DEPARTMENT OF MECHANICAL ENGINEERING - · PDF file · 2015-06-17measurement of flow...

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MECHANICS OF FLUIDS AND HYDRAULIC MACHNIES educateindiaprashanth.wordpress.com M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI DEPARTMENT OF MECHANICAL ENGINEERING MECHANICS OF FLUIDS AND HYDRAULIC MACHINES LAB PRASHANTH REDDY M [Type the company name]

Transcript of DEPARTMENT OF MECHANICAL ENGINEERING - · PDF file · 2015-06-17measurement of flow...

MECHANICS OF FLUIDS AND HYDRAULIC MACHNIES educateindiaprashanth.wordpress.com

M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

DEPARTMENT

OF

MECHANICAL ENGINEERING

MECHANICS OF FLUIDS AND HYDRAULIC MACHINES LAB

PRASHANTH REDDY M

[Type the company name]

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

LIST OF EXPERIMENTS

S.NO. NAME OF EXPERIMENT PAGE

NO. DATE REMARK SIGN

1 IMPACT OF JETS ON VANES

(flat, curved ,semi sphere)

2 CALIBRATION OF VENTURIMETER

3 CALIBRATION OF ORIFICEMETER

4 FRICTION FACTOR

5 VERIFICATION OF BERNOULIS

THEOREM

6 CENTRIFUGAL PUMP

7 RECIPROCATING PUMP TEST RIG

8 FRANCIS TURBINE

9 PELTON WHEEL TURBINE

10 KAPLAN TURBINE

11 MULTI STAGE CENTRIFUGAL

PUMP

12

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

IMPACT OF JETS ON VANES

AIM:

To determine the Co-efficient of Impact for vanes.

APPARATUS:

Impact of Jet experimental Set up, stop watch & Vane of different shape (flat and

curved)

DESCRIPTION:

Momentum equation is based on Newton’s second law of motion which states that

the algebraic sum of external forces applied to control volume of fluid in any direction is

equal to the rate of change of momentum in that direction. The external forces include the

component of the weight of the fluid & of the forces exerted externally upon the

boundary surface of the control volume. If a vertical water jet moving with velocity is

made to strike a target, which is free to move in the vertical direction then a force will be

exerted on the target by the impact of jet, according to momentum equation this force

(which is also equal to the force required to bring back the target in its original position)

must be equal to the rate of change of momentum of the jet flow in that direction.

PROCEDURE:

1. Fix a required vane (suppose a Flat Plate) to the lever.

2. Adjust the Balancing Weight so that the lever becomes horizontal.

3. Add Known weight to the pan attached to one end of the lever. This will disturb

the initial balance of the lever.

4. Start the supply. The jet of water through the nozzle will impinge on the vane. The

force due to impact of water will be acting on the vane in the upward direction.

5. Adjust the Supply valve so that the lever will be back to its initial balance.

6. Adjust the weights in the pan and take few more readings.

7. With different vanes attached, and weights, repeat the procedure.

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DIAGRAM:

Nozzle

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TABULAR COLUMS:

S.NO Fa

(Actual

Force in Kg)

Ft

(Theoretical

force in Kg)

t

(Time taken for

h cm raise of

water in tank)

Q= 𝑨×𝒉

𝒕 K = Fa/Ft

(coefficient of

Impact)

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CALCULATIONS:

Ft= ρav2(1+cos φ)

Fa= Load added to the pan

ρ= density of the water=1000 kg/m3

A= area of nozzle

V= velocity of jet in m/sec

φ= angle made by the jet after impact (0˚ for flat wane and 135˚ for curved wane)

Q= discharge of water m3/sec

Q= [Volume of tank / Time taken for h cm raise of water in tank]

= [Area of Tank × h]/t m3/sec

Area of tank=50×50 cm2

Q=area of nozzle × velocity of jet = a×v m3/sec

K = 𝐹𝑎

𝐹𝑡

Dia of nozzle = 1 cm

RESULT:

PRECAUTIONS:

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CLAIBRATION OF VENTURIMETER

AIM:

To determine the coefficient of discharge of venturi meter.

APPARATUS:

A pipe provided with inlet and outlet and pressure tapping and venturi in between

them, Differential u-tube manometer, Collecting tank with piezometer, Stopwatch,

Scale

THEORY:

Venturi, the Italian engineer, discovered in 1791 that a pressure difference related

the rate of flow could be created in pipe by deliberately reducing its area of cross-section.

The modern version of the venturi meter was first developed and employed for

measurement of flow of water by Clemens Herschel in 1886. Venturi meter continues to

be the best and most precise instrument for measurement of all types of fluid flow in

pipes. The meter consists of a short length of gradual convergence throat and a longer

length of gradual divergence. The semi-angle of convergence is 8 to 10 degrees and the

semi-angle of divergenceis3 to 5 degrees. By measuring the difference in fluid pressure

becore and after throt the flow rate can be obtained from Bernoulli's equation.

PROCEDURE:

1. The pipe is selected for doing experiments

2. The motor is switched on, as a result water will flow

3. According to the flow, the ccl4 level fluctuates in the U-tube manometer

4. The reading of H1 and H2 are noted

5. The time taken for 5 cm rise of water in the collecting tank is noted

6. The experiment is repeated for various flow in the same pipe

7. The co-efficient of discharge is calculated

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SCHEMATIC DIAGRAM:

Venturimeter

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TABULARCOLUMN:

S.NO Manometric head Time taken for h

cm raise of water

in tank t

Theoretical

Discharge

(Qt) m3/sec

Actual

Discharge

(Qa) m3/sec

Coefficient of

discharge

Cd = Qa/Qt h1 h2 hw

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CALCULATIONS:

t = Time taken for h cm raise of water in tank

h1= Manometric head in first limb m

h2 = Manometric head in second limb m

hw = Venturi head in terms of flowing liquid m

= (h2-h1) × {Specific gravity of ccl 4

specific gravity of water-1}

Specific gravity of carbontetracloride (ccl4) = 1.6

Specific gravity of water = 1

Diameter of the pipe = 4 cm

Diameter of the throat = 2.4 cm

Area of collecting tank = 50×50 cm2

Theoretical Discharge (Qt) = K× ℎ m3/sec

K=𝑎1×𝑎2× 2𝑔

𝑎12−𝑎22

a1=area of cross section of pipe

a2=area of cross section of pipe at throt

Actual Discharge (Qa) = [Volume of water collected in tank/time taken to collect water]

= [Area of tank × height of water collected in tank]/ t m3/sec

Coefficient of discharge (Cd) = Qa / Qt

RESULT:

PRECAUTIONS:

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CLAIBRATION OF ORIFICEMETER

AIM:

To determine the coefficient of discharge of orifice meter.

APPARATUS:

A pipe provided with inlet and outlet and pressure tapping and Orifice in between

them, Differential U-tube manometer, Collecting tank with piezometer, Stopwatch, Scale

THEORY:

An orifice plate is a thin plate with a hole in it, which is usually placed in a pipe.

When a fluid passes through the orifice, its pressure builds up slightly upstream of the

orifice, but as the fluid is forced to converge to pass through the hole, the velocity

increases and the fluid pressure decreases. A little downstream of the orifice the flow

reaches its point of maximum convergence, afterd that, the flow expands, the velocity

falls and the pressure increases. By measuring the difference in fluid pressure across

tappings upstream and downstream of the plate, the flow rate can be obtained from

Bernoulli's equation.

PROCEDURE:

1. The pipe is selected for doing experiments

2. The motor is switched on, as a result water will flow

3. According to the flow, the ccl4 level fluctuates in the U-tube manometer

4. The reading of H1 and H2 are noted

5. The time taken for 5cm rise of water in the collecting tank is noted

6. The experiment is repeated for various flow in the same pipe

7. The co-efficient of discharge is calculated

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SCHEMATIC DIAGRAM:

Orifice meter

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TABULARCOLUMN:

S.NO Manometric head Time taken for h

cm raise of water

in tank t

Theoretical

Discharge

(Qt) m3/sec

Actual

Discharge

(Qa) m3/sec

Coefficient of

discharge

Cd = Qa/Qt h1 h2 hw

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CALCULATIONS:

t = Time taken for h cm raise of water in tank

h1= Manometric head in first limb m

h2 = Manometric head in second limb m

hw = Venturi head in terms of flowing liquid m

= (h2-h1) × {Specific gravity of ccl 4

specific gravity of water-1}

Specific gravity of carbontetracloride (ccl4) = 1.6

Specific gravity of water = 1

Diameter of the pipe = 4 cm

Diameter of the throat = 2.4 cm

Area of collecting tank = 50×50 cm2

Theoretical Discharge (Qt) = K× ℎ m3/sec

K=𝑎1×𝑎2× 2𝑔

𝑎12−𝑎22

a1=area of cross section of pipe

a2=area of cross section of pipe at throt

Actual Discharge (Qa) = [Volume of water collected in tank/time taken to collect water]

= [Area of tank × height of water collected in tank]/ t

Coefficient of discharge Cd = Qa/Qt

RESULT:

PRECAUTIONS:

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

FRICTION FACTOR

AIM:

To determine the Darcy's friction factor (f) of the given pipe

APPARATUS:

A pipe provided with inlet and outlet and pressure tapping, Differential u-tube

manometer, collecting tank with piezometer, Stopwatch, Scale.

DESCRIPTION:

When the fluid flows through a pipe the viscosity of the fluid and the inner surface

of the pipe offer resistance to the flow. In overcoming the resistance some energy of the

flowing fluid is lost. This is called the major loss in pipe flow. Boundary roughness,

which has little significance in laminar flow, plays an important role in turbulence. This,

together with transverse momentum exchange of fluid particles due to the perpetual

turbulent intermixing, are the main sources of tangential or shear stresses in turbulent

flow. Various equations have been proposed to determine the head losses due to friction.

These equations relate the friction losses to physical characteristics of the pipe and

various flow parameters.

PROCEDURE:

1. The pipe is selected for doing experiments

2. The motor is switched on, as a result water will flow

3. According to the flow, the mercury level fluctuates in the U-tube manometer

4. The reading of H1 and H2 are noted

5. The time taken for 5cm rise of water in the collecting tank is noted

6. The experiment is repeated for various flow in the same pipe

7. The co-efficient of discharge is calculated

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SCHEMATIC DIAGRAM:

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

TABULARCOLUMN:

S.NO Manometric head Time taken for h

cm raise of water

in tank t sec

Discharge

(Q)

m3/sec

Velocity

(v) m/sec

Friction

factor (f) h1 h2 hf

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CALCULATIONS:

Friction factor (f) = 2 x g x D x hf

l x v2 Where,

g = Acceleration due to gravity (m / sec2)

D for circular pipe =4x𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎

𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 =4x

𝜋𝑟2

𝜋𝑑=d

d= Diameter of the pipe = 2cm

D for squarer pipe = 4x𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎

𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟=4x

𝑤xh

2x w+h

w= width of pipe, h= height of pipe

l = Length of the pipe = 200cm

v = Velocity of liquid following in the pipe (m / s)

hf = Loss of head due to friction (m)

= (h2-h1) × { Specific gravity of Hg

specific gravity of water - 1} Where

h1 = Manometric head in the first limbs

h2 = Manometric head in the second limbs

Actual Discharge Q = A x h

t (m

3 / sec)

Where

A = Area of the collecting tank (m2)

h = Rise of water for 5 cm (m)

t = Time taken for 5 cm rise (sec)

Also

Q=Velocity in the pipe X Area of the pipe

= v x a

v= Q/a

RESULT:

PRECAUTIONS:

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

VERIFICATION OF BERNOULIS THEOREM

AIM:

To verify the Bernoulli’s theorem.

APPARATUS:

A supply tank of water, a tapered inclined pipe fitted with no. of piezometer tubes

point, measuring tank, scale, and stop watch.

THEORY:

Bernoulli’s theorem states that when there is a continues connection between the

particle of flowing mass liquid, the total energy of any sector of flow will remain same

provided there is no reduction or addition at any point. I.e. sum of pressure head and

velocity head is constant.

PROCEDURE:

1. Open the inlet valve slowly and allow the water to flow from the supply tank.

2. Now adjust the flow to get a constant head in the supply tank to make flow in and

outflow equal.

3. Under this condition the pressure head will become constant in the piezometer

tubes. Note down piezometer readings.

4. Note down the quantity of water collected in the measuring tank for a given

interval of time.

5. Compute the area of cross-section under the piezometer tube.

6. Compute the values of velocity head and pressure head.

7. Change the inlet and outlet supply and note the reading.

8. Take at least three readings as described in the above steps.

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SCHEMATIC DIAGRAM:

Throt

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TABULARCOLUMN:

S.NO Pizeometer

Reading

time for

5cm rise

Discharge

Q m3/sec

Pressure

Head m

Velocity

Head m

Datum

head m

Total Head

S.NO Pizeometer

Reading

time for

5cm rise

Discharge

Q m3/sec

Pressure

Head m

Velocity

Head m

Datum

head m

Total Head

S.NO Pizeometer

Reading

time for

5cm rise

Discharge

Q m3/sec

Pressure

Head m

Velocity

Head m

Datum

head m

Total Head

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CALCULATIONS:

Pressure head = P

ρg m

Velocity head = v2

2g m

Datum head = Z = 0 m (for this experiment)

Velocity of water flow = v

Q (Discharge) = [Volume of water collected in tank/time taken to collect water]

= [Area of tank × height of water collected in tank]/ t m3/sec

Also

Q= velocity of water in pipe × area of cross section = v × Ax m3/sec

Area of cross section (Ax) = At + [(Ai−At)×Ln

L] m

2

At = Area of Throt

Ai = Area of Inlet

Dia of throt = 25mm

Dia of inlet = 50mm

Ln= distance between throt and corresponding pizeometer

L=length of the diverging duct or converging duct = 300mm

Distance between each piezometer = 75mm

Total Head = 𝑃

ρg +

v2

2g + Z

RESULT:

PRECAUTIONS:

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CENTRIFUGAL PUMP

AIM:

To study the performance characteristics of a centrifugal pump

APPARATUS:

1. centrifugal pump test setup

2. stop watch

DESCRIPTION:

The operation of filling water in the suction pipe casing and a portion delivery pipe for

the removal of air before starting is called priming.

After priming the impeller is rotated by a prime mover. The rotating vane gives a

centrifugal head to the pump. When the pump attains a constant speed, the delivery valve

is gradually opened. The water flows in a radially outward direction. Then, it leaves the

vanes at the outer circumference with a high velocity and pressure. Now kinetic energy is

gradually converted in to pressure energy. The high-pressure water is through the

delivery pipe to the required height.

PROCEDURE:

1) Prime the pump, close the delivery valve and switch on the unit.

2) Open the delivery valve and maintain required head. Note the reading of pressure

gauge.

3) Note the corresponding reading from vacuum gauge.

4) Measure the area of the collecting tank

5) Close the drain valve and note the time for 10 cm rise of water level in the collecting

tank.

6) For different delivery head repeat the experiment.

7) For every set of reading note the time taken for 3 rev. Energy meter.

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SCHEMATIC DIAGRAM:

Centrifugal pump

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

TABULARCOLUMN:

S.NO Pressure

gauge

reading

Pd

(Kg/cm2)

Vacuum

gauge

reading

mm of

Hg(Ps)

Time for

3 rev of

Energy

meter

seconds

(te)

Time for 10

cm rise in

collecting

tank (t)

seconds

Discharge

(Q)

m3/sec

Input

Power

Pi

Output

Power

Po

η%

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CALCULATIONS:

1. The total effective head H in meters of water column = Hd + Hs + Z

* since the delivery pressure is in Kg/cm2 and suction gauge readings are in mm

of Hg, the total head developed by the pump to be converted in meters of water

column.

Where Hd = delivery head = Pd/ρ kg/cm2

Hs = suction head= 𝑃𝑠×13600

𝜌 mm of Hg

Z= datum level difference = 2.8 m

Note: the velocity head and the loss of the head in the suction pipe are neglected.

2. Knowing the discharge Q = 𝑨×𝒉

𝒕 m

3/sec.

Where

A = Area of the collecting tank (m2) = 70 X 70 cm

2

h = 10 cm rise of water level in the collecting tank

t = Time taken for 10 cm rise of water level in collecting tank.

2. The work done by the pump is given by Po = ρ × g × Q × H

1000 Kw

Where,

ρ = Density of water = 1000 (kg / m³)

g = Acceleration due to gravity = 9.81 (m / s2)

H = Total head of water (m)

H = suction head (Hs) + delivery Head (Hd) + Datum Head

3. The input power Pi = 3600 × Nr

E × te Kw

Where

Nr = Number of revolutions of energy meter disc

E = Energy meter constant = 150 (rev / Kw hr)

T = time taken for ‘Nr’ revolutions (seconds)

4. The efficiency of the pump = (Po/ Pi) ×100 %

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GRAPH:

1. Actual discharge Vs Total head

2. Actual discharge Vs Efficiency

3. Actual discharge Vs Input power

4. Actual discharge Vs Output power

RESULT:

PRECAUTIONS:

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RECIPROCATING PUMP TEST RIG

AIM:

To study the characteristics of Reciprocating pump.

APPARATUS:

1) Reciprocating pump test setup

2) Stop watch

DESCRIPTION:

Reciprocating pumps also classified as positive displacement pumps as a definite

volume of liquid is trapped in a chamber which is alternatively filled from the inlet and

emptied at a higher pressure through the discharge. Most piston pumps are acting with

liquid admitted alternatively on each side of the piston so that one part of the cylinder is

being filled while the other is being emptied to minimize fluctuations in the discharge.

It consists of a double action Reciprocating pump of size 25×20 mm with air

vessel coupled to a 1 HP, 1440 rpm single phase motor, piping system consisting of

pipes, gate valve, foot valve, pressure and vacuum gauges. Collecting tank with gauge

glass scale fittings and drain valve. Panel with switch, starter and energy meter.

PROCEDURE:

1. Keep the delivery valve open and switch on the pump. Slowly close the delivery

valve and maintain a constant head.

2. Note the delivery and suction gauge reading.

3. Note the time for 10 rev of Energy meter.

4. Note the time for 10 cm rise in water level in the collecting tank.

5. Note the speed of the pump (N) rpm.

6. Repeat the procedure for various openings of the delivery valves.

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SCHEMATIC DIAGRAM:

Reciprocating pump

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TABULARCOLUMN:

S.NO Pressure

gauge

reading

Pd

(Kg/cm2)

Vacuu

m

gauge

reading

mm of

Hg(Ps)

Time

for 3

rev of

Energy

meter

(te)sec

Time for

10 cm rise

in

collecting

tank

(t)sec

Speed

NP

rpm

Discharge

(Q)

m3/sec

Input

Power

Pi

Output

Power

Po

η%

`

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CALCULATIONS:

Stroke length of the pump (L) = 0.045 m

Bore (d) = 0.04m

Piston area (a) = (π/4) × (0.04)2

Area of the collecting tank (A) = 50 X 50 cm2

NP = speed of mortar in rpm

To find the percentage of slip = Qt – Qa

Qt× 100

Qt = theoretical discharge = 2 L×a×Np

60 m/sec

Qa = Actual discharge = Q= 𝑨×𝒉

𝒕 m/sec

A = Area of the collecting tank

t = time for (h) rise in water level.

To find the overall efficiency of the pump = Po/Pi

The input power Pi = 3600 × N

E × te Kw

Where

N = Number of revolutions of energy meter disc

E = Energy meter constant = 1600 (rev / Kw hr)

T = time taken for ‘Nr’ revolutions (seconds)

Output power Po = ρ × g × Q × H

1000 Kw

Where,

ρ = Density of water = 1000 (kg / m³)

g = Acceleration due to gravity = 9.81 (m / s2)

H = Total head of water (m)

H = suction head (Hs) + delivery Head (Hd) + Datum Head

Where Hd = delivery head = Pd/ρ kg/cm2

Hs = suction head= 𝑃𝑠×13600

𝜌×1000 mm of Hg

Z= datum level difference = 2.8 m

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

GRAPH:

1. Actual discharge Vs Total head

2. Actual discharge Vs Efficiency

3. Actual discharge Vs Input power

4. Actual discharge Vs Output power

RESULT:

PRECAUTIONS:

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

FRANCIS TURBINE

AIM:

To study the characteristic of a Francis turbine.

APPARATUS:

1. Francis turbine test setup

2. Tacho meter

3. stop watch

STARTING UP:

Modern Francis turbine in an inward mixed flow reaction turbine it is a medium

head turbine. Hence it required medium quantity of water. The water under pressure from

the penstock enters the squirrel casing. The casing completely surrounds the series of

fixed vanes. The guides’ vanes direct the water on to the runner. The water enters the

runner of the turbine in the dial direction at outlet and leaves in the axial direction at the

inlet of the runner. Thus it is a mixed flow turbine.

The unit essentially consists of a spiral casing and rotor assembly with runner

shaft and break drum, all mounted on suitable study base frame. An elbow fitted draft

tube is provided for the purpose of regaining the kinetic energy from the exit water and

also facilitating easy accessibility of the turbine due to its location at higher level than the

tail race. A transparent hollow Perspex cylinder is provided in between the draft bend and

the casing for the purpose of observation of flow at exit of runner. A rope brake

arrangement is provided to load the turbine. The output of the turbine can be controlled

by adjusting the guide vanes for which a hand wheel and a suitable link mechanism is

provided. The net supply head on the turbine is measured by a pressure and vacuum

gauge.

Make sure that before starting the pipeline are free from foreign matter. Also note

whether all the joints are watertight and leak proof. Prime the pump and start it with

closed gate valve. The guide vane in the turbine should also be in the closed position

while starting the pump. See that all the ball bearings and bush bearings in the units are

properly lubricated.

Then slowly open the gate valve situated above the turbine and open the cock fitted to the

pressure gauge and see that the pump develops the rated head. If the pump develops the

required head, slowly open the turbine guide vanes by rotating the hand wheel (which

operates the guide vanes through suitable link mechanism) until the turbine attains the

normal rated speed. Run the turbine at the normal speed (1250 rpm) for about one hour

and carefully note the following:

1. Operation of the bearings, temperature rise, noise etc.

2. Vibration of the unit.

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SPECIFICATION:

1. Spiral casing: made of cast iron with smooth inner surface.

2. Runner: made of gunmetal casting designed for efficient operation.

Accurately machined and smoothly finished.

3. Guide vane :consists of guide vanes rotating in gunmetal bushes

Mechanism operated by hand wheel through a link mechanism.

4. Shaft : stainless steel accurately machined

5. Bearing: one number ball bearing and one number taper roller bearing.

6. Draft tube bend: provided at the exit of the runner with a transparent

cylindrical window for observation of flow past the runner to the bend is

connected a draft tube of mild steel fabrication.

7. Brake arrangement: consists of a machined and polished cast iron brake drum,

cooling water pipes, internal water scoop discharge pipe, spring balances, screw

rod, and belt brake arrangement.

PROCEDURE:

1. keep the guide vane in full open position and slowly open the gate valve so that

the turbine runs at 1200 rpm

2. apply load supply by adjusting the screw rod connected to the belt, for loading

the brake drum and tighten the locknut

3. the speed gets reduced open the gate valve to an extent the speed is 1200 rpm

4. note the spring balance reading, pressure gauge reading & manometer reading

5. Repeat the above procedure for different loads & for different guide vane

opening.

SUPPLY PUMP:

1. Rated head : 20 m

2. Discharge : 2000lpm

3. normal speed : 1440 rpm

4. Power required : 15hp (11.2 Kw)

5. Size of pump : 100×100 mm

6. Type : centrifugal high speed single suction volute.

FRANCIS TURBINE:

1. rated supply head : 15.0 m

2. discharge : 2000lpm

3. rated speed : 1250 rpm

4. unit speed : 51.5 rpm

5. specific speed : 95.5 rpm

6. runner diameter : 150 mm

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7. no. of guide vanes : 8

8. brake drum diameter : 300 mm

FLOW MEASURING UNIT:

Size of venturi meter 100 mm

Throat diameter for venturi meter 60 mm

Manometer Double column deferential type.

TABULARCOLUMN:

S.no Gate

opening

Pressure

gauge

Vacuum

pressure

Manometer

reading

Speed of

rotation

Spring

balance

Pi Po η%

h1 h2 T1 T2

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

SCHEMATIC DIAGRAM:

Francis turbine

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CALCULATIONS:

Input power (Pi) = (ρ×g×Q×h) watts

Flow rate of water Q = Cd ×𝑎1×𝑎2× 2𝑔𝐻

𝑎12−𝑎22 m

3/sec

d1 = dia. Of venture inlet = 100 mm

d2 = dia. Of venture throught = 60 mm

Cd = coefficient of discharge of venturimeter = 0.9

Where a1 = area of inlet of the venturimeter.

a2 = area of the venturimeter throat.

H = h1-h2 [𝑠1

𝑠2-1]

h = Total head of water (m)

h = suction head (hs) + delivery Head (hd) + Datum Head

Where hd = delivery head = Pd/ρ kg/cm2

hs = suction head= 𝑃𝑠×13600

𝜌×1000 mm of Hg

Output power (Po) = 2π×N×T

60 watts

T = (T1-T2) ×g × radi. Of break drum

radi. Of break drum = 0.15m

N = speed in tacho meter

Efficiency of the turbine ηm%= Po/Pi

Electrical efficiency = ηe% = po / Pi

po= electrical output = V × I watts

GRAPHS:

1. speed vs. output power

2. speed vs. efficiency

RESULT:

PRECAUTIONS:

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

PELTON WHEEL TURBINE

AIM:

To conduct load test on Pelton wheel turbine and to study the characteristics of

pelton wheel turbine.

APPARATUS REQUIRED:

Venturi meter, Stopwatch, Tachometer, Dead weight.

DESCRIPTION:

Pelton wheel turbine is an impulse turbine, which is used to act on high loads and

for generating electricity. All the available heads are classified in to velocity energy by

means of spear and nozzle arrangement. Position of the jet strikes the knife-edge of the

buckets with least relative resistances and shocks. While passing along the buckets the

velocity of the water is reduced and hence an impulse force is supplied to the cups which

in turn are moved and hence shaft is rotated.

PROCEDURE:

1. The Pelton wheel turbine is started.

2. All the weight in the hanger is removed.

3. The pressure gauge reading is noted down and it is to be maintained constant for

different loads.

4. The Venturi meter readings are noted down.

5. The spring balance reading and speed of the turbine are also noted down.

6. A 5Kg load is put on the hanger, similarly all the corresponding readings are

noted down.

7. The experiment is repeated

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SCHEMATIC DIAGRAM:

Pelton Wheel Turbine

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

TABULARCOLUMN:

S.no Gate

opening

Pressure

gauge

Vacuum

pressure

Manometer

reading

Speed of

rotation

Spring

balance

Pi Po η%

h1 h2 T1 T2

Electrical output

Load (kw)

Voltage

V

Current (I) A Speed(N)

rpm

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CALCULATIONS:

Input power (Pi) = (ρ×g×Q×h) W

Flow rate of water Q = Cd 𝑎1×𝑎2× 2𝑔𝐻

𝑎12−𝑎22 m

3/sec

d1 = dia. Of venture inlet = 65 mm

d2 = dia. Of venture throught = 39 mm

Cd = coefficient of discharge of venturimeter = 0.9

Where a1 = area of inlet of the venturimeter.

a2 = area of the venturimeter throat.

H= h1-h2 [𝑠1

𝑠2-1]

h = Total head of water (m)

h = suction head (hs) + delivery Head (hd) + Datum Head

Where hd = delivery head = Pd/ρ kg/cm2

hs = suction head= 𝑃𝑠×13600

𝜌×1000 mm of Hg

Output power (Po) = 2π×N×T

60 watts

T = (T1-T2) ×g × radi. Of break drum

radi. Of break drum = 0.15m

N = speed in tacho meter

Efficiency of the turbine ηm%= Po/Pi

Electrical efficiency = ηe% = po / Pi

po= electrical output = V × I watts

GRAPHS:

1. speed vs. efficiency

2. speed vs. power input

3. speed vs. power out put

RESULT:

PRECAUTIONS:

MECHANICS OF FLUIDS AND HYDRAULIC MACHNIES educateindiaprashanth.wordpress.com

M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

KAPLAN TURBINE

AIM:

To study the characteristics of a Kaplan turbine

APPARATUS REQUIRED:

Venturimeter, Stopwatch, Tachometer, Dead weight.

DESCRIPTION:

Kaplan turbine is an axial flow reaction turbine used in dams and reservoirs of low

height to convert hydraulic energy into mechanical and electrical energy. They are best

suited for low heads say from 10m to 5 m. the specific speed ranges from 200 to 1000

The flow through the pipelines into the turbine is measured with the office meter fitted in

the pipeline. A mercury manometer is used to measure the pressure difference across the

orifice meter. The net pressure difference across the turbine output torque is measured

with a pressure gauge and vacuum gauge. The turbine output torque is determined with

the rope brake drum. A tachometer is used to measure the rpm.

PROCEDURE:

1. Keep the runner vane at require opening

2. Keep the guide vanes at required opening

3. Prime the pump if necessary

4. Close the main sluice valve and they start the pump.

5. Open the sluice valve for the required discharge when the pump motor switches

from star to delta mode.

6. Load the turbine by adding weights in the weight hanger. Open the brake drum

cooling water gate valve for cooling the brake drum.

7. Measure the turbine rpm with tachometer

8. Note the pressure gauge and vacuum gauge readings

9. Note the orifice meter pressure readings.

Repeat the experiments for other loads

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

SCHEMATIC DIAGRAM:

Kaplan turbine

MECHANICS OF FLUIDS AND HYDRAULIC MACHNIES educateindiaprashanth.wordpress.com

M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

TABULARCOLUMN:

S.no Gate

opening

Pressure

gauge

Vacuum

pressure

Manometer

reading

Speed of

rotation

Spring

balance

Pi Po η%

h1 h2 T1 T2

Electrical output

Load (kw)

Voltage

V

Current (I) A Speed(N)

rpm

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CALCULATIONS:

Input power (Pi) = (ρ×g×Q×h) kW

Flow rate of water Q = Cd 𝑎1×𝑎2× 2𝑔𝐻

𝑎12−𝑎22 m

3/sec

d1 = dia. Of venture inlet = 0.13 m

d2 = dia. Of venture throught = 0.078 m

Cd = coefficient of discharge of venturimeter = 0.9

Where a1 = area of inlet of the venturimeter.

a2 = area of the venturimeter throat.

H= h1-h2 [𝑠1

𝑠2-1]

h = Total head of water (m)

h = suction head (hs) + delivery Head (hd) + Datum Head

Where hd = delivery head = Pd/ρ kg/cm2

hs = suction head= 𝑃𝑠×13600

𝜌×1000 mm of Hg

Output power (Po) = 2π×N×T

60 watts

T = (T1-T2) ×g × radi. Of break drum

radi. Of break drum = 0.15 m

N = speed in tacho meter

Efficiency of the turbine η%= Po/Pi

Electrical efficiency = ηe% = po / Pi

po= electrical output = V × I watts

GRAPHS:

1. speed vs. efficiency

2. speed vs. power input

3. speed vs. power out put

RESULT:

PRECAUTIONS:

MECHANICS OF FLUIDS AND HYDRAULIC MACHNIES educateindiaprashanth.wordpress.com

M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

MULTI STAGE CENTRIFUGAL PUMP

AIM:

To study the performance characteristics of a centrifugal pump

APPARATUS:

1. centrifugal pump test setup

2. stop watch

DESCRIPTION:

A centrifugal pump with a single impeller that can develop a differential pressure

of more than 150 psi between the suction and the discharge is difficult and costly to

design and construct. A more economical approach to developing high pressures with a

single centrifugal pump is to include multiple impellers on a common shaft within the

same pump casing. Internal channels in the pump casing route the discharge of one

impeller to the suction of another impeller. The water enters the pump from the top left

and passes through each of the four impellers in series, going from left to right. The water

goes from the volute surrounding the discharge of one impeller to the suction of the next

impeller.

A pump stage is defined as that portion of a centrifugal pump consisting of one impeller

and its associated components. Most centrifugal pumps are single-stage pumps,

containing only one impeller. A pump containing seven impellers within a single casing

would be referred to as a seven-stage pump or, or generally, as a multi-stage pump.

PROCEDURE:

1. Prime the pump, close the delivery valve and switch on the unit.

2. Open the delivery valve and maintain required head. Note the reading of pressure

gauge.

3. Note the corresponding reading from vacuum gauge.

4. Measure the area of the collecting tank

5. Close the drain valve and note the time for 10 cm rise of water level in the

collecting tank.

6. For different delivery head repeat the experiment.

7. For every set of reading note the time taken for 3 rev. Energy meter.

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SCHEMATIC DIAGRAM:

Centrifugal pump

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

TABULARCOLUMN:

S.

N

O

Pressure

readings

Pressure

gauge

reading

Pd

(Kg/cm2)

Vacuum

gauge

reading

mm of

Hg(Ps)

Time for 3

rev of

Energy

meter

seconds

(te)

Discharge

(Q)

m3/sec

Input

Power

Pi

Output

Power

Po

η% h1 h2

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M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI

CALCULATIONS:

Flow rate of water Q = Cd 𝑎1×𝑎2× 2𝑔𝐻

𝑎12−𝑎22 m

3/sec

d1 = dia. Of venture inlet = 65 mm

d2 = dia. Of venture throught = 39 mm

Cd = coefficient of discharge of venturimeter = 0.9

Where a1 = area of inlet of the venturimeter.

a2 = area of the venturimeter throat.

H= h1-h2 [𝑠1

𝑠2-1]

h = Total head of water (m)

h = suction head (hs) + delivery Head (hd) + Datum Head

Where hd = delivery head = Pd/ρ kg/cm2

hs = suction head= 𝑃𝑠×13600

𝜌×1000 mm of Hg

1. The work done by the pump is given by Po = ρ × g × Q × H

1000 Kw

Where,

ρ = Density of water (kg / m³)

g = Acceleration due to gravity (m / s2)

H = Total head of water (m)

2. The input power Pi = 3600 × N

E × te Kw

Where

N = Number of revolutions of energy meter disc

E = Energy meter constant = 150 (rev / Kw hr)

T = time taken for ‘Nr’ revolutions (seconds)

3. The efficiency of the pump = (Po/ Pi) ×100 %

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GRAPH:

1. Actual discharge Vs Total head

2. Actual discharge Vs Efficiency

3. Actual discharge Vs Input power

4. Actual discharge Vs Output power

RESULT:

PRECAUTIONS:

MECHANICS OF FLUIDS AND HYDRAULIC MACHNIES educateindiaprashanth.wordpress.com

M PRASHANTH REDDY [email protected] RAM PRASHANTH REDDY MAMIDIPALLI