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ANNAMALAI UNIVERSITY

DEPARTMENT OF MECHANICAL ENGINEERING

FACULTY OF ENGINEERING AND TECHNOLOGY

III SEM B.E MANUFACTURING ENGINEERING

MECHANICAL ENGINEERING LABORATORY INSTRUCTION MANUAL CUM OBSERVATION NOTEBOOK

NAME : ___________________________

ROLL No : ____________________________

SUBJECT : ____________________________

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ANNAMALAI UNIVERSITY

DEPARTMENT OF MECHANICAL ENGINEERING

IIIIII SSeemm.. BBEE MMaannuuffaaccttuurriinngg EEnnggiinneeeerriinngg 22001166--22001177..

MECHANICAL ENGG. LABORATORY

AA.. IINNTTEERRNNAALL CCOOMMBBUUSSTTIIOONN EENNGGIINNEESS LLAABB

1. Study and Performance on Kaeser Air compressor Test rig. 06

2. Load test on Kirloskar AV I engine. (Single arm type) 18

3. Load test on PSG 5HP engine. 28

4. Study and valve timing on Field marshal 8 HP engine. 38

5. Study and Port timing on Petter 3 HP engine 44

GIVEN Calorific value of Diesel : 42,000 kJ/kg. Specific Gravity of Diesel : 0.835 Density of water : 1000 kg/m3

REFERENCE

Extrapolation (or) Willan’s line method 26 The general shape of characteristics curve 27

BB.. DDYYNNAAMMIICCSS LLAABB [[VENUE: I C Engines Laboratory Annex (New building)] 1. Determination of characteristics curves of Watt governor and Hartnell governor. 47

2. Determination of mass moment of inertia of fly wheel. 53

3. Determination of mass moment of inertia of connecting rod. 58

4. Experimental verification of natural frequency of Torsional vibration of a

single rotor shaft system. 62

5. Study and experiments on cam and follower mechanism. 64

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Instructions to the students 1. Be regular and be punctual to classes

2. Come in proper uniform stipulated

3. Ensure safety to your body organs and laboratory equipment

– SAFETY FIRST DUTY NEXT

4. Read in advance the contents of the instruction manual pertaining to the experiment

due and come prepared. Understand the related basic principles.

5. Maintain separate observation and record note books for each laboratory portion of

the course wherever justified.

6. Though you work in a batch to conduct experiment, equip yourself to do

independently. This will benefit you at the time of tests and university examinations.

7. Independently do the calculations and sketching. If there is difficulty, consult your

batch mate, classmate, teacher(s) and Laboratory in-charge.

Do not attempt to simply copy down from others. You may fulfill the formalities but

you stand to loose learning and understanding

8. Obtain the signature of teacher (s) in the laboratory observation note book and record

note book then and there during class hours (with in a week subsequent to

experimentation).This will relieve the teacher (s) from giving reminder.

9. Help to maintain neatness in the laboratory.

10. Students are advised to retain the bonafide record notebook till they successfully

complete the laboratory course.

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CONTENTS

Sl No Date Name of the Experiment Staff Signature

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CONTENTS

Sl No Date Name of the Experiment Staff Signature

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Expt.No : Date : STUDY AND PERFORMANCE TEST ON KAESER AIR COMPRESSOR

(A)Study on Kaeser Air compressor

Introduction:- Air Compressors are used to raise the pressure of air with the minimum expenditure of energy. An air-compressor sucks the air from the atmosphere, compresses it and delivers the same under high pressure to a storage tank. Since the compression of air requires some work to be done on it, some form of prime mover must drive a compressor. The compressed air is used for many purposes such as for operating pneumatic drills, rivets, road drills, paint spraying, air motors and in starting and supercharging of I.C. Engines etc. It is also utilized in the operation of lifts, rams, pumps and a variety of other devices. In heavy vehicle automobile, compressed air is also used for power brakes.

Air Compressors are classified into: a) Reciprocating air compressors b) Rotary air compressors.

Classification of Reciprocating Air Compressor:

(i) Single acting compressor, (ii) Double acting compressor, (iii) Single stage compressor, (iv) Multi stage compressor

Single acting reciprocating compressor: In single acting compressor the air is compressed in the cylinder on one side of the piston. Double acting compressor: In double acting compressor the air is compressed on both sides of the piston. Single stage compressor: In single stage compressor, the air is compressed in a only one cylinder. Multi stage compressor: In multistage compressor, the air is compressed in two or more cylinders. Multi stage compression is done to achieve high pressure ratio. In a compressor when compression ratio exceeds 5, generally multistage compression is adopted. The following arrangements are generally in practice for reciprocating compressors. No. of stages Delivery press

One up to 5 bars Two 5 to 35 bar Three 35 to 85 bar

Four above 85 bars

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According to pressure range the compressors are also classified as: Fans : Pressure ratio is 1 to 1.1

Blowers : Pressure ratio is 1.1 to 4.0 Compressors : Pressure ratio is above 4

In this experiment the compressor given for study is Keaser air compressor, which is a single acting, two stage, air cooled reciprocating air compressor .The specifications of the compressor is also given below.

Some Important parts and their functions Low pressure cylinder: Air is compressed from atmospheric pressure to intermediate pressure in L.P. Cylinder. High Pressure Cylinder: Air is compressed from intermediate pressure to delivery pressure in H.P Cylinder. Inter Cooler: Air is cooled in between the two compression stages at constant pressure. After Cooler: Air is cooled after the compression is over to accommodate more air in the receiver tank. Air Filter: It filters dust particles from the air. Otherwise the dust particles will adhere the inner surface of the cylinder and thereby increases the friction between the cylinder and piston. Due to this more power loss, wear and tear will be taking place. Orifice Meter: It is used to measure the actual flow of air for compression by measuring pressure difference across the orifice using manometer. Air Stabilizing Tank: During suction stroke, the air from atmosphere is sent into the LP cylinder. During compression, air is sent to HP cylinder through inter cooler. The flow of air in the pipe line from atmosphere to the LP cylinder is not uniform (i.e. intermittent) due to the suction of the air taking place in the alternative strokes. To measure the flow rate of air, the flow must be uniform across the orifice. Otherwise the manometer reading will fluctuate. Hence an air stabilizing tank is introduced between orifice meter and LP cylinder. This stabilizes the flow of air between the air filter and stabilizing tank. While connecting the pipe line and the stabilisingtank,see, that these are connected in diametrically opposite. However, air stabilizing tank, is fitted only in the experimental air compressors to measure the flow rate of air. Safety Valves: It releases the air when the pressure of air exceeds the desired limit. Pressure Cut off Switch: It is a device used to disconnect the electrical circuit, when the pressure of air in the receiver tank reaches the desired pressure. This disconnects the circuit of no volt coil in the Star-Delta Starter; thereby it switches off the motor.

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Prime mover and Dynamometer:The prime mover used for the compressor is a trunnion type electrical motor. This motor itself is acts as dynamometer to measure the input power of the compressor. Work done on P-V diagram of compressor:

Fig.2 P-V diagram for Single stage air compressor. The P-V diagram of a single stage reciprocating air compressor with zero clearance is shown in Fig.2 The air is sucked in from the atmosphere during the suction stroke AB at pressure Pa(i.e. at atmospheric pressure). At the end of suction stroke the air is compressed polytropically during the part of its return stroke (process BC). During compression stroke the pressure and temperature of air increases and volume decreases. This happens until the pressure Pd (delivery pressure) in the cylinder is sufficient to force open the delivery valve at C after which no more compression takes place. The delivery occurs during the remainder of the return stroke CD. The work done on the air per stage is area ABCDA. In the case of air compressors, the inlet and outlet valves are operated by pressure difference only. Not by any external means.

Fig.3 P-V diagram for two stage air compressor with inter cooling.

The air is sucked in LP cylinder during the suction stroke at intake pressure Pa and Temperature Ta. After compression in the first stage from B to E it is delivered to the intercooler, at a constant pressure Pi. The air is cooled in an intercooler, at a constant pressure Pi before passing it to second stage. The process of inter cooling is represented by the line EF. The air from the intercooler is then directed to the second stage ofcompression FG to the delivery pressure Pd. Then the air delivered to the receiver tank at constant pressure Pd. This process is represented by GD in the P-V diagram. The shaded area CEFGC shows the amount of work saved due to two stage compression with intercooling per cycle.

D C

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(B)Performance test Aim: To determine the volumetric efficiency of the low pressure cylinder of Keaser Air Compressor at NTP condition, and to draw the performance characteristics curve of a compressor . Specifications: Type :

Speed :

Type of cooling :

Bore

L.P. cylinder :

H.P. cylinder :

Stroke :

Maximum pressure :

Motor output :

Free air delivered :

Precautions:

1. Before starting the experiment, the air which is already compressed if any in the reservoir is released out so that initial gauge pressure in compressor reservoir is zero/atmospheric.

2. The initial load on the motor while starting should be avoided by opening the valves provided at the top of the L.P. and H.P. cylinders.

3. The pin is inserted on the torque arm hole to prevent jerk while starting.

4. After starting the motor the pin is removed to get the correct load on the motor. Procedure:

1. The motor is started using the automatic star-delta starter by pressing the green colour button.

2. The valves provided at the top of the LP and HP cylinders, water drain cock and the air outlet valves are closed after the motor has gained its speed and attains 960 rpm. The increase in pressure of air in the receiver tank is indicated by the pressure gauge.

3. The pressure of air is maintained constant to the desired value (say 2 kgf /cm2) by adjusting at the opening of the compressed air outlet valve in the reservoir manually.

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4. The following observations are to be taken by keeping reservoir pressure constant. Say (2 kgf / cm2) a) Speed (N) b) Manometer reading (ℎ𝑤𝑤)(Pressure difference across orifice) c) Temperature of air entry to LP cylinder [T1] d) Temperature of air entry to HP cylinder [T2’] afterattaining the steady state condition. e) Load on the motor in kgf. [Tm1-Tm2]

5. The same procedure of observations are to be made for the other reservoir pressures [4,6,8,10,12 Kgf / cm2]

6. Then the motor is switched off by pressing the red color ( or off) button of the starter after removing the load on the motor. This can be done by opening the valves provided at the top of LP and HP cylinders.

. Note : The load on the motor[Tm1-Tm2] has to be taken by keeping the torque arm

horizontal and this position can be verified by inserting the pin into the torque arm hole. Observations: Room Temperature [Ta] :_________oC Orifice dia. [do] : _____mm.

Sl. No

Receiver pressure (P3)

Speed N

Entry temp Manometer reading(hw) in mm

Load on motor

(𝑇𝑇𝑇𝑇1 − 𝑇𝑇𝑇𝑇2)

kgf/cm2 (gauge)

Bar (abs.) Rpm LPC

𝑇𝑇1 HPC 𝑇𝑇2

1

L (left arm)

R (right arm)

L+R 𝑁𝑁 𝐾𝐾𝐾𝐾𝑓𝑓

Note :1 kgf / cm2 (gauge) = (1 x 0.981) + 1.013 bar [Absolute] Bar

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Specimen calculations: (For gauge pressure = kg/cm2)

1. Density of air (ρa)�𝑘𝑘𝐾𝐾𝑇𝑇3� = 𝑃𝑃𝑎𝑎

𝑅𝑅𝑎𝑎𝑇𝑇𝑎𝑎 = 1.013 x 105

287 ×(273+ ) =

Where Pa = Atmospheric pressure, 1.013 x 105N/m2

Ra = Universal Gas constant, 287 J/kgK

Ta = Room Temperature in Kelvin

2. Pressure head in terms of air (ha), m ρa ha = ρw hw

ℎ𝑎𝑎 = ℎ𝑤𝑤𝜌𝜌𝑤𝑤𝜌𝜌𝑎𝑎

= × 1000

=

where ρw = density of water = 1000 kg/m3

hw = head of water in m=( × 10−3 m) 3. Velocity of air through orifice, (Va), m/sec. = �2𝐾𝐾ℎ𝑎𝑎

4. Area of orifice, (m2) = (Ao) =𝜋𝜋𝑑𝑑02

4where do = dia of orifice= 18 mm = 0.018 m

= 3.14 ×(0.018)2

4= 2.54 × 10−4 𝑇𝑇 (Const)

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5. Volume flow rate of air at inlet condition (Qa), (m3/sec) = 𝐶𝐶𝑑𝑑 𝑥𝑥 𝐴𝐴𝑜𝑜 𝑥𝑥 𝑉𝑉𝑎𝑎 Where Cd = Coefficient of discharge of orifice = 0.6

6. Vol.of air compressed at NTP (Normal Temp. and Press.),( m3/sec.) Qa at NTP = 𝑄𝑄𝑎𝑎

273𝑇𝑇𝑎𝑎

7. Theoretical Volume of air, m3 /sec Qth = 𝜋𝜋𝐷𝐷2𝐿𝐿𝑁𝑁

4 × 60

where

L = Stroke length of LP cylinder = 0.076m N = Speed in rpm D = Diameter of LP Cylinder = 0.110 m 8. Volumetric efficiency of the L.P. Cylinder, % = 𝑄𝑄𝑎𝑎𝑎𝑎𝑎𝑎𝑁𝑁𝑇𝑇𝑃𝑃

𝑄𝑄𝑎𝑎ℎx100

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9. Input power , (kW) =2 x π x R x N x (Tm 1−Tm 2)60x1000

where, R = Torque arm length = Half the distance between the spring balance centers.

𝑆𝑆𝑆𝑆𝑆𝑆 𝑎𝑎ℎ𝑒𝑒 𝑣𝑣𝑎𝑎𝑣𝑣𝑆𝑆𝑒𝑒 (𝑇𝑇𝑇𝑇1 − 𝑇𝑇𝑇𝑇2)𝑖𝑖𝑖𝑖 𝑁𝑁𝑒𝑒𝑤𝑤𝑎𝑎𝑜𝑜𝑖𝑖 ;𝑁𝑁 = 𝑆𝑆𝑆𝑆𝑒𝑒𝑒𝑒𝑑𝑑 𝑜𝑜𝑓𝑓 𝑎𝑎ℎ𝑒𝑒 𝑇𝑇𝑜𝑜𝑎𝑎𝑜𝑜𝑚𝑚 = 960 𝑅𝑅𝑆𝑆𝑇𝑇 10. Mass flow rate of air [Ma], kg/sec. = 𝑄𝑄𝑎𝑎 × ρ𝑎𝑎

11.Isothermal power ,(kW) =Ma xRa xT1

1000 x 𝑣𝑣𝑖𝑖 �P3

P1�

Where P3 = Receiver pressure in bar

P1= atmospheric pressure = 1.01325bar Ra = 287 J/kgk T1 = Atmospheric temperature in K 12. Isothermal efficiency = Isothermal power

Inputpower x 100

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13. Adiabatic power in kW = � 𝛾𝛾𝛾𝛾−1

� × �Ma x Ra1000

� x �T1 + T2′ � × ��P3

P!�γ−12γ − 1�

Where γ = 1.4 [Ratio of Specific heats] 𝑇𝑇1 R , 𝑇𝑇2

1P

are in Kelvin 14. Adiabatic efficiency,(%) = Adiabatic power

Input power

15. Free air delivered,� m3

kW S� = Qa

Input power

16. Heat rejected in inter cooler,( kJ/sec). = �𝑇𝑇𝑎𝑎 × 𝐶𝐶𝑆𝑆𝑎𝑎 × �𝑇𝑇2 − 𝑇𝑇2

1� �

Where T2 = Temperature of air entry to intercooler

ie T2= T1x �P2P1�

n−1n

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Where n = 1.35

T1 = Atmospheric temperature in k.

Where P2 = Intermediate pressure assuming perfect intercooling, P1 = Atmospheric pressure. P2 = �P1xP3 bar.

Cp of air = 1.005

Draw the following performance characteristic curves graphs:

Receiver Pressure (P3) 𝑉𝑉𝑉𝑉 Volumetric efficiency. 𝑉𝑉𝑉𝑉 Isothermal efficiency. 𝑉𝑉𝑉𝑉 Adiabatic efficiency. 𝑉𝑉𝑉𝑉 FAD. 𝑉𝑉𝑉𝑉 Heat lost in the intercooler. RESULT

The performance test on Keaser air compressor was conducted and the volumetric efficiency of L.P cylinder at NTP condition was determined at various receiver pressure , and the various performance characteristic curves of the were also drawn

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Result Tabulation: Sl.No Reci. Press.

P3, (bar)

(abs.)

𝑄𝑄𝑎𝑎 𝑇𝑇3

𝑆𝑆𝑒𝑒𝑆𝑆

𝑄𝑄𝑎𝑎ℎ 𝑇𝑇3

𝑆𝑆𝑒𝑒𝑆𝑆

Vol.η %

I.P kW

Adiabatic Isothermal Heat rejected

𝑘𝑘𝑘𝑘/𝑉𝑉

FAD

𝑇𝑇3 𝑘𝑘⁄ 𝑊𝑊 𝑆𝑆𝑒𝑒𝑆𝑆⁄ Power kW

η in %

Power kW

η in %

FAD – Free Air Delivered.(The FAD is actual volume delivered at the stated pressure reduced to intake temperature and pressure and is expressed in cubic meter per min).

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Expt.No: Date:

LOAD TEST ON KIRLOSKAR AV-I (Single arm) ENGINE Aim: To conduct a load test on Kirloskar AV-I(Single arm) engine by running the engine at different loads at 1500 rpm and to find the economic load. Also determine the load characteristics of the engine and generator. Instruments required:

Swing field [DC] generator type dynamometer, Bank of electrical resistances to apply load, Stop-watch to note the time for a definite volume of fuel consumption. Tachometer to measure the speed of the engine.

Specifications: Type :

No. of strokes per cycle :

Type of cooling :

Fuel used :

Speed :

Power :

Bore :

Stroke :

Preliminary calculations:

While conducting load test the speed is retained constant and the load is varied (viz, no load, min load, medium load & full load). Normally six equi-distributed loads are chosen in the range from no load to full load. Preliminary calculations are made to determine the tension T or a tangential force (W) needed at the dynamometer to effect the desired power output (Load).

In this test setup the engine is coupled to a swing field DC generator type dynamometer. There are provisions to measure the electrical output of the generator .

When the electrical resistances are included in the output circuit, the generator and

hence the engine gets loaded. The generator swings about the trunnion and force required to keep the torque arm horizontal is the measure of engine output.

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The tension T corresponding to the rated power output is calculated using the expression for brake power. Then the values of net tension for other part load outputs (0%,20%,40%,60%,80%) are calculated and tabulated in the observation table. We know that Brake power (BP) = Torque x Angular Velocity

BP = 𝑇𝑇 × 𝑅𝑅𝑒𝑒 × 2π N 60

Where, BP = Brake Power in kilowatts

T = Load or net tensions in Newton

N = Speed of the engine in rpm

Re = Torque arm length

{Torque arm length = Distance between the center lines of the spring balance to generator shaft. }

Hence Re=________ cm = ________ m

The equation to find BP is: BP in kW = 2 πRe N(T)60 x 1000

Hence (T) = B.P x 60 x 1000

2 π Re N in Newton

The value for ‘B.P’ and ‘N’ are to be taken from the specifications of the engine. Substitute the value of Re in “Meters” The value obtained for (T) from the above equation is the full load of the engine in

Newton.

(T) = ---------------------------------------------- = in N (T) = N

(T)= (𝑇𝑇)𝑖𝑖𝑖𝑖 𝑁𝑁

9.81 in kgf

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Procedure: 1. After evaluating the loads to be applied on the engine, check the fuel level in fuel tank, flow of cooling water to the engine, switches on the panel board are in off position, level of lubricant in the sump as indicated by the dipstick and no-load on the engine as indicated by the loading device etc. The DC machine coupled to the engine is run as a generator and the applied load is varied by altering the load resistances.

2. After observing all the precautions, the engine has to be started. Starting Procedure:-

i. The decompression lever located at the rocker box is turned to the vertical upward position. [Decompression On]

ii. The starting handle is inserted on the cam shaft iii. The cam shaft is manually rotated faster in clockwise direction. iv. After the engine shaft has gained sufficient momentum, the decompression lever is

brought to the horizontal position [Decompression Off]. Then the engine gets started. 3. Allow the engine to run for 5 to 10 minutes to attain steady condition at its rated speed of 1500 rpm. 4. Now, load the engine approximately to 20 % of full-load by switching on the load mains and suitable electrical resistances (Trial and error method). 5. The speed of the engine is to be maintained at 1500 rpm by adjusting speed governor. 6. The actual spring balance readings are noted after the generator torque arm to its horizontal position by adjusting the hand wheels located above the spring balance 7. Note the time for 10 cc of fuel consumption twice at this load and average the values. 8. In the same way, maintain the speed at rated value and note the time for 10cc of fuel consumption at 40 %, 60 %, 80 % of full load, full load and at no load. 9. After taking all the readings, the engine is switched off at no load by pulling the control rod of the fuel injection pump.

Engine

Generator

Load

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Observations:- Speed to be maintained: 1500 rpm

Sl. No. %𝑜𝑜𝑓𝑓 𝑣𝑣𝑜𝑜𝑎𝑎𝑑𝑑

Cal. load (T)

App. load (T)

Time taken for 10 cc of fuel consumption

in sec. 𝑁𝑁 𝑘𝑘𝐾𝐾𝑓𝑓 𝑘𝑘𝐾𝐾𝑓𝑓 𝑁𝑁

𝑎𝑎R1 𝑎𝑎R2 𝑎𝑎𝑎𝑎𝑣𝑣𝑒𝑒

Note: At the beginning of the experiment, the conditions of viscous friction will be more than that at steady running. Hence observation at no load if made at the beginning will result in arriving of incorrect higher fuel consumption value. In order to overcome this error observations at no load are to be made at the end of the experiment

Specimen calculations: (For ……………% of load) 1. Fuel Consumption [FC], kg/hr

= Vol. flow rate, cc/s x density of fuel, kg/cc x 3600 s/hr.

=� 10tave

� x Sp. gravity of diesel x density of water in kg/cc x 3600

FC = � 10

�x 0.835 x 1000 x 10−6 x 3600 = _____________kg/hr

2. Brake Power [BP], (kW) = 2 πRe N(T) 60 x 1000

𝑆𝑆𝑆𝑆𝑆𝑆𝑉𝑉𝑖𝑖𝑎𝑎𝑆𝑆𝑖𝑖𝑎𝑎𝑒𝑒 (𝑇𝑇) 𝑉𝑉𝑎𝑎𝑣𝑣𝑣𝑣𝑒𝑒 𝑖𝑖𝑖𝑖 "N𝑒𝑒𝑤𝑤𝑎𝑎𝑜𝑜𝑖𝑖"

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BP = 2 π x x 1500 ( )

60 x 1000=

3. Specific Fuel Consumption [SFC], (kg/kW-hr) = FC

BP

SFC = --------------- = …………. kg/kW-hr

4. Frictional Power [Fr. P], kW

To be obtained from the graph of BP Vs FC by extrapolation method. Frictional power is assumed to be constant at all loads.

Fr.P =

5. Indicated Power [IP], (kW) = BP+ Fr.P

IP =

6. Mechanical Efficiency = Brake powerIndicated power

x 100 %

ηMech =

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7. Fuel Power [Fu.p] (or) Heat Input, (kW) = FC ,�kg

hr �xC .V,(kJkg )

3600 [C.V of Diesel 42,000(kJ

kg)]

Fu P =

8. Brake Thermal Efficiency = BPfuel power

x100 %

ηBth =

9. Indicated Thermal Efficiency = IPFuel power

x100 %

ηIth =

10.Brake Mean Effective Pressure [BMEP],(bar) = BPx 103 x60

105 x L x A x N ′ x n

= BP x 60

100 L A N ′ n in bar

BMEP =

N’ = Number of power strokes per minute.

For 4 stroke engine N' =𝐍𝐍𝟐𝟐; For 2 stroke engine N' = N; n = Number of cylinders

A = Area of bore(m2)= 𝜋𝜋𝑑𝑑2

4(d = Dia of bore in m)= …………….m2; L = Length of stroke,m

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11. Indicated Mean Effective Pressure [IMEP] (bar) = IPx 60100LA N ′ n

IMEP =

12. Torque, Nm = (T) x Re

Torque = Draw the following Graphs:

1. BP Vs SFC, B.Th. efficiency,

I.Th.efficiency, Mech. efficiency, IMEP, BMEP , Torque .

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RESULT TABULATION

Result: The Load test on KIRLOSKAR AV I(Single arm type) Engine was conducted at different load condition, the economic load of engine was found at ___________rpm is ________kW , and the corresponding performance characteristic curve were also drawn.

l. No.

𝐾𝐾𝑤𝑤

B.P

𝑘𝑘𝐾𝐾/ℎ𝑚𝑚

F.C

𝑘𝑘𝐾𝐾/𝑘𝑘𝑤𝑤. ℎ𝑚𝑚

SFC

𝑘𝑘𝑤𝑤

Fu.P

𝑘𝑘𝑤𝑤

I.P

%

I.Th.η

%

B.Th.η

%

Mech.η

𝑆𝑆𝑎𝑎𝑚𝑚

BMEP

𝑆𝑆𝑎𝑎𝑚𝑚

IMEP

𝑁𝑁𝑇𝑇

Torque

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Extrapolation (or) Willan’s line method:

This is a method of determining the friction power and hence the indicated power of C.I engine. It is based on the fact that at part loads the combustion is completed within the engine cylinder. Hence at a given speed the rate of fuel consumption bears a linear relationship with power output/torque.

A plot therefore of rate of fuel consumption versus power output/torque at a particular speed will be a straight line in the light load region. This straight line is Willan’s line. The amount of negative power output/torque obtained by extrapolation of the plot at zero rate of fuel consumption represents the frictional power (power required to over come friction) of the engine at the specified speed.

The rapid increase in the slope of Willan’s line at high load denotes a reduction in combustion efficiency as more and more fuel is pumped in to the given volume of air.

Since petrol engine is throttled to maintain a high fuel/air ratio with load, combustion is not complete with in the cylinder. In this case a plot of power output versus rate fuel consumption does not yield a straight line. Hence extrapolation is difficult and not suitable for use with petrol engines. Note: Finding friction power in this method directly depends on the flow rate of fuel consumed. It has to be noted that any fuel leak in the fuel line will result in incorrect frictional power.

BP in kW

BP Vs. FC (Extrapolation Method)

0 0.2 0.4 0.6 0.8

1 1.2 1.4 1.6 1.8

2

-1.5 -1 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.5 Friction Power

27

The general shape of characteristics curve are shown – For reference

ECONOMIC LOAD CURVE

28

Expt.No : Date:

LOAD TEST ON PSG 5 HP ENGINE Aim: To conduct a load test on PSG 5 HP engine by running the engine at different loads at 610 rpm and to find the economic load. Also to determine the load characteristics of the engine. Instruments required:

Friction brake to indicate the load, Digital tachometer to measure the speed of the engine, Stop-watch to note the time for a definite volume of fuel consumption.

Specifications:

Type :

No. of strokes per cycle :

Type of cooling :

Fuel used :

Speed :

Power :

Bore :

Stroke :

Procedure: 1. After evaluating the full-load of the engine in kgf with the help of specifications given, check the fuel level in fuel tank, flow of cooling water to the engine, level of lubricant in the sump as indicated by the dipstick, and no-load on the engine as indicated by the loading device etc., before starting the engine. 2. Start the engine by cranking and allow it to run for 5 to 10 minutes to attain steady condition at its speed of 610 rpm. Allow the cooling water for the brake drum in order to cool it while applying loads. 3. Now, load the engine to 20 % of full-load. Then check for its speed and note down the actual load on the engine. 4. Note the time for 10 cc of fuel consumption twice at this load and average the values. 5. In the same way, maintain the speed at its rated value and note the time for 10cc of fuel consumption at 40 %, 60 %, 80 % of full load, full load and at no load. After taking no load reading the engine is stopped by engaging the fuel cut off lever.

29

Preliminary calculations: The loading device attached to this Engine is Mechanical Brake drum type.

We know that ,

Brake power (BP) = Torque x Angular Velocity

BP =(T1 –T2) Re ×2π N 60

Where, BP = Brake Power in kilowatts

(T1 - T2) = Load or net tensions in Newton

N = Speed of the engine in rpm

Re = Effective radius of brake drum in meters = (R +t2 )

(Where ‘R’ is the radius of the brake drum , obtained from its circumference and ‘t’ is the thickness of the belt = 8 mm = 0.008 m) Circumference of the brake drum, 2πR = (To be obtained from Engine brake drum)

R =

Hence Re =

The equation to find BP is: BP in kW = 2 πRe N(T1−T2)

60 x 1000

Hence (T1- T2) = B.P x 60 x 1000

2 π Re N in Newton's

The value for ‘B.P’ and ‘N’ are to be taken from the specifications of the engine. The value obtained for (T1- T2) from the above equation is the full load of the engine in

Newton.

(T1- T2) = N

(T1- T2)= (𝑇𝑇1− 𝑇𝑇2)𝑖𝑖𝑖𝑖 𝑁𝑁9.81

in kgf

30

Observations:- Speed to be maintained :__________ rpm

Sl. No

%of load

Calculated load (T1-T2)

Applied load (T1-T2)

Time taken for 10cc of fuel consumption in sec

N kgf kgf N t1 t2 tave

Note: At the beginning of the experiment, the conditions of viscous friction will be more from that at steady running. Hence observation at no load if made at the beginning will result in arriving of incorrect higher fuel consumption value. In order to overcome this error observations at no load are to be made at the end of the experiments.

Specimen Calculations: (For ……………% of load) 1. Fuel Consumption [FC], kg/hr

= Vol. flow rate, cc/s x density of fuel, kg/cc x 3600 s/hr.

=� 10tave

� x Sp. gravity of diesel x density of water in kg/cc x 3600

FC = � 10

�x 0.835 x 1000 x 10−6 x 3600 = ________________ kg/hr

2. Brake Power [BP], (kW) = 2 πRe N(T1−T2)

60 x 1000

𝑆𝑆𝑆𝑆𝑆𝑆𝑉𝑉𝑖𝑖𝑎𝑎𝑆𝑆𝑖𝑖𝑎𝑎𝑒𝑒 (𝑇𝑇1 − 𝑇𝑇2) 𝑉𝑉𝑎𝑎𝑣𝑣𝑣𝑣𝑒𝑒 𝑖𝑖𝑖𝑖 ""Newton" "

BP = 2 π x x 610 ( )

60 x 1000 =

31

3. Specific Fuel Consumption [SFC], (kg/kW-hr) = FCBP

SFC =

4. Frictional Power [Fr. P], kW

To be obtained from the graph of BP Vs FC by extrapolation method. Frictional power is assumed to be constant at all loads.

Fr.P =

5. Indicated Power [IP], (kW) = BP+ Fr.P

IP =

6. Mechanical Efficiency = Brake powerIndicated power

x 100

ηMech =

7. Fuel Power [Fu.p] (or) Heat Input, (kW) = FC ,�kg

hr �xC .V,(kJkg )

3600 [C.V of Diesel 42,000(kJ

kg)]

Fu P =

8. Brake Thermal Efficiency = BPfuel power

x100

ηBth =

9. Indicated Thermal Efficiency = IPFuel power

x100

ηIth =

32

10.Brake Mean Effective Pressure [BMEP],(bar) = BPx 103 x60

105 x L x A x N ′ x n

= BP x 60

100 L A N ′ n in bar

BMEP =

11. Indicated Mean Effective Pressure [IMEP] (bar) = IPx 60100LA N ′ n

IMEP =

12. Torque, Nm = (T1 - T2) x Re

Torque =

Draw the following Graphs:

1. BP Vs SFC, B.Th. efficiency,

2. BP Vs I.Th.efficiency,

Mech. efficiency, IMEP, BMEP and Torque .

N’ = Number of power strokes per minute.

For 4 stroke engine N' =𝐍𝐍𝟐𝟐; For 2 stroke engine N' = N; n = Number of cylinders

A = Area of bore(m2)= 𝜋𝜋𝑑𝑑2

4(d = Dia of bore in m)= …………….m2; L = Length of stroke,m

33

RESULT TABULATION

Sl no Appl.load BP F C S F C Fu.P I P 𝜂𝜂𝐵𝐵 𝑎𝑎ℎ 𝜂𝜂𝐼𝐼 𝑎𝑎ℎ 𝜂𝜂𝑇𝑇𝑒𝑒𝑆𝑆 ℎ BMEP IMEP Torque

N kW Kg/hr Kg/kW-hr kW kW % ``% % bar bar Nm

1

2

3

4

5

6

Result : The Load test on PSG 5 HP Engine was conducted at different load condition, the economic load of engine was found at ___________rpm is ________kW and the corresponding performance characteristic curve were also drawn.

34

STUDY AND VALVE TIMING ON FOUR STROKE IC ENGINES

Aim:-

To determine the valve timings/settings of a four stroke IC engine and to draw the valve timing diagram.

Valve timing Diagram:- Valve timing diagram is the graphical form of representing the timings (instants) in terms of the corresponding crank positions at which the inlet and exhaust valves open and close. The crank positions and the sequences are described with reference to the nearest dead center.

Purpose:- At the time of every overhauling and repairing, the engine parts are dismantled and reassembled. After such reassembling, the valve timings are to be verified with the manufacturers recommendations, i.e, the original settings, and adjusted using the valve tappet if required. The valve timings are also to be periodically verified with original settings because of the wear and tear of the elements of the valve mechanism.

Procedure:-

Identification of valves:- The engine cylinder head cover is removed to facilitate the observation of valves, valve mechanisms and their workings in case of multi cylinder engine. Incase of single cylinder engine the rocker arm box cover is removed. The engine crank shaft is manually turned in the correct direction (clockwise from the cranking end) and the functioning of valves is noted. Both inlet and exhaust valves will be in simultaneous operation at a particular period. This period represents the overlapping between two cycles (the suction process of later cycle overlapping with the exhaust process of the current cycle.) The valve which closes during this period is the exhaust valve while that opens is the inlet valve.

Rotating member for markings:- A circular member (flywheel /pulley/brake drum) mounted on the crankshaft is used for making the markings related to the various events (IVO,IVC,EVO & EVC) with reference to the nearest dead center IDC/ODC for horizontal engine or TDC/BDC for vertical engine. The circumference of the circular member used will be useful in determining the various crank angles. A pointer or an indicator fixed on the engine frame serves as the reference against which all the markings are made on the rim of the circular member.

35

Dead Centers:- When the engine runs, the piston reciprocates between two extremities of the stroke. These extreme positions are called as dead centers, the velocity of reciprocation being zero at these instants. The crank positions corresponding to the dead center positions of the piston are termed TDC/ BDC or IDC /ODC positions. The crank is turned manually and the marking is made on the rim of the circular member when the piston is at a definite position (piston bottom coincides with the bottom of the cylinder) during its downward journey. On further rotation of the crank, the piston moves down, reaches the BDC position and reverses its direction to move up. During the upward journey the piston will reach the same definite position at which a marking was done earlier on the rim of the circular member. Now another marking is done on the rim of the circular member. Mid point of the two markings (on the shorter arc length) is the BDC marking on the rim. TDC marking is made diametrically opposite to the BDC marking. At the commencement of valve just moving away from its seat due to the actuation by cam through push rod, the valve is said to have opened. When the valve has completed the movement against its seat by spring action, it is said to have closed. When the crank is turned, the instant at which a piece of paper inserted in-between the valve stem and tappet/rocker arm is just gripped is the valve opening. IVO and EVO markings are made on the circular member corresponding to the openings of the inlet and exhaust valves. Similarly the instant at which the paper just looses the grip in between the valve stem and tappet/rocker arm is the valve closing. IVC and EVC markings are made on the circular member.

The distances of the IVO,IVC,EVO & EVC markings on the circular member, from the nearest dead center are measured.[Fuel injection timing can also be noted by noting down the starting and ending of fuel injection through nozzle( i.e., fuel injector)] The observed diagram is drawn using the dimension details of markings on the circular member and the direction of rotation of the flywheel. The circle, drawn of any convenient radius, represents the circular member (flywheel/pulley/brake drum). Inferring from the observed diagram, the events, their sequences and timings in relation to the nearest dead center are tabulated as shown. The arc lengths are expressed as crank angles in the last column of the table. The valve timing diagram is drawn using the details (event, sequence and crank angle) in the table.

36

Observation tabulation:- Circumference of flywheel = ............cm.

Sl. No.

Events

Sequence of Operation

Distance from the nearest dead center ‘x’ in cms

Crank angle ‘θ’ in degrees

1 2 3 4

IVO

IVC

EVO

EVC

Crank angle, 𝜃𝜃 = 𝑋𝑋 ×360

𝐶𝐶𝑖𝑖𝑚𝑚𝑆𝑆𝑆𝑆𝑇𝑇𝑓𝑓𝑒𝑒𝑚𝑚𝑒𝑒𝑖𝑖𝑆𝑆𝑒𝑒 𝑜𝑜𝑓𝑓 𝑎𝑎ℎ𝑒𝑒𝑓𝑓𝑣𝑣𝑒𝑒𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑣𝑣/ 𝑆𝑆𝑚𝑚𝑎𝑎𝑘𝑘𝑒𝑒 𝑑𝑑𝑚𝑚𝑆𝑆𝑇𝑇/𝑆𝑆𝑆𝑆𝑣𝑣𝑣𝑣𝑒𝑒𝑒𝑒

Model OBSERVED DIAGRAM

X1

X2 X3

X4

37

Procedure for drawing V.T.D. Draw a vertical line to represent the line of stroke. Mark a point 1 on it. Draw a semi-circle of any convenient radius taking point 1 as center to its RHS of the vertical line from ‘a’ to ‘b’. Then mark another point 2 little above the point 1 on the vertical line (say 5 to 10 mm). Then draw another semi-circle from ‘b’ to ‘c’ taking 2 as center on the LHS of the vertical line. Similarly draw semi-circles from ‘c’ to ‘d’ on the RHS taking 1 as center and from ‘d’ to ‘e’ on the LHS taking 2 as center. Then extend the arcs to the desired length from ‘a’ as well as from ‘e’. Then locate all the events on the circular arcs as per the tabulation . Join these points with the center of the circle (Take point 1 as center) by radial lines. This radial line represents the crank positions corresponding to the events.

Result: The angle through which the inlet valve remains open : The angle through which the Exhaust valve remains open : The angle of overlap :

-------------------------------------------------------------------------------------------------- Note: In the place of θ1, θ2, θ3 & θ4 , the actual values of the crank angles in degrees are to be substituted

c

b

a

2 1

d

e

θ4

θ3 θ2

θ1

Model VALVE TIMING DIAGRAM

S – Suction C – Compression P – Power E – Exhaust

38

Expt.No : Date:

STUDY AND VALVE TIMING ON FIELD MARSHAL 8 HP ENGINE

Aim:-

To study and determine the valve timings/settings on field marshal 8 HP engine and to draw the valve timing diagram.

Requirements:-

Measuring tape A circular member attached to crank shaft A reference point

Specifications:- Type :

Fuel used :

Power :

Speed :

Bore :

Stroke :

Type of Cooling :

Valve mechanism :

OBSERVED DIAGRAM

39

Observation tabulation:- Circumference of flywheel = cm.

Sl. No.

Events


Distance from the nearest dead centre ‘x’ in cms


1 2 3 4

IVO

IVC

EVO

EVC

SPECIMEN CALCULATIONS:

θ1 = 𝑋𝑋1𝐱𝐱 𝟑𝟑𝟑𝟑𝟑𝟑

Circumference of the flywheel / pulley / brake drum =

θ2 = 𝑋𝑋2 𝐱𝐱 𝟑𝟑𝟑𝟑𝟑𝟑


θ3 = 𝑋𝑋3 𝐱𝐱 𝟑𝟑𝟑𝟑𝟑𝟑


θ4 = 𝑋𝑋4 𝐱𝐱 𝟑𝟑𝟑𝟑𝟑𝟑


40

VALVE TIMING DIAGRAM

RESULT: The experiment was conducted and the Valve Timing Diagram for field marshal 8 HP

engine was drawn. From the valve timing diagram the following values are obtained

Crank angle for which the Inlet Valve remains open( θ1+180+ θ2) =

Crank angle for which the Exhaust valve remains open(θ3+180+ θ4) =

Angle of over lap (θ1+ θ4) =

41

General Instruction and procedure for

STUDY AND PORT TIMING ON TWO STROKE IC ENGINES

Aim : To determine the port timings of Two stroke IC engines and to draw the port timing

diagram.

Ports: The ports in a two stroke engine are small openings in the cylinder walls diametrically

opposite to each other. The piston reciprocating inside the cylinder uncovers (opens) and covers

(closes) the ports, thus performing the function of valves in 4 - stroke engines. The exhaust port

lies slightly above the inlet (transfer) port. Hence during the ascending of the piston in a vertical

engine, the closing of exhaust port precedes the closing of transfer port. Similarly in the

descending of the piston, the exhaust port is uncovered first and then the inlet (transfer) port.

The above said facts help in identifying the inlet and exhaust ports.

Procedure: Dead centers: On removal of the cylinder head, the piston top can be visualized. Rotation of the

flywheel makes the piston reciprocate. Make a mark on the flywheel against the fixed

pointer corresponding to a specific position of the piston in its upward stroke. On further

rotation of the flywheel piston reaches the TDC position (maximum height) and reverses

its direction of motion down wards. The position at which a mark was made earlier during

upward stroke will be reached again during the downward stroke. Now make a mark on the

flywheel. The mid point of the markings is the TDC point on the flywheel. As the direct

location of TDC point requires visual judgment and is difficult, the procedure mentioned is

suggested. The BDC point is marked in the flywheel diametrically opposite to TDC point. A

port is said to open when the piston just starts uncovering it. It is said to close only when piston

has totally covered it

After marking the dead center positions on the flywheel, markings are made

corresponding to the opening and closing of the inlet and exhaust ports.

42

Observed diagram: A circle of any convenient radius is drawn to represent the flywheel. The markings made on the flywheel are also shown in the circle and the distance of them from the nearest dead center are marked. When the sense of rotation of the flywheel is added, the observed diagram becomes complete.

The details of the observed diagram ie. the crank positions (angle and sequence) in relation to the nearest dead center, corresponding to the opening and closing of the inlet and exhaust valves are tabulated. The respective crank angles are calculated from the arc lengths in observed diagram.

Observation tabulation:-

Circumference of flywheel = ............cm. Sl. No.

Events




1 2 3 4

IPO IPC EPO EPC

Crank angle, θ = 𝐗𝐗 𝐱𝐱 𝟑𝟑𝟑𝟑𝟑𝟑

Circumference of the flywheel / pulley / brake drum

Model OBSERVED DIAGRAM

IPO IPC

EPO EPC

X1 X2 X3 X4

TDC

BDC

43

Port Timing Diagram:

The port timing diagram is drawn using the inferences from the table. The crank angles,

the sequences in relation to the nearest dead center position of the crank and the required

direction of travel of the observer along the circular diagram are furnished in the port timing

diagram.

Result: The angle through which the inlet port remains open : The angle through which the Exhaust port remains open :

-------------------------------------------------------------------------------------------------- Note: In the place of θ1, θ2, θ3 & θ4 , the actual values of the crank angles in degrees are to be substituted

Model PORT TIMING DIAGRAM

θ1 θ2 θ3 θ4

44

Expt. No : Date:

STUDY AND PORT TIMING ON PETTER 3 HP ENGINE Aim :

To study and determine the port timings of Petter 3 HP engine and to draw the port

timing diagram.

Requirements:- Measuring tape A circular member attached to crank shaft A reference point

Specifications:- Type :

Fuel used :

Power :

Speed :

Bore :

Stroke :

Type of Cooling :

OBSERVED DIAGRAM

45

Observation tabulation:- Circumference of flywheel = cm.

Sl. No.

Events




1 2 3 4

IPO

IPC

EPO

EPC

SPECIMEN CALCULATIONS

θ1 = 𝑋𝑋1𝐱𝐱 𝟑𝟑𝟑𝟑𝟑𝟑Circumference of the flywheel / pulley / brake drum

=

θ2 = 𝑋𝑋2 𝐱𝐱 𝟑𝟑𝟑𝟑𝟑𝟑


θ3 = 𝑋𝑋3 𝐱𝐱 𝟑𝟑𝟑𝟑𝟑𝟑


θ4 = 𝑋𝑋4 𝐱𝐱 𝟑𝟑𝟑𝟑𝟑𝟑


46

PORT TIMING DIAGRAM

RESULT:

The experiment was conducted and the Port Timing Diagram for the Petter 3 HP engine

was drawn

Crank angle for which the Inlet port remains open ( θ1 + θ2) =

Crank angle for which the Exhaust port remains open (θ3 + θ4) =

47

DYNAMICS LABORATORY Expt. No. Date:

DETERMINATION OF CHARACTERISTICS CURVES OF WATT GOVERNOR AND HARTNELL GOVERNOR

Watt Governor

Aim: To determine the characteristic curves of Watt governor. Introduction:

The function of a governor is to maintain the mean speed of a machine/prime mover, by regulating the input to the machine/prime mover automatically, when the variation of speed occurs due to fluctuation in the load.

SPECIFICATIONS: Length of each link ‘l ‘ = 125 mm Initial height of governor (ho) = 95 mm Mass of each ball (m) = 0.306 kg Description:

The drive unit consists of a small electric motor connected through the belt and pulley arrangement. A DC variac effects precise speed control and an extension of the spindle shaft allows the use of hand held tachometer to find the speed of the governor spindle. A graduated scale is fixed to the sleeve and guided in vertical direction. Procedure:

Mount the watt governor mechanism on the drive unit of the governor apparatus. Vary the governor spindle speed by adjusting the variac. The speed can be determined by the hand tachometer.

Increase the speed of the governor spindle gradually by adjusting the variac and note down the speed at which the sleeve just begins to move up. Take four or five sets of readings by increasing the governor speed in steps and note down the corresponding sleeve displacement within the range of the governor and tabulate the observations. Observation table:

Sl.No. Speed (N) in rpm

Sleeve displacement x in cm m

1

2 3 4

5

48

Specimen Calculation: (for …………….reading) Height of the governor h = [ho - (x/2)] = From the figure we can write Cos α = h / l ∴ α = Cos-1 (h / l ) = The controlling force Fc = m ω2 r Where m = mass of fly ball in kg =0.306kg. ω = Angular velocity of spindle in rad/sec = 2 π N / 60

= r = radius of rotation of the balls.

`r' = l Sin α + 50 mm

=

The controlling force Fc = m ω2 r

FC = Result tabulation:

Sl.No. Speed (N) in rpm Radius of rotation (r) in

meter. Controlling force Fc in N

1

2

3 4 5

Graph: Draw Speed Vs Displacement Radius Vs Controlling force Result:

49

Scale

DriveMotorUnitSleeve

Flyball

Spindle

Link

EXPERIMENTAL SETUP OF WATT GOVERNOR

C

B

A100

r

α

α

100

ho=95

l=125

All dimensions are in mm

WATT GOVERNOR CONFIGURATION

50

Hartnell Governor

Aim: To determine the characteristic curves of the given Hartnell Governor. Introduction:

This governor comes under the spring loaded type centrifugal governors. The control of the speed is affected either wholly or in part by means of springs. The centrifugal governors are based on the balancing of centrifugal force on the rotating balls by an equal and opposite radial force, known as the controlling force. It consists of two balls of equal mass, which are attached to the arms as shown in fig. These balls are known as governor balls or fly balls. The balls revolve with a spindle, which is driven by the engine through bevel gears. The upper ends of the arms are pivoted to the spindle, so that the balls may rise up or fall down as they revolve about the vertical axis. The arms are connected by the links to a sleeve, which is keyed to the spindle. This sleeve revolves with the spindle but can slide up & down. The balls and the sleeve rises when the spindle speed increases and falls when the speed decreases. The sleeve is connected by a bell crank lever to a throttle valve. The supply of the working fluid decreases when the sleeve rises and increases when it falls. Description:

The drive unit of the governor consists of a small electric motor connected through a belt and pulley arrangement. A D.C. Variac affects precise speed control. A photoelectric pick up is used to find speed of the governor spindle. The set up is designed to produce pulses proportional to r.p.m of shaft using phototransistor as the sensing element. A graduated scale is fixed to the sleeve and guided in vertical direction. Procedure:

Mount the Hartnell governor mechanism on the drive unit of the governor apparatus. Vary the governor spindle speed by adjusting the variac. Increase the speed of the governor spindle gradually by adjusting the variac and note down the speed at which the sleeve just begins to move up. Take four or five sets of readings by increasing the governor speed gradually in steps and note down the corresponding sleeve movement within the range of the governor. Specifications: Observation table: Mass of the fly ball = 0.700 kg Length of ball arm (a) = 75 mm Length of sleeve arm (b) = 115.5 mm Initial radius ro = 165 mm

Sl. No. Speed(N)

in rpm

Sleeve displacement ‘x’ in

cm m

1

2

3

4

5

51

Specimen calculation: The controlling force Fc = m ω

2 r

Where m = mass of fly ball in kg ω = Angular velocity of spindle in rad/sec

ω = 2 π N / 60

=

Radius of rotation (r) = ro + (a / b) x

=

The controlling force Fc = m ω2 r

FC =

Result tabulation:

Sl. No.

Speed (N) in rpm

Radius of rotation (r) in meter.

Controlling force (Fc) in N

1 2 3 4 5

Graph: Draw Speed Vs Displacement Radius Vs Controlling force Result:

52

ScaleDriveMotorUnit

Bell Crank Lever

LocknutSpringFlyball

NutFrame

EXPERIMENTAL SETUP OF HARTNELL GOVERNOR

300

r

b

G

a

80C

Spring

HARTNELL GOVERNOR CONFIGURATION

53

Expt. No. : Date :

DETERMINATION OF MASS MOMENT OF INERTIA OF FLY WHEEL

Aim: Experimentally determine the mass moment of inertia of the flywheel along with the shaft and verify the same theoretically. Theory: Let the mass descend under the force of gravity starting from rest with uniform acceleration. As per Newton's second law of motion of the falling mass m g - T = m a ……….. (1) Where m = Mass of the falling body T = tension in the string a = acceleration of the falling mass therefore T = m g - m a = m (g - a) ………….. (2) Considering the motion of the flywheel the equation of motion is Tq = I α But net torque, Tq = Torque due to tension T - Frictional torque = (T.r) - ( Ff . r ) = ( T - Ff ) r Where Ff is Frictional force in Newton Therefore ( T - Ff ) r = I.α .. ….. (3) Substituting the value of T in equation (3) [m (g - a) - Ff] r = I. α But α = a / r Therefore [m (g - a) - Ff] r = I × (a/r)

or I = (r2 / a) [ m(g - a ) - Ff ] …….. (4)

To find ` a': Let `t’ is the time taken by the falling mass to travel the distance `h" u = 0; S = h ; t = t We know that S = u t +

21 a t

2

h = 0 + 21 a t

2

Therefore a = 2

2th

Moment of inertia of the flywheel ‘I’ can be found by substituting the value of ‘a’ in eqn. (4)

54

Procedure:

First of all find out the force needed to overcome the friction present on the bearings when it just begins to rotate by gradually adding the weight to the weight pan which is attached to the one end of the string.

Then some more known weight is added and allow the mass to fall under the force of gravity. Note down the time for first 10 revolutions of the flywheel starting from rest. Conduct the experiment two or three times with the same mass and take the average time value. Repeat the experiment with different masses and tabulate the observations. Observations: Frictional force Ff = mf × g = (Where m

f = mass added to overcome the friction)

Mass of falling body (m)

in Time for 10 revolutions in seconds

grams kg. t1 t2 t3 taverage

Specimen Calculations: m - mass added r - radius of the shaft in mtrs. h - distance travelled for 10 revolutions h = d××π10 in meter Where d is diameter of the shaft h =

a = 2

2th

= ……….. m/s2 Therefore

Iept. =

( )[ ] 22

kgmFagmarI f−−=

55

Mass Moment of Inertia of Flywheel (Theoretical ) CALCULATIONS: Mass Density of the material (ρ) of the flywheel and shaft = 6840 kg/m3 Radius of gyration k: k2 for solid cylinder of diameter. `d' = d2 / 8 k2 for hollow cylinder = (do2 + di2) / 8 1) Moment of Inertia of shaft: Dia. of the shaft (d) = Length of the shaft = Area of cross section = (π/4)d2

= Volume of the shaft = Area of cross section x Length of the shaft = Mass of the shaft (m) = Volume x mass density

=

(Radius of gyration)2 = k2 = d2 / 8 I shaft = m. k2 = 2) Moment of Inertia of Hub: Outer dia. of the hub, do = Inner dia. of the hub, di = Width of the hub = Area of cross section = (π/4) (do2 - di2)

=

Volume of the Hub = Area of cross section x width of the Hub = Mass of the Hub(m) = Volume x mass density

=

56

(Radius of gyration)2 = k2 = (do2 + di2) / 8 = I Hub = m. k2 = 3) Moment of Inertia of Disk: Outer dia. of the disk, do = Inner dia. of the disk, di = Width of the disk = Area of cross section = (π/4) (do2 - di2)

=

Volume of the Disk = Area of cross section x width of the Disk = Mass of the Disk (m) = Volume x mass density

=

(Radius of gyration)2 = k2 = (do2 + di2) / 8 = I shaft = m. k2 = 4) Moment of Inertia of Rim: Outer dia. of the rim, do = Inner dia. of the rim, di = Width of the rim = Area of cross section = (π/4) (do2 - di2)

=

Volume of the Rim = Area of cross section x width of the Rim = Mass of the rim (m) = Volume x mass density

=

57

(Radius of gyration)2 = k2 = (do2 + di2) / 8 I rim = m. k2 = The mass moment of inertia of flywheel , I = ISHAFT + IHUB +IDISC + IRIM (Theoretical ) = Result: Mass moment of inertia of flywheel : Theoretical =

Experimental =

All Dimensions are in mm

SECTIONAL FRONT VIEW OF FLYWHEEL

Shaft

Hub

Disc

Rim

Bearing

245

68

55

18

Ø252Ø194Ø44Ø25

58

Expt. No.: Date :

DETERMINATION OF MASS MOMENT OF INERTIA OF CONNECTING ROD Aim:

To determine the mass moment of inertia of the given connecting rod about an axis through the centre of gravity perpendicular to the plane of oscillation. Materials required:

Connecting rod, stop clock, scale, weighing balance Theory:

The mass moment of inertia of a connecting rod is the product of the mass of the connecting rod and the square of the radius of gyration. In this experiment the connecting rod is assumed as a compound pendulum and the relationship established between the length of simple pendulum and the frequency of oscillation. This relationship is used to find the radius of gyration and hence the moment of inertia of the connecting rod. Procedure:

First of all, find the mass of the given connecting rod using weighing balance. Then allow oscillating the connecting rod as a compound pendulum gently on a knife edged pivot about its small end as shown in fig. Note down the time for 20 oscillations. Repeat this for two or three times and take the average time.

Similarly repeat the same procedure with big end of the connecting rod and note down the timings. Measure the distance between the points of suspension of small and big end with the pivot. This distance represents the length of the connecting rod. Tabulate all the observations. Observations : Connecting rod No. : Length of the connecting rod (l) : Mass of the connecting rod(m) :

Connecting Rod No.

Point of suspension

Time for 20 oscillations in sec Time for one

oscillation ‘tp’ in sec

Frequency fn = 1/tp

in Hz t1 t2 t3 tmean

Small end fn1 =

Big end fn2 =

59

CONNECTING ROD

h1, h2 = distance of centre of gravity from the point of suspension of small end and big end respectively

l = length of connecting rod G = centre of gravity Theory & Calculations: (Assume connecting rod as a Compound pendulum) We know that the frequency of oscillation of a compound pendulum f n = 1

2𝜋𝜋 �𝐾𝐾ℎ

𝑘𝑘𝐾𝐾2 +ℎ2

This equation can be written as

60

f n = 12𝜋𝜋 �

𝐾𝐾𝑣𝑣𝑒𝑒𝑒𝑒

where leq = 𝑘𝑘𝐾𝐾2 +ℎ2

ℎ

`h ‘ is the distance of centre of gravity from the point of suspension. kg is the radius of gyration about its centre of gravity. Let leq1 = length of equivalent simple pendulum when suspended from the small end leq2 = length of equivalent simple pendulum when suspended from the big end h1, h2 = distance of centre of gravity from the point of suspension of small end and big end respectively f n1 = 1

2𝜋𝜋 �𝐾𝐾

𝑣𝑣𝑒𝑒𝑒𝑒 1

Squaring both sides

f n12 = 1

4𝜋𝜋2 𝐾𝐾𝑣𝑣𝑒𝑒𝑒𝑒 1

∴ leq1 = 𝐾𝐾

4𝜋𝜋2 𝑓𝑓𝑖𝑖12 =

9.814𝜋𝜋2 ×( )2 =

Similarly we can write

leq2 = 𝐾𝐾

4𝜋𝜋2 𝑓𝑓𝑖𝑖22 =

9.814𝜋𝜋2 ×( )2 =

But leq1 = 𝑘𝑘𝐾𝐾2 +ℎ1

2

ℎ1

h1 leq1 = kg2 + h1

2

kg

2 = h1 ( leq1 - h1) …….(1)

Similarly we can write kg

2 = h2 ( leq2 - h2) ……..(2)

Equating (1) & (2) h1 (leq1 - h1) = h2 ( leq2 - h2) …..(3) Also we know that h1 + h2 = l ……….(4) Where ‘l’ is the actual length of the connecting rod between the points of suspension. Solving the equation (3) and (4) we can get h1 and h2 Then substituting the value of h1 or h2 in equation (1) or (2) we can get ‘ kg

’

61

Then Moment of Inertia of the connecting rod I = m × kg2

Where m is the mass of the connecting rod. I = m × kg

2 = I = Result: The mass moment of inertia of the connecting rod = …….. kg.m2

62

Expt. No.: Date : EXPERIMENTAL VERIFICATION OF NATURAL FREQUENCY OF TORSIONAL

VIBRATION OF SINGLE ROTOR SHAFT SYSTEM Aim:

To study the Torsional vibration of single rotor- shaft system

Description: Figure shows the general arrangement for carrying out the experiments. One end of the shaft is gripped in the chuck and heavy rotor free to rotate in ball bearing

is fixed at the other end of the shaft. The bracket with fixed end of shaft can be clamped at any convenient position along

lower beam. Thus length of the shaft can be varied during the experiments. Specially designed chucks are used for clamping the ends of the shaft. The ball bearing support to the rotor provides negligible damping during experiment. The bearing housing is fixed to side member of mainframe.

Procedure: 1. Fix the bracket at convenient position along the lower beam 2. Grip one end of the shaft at the bracket by chuck. 3. Fix the rotor on the other end of the shaft. 4. Twist the rotor through some angle and release. 5. Note down the time required for 20 oscillations 6. Repeat the procedure for different lengths of shaft 7. Make the following observations in the table:

a. Shaft diameter d = 4.6 mm b. Diameter of rotor D = 225 mm c. Mass of rotor m = 3.088 kg d. Modulus of rigidity for shaft material (MS) G = 8 × 1010 N/m2

Observation table:

Effective Length of Shaft (L)

Time for 20 oscillations in seconds

Periodic time (tp) in seconds

Frequency fn expt = 1/tp

Hz cm m t1 t2 t3 tave

Theoretical Calculations: Torsional stiffness of the shaft : kt = 𝐺𝐺 𝐼𝐼𝑆𝑆

𝐿𝐿

Where G = Modulus of rigidity for shaft : G = 8 x1010 N/m2

L = Effective length of shaft

63

Ip = Polar moment of inertia of shaft = 𝜋𝜋𝑑𝑑4

32 in m4

d = Diameter of the shaft in m

L

d

Frame

Rotor

Bracket

Chuck

Shaft

EXPERIMENTAL SETUP FOR TORSIONAL VIBRATION OF SINGLE ROTOR - SHAFT

SYSTEM Torsional stiffness of the shaft : kt = 𝐺𝐺 𝐼𝐼𝑆𝑆

𝐿𝐿 = ……. N.m

We know that 𝑓𝑓𝑖𝑖 𝑎𝑎ℎ𝑒𝑒𝑜𝑜 = 12𝜋𝜋�𝑘𝑘𝑎𝑎

𝐼𝐼 Hz

Where

kt = Torsional stiffness of the shaft

I = mass moment of inertia of the rotor = m k2

k = radius of gyration = �𝐷𝐷2

8

fntheo = Result tabulation:

Length of shaft in m. fexpt. (Hz) ftheo (Hz)

64

Expt. No. Date:

STUDY AND EXPERIMENTS ON CAM AND FOLLOWER MECHANISM

Aim: To study various types of cams & followers and to draw displacement diagram of the follower for the given four cam profiles. Description: The cam is a reciprocating, oscillating or rotating body, which imparts reciprocating or oscillating motion to a second body called the follower with which it is in contact. The cam mechanisms are commonly used in printing machinery, in automatic machines, machine tools, internal combustion engines, control mechanisms etc., There are at least three members in a cam mechanism 1. The cam, which has a contact surface either curved or straight. 2. The follower whose motion is produced by contact with the cam surface. 3. The frame which supports and guides the follower and cam

The cam rotates usually at constant angular velocity and drives the follower whose motion depends upon the shape of the cam.

The apparatus consists of roller follower and provision for mounting disc cams. The different cams are mounted one after another and rotated through the handle. The translator motion of the follower can be determined by the arm attached to the follower.

The displacement of the follower at various angular position of the cam is determined by attaching a paper over the plate. (On which the projecting arm moves). Using this observation the displacement diagram of the follower for the given cam can be drawn. Procedure:

Mount one of the cam profiles (say A) on the apparatus. Fix a white paper on the plate (on which the projecting arm moves). Rotate the cam using the handle through a known angular displacement (i.e., coinciding the follower with the division made on the cam). Now the position of the projecting arm on the paper can be marked. Similarly subsequent positions of the follower at other known angular positions can be determined for one full rotation of the cam. The same procedure has to be repeated for other cams. Tabulate the observations.

65

Observation table:

Angular displacement of cams in degree

Linear displacement of the follower of the cam

A B C D

30

60

90

120

150

180

188 - - -

210

240

270

285 - - -

300

315 - - -

330

345 - - -

360

Graph: To draw angular displacements of the cam Vs the linear displacements of the follower Result:

66

Graphical Method

Polar Chart

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 100 200 300 400

Circular Arc Cam

Offset Cam

0

1

2

3

4

5

Offset Cam

Circular Arc Cam

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