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The Physics of Electric Vehicles. Circle Calibrate your video screen or projector. The next slide must show as a circle for the pictures to have the correct.

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Presentation on theme: "The Physics of Electric Vehicles. Circle Calibrate your video screen or projector. The next slide must show as a circle for the pictures to have the correct."— Presentation transcript:

1 The Physics of Electric Vehicles

2 Circle Calibrate your video screen or projector. The next slide must show as a circle for the pictures to have the correct aspect ratio.

3

4 The Physics of Electric Vehicles By Russ Lemon Russ@FarTooMuch.Info

5 How they work … And why

6 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.

7 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.”

8 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.

9 Energy 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

10 Power 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

11 Rolling Resistance Rolling resistance of an average Radial Ply Passenger Tire inflated to 32 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.

12 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.

13 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 is about.11

14 Air Resistance 2 For a vehicle with a frontal area of 30 ft 2, traveling at 60 mph at sea level with a drag coefficient of 0.4, the drag would be about 110 lb. That would be about 17.7 horsepower or about 13.2 kW.

15 Air Resistance 3 Power needed to overcome air resistance increases with the cube of the vehicles velocity. Going from 60 to 75 mph is an air resistance power increase of 95% Energy to overcome air resistance to go a fixed distance, increases with the square of the vehicles velocity. (GPM or KW-hr)

16 Very High Drag

17 Cd between 0.4 & 0.5?

18 Cd about 0.19

19 Cd about 0.11

20 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 29 kW more.

21 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.

22 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 48 kW more.

23 Losses Roughly 4.8 + 13.2 or 18.0 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 30 ft 2 with a Cd of 0.4. Roughly 18.0 + 28.6 or 46.6 kW would be needed to maintain 60 mph up a 6% grade. Roughly 47 + 36 or 83 kW would be needed to accelerate at 2.2 mph/sec up the 6% grade at 60 mph.

24 Losses 2 Running 18 kW for 40 minutes run would be 12 kW-hr of energy for a distance of 40 miles at 60 mph. With a battery pack of 144 volts, this would be about 90 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 120 amp-hr battery for the described vehicle.

25 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 )

26 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.

27 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 min @ 75 A. (178 amp-hr @ 20 hr rate)

28 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.

29 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.]

30 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.

31 Drive Train 2 Selected tire size is P185/60R14 Tire will make 888 revolutions per mile Each tire will hold 1047 lb at 35 psi Total gear ratio is 3.75:1 Motor RPM @ 60 mph is 3330 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.

32 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

33 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.]

34 Motor Characteristics Torque increases with current. Back voltage increases with current and motor speed [rpm]. [Motors are also a generator].

35 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.

36 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.

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38 Assumptions 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 30 ft 2. Vehicle is at Sea Level.

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40 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.

41 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, I 2 R 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]

42 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.

43 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 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.

44 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.

45 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.

46 Available Current The 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

47 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.

48 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.

49 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.

50 ... 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.

51 San Diego Car Conversion Project July-August 2008 Physics of Electric Vehicles by Russ Lemon Russ@FarTooMuch.Info

52 The End

53 To Return http://EVAoSD.com


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