applied fluid mechanics...applied fluid mechanics (2160602) department of mechanical engineering...

Department of Mechanical Engineering

Darshan Institute of Engineering. & Technology, Rajkot

APPLIED FLUID MECHANICS Lab Manual

6th SEM Civil Engineering

DARSHAN INSTITUTE OF ENGINEERING AND

TECHNOLOGY, RAJKOT

APPLIED FLUID MECHANICS

Sr.

No. Experiment

Start

Date

End

Date Sign Remark

1. To determine Fluid friction factor for the

given pipes.

2. To study Laminar and Turbulent Flow

and It’s visualization on Reynolds’s

Apparatus.

3. To Study performance characteristics of a

Pelton wheel Turbine

4. To study performance characteristics of a

Francis Turbine

5. Study of centrifugal pump characteristics.

6 To calibrate the given Rectangular,

Triangular and Trapezoidal Notches

Applied Fluid Mechanics (2160602) Department of Mechanical Engineering Darshan Institute of Engineering and Technology, Rajkot 1.1

EXPERIMENT 1

1. Objective

To determine Fluid friction factor for the given pipes.

2. Aim

To determine friction loss of head

3. Introduction

The flow of liquid through a pipe is resisted by viscous shear stresses within the liquid

and the turbulence that occurs along the internal walls of the pipe, created by the

roughness of the pipe material. This resistance is usually known as pipe friction and is

measured is meters head of the fluid, thus the term head loss is also used to express the

resistance to flow.

Many factors affect the head loss in pipes, the viscosity of the fluid being handled, the

size of the pipes, the roughness of the internal surface of the pipes, the changes in

elevations within the system and the length of travel of the fluid.

The resistance through various valves and fittings will also contribute to the overall head

loss. In a well-designed system the resistance through valves and fittings will be of minor

significance to the overall head loss and thus are called Major losses in fluid flow.

4. The Darcy-Weisbach equation

Weisbach first proposed the equation we now know as the Darcy-Weisbach formula or

Darcy-Weisbach equation.

hf = 4fLV2

2gD

Where, hf = Head loss in meter

f = Darcy friction factor depends

L = length of pipe work in meter

V = Velocity of fluid in meter/sec

D = inner diameter of pipe in meter

g = Acceleration due to gravity meter /second2

Pipe Friction Losses


The Darcy Friction factor used with Weisbach equation has now become the standard

head loss equation for calculating head loss in pipes where the flow is turbulent.

5. Apparatus Description

The experimental set up consists of a large number of pipes of different diameters. The

pipes have tapping at certain distance so that a head loss can be measure with the help of

a U – Tube manometer. The flow of water through a pipeline is regulated by operating a

control valve which is provided in main supply line. Actual discharge through pipeline is

calculated by collecting the water in measuring tank and by noting the time for collection.

6. Technical Specification

Pipe: MOC = P.U.

Test length = 1000 mm

Pipe Dia. Pipe 1: ID: 16 mm

Pipe 2: ID: 21 mm

Pipe 3: ID: 26.5 mm

7. Experimental Procedure

1. Fill the storage tank/sump with the water.

2. Switch on the pump and keep the control valve fully open and close the bypass

valve to have maximum flow rate through the meter.

3. To find friction factor of pipe 1 open control valve of the same and close other to

valves

4. Open the vent cocks provided for the particular pipe 1 of the manometer.

5. Note down the difference of level of mercury in the manometer limbs.

6. Keep the drain valve of the collection tank open till it’s time to start collecting the

water.

7. Close the drain valve of the collection tank and collect known quantity of water

8. Note down the time required for the same.

9. Change the flow rate of water through the meter with the help of control valve and

repeat the above procedure.

10. Similarly for pipe 2 and 3. Repeat the same procedure indicated in step 4-9

11. Take about 2-3 readings for different flow rates.



8. Observations Table

Length of test section L = 1000 mm = 1 m

Pipe 1 Internal Diameter of Pipe, D= 16 mm

Cross Sectional Area of Pipe = 200.96 mm2 = 2.0 x 10-4 m2

Sr. No. Qty (liters) t (sec) h1 – h2 (m)

1

2

3

4

5

Pipe 2 Internal Diameter of Pipe, D= 21 mm



1

2

3

4

5

Pipe 3 Internal Diameter of Pipe, D= 26.5 mm



1

2

3

4

5



9. Calculations

1. Q = L

time required to collect L ltrs in m3/sec

2. Mean velocity, V = Q

A in meter/sec

According to Darcy- Weisbach Equation for frictional loss of head due to pipe friction

hf =4fLV2

2gD

In the above equation, everything is known to us except “f”

Conversion Factor: 1 mm of Hg = 0.0136 m of water

10. Result table:

Sr. No. Actual Discharge

(m3/s)

Actual velocity

(m/s)

Head loss due

to friction (hf)

Friction factor

(f)

Pipe -1

Pipe -2

Pipe-3



11. Conclusion


EXPERIMENT 2

1. Objective

To study Laminar and Turbulent Flow and It’s visualization on Reynolds’s Apparatus

2. Aim

To determine Reynolds number and type of flow

3. Introduction

The properties of density and specific gravity are measures of the “heaviness” of fluid.

These properties are however not sufficient to uniquely characterize how fluids behave

since two fluids (such as water and oil) can have approximately the same value of density

but behaves quite differently when flowing. There is apparently some additional property

that is needed to describe the “fluidity” of the fluid.

Viscosity is defined as the property of a fluid which offers resistance to the movement of

one layer of fluid over another adjacent layer of the fluid. It is an inherent property of

each fluid. Its effect is similar to the frictional resistance of one body sliding over other

body. As viscosity offers frictional resistance to the motion of the fluid consequently. In

order to maintain the flow, extra energy is to be supplied to overcome effect of viscosity.

The frictional energy generated comes out in form of heat and dissipated to the

atmosphere through boundary surfaces.

4. Types of flow

Laminar flow

Laminar flow is that type of flow in which the particle of the fluid moves along well

defined parts or streamlines. In laminar flow all streamlines are straight and parallel. In

laminar flow one layer of fluid is sliding over another layer, whenever the Reynolds

number is less than 2000, the flow is said to be laminar. In laminar flow, energy loss is

low and it is directly proportional to the velocity of the fluid. The following reasons are

for the laminar flow, fluid has low velocity, fluid has high viscosity and diameter of pipe

is large.

Turbulent Flow:

The flow is said to be turbulent flow it he flow moves in a zigzag way. Due to movement

of the particles in a zigzag way the eddies formation take place which are responsible of

high-energy losses.

Reynolds’s Apparatus


In turbulent flow, energy loss is directly proportional to the square of velocity of fluid. If

Reynolds number is greater than 4000, then flow is said to be turbulent

Reynolds’s Number

Reynolds was first to determine the translation from laminar to turbulent depends not

only on the mean velocity but on the quality

Re =VD

Where, = Density of Fluid

D = Diameter of pipe

=Dynamic Viscosity

The term is dimensionless and it is called Reynolds Number (Re). It is the ration of the

inertia force to the viscous force

Re =Intertia Force

Viscous Force

Re =V2

(VD)

Re =VD

This indicates that it is non-dimensional number.


The apparatus consists of

1. A tank containing water at constant head

2. Die container

3. A glass tube

4. The water from the tank is allowed to flow through the glass tube. The velocity of

flow can be varied by regulating valve. A liquid die having same specific weight

as that of water has to be introduced to glass tube.

Additional materials or Equipments required are

1. Stop Watch

2. Measuring Flask

3. Color Dye

4. Water Supply




1. Switch on the pump and fill the head tank. Manually also fill the dye tank with

some amount of bright dye liquid provided.

2. Open the control valve slowly at the bottom of the tube and release small flow of

dye.

3. Observe the flow in the tube.

4. Note down the time for 1 liter of discharge with the help of stopwatch and

measuring flask.

5. Repeat the above process for various discharges

7. Observations

The following observations are made:

1. When the velocity of flow is low, the die filament in the glass tube is in the form

of a straight line of die filament is parallel to the glass tube which is the case of

laminar flow as shown in fig.

Laminar flow

2. With the increase of velocity of flow the die filament is no longer straight line but

it becomes wavy one as shown in fig. this is shown that flow is no longer laminar.

This is transition flow.

Transition flow

3. With further increase of velocity of the way die filament is broken and finally

mixes in water as shown in fig.

Turbulent flow



8. Observation Table

Sr.

No.

Time for 500 ml

discharge in (Sec)

Discharge

Q (m3/s)

Velocity V

(m/s)

Reynolds

No. Re

Observe the flow

(Laminar,

Transition,

Turbulent)

1

2

3

4

5

Conversion Factors

1 liter/sec = 0.001 m3/sec

0.5 liter/sec = 0.0005 m3/sec

9. Calculations

Since D = 0.02 m

Area = A =π

4d2 =

π

4 x 0.022 = 0.000314 m2

Ambient Temperature is 300 C, = 0.801 x 10-6

Q =0.0005

time required to collect 0.5 ltrs of water in m3/sec

V =Q

A=

Q

0.000314 in m/sec

Re =VD

=

VD

v (because

)

Where, = Kinematic viscosity of water which in m2/s

V = Velocity of Water in m/s

D = Diameter of pipe is 0.030 m



10. Appendix

Dynamic and Kinematic Viscosity of Water in SI Units:-

Temperature t

(0C)

Dynamic Viscosity µ (N.s/m2) x

10-3

Kinematic Viscosity ν (m2/s) x

10-6

0 1.787 1.787

5 1.519 1.519

10 1.307 1.307

20 1.002 1.004

30 0.798 0.801

40 0.653 0.658

50 0.547 0.553

60 0.467 0.475

70 0.404 0.413

80 0.355 0.365

90 0.315 0.326

100 0.282 0.294

11. Conclusion


EXPERIMENT 3

1. Objective

To Study performance characteristics of a Pelton wheel Turbine

2. Aim

1. To determine the output power of Pelton turbine.

2. To determine the efficiency of the Pelton turbine.

3. Introduction

A turbine is a machine which converts the fluid energy into mechanical energy which is

then utilized to run the electric generator of a power plant. Fluid used can be water or

steam. The Pelton wheel is a tangential flow impulse turbine. The water strikes the bucket

along the tangent of the runner. The energy available at the inlet of the turbine is only

kinetic energy. The pressure at the inlet and outlet of the turbine is atmosphere. The

turbine is used for high head.

4. Nomenclature

A Cross section area of pipe m2

Cv Co-efficient of pitot tube.

D Diameter of pipe M

dB Diameter of brake drum m

dR Diameter of rope m

Ei Input power kW

Eo Output power kW

g Acceleration due to gravity m/sec2

H Total head m

h Manometer difference m

h1,h2 Manometer reading at both points cm

N RPM of runner shaft RPM

P Pressure gauge reading kg/cm2

Re Equivalent Radius m

Pelton Wheel Turbine


Q Discharge m3/sec

T Torque N m

V Velocity of water m/sec

W1 Spring balance weight kg

W2 Adjustable weight Kg

W3 Weight of Rope Kg

w Density of Water kg/m3

m Density of Manometer fluid i.e. Hg kg/m3

ηt Turbine efficiency %

5. Block Diagram

Figure 3.1 Pelton wheel turbine test rig

(V1 – bypass valve, V2 – valve for cooling water or brake drum, V3 – drain valve for

sump tank, V4 & V5 – valve for manometer for pressure tapping,

V6 & V7 – valve on manometer for air vent)



Fig 3.2 Experimental apparatus

6. Theory

Pelton turbine is an impulse turbine. In an impulse turbine, all the available energy of

water is converted into kinetic energy or velocity head by passing it through a contracting

nozzle provided at the end of the penstock. The water coming out of the nozzle is formed

into a free jet, which strikes on a series of buckets of the runner thus causing it to revolve.

The runner revolves freely in air. The water is contact with only a part of the runner at a

time, and throughout its action on the runner.

7. Description

The set up consists of centrifugal pump, turbine unit, and sump tank, arranged in such a

way that the whole unit works as re-circulating water system. The centrifugal pump

supplies the water from sump tank to the turbine. The loading of the turbine is achieved

by rope brake drum connected with weight balance. The turbine unit can be visualize by a



large circular transparent window kept at the front. A bearing pedestals rotor assembly of

shaft, runner and brake drum, all mounted on suitable cast iron base plate.

8. Utilities Required

➢ Electricity Supply: Single Phase, 220 V AC, 50 Hz, 5-15 Amp. Combined socket

with earth connection.

➢ Water supply (Initial fill).

➢ Drain Required.

➢ Floor Area required: 1.5𝑚 × 0.75𝑚

➢ Mercury (Hg) for manometer: 250 gms

➢ Tachometer for RPM measurement.


Starting Procedure:

➢ Close all the valves provided.

➢ Fill sump tank ¾th with clean water and ensure that no foreign particles are there.

➢ Fill manometer fluid i.e. Hg. in manometer by opening the valves of manometer

and one PU pipe from pressure measurement point of pipe.

➢ Connect the PU pipe back to its position and close the valves of manometer.

➢ Open the by-pass valve and ensure that there is no load on the brake drum.

➢ Switch on the pump with the help of starter.

➢ Close the by-pass valve.

➢ Open pressure measurement valves of the manometer.

➢ Open the air release valve provided on the manometer, slowly to release the air

from manometer. (This should be done very carefully)

➢ When there is no air in the manometer, close the air release valves.

➢ Now turbine is in operation.

➢ Load the turbine with the help of hand wheel attached on the top of weight

balance.

➢ Note the manometer reading and pressure gauge reading.

➢ Measure the load applied and RPM of the turbine.

➢ Repeat the experiment at different load.



➢ Repeat the experiment for different discharge by regulating the nozzle position by

the hand wheel provided for same.

Closing Procedure:

➢ When the experiment is over, first of all remove the load on dynamometer.

➢ Open the by-pass valve.

➢ Close the ball valves provided on manometer.

➢ Switch OFF Pump with the help of starter.

➢ Switch OFF main power supply.

➢ Drain the sump tank by the drain valve provided.

10. Observation & Calculation

Given Data:

➢ Acceleration due to gravity g =9.81m/sec2

➢ Diameter of pipe, D = 0.052 m

➢ Density of water, w = 1000 kg/m3

➢ Diameter of brake drum, dB = 0.2 m

➢ Density of Manometer fluid Hg, m = 13600 kg/m3

➢ Diameter of rope, dR = 0.012 m

➢ Co-efficient of pitot tube, Cv = 0.98

➢ Weight of Rope, W3 = 0.116 kg

Observation Table:

Sr.

No

N

RPM

P

kg/cm2

h1

(cm)

h2

(cm)

W1

(kg)

W2

(kg)

Set 1

Set 2



Calculations:

𝐻 = 10𝑃

𝐴 = 𝜋

4𝐷2

ℎ = (ℎ1 − ℎ2)

100 𝑚

𝑉 = 𝐶𝑣√2𝑔ℎ (𝜌𝑚

𝜌𝑤− 1)

𝑄 = 𝐴𝑉

𝐸𝑖 = 𝜌𝑤𝑔𝑄ℎ

1000 𝑘𝑊

𝑅𝑒 = 𝑑𝑏 + 2𝑑𝑟

2

𝑇 = (𝑊1 + 𝑊2 + 𝑊3)𝑔𝑅𝑒

𝐸𝑜 = 2𝜋𝑁𝑇

60000 𝑘𝑊

𝜂𝑡 = 𝐸𝑜

𝐸𝑖 × 100 %



Result Table:

Sr.

No

H

(m of WC)

Q

(m3/sec)

Ei

(KW)

Eo

(KW)

ηt

(%)

11. Conclusion


EXPERIMENT 4

1. Objective

To study performance characteristics of a Francis Turbine

2. Aim

1. To determine the output power of Francis Turbine.

2. To determine the efficiency of the Francis Turbine.

3. Introduction

Francis Turbine, named after James Bichens Fransis, is a reaction type of turbine for

medium high to medium low heads and medium small to medium large quantities of

water. The reaction turbine operates with its wheel submerged in water. The water before

entering the turbine has pressure as well as kinetic energy. The moment on the wheel is

produced by both kinetic and pressure energies. The water leaving the turbine has still

some of the pressure as well as kinetic energy.

4. Nomenclature

A Cross section area of pipe m2

Cv Co-efficient of pitot tube.

D Diameter of pipe m

dB Diameter of brake drum m

dR Diameter of rope m

Ei Input power kW

Eo Output power kW

g Acceleration due to gravity m/sec2

H Total head m

h Differential pressure of manometer m

h1,h2 Manometer reading at both points cm

N RPM of runner shaft RPM

Pd Delivery pressure kg/cm2

PS Suction pressure mmHg

Francis Turbine


Q Discharge m3/sec

Re Equivalent Radius m

T Torque N m

V Velocity of water m/sec

W1 Applied weight kg

W2 Dead weight (obtain from spring balance) kg

W3 Weight of hanger kg

W4 Weight of Rope kg

w Density of Water kg/m3

m Density of Manometer fluid i.e. Hg kg/m3

ηt Turbine efficiency %

5. Block Diagram

Figure 4.1 - Francis turbine test rig

(V1 - valve for discharge pressure, V2 - valve for suction pressure, V3 & V4 - valve for

pitot tube, V5 & V6 – Air bleeding valve, V7 – Drain valve for sump tank,

V8 – valve for cooling water of brake drum)

Francis Turbine


Figure 4.2 – Experimental apparatus

6. Theory

Originally the Francis turbine was designed as a purely radial flow type reaction turbine

but modern Francis turbine is a mixed flow type in which water enters the runner radially

inwards towards the centre and discharges out axially. It operates under medium heads

and requires medium quantity of water.

Francis Turbine


7. Description

The present set-up consists of a runner. The water is fed to the turbine by means of

Centrifugal Pump, radially to the runner. The runner is directly mounted on one end of a

central SS shaft and other end is connected to a brake arrangement. The circular window

of the turbine casing is provided with a transparent acrylic sheet for observation of flow

on to the runner. This runner assembly is supported by thick cast iron pedestal. Load is

applied to the turbine with the help of brake arrangement so that the efficiency of the

turbine can be calculated. A draught tube is fitted on the outlet of the turbine. The set-up

is complete with guide mechanism. Pressure and vacuum gauges are fitted at the inlet and

outlet of the turbine to measure the total supply head on the turbine.


➢ Electricity Supply: Three Phase, 440 V AC, 50 Hz, 5kW with earth connection.

➢ Water supply (200 liters.)

➢ Drain required.

➢ Floor Area required: 2 m x 1 m

➢ Mercury for manometer, 250 gm.

➢ Tachometer to measure RPM


Starting Procedure:

➢ Clean the apparatus and make tank free from Dust.

➢ Close the drain valve provided.

➢ Fill Sump tank ¾ with Clean Water and ensure that no foreign particles are there.

➢ Fill manometer fluid i.e. Hg. in manometer by opening the valves of manometer

and one PU pipe from pressure measurement point of pipe.

➢ Connect the PU pipe back to its position and close the valves of manometer.

➢ Ensure that there is no load on the brake drum.

➢ Switch on the Pump with the help of Starter.

➢ Open the Air release valve provided on the Manometer, slowly to release the air

from manometer. (This should be done very carefully.)

➢ When there is no air in the manometer, close the air release valves.

Francis Turbine


➢ Now turbine is in operation.

➢ Apply load on hanger and adjust the spring balance load by hand wheel just to

release the rest position of the hanger.

➢ Note the manometer reading, pressure gauge reading and vacuum gauge reading.

➢ Measure the RPM of the turbine.

➢ Note the applied weight and spring balance reading.

➢ Repeat the same experiment for different load.

➢ Regulate the discharge by regulating the guide vanes position.

➢ Repeat the experiment for different discharge.

Closing Procedure:

➢ When the experiment is over, first remove load on dynamometer.

➢ Open the by-pass valve.

➢ Close the ball valves provided on manometer.

➢ Switch OFF Pump with the help of starter.

➢ Switch OFF main power supply.


Given Data:

➢ Acceleration due to gravity g =9.81m/sec2

➢ Diameter of pipe, D = 0.08 m

➢ Density of water w = 1000 kg/m3

➢ Diameter of brake drum, dB = 0.2 m

➢ Density of Manometer fluid Hg, m = 13600 kg/m3

➢ Diameter of rope, dR = 0.012 m

➢ Co-efficient of pitot tube, Cv = 0.98

➢ Weight of hanger, W3 = 0.246 kg

➢ Weight of Rope, W4 = 0.104 kg

Francis Turbine


Observation table:

Sr.

No.

N

RPM

Ps

kg/cm2

Pd

mmHg

h1

(cm)

h2

(cm)

W1

(kg)

W2

(kg)

Set 1

1.

2.

Set 2

3.

4.

5.

Set 3

6.

7.

8.

9.

10.

Calculations:

𝐻 = 10 (𝑃𝑑

760+ 𝑃𝑠) 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

𝐴 = 𝜋

4𝐷2

ℎ = (ℎ1 − ℎ2)

100 𝑚

𝑉 = 𝐶𝑣√2𝑔ℎ (𝜌𝑚

𝜌𝑤− 1)

Francis Turbine


𝑄 = 𝐴𝑉

𝐸𝑖 = 𝜌𝑤𝑔𝑄ℎ

1000 𝑘𝑊

𝑅𝑒 = 𝑑𝑏 + 2𝑑𝑟

2

𝑇 = (𝑊1 + 𝑊3 + 𝑊4 − 𝑊2 )𝑔𝑅𝑒

𝐸𝑜 = 2𝜋𝑁𝑇

60000 𝑘𝑊

𝜂𝑡 = 𝐸𝑜

𝐸𝑖 × 100 %

Francis Turbine


Result Table:

Sr.

No

H

(m of WC)

Q

(m3/sec)

Ei

(kW)

Eo

(kW)

ηt

(%)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. Conclusion


EXPERIMENT 5

1. Objective

Study of centrifugal pump characteristics.

2. Aim

1. To determine : (i) Power input, (ii) Shaft output, (iii) Discharge, (iv) Total head,

(v) Pump Output, (vi) Overall efficiency, (vii) Pump efficiency

2. To plot the following performance characteristics: (i) Head Vs Discharge, (ii)

Pump efficiency Vs Discharge

3. Introduction

The hydraulic machines, which convert the mechanical energy into hydraulic energy, are

called pumps. The hydraulic energy is in the form of pressure energy. If the mechanical

energy is converted into pressure energy by means of centrifugal force acting on the fluid,

the hydraulic machine is called centrifugal pump.

4. Nomenclature

A Area of measuring tank m2

EMC Energy meter constant Pulses/kWh

Ei Pump input kW

ES Shaft output kW

Eo Pump output kW

g Acceleration due to gravity m/s2

H Total head M

hpg Height of pressure gauge from vacuum gauge M

N Speed of pump RPM

P Pulses of energy meter

Pd Delivery pressure kg/cm2

PS Suction pressure mmHg

Q Discharge m3/s

R Rise of water level in measuring tank M

Centrifugal Pump


R1 Final level of water in measuring tank Cm

R2 Initial level of water in measuring tank Cm

t Time taken by R Sec

tp Time taken by P Sec

Density of fluid kg/m3

ηm Motor efficiency %

ηo overall efficiency %

ηp Pump efficiency %

5. Block Diagram

Figure 5.1 Centrifugal pump test rig

(V1 – flow control valve at discharge of pump, V2 - control valve at suction of pump, V3

– valve for delivery pressure, V4 – valve for suction pressure, V5 – drain valve of

measuring tank, V6 – Drain valve for sump tank)

Centrifugal Pump


Fig 5.2 Experimental apparatus

6. Theory

The centrifugal pump acts as a reversed of an inward radial flow reaction turbine. This

means that the flow in centrifugal pumps is in the radial outward directions. The

centrifugal pump works on the principle of forced vortex flow, which means that an

external torque rotates a certain mass of liquid, the rise in pressure head of the rotating

liquid takes place. The rise in pressure head at any point of the rotating liquid is

proportional to the square of tangential velocity of (i.e. rise in pressure head = V2/ 2g or

2r2/2g) the liquid at that point. Thus at the outlet of the impeller where radius is more,

the rise in pressure head will be more and the liquid will be discharged at the outlet with a

high- pressure head. Due to this high-pressure head, the liquid can be lifted to a high

level.

Centrifugal Pump is a mechanical device, which consists of a body, impeller and a

rotating mean i.e. motor, engine etc. Impeller rotates in a stationary body and sucks the

fluid through its axes and delivers through its periphery. Impeller has an inlet angle,

outlet angle and peripheral speed, which affect the head and discharge. Impeller is

rotated by motor or i.c. engine or any other device.

Centrifugal Pump


7. Description

Centrifugal Pump Test Rig consists of a sump, a centrifugal pump, a DC motor and

measuring tank. To measure the head, pressure and vacuum gauges are provided. To

measure the discharge, a measuring tank is provided. Flow diversion system is provided

to divert flow from sump tank to measuring tank and from measuring tank to sump tank.

A valve is provided in pipeline to change the rate of flow.


➢ Electricity Supply: Single Phase, 220 V AC, 50 Hz, 5-15 Amp.

➢ Combined socket with earth connection.

➢ Water Supply (Initial Fill).

➢ Floor Drain Required.

➢ Floor Area Required: 1.5𝑚 × 0.75𝑚


Starting Procedure:

➢ Clean the apparatus and make tanks free from dust.

➢ Close the drain valves provided.

➢ Fill sump tank ¾ with clean water and ensure that no foreign particles are there.

➢ Open flow control valve given on the water discharge line and control valve given

on suction line.

➢ Ensure that all On/Off switches given on the panel are at OFF position.

➢ Set the desired RPM of motor / pump with the speed control knob provided at the

control panel.

➢ Operate the flow control valve to regulate the flow of water discharged by the

pump.

➢ Operate the control valve to regulate the suction of the pump.

➢ Record discharge pressure by means of pressure gauge, provided on discharge

line.

➢ Record suction pressure by means of vacuum gauge, provided at suction of the

pump.

➢ Record the power consumption by means of energy meter, provided in panel with

the help of stop watch.

Centrifugal Pump


➢ Measure the discharged by using measuring tank and stop watch.

➢ Repeat the same procedure for different speeds of pump.

➢ Repeat the same procedure for different discharge with constant speed.

Closing Procedure:

➢ When experiment is over, open gate valve properly provided on the discharge line.

➢ Reduce the RPM of the pump with the help of DC drive.

➢ Switch OFF the pump first.

➢ Switch OFF power supply to panel.


Given Data:

➢ Area of measuring tank A = 0.125 m2

➢ Acceleration due to gravity g = 9.81 m/sec2

➢ Motor Efficiency, ηm = 80 % (assumed)

➢ Density of water = 1000 kg/m3

➢ Energy Meter Constant, EMC = 3200 Pulses / kW hr

➢ Height of pressure gauge from vacuum gauge, hpg = 1 m

Observation Table:

Sr.

No

N

(RPM)

Pd

(kg/cm2)

PS

(mmHg)

R1

(cm)

R2

(cm)

t

(sec)

tP

(sec) P

Set 1

1.

2.

3.

4.

Set 2

5.

6.

7.

8.

9.

Centrifugal Pump


Calculations:

𝐸𝑖 = 𝑃

𝑡𝑝 ×

3600

𝐸𝑀𝐶 (𝑘𝑤)

𝐸𝑠 = 𝐸𝑖 × 𝜂𝑚

𝑅 = (𝑅1 − 𝑅2)

100 𝑚

𝑄 =𝐴 × 𝑅

𝑡

𝐻 = 10 (𝑃𝑑 + 𝑃𝑠

760) + ℎ𝑝𝑔 (𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟)

𝐸𝑜 = 𝜌𝑤𝑔𝑄𝐻

1000 𝑘𝑊

𝜂𝑡 = 𝐸𝑜

𝐸𝑖 × 100 %

𝜂𝑡 = 𝐸𝑜

𝐸𝑠 × 100 %

Centrifugal Pump


Result Table:

Sr.

No

N

(RPM)

H,

(m of

water)

Q

(m3/sec)

Ei

(kW)

Es

(kW)

Eo

(kW)

ηoverall

(%)

ηpump

(%)

11. Conclusion

Notches


EXPERIMENT NO. 6

1. Objective

To calibrate the given Rectangular, Triangular and Trapezoidal Notches

2. Introduction

Measurement of flow in open channel is essential for better management of supplies of water.

Notches and Weirs are used to measure the rate of flow of liquid (discharge) indirectly from

measurements of the flow depth.

Notch is a device used for measuring the rate of flow of liquid through a small channel or a

tank. It is an opening in the side of a measuring tank or reservoir extending above the free

surface. Weir is a concrete or masonry structure, placed in open channel over which the flow

occurs like a river. A notch is small in size whereas weir is a notch on a large scale.

A weir/notch is an orifice placed at the water surface so that the head on its upper edge is

zero. Hence, the upper edge can be eliminated, leaving only the lower edge named as weir

crest. A weir/notch can be of different shapes - rectangular, triangular, trapezoidal etc. A

triangular weir is particularly suited for measurement of small discharges.

Equation of discharge for notch and weir will remain same.

1. Rectangular Notch

The discharge over an un-submerged rectangular sharp-crested notch is defined as:

𝑄𝑡ℎ = 2

3 × 𝐿 × √2𝑔 × 𝐻

32

H = Head of water over the crest

L = Length of notch or weir

Figure 3.1 Rectangular notch

Notches

Applied Fluid Mechanics (2160602) Department of Mechanical Engineering

Darshan Institute of Engineering and Technology, Rajkot 6.2

2. Triangular Notch (V-Notch)

The rate of flow over a triangular weir mainly depends on the head H, relative to the crest of

the notch; measured upstream at a distance about 3 to 4 times H from the crest. For triangular

notch with apex angle , the rate of flow Q is obtained from the equation,

𝑄𝑡ℎ = 8

15 √2𝑔 𝑡𝑎𝑛

𝜃

2 𝐻

52

Figure 3.2 Triangular notch

3. Trapezoidal Notch

Also known as Cipolletti weirs are trapezoidal with 1:4 slopes to compensate for end

contraction losses. The equation generally accepted for computing the discharge through an

unsubmerged sharp-crested Cipolletti weir with complete contraction is:

𝑄𝑎𝑐𝑡 =2

3× 𝐶𝑑 × 𝐿 × √2𝑔 × 𝐻

32

Where, Q = Discharge over notch (m3/sec)

L = Bottom of notch width

H = Head above bottom of opening in meter

Figure 3.3 Trapezoidal notch

Notches



The pump sucks the water from the sump tank, and discharges it to a small flow channel. The

notch is fitted at the end of channel. All the notches and weirs are interchangeable. The water

flowing over the notch falls in the collector. Water coming from the collector is directed to

the measuring tank for the measurement of flow.

The following notches are provided with the apparatus:

1. Rectangular notch (Crest length L = 0.050m)

2. Triangular notch (Notch Angle – 600)

3. Trapezoidal notch (Crest length L = 0.075m; Slope = 4V:1H)

(1) Rectangular notch (2) V- notch (3) Trapezoidal notch

Figure 3.4 Different types of notches used in apparatus


1. Fit the required notch in the flow channel.

2. Fill up the water in the sump tank.

3. Open the water supply gate valve to the channel and fill up the water in the channel

up to sill level.

4. Take down the initial reading of the crest level (sill level).

5. Now start the pump and open the gate valve slowly so that water starts flowing over

the notch.

6. Let the water level become stable and note down the height of water surface at the

upstream side by the sliding depth gauge.

7. Close the drain valve of measuring tank, and measure the discharge.

8. Take the reading for different flow rates.

9. Repeat the same procedure for other notch also.

Notches



5. Observations

Notch Type: Rectangular

Sr.

No.

Still level reading,

s, (m)

Water height on upstream side,

h, (m)

Discharge time for 10 liters,

t, (sec)

1

2

3

Notch Type: Triangular

Sr.

No.


s, (m)


h, (m)


t, (sec)

1

2

3

Notch Type: Trapezoidal

Sr.

No.


s, (m)


h, (m)


t, (sec)

1

2

3

6. Calculations

Rectangular Notch

1. Head over the notch, 𝐻 = (ℎ − 𝑠), 𝑚 = _______________, 𝑚

2. Actual Discharge , 𝑄𝑎𝑐𝑡 = 𝑙

𝑡=

0.01

𝑡, 𝑚3/𝑠𝑒𝑐 = _____________, 𝑚3 𝑠𝑒𝑐⁄

3. Crest length of notch = 0.05 m

4. Theoretical discharg, 𝑄𝑡ℎ = 2

3 𝐿. √2𝑔 𝐻

3

2, 𝑚3 𝑠𝑒𝑐⁄ = _______________, 𝑚3 𝑠𝑒𝑐⁄

5. Coefficient of discharge 𝐶𝑑 = 𝑄𝑎𝑐𝑡

𝑄𝑡ℎ= ________________

Triangular notch

Notches




𝑡=

0.01

𝑡, 𝑚3/𝑠𝑒𝑐 = _____________, 𝑚3 𝑠𝑒𝑐⁄


4. Theoretical discharg, 𝑄𝑡ℎ = 8

15 √2𝑔 𝑡𝑎𝑛

60

2 𝐻

5

2, 𝑚3 𝑠𝑒𝑐⁄ = _______________, 𝑚3 𝑠𝑒𝑐⁄


𝑄𝑡ℎ= _______________

Trapezoidal Notch (or Cipolletti Weir)



𝑡=

0.01

𝑡, 𝑚3/𝑠𝑒𝑐 = _____________, 𝑚3 𝑠𝑒𝑐⁄


4. Theoretical discharg, 𝑄𝑡ℎ = 1.84. 𝐿. 𝐻3

2, 𝑚3 𝑠𝑒𝑐⁄ = _______________, 𝑚3 𝑠𝑒𝑐⁄


𝑄𝑡ℎ = _______________

7. Result Tables

Notch Type: Rectangular

Sr.

No.

Theoretical Discharge,

Qth,. (m3/sec)

Actual Discharge,

Qact,. (m3/sec) Cd

1

2

3

Notch Type: Triangular

Sr.

No.

Theoretical Discharge,

Qth,. (m3/sec)

Actual Discharge,

Qact, (m3/sec) Cd

1

2

3

Notch Type: Trapezoidal

Sr. Theoretical Discharge, Actual Discharge, Cd

Notches



No. Qth,. (m3/sec) Qact,. (m3/sec)

1

2

3

8. Conclusion

Notches


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