan, M Against

Flow rate, Q2).

1.10 CONCLUSION :

From the graph above, we found that the flow rate increased when mass of weight pan

increased in according time.

After this experiment, we can accept and prove the validity of theoretical expressions for the

force exerted by a jet on targets of various shapes. The force will be an equal and opposite

reaction force exerted on the jet by the target.



2.0 EXPERIMENT :

Bernoulli’s Theorem Demonstration

2.1 INTRODUCTION :

The flow of a fluid has to confirm with a number of scientific principles in particular the

conservation of mass and the conservation of energy.

The first of these when applied to a liquid flowing through a conduit requires that for steady

flow the velocity will be inversely proportional to the flow area. The second requires that if

the velocity increases then the pressure must decrease.

The test section consists of a classical Venturi machined in clear acrylic. A series of wall

tapings allow measurement of the static pressure distribution along the converging pipe.

These tapings are connected to manometer tubes.

Water is fed through a hose connector and is controlled by a flow regulator valve at the

outlet of the test section.

The venturi can be demonstrated as a means of flow measurement and the discharge

coefficient can be determined.

2.2 OBJECTIVE :

To demonstrate the Bernoulli’s Theorem and to investigate the validity of Bernoulli’s

Theorem as applied to the flow of water in a conical pipe.

2.3 APPARATUS :

Bernoulli’s theorem demonstration unit.



2.4 LABORATORY EQUIPMENT :

FIGURE 2.1 : BERNOULLI’S THEOREM DEMONSTRATION UNIT


Hose connector

Flow regulator valve

Manometer-tubes

Venturi

Wall tapping

8


FIGURE 2.2 : ILLUSTRATION OF BERNOULLI’S THEOREM DEMONSTRATION UNIT


Air bleed screw

Test section

manometer-tubes

unions

Water inlet

Water outlet

adjustable feet

9


2.5 PROCEDURES :

1. Ensure that the clear acrylic is installed with the converging section upstream. Also check

that the unions are tightened.

2. The apparatus located on the flat top of the bench.

3. A spirit level attached to baseboard and level the unit on top of the bench.

4. Water filled into the volumetric tank of the hydraulic bench until approximately 90%.

5. Connect a flexible hose to the outlet and make sure that it is directed into the channel.

6. Partially open the outlet flow control valve at Bernoulli’s theorem demonstration unit.

7. All manometer tubing are checked properly connected to the corresponding pressure taps

and are air bubble free.

8. The discharge valve adjusted to a high measurable flow rate.

9. The water flow rate measured using volumetric method after the discharge.

10. The hypodermic tube slide gently connected to manometer #g, so that its end reaches the

cross section of the Venturi tube at #A. Wait for some time and note the readings from

manometer #G is the sum of the static head and velocity heads.

11. Step 10 repeated for the other sections.

12. Step 8 to 11 repeated with three other decreasing flow rates by regulating the venture

discharge valve.

13. The velocity VIB calculated using the Bernoulli’s equation where,

ViB= √2 x g x (hG – hi)

14. The velocity ViC calculated using the continuity equation where

ViC= Qave / Ai

15. The difference between two calculated velocities determined.

16. Water supply valve and venturi discharge valve have to be closed.

17. Water supply pump have to be turned off.

18. Water drained off from the unit when not in use.

2.6 RESULT AND DATA :

The example of calculation for experiment 1,2 and 3 are shown as below:

1. Qave = 10 lmin-1 = 10/60000 = 0.0002 m3/s

2. DiA = 25.0 mm = 25/1000 = 0.025 m

3. AiA = π DiA2/4

= π (0.025)2 / 4

= 0.00049 m2



4. ViB = √2 x g x (h*G – hi)

= √2 x (9.81) x (0.225-0.230)

= 0.313 ms-1

5. ViC = Qave/AiA

= 0.0002 / 0.00049

= 0.408 ms-1

6. ViB-Vic = 0.313 – 0.408

= -0.095 ms-1

Table 2.1 : Result For Experiment 1 below showing the data needed to show that when the

velocity of water increase, then the pressure must increase.

TABLE 2.1 : RESULT FOR EXPERIMENT 1CROSS SECTION USING BERNOULLI

EQUATIONUSING CONTINUITY

EQUATIONDIFFERENCE

i Qave

lmin-

1

Di

mmh*=hG

mm

hi

mm

ViB= √2 x g x (h* – hi)

ms-1

Ai= πDi2/4

m2

ViC=Qave/Ai

ms-1

ViB-Vic

ms-1

A 10 25.0 177 177 0.00 0.00049 0.408 -0.408B 10 13.9 174 170 8.86 0.00015 1.333 7.527C 10 11.8 169 173 8.86 0.00011 1.818 7.042D 10 10.7 166 172 10.85 0.00009 2.222 8.628E 10 10.0 164 171 10.88 0.00008 2.500 8.380F 10 25.0 164 167 7.13 0.00049 0.408 6.722

Table 2.2 and 2.3 below are showing the second and third test done in the laboratory:



EQUATIONDIFFERENC

Ei Qave

lmin-1

Di

mmh*=hG

mm

hi

mm

ViB= √2 x g x (h* – hi)

ms-1

Ai= πDi

2/4

m2

ViC=Qave/Ai

ms-1

ViB-Vic

ms-1

A 10 25.0 253 256 7.67 0.00049 0.408 7.262B 10 13.9 202 256 32.55 0.00015 1.333 31.217C 10 11.8 142 254 46.88 0.00011 1.818 45.059D 10 10.7 102 253 54.43 0.00009 2.222 52.208E 10 10.0 61 249 60.74 0.00008 2.500 58.240F 10 25.0 129 140 14.69 0.00049 0.408 14.282



EQUATIONDIFFERENC

Ei Qave

lmin-

1

Di

mmh*=hG

mm

hi

mm

ViB= √2 x g x (h* – hi)

ms-1

Ai= πDi2/4

m2

ViC=Qave/Ai

ms-1

ViB-Vic

ms-1

A 10 25.0 225 230 9.91 0.00049 0.408 9.502B 10 13.9 179 229 31.32 0.00015 1.333 29.987C 10 11.8 122 228 45.60 0.00011 1.818 43.782D 10 10.7 86 225 52.22 0.00009 2.222 49.998E 10 10.0 43 222 59.27 0.00008 2.500 56.77



F 10 25.0 121 134 15.97 0.00049 0.408 15.562

2.7 DISCUSSION :

The difference between the velocities is caused due to the difference in the cross sectional

area of the water channels. The difference of the velocity also caused due the inner surface

frictional losses. The smaller cross sectional area have a larger velocity value.

2.8 CONCLUSION :

By doing these lab experiment, we observe the theory that stated in Bernoulli theorem, we

see practically in this demonstration experiment. This experiment also proves that the

smallest section in area having a largest velocity which is in Bernoulli’s theorem.

2.9 REFERENCES :

1. Robert L.Mott Applied Fluid Mechanics (6th edition), 2001, London

2. Joseph B. Franzini Water Resources Engineering, 1999, NewYork


A

electronic device

framework

test pipe

valve

length of test pipe=0.5 m

tappingpoint

tappingpoint


3.0 EXPERIMENT :

Osborne Reynold’s Demonstration

3.1 INTRODUCTION :

The Energy Losses in Pipes accessory consists of a test pipe, orientated vertically on the

side of the equipment, which may be fed directly from the Hydraulics Bench supply.

The flow rate may be controlled by a valve at the discharge end of the test pipe. Head loss

between two tapping points in the test pipe is measured using an electronic pressure

device.

Water discharging from the head tank is returned to the sump tank of the Hydraulics Bench.

3.2 OBJECTIVE :

To compute Reynold’s number, R. To observe the laminar, transitional and turbulent flow

and to investigate the variation of friction head along a circular pipe and to calculate the

roughness of the pipe.

3.3 APPARATUS :

Osborne Reynold’s Demonstration ( Model : FM 11)


FIGURE 3.1 : ILLUSTRATION OF OSBORNE REYNOLD’S DEMONSTRATION ( MODEL : FM 11)



FIGURE 3.2 : OSBORNE REYNOLD’S DEMONSTRATION ( MODEL : FM 11)


Reynold’s Number Formula, R = VL/

Where, R = Reynolds Number

V = Fluid Velocity (ms-1)

L = Characteristic Length or Diameter (m)

= Kinematics viscosity (m2s-1)

Flow Rate Formula, Q = V/T



Where, Q = Flow Rate (Ls-1)

V = Volume of fluid (L)

T = Time Taken (s)

3.6 PROCEDURE :

1. The dye injector is lowered until it is just above the bell mouth inlet.

2. The inlet valve is opened and water is allowed to enter stilling tank.

3. A small overflow spillage through the over flow tube is ensured to maintain a constant level.

4. Water is allowed to settle for a few minutes.

5. The flow control valve is opened fractionally to let water flow through the visualizing tube.

6. The dye control needle valve is slowly adjusted until a slow flow with dye injection is

achieved.

7. The water inlet and outlet valve is regulated until a straight identifiable dye line is achieved.

The flow will be laminar.

8. The flow rate is measured using volumetric method.

9. The experiment is repeated by regulating water inlet and outlet valve to produce transitional

and turbulent flow.

3.7 RESULT AND CALCULATION :

Reynolds number, R = VL/

Given,

d= 10 mm

= 0.893 × 10-6 mm

V = ?

1. For laminar flow:

Flow Rate, Q = VA

V = Q/A = (0.018 × 10-3)/ [π (10 × 10-3)2/4]

= 1.41 × 10 -9 ms -1

Reynolds number, R = [(1.41 × 10-9ms-1)( 10 × 10-3 m)] / ( 0.893 × 10-9 m2s-1)

= 0.016

2. Transitional flow:

Flow Rate, Q = VA

V = Q/A = (0.019 × 10-3)/ [π (10 × 10-3)2/4]

= 1.49 × 10 -9 ms -1

Reynolds number, R = [(1.49 × 10-9ms-1)( 10 × 10-3 m)] / ( 0.893 × 10-9 m2s-1)

= 0.017

3. Turbulent flow :

Flow Rate, Q = VA



V = Q/A = (0.018 × 10-3)/ [π (10 × 10-3)2/4]

= 1.41 × 10 -9 m/s

Reynolds number, R = [(1.41 × 10-9 ms-1)( 10 × 10-3 m)] / ( 0.893 × 10-9 m2s-1)

= 0.016

Tables below are showing the data for laminar flow, transitional flow and turbulent flow:

TABLE 3.1 : DATA FOR LAMINAR FLOWLAMINAR FLOWVolume T1

sT2s

T3s

Tavg

sQ ls-1

17.34 6.78 8.81 10.98 0.018

TABLE 3.2 : DATA FOR TRANSITIONAL FLOWTRANSITIONAL FLOWVolume T1

sT2s

T3s

Tavg

sQ ls-1

16.84 7.03 8.44 10.77 0.019

TABLE 3.3 : DATA FOR TURBULENT FLOWTURBULENT FLOWVolume T1

sT2s

T3s

Tavg

sQ ls-1

16.56 7.34 9.06 10.99 0.018

3.8 DISCUSSION :

From the calculation done, we observed that flowrates for laminar flow and turbulent flow

are 1.41×10-9 ms-1. So that, the Reynold’s number for both situation is same which is 0.016.

However, the result for transitional flow is almost same but the flowrate is higher than both

laminar and turbulent flow.

The flow rate may be controlled by a valve at the discharge end of the test pipe. Head loss

between two tapping points in the test pipe is measured using an electronic pressure

device.

3.9 CONCLUSION :

From the laboratory test result, we can conclude that the Reynold’s number, R can be

computed by this test. The laminar, transitional and turbulent flow can be specified, and the

variation of friction head along a circular pipe is possible to be investigated, and the

roughness of the pipe should be found by the calculation.



4.0 EXPERIMENT :

Experiment B (Hydrostatic Pressure)

4.1 INTRODUCTION :

The hydrostatic force on any surface is due to the fluid pressure acting on that surface.

Pressure is a normal stress which is positive when in compression. Since the pressure is

everywhere normal to the surface, the resultant pressure force (Fp) is also normal to the

surface. The magnitude of Fp is

where p = ysin , = specific weight of the fluid, y = distance measured from level of zero

pressure and measured in the plane of the surface, and = angle which the plane of the

surface makes with the horizontal.There are 2 types of experiment should be done, such as

partially submerged vertical surface and fully submerged vertical surface. However, this

report only contained of the experiment for partially submerged vertical surface.

4.2 OBJECTIVE :

To determine the centre of pressure on a partially submerged plane surface.

4.3 APPARATUS :

Hydraulics Bench F1-10 and Hydrostatic Pressure Apparatus F1-12


FIGURE4.1 : HYDRAULICS BENCH F1-10 AND HYDROSTATIC PRESSURE APPARATUS F1-12



FIGURE4.2 : ILLUSTRATION OF HYDRAULICS BENCH F1-10 AND HYDROSTATIC PRESSURE

APPARATUS F1-12


For Partial Immersion,



mgL = (g/2)by2 [ (a + d) – y/3]

... m/y2 = (ρb/2L)[a + d] – (ρb/2L)(y/3)

4.6 PROCEDURE :

1. The quadrant is placed on the two dowel pins and the clamping screw is used to fasten the

balance arm.

2. a, L, depth d and width b of the quadrant is measured with the Perspex tank on the bench.

The balance arm is positioned on the knife edges (pivot).

3. The balance pan is hanged from the end of the balance arm. A length of hose is connected

from the drain cock to the sump and a length from the bench feed to the triangular aperture

on the top of the Perspex tank.

4. The tank is leveled using the adjustable feet and spirit level.

5. The counter balance weight is moved until the balance arm is horizontal.

6. The drain cock is closed and water is admitted until the level reaches the bottom edge of the

quadrant.

7. A weight is placed on the balance pan , slowly adding water into the tank until the balance

arm is horizontal.

8. The water level is recorded on the quadrant and the weight on the balance pan.

9. Fine adjustment of the water level can be achieved by overfilling and slowly draining using

the stop cock.

10. The above step is repeated for each increment of weight until the water level reaches the

top of the quadrant and face. Then remove each increment of weight noting weights and

water levels until the weights have been removed.



4.7 RESULTS AND CALCULATIONS :

The dimension of the equipment is:

a = 26.5 mm, b = 74 mm, d = 167 mm, L = 275 mm

Plot m/y2 vs y graph. The slope of this graph should be

− pb6 L

and the intercept should be

pb2L

(a+d )

. The

− pb6 L

and

pb2L

(a+d )

is theory part.

For the experiment part:

Let a1 = slope of the graph , let b1 = intercept for the graph

= - 0.0015/20 = 0.039 g/mm2

= - 7.5 x 10-5 g/mm3

For the theory part:

let a2 = − pb6 L let b2 =

pb2L

(a+d )

− pb6 L =

1000000 x746 x275

pb2L

(a+d )=

1000000 x742x 275

(26 .5+167 )

a2 = −4 .485 x10−5

g/mm3 b2 = 0.026 g/mm2

Since the theory and experiment result are different, it should to find out the percentage

different between the both result by using the following formulas:

a1−a2a2

x100%

= 67.22% and

b1−b2b2

x100%

= 50%

TABLE 4.1 : EXPERIMENT RESULT FOR HYDRAULICS WATER PRESSUREFILLING TANK DRAINING TANK AVERAGE

Weight, m

g

Height of water, y

mm

Weight, mg

Height of water, y

mm

m y y2 m/y2

10 2.30 10 2.20 10 2.25 5.06 1.9820 2.50 20 3.00 20 2.75 7.56 2.6430 3.50 30 3.60 30 3.55 12.60 2.3840 4.10 40 4.10 40 4.10 16.81 2.3850 4.60 50 4.70 50 4.65 21.62 2.3160 5.10 60 5.00 60 5.05 25.50 2.3570 5.50 70 5.50 70 5.50 30.25 2.3180 5.90 80 5.90 80 5.90 34.81 2.30



90 6.30 90 6.30 90 6.30 39.69 2.27100 6.60 100 6.60 100 6.60 43.56 2.30

FIGURE 4.3 : GRAPH OF m/y2 vs yError! Not a valid link.

LEGEND:

Slope line for hydraulics pressure lab test values

Theoretical linear slope line for hydraulics pressure

4.8 DISCUSSION :

From the results obtained, the different percentage between the theoretical part and

experiment part is not obvious. The different between both sides is less than 10%, this lab

results is acceptable.

There are so many differences between theoretical and experimental part caused by human

error. This can be occurred because when taking the reading, it only use eyes look at the

reading directly and not by some apparatus which is more accuracy. This laboratory results

is among the range from normal situation.

4.9 CONCLUSION :

From the graph, we almost get the linear line. Here we determine the static thrust exerted

by a fluid on a partially submerged surface and allow comparison of the measured

magnitude and position.

The higher the mass, the higher the stability force and point we needed. For example like

the ship, they design the center of gravity at the bottom of the ship. Hence it prevents the

ship from sunk.

In the hydrostatic state, the pressure is constant along a horizontal plane, pressure at a

point is independent of orientation and pressure change in any direction is proportional to

the fluid density and vertical change in depth.

Here we also found out that two points at the same depth have the same pressure and

orientation of a surface has no bearing on the pressure at a point in a static fluid.



5.0 EXPERIMENT :

Demonstration Of Pelton Turbine

5.1 INTRODUCTION :

Turbines are classified into 2 general category impulse and reaction. In both types the fluid

passes through a runner having blades. The momentum of fluid in the tangential direction is

changed and so tangential force on the runner is produced the runner therefore rotates and

performs useful work, while the fluid leaves with reduced energy.

5.2 OBJECTIVE :

To determine the operating characteristics of a pelton turbine such as power, efficiency and

torque at various speeds, and to determine the typical Turbine Curve. To plot the

characteristic curve for a turbine operated at a different fluid flow rates and to compare the

spear valve control in turbine performance.

5.3 APPARATUS :

Basic hydraulic bench, Demonstration Of Pelton Turbine Apparatus(Model : FM 41), stops

watch.



5.4 LAB EQUIPMENT :

FIGURE 5.1 : DEMONSTRATION OF PELTON TURBINE APPARATUS(MODEL : FM 41)-FRONT VIEW


Spring balance

Pelton Turbine

Tensioning screw

Pressure Gauge

Spear valve (nozzle)

23


FIGURE 5.2 : DEMONSTRATION OF PELTON TURBINE APPARATUS(MODEL : FM 41)-REAR VIEW


Pulley

24


5.5 PROCEDURES :

1. The turbine throttle valve must close before start to pumping the water. The motor pump is

started at minimum load. The valve are fully open and then allowing the water to circulate

until all the air-bubble are disappeared.

2. All the tensioning screw at pulley wheel tighten up and its adjust until the turbine is almost

start to turning.

3. The force value that applied at the pulley, starting with point 0 (Fb=0) until the point 8

(Fb=x.).

4. Every time force applied at pulley brake, the load gauge is tighten up then record down the

value. This step repeated until the 8 point.

5. Each point the force applied at pulley brake, the value are recorded and the value recorded

at the given table.


Given data :

g= 9.81ms-1

= 3.142

r= 0.04m

= 1000kgm-3

Below is the flow rate measurement data table:

TABLE 5.1 : FLOW RATE MEASUREMET DATA TABLE

NO OF V1 TI T2 Tave Q Q

EXPERIMENT L min min min lpm m3s-1

01 5 1.083 1.067 1.075 4.65 0.00007802 5 0.650 0.650 0.650 7.69 0.00012803 5 0.700 0.650 0.675 7.41 0.00012304 5 0.720 0.730 0.725 6.90 0.00011505 5 0.850 0.750 0.800 6.25 0.00010406 5 0.470 0.530 0.500 10.00 0.00016707 5 0.620 0.880 0.750 6.67 0.00011108 5 0.710 0.930 0.820 6.10 0.000102

Pelton Turbine experimental data is expressed in the Table 5.2 : PELTON TURBINE

EXPERIMENTAL DATA below:

TABLE 5.2 : PELTON TURBINE EXPERIMENTAL DATANO OF m1 m2 Fb1 Fb2 Fb N1 N2 N N

EXPERIMENT g g N N N rpm rpm rpm Hz

01 0 0 0.000 0.000 0.000 1457 1449 1453 24.2202 10 50 0.098 0.491 0.294 1056 1100 1078 17.9703 30 100 0.294 0.981 0.638 1538 1544 1541 25.6804 50 130 0.491 1.275 0.883 1444 1382 1413 23.55



05 50 160 0.491 1.570 1.030 1188 1151 1170 19.5506 50 200 0.491 1.962 1.226 784 765 775 12.2907 60 220 0.589 2.158 1.373 627 444 536 8.9308 60 240 0.589 2.354 1.472 216 218 217 3.62

The calculation for all data in Table 5.3 : Pelton Turbine Experimental Result Summary are

shown as below:

Hi = P1/Pg = 1.80/(1000 x 9.81) = 0.00018m

Ph = x g x Hi x Q = 1000 x 9.81 x 0.00018 x 0.000078 = 0.00023W

T = Fb x r = 0.294 x 0.04 = 0.012 Nm

Pb = 2 x x N x T = 2 x 3.142 x 17.97 x 0.012 = 1.329 W

TABLE 5.3 : PELTON TURBINE EXPERIMENTAL RESULT SUMMARYNO OF Q Fb N P1 Hi Ph T Pb Et

EXPERIMENT m3s-1 N Hz bar m W Nm W %

010.00007

80.00

024.2

2 1.700.0001

70.0001

3 0.000 0.000 0

020.00012

80.29

417.9

7 1.800.0001

80.0002

3 0.012 1.329 5760

030.00012

30.63

825.6

8 1.800.0001

80.0002

2 0.026 4.116 18522

040.00011

50.88

323.5

5 1.850.0001

90.0002

1 0.035 5.226 24578

050.00010

41.03

019.5

5 1.850.0001

90.0001

9 0.041 5.062 26266

060.00016

71.22

612.9

2 1.700.0001

70.0002

8 0.049 3.982 14055

070.00011

11.37

3 8.93 1.700.0001

70.0001

9 0.055 3.083 16321

080.00010

21.47

2 3.62 1.700.0001

70.0001

7 0.059 1.339 7750

From the result in the tables above, we can plot the Pelton Turbine Characteristics Curve.

The operating characteristics of a turbine are often conveniently shown by plotting torque,

T, brake power, Pb and turbine efficiency Et, against turbine rotational speed, N for a series

of volume flow rates, Qv such as shown in Figure 5.3,5.4 and 5.5 below:

FIGURE 5.3 : TORQUE, T AGAINST TURBINE ROTATIONAL SPEED, N



TORQUE, T AGAINST TURBINE ROTATIONAL SPEED, N

0.000

0.010

0.020

0.030

0.040

0.050

0.060

0.070

3.62 8.93 12.92 19.55 23.55 25.68 17.97 24.22

Rotational Speed, N

Tor

que,

T

LEGEND:

Slope line for torque against rotational speed according to lab

test values

Theoretical linear slope line

FIGURE 5.4 : BRAKE POWER, Pb AGAINST TURBINE ROTATIONAL SPEED, N



BRAKE POWER, Pb AGAINST TURBINE ROTATIONAL SPEED, N

0.000

1.000

2.000

3.000

4.000

5.000

6.000

3.62 8.93 12.92 19.55 23.55 25.68 17.97 24.22

Rotational Speed, N

Bra

ke P

ower

, P b

LEGEND:

Slope line for brake power lab test values

Theoretical linear slope line for brake power against rotational

speed

FIGURE 5.5 : TURBINE EFFICIENCY, Et AGAINST TURBINE ROTATIONAL SPEED, N

TURBINE EFFICIENCY, Et AGAINST TURBINE ROTATIONAL SPEED, N

0

5000

10000

15000

20000

25000

30000

3.62 8.93 12.92 19.55 23.55 25.68 17.97 24.22

Rotational Speed, N

Tur

bine

Effi

cien

cy, E t

LEGEND:

Slope line for turbine efficiency lab test values

Theoretical linear slope line



5.7 DISCUSSION :

From the pelton experiment, we found that the pelton turbine bucket rotate when the water

from spear valve heat at the bucket. When bucket rotate its runs very high rotation and the

rotor that fixed at turbine start too rotated at sane rate of turbine rotates .The rotation of

the bucket depend on the pressure from the spear valve.

5.8 CONCLUSION:

From the calculation, we can plot the Pelton turbine characteristics curve which is consist of

torque, T, brake power, Pb and turbine rotational speed, N for a series of volume flow rates

Qv as shown in 5.1, 5.2 and 5.3.

From the graph, we observe that the efficiency reaches a maximum and then falls. The

torque curve falls constantly and linearly.

Since the speed of the generator is fixed to produce a given frequency of electricity, the

optimum conditions for operation occur when the maximum turbine efficiency coincides with

the rotational speed of the generator.

So that, the load on the generator increases then the flow of water to the turbine must

increase to maintain the required operating speed.



6.0 EXPERIMENT :

Experiment J (Determination Of Coefficient Of Velocity)

6.1 INTRODUCTION :

An orifice is an opening usually circular in the side or base of a tank or reservoir through

which fluid is discharged in the form of a jet, usually into the atmosphere. The volume rate

of flow discharged through an orifice will depend upon the head of the fluid above the level

of the orifice and it can therefore, be used as a means of flow measurements.

The term ‘small orifice’ is applied to an orifice with the head producing flow so that it can be

assumed that this head does not vary appreciably from point to point across the orifice.

By taking the datum for potential energy as the center of the orifice and by applying

Bernoulli’s equation, assuming that there is no loss of energy, then the velocity of jet,

v=(2gh)-2. This is a statement of Torricelli’s theorem that the velocity of the issuing jet is

proportional to the square root of the head producing flow. However, in practice, the actual

discharge is considerably less than the theoretical discharge.

6.2 OBJECTIVE :

To measure the coefficient of velocity for a small orifice opening.

6.3 APPARATUS :

Hydraulic bench F1-10, orifice and the jet apparatus F1-17, stop watch




FIGURE 6.0 : ORIFICE AND THE JET APPARATUS F1-17

6.5 PROCEDURE :

1. The apparatus connected to the bench and the overflow pipe hose are drained to the sump

tank.

2. The apparatus leveled by adjusting the feet and ensure the path of the jet coincide with the

row of measuring needles.

3. The blank sheet graph paper are placed on the backboard and raise the needles to clear the

path of the water jet.

4. The overflow pipes is raised up and then the flow control valve opened , the water flow into

the head tank.

5. The valve adjusted until the water just spilling into the overflow pipe. The head values taken

down from scale point up to the outlet of vena contract (h).


Baffle

Inlet pipe

Thumb Nut

Needl

Orifice Plate with ‘O’ Ring

Backboar

Screw

Locknut

Head

31


6. The needles adjusted up to the just top of water and mark level at top of the needles at

sheet paper provided.

7. The step repeated from the head pressure 360 to 280.


There are 5 head used for these test, which is shown in Table 6.1, 6.2, 6.3, 6.4 and 6.5. The

distance over head correspond to height at each head is expressed by plotted graph in

Figure 6.1, 6.2, 6.3, 6.4 and 6.5 below:

TABLE 6.1 : RESULT FOR ORIFICE AT HEAD OF 360mm

HEAD, h NEEDLE HEIGHT, y DISTANCE, x x2 x2/h

mm NO. mm mm mm2 mm360 1 0 95 9025 25

2 3 95 9025 25 3 16 145 21025 58 4 28 195 38025 106 5 45 245 60025 167 6 61 295 87025 242 7 86 345 119025 331 8 111 395 156025 433

FIGURE 6.1 : DISTANCE OVER HEAD CORRESPOND TO HEIGHT AT HEAD OF 360mm

DISTANCE OVER HEAD CORRESPOND TO HEIGHT AT HEAD OF 360mm

25 2558

106

167

242

331

433

0

50

100

150

200

250

300

350

400

450

500

0 3 16 28 45 61 86 111

y (mm)

x2/h (

mm

)



mm NO. mm mm mm2 mm



340 1 0 95 9025 27 2 3 95 9025 27 3 12 145 21025 62 4 23 195 38025 112 5 39 245 60025 177 6 66 295 87025 256 7 94 345 119025 350 8 127 395 156025 459




mm NO. mm mm mm2 mm

320 1 0 95 9025 28 2 3 95 9025 28 3 11 145 21025 66 4 30 195 38025 119 5 51 245 60025 188 6 70 295 87025 272 7 100 345 119025 372 8 133 395 156025 488




TABLE 6.4 : RESULT FOR FOR ORIFICE AT HEAD OF 300mm


mm NO. mm mm mm2 mm

300 1 0 95 9025 30 2 3 95 9025 30 3 18 145 21025 70 4 34 195 38025 127 5 54 245 60025 200 6 75 295 87025 290 7 108 345 119025 397 8 143 395 156025 520




TABLE 6.5 : RESULT FOR FOR ORIFICE AT HEAD OF 280mm


mm NO. mm mm mm2 mm

280 1 0 95 9025 32 2 7 95 9025 32 3 21 145 21025 75 4 33 195 38025 136 5 57 245 60025 214 6 81 295 87025 311 7 117 345 119025 425 8 154 395 156025 557




6.7 DISCCUSION :

The experiment show that the water pressure gradually decrease when the head pressure

decreased from high value up certain low value from high head values (refer to table

provided). It can be prove by seeing the graph in Figure 6.1 to 6.5, shows that the curves is

slowly increasing from the origin line up to the certain value(y) in the graph.

6.8 CONCLUSION :

From the experiment, we found out that when the head pressure gradually reduced from

360 up to 280, the water jet power will reduce as well. Therefore the pressure of water is

control by the height of the head pressure up to the outlet of water flows.


04-fluid report-group 3 - copy

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