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The Physics of
Electric Vehicles
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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• 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
• 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
• power = force x speed• watts = newtons x
meters/second• watts = volts x amps• one horsepower = 746 watts
Other
• 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.• (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• 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• 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• 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.
RangeNow 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 2To 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 CurrentThe total capacity of the battery is non-linear. 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.)
NOTICEThe 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 motor-generators include gasoline, diesel, propane, etc. Be sure the controller can take the higher voltage. Voltage should not exceed the maximum battery charging voltage.
The End
