chapter 2.ppt

Dr. B DayalDr. B Dayal

PERFORMANCE CHARACTERISTICS PERFORMANCE CHARACTERISTICS OF ENGINESOF ENGINES

gasolinegasoline dieseldiesel

ACCELERATION PERFORMANCE ENGINE POWER TRACTION LIMITS ON THE TYRE WHEELS.POWER LIMITED ACCELERATION.ENGINE.POWER = TORQUE X SPEED = 2πTNPower = rate of work done = force x dS / dt = force x V

Force = Fx = power / V

Fx = Max = power / V

ax = power / MV = power . g /(W . V)

POWER TRAIN IN AUTOMOBILE SYSTEM

ENGIENGINENE

CLUTCCLUTCHH

GEAR GEAR BOXBOX

PROPELLOR PROPELLOR SHAFTSHAFT

DIFFERDIFFERENTIALENTIAL

AXLE AXLE SHAFTSHAFT

AXLE AXLE SHAFTSHAFT

FINAL DRIVEFINAL DRIVEWHEELWHEEL

WHEEWHEELL

POWER TRAINTc = Te – Ie αe

Where, Tc = torque at the clutchTe = engine torque at a given speedIe = engine rotational inertiaαe = engine rotational acceleration

TD = (Tc – Itαt) . GWhere, TD = torque output to the propeller shaft

G = gear ratioIt = rotational inertia of the transmission

Ta = Fxr + Iw αw = (TD – Id αd)GFD

Where, Ta = torque on the axlesFx = tractive force at the groundr = radius on the wheelIw = rotational inertia of the wheels and axle shaftsαw = rotational acceleration of the wheels

POWER TRAINId = rotational inertia of the drive shafts

αd = rotational acceleration of the drive shaft

GFD = gear ratio of the final drive

α d = GFD . αW

And αe = G . Αd = G . Gfd . Αw

Fx = Te . Goverall / r – [(Ie + It)G2overall + IdG2

FD + Iw]ax / r2

Where, Goverall = combined ratio of transmission and final drive.Taking efficiency of transmission and final drive into consideration:

Fx = [Te . Goverall . ƞtf / r] – [(Ie + It)G2overall + IdG2

FD + Iw] ax / r2

Where, ƞtf = combined efficiency of transmission and final drive.• First term on the right side represents the steady state tractive

force available.• The second term represents the “loss” of tractive force due to

inertia of the engine and driveline components.

POWER TRAINKnowing the tractive force, it is now possible to predict the

acceleration performance of the vehicle.M . ax = W . Ax / g = Fx – Rx – DA – Rhx – W sin ƟWhere, Rx = rolling resistance forces

DA = aerodynamic forcesRhx = hitch (towing forces)

Since Fxincludes the tractive force and rotational terms. For rotational terms, taking lumped mass into consideration:

(M + Mr)ax = (W + Wr)ax / g = [Te . Goverall . ƞtf / r] - Rx – DA – Rhx – W sin Ɵ

Where, Mr = equivalent mass of rotating componentsM + Mr = “effective mass”

And (M + Mr) / M = “mass factor”• The mass factor will depend upon the operating gear. The

representative number is often taken as:Mass factor = 1 +0.04 + 0.0025 G2

overall

TRACTIVE EFFORT – SPEED CHARACTERISTICS FOR A MANUAL TRANSMISSIONTRACTIVE EFFORT – SPEED CHARACTERISTICS FOR A MANUAL TRANSMISSION

The constant engine power line is equal to the maximum power of the The constant engine power line is equal to the maximum power of the engine, which is upper limit of the tractive effort available, less any losses in engine, which is upper limit of the tractive effort available, less any losses in the drive line.the drive line.

TORQUE CONVERTORTORQUE CONVERTOR

CHARACTERISTICS OF A CHARACTERISTICS OF A TYPICAL TORQUE CONVERTERTYPICAL TORQUE CONVERTER

At 0 output speed, At 0 output speed, output torque is more output torque is more than two times the than two times the engine torque, engine torque, providing for good providing for good take off.take off.As speed builds up As speed builds up the transmission the transmission input approaches the input approaches the engine speed and engine speed and torque ratio drops to torque ratio drops to 1.1.

TRACTIVE EFFORT – SPEED CHARACTERISTICS FOR AN AUTOMATIC TRANSMISSION

In first gear the tractive In first gear the tractive effort rises significantly.effort rises significantly.Road load forces are Road load forces are shown arising from rolling shown arising from rolling resistance, aerodynamic resistance, aerodynamic drag and road grade.drag and road grade.The intersection between The intersection between the road load curve and the road load curve and tractive effort curves is the tractive effort curves is the maximum speed that can maximum speed that can be sustained in that gear.be sustained in that gear.

SELECTION OF GEAR RATIOS BASED ON GEOMETRIC PROGRESSION

The actual ratios selected The actual ratios selected for a transmission may be for a transmission may be tailored for performance in tailored for performance in specific modes – an optimal specific modes – an optimal first gear for starting, a first gear for starting, a second or third gear for second or third gear for passing, and a high gear for passing, and a high gear for fuel economy.fuel economy.The best gear ratios usually The best gear ratios usually fall close to a geometric fall close to a geometric progression, in which the progression, in which the ratios change by a constant ratios change by a constant percentage from gear to gear.percentage from gear to gear.

GEAR RATIOS ON A TYPICAL GEAR RATIOS ON A TYPICAL PASSENGER CARPASSENGER CAR

EXAMPLE PROBLEMSProb 1. we are given the following information about the engine and

drive train components for a passenger car:Engine inertia 0.8 in – lb - sec2

Torque available:

Transmission data:

Final drive: inertia = 1.2 in – lb – sec2; ratio = 2.92; efficiency = 0.99; wheel size = 12.59 inches

Calculate the effective inertia of the drive train components in first gear

Rpm 800 1200 1600 2000 2400 2800 3200 3600 4000 4400 4800 5200

torque 120 132 145 160 175 181 190 198 200 201 198 180

Gear 1 2 3 4 5

Inertia (in – lb – sec2) 1.3 0.9 0.7 0.5 0.3

ratio 4.28 2.79 1.83 1.36 1.00

efficiency 0.966 0.967 0.972 0.973 0.970

EXAMPLE PROBLEMS

Prob 2. calculate the maximum tractive effort and corresponding road speed in first gear and fifth gear of the car given in previous slide when inertial losses are neglected.

TRACTION LIMITED ACCELERATION

Presuming there is adequate power from the engine, the acceleration may be limited by the coefficient of friction between the tyre and road.

Fx = μWWhere, μ = coefficient of friction

W = weight of the vehicleThe weight on a drive wheel then depends on the static

plus the dynamic load due to acceleration, and on any traverse shift of load due to drive torque.

TRANSVERSE WEIGHT SHIFT DUE TO DRIVE TORQUE

Transverse weight shift occurs on all solid drive axles. The drive shaft into the differential imposes a torque Td on the

axle.A torque due to suspension roll A torque due to suspension roll stiffness, Tstiffness, Tss, is produced., is produced.Any difference between these Any difference between these two must be absorbed as a two must be absorbed as a difference in weight on the two difference in weight on the two wheels.wheels.If the axle is of non-locking If the axle is of non-locking type, then the torque delivered to type, then the torque delivered to both wheels will be limited by the both wheels will be limited by the traction limit on the most lightly traction limit on the most lightly loaded wheel.loaded wheel.

TRANSVERSE WEIGHT SHIFT DUE TO DRIVE TORQUE∑To = (Wr / 2 + Wy – Wr / 2 + Wy)t/2 + Ts – Td = 0

Or Wy = (Td – Ts) / t

In the above equation, Td can be related to the drive forces because

Td = Fx . R / GFD

Where, Fx = total drive force from the two rear wheels.r = tyre radius

GFD = final drive ratioRoll torque produced by suspension requires analysis of the

whole vehicle because the reaction of the drive torque on the chassis attempts to roll the chassis on both the front and rear suspensions.

DRIVE TORQUE REACTIONS ON THE CHASSIS

• The drive torque reaction at the engine / transmission is transferred to the frame and distributed between the front and rear suspensions.

• It is generally assumed that the roll torque produced by a suspension is proportional to roll angle of the chassis


Then:

Tsf = KΦf . Φ

Tsr = KΦr . Φ

KΦ = KΦf + KΦr

Where, Tsf = roll torque on the front suspension

Tsr = roll torque on the rear suspension

KΦf = front suspension roll stiffness

KΦr = rear suspension roll stiffness

KΦ = total roll stiffnessΦ = roll angle

Φ = Td / KΦ = Td / (KΦf + KΦr)

Therefore, Tsr = KΦr . Td / (KΦf + KΦr)

Thus, Wy = [Fxr / GFDt] . [1 – KΦr / (KΦf + KΦr)]

Or Wy = [Fxr / GFDt] . [KΦf / KΦ]


The net load on the rear axle during acceleration will be its static plus dynamic components.

For a rear axle:Wr = W . (b/L + axh/gL)

Or Wr = W . (b/L + Fxh/MgL)The weight on the right rear wheel Wrr = Wr / 2 – Wy

Or Wrr = Wb/2L + Fxh/2L – (FxrKΦf / Nft KΦ) Fx = 2μWrr = 2μ[Wb/2L + Fxh/2L – (Fxr KΦf/ Nft KΦ)

Traction limitsFxmax = (μWb/L) / [1 – μh/L + 2μrKΦf / Nft KΦ]

For a solid rear axle with a locking differential and in the case of an independent rear suspension:

Fxmax = (μWb/L) / [1 – μh/LFor the solid front drive axle with non-locking differential:

Fxmax = (μWc/L) / [1 + μh/L + 2μrKΦr / Nft KΦ]


Traction limits (CONTD)For the solid front drive axle with

locking differential and in the case of independent front drive axle:

Fxmax = (μWc/L) / [1 + μh/L]

EXAMPLE PROBLEMSProb 1. find the traction limited acceleration for the rear drive

passenger car with and without a locking differential on a surface of moderate friction level. The following information is available:

Weight: front = 2100 lb; rear = 1850 lb; total weight = 3950 lbCG height = 21.0 in; wheel base = 108 in; coefficient of

friction = 0.62; tread = 59.0 in; final drive ratio = 2.90; tyre size = 13.0 in; roll stiffness = front = 1150 ft-lb/deg; rear = 280 ft – lb/deg

Prob 2. find the traction limited performance of a front wheel drive vehicle under the same road conditions as the problem above. The essential data are:

Weight: front = 1950 lb; rear = 1150 lb; CG height = 19.0 in; wheel base = 105 in; coefficient of friction = 0.62; tread = 60 in; final drive rastio = 3.70; tyre size = 12.59 in; roll stiffness: front = 950 ft – lb/deg; rear = 620 ft – lb/deg.

chapter 2.ppt

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