Aerospace and Mechanical Systems

8/8/2019 Aerospace and Mechanical Systems

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University of Manchester

School of Mechanical, Aerospace and Civil Engineering

Aerospace and Mechanical SystemsDr D.A. Bond

Pariser Bldg. C/21

e-mail: [email protected]: 0161 306 8733

UNIVERSITY OFMANCHESTER

1st YEAR LECTURE NOTES

AEROSPACE AND MECHANICAL SYSTEMS

1: ENGINEERING SYSTEMS

2: AUTOMOTIVE POWER TRANSMISSION SYSTEM

3: AEROSPACE POWER TRANSMISSION SYSTEM


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Aerospace and Mechanical Systems - What is an Engineering System?

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1. What is an Engineering System?

An engineering system is a set of components that function together to perform a required duty.Systems may be large or small and are defined by precise boundaries of theoretical or physicalform. System boundaries are chosen to concentrate attention on the system of interest. Two typesof information cross the system boundaries; input and output. In the simplest situation there willbe a single system input and single output however most engineering systems are complex andmay involve multiple inputs and outputs.

Consider the following examples

Chemical plant: The system may be the production process required to produce plutoniumfuel for a power generating reactor. The components of the system would be dissolvingvessels, mixers, precipitation vessels, filtration and other components.

Aircraft: The total transportation system of an aircraft may comprise, wings, fuselage,

power-plant, stabilisers, rudder undercarriage, environmental control etc.

Automobile: As with aircraft an automobile is comprised of many interoperating systemseg: power-plant, brakes, wheels, suspension drive train, fuel system

Within these higher level systems there may be lower level systems. Whilst at a lower level these

sub-systems still fall into the description of a system provided thay are comprised of multiplecomponents. Consider the following examples of sub-systems within an automobile propulsion

system:

Engine - which has further sub-systems including ignition, exhaust, air intake, cooling.

Gearbox with sub-systems such as driveshafts and gears, mechnical levers.

Clutch comprised of clutch plates, springs, hydraulic actuators, mechanical levers.

Differential/Final-drive with planet gears, drive gears, bearings, housings.

Propshaft including bearings, couplings and possibly CV or universal joints.

Axle including bearings, couplings and possibly CV or universal joints. Wheels which includes bearings, tyres, rims.

Brakes which may comprise brake levers, pads, rotation sensors, hydraulic cylinders.

The above systems and sub-systems are not an exhaustive set for each case but are indicative ofthe breadth of mechanical or electrical hardware that may comprise a system. It is thereforepossible to define systems of very different types. The essential principle is that the boundary of asystem is precisely and clearly defined.

Size does not define a system; in fact large systems may be formed by a set of smaller systems, asin the case of transport or in the chemical plant above. In most systems it is possible to identifylevels of increasing engineering detail until a fundamental component is reached. However theobjective of this course is not to discuss at length the relationship between systems, level of sub-

systems and components, but rather to study how components make a contribution to the functionof systems and how systems perform.

The course is a mixture of quantitative and qualitative topics which shall explore a number ofaspects of different engineering fields. Engineering analysis will be limited to simple cases orsimplified components but will be the starting point for more detailed analyses which will becovered in future years.


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1.2 Relationship to Systems Engineering

The concepts of Systems Engineering are followed to some extent in this course in that we lookat the performance of a system rather than the individual parts that comprise the system.

"Systems Engineering is an interdisciplinary approach and means to enable the realization of

successful systems. It focuses on defining customer needs and required functionality early in thedevelopment cycle, documenting requirements, then proceeding with design synthesis and system

validation while considering the complete problem: Operations, Performance, Test,

Manufacturing, Cost & Schedule, Training & Support, and Disposal."

International Council On Systems Engineering

"A branch of engineering which concentrates on the design and application of the whole as

distinct from the parts looking at a problem in its entirety, taking account of all the facets and

all the variables and linking the social to the technological."

Ramo S. in Stevens et al, 1998, Systems Engineering: Coping With Complexity', Prentice Hall Europe

"An interdisciplinary approach encompassing the entire technical effort to evolve and verify an

integrated and life-cycle balanced set of system people, product and process solutions that satisfy

customer needs."

Shishko, R, NASA Systems Engineering Handbook, 1995.

1.3 Course examples

This course draws heavily on engineering examples from Transport systems and will coverexamples relevant to both Mechanical and Aeronautical applications. Where the dual purpose of asystem is not obvious please ASK where such a system may be employed in the application ofyour interest.

Figure 1-1: Aircraft Structural, Propulsion and Control Systems

The two most obvious and fundamental industries to the Mechanical and Aeronautical courseswill be used to provide the majority of the examples for analysis; Aircraft and Automobiles.


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Before we start into the analysis and consideration of systems within these transport devices it isworth considering at a higher level the similarity of systems used in the two devices.

Generic

System

Generic Purposes Automobile Examples Aero Examples

Structure Hold other systems together. Provide an environment for payload.

Crash protection

Support vechicle in/on medium oftravel

Body/ChassisWheels

FuselageWingsUndercarriage

Propulsion Propels vehicle

Provides power for otherapplications

IC Engines IC EnginesGas Turbine EnginesRockets

Fuel Supply an appropriate quantity of fuelto the propulsion system

Maintain efficiency of propulsionsystem

Monitor usage of fuel source

Fuel Injection/Carburettor Fuel Injection/Carburettor

Hydraulic/Pneumatic

To control motion of structure viaactuators

Brake System/MasterCylinderSome clutch systems

Control Surface DeflectionBrakesG-Suit

Electrical Power distribution in the vehicle LightingHeating

LightingHeatingSome component actuation

Instrumentation

Provide information on performance TachometerFuel gauges

HSIAirspeed IndicatorNav Instruments

These are all very generic systems within not only transport systems but any operation thatrequires power to be supplied from a fuel source, control of a process or actuation of components.The systems breakdown is not limited to mechanical devices but also can be used to consider

biological systems such as in a human.

This course will focus on the propulsion systems as they contain many examples of complexsystem applications. In particular the course will look at:

Auto Power Transmission Systems including engines, turbochargers, clutches, gearboxes,differentials

Aero propulsion systems propeller driven, gas turbine, turbocharged engines

Braking Systems Disc and drum brake mechanisms

Hydraulic and Pneumatic Systems Hydraulic actuators and G-suits

Questions

Try to think of equivalent systems in a human that match onto the generic systems listed in thetable above.


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Aerospace and Mechanical Systems Automotive Power Transmission System

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2. Automotive Power Transmission System

2.1 System elements

The power transmission system of a truck, car or motorcycle is an integrated set of carefullymatched components from engine to wheels. The principal elements for a vehicle with two

driving wheels are:

ENGINE CLUTCH GEARBOX DIFFERENTIAL WHEELS

Figure 2-1: Principal Elements in Power Transmission System

2.2 Engines

2.2.1 Principle of Operation of Automotive Internal Combustion (IC) engines

2.2.1.1 Four stroke cycle

Figure 2-2: Four Stroke Engine Cycle

(after Anderson J.D., Aircraft Performance and Design, McGraw Hill,1999)


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Aerospace and Mechanical Systems - Automotive Power Transmission System

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Prior to arriving at the engine cylinder air is drawn into the engine through the carburettor, wherefuel is added. The resultant fuel-air mixture is then ducted to the top of the engine, ready to bedrawn into the cylinder through the intake (see Figure 2-2). Once at this stage the four stoke cycle

begins. The four stages (which correspond to an up or down motion of the piston also known asa stroke) are:

a. Induction/Intake Stroke where the piston moves downwards, sucking the available fuel-air mixture through an opened valve into the area at the top of the cylinder known as thecombustion chamber. The sucking process continues all the time that the piston movesdown the cylinder.

b. Compression Stroke where the inlet valve closes and the piston moves upwards,compressing the fuel-air mixture (consequently increasing the temperature of the fuel airmixture). As the piston approaches the top of the stroke, known as Top Dead Centre (TDC),spark from the spark plug ignites the fuel-air mixture, causing it to combust. Thecombustion process increases the pressure of the mixture and the burning mixture wants toexpand.

c. Power/Work Stroke - the burning/expanding gases force the piston downwards providing

power for the engine to supply less a component that is used to power the other threestrokes in the other cylinders of the engine.

d. Exhaust Stroke- where the exhaust valve opens and the piston moves upwards expelling theburnt gases out of the cylinder at high velocity.

Less formally, the four strokes are sometimes called Suck, Squeeze, Bang, Blow.

A four-stroke engine may have any number of cylinders from one upwards. However, most arearranged in pairs as paired cylinders make an engine easier to balance. Most engines use part ofthe power generated in the power stroke to rotate a heavy flywheel to keep the rotation smoothand to keep the crankshaft turning during the "non-power" strokes.

Questions

What is the purpose of the carburettor and how does it work?

Consider the Perfect Gas Laws (PV=nRT) and understand why and to what extent thetemperature must rise when the fuel/air gas mixture is compressed.

How does a camshaft operate and what is its main purpose?

2.2.1.2 Two stroke

The mechanics of a two-stroke engine are more simple than those of a 4 stroke engine. Unlike thefour-stroke engine, the complete cycle takes only one upstroke and one downstroke of the piston,so some of the four stages of operation must occur simultaneously.

a. Induction/Intake Stage The fuel and air are mixed in the carburettor in the usual way, butinstead of being sucked into the top of the cylinder, the fuel/air mixture is drawn into the

crankcase by the vacuum created during the upward stroke of the piston. The pressureimbalance caused by the vacuum being created causes the inlet poppet valve to openallowing fuel into the crankcase.

b. Compression Stage while the upwards motion of the piston is drawing in a new fuel airmixture, a previous mixture is being compressed above the piston. The upwards motion ofthe piston in a single cylinder two-stroke engine is driven by a flywheel. The crankcase isconnected to the combustion chamber in the cylinder by an inlet port (also known as a


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transfer port) and to the exhaust manifold by the exhaust port. In the induction andcompression stages of the cycle both ports are covered, sealing the cylinder and allowingthe new charge of fuel-air mix to be drawn in and the previous mix to be compressed.

c. Power/Work Stage - At the top of the upstroke the spark plug ignites the fuel mixture andthe burning fuel expands, driving the piston downward. During the downward stroke the

poppet valve is forced closed by the increased crankcase pressure. As the piston movesdown the inlet and exhaust ports are uncovered (replicating the function of the valves in thefour-stroke engine).

d. Exhaust Stage as the inlet port is now open the compressed fuel/air mixture in thecrankcase moves through the port (around the piston) and into the cylinder. This addition ofthe new fuel/air mixture expels the burnt exhaust gasses out the exhaust port. Unfortunatelythough, some of the fresh fuel mixture is usually expelled as well.

Figure 2-3: Two Stroke Cycle

Questions

How is the lubrication of a two-stroke engine different to that of a four-stroke engine andwhy does that mean you should not use two-stroke fuel in a four-stroke engine?

What would be more efficient a two-stroke or four stroke engine? Why?

2.2.2 Turbocharger

Turbochargers compress the air being fed into an engine thus forcing more air into a cylinderduring the induction stroke. The increased amount of oxygen in the compressed air means thatadditional fuel may be added (the fuel/air mix ratio being fixed at a ratio such that the fuel aircombustion process consumes all or nearly all of the fuel).This allows more power to begenerated during each power stroke and hence more power will be available from the engine. Aturbocharged engine produces more power overall than the same engine without a turbocharger.A parameter used to assess the impact of turbocharger is the power-to-weight ratio for an engine.The weight of a transport engine is directly related to the amount of energy required to propel it.


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So as high as possible a power-to-weight ratio for an engine is ideal. A turbocharger is quite lightbut adds quite a lot of power to an engine and so helps increase an engines power-to-weight ratio.

The turbocharger is connected to the engine exhaust manifold and uses the high velocity exhaustgas flow from the engine to spin a turbine, which then drives an air compressor. Turbochargerturbines operate at speeds over 120,000 rpm.

Figure 2-4: The Turbocharger System (from www.howsttuffworks.com)

Figure 2-5: A Turbocharger Operation (from www.howsttuffworks.com)

A Turbocharger may increase the pressure of the engine inlet air by around 7psi (this isapproximately 50% of the normal atmospheric air pressure 14.7psi). This then allows almost 50%more air and fuel to be added to the cylinder. However 50% more energy is not produced due toinefficiencies in the system and restricted exhaust flow. Performance increases of 30-40% aremore typical.

As will be examined in future lectures on gas turbine engines, the power to spin the turbine andcompress the inlet air must come out of the additional power provided. This is seen through theturbine making it more difficult to force the exhaust gases out so the engine has to use part of itspower stroke energy to drive the exhaust gases out.

A turbocharger also helps at high altitudes where the air is less dense. Normally aspirated engineswill have reduced power at high altitudes for exactly the opposite reason why turbochargers


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increase power. The air-pressure is lower, the air density less hence less air is sucked into acylinder during an induction stroke. A turbocharger overcomes this problem and although aturbocharged engine may also have reduced power at high altitudes (as the same inlet fuel/air

pressure cannot be achieved from the lower starting air pressure), the reduction will besignificantly less as the less dense air is easier for the turbocharger to compress thus requiring

less energy from the engine.2.2.3 Supercharger

A supercharger is similar to a turbocharger in that they are both forced induction systems thatcompress the air flowing into an engine. Also like a turbocharger the supercharger relies upon arotating air compressor to provide the compression. Where the turbocharger uses the engineexhaust gases to spin a turbine to drive the compressor, a supercharger is connected via a beltdirectly to the engine. A supercharger provides a similar boost in inlet fuel/air mix pressure(approx 7psi) to that of a turbocharger.

Theoretically a turbocharger is more efficient because it makes use of the energy contained in theexhaust gas flow that would otherwise simply be lost to the environment. However the exhaustback-pressure that this creates reduces the efficiency of a turbocharger such that the net additionalpower provided is slightly less than that of a supercharger when installed on an engine operatingat relatively low speeds (rpm). At higher engine speeds (rpm) a turbocharger becomes moreefficient and is the preferred method for boosting the performance of high-speed engines.

Questions.

How do the operating speeds of a turbine compare to those of the overall engine?

What operating temperature would you expect a turbocharger turbine to operate at and howwould that affect the design and manufacture of a turbine?

What might be some of the critical design issues with a turbine due to its high operatingspeeds?

What happens when a turbo blows-up and how might the driver recognise it?

Can a turbocharger be added to a normal road vehicle?

Why are aero engines turbocharged?

2.2.4 Measuring Power, Torque and Speed of an Engine

The power (P) transmitted by an engine shaft is the product of the torque (T) and the rotationalspeed ( ).

TP = (2.1)

where Power has SI units of Watts (W= Nm/s), Torque SI units of Newton Metres (Nm) androtational speed SI units of Radian per Second (rad/sec).


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Example:

If an engine develops 70Nm of Torque at 3000 rpm the power generated by the engine is:

kWhp

RPM

where

hpkWP

7457.01

60

2

5.290.2260

2300070

=

=

==

=

The Torque generated at the output of the crankshaft can be measured at different engine speedsto produce what are known as Torque curves for an engine. Consider the example below for a1996 1.5l turbo-charged Honda F1 engine. This data may be used to calculate the maximumpower output of the engine over the operating engine speeds.

300

400

500

600

700

800

6 7 8 9 10 11 12 13

Engine Speed (x1000 rpm)

Torque(Nm)

300

400

500

600

700

800

Power(kW)

Figure 2-6: Torque curve for a Honda F1 engine

Note the speed range of the engine shown is much higher than for a normal road vehicle.

Questions:

Why is a high engine speed attractive?

What are the disadvantages of high engine speeds?


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2.3 Clutch

2.3.1 Introduction

The clutch provides the means of separating the engine from the drive train when a vehicle needsto be at rest or when changing gear. A very common design consists of two plates pressedtogether by a spring. Torque is transmitted from one plate to the other through friction. The clutchplates are separated by forcing one plate back against the spring and separating the connectionbetween plates (this is of course what you do when you depress the clutch pedal in a car).

Figure 2-7: Diaphragm Spring Clutch

The above described clutch is known as a single plate dry clutch. In industrial applications, adifferent arrangement is often used. The clutch plates can be immersed in oil. The oil reduceswear as the clutches engage and disengage machinery quickly. Clearly the coefficient of frictionis reduced considerably therefore multiple clutch plates are used to improve torque capacity.There are clutch designs that also use multiple clutch plates without being immersed in oil.

Figure 2-8: Multiple Plate Clutch


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Note the multiple clutch plates must be free to move axially on their shaft in order that each platehas the contact force of the spring. If the plates were not free to move axially then the springcontact force would be shared between the plates thus reducing the torque capacity.

2.3.2 Clutch Analysis

Figure 2-9 shows a friction disk having an outside diameter D and an inner diameter d. Asimplified analysis may be used to determine the axial force F necessary to transfer a certainTorque T.

FdDr dr

dA

Figure 2-9: Simplified Sketch of Clutch Plate

Two methods of solving this problem exist:

Uniform Wear If the disk is assumed to be rigid then the greatest amount of wear willfirst occur in the outer areas of the plate where the work done by friction is greatest (as the

outer rotates at a higher linear velocity). After a certain amount of wear the pressuredistribution will change so as to permit wear to be uniform.

Uniform Pressure A simpler method of analysis but which requires a complex design forthe clutch assumes that the pressure distribution across the plate is kept uniform.

2.3.2.2 Uniform wear analysis

After initial wear the pressure exerted on the plate will be greatest at r = d/2 in order for the wearto be uniform. Define this maximum pressure as Pmax. Wear is proportional to work, work isequivalent to a force multiplied by a distance and the distance a point at radius r travels in a giventime relative to the inside radius is equivalent to the ratio of the radii (as Vr = r). This allows thepressure at any point to be written as:

rdPP 2/max= (2.2)

The elemental Force (dF) being applied on elemental area dA(= 2 rdr) is simply the pressure atradius r multiplied by dA. Thus the total Force may be found by integrating of the range r = d/2 to

r = D/2:


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( )dDdP

drdP

drrr

dP

drrP

dFF

D

d

D

d

D

d

D

d

=

=

=

=

=

2

.22/

.2.

max

2/

2/

max

2/

2/

max

2/

2/

2/

2/

(2.3)

The torque that this clutch is able to transfer is found by integrating the product of the elementalFriction Force (dFf= dF with = coefficient of Friction between the clutch plates) and theradius that it acts at (r):

( ) ( )dDFdDdP

drrdP

drrr

drP

rdFT

D

d

D

d

D

d

+==

=

=

=

48

.

.22/

22max

2/

2/

max

2/

2/

max

2/

2/

(2.4)

2.3.2.3 Uniform Pressure AnalysisA similar approach may be used considering the pressure to be uniformly distributed to give:

( )224

dDP

Funiform

=

(2.5)

( )3312

dDP

Tuniform

=

(2.6)

Questions

What effect would using multiple clutch plates have on the size of a clutch required totransfer a given torque?

What effect would using a stiffer clutch spring have on the Torque able to be transferred bya clutch?

Is the simplified clutch described above any different to a simple disc brake? If not could asimilar analysis be used for a disc brake?


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2.4 Gearbox

Figure 2-10: Four-speed Gearbox

The detail design of gearboxes is beyond the scope of this course and their study will be confinedto gear ratios to match engine power to vehicle performance.

The power transmission system provides the drive from the engine to the wheels. An engine mayoperate at speeds of 1000 rpm to 7000rpm for a road vehicle (and as shown earlier even higherfor performance vehicles). Typically a road wheel has a rolling radius (r) of approximately 0.25m(measured from the ground to the centre of the wheel). The speed of the wheel is therefore:

rV/= (2.7)

where V is the road speed.


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Therefore for road speeds up to 120mph, varies from 0 to 215rad/sec or 2049 rpm. Thetransmission system must match 1000-7000rpm for the engine to 0-2000rpm for the wheels.Disengaging the clutch enables the engine to run whilst the rest of the transmission system isstationary.

The possible gear ratio range varies from approximately 15:1 (= 5000:341 considering 20mph infirst gear say at 5000rpm engine speed) to approximately 3.5:1 (= 3800:1170 considering 70mphat 3800 rpm engine speed).

Gearboxes generally operate most efficiently when they do not have to perform large speedvariations. If the final drive has a speed reduction of about 4:1 then the gearbox does not have to

operate at high gear ratios. Typically differential gears have gear reduction rations of about 4:1for automobiles.

Let the engine speed (also the input speed to the gearbox) be 1, the output speed of the gearboxbe 2 (also the input speed to the final-drive) and the output speed of the final drive be 3 whichis also the speed of the wheels.

The relationship between the engine speed and the wheel speed is then:

r

Vratiodrivefinalratiogearbox

speedwheelratiodrivefinalratiogearbox

=

=

= 3

3

2

2

11

(2.8)

From this equation the gearbox ratios may be calculated:

Consider a typical road car

Gearbox Ratios

First 3.25:1

Second 1.90:1Third 1.25:1

Fourth 0.937:1

Fifth 0.771:1

Final-drive 4.25:1


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2.5 Differential/Final-drive

Figure 2-11: Final-drive (Differential) Locations on modern automobiles (from www.howsttuffworks.com)

The detail design of the final drive is also beyond the scope of this course but we shall considerits functionality. The function of components is, of course, an essential aspect to theunderstanding of system performance.

The 'final drive' is a differential gearbox that distributes the drive to the two driven wheels. On afront engine, rear wheel car drive the differential is often incorporated into a back axle or as aseparate gearbox at the back if the car has independent rear suspension. On a front engine, frontwheel drive car the final drive is usually incorporated into the gearbox housing but it can belocated at the points where the drive shaft couplings 'emerge' from the gearbox.

Figure 2-12: Transfer of Torque by a Differential Gearbox

The differential serves several purposes.

It is a final, fixed reduction gear in the drive system to match engine speed to wheel speed.

It creates two drive shafts to the wheels from a single drive input from the gearbox.

It allows the driven wheels to rotate at different speeds. Travelling in a straight line bothwheels must rotate at the same speed. Around a bend though, the inside wheel must rotatemore slowly than the outside wheel.


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Figure 2-13: Variation in distance travelled by internal and external wheels:

If the differential did not perform this last task the wheels would be locked together and rotate at

the same speed all of the time. For a vehicle to turn, one tyre would have to slip. With moderntyres and sealed roads, a great deal of force is required to make a tire slip. That force would have

to be transmitted through the axle from one wheel to another, putting a heavy strain on the axlecomponents and greatly increasing the wear rate on the tyres.

Figure 2-14: Rear Differential

On front wheel drive vehicles, the differential unit is normally part of the transaxle assembly. Onrear wheel drive vehicles, it is part of the rear axle assembly. A differential unit is usuallycontained within a cast iron casting known as the differential case. Located inside the case are thedifferential pinion shafts and gears and the axle (or wheel) side gears.


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When an automobile is moving straight ahead, both wheels are free to rotate. Engine power isapplied to the pinion gear, which rotates the ring gear. Beveled pinion gears are carried around bythe ring gear and rotate as one unit. Each axle receives the same power, so each wheel turns at thesame speed.

Figure 2-15: Operation of a Differential

When the car turns a sharp corner, only one wheel rotates freely. Torque still comes in on thepinion gear and rotates the ring gear, carrying the bevelled pinions around with it. However, oneaxle is held stationary (or rotate more slowly) and the bevelled pinions are forced to rotate ontheir own axis and "walk around" the slower rotating axle side gear gear. The other side is forcedto rotate more rapidly because it is subjected to the sum of the of the ring gear rotation and therotation of the pinion gears walking about the other axle axis.

To prevent a loss of power on slippery surfaces, a differential lock is often used to lock the twoaxles together until the slippery spot is passed, at which point they are released. Thesedifferentials are referred as to limited-slip or traction-lock differentials. When the car isproceeding in a straight line, the differential gears are locked against rotation due to gear reaction.When the vehicle turns a corner or a curve, the differential pinion gears rotate around thedifferential pinion shaft. The differential pinion gears allow the inside axle shaft and drivingwheels to slow down. On the opposite side, the pinion gears allow the outside wheels to


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accelerate. Both driving wheels resume equal speeds when the vehicle completes the corner orcurve. This differential action improves vehicle handling and reduces driving wheel tire wear.

Example

The tooth numbers for the automotive differential shown below are N2=17, N3=54, N4=11,

N5=N6=16. If the drive shaft turns at 1200rpm what is the speed of the right hand wheel if it isjacked up and free to rotate and the left wheel remains on the road surface?

Figure 2-16: Schematic of a rear differential (after Shigley, J.E., Mechnical Engineering Design, McGraw Hill, 1975)

The ring gear rotates 17/54

th

of a rotation for every rotation of the drive shaft =120017/54=378rpm.

As the Left Hand wheel is locked gear 5 does not move.

During one rotation of the ring gear the planet gears will rotate about their axis 16/11 timesbecause of their interaction with the stationary left hand axle gear (No 5) =37816/11=550rpm.

If the planet gears were to remain stationary about the wheel axis their rotation about theirown axis would cause the right hand axle gear (No 6) to rotate at a speed 11/16

thof the

planet gears i.e. = 55011/16=378rpm.

Added to this planet gear walking rotation is the overall ring gear rotation of 378rpm

therefore the total rotational speed of the right hand axle = 378 + 378 = 756rpm.

Questions

Why might the trait of differential gearboxes illustrated in the example above not bebeneficial for car performance (especially in the wet or in boggy conditions)

How do modern vehicle overcome this problem


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2.6 Performance

2.6.1 Power

The engine develops a certain amount of power and torque depending on its speed of operation(and the fuel input).

The torque T produced at the wheels gives a force F = T/r where r is the rolling radius of the

wheel. F provides:

a means of accelerating the car (F = ma), and

a means of climbing hills, F>mg.sin for the vehicle to move at all up the slope, the extraforce is available for acceleration ( is the inclination of the slope).

The power produced by the engine determines the rate at which a vehicle can work. At constantspeed, the power of the engine determines the maximum vehicle speed considering the rate ofwhich work is done:

raising the mass of the vehicle (climbing a hill), overcoming drag from the air,

overcoming rolling resistance, and overcoming losses in the power transmission systems.

Example:

Calculate the power required to drive a car of mass 1300kg up a hill of inclination of 10 at 18m/s (approx 40 mph).

53hp39.86kW

18sin109.811300

VelocityForce

imeDistance/TForce

timeWorkPower

=

=

=

=

=

= /

So, unless a vehicle has a power output greater than 53hp it cannot climb a 10 slope at 40 mph.

Question

How does power to weight ratios of vehicles affect their hill climbing capabilities.

2.6.2 Resistance to motion

The drag D due to wind resistance is given by

2d V.ACD = (2.9)

where:

Cd = drag coefficient (no units, derived by experiment or Computational Fluid Dynamicsand depends on detail shape),

A = frontal area of the vehicle,

= density of air (1.2 kg/m3 at sea level) and

V = velocity (m/s) of air over the vehicle.


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Example

Calculate the power required to overcome drag on a car with a Cd of 0.35 travelling at 100 mph(45 m/s) in still air if the frontal area of the car is 2.5m

2.

D = (0.35).(0.5).(2.5).(1.2).(45)2

= 1063N

The power required to overcome the drag force = D V

Power = 1063.(45) = 47.8kW = 64hp

Think about the size of this answer, it is a large amount of power for a car just to overcome drag.Remember that there must be an excess of power for rapid acceleration or towing applications.

If the rolling resistance of the vehicle is 120N and is independent of speed, then the power

required to overcome rolling resistance is 120.(45) = 5.4kW = 7.2 hp. Which means the totalpower required to travel at 100mph is approximately 71hp.

2.7 Web References:

a. http://www.howstuffworks.com/sc-engines-automotive.htm

b. http://www.keveney.com/Engines.htmlc. http://www.familycar.com/auto101.htm

d. http://www.autoshop-online.com/auto101

e. http://www.ogura-clutch.com/whatsnew/feature5.html

f. http://www.turbointernational.com

g. http://www.mercedes-benz.com/e/cars/e-class-t/facts_m2.htm


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Tutorial Sheet 1: Automotive Power Transmission System

i. Explain the 4T and 2T IC engine cycles.

ii. Over what speed range does a modern car engine operate? Explain briefly two principalfactors that limit the speed of engines.

iii. Name the elements in an automotive power transmission system. What functions does the

final drive ('differential' gearbox) perform in the power transmission system of a car?

iv. A Harley-Davidson Sportster 883 develops a maximum torque of 6INm at 2500rpm.Calculate the power output at this speed. If the torque reduces to 50Nm at 5000rpni,calculate the power developed at 5000 rpm.

Answers (16kW/21.8hp, 26.2kW/35. 7hp)

v. A Kawasaki GPZ 500S develops a maximum torque of 46.1Nm at 8500rpm. Calculate the

power developed at 8500 rpm. The maximum power developed by the engine is 44.1 kW at9800rpm, calculate the torque developed at this speed.

Answers (41kW/55.8hp, 43Nm)

vi. A typical ' hot hatch' saloon car might develop 130 bhp (Brake horse power, the poweroutput at the dynamometer brake) and have a mass of 900 kg. A typical sports/all roundermotorcycle, eg Honda CB 500, has a power output of 55bhp and a mass of 170 kg.Calculate the power to weight ratio of the two machines and comment on their expectedperformances.

vii. Calculate the power - engine speed characteristic from the torque curves for the followingtwo engines, Explain why one of the engines will need more gears than the other foreffective driving.

60

70

80

90

100

110

120

130

140

150

160

2 3 4 5 6 7 8 9

Engine Speed (x1000 rpm)

Torque(Nm)

50

60

70

80

90

100

110

120

Power(kW)

Figure 2-17: Torque Curves


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viii. Derive the uniform pressure expressions for Force and Torque for a single plate dry clutch.Answers (see eqns 2.5 and 2.6

ix. A single dry plate clutch has a single pair of mating surfaces 300mm outer diameter by 225mm internal diameter. The coefficient of friction in 0.25 and the maximum pressure is825kPa. Find the torque capacity using:a) the uniform wear assumptionb) the uniform pressure assumption.

Answers (718Nm, 842Nm)

x. A single dry plate clutch has a single pair of mating friction surfaces, is 200m outsidediameter and 100mm inside diameter and has a coefficient of friction of 0.30. What is themaximum pressure corresponding to an actuation force of 15kN? Use both the uniform

wear and uniform pressure assumption.Answers (955kPa, 637kPa)

xi. A car has a gear ratio of 0.8:1 in fifth gear, a final drive ratio of 3.9:1 and a wheel rollingradius of 0.29. Calculate the engine revs per 10mph of road speed in fifth gear. If the topspeed is 115 mph, at what is the engine speed?

Answers (459.2rev/10mph, 5281rpm)

xii. Calculate the required gearbox ratios for the following conditions.

Gear Road Speed

(mph)

1 18

2 253 33

4 46

for an engine speed in each case of 3000rpm, a final drive ratio of 4:1 and a wheel rollingradius of 0.25m

Answers:(2.44:1, 1.757.1, 1.33:1, 0.955:1)

xiii. Plot a graph of the drag force on a vehicle against the vehicle speed (cover the values at 10mph, 30mph, 50mph, 70mph, 90mph, 110mph, 130mph, 150mph). Assume that the frontalarea of the vehicle is 2.0m

2and that the vehicle has a Cd of 0.35. The density of air is

1.2kg/m3.Answers (8.9, 75.5,210, 411, 680, 1015, 1418, 1888 N)

Take note of this result, gain a feel for the numbers.

xiv. Calculate the power required to overcome the drag of the vehicle in Q(xiii) at 70 and 110

mph. If the rolling resistance of the vehicle is 300N (assume independent of speed),calculate the power to overcome rolling resistance.

Answers (12.9, 49.9, 9.4, 14.8).

Note how the rolling resistance is less important at high speed.


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Aerospace and Mechanical Systems