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
• A decimeter is a tenth of a meter, or about
3 & 15/16 inches
• 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 hp-hr about 2.7 MJ
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
one horsepower = lbf x mph / 375
(i.e. 1 hp = 5 lbf @ 75 mph)
1 hp = ft-lb x rpm / 5252
Rolling Resistance
• Rolling resistance of an average Radial Ply
Passenger Tire inflated to 44 psi 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 vehicle weighing 4000 lb, a rolling
resistance of 1% of load represents a
drag of 40 lb.
• At 60 mph, a drag of 40 lb represents a
loss of 6.4 horsepower or about 4.8 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 1st Aptera was about .11
Air Resistance 2
• For a vehicle with a frontal area of 20
ft2, traveling at 60 mph at sea level
with a drag coefficient of 0.4, the drag
would be about 74 lb.
• That would be about 11.1 horsepower or
about 8.3 kW.
Air Resistance 3
• Power needed to overcome air
resistance increases with the cube of
the vehicles velocity.
• Going from 50 to 63 mph will cause
air resistance power to double
• Energy to overcome air resistance to
go a fixed distance, increases with
the square of the vehicles velocity.
(GPM or 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 4000 lb vehicle
to climb a 6% grade is 240 lb.
• To climb a 6% grade at 60 mph, A 4000
lb vehicle needs an additional 38.4
horsepower or about 28.6 kW more.
Acceleration
• If you drop something, it will accelerate
at the rate of about 22 mph/sec. This
is know as a 1 g acceleration.
• An horizontal acceleration of half that,
or about 10 mph/sec would be an
‘aggressive’, typical of a sports car with
a very fast driver.
• An acceleration of 2 mph/sec would be
‘conservative’, typical of an older driver,
or of a Honda Civic or VW GT.
Acceleration 2
• An acceleration of 2.2 mph/sec, or 0.1 g,
of a 4000 lb vehicle would require a
force of 400 lb.
• At 60 mph, this would require an
additional 64 horsepower or about 47.7
kW more.
Losses
• Roughly 4.8 + 8.3 or 13.1 kW would be
needed to maintain 60 mph on a level
road with a 4000 lb vehicle with typical
radial tires and a cross section of 20 ft2
with a Cd of 0.4.
• Roughly 13.1 + 28.6 or 41.7 kW would be
needed to maintain 60 mph up a 6%
grade.
• Roughly 41.7 + 47.7 or 89.4 kW would be
needed to accelerate at 2.2 mph/sec up
the 6% grade at 60 mph.
Losses 2
• Running 13.1 kW for 40 minutes run would be
8.7 kW-hr of energy for a distance of 40 miles
at 60 mph.
• With a battery pack of 144 volts, this would be
about 61 amp-hr of usage.
• 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 76 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 22 mph/sec
• Measure how long it takes on a level road
to coast from 65 to 55 mph in sec (t)
• Deceleration (d) = (65-55)/t mph/sec
• Force (f) is vehicle weight * d/22 lbs
• Loss is about f * 60 /375 horsepower
• ( 1 hp = 375 lb-mph = 746 watts )
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 weight of about 65 lb.
• 18 batteries in series will supply 144 volts.
• 18 batteries will weight about 1170 lb.
• 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 more than
40 miles 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 865 revolutions per mile
Each tire will hold 1201 lb at 44 psi
Total gear ratio is 3.75:1
Motor RPM @ 60 mph is 3244
Maximum gross vehicle weight
(including 143 lb motor, 1134 lb of
batteries, 50 lb of controller & wiring,
two 250 lb occupants and 250 lb of
‘stuff’) is 3700 lb.
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”
• Weight is 143 lb
• 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 4000 lb.
Tire drag is 1% of vehicle weight.
Aerodynamic Cd is 0.4.
Cross sectional area is 20 ft2.
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 [70 mph?] 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 overcome losses to cause a 1 g acceleration.
• Measure how long it takes on a level road
to accelerate from 55 to 65 mph in
seconds (t).
• Acceleration (a) at 60 mph is about 10/t.
• Force (f) is about weight * a/22 lbs.
• Acceleration Hp is about f * 60 / 375.
• Total Hp is Acceleration Hp + Loss Hp.
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 (lb) x
speed (mph) / 375 is the hp needed to maintain
that speed. Multiply hp by .746 to get kW.
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
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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