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iisjsmith said:
Hi guys I just registered here because this thread is annoying the heck out of me and I wanna clear some things up
iisjsmith: are you an EE? I know EE's need to take thermodynamics, and some basic mechanics classes... you are completely ignoring some very important factors.
- first of all, WarpedOne's calculations(which are fairly sound and logical, will address in a moment) have nothing to do with, and need not have anything to do with the motor, voltage, frequency or anything in the electrical system.
** so basically, leave your electrical power calculations out of it. they are not nessicary. all we need is the efficiency of the system, and mechanics.
**note; actually WarpedOne's calculations rely on statistics mostly... which is perfectly fine. his linear/square and cube rule come from mechanics.
Michael: his speculation is quiet logical, and really only needs some fine tuning for a more accurate prediction.
Correct, as there is more drag, you need more power from the motor to over come it, WarpedOne has been using this principle.
Ill try to help you guys understand.
Enery used, or work done is: W = F * D (Work = Force times Distance) where the Distance is obvious (miles...) Force is the force fighting you (rolling friction, viscious drag from bearings/transmission, tire drag, tire deformation, and aerodynamic drag) and W is energy used (joules, gallons of gas, KWh (kilowatt-hours), whatever)
what would be more insightful would be to use P = F * V (Power = Force times Velocity) where the force is same as above, Power is amount of energy expended per unit time (i.e. HP, KW(kilowatts), watts, gallons of gas per hour, etc etc...)
now, if frictional force was constant... then your range would not change based on cruising speed. (this is ALL assuming we accelerate to cruising speed on a full charge, and stay there until the battery dies, NO stop and go driving.. that changes everything.)
P = F * V ---> ie if you double the velocity, you double the power used, but since your actually traveling twice as fast, you end up using the same amout of energy to travel a certain distance.
however, the frictional force, F is not a constant; it is actually (for the most part) a function of velocity itself. basically, as velocity goes up, so does the friction force... so your Power goes WAY up.
if we want to get more specific, we have to analyze all the forces seperately.
I previously stated the drag force, F is comprised of many components:
-General kinetic friction
-Tire drag
-Tire deformation
-viscius drag from oil/grease/tranny/bearings
-aerodynamic drag.
Some of these forces (general & tire drag) are constants. This is the linear part of WarpedOne's calculations.
Some of these forces vary linearly with velocity, (i.e. double the velocity, double the drag force) (tire deformation, viscious drag) this is the squared part of his calculations (i.e. double the speed, quadrouple the power required.)
but it gets more complicated then that... as you approach high speeds, the aerodynamic drag will approace a square of the velocity (thiis is the cubed component WarpedOne speaks of.)
all we need is the power required to tranverse at some speed, the battery pack capacity, and the motor/battery efficiency to do some basic estimations. for example if were traveling at a constant velocity, the energy used would be P*t and distance traveled would be V*t that would give us the range at a specifc speed.
WarpedOne: I'm going to throw your some ideas/thoughts to try and help improve your calcs:
1.) you didn't include motor effiency in any of your estimations. Just use 90% across the board. this will lower your range calcs.
--I.E. 0.9*power sucked from battery = power put out by motor(250 peak)
2.) you estimate at top speed, 135 that the motor is producing peak HP at that rpm (max rpm) when in fact the top speed is gear limited and the car is still accelerating when it hits redline at 135mph. so infact to travel 135mph it needs LESS then the peak hp at 13500 rpm (or whatever peak rpm is, dont feel like looking it up) try assuming it uses 80-90% of that power. this will increase your range calcs.
3.) I think that going to a power reqrment that's a cube of the speed at such a low speed is a bit much. say perhaps it doesnt become a cube until 80+mph. then scale evenly.
this is fun stuff, good luck.
apologies for grammer, spelling, wording, organization etc. dont feel like proofreading. later.
okashira said:
re 1.): You are correct. Published power curve is very likely to be output power. Input power is somewhat higher (efficiency). This paints my numbers as some 10% to 20% too optimistic.
re 2.): I'm not sure you understood correctly. I assumed "max speed @ max power" in my first post, in my second one I corrected that as I've seen that max speed is indeed gear limited. Max power in second gear is availble around 80mph. I also came up with a "power max" speed of around 160mph using third gear with right ratio.
re 3.): My calculations are based on top speed where (upper limit of available) power is known and then 'deduced' down to lower speeds using cube / square / linear power-speed relations. I've chosen 55mph (90km/h) as a transition point between cube and square power-speed relation. If we choose 80mph for that transition point we'll get somewhat lower range results at lower speeds. I will do that at later time to see how much difference there is.
There is another thing to point out again. Published power curve puts an upper limit on available power at some motor revs. You cannot draw more power, only less. We know that roadster is capable of 135mph and that it hits this at redline where maximum available power is only 105kW. My calculations are based on assumption that for 135mhp the car uses all of availabe 105kilowats. This is probably not true, but we do not know to what degree. Again, this puts my numbers on conservative side.
2.) I'm saying the car needs less then the published 105kw to transverse at 135mph. (the car is still accelerating, the only reason it stops at 135 is the motor rev limiter) however the amt below 105kw is problably low.
3.) yes I know your using linear square cube power relations, if you read my post you may understand why you've been doing that.
the cube 'law' is more of a rule of thumb for estimating vehicle top speed. the power dependence on speed does not become a 'cube' until a higher speed then what you have been using. again aerodynamic drag is not so predictable. it may be a square relationship with power(force is direclty speed-dependent) or a cube relationship with power(force is square of speed), you must determine it experimentally.
here's what were looking at:
P(power) = F(force) * V(velocity), where F = f(V) and cosists of the components i outlined before.
and thus P = V*f(v)
we can assume f(v) does not have an order higher then a square (the aerodynamic component)
we get that P = V*f(v) = F(V) = some 3rd order polynomial.
P = A + Bv + Cv^2 + Dv^3; A, B, C, D constants.
D is a scaling component of the aero drag
C is a scaling component of the viscios drags and the aero drag
B is a scaling component of the basic kinetic friction
A is a constant power draw needed to operate the vehicle (ie computer system, lighting, a/c, heater, cigarrette lighter etc.)
v is the velocity and assuming steady state (vehicle is not accelerating but transversing at a constant speed.)
A,B,C,D are best found using hopefully at minimun four data points (is power req'd to tranverse four different velocities, then we can solve for A,B,C,D with some simple math.)
however since we only have one, mabye two data points you just gotta do it by ear.
I'm just saying that it likely doesnt go to a cube until a higher speed then what you have been using.
3.) yes I know your using linear square cube power relations, if you read my post you may understand why you've been doing that.
the cube 'law' is more of a rule of thumb for estimating vehicle top speed. the power dependence on speed does not become a 'cube' until a higher speed then what you have been using. again aerodynamic drag is not so predictable. it may be a square relationship with power(force is direclty speed-dependent) or a cube relationship with power(force is square of speed), you must determine it experimentally.
here's what were looking at:
P(power) = F(force) * V(velocity), where F = f(V) and cosists of the components i outlined before.
and thus P = V*f(v)
we can assume f(v) does not have an order higher then a square (the aerodynamic component)
we get that P = V*f(v) = F(V) = some 3rd order polynomial.
P = A + Bv + Cv^2 + Dv^3; A, B, C, D constants.
D is a scaling component of the aero drag
C is a scaling component of the viscios drags and the aero drag
B is a scaling component of the basic kinetic friction
A is a constant power draw needed to operate the vehicle (ie computer system, lighting, a/c, heater, cigarrette lighter etc.)
v is the velocity and assuming steady state (vehicle is not accelerating but transversing at a constant speed.)
A,B,C,D are best found using hopefully at minimun four data points (is power req'd to tranverse four different velocities, then we can solve for A,B,C,D with some simple math.)
however since we only have one, mabye two data points you just gotta do it by ear.
I'm just saying that it likely doesnt go to a cube until a higher speed then what you have been using.
I am in no way an EE. But I have spent several weeks researching PWM and other VFD technologies for a pet project of mine, and I finally understand how they work.
You can spout equations all day long but it doesn't change that fact that at some point you are only increasing the frequency of the power, not the voltage. You cannot supply a voltage greater than the line voltage, which in this case is a nominal 375V. I believe I calculated that the frequency of the power at 8000rpm would be 266Hz. I admit I don't remember the reference...I'll have to check my links.
So once you are supplying the inverter with 375V at 266Hz you are at the max of the volts/hertz curve. Anything above that means only increasing frequency, which does not consume more power. So while drag might increase because you are going faster, you do NOT draw more power from the battery pack.
This is just the way PWM inverters work.
You can spout equations all day long but it doesn't change that fact that at some point you are only increasing the frequency of the power, not the voltage. You cannot supply a voltage greater than the line voltage, which in this case is a nominal 375V. I believe I calculated that the frequency of the power at 8000rpm would be 266Hz. I admit I don't remember the reference...I'll have to check my links.
So once you are supplying the inverter with 375V at 266Hz you are at the max of the volts/hertz curve. Anything above that means only increasing frequency, which does not consume more power. So while drag might increase because you are going faster, you do NOT draw more power from the battery pack.
This is just the way PWM inverters work.
martin eberhard wrote in his october blog:
Motor frequency: The motor makes one revolution for every two cycles of its three-phase AC input. So when the motor is turning at 13,500 RPM (which is 225 revolutions per second) the AC frequency is 450 cycles per second.
Your 266Hz @ 8k rpm agrees with this.
Also if you observe the power curve it is quite peculiar - it has somewhat triangular shape. At lower revs (0 - 8k) it grows almost linearily with motor revs. This is logical as power is simply torque multiplied with revs and torque is almost constant in this range. But over 8000k revs power suddenly starts to drop. And again almost linearily with increasing revs. Torque drop is even more irregular.
You might be onto something, I just find it hard to believe that car would consume energy at the same rate at 80mph and 135mph. That would mean that at 80mph the motor / inverter would produce much more heat than at 135mph. all that surplus energy must go somewhere. Going over 80mph would be free lunch regarding range and less of a burden on internals ???
I looked over your previous post where you've stated:
As we accelerate past 80mph the inverter continues to increase the frequency supplied to the motor...but the voltage no longer increases. The volts/hertz ratio starts to change from the baseline and our available torque starts to suffer. But since the voltage is not increasing we are not draining the battery pack any faster.
It is true that AC frequency does not affect power, but only as long as you have perfect sine waves or at least continous waves.
If output is not like sine waves but more like isolated spikes or bursts of voltage, increasing the frequency of these bursts would increase awerage power although its voltage stayes constant.
Motor frequency: The motor makes one revolution for every two cycles of its three-phase AC input. So when the motor is turning at 13,500 RPM (which is 225 revolutions per second) the AC frequency is 450 cycles per second.
Your 266Hz @ 8k rpm agrees with this.
Also if you observe the power curve it is quite peculiar - it has somewhat triangular shape. At lower revs (0 - 8k) it grows almost linearily with motor revs. This is logical as power is simply torque multiplied with revs and torque is almost constant in this range. But over 8000k revs power suddenly starts to drop. And again almost linearily with increasing revs. Torque drop is even more irregular.
You might be onto something, I just find it hard to believe that car would consume energy at the same rate at 80mph and 135mph. That would mean that at 80mph the motor / inverter would produce much more heat than at 135mph. all that surplus energy must go somewhere. Going over 80mph would be free lunch regarding range and less of a burden on internals ???
I looked over your previous post where you've stated:
As we accelerate past 80mph the inverter continues to increase the frequency supplied to the motor...but the voltage no longer increases. The volts/hertz ratio starts to change from the baseline and our available torque starts to suffer. But since the voltage is not increasing we are not draining the battery pack any faster.
It is true that AC frequency does not affect power, but only as long as you have perfect sine waves or at least continous waves.
If output is not like sine waves but more like isolated spikes or bursts of voltage, increasing the frequency of these bursts would increase awerage power although its voltage stayes constant.
iisjsmith said:
HAHAHHAA.. OK i get it now. iisjsmith, we are not talking about full throttle acceleration. we are talking about steady state cruising.
the voltage is most certantly not constantly 375 volts at 266hz(8000rpm) if that was the case, the car would be UNDRIVABLE. you would accelerate with maxminum force AT ALL TIMES. the throttle would be and ON OFF SWITCH and would make cruising at a constant speed IMPOSSIBLE.
in order to get the desired power, the computer/inverter/whatever varies the voltage based off 'throttle' position.
so at 90mph it may be 95 volts
100mph 120 volts
110 mph 160 volts
etc etc etc...
since we have NO IDEA what the crusing voltages at any time are.. we cannot use that to calculate the car's range.
so we have the voltage at peak acceleration at 8000rpm. which doesn't help us at all.
the only thing we know is how much power the car needed to go a certain speed. (105kw at 135mph) from that we can use mechanics to extrapolate power at other speeds, thus giving us the vehicles range.
seriously, eletronics, voltage, ac, inverters have no place in determining this.
all we need is: effiency and power and velocity.
I'm sorry, but you should really read my previous posts again. This is not a potentiometer system, like those found in DC motor systems. The throttle does not vary the voltage to the motor, thus altering its speed. The throttle position tells the inverter what frequency to make the electric power that is fed to the motor. The voltage adjustment is made by the inverter, and that has nothing to do with speed...only torque.
There are some very helpful online resources for explaining the way pulse width modulation inverters work in relation to AC motors. The one I like the best is the Siemens website...they have some online courses at http://www.sea.siemens.com/step/default.html. If you just want to download the PDFs you can go to http://www.sea.siemens.com/step/downloads.html. Focus on the sections titled "Field Weakening" and "Volts per Hertz". They explain how the voltage is adjusted.
Saying that electronics and inverters have no place in this discussion is like saying carburetors and fuel injection systems have no place in a discussion about ICE vehicles.
iisjsmith said:
I have read your posts, and I understand them perfectly, however you are still a little confused, and missing some important factors.
I should have been more clear in the above quote, the throttle does not control power, it does infact control torque (via voltage to the motor.)
The throttle absolutly does not control frequency. the frequency is dependent on the motor's speed, NOT the other way around. you cannot control speed, only the acceleration to acheive that speed, after a period of time(through torque).
this is not really a discussion about electric cars, ICE cars or whatever. it's just about the driving range of this particular vehicle.
it doesnt matter what kind of car it is, we can express the energy used (50KWh in this car) as a function of the power used at a certain velocity, to derive the driving range. and the power used at a certain constant speed just happens to be the power needed to overcome drag
(P=F*V) divided by the efficiency of the system.
--a gallon of gas, a charged battery pack, a tank of hydrogen all contain a certain amount of ENERGY.
to expend a certain amount of enery we produce a certain amout of POWER over a period of TIME(and driving distance related to time by speed).
E = P * t or more accuratly E = integral(P*dt), where P = F * V... F is a function of the frictional forces descirbed above, and also the intertial force of the vehicle when under acceleration.
this is why even EE's have to take basic mechanics, thermodynamics and physics courses. you need to look at the big picture.
I'm sorry, but this is completely incorrect. Any basic understanding of AC induction motors and variable frequency drive systems should tell you that. Just read any document that explains AC motor theory and you will see that the speed of the motor is controlled by only two factors: the number of poles and the frequency of the power. The number of poles is obviously a static number, since you can't just swap poles in and out. But the frequency can be adjusted by using an inverter. Here is some info I found in two minutes of googling:
- From http://en.wikipedia.org/wiki/Variable_frequency_drive:
"A variable-frequency drive (VFD) is a system for controlling the rotational speed of an alternating current (AC) electric motor by controlling the frequency of the electrical power supplied to the motor. A variable frequency drive is a specific type of adjustable-speed drive. Variable-frequency drives are also known as adjustable-frequency drives (AFD), variable-speed drives (VSD), AC drives or inverter drives."
- From the Joliet Technologies website (http://www.joliettech.com/what_is_a_variable_frequency_drive.htm):
"With one pole pair isolated in a motor, the rotor (shaft) rotates at a specific speed: the base speed. The number of poles and the frequency applied determine this speed"
- From the Drive Systems Inc. website (http://www.drivesys.com/) section on "Learn More About AC Induction Motors":
"Induction motors work by electrically inducing an electro-magnetic pole into the rotor. The magnetic field that surrounds the rotor appears to rotate which has the effect of pulling the rotor in the direction of rotation. The speed of the rotation is determined by the frequency of the applied alternating current - change the frequncy and the rotor speed is changed. This is the function of the variable frequency drive (VFD or AC drive)."
- From the Tesla Motors blog "Motor City" (http://www.teslamotors.com/blog1/?p=30):
"As noted above, AC motors designed for appliances usually run at one speed. Some of you have commented that we should use a Continuously Variable Transmission (CVT) to match our motor speed to the desired speed of the car. This would be true if we ran our motor on a fixed frequency. But we don’t. Like the GM cars, and like other AC electric car motors, we feed the motor with a variable frequency AC waveform, using frequency to regulate torque and therefore speed."
Again, this is completely incorrect. In a potentiometer system you would be right...the throttle controls the voltage supplied to the motor. In a DC motor the voltage controls the speed. The more voltage supplied the faster it turns. This is NOT how the Tesla Roadster and other AC motor drive systems work. It uses a variable frequency drive system to alter the frequency of the power. From the blog "Motor City" on the Tesla Motors website they say "Tesla Motors’s Power Electronics Module (PEM), in turn, uses a similar kind of variable frequency, IGBT inverter..."
So you cannot discuss range of the Tesla Roadster without factoring in the way their drive system works. And I stand by my opinion that you will see a maximum battery drain at maximum power (80mph) and then NO additional battery drain as you go faster.
Instead of reading more about ac motors, you should pick a physics book.iisjsmith said:
So the they use a phase shift to control the torque of an A/C motor, that's cool stuff.. I can see how it can get very complicated (from a smooth control standpoint.) yeah, I can now see how the voltage stays constant, and the frequency sets the speed of the motor(dumbing it down a bit)
when there is a large load on the motor (such as the weight of a car) you cant just change the frequency to set the speed of the motor. that is, if you want the car to drive smoothly. you carefully introduce a phase shift, to control the torque.
but that doesnt really help you case... If I get the time, ill read up on AC motors some, so I can give a clear explination on how the power is not constant for a particualr RPM. but I should be stufying for finals this week.
but, allow me to speculate again: The frquency holds a particular speed for the motor. the phase shift is dependent on the torque output(and the other way around.) if your holding a constant speed/frequency, and then you introduce a sudden load on the motor, it will slow the motor for an instant, a very small amout... just enough to introduce a phase shift that helps the motor create a torque large enough to match that load.
Ill leave you with this: at 80mph, it only takes about 40-50 horsepower to keep the car in motion (probably even less for the tesla). if your car is puttering along at 80mph, and the motor is producing 250hp, where is all that leftover energy going? (serious question)
If the Tesla used 250hp to cruise at 80mph with a 50KWh battery, it will have a range of 21.5 miles at that speed before the battery runs complely dry. Good thing it obeys the laws of physics.
Here are some numbers I worked up, after figuring out the Roadster ESS specs:
The ESS is 324V and 152Ah. To drain the ESS in one hour you discharge at a 1C rate, which is 152A. 152A * 324V = 49248W or 49.3kW or 66.1hp. So if we supply 49.3kW of power for one hour we will completely drain the ESS. If you look at the the torque/power curve on the TM website (http://www.teslamotors.com/performance/performance.php) then you will see that 49.3kW is about 2000RPM.
The Roadster can go 250 miles on the EPA Highway cycle, which is an average speed of 48mph. That is 5.2 hours of driving to completely drain the ESS.
5.2 hours is a .19C discharge rate (1 / 5.2 = .19), which is 152A * .19 = 28.88A. This is 28.88A * 324V = 9357W or 9.4kW or 12.6hp. So if we supply 9.4kW of power for 5.2 hours we will completely drain the ESS. The torque/power curve shows 9.4kW to be in the few hundreds of RPM.
At a maximum power of 189kW we need 189000W / 324V = 583A. This is a 583/152 = 3.84C discharge rate. If 1C is one hour then 3.84C is 1h / 3.84C = .26hour, which is .26h * 60min = 15.6minutes. So if you drive at full power, which is 189kW and 8000RPM, you will drain the ESS in 15.6 minutes.
If full speed is at 80mph, then we are looking at a full power range of 80mph / 60min = 1.33miles per min * 15.6 min = 20.8 miles.
The ESS is 324V and 152Ah. To drain the ESS in one hour you discharge at a 1C rate, which is 152A. 152A * 324V = 49248W or 49.3kW or 66.1hp. So if we supply 49.3kW of power for one hour we will completely drain the ESS. If you look at the the torque/power curve on the TM website (http://www.teslamotors.com/performance/performance.php) then you will see that 49.3kW is about 2000RPM.
The Roadster can go 250 miles on the EPA Highway cycle, which is an average speed of 48mph. That is 5.2 hours of driving to completely drain the ESS.
5.2 hours is a .19C discharge rate (1 / 5.2 = .19), which is 152A * .19 = 28.88A. This is 28.88A * 324V = 9357W or 9.4kW or 12.6hp. So if we supply 9.4kW of power for 5.2 hours we will completely drain the ESS. The torque/power curve shows 9.4kW to be in the few hundreds of RPM.
At a maximum power of 189kW we need 189000W / 324V = 583A. This is a 583/152 = 3.84C discharge rate. If 1C is one hour then 3.84C is 1h / 3.84C = .26hour, which is .26h * 60min = 15.6minutes. So if you drive at full power, which is 189kW and 8000RPM, you will drain the ESS in 15.6 minutes.
If full speed is at 80mph, then we are looking at a full power range of 80mph / 60min = 1.33miles per min * 15.6 min = 20.8 miles.
jsmith, I agree with okashira that your are clearly confused. Perhaps the root of your confusion is that your calculations (referring to power graphs etc) determine the maximum power output available at various speeds. What we are trying to determine is the maximum range of the vehicle at various speeds, which is accomplished by using only a small fraction of the maximum power output.
In a nutshell: For a given RPM, different loads on a motor will cause correspondingly different electrical power consumption.
The power output required to maintain a constant speed will be determined by the sum of the wind resistance, rolling resistance etc. We know that at high speeds wind resistance dominates, so a lot more power is required to maintain 135mph than 80mph. At very low speeds the static power consumption (A/C) will no longer be negligible, and also the efficiency of an induction motor will suffer at low speeds.
In a nutshell: For a given RPM, different loads on a motor will cause correspondingly different electrical power consumption.
The power output required to maintain a constant speed will be determined by the sum of the wind resistance, rolling resistance etc. We know that at high speeds wind resistance dominates, so a lot more power is required to maintain 135mph than 80mph. At very low speeds the static power consumption (A/C) will no longer be negligible, and also the efficiency of an induction motor will suffer at low speeds.
jjsmith:
I think what you are missing is that power <> voltage.
Yes, the engine might be running at constant voltage, but that doesn't mean it is consuming constant power. Power on an electric circuit is I.V, that is voltage times current. If voltage stays constant but current increases, power increases.
The fact that the controller uses frequency to control power does not tell you that current is constant. It just tells you that it uses another parameter, not voltage, to control current. It is not a simple resistor, where I=V/R. It is a much more complex calculation that yelds I in an AC motor.
In the case of an AC motor, when frequency increases above rotational speed, current increases. Thus, the engine consumes more power.
But as others have posted, you don't need to know that. No electronics can modify the laws of physics. If the car needs a certain amount of energy to beat resistance (at constant speed), then that's the exact amount of power the engine is developing. And a car, or any moving mass not floating in vacuum for that matter, needs an increasing amount of energy to move as speed increases.
I think what you are missing is that power <> voltage.
Yes, the engine might be running at constant voltage, but that doesn't mean it is consuming constant power. Power on an electric circuit is I.V, that is voltage times current. If voltage stays constant but current increases, power increases.
The fact that the controller uses frequency to control power does not tell you that current is constant. It just tells you that it uses another parameter, not voltage, to control current. It is not a simple resistor, where I=V/R. It is a much more complex calculation that yelds I in an AC motor.
In the case of an AC motor, when frequency increases above rotational speed, current increases. Thus, the engine consumes more power.
But as others have posted, you don't need to know that. No electronics can modify the laws of physics. If the car needs a certain amount of energy to beat resistance (at constant speed), then that's the exact amount of power the engine is developing. And a car, or any moving mass not floating in vacuum for that matter, needs an increasing amount of energy to move as speed increases.
iisjsmith said:
I registered here to try and clarify this issue, because iisjsmith has half-understood how AC drive works, so with additional knowledge he can see how WarpedOne's calculations indeed match with the operation of an AC motor drive (and reality) and I can take a stab at the issues that okashira raises, in which he is absolutely correct, but hear me out:
- iisjsmith is correct when stating that the speed of an AC motor is controlled by the frequency applied.
his following reasoning is flawed however, because no AC motor can "jump" in frequency, expecially when attached to the wheels of a car.
If you would try to make a sudden change in frequency, either one of two things will happen:
1. the motor follows the change in frequency if it can, so the wheels will change speed suddenly - this means loss of control: spinning wheels.
2. the motor fails to follow the jump in frequency, now the field in the AC motor is rotating at a different speed than the rotor, which will give the effect of alternating pushes and pulls with the frequency of the difference between field and rotor speed. The motor will take a long time to get to the same speed of the field if it fails to follow the field changes and it is very uncomfortable to be present in a vehicle that is pushed forward and back several times per second at maximum power while rolling down the road.
To avoid these problems, the AC drive in a car uses the technique of phase control. This means that the inverter is *locked* to the frequency of rotation of the AC motor, but it can slow down or speed up a little bit, maximum is a 90 degree phase difference with the field of the rotor.
This is why Okashira said that the frequency is determined by the motor: it is. The inverter *has* to follow the frequency of the motor, it can only try to change it s-l-o-w-l-y by applying a field that is ahead or lagging with respect to the motor, to increase or decrease the frequency.
To make this even more obvious, the following example: what if a car with AC drive is riding up a hill so steep that it is losing speed, even with the AC drive at full throttle?
You guessed it: the inverter applies a maximum field to increase motor speed but the motor slows down, so the inverter has to keep slowing down to stay within 90 degrees of the motor's field or it risks losing control over the motor completely. The speed of the car determines the motor speed and the motor speed determines the inverter's speed. All the inverter can do is generate a _torque_ by applying a field that leads or that lags the motor's field, to try and increase or decrease the motor's frequency. Just like a gas engine pushes against a cylinder to create a torque, which may speed the crankshaft up if the torque is larger than the resulting friction of the wheels.
Now about consumption:
An AC motor drive is an inverter, which is capable of generating an AC waveform with variable frequency *and* variable voltage!
In commercial VFD's the latter is often hardly used, maybe only to limit startup current surges, so a motor always runs at max voltage. But in an AC drive in a car the voltage is certainly a part that is necessary, because the motor is not running in a certain range of speeds all the time, it varies from zero (actually negative, to back up) to redline all the time and the torque requirements vary also all the time.
Varying the voltage in an inverter is simple: the same way that it approximates a sine wave by switching on/off in a rising and falling pattern to create and average voltage that follows the sine waveform, it can also reduce the on-time by a certain percentage, which will reduce the AC voltage by that percentage. If it switches on only 50% of how much it should to make a max AC voltage, then the output is half as high, for example 150 Volts instead of 300V. If it switches on for only 10%, the output is only 30V.
Now why would an inverter do that? Many reasons:
- when the AC motor is at standstill and the inverter starts to push against it with a very low frequency field, the current is very high for a low voltage due to the lack of impedance (which is linear with frequency for an inductor) so the inverter needs to protect itself from over-current by applying only a low voltage and monitor current - voltage can steadily rise with increasing frequency, this is exactly why the motor *power* curve has an almost linear ascend up to its maximum power point: it is current-limited.
- partial throttle: when you want to maintain speed, only a low torque is necessary to keep the motor (and the car) moving at flat road.
It would be a waste to generate a maximum voltage waveform by the inverter, create a strong field and then not use it to generate much power. So the inverter reduces the voltage waveform and phase difference with the motor field, so it only generates as much torque as necessary while consuming as little power as necessary. There is a certain minimum voltage level that the inverter must generate, because the field of the motor will automatically already create a voltage in the motor windings (back EMF) so in order to apply power, the inverter must create a voltage that is larger than this back EMF, otherwise there would be no current and thus no power to the motor.
- partial braking: one of the interesting things of an AC motor is that it is just as good a generator as a motor and the inverter will automatically do either one, so when creating a field that is lagging the motor, the current will be in the opposite direction and flow from the inverter into the battery pack (recharging). The energy from the moving car is converted back into electrical power, slowing down the car. I like this part the best of the whole AC drive: put the power back in the battery instead of heating the brake pads. When did you see a gas car pump gas back into the tank while braking?
- battery sag: when pulling a large current from the pack and towards the end of the charge of the pack, the DC battery voltage will drop. Since the AC voltage cannot be larger than the DC input voltage (in motoring mode - regeneration is a different story) the AC voltage will need to be reduced when the DC voltage sags. This results in a reduced power but the alternative would be a distorted AC waveform with the top cut off, which is hard on inverter and motor.
Now this only leaves us the reduction of power beyond the maximum power point of the AC motor. As already indicated, frequency increases and this causes increased impedance of the motor, the result is that even when the inverter applies the full voltage to the motor, above a certain frequency the motor will not draw the maximum current because the current does not have the time to ramp up due to the fast change of voltage (high frequency). The result is a somewhat linear drop of power with increasing frequency.
Note that also the back EMF rises with increasing speed, so either this or the impedance will take care of limiting current.
So, how will the range of Tesla's Roadster be determined? By the power that the inverter needs to feed to the AC motor to create torque that overcomes the drag of the total car.
How could we have avoided all the theory above? By realizing that the energy loss from friction by the total car must be replenished by the energy coming from the batteries in DC form, which is exactly the calculation that WarpedOne did.
Knowing that the DC energy is converted by an inverter into an AC waveform with varying frequency and voltage to create a varying AC current which causes a magnetic field in an AC motor that causes a torque proportional to the current, which should cancel out the torque resulting from the friction to keep the car running at the same speed.... Well, that part can be skipped. (Although I am a techie, so I take things apart to learn about how they work)
Note that although I am an EE, I have not majored in power electronics or motor technology, it is mainly from practical experience and reading up on AC drive that I came to the above understanding. Please correct any errors or omissions.
BTW - In practice, I use my AC drive all the time, because my truck has an AC drive alike (but of lower power) than the Tesla.
You can see it here: http://evalbum.com/694
W8MM
R1.5 #325 + Mdl S #01380
cor_van_de_water said:
Thanks so much for the tutorial.
I am an EE as well, but in the RF area.
Much appreciated.
Anatoly Moskalev posted on Tesla's blog regarding range. I am copying his posts here as it would be a big waste for them to disappear with next blog update.
Anatoly Moskalev wrote on March 27th, 2007 at 10:54 am on Tesla's blog:
Anatoly Moskalev wrote on March 27th, 2007 at 11:26 am on Tesla's blog:
Anatoly Moskalev wrote on March 28th, 2007 at 10:14 am
Anatoly Moskalev wrote on March 27th, 2007 at 10:54 am on Tesla's blog:
Anatoly Moskalev wrote on March 27th, 2007 at 11:26 am on Tesla's blog:
Anatoly Moskalev wrote on March 28th, 2007 at 10:14 am
Thanks, cor_van_de_water, for your explanation.
I had some limited understanding of how the AC motor works from bit and pieces seen elsewhere and your explanation ties together well what I thought was happening.
After reading the previous postings from others I was starting to doubt myself, so you got me back to trusting my instincts.
("Frequency must be tied closely to motor speed! Up to 90 degrees off is what I needed to hear.")
Now I have another question...
On the teslamotors.com blogs was a discussion of why the power curve starts to drop off at higher RPMs.
I theorized losses such as controller transitor switching losses (at higher frequencies), bearing drag, and wind resistance against the rotor.
Anatoly seemed to confirm these theories.
...Then Wolfgang came along and said that those are incidental and the real problem is insulation limits in the motor windings.
I accepted this as an overlooked truth, but now I am starting to wonder. Could the "IGBT switching losses" actually be more of a problem that he suggests? Is motor heat or battery current output the real issue?
Is it a given that voltage applied to the eMotor is at most the ESS pack voltage? I heard that the ESS can have up to 411V (when fully charged) so shouldn't the windings only ever see 411V max? Now wouldn't max voltage already be applied at less than highest RPMs? If so, I would think that the winding insulation would need to be able to handle 411V, and would never get more even at the higher RPMs. So, (assuming that rotor aerodynamic drag, & bearing lubricant viscosity are incidental concerns), why does Tesla eMotor power drop off above 6500RPM?
At this point I am starting to think it is all because heat and/or limits of the batteries. Perhaps they have to reduce the voltage or max phase slip angle above 6500RPMs because the ESS would start to damage the batteries too much or the eMotor would overheat?
When you look at the torque curve it appears to start a linear (intentional?) downslope right at 6500RPMs which (when using HP=Torque*RPMs) causes the horsepower to peak and then drop off slowly towards redline.
I would think that they could make the torque dropoff less gradual and maintain 248hp (200KW) horsepower from 8000-13.500RPMs. So why not?
Perhaps the batteries get tired of giving out 4C too long and need a break? Perhaps the motor starts to overhead? Anyone know?
=========================
# cor_van_de_water wrote:
## "Since the AC voltage cannot be larger than the DC input voltage"
Is that absolutely true?
From a Toyota Prius whitepaper:
"Power Control Unit The power control unit contains an inverter that converts the DC from the battery into an AC for driving the motor and a DC/DC converter for conversion to 12V. In THS II, a high-voltage power circuit that can INCREASE THE VOLTAGE from the power supply to 500V, has been added. Based on the relationship of Power = Voltage x Current, increasing the voltage makes it possible to reduce the current, which in turn makes it possible to reduce the size of the inverter. Also, because the control circuits have been integrated, the size of the power control unit itself has remained almost the same as before."
The Prius traction battery pack outputs less than 500V, so they have a way to boost the voltage...
====================================================
When I look at Tesla eMotor torque output, torque is constant (~200ft-lb) from 0-6500RPMs, then drops off in a very linear fashion at a near perfect 45 degree angle.
As I look at a description of "Variable Frequency Drive" here:
http://en.wikipedia.org/wiki/Variable-frequency_drive
I see mention of having a specific "synchronous" speed based on AC frequency. If max frequency of the inverter corresponds to synchronous speed of 6500RPM, then perhaps higher RPMs are operating in "field weakening" mode which would readily explain why torque drops off like that and power peaks then drops off slowly.
Now that brings me back to the idea that IGBT switching limits may be what causes the dramatic shift in torque curve. Am I wrong?
The overall effect seems well engineered to other limits within the vehicle system.
For instance, the ESS batteries can perhaps give 4C (~200KW) of current for a short time, but then they need a break, so it would be wise to let the eMotor power start to drop off as the motor heads toward redline. The same thing could be said for eMotor temperature. I think it has a "duty cycle" where you can push it over the sustained limit temporarily then you need to give it a rest.
It seems like there is a precarious relationship between battery output limits, IGBT switching limits, motor winding insulation limits, motor temperature rejection limits, and gearing. It all seems very elegant the way all the limits line up just right so that battery, eMotor and gearing are highly optiimized to do 0-60 in 1st gear, and get just to the 1/4 mile in 2nd gear.
Maybe I am wrong, but I am thinking that battery winding insulation is really driven by IGBT switching limits and max HP requirements. The PEM runs up to as high a frequency as possible (indicating 6500RPMs), and this determines max motor power output based on voltage. The ESS voltage is tuned to provide the desired HP at syncronous speed, with anticipated drop off at higher RPMs. To achieve the performance goals of the vehicle, 250hp is selected, and so ~375V comes up as the needed voltage to get that power at 6500RPMs. The winding insulation is then designed to handle about 400V.
Before I considered the above, I thought "why not just have higher voltage ESS", but then it seems that there would be no point because you can't get more power out of the eMotor just by adding more voltage once you are past the synchronous speed limit determined by the max frequency produced by the PEM. So, one could argue that the motor windings are the limit based on voltage, but it seems to me that is just a side effect of other design considerations that revolve around the IGBTs.
After stewing all the above around in my head, I stumble on this:
http://www.plantservices.com/articles/2002/48.html?page=2
Which discusses "Vector and Direct Torque Control" techniques to manage motor power based on "torque and flux" rather than voltage and frequency. Well that is more than I can handle at this point, but the "old school" rules using voltage and frequency still seem to line up with Tesla's eMotor torque/power graph so I am just going to have to guess that Vector/DTC control is just another way of controlling working under the same constraints.
I had some limited understanding of how the AC motor works from bit and pieces seen elsewhere and your explanation ties together well what I thought was happening.
After reading the previous postings from others I was starting to doubt myself, so you got me back to trusting my instincts.
("Frequency must be tied closely to motor speed! Up to 90 degrees off is what I needed to hear.")
Now I have another question...
On the teslamotors.com blogs was a discussion of why the power curve starts to drop off at higher RPMs.
I theorized losses such as controller transitor switching losses (at higher frequencies), bearing drag, and wind resistance against the rotor.
Anatoly seemed to confirm these theories.
...Then Wolfgang came along and said that those are incidental and the real problem is insulation limits in the motor windings.
I accepted this as an overlooked truth, but now I am starting to wonder. Could the "IGBT switching losses" actually be more of a problem that he suggests? Is motor heat or battery current output the real issue?
Is it a given that voltage applied to the eMotor is at most the ESS pack voltage? I heard that the ESS can have up to 411V (when fully charged) so shouldn't the windings only ever see 411V max? Now wouldn't max voltage already be applied at less than highest RPMs? If so, I would think that the winding insulation would need to be able to handle 411V, and would never get more even at the higher RPMs. So, (assuming that rotor aerodynamic drag, & bearing lubricant viscosity are incidental concerns), why does Tesla eMotor power drop off above 6500RPM?
At this point I am starting to think it is all because heat and/or limits of the batteries. Perhaps they have to reduce the voltage or max phase slip angle above 6500RPMs because the ESS would start to damage the batteries too much or the eMotor would overheat?
When you look at the torque curve it appears to start a linear (intentional?) downslope right at 6500RPMs which (when using HP=Torque*RPMs) causes the horsepower to peak and then drop off slowly towards redline.
I would think that they could make the torque dropoff less gradual and maintain 248hp (200KW) horsepower from 8000-13.500RPMs. So why not?
Perhaps the batteries get tired of giving out 4C too long and need a break? Perhaps the motor starts to overhead? Anyone know?
=========================
# cor_van_de_water wrote:
## "Since the AC voltage cannot be larger than the DC input voltage"
Is that absolutely true?
From a Toyota Prius whitepaper:
"Power Control Unit The power control unit contains an inverter that converts the DC from the battery into an AC for driving the motor and a DC/DC converter for conversion to 12V. In THS II, a high-voltage power circuit that can INCREASE THE VOLTAGE from the power supply to 500V, has been added. Based on the relationship of Power = Voltage x Current, increasing the voltage makes it possible to reduce the current, which in turn makes it possible to reduce the size of the inverter. Also, because the control circuits have been integrated, the size of the power control unit itself has remained almost the same as before."
The Prius traction battery pack outputs less than 500V, so they have a way to boost the voltage...
====================================================
When I look at Tesla eMotor torque output, torque is constant (~200ft-lb) from 0-6500RPMs, then drops off in a very linear fashion at a near perfect 45 degree angle.
As I look at a description of "Variable Frequency Drive" here:
http://en.wikipedia.org/wiki/Variable-frequency_drive
I see mention of having a specific "synchronous" speed based on AC frequency. If max frequency of the inverter corresponds to synchronous speed of 6500RPM, then perhaps higher RPMs are operating in "field weakening" mode which would readily explain why torque drops off like that and power peaks then drops off slowly.
Now that brings me back to the idea that IGBT switching limits may be what causes the dramatic shift in torque curve. Am I wrong?
The overall effect seems well engineered to other limits within the vehicle system.
For instance, the ESS batteries can perhaps give 4C (~200KW) of current for a short time, but then they need a break, so it would be wise to let the eMotor power start to drop off as the motor heads toward redline. The same thing could be said for eMotor temperature. I think it has a "duty cycle" where you can push it over the sustained limit temporarily then you need to give it a rest.
It seems like there is a precarious relationship between battery output limits, IGBT switching limits, motor winding insulation limits, motor temperature rejection limits, and gearing. It all seems very elegant the way all the limits line up just right so that battery, eMotor and gearing are highly optiimized to do 0-60 in 1st gear, and get just to the 1/4 mile in 2nd gear.
Maybe I am wrong, but I am thinking that battery winding insulation is really driven by IGBT switching limits and max HP requirements. The PEM runs up to as high a frequency as possible (indicating 6500RPMs), and this determines max motor power output based on voltage. The ESS voltage is tuned to provide the desired HP at syncronous speed, with anticipated drop off at higher RPMs. To achieve the performance goals of the vehicle, 250hp is selected, and so ~375V comes up as the needed voltage to get that power at 6500RPMs. The winding insulation is then designed to handle about 400V.
Before I considered the above, I thought "why not just have higher voltage ESS", but then it seems that there would be no point because you can't get more power out of the eMotor just by adding more voltage once you are past the synchronous speed limit determined by the max frequency produced by the PEM. So, one could argue that the motor windings are the limit based on voltage, but it seems to me that is just a side effect of other design considerations that revolve around the IGBTs.
After stewing all the above around in my head, I stumble on this:
http://www.plantservices.com/articles/2002/48.html?page=2
Which discusses "Vector and Direct Torque Control" techniques to manage motor power based on "torque and flux" rather than voltage and frequency. Well that is more than I can handle at this point, but the "old school" rules using voltage and frequency still seem to line up with Tesla's eMotor torque/power graph so I am just going to have to guess that Vector/DTC control is just another way of controlling working under the same constraints.
You might be on to something. In Tesla' blog Never a dull moment (http://www.teslamotors.com/blog4/?p=38) they state:
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