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Approximate Rating of Powertrain

•To determine the forces needed for a 3000lb vehicle to accelerates at 10mph/second; assume aerodynamic, rolling and hillclimbing force counts extra 10% of the forces needed

Force = 1.1*mass*acceleration

=1.1*3000lb*10mph/second*5280ft/3600second

=48400 lbm =1503 lbf

Force = 1.1*mass*acceleration

=1.1*3000lb/2.2*10mph/s*1609/3600second

=6704 N

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Rating of Powertrain

•Determine the average power needed to accelerate the vehicle from zero speed to 60mph

energy = mass*V*V / 2

=3000lb/2.2*[60mph*1609/3600second] 2 / 2

=490318 joules

=[60mph*1609/3600] / [10mph/second*1609/3600]

=[26.8 m/s] / [4.47m/s2]

=6 seconds

Approximate Rating of Powertrain

•Alternatively, determine the forces needed for a 3000lb vehicle to accelerate to 60mph in 10 seconds; assume aerodynamic, rolling and hillclimbing force counts extra 10% of the forces needed, and a constant acceleration

–Final speed V= 60mph*1609/3600=26.8 m/second

–Acceleration a = V / t =26.8 / 10 = 2.68 m/S2

–Force F = m*a = 3000/2.2*2.68 m/S2 = 3657 N

–Power = F V = 3657 * 26.8 = 98 kW (at 60mph speed)

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Rating of Powertrain

•The above assumed a constant acceleration. In real life, the acceleration near 60mph will be greatly reduced. Therefore, the actual power needed to accelerate the vehicle is much less than 90kW

Average power = Final power / 2 = 49 kW

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Rating of Motor

•Assume the effective tire radius is R

•Torque at wheel is

Tw=FR

Tmotor=Tw / rg

Where rg is gear ratio

• Alternatively, motor torque is

T=P/wm

Where wm is motor angular speed

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Size of Drive Train

•Motor size is determined by

D 2l = 6.1×C108 ABP 1n

Where P is motor input power, in kW, P=Pmax / efficiency A is airgap current density

B is airgap magnetic flux density

C is a constant, between 0.5 and 0.9 n is motor speed in rpm

D is inner diameter of stator or inner diameter of rotor

L is effective length of stator/rotor

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Size of Motor

•Note that the power required to cruise a vehicle on highway at 60mph is only 6% of the power needed to accelerate the vehicle from 0 to 60mph in 10 seconds.

•Since most motors can be designed to overload for a short time, a motor can be designed at much lower ratings. Example:

–30kW rated power (13.8kW dragging at 60mph, 1/3 rated)

–2 times overload for 60 seconds (60 kW)

–3 times overload for 30 seconds (90 kW)

–4 times overload for 20 seconds (120 kW)

–5 times overload for 10 seconds (150 kW)

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Efficiency

•Note also that a motor can have efficiency (including controller) of over 90%, while an engine only has efficiency less than 30%

•An ICE does not have the overload capability as that of a motor. That’s why the rated power of ICE is usually much higher than required for highway cruising

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Vehicle Power Plant Characteristics

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Engine Performance at Full Throttle

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Motor Performance at Full Load

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Tractive Effort of Internal

Combustion Engine

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Tractive Effort of EV with Single Gear

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Continuously Variable Transmissions

(CVT)

•Provide infinite gear ratios

•Virtually matching any engine speed with vehicle speed

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Vehicle Performance – Speed and

Gradeability of ICEV

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Vehicle Performance – Speed and

Gradeability of EV

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Driving Cycles

Driving time, sec.

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Braking Performance

•Energy wasted during braking in conventional vehicles

•Can be partially recovered in EV and HEV

•ABS performance can be improved in HEV/EV

•Traction control is easier to achieve in HEV/EV

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