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Department of Petroleum Engineering NRIH

Department

Of

PETROLEUM ENGINEERING

Lab Manual for

BASIC ENGINEERING MECH

LABORATORY (II-Year PET I-Sem)

Prepared by

Mr. D.U.M.Manikanta M.E

Asst. Professor


LIST OF EXPERIMENTS

S.NO:

NAME OF THE EXPERIMENT

PAGE NO:

1.

To Calibrate pressure gauge using standard pressure and weights

2.

Valve timing diagram of a 4-stroke diesel engine

3. Port timing diagram of a 2-stroke petrol engine

4.

Perform load test at full load, half load,1/4th

load on a 4-stroke Ruston

engine and draw the performance curves

5.

Find the volumetric efficiency and isothermal efficiency of the given

compressor

6.

To determine the moment of inertia of a fly wheel and shaft

experimentally and compare the values with the calculated values

7.

To determine the modulus of rigidity of the material of the wire by

torsional oscillators

8.

Brinnels hardness test


1. TO CALIBRATE PRESSURE GAUGE USING STANDARD

PRESSURE AND STANDARD WEIGHTS.

Aim: To calibrate pressure gauge using standard pressure and standard weights.

APPARATUS: Bourdon gauge, dead weight tester.

DIAGRAM:


Dead weight tester and Bourdon tube pressure gauge

STEPWISE PROCEDURE:

Theory: Pressure gauge, especially Bourdon’s gauge is calibrated by means of dead weight tester. The

essential components of such a tester is reservoir ‘R’, cylinder ‘C’, barrel ‘B’ and passage up to ‘C’ to

hold up clean dry oil. A spindle ‘S’ with highly finished surface and precise cross-sectional area slides

vertically in the barrel ‘B’ through close fitting, highly polished bearing ‘b’ and carries the table ‘T’ at

its upper end. Its lower end rests on ‘HP’ piston. Screws ‘Ls1’ and ‘Ls2’ lock the passage of oil when

required. The tester is mounted on a stand ‘ST’. It is provided with special precise weights marked in

terms of pressure. A double piston ‘DP’ can be moved forward and backward by rotating the handle ‘H’.

The double acting piston works both ways and does not allow the oil to leak. The gauge under

calibration can be connected at ‘D’ such that the connection is leak tight. With ‘Ls1’ and ‘Ls2’ open, the

handle is rotated such that the oil is just in level with the gauge connecting points ‘D’, and ‘Ls2’ is now

locked. The gauge ‘G’ is mounted carefully. ‘H’ is rotated until the table ‘T’ is at raised position in line

with the upper edge of colour band. ‘Ls1’ is then locked. ‘Ls2’is opened and the handle H is rotated such

that the gauge needle just moves and reads some minimum pressure characteristics of the tester because

of the weight of the unloaded table acting through the piston’ LP’. A weight is placed on the table

increasing the pressure on the oil in the tester. The gauge pressure reading should give a reading

corresponding to the amount scribed on the weight if it is operating correctly. If not, the dial is rotated so

that the needle points to be correct pressure. Another weight is added and another gauge reading is noted

and so on. A combination of weights can be used with thinner piston ‘HP’, with a multiplying factor

given by the manufacturer, Say 20. Such a tester can give pressure values accurate up to + 0.05% of the

pressure being measured. The range of pressure is typically 0.5 to 10000 kg / cm2 using dual spindles.

The pressure range that dead weight tester can measure is limited by the area of cross –section of the

spindle S and the number of weights that can be safely placed on the table. In order to increase the range,

another spindle of smaller cross section can be provided increasing the pressure range for the same

weight placed on the table by a factor equal to the ratio of the cross-sectional area of the two spindles.

The weights are typically for 0.05,0.1,0.5,1,2,5,9,10 kg / cm2 .With thinner spindle and factor of 20these

can give 1,2,10,20,40,100,130,200 kg / cm2.


1. Assemble Bourdon’s pressure gauge with dead weight tester.

2. Apply pressure in terms of weights on table and measure the value of it on pressure gauge.

3. Repeat the procedure for ascending and descending weights on table and corresponding

measurement on Bourdon pressure gauge.

4. Plot the graph of applied weights (x-axis) versus pressure on gauge (y-axis)

OBSERVATIONS:

S.No Applied weight Measured Pressure on

Gauge


2. VALVE TIMING DIAGRAM OF A 4-STROKE DIESEL ENGINE

Aim: To draw the Valve Timing Diagram for given 4-stroke engine and calculate different periods.

Apparatus: Scale, Thread and Chalk

Engine Details: 4-Stroke, Single Cylinder, Constant Speed, Water-Cooled, Vertical Diesel

Engine, 5 BHP, 1500 rpm.

Theory & Definitions: A 4-stroke diesel engine works on diesel cycle which involve suction stroke,

compression stroke, power stroke (expansion), and exhaust stroke. During the suction stroke fresh air is

drawn into the cylinder by moment of piston from TDC to BDC. During this period the exhaust valve

should be closed theoretically as the piston reaches BDC the suction stroke is completed and air drawn

must be compressed during the moment of the piston from BDC to TDC. The air in cylinder must under

go compression. Both inlet valve and exhaust valve will remains closed during compression stroke diesel

fuel is injected into the cylinder nearly at the end of the compression stroke and it will get ingoted due to

high pressure and temperature in the cylinder. The piston starts moving from TDC to BDC due to high

pressure inside the cylinder resulting in power stroke. During the power stroke both valves are closed.

The expanded gases must be explained from a cylinder after completion of expansion stroke so that fresh

air can be drawn suction stroke. During the momentum of piston from BDC to TDC the exhaust valve is

kept open when inlet valve is kept closed so that gas will escaped from exhaust valve. The moment from

BDC to TDC is called are exhaust stroke, suction stroke the cycle is repeated.

Lead: Valve is said to be given ‘Lead’ when it opens before the piston has recharged the dead center.

Lag: Valve is said to have lag when it closes after the piston have recharged dead center.

Overlap: Overlap is a period during which both Inlet and Exhaust Valves are OPEN.

Inlet Valve Period: Inlet valve opens before top dead center (TDC) and closed after BDC with reference

to vertical engine. The reason for giving lead to inlet valve is to give sufficient time for full opening of

the valve and avoid throttling of incoming air. Also the depression in the cylinders at the end of the

exhaust stroke caused by the momentum of outgoing gases assist the fresh charge to be drawn into the

cylinder occupying its inertia. But the valves can’t be opened to early. For the same period, known as

own lap both inlet and exhaust valve are open that in sequencing the cylinder of burnt gases. The piston

moves from TDC to BDC to facilitate drawing of air into cylinder however the inlet valve is not closed

at BDC, but after piston crosses BDC and I moving up there is a log in the closing of the inlet valve. The

inlet valve is closed after BDC the log is given about 30-40 degrees. This done so as to induce as large a

fresh charge as possible due to high speed of piston. The air does not keep place with the speed of the

piston and if the inlet valve is closed at BDC the cylinder would not be completely filled with fresh air.

Advantage is taken of the high momentum of air due to which the suction continues an after piston richly

BDC. Therefore closing the inlet valve a little later then BDC induces more charge inlet the cylinder as

piston is relation stationary and crank can swing through a wide angle with little motion of the piston.

The compression will take place inside the cylinder when both valves are closed.

Exhaust Valve Period: After the power stroke the combustion gases are to be exhausted. The exhaust

valve opens with a lead of 40-45 degrees i.e., before piston reaches BDC in power stroke this will

facilitate escape of large qualities of exhaust gases to learn the cylinder even before BDC is reaches

momentum of piston from BDC to TDC leads to pushing out of gases from cylinder. The exhaust valve

is not closed at end of the exhaust stroke but closes 15 degrees of crank angle after TDC. The only

opening of exhaust valve is associated will loss of power due to the shunting of power stroke but is

compensated by better sequencing and less negative work is expelling the exhaust gases. Even after the

piston has reached TDC some exhaust gases left in combustion chamber and allowed to escape under

influence of momentum of incoming air.


In a four-stroke engine opening and closing of valves and fuel injection do not take place exactly at the

end of dead centre positions. The valves open slightly earlier and close after that respective dead centre

position. The injection (Compression ignition) also occurs prior to the full compression and piston

reaches the dead centre position. All the valves operated at some degree on either side in terms of crank

angles from dead centre position.

Procedure:

1. Rotate flywheel freely by hand.

2. Now while rotating observe the piston at Top Dead Centre (TDC) and mark with chalk on

flywheel.

3. Similarly by rotating, mark the position of Bottom Dead Centre (BDC).

4. It is to be observed that it takes 2 rotation of flywheel to complete one cycle of operation.

5. Now identify inlet and exhaust valves.

6. Find out direction of rotation of flywheel (crank shaft).

7. Bring flywheel to TDC position (pointer).

8. Go on rotating flywheel slowly and observe the position (functioning) of all valves.

9. Now observe when inlet valves opens mark it on flywheel Inlet Valve Open (IVO).

10. Slowly rotate flywheel, and observe when Inlet Valve Close (IVC).

11. Rotate further observe when Exhaust Valve Opens (EVO).

12. Rotate further & observe when Exhaust Valve Closes (EVC).

13. Same time note down IVO & mark all these on flywheel.

14. With small thread & scale find out circumference of flywheel.

15. With marking of IVO, IVC, EVO & EVC find out lengths with thread & scale.

16. Then draw spiral diagram with data in marking on flywheel.

Tabular Column:

Sl. No

Valve

Event

Distance in ‘cm’ from nearest Dead Center

Crank

Angle θ

(degree) Before

TDC

After

TDC

Before

BDC

After

BDC

1 IVO

2 IVC

3 EVO

4 EVC


Formulae & Calculations:

(1) Let, Arc length = X cm

(2) Angle ‘θ‘ = 360 x X

---------------

2 x π x R Where,

R - Radius of flywheel in ‘cm’

P - Perimeter of flywheel in ‘cm’

(3) Angle of Overlap (θ) = IVO angle + EVC angle

Precautions:

1. The valve opening should be taken as a point where first begins to open.

2. The valve closing should be taken as a point where close completely.

3. The fly wheel should be rotated in proper direction.

Result: The timing of Valve Opening and Closing for the given 4-stroke engine are determined and

Valve Timing Diagram has been drawn.


3. PORT TIMING DIAGRAM OF A 2-STROKE PETROL ENGINE

Aim: To find out the timing of the inlet port and exhaust port operation of the given petrol engine and

to represent the result through a diagram.

Apparatus: Scale, Thread and Chalk

Engine Details: 2-Stroke, Single Cylinder, Constant Speed, Air-Cooled, Horizontal Petrol

Engine, 3 BHP, 1500 rpm.

Theory: The timing of the sequence of two stroke petrol engine is represented graphically. The events

such as opening and closing of inlet ports, transfer ports and exhaust port are shown graphically with

respect to cram angles from dead centre positions. This diagram is known as port timing diagram.

The inlet port is uncovered by the piston 450 – 55

0 before the top dead centre position. The inlet

port is covered 450 – 55

0 after the top dead centre position. The exhaust port is uncovered and covered

650 and 75

0 before and after bottom dead centre respectively. The transfer port is uncovered and covered

550 and 65

0 before and after bottom dead centre respectively. Ignition occurs 15

0 – 25

0 before top dead

centre.

Procedure: Remove the port covers, if necessary to see the ports. Throughout the experiments the rotation of

the flywheel has to be in one direction – Clockwise or Anti-clockwise direction. Mark the fixed or

reference point on the frame or note the pointer attached to the frame. Rotate the flywheel and before the

piston reaches the top dead centre coincide the piston top or one of the piston ring edges with the exhaust

port top edge. Have a mark on the flywheel with respect to the fixed point (Say TDC 1). Rotate the

flywheel again and the piston moves towards the TDC and towards the BDC. When the piston or one of

the piston ring edge again coincides with the same exhaust port edge, mark this point on the flywheel

with respect to the fixed point (Say TDC 2).

INLET PORT (CRANK CASE COMPRESSION ON PETROL ENGINE):

1. When the piston just opens the inlet port, mark a point on the flywheel with respect to the

fixed point (TPO.)

2. When the piston completely closes the exhaust port, mark a point on the flywheel with

respect to the fixed point (EPC).

Measure the circumference of the flywheel, measure the peripheral length from TDC 1 to TDC 2

along the direction of rotation. Take half of this timing length and mark a line from TDC 1 along the

directions of rotation. Indicate the line as TDC – BDC line.

Measure the timing length from TDC to IPO and IPC. Measure the timing length from BDC to

TPO, TPC, EPO and EPC. Tabulate the readings as below:

Tabular Column:

S.NO PORT

EVENT

Distance in ‘cm’ from Crank

angle Before

ODC

After

ODC

Before

ODC

After

ODC

1) EPO

2) EPC

3) TPO

4) TPC


Port Timing Diagram of Single Cylinder 2-Stroke Petrol Engine:

Precautions:

1. Lubricate all the parts for smooth operation before doing the experiments.

2. Note the correct the direction of the crank shaft and mark the direction of rotation of flywheel.

3. Rotate the crank shaft always in the correct direction.

4. The valve opening should be taken as a point where first begins to open.

5. The valve closing should be taken as a point where close completely.

6. The fly wheel should be rotated in proper direction.

Result: The timing of Valve Opening and Closing for the given 4-stroke engine are determined and

Valve Timing Diagram has been drawn.


4. PERFORM LOAD TEST AT FULL LOAD, HALF LOAD,1/4TH

LOAD ON A 4-STROKE RUSTON ENGINE AND DRAW THE

PERFORMANCE CURVES

Aim: To conduct performance test on single cylinder, vertical, water –cooled diesel engine and hence to

determine frictional power and draw the performance characteristic curves.

Apparatus: Single cylinder diesel engine test rig coupled with rope brake dynamometer, stop watch.

Engine Specification: Type : 4-Stroke Diesel Engine (Water cooled)

Make : Field-Marshal

Bore : 85 mm

Stroke : 110mm

Speed : 1500 rpm

Output : 5HP

Compression Ratio : 16.5 : 1

Brake Drum Radius : 0.185 m

Orifice Diameter : 15MM

Specific Gravity of H.S.D.OIL : 0.85 gm/ml

Calorific Value : 10,000 K cal/kg

Description: The water-cooled single cylinder diesel engine is coupled with a rope brake dynamometer. Separate

cooling lines are provided for the drum and the engine. Thermocouples are arranged for sensing the

temperature of cooling water consisting of fuel tank mounted on stand, burette with 3-way cock

arrangement is provided.

Theory: Load test is conducted to study the performance characteristics of the engine. The single cylinder diesel

engine is run at a constant speed of 1500 rpm. The engine is loaded in steps of constant interval loads i.e.

0kg, 2kgs, 4kgs etc. At each load fuel consumed is determined. The output of the engine is calculated as

follow.

BP = (П W D N)/60000 kW , W=( W1 –W2 ) Kg

A graph with BP on X- axis and fuel consumed per hour (FCH) on Y-axis is plotted. The line joining the

all data points when extended back, it intercepts the – ve X-axis. The negative intercept magnitude gives

the Frictional Power of the engine. The line connecting the data points is known as the WILAN’S LINE.

The other performance parameters like Brake Mean Effective Pressure (Bmep), indicated

thermal efficiency (ith), Brake thermal efficiency (bth), Mechanical efficiency (mech), Specific Fuel

Consumption (SFC) are determined and graphs are plotted.

Maximum load on the engine (Wmax ) can be calculated as follows

Wmax = {3.68 x 60000} / {ПDN x 9.81}

Procedure: 1. The fuel level in the tank is checked.

2. Lubricating oil level is checked.

3. The engine is started at no load condition and the time taken for 10 ml fuel consumption is noted.

4. A load of 2 kgs is applied on the engine, the spring balance reading w2 , applied load w1 , time

taken for 10 cc of fuel consumption are noted down.

5. The above procedure is repeated at different loads like 4kgs, 6kgs, ----- 15 kgs.

6. Frictional Power is obtained from the WILAN’S LINE graph.

7. The other parameters like SFC, Bmep, IP, ith, bth, mech , are calculated.

8. Graphs are plotted as given below.


i) BP Vs FCH

ii) BP VS SFC

BP VS Bmep

iii) BP VS mech

iv) BP VS bth

v) BP VS ith

Observations:

S.No. Dynamometer reading Speed

(rpm)

Time for 10 cc of

Fuel Consumption

(sec) W1 (kg) W1 (kg) W = W1 - W2

(kg)

Model Calculations:

1. Break Power:

BP = (П WDN x 9.81) / 60000 kW

2. Fuel consumption per hour (FCH):

FCH = (10 x 3600 x 0.8275) / (t x 1000) kg/hr

3. Specific Fuel Consumption:

SFC = BP

FCH kg / kW.hr

4. Indicated Power (IP):

IP = BP + FP (FP is obtained from WILAN’S LINE graph)

5. Mechanical Efficiency (mech)

mech = BP/IP %

6. Break Mean Effective Pressure:

Bmep = (60000 x BP) / (LA(N/2) x 105

x n) bars

7. Indicated Thermal Efficiency

ith = (IP x 3600 x 100) / (FCH x CV) %

8. Break Thermal Efficiency:

bth = (BP x 3600 x 100) / (FCH x CV) %

Where

i) IP and BP are in kilo watts

ii) CV- calorific value of the fuel in kJ/kg

Result Table:

Speed N = 1500 rpm

S.No Load

(kg)

BP

(KW)

FCH

(KG/hr)

FP

(Kw)

IP

(Kw)

SFC

(Kg/kw-

hr)

Bmep

Bar mech

%

bth

%

ith

%


Precautions: i) The engine should be started and stopped at No Load condition.

ii) Cooling water supply must be ensured throughout the experiment.

iii) The readings should be noted without parallax error.

iv) Lubricant oil level to be checked.

Review Questions: 1. Define mean effective pressure

2. Briefly discuss the various efficiency terms associated with an engine

3. Mention the basic aspects covered by the engine performance

4. What are the methods available for improving the performance of an engine

5. List the types of exhaust temperatures measured

Trouble Shooting: 1. Engine will not start due to air lock in the fuel system-

i) Open the bleed- off valve and release the air lock.

2. Engine will not start due to diesel filter choked –

i) Remove the filter and clean it.

3. Engine will not start if the holding bolts are loose –

i) Tighten the bolts so that required injecting pressure occurs.

4. Abnormal noise -

i) Check the engine jacket cooling system.

ii) Check the bearings condition.

iii) Check the level and condition of lubricating oil / lubricating filter.

Inference: Brake Thermal efficiency around 25%

Indicated Thermal efficiency around 35%

Friction Power loss around 16%

Mechanical efficiency around 75%

Specific Fuel Consumption for diesel engine is around 0.834

Applications: Understanding of speed Vs Load

Diesel consumption Vs Load per unit time


5. FIND THE VOLUMETRIC EFFICIENCY AND ISOTHERMAL

EFFICIENCY OF THE GIVEN COMPRESSOR

AIM:

To conduct performance test on two stage air compressor and to determine the volumetric & isothermal

efficiency at various discharge pressure.

Equipment Data:

Diameter of low pressure cylinder = 70 mm

Diameter of high pressure cylinder = 60 mm

Length of stroke = 69 mm

Orifice Diameter = 10.00 mm

Coefficient of discharge of orifice = 0.65

Area of Orifice = 1.1304 x 10-4

m2

THEORY:

An air compressor is a machine which takes the air (ambient pressure and temperature condition) from

the atmosphere during its suction stroke and compresses that air to high pressure with the help of a

piston and cylinder arrangement, is known as Reciprocating air compressor. When this process carried

out with more than one cylinder then such type of compressor is called Multi stage air compressor.

DESCRIPTION:

Two stage air compressors is a reciprocating type, driven by a prime moveri.e a 3 phase AC motor

through belt. The test rig consists of a base on which the tank is mounted. The pressure & temperature of

the air at different points are measured by pressure gauges and thermo couple with digital temperature

indicator respectively. An electrical pressure safety valve is provided as additional safety .The suction is

connected to an air tank with a calibrated orifice through the water manometer to measure pressure head

of air. The input of the motor can be measured by an energy meter.

OBSERVATIONS:

Manometer diff. = hw………mm

Speed of compressor = Nc…………rpm

Ambient pressure = P1………..kg/cm2

L P cyl. Pressure = P2……..kg/cm2

H P cyl. Pressure = P3………..kg/cm2

Head of air = Ha……………m

Actual volume of air = Va………..m3/min

Theoretical volume of air = Vthe………..m3/min

Volumetric efficiency =v…………..%

Output power of motor =BP………….KW

Power of compressor = BPc………..KW


PROCEDURE:

1. Connect the power supply to compressor

2. Close the outlet valve

3. Switch on the compressor & note down the readings

a. Inlet air pressure P1

b. Discharge pressure of first stage P2

c. Discharge pressure of second stage P3

d. Manometer reading

e. Also note down different temperatures.

T1 = First stage Inlet temperature.( amb.air temp.)

T2 = First Stage Outlet temperature.

T3 = Second Stage Inlet temperature. (Intercooler)

T4 = Second Stage Outlet temperature

4. Note down the energy meter No. of impulse with respect to time.

5. Repeat the experiments for different delivery pressure.

TABULAR COLUMN

SL.NO. Speed

Nc

Time

in sec

Mano

Meter

Reading

hw

P2 P3 T1 T2 T3 T4

FORMULAE:

1. Head of Air at NTP:

Diff in Manometer (hw) x Density of water (ρw)

Ha = ----------------------------------------------- ---------… m of water.

1000 x Density of Air (ρa)

Ha = (hw) X ρw……m of water.

1000 ρa

ρw = 1000 kg/m3

ρa = 1.29273 kg/m3

2. Volume of air compressed:

Va = Cd ao (2g Ha) x 3600…….. m³/hr

Where, Cd = co-efficient of discharge=0.62

ao = area of the orifice.

d = dia of orifice = 10mm.

g = gravitational constant.

3. Theoretical volume of air compressed:

Vth = d1²/ 4 x L x Nc x 60 …. m³/hr


Nc = Compressor speed

d1 = LP cyln. Dia = 60.33mm

d2 = HP cyln. Dia = 57mm

L1&L2= stroke length=62mm

4. Volumetric Efficiency:

% ηvol= Va x 100 …….%

Vth

5.Compression Ratio: P3

Rp = -----

P1

6.Isothermal power: Piso = Ma Ra Ta loge (Rp) x 0.736 …………KW

1000

Ma = Va x ρa

Va = ……m3/hr

Ra = 0.287 kJ/kg deg K

Ta = Ambient temp. in deg K

7. Isothermal Efficiency: P iso

η iso = ---------------

P1

8. Power input to the motor:

P = no of unit consumed in energy meter x 3600 …KW

Time taken

9. Power of compressor

Pc = power of the motor x transmission efficiency…..KW

Where, transmission efficiency = 80%


6. TO DETERMINE THE MOMENT OF INERTIA OF A FLY WHEEL AND SHAFT

EXPERIMENTALLY AND COMPARE THE VALUES WITH THE CALCULATED VALUES

Aim:

To determine the moment of inertia of a flywheel.

Apparatus:

Fly wheel, weight hanger, slotted weights, stop watch, meter scale.

Theory:

The flywheel consists of a heavy circular disc/massive wheel fitted with a strong axle projecting on

either side. The axle is mounted on ball bearings on two fixed supports. There is a small peg on the axle.

One end of a cord is loosely looped around the peg and its other end carries the weight-hanger.

Let "m" be the mass of the weight hanger and hanging rings (weight assembly).When the mass "m"

descends through a height "h", the loss in potential energy is

The resulting gain of kinetic energy in the

rotating flywheel assembly (flywheel and axle) is

Where

I -moment of inertia of the flywheel assembly

ω-angular velocity at the instant the weight assembly touches the ground.

The gain of kinetic energy in the descending weight assembly is,

Where v is the velocity at the instant the weight assembly touches the ground.

The work done in overcoming the friction of the bearings supporting the flywheel assembly is


Where

n - number of times the cord is wrapped around the axle

Wf - work done to overcome the frictional torque in rotating the flywheel assembly completely once

Therefore from the law of conservation of energy we get

On substituting the values we get

Now the kinetic energy of the flywheel assembly is expended in rotating N times against the same

frictional torque. Therefore

and

If r is the radius of the axle, then velocity v of the weight assembly is related to r by the equation

Substituting the values of v and Wf we get:

Now solving the above equation for I

Where, I = Moment of inertia of the flywheel assembly

N = Number of rotation of the flywheel before it stopped

m = mass of the rings

n = Number of windings of the string on the axle

g = Acceleration due to gravity of the environment.

h = Height of the weight assembly from the ground.

r = Radius of the axle.

Now we begin to count the number of rotations, N until the flywheel stops and also note the duration of

time t for N rotation. Therefore we can calculate the average angular velocity in radians per

second.

Since we are assuming that the torsional friction Wf is constant over time and angular velocity is simply

twice the average angular velocity

Procedure for doing Simulator

Choose any desired environment by clicking on the ‘combo box’.

Adjust the sliders to have suitable dimensions for flywheel arrangement.

Click on ‘Release fly wheel’ to start the experiment.


No of revolutions (N) of the flywheel, after the loop slips off from peg is indicated on the side of

axle.

The time taken by flywheel to come to rest is noted from stop watch.

Repeat the experiment for different values of variables.

Procedure for doing Real Lab

The length of the cord is carefully adjusted, so that when the weight-hanger just touches the

ground, the loop slips off the peg.

A suitable weight is placed in the weight hanger

A chalk mark is made on the rim so that it is against the pointer when the weight hanger just

touches the ground.

The other end of the cord is loosely looped around the peg keeping the weight hanger just

touching the ground.

The flywheel is given a suitable number (n) of rotation so that the cord is wound round the axle

without overlapping.

The height (h) of the weight hanger from the ground is measured.

The flywheel is released.

The weight hanger descends and the flywheel rotates.

The cord slips off from the peg when the weight hanger just touches the ground. By this time the

flywheel would have made n rotations.

A stop clock is started just when the weight hanger touches the ground.

The time taken by the flywheel to come to a stop is determined as t seconds.

The number of rotations (N) made by the flywheel during this interval is counted.

The experiment is repeated by changing the value of n and m.

From these values the moment of inertia of the flywheel is calculated using equation

.

Observations

Mean value of moment of inertia, I =.........kgm2


Result

Moment of inertia of the fly wheel =.........kgm2

7. TO DETERMINE THE MODULUS OF RIGIDITY OF THE

MATERIAL OF THE WIRE BY TORSIONAL OSCILLATORS

Aim:

To determine the rigidity modulus of the suspension wire using torsion pendulum.

Apparatus:

The given torsion pendulum, two identical cylindrical masses, stop watch, meter scale, etc.

Theory:

What is Torsional Oscillation?

A body suspended by a thread or wire which twists first in one direction and then in the reverse

direction, in the horizontal plane is called a torsional pendulum. The first torsion pendulum was

developed by Robert Leslie in 1793.

A simple schematic representation of a torsion pendulum is given below,

The period of oscillation of torsion pendulum is given as,

Where I=moment of inertia of the suspended body; C=couple/unit twist


But we have an expression for couple per unit twist C as,

Where l =length of the suspension wire; r=radius of the wire; n=rigidity modulus of the suspension

wire

Substituting (2) in (1) and squaring, we get an expression for rigidity modulus for the suspension wire

as,

We can use the above formula directly if we calculate the moment of inertia of the disc, I as (1/2)MR

2.

Now, let I0 be the moment of inertia of the disc alone and I1 & I2 be the moment of inertia of the disc

with identical masses at distances d1&d2 respectively. If I1 is the moment of inertia of each identical

mass about the vertical axis passing through its centre of gravity, then

But from equation (1) ,

Where T0,T1,T2 are the periods of torsional oscillation without identical mass, with identical pass at

position d1,d2 respectively.

Dividing equation (6) by (9) and using (5),

Therefore, The moment of inertia of the disc,

Now substituting equation (2) and (5) in (9),we get the expression for rigidity modulus 'n' as,


Applications of Torsional Pendulum:

1. The working of " Torsion pendulum clocks " ( shortly torsion clocks or pendulum clocks), is based on

torsional oscillation.

2.The freely decaying oscillation of Torsion pendulum in medium(like polymers),helps to determine

their characteristic properties.

3.New researches, promising the determination of frictional forces between solid surfaces and flowing

liquid environments using forced torsion pendulums.

Procedure: (for performing in lab)

PART 1: Determination of Rigidity modulus using Torsion pendulum alone

The radius of the suspension wire is measured using a screw gauge.

The length of the suspension wire is adjusted to suitable values like 0.3m,0.4m,0.5m,.....0.9m,1m

etc.

The disc is set in oscillation. Find the time for 20 oscillations twice and determine the mean

period of oscillation ' T0 '.

Calculate moment of inertia of the disc using the expression, I = (1/2)MR2

.

Determine the rigidity modulus from the given mathematical expression.

PART 2: Determination of rigidity modulus and moment of inertia using torsion pendulum

with identical masses

The radius of the suspension wire is measured using a screw gauge.

The length of the suspension wire is adjusted to suitable values like 0.3m,0.4m,0.5m,.....0.9m,1m

etc.

The disc is set in oscillation. Find the time for 20 oscillations twice and determine the mean

period of oscillation ' T0 '.

The two identical masses are placed symmetrically on either side of the suspension wire as close

as possible to the centre of the disc, and measure d1 which is the distance between the centers of

the disc and one of the identical masses.

Find the time for 20 oscillations twice and determine the mean period of oscillation ' T1 '.

The two identical masses are placed symmetrically on either side of the suspension wire as far as

possible to the centre of the disc, and measure d2 which is the distance between the centers of the

disc and one of the identical masses.

Find the time for 20 oscillations twice and determine the mean period of oscillation ' T2 '.

Find the moment of inertia of the disc and rigidity modulus of the suspension wire using the

given formulae.

Observations:

Length of the suspension wire=................m

Radius of the suspension wire=..............m

Mass of each identical masses=.............kg


d1=...........m

d2=...........m

For Part 1:

For Part 2:


Calculations:

T0 = ........s

T1 = .........s

T2 = .........s

For Part 1:

For Part2:

Moment of inertia of the given disc,

And the rigidity modulus of the suspension wire,


8. BRINNELS HARDNESS TEST

AIM: To measure the Brinnel hardness number for the given material.

APPARATUS: Brinnel’s hardness testing machine with accessories, emery paper,

microscope, specimen.

THEORY:

Hardness is the property exhibited by a material. It can be defined as the property of a

material by virtue of which it resists scratch, wear, abrasion or indentation.

DESCRIPTION:

For a number of engineering materials which are subjected to friction such as steel,

cast iron etc. it is necessary to find out their resistance to wear and tear (hardness). Hardness

of a surface can be increased by heat treatment or by chemical treatment and finding out the

hardness can check the efficiency of the process. The Brinnel’s hardness test is carried out by

forcing a hardened steel ball of diameter D under a load of P into a test specimen and

measuring the mean diameter d of the indentation left on the surface after removal of the load.

Normally for hard materials a ball of 10 mm diameter should be used. For soft material 5mm,

2.5mm, 2mm and 1mm are to be used depending upon the softness of the surface.

The hydraulic pump applies the load required for specified time. A Brinnel

Microscope is used to measure the Indentation.


BHN =

2P

∏D[D -√( D 2 - d

2) ]

Where P is the load adjusted in the machine in N

D is the diameter of indenter and

d is the diameter of impression

PROCEDURE:

1. Polish the surface with emery paper.

2. Place the specimen on the work table and raise it by turning the elevating screw till the

small pointer on the dial reaches the set position. Now the specimen is subjected to the

preliminary the load 10 kgf

3. Adjust the diaphragm the required weight, that is, if the penetrate diameter is 25mm,

and P/D2 ratio is 30, then the load to be adjusted to 187.5 Kg. If the diameter of

penetrator is 10 mm, then the load is 30 Kg (300N). Apply the

load by operating the lever arm.

4. Wait for 30 Sec for soft materials and 15 sec for hard material so as to make the load

reach the specimen fully. Wait till the pointer stops moving.

5. Remove the specimen and measure the diameter of the indentation correct to 0.1mm

with Brinnel microscope. To do this, keep the specimen at microscope adjusted

indentation to the scale of the microscope and measure the diameter of the indentation.

6. Repeated the process to obtain at least 4 different sets of observation for the same

material.

7. Brinnel Hardness number B.H.N =

2P

∏D[D -√( D 2 - d

2)

]

In Brinnel’s Machine the surface area of the Indentation is calculated and is used as

an index of hardness of the metal.

The surface area of Indentation is dependent upon the depth of penetration. The load

applied (in kgf) divided by the spherical area of Indentation in square mm is taken as the

Brinnel’s Hardness number.


OBSERVATIONS:

Diameter of the indenter = mm

Load = kgf

TABULAR FORM

Diameter Diameter of impression

Load P B.H.N

S.No Material of indentor kgf

mm

Trail I

Trail II Average

1

2

3

4

5


CALCULATION:

B.H.N =

2P

∏D[D -√( D 2 - d

2) ]

=

RESULT:

Brinnel Hardness Number for the given material = _________ BHN

Documents

Department Of - oss.nri.edu.inoss.nri.edu.in/eLearning/R13-BTech-2-1/LAB_Manuals/... · Dead weight tester and Bourdon tube pressure gauge ... calibration can be connected at ‘D