Transcript Electric Vehicles

The Physics of
Electric Vehicles
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The Physics of
Electric Vehicles
By
Russ Lemon
[email protected]
How they work …
And why
Electric Vehicles
• What follows is a discussion of how
chemical energy is converted into electric
energy and then into mechanical energy to
propel a vehicle. The discussion includes
how mechanical energy is used to overcome
the vehicle losses of tire and aerodynamic
drag, and yet have enough energy left over
to climb hills and accelerate the vehicle.
• To go fast and far, minimize your losses.
William Thomson
[aka Lord Kelvin]
1824-1907
“When you can measure what you are
speaking about, and express it in numbers,
you know something about it; but when you
cannot measure it, when you cannot
express it in numbers, your knowledge is
of a meager and unsatisfactory kind: it
may be the beginning of knowledge, but
you have scarcely, in your thoughts,
advanced to the stage of science.”
Basic Units
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A decimeter is a tenth of a meter
A cubic decimeter is a liter
A liter of cold water has about 1 kg of mass
In San Diego that 1 kg of mass has a
weight of about 9.8 newtons (force)
• Raise it up 1 meter and you have done 9.8
joules of work.
• Raise it up 1 meter in one second requires a
power of 9.8 watts.
Energy
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energy = force x distance
joules = newtons x meters
joules = volts x coulombs
1 kW-hr = 3.6 MJ [megajoules]
1 BTU about 1054.8 J
Power
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power = force x speed
watts = newtons x meters/second
watts = volts x amps
one horsepower = 746 watts
Other
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1 kph = 1 kilometer per hour
1 kph = 0.278 m/s
1 t = 1 Mg = 1000 kg = 1000000 g
Air Density at 0 C & at sea level is
about 1.293 Kg/m3
Rolling Resistance
• Rolling resistance of an good Radial Ply
Passenger Tire inflated to 300 kPa is
about 1% of the weight on the tire.
• Rolling resistance of an average Bias Ply
Tire can be more than double that of a
radial ply tire with the same load and
pressure.
• Rolling resistance is measured at
maximum inflation pressure and increases
as tire pressure decreases.
Rolling Resistance 2
• For a 1500 kg vehicle, a weight of about
14.7 kN, a rolling resistance of 1% of load
represents a drag of about 147 N.
• At 100 kph, a drag of about 147 N
represents a loss of about 4.1 kW.
• There are now special low rolling
resistance passenger tires with a rolling
resistance as low as 0.6% of load.
Air Resistance
• Air Resistance is proportional to the
density of the air, the drag coefficient
of the vehicle, the frontal area of the
vehicle, and the speed of the vehicle
squared.
• Typical Coefficient of Drag (Cd) for a
modern passenger vehicle [with windows
rolled up] is about 0.4. The EV1 was
about .19. The Aptera 2 is about .15
Air Resistance 2
• For a vehicle with a frontal area of 1.5
m2, traveling at 100 kph at sea level with
a drag coefficient of 0.4, the drag would
be almost 300 N.
• That would be about about 8.3 kW.
Air Resistance 3
• Power needed to overcome air
resistance increases with the cube of
the vehicles velocity.
• Going from 100 to 126 kph doubles
the power needed to overcome air
resistance.
• Energy to overcome air resistance to
go a fixed distance, increases with
the square of the vehicles velocity.
(KW-hr)
Very High Drag
Cd about 0.4?
Cd about 0.19
Cd about 0.11
Climbing Hills
• The maximum freeway grade is 6%
• Some San Diego roads have grades as
high as 24%.
• The force needed for a 1500 kg vehicle
to climb a 6% grade is 880 N.
• To climb a 6% grade at 100 kph, A 1500
kg vehicle needs an additional 24.5 kW.
• [Note: a 100% grade is 45 degrees.]
Acceleration
• If you drop something, it will accelerate
at the rate of about 9.8 m/sec2. This is
know as a 1 g acceleration.
• An horizontal acceleration of half that,
or about 5 m/sec2 would be ‘aggressive’,
typical of a sports car with a fast driver.
• An acceleration of 1 m/sec2 would be
‘conservative’, typical of an older driver
with a Honda Civic or VW GT.
Acceleration 2
• An acceleration of 1 m/s2, 3.6 kph/s or
about 0.1 g, of a 1500 kg vehicle would
require a force of about 1.47 kN.
• At 100 kph, this would require an
additional power of about 40.8 kW.
Losses
• Roughly 4.1 + 8.3 or 12.4 kW would be
needed to maintain 100 kph on a level
road with a 1400 kg vehicle with typical
radial tires and a cross section of 1.5 m2
and a Cd of 0.4.
• Roughly 12.4 + 24.5 or 36.9 kW would be
needed to maintain 100 kph up a 6%
grade.
• Roughly 36.9 + 40.8 or 77.7 kW would be
needed to accelerate at 3.6 kph/sec up
the 6% grade at 100 kph.
Losses 2
• Running 12.4 kW for 40 minutes run would be
8.27 kW-hr of energy for a distance of 67 km
at 100 kph.
• With a battery pack of 144 volts, this would be
about 57.4 amp-hr of usage. [86 A for 40 min]
• For long life of a Lead-Acid battery, the depth
of discharge should be less then 80%. Even an
80% DOD would shorten the life. A 100%
DOD would give a very short life.
• Thus the need for at least a 72 amp-hr
battery for the described vehicle.
Measure Losses
• It takes a force equal to the weight of
the vehicle to cause a 1 g deceleration
• A 1 g deceleration is about 9.8 m/s2
• Measure how long it takes on a level road
to coast from 108 to 90 kph in sec (t)
[30 to 25 m/s]
• Deceleration (d) = (30-25)/t m/s2
• Force (f) is vehicle mass * d
Battery 1
• The source of energy for an electric
vehicles is its battery.
• The battery must supply enough current
to the electric motor in order for it to
supply the needed torque.
• The battery must have enough voltage to
force the needed current through the
electric motor for the desired speed.
• The battery must have enough energy to
supply the needed power for the needed
amount of time.
Battery 2
• U. S. Battery makes an 8 volt battery with
a 75 amp discharge time of 85 minutes
called the US-8VGC.
• It has a mass of about 30 kg.
• 18 batteries in series will supply 144 volts.
• 18 battery mass will be about 531 kg.
• Amp-Hr rating of about 106 @ 75 A.
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(178 amp-hr @ 20 hr rate)
Battery 3 – [rules of thumb]
• Lead-acid batteries in an electric vehicle
need to be at least 33% of a good vehicles
gross weight to get a range of about 64
km with conservative driving.
• To get good performance, you need at
least 33% of the vehicles gross weight to
be active, on-line battery.
Battery 4 –
[lead-acid battery life]
• Do not exceed 80% depth of discharge.
• Keep battery voltage within normal range.
[For 144 V pack, keep pack above 120 V
and below 185 V at all times.]
• Limit maximum current. [Excessive
current leads to short life and even
battery failure.] [Keep maximum current
below the current that gives a full charge
to 80% Discharge time of 20 minutes.]
Drive Train
• The electric motor must have enough
torque to overcome the losses, climb hills
and accelerate the vehicle to a useful
speed.
• The electric motor must have enough
speed for the vehicle.
• Gears are used to match the electric
motor characteristics to the vehicle
requirements.
Drive Train 2
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Selected tire size is P185/70R14
Tire will make 537 revolutions per km
Each tire will hold 475 kg at 300 kPa
Total gear ratio is 3.75:1
Motor RPM @ 100 kph is 3356
Maximum gross vehicle weight
(including 65 kg motor, 515 kg of
batteries, 23 kg of controller & wiring,
two 100 kg occupants) is 1539 kg.
Electric Motor
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Series wound direct current motor
In any gear, speed is proportional to RPM
Constant torque for even acceleration
Torque roughly proportional to current
Increasing voltage is necessary to
maintain current to maintain torque as
vehicle speed and motor RPM increase
• Batteries must have enough voltage and
current to maintain desired speed
Electric Motor 2
• The selected electric motor is the Advanced DC
FB1-4001
• Diameter is 9.1”
• Mass is 65 kg
• Max continuous rated current is 180 A
• Max 1 hour rated current is 200 A
• Max 5 minute rated current is 340 A
• Current is limited by motor temperature
• Motor speed should be kept under 6000 rpm
[High rpm causes rapid brush and bearing wear.]
Motor Characteristics
• Torque increases with current.
• Back voltage increases with current
and motor speed [rpm].
• [Motors are also a generator].
Vehicle Characteristics
You select with your foot the current sent to
the electric motor. With a constant current
you have a constant torque. As the vehicle
accelerates from a stop, the controller
increases the voltage on the motor to maintain
that current until there is no more voltage.
[battery voltage reached] As the vehicle
continues to accelerate, current and therefore
torque decrease, causing acceleration to also
decrease until torque is just enough to match
losses and you maintain a constant speed.
Vehicle Characteristics 2
In the following graph, for a given foot
setting, you follow a constant torque line
up to the battery voltage and then follow a
horizontal line to the right as rpm and
vehicle speed increase. Note the
corresponding decrease in torque.
You must have enough battery voltage to
push the current you need to get the
torque you need to go the speed you need.
Assumptions
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Battery voltage is 144 volts.
Maximum controller current is 500 amps.
Motor is Advanced DC FB1.
Vehicle gross weight is 1500 kg
Tire drag is 1% of vehicle weight.
Aerodynamic Cd is 0.4.
Cross sectional area is 1.5 m2.
Vehicle is at Sea Level.
Warning
Note that the highest force in the
previous slide is for a current of almost
500 A that will quickly overheat the
motor. The continuous current must be
less then 180 A and that means that the
continuous force must be less than 1/3 of
the maximum force shown.
Motor Comment
Remember that power is the product of
torque and rpm. With the ADC FB1-4001,
the 200 A continuous rating is a torque
limit of about 30 ft-lbs. At 30 ft-lbs, it
takes about 144 V for a motor speed of
5500 rpm. This is about 31 hp. Actually,
I2R losses in battery, controller and wiring
will reduce the actual voltage available to
the motor. At 80% DOD with a 200 A load,
the maximum voltage at the motor may be
as low as 120 V for only 4500 rpm. [25 hp]
Motor Comment 2
Gearing the motor for 4500 rpm at the top
vehicle speed [120 kph?] will take full
advantage of the capability of the battery,
controller and motor in the real world.
Too many car conversions fail to take into
account worst case conditions. [The last
hill to climb with batteries at 80% DOD.]
Of course some have the option to shift to
a lower gear and struggle at a lower speed.
Measure Performance
• It takes a force equal to the weight of the
vehicle in addition to the force to over-come
losses to cause a 1 g acceleration.
• Measure how long it takes on a level road to
accelerate from 90 to 108 kph in seconds (t).
[25 to 30 m/s]
• Acceleration (a) at 100 kph is about 5/t m/s.
• Force (f) is about mass * a. [f in Newtons]
• Acceleration kW is about f * 27.5.
• Total kW is Acceleration kW + Loss kW.
Range
Now that we have a rough idea of the
vehicle’s performance, the next question is
how far will it go on a charge. In other
words, what is its range? Range should
really be determined by how far it will go
on 80% of a charge since completely
discharging a battery will ruin it. Note
that the capacity [amp-hr] decreases as
the current increases. Also note that the
voltage decreases as the charge is used up.
Range 2
To estimate range at a given speed, determine
the force needed at that speed. The force (N) x
speed (kph/3.6) is the kW needed to maintain
that speed. Divide kW by the battery voltage to
get battery current. Estimate battery amp-hr at
that current and divide by the current. Multiply
hr by 0.8 to get the approximate number of
hours. Multiply hours by the speed to get an
estimate of range.
Available Current
The total capacity of the battery is nonlinear. The minutes the battery can
provide power decreases faster then the
amps supplied by the US 8V GC battery:
1041 minutes @ 10 amps
341 minutes @ 25 amps
146 minutes @ 50 amps
94 minutes @ 75 amps
66 minutes @ 100 amps
50 minutes @ 125 amps
40 minutes @ 150 amps (est.)
NOTICE
The numbers used on the previous slides
were taken from the best information and
estimates available. Exact measured
numbers were not available. Therefore,
notice is given that the conclusions are
approximate ballpark estimates. Actual
performance to be determined.
Charging
U.S. Battery recommends that:
• Voltage not exceed 2.585 V per cell
• Current not exceed AH/10
• Time not exceed 10 hours
• http://www.usbattery.com/pages/usbspecs.htm
In other words, for a “144 volt” pack, the
charging current should not exceed 165/10 or
about 16.5 amps until limited by the total voltage
that must not exceed 186 volts. Maximum charge
time is 10 hours. Check water level after charge.
Charging
186 volts times 16.5 amps is 3065 watts.
That would be about 26 amps from a 120
volt source, or 13 amps from a 240 volts
source, not taking into account efficiency
of the charger. If time were short, the
batteries could be charged at 25 amps.
That would be 4650 watts, or almost 20
amps from a 240 volt source. A 30 amp
240 volt service is best for charging.
. . . ie Hybrid
For long trips a small motor-generator can
be added to extend range. Motor
generators are made to run on a variety of
different fuels. Commercial motorgenerators include gasoline, diesel, propane,
etc. Be sure the controller can take the
higher voltage. Voltage should not exceed
the maximum battery charging voltage.
San Diego
Car Conversion Project
Physics of Electric Vehicles
by
Russ Lemon
[email protected]
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