234 comments on The energy efficiency of cars
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"The efficiency of an electric motor is a curve"
True for induction machines but not for modern permanent magnet machines, these exhibit a very high part-load efficiency.
EDIT: Of course a PM machine will be more expensive!
crobar, I agree totally, though copper losses in PM machines will be fixed for a given torque so the efficiency will still fall, but not as much. When I suggested permanent magnet machines would be used (on another post) I got slated for it by Engineer_Poet, You just can't win!
The I sq R law applies to everything electrical including PM motors. I represents amps and R is resistance measured in ohms. Ohms are mostly constant while amps are a function of torque. Therefore a motor's efficiency is drops at the square of load. Light loads are very efficient while heavy loads are only a fraction of peak efficiency. The same law applies to fuel cells which only match their high efficiency claims at very light loads.
You've got that almost entirely wrong.
1) I sq R applies, in synchronous motors, only to that part of the impedance presented by the motors which is due to resistance. Most impedance in synchronous motors occurs due to inductance. Also in typical high-efficiency synchronous motors, the field is created by permanent magnets which make no contribution to energy use. In induction motors, same except must also contend with high currents at low voltages, therefore high I sq R losses in cheap aluminum conductors in rotors, often reduced by using costly copper bars in high-Q induction motors eg. Tesla.
2) A typical electric motor's efficiency INCREASES with the load up to rated output, heavily dependent on issues such as a) synchronous or inductive? b) if inductive, what rotor resistance? c) what tradeoffs has the manufacturer made regarding i) winding design and conductor cross-section ii) what quality of magnetic steel chosen? iii) what cooling method chosen, if air, what fan design etc? iv) what engineering tradeoffs made regarding bearings vs. cost etc?
Most reasonable size electric motors will achieve peak efficiency at full rated power, and effic. will drop off dramatically at light loads.
Lengould,
Just to expand on what you say, (have said this above) maximum efficiency occurs when the "load dependant loss = Non load dependant loss". This can be simplified to saying "copper (I*I*R) = Iron (EDDY+Hysterisis)". This applies to transformers as well, and their will be a mechanical equivalent I expect.
At no load, there are no copper losses on a PM motor if electronically commutated since current is prop. to torque.
An induction motor has to draw magnetising current so it has a "no load" copper loss, though less than at full load. For a given frequency and excitation (flux density) Stator iron losses are fixed and rotor iron losses are often ignored altogether, because the rotor slip frequency is so low under any permitted operating conditions. All copper losses are I*I*R (irrespective of motor type), and so increase with load current and when these = iron loss (and windage) the point of maximum efficiecy is reached. The rotor resistance and inductance can be "referred" back to the stator in the equivalent circuit and can be established by stall and off load tests, in a similar way to short circuit tests etc on transformers.
As long as current must flow through a (non-superconducting) wire there will be I sq R losses. Even at no load, current must flow if motor is to rotate. Still needs to produce torque req'd to overcome windage, bearing, etc. losses.
Your [q]maximum efficiency occurs when the "load dependant loss = Non load dependant loss".[/q] Not sure about that, never heard of it before, but that doesn't make it wrong. Sounds a bit iffy regarding motor theory though.
I should have said negligable. Yes there will be a small loss, but I assumed no load, no torque so no current, but friction and windage yes so a small loss, its one of my famous approximations.
Its a law, of similar standing the maximum power transfer theory (maximum power tranfer occurs when the source impedance = load impedance) under this condition efficency is 50%. It comes from differentiation and finding the turning points (max and min) of the equation that describes efficiency in terms of iron and copper loss. Its a long time ago since I was forced to derive it, but if I can find a reference to it I will. (if your interested) It may be an approximation for a motor becuase losses are more complex, but for a transformer it holds true.
Lengould, I don't know if this link will work, but illustrates the theory and how to derive it. Its one of those laws I suspect will apply to any energy transfer system, though as an electrical engineer, I have only seen the proof for electrical machines. This document is for dc motors, but the maths does not care what the motor type is.
If you are going to comment on motor theory in the future, please don't question those who understand it unless you do yourself.
As the saying goes, those who think they know everything are anoying to those that do.
http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT%20Kharagpur/Basic%20Electrical%20Technology/pdf/L-40(TB)(ET)%20((EE)NPTEL).pdf
Edit:
It does not work as a link, but cut and paste it into google and it does. Perhaps one of the oil drum gurus can tell me how to create a working link from it.
I would just recommend that you review theory of efficiency calculations. A motor "at no load" by definition is doing zero effective work, but still drawing some power, therefore has zero (eg. lowest possible) efficiency.
That's why peak efficiecy occurs at some load much greater than zero, because at zero load, unless you have designed your motor very badly, the fixed loss (iron) is much higher than the variable loss (copper), hence the efficiency is not maximum. Check it out for your self, i gave you a link.
Sorry, I don't think I'll be listening any further to anyone who applies this sort of logic to electric motors. 50% is maximum efficiency? Ha.
For example, this quote from EnergyStar.gov presentation, http://www.energystar.gov/ia/business/networking/presentations/feb_05_mo...
“Right-size” the Motor
• Choose the correct rating for the application
– Oversized motors have lower efficiency andpower factor
– Highest efficiency 75 -100% of rated load
I can only conclude you very little about electrical engineering. The maximum power transfer theory is just that. maximum efficiency does not necessarily, and usually does not, occur at maximum power transfer, that is obvious. If you drew maximum power from the grid, 50% of the energy would be lost in the grid and it would probably burn out.
I was simply saying the maximum power transfer theory is well known, and so is the condition for maximum efficiency, but they are two different conditions. You really need some additional training on these very basic and fundamental ideas. Any good electrical engineering text book will help you.
In that case these motors have been designed to have variable and fixed loss equal at 75%. Dead simple, the law applies.
Over sized motors do have low efficiency, because the iron loss is greater than the variable loss at below optimum load. I can't understand why you can't grasp it.
Whether you like it or not the theory holds true.
Perhaps your problem is one of definition. Define "maximum power transfer" in your terms. Is that the amount of power one can put into a locked-rotor condition motor for the few seconds before it burns out, or the maximum manufacturer rated power input? If the second, then you're clearly wrong because every properly designed industrial motor is designed to achieve its maximum efficiency at or very near its maximum rated power. Check any other manufacturer's publications if you don't like the Baldor / Reliance website which I referenced.
I think you may be badly confused, mixing motor theory with high-frequency RF circuit theory, where the impedance to power transfer is almost entirely due to the very high frequency effects (often stray capacitances and inductances developed in the conductors and feeder circuit elements themselves and not the load circuit). In those cases, yes, I'll agree that such issues as that which you describe can limit efficiency to a percentage much lower than that of a simple effectively unlimited power source feeding into an industrial motor at very low frequency and with only rated load or less on the motor.
?? Perhaps further definition of terms would make this statement appear less obviously wrong? I don't care what power level you're going to draw from the grid you're never going to see 50% of it dissipated in the grid itself. Though that must be some scary sort of test situation. You haven't lived until you've observed a dead short on a 2 MW bussbar in a switch room (as I have), and the results, effectively a lot of vapourized copper and steel. The formulae for calculating exactly the short-circuit capability of a service feeder are well-known. The following quote from Federal Pacific's Transformer Basics details it http://www.federalpacific.com/university/transbasics/chapter5.html
So the maximum draw one can get from the grid is entirely dependent on the size and impedance of the nearest upstream transformer to the point of the short, and it can be impressive even though eg. a short on a building feeder behind a 2 MW transformer which can drive molten copper right through steel doors (which I've seen) is still only drawing 40 MW from a grid capable of delivering 25,000+ MW in normal operation here. Look up BIL ratings reference switchgear. So what losses the grid would experience delivering its total 25,000+ MW to a single point depends entirely on the design of the nearest upstream equipment, but if that's a code rated transformer, it will suffer no more than 5% losses, not 50%.
But enough of this silly correcting of your errors. Its way off the websites topics of interest.
http://en.wikipedia.org/wiki/Maximum_power_theorem
Thats why I said is you apply it to power circuits, you would burn them out (blow your self up if you please), but it still applies regardless.
A transformer of 5% impedance would supply maximum power into a similar load impedance. After that point more power would be dissipated in the transformer (you do not operate power circuits under these conditions, not for long in any case). If the load impedance is infinite, no power transfer, if the load impedance is zero, no power transfer. Dead simple, again your theory shows serious lacking.
You keep making a mistake common to people without education in electromagnetics, which is confusing impedance with resistance. Resistance dissipates power, but inductive and capacitive impedances do not. Most components of the electric power grid (transformers and transmission lines) have a net inductive impedance and will limit current to a maximum without dissipating anything close to the power that a naïve calculation would imply. There are even more non-intuitive consequences of this which must be taken into account by power engineers, but I won't go into those.
Maximum power transfer theorem applies to complex (R+jX) circuits. It is used to derive maximum torque for a squirrel cage induction motor for example, which is only a "special" power transformer after all. The mechanical output is represented by a variable resistance and the loss by a fixed resistance. Maximum torque occurs when maximum power is dissipated into the total equivalent resistance. As slip increases beyond a certain value, rising XL of the rotor limits this power and the turning point (pull out torque) is reached and this coincides with maximum power transfer into the resistance of secondary cicuit of the transformer (motor). By adding additional resistance (slipring motor) the maximum torque condition can be recovered, though at a greater slip than with a lower rotor resistance. The peak torque is always the same, it just occurs at an ever greater value of slip. This is all basic well documented stuff, I can't understand the problem.
So no confusion between impedance and resistance on my behalf. I also understand electromagnetics reasonably well.
You're obfuscating the issue (how trollish of you).
You said above, and I copy-and-paste:
No, they would not even necessarily burn out; they could easily self-limit to currents less than what would cause physical or thermal failure. Of course, a properly protected system will shut down before it fails physically.
Not in the slightest,
There is no way a power transformer with 5% impedance operating at rated voltage would limit the current to survivable levels. It would be an exciting experiment though if you have one at hand.
O RLY? Do you think the transformer would fail from physical overstress, or would overheat faster than the fuses upstream would blow? Do you think that the network it's connected to has zero source impedance?
Things like shorted outputs happen, and the hardware usually survives (that's the design criterion of the protection systems). That's an existence proof.
Fault conditions are an "instantaneous" event and its obvious (even to you I guess) that the protection employed passes a value of I*I*t much less than that required to damage a power transformer, otherwise it would not be protection would it? your just being a clown. Your argument has nowt to do with the fact you cannot operate a power transformer into a load under maximum power transfer (into the load) conditions for long without burning it out. The same goes for operating an induction motor at maximum torque, you will burn it out, but not straight away!!!!!! It has thermal capacity that can absorb some overload and hopefully more than the I*I*t that the overload will allow through, though not always because some folk turn up the overload too high. Its so simple it hurts that you can't grasp it, God help us.
From what I recall from my last installing a 2MVA transformer, the 11kV side has negligable impedance when referred to the 415 secondary. The 40,000 kA or so of fault current is limited almost totally by the transformer leakage inductance and if not protected, there would soon be a mess. The 11kV impedance would not protect very much at all. There is a GEC film showing all this, it's great fun and a big bang at the end.
So you admit that maximum power transfer isn't a design load condition? That it's neither necessary nor desirable to operate anywhere near it? And you've been raising this as an issue, why?
You have descended fully into trolldom. Get lost.
I never said it was. This began when I said to lengould that maximum efficiency occurred when fixed loss equalled variable loss and this idea was of similar standing in electrical engineering to the maximum power transfer theory. Both are well established laws. He took this as me claiming maximum efficiency is 50%. Well, this mistake was made over a 100 years ago in Edison's era. These two are separate conditions and for some reason you stepped in halfway through and totally misinterpreted what I was saying. so I will clarify,
Maximum efficiency occurs when fixed loss (iron windage)=variable loss (Copper). This applies (as far as I am aware) to all electrical machines. It puts limits on operating efficiency, since you can't simply assume using a larger machine at low load (hence reducing resistive I*I*R losses) will improve efficiency. I strongly suspect there is a mechanical analogy for hydraulic motors, ICE etc etc.
The maximum power transfer theory states that maximum power can be transferred from an emf source when the load resistance = the EMF source resistance. At this point efficiency is 50%, not desirable for a power system, obviously because it will you cook the source (Battery) due to loss, but it may be desirable if you want maximum power transfer for a signal. It was widely applied during the days of valve amplifiers, but not transitor amplifiers, because their source impedance is so low they would be destroyed if it was applied. It also applies to power systems, but is not an operating conditon (I never ever said it was) for obvious reasons. It is more complex because the loss in the source may not equal the load because the source impedance is complex (R+jX). However, reducing the load resistance will eventually cause the condition of maximum power because Z will cause the voltage to drop faster than the current rises. This is the point of maximum power transfer and at this point (probably well before) most transformers will be operating in an overload condition.
Its late at night here in the uk, I have tried to put this into words and reason with you as best I can. If you can't agree now I just assume you just want to pick fault regardless of logic. You are getting offensive now, which helps no one. I am disappointed in you telling a fellow engineer to get lost. Calling me a troll, well I see that as a joke, telling me to get lost, can't you do better than that?
I note that you've still not responded to the electric motor manufacturer's efficiency rating vs power curves I referenced at the beginning of this discussion, which clearly put paid to your ridiculous "Maximum efficiency occurs when fixed loss (iron windage)=variable loss (Copper). This applies (as far as I am aware) to all electrical machines."
This whole discussion has been about your refusal to accept taht your hypothesis does not apply to electric motors or power grids, as I keep proving repeatedly.
C'mon back if you ever get out of first year.
I did look at your information and its basic "sales brochure" performance curves with a few elementry calclations on cost saving, nothing I did not know already prior to reading.
I'm at a loss by how you draw your conclusions from the graphs, which don't give any information about specific loss. For instance you cannot conclude at peak efficiency for each curve what either iron or copper losses are. I don't know what your on, but I suggest you go and get treatment for it. I can only repeat, buy yourself a good book and will will find out I was right all along and your tantrums can cease.
EDIT,
Here is the mathematical proof that max eff. occurs when fixed=variable.
c=constant loss
b=coefficient for I loss
a=coefficent for I*I loss
n=efficiency
so loss=aI2+bI+c
n=(powerin-loss)/powerin (by a bit of manipulation can rearrange to)
n=powerout/(powerout + Loss)
powerout=VI cos(phi)
By substitution you get
n=V*I*cos(phi)/((V*I*cos(phi))+aI2+bI+c)
to find max (or min), differentiate with respect to I and make equal to zero. This is a quotient because both numerator and denominator are functions of "I", so one has to use the quotient rule VIZ
dn/di= v*du/di - u*dv/di / v2
where v=((V*I*cos(phi))+aI2+bI+c)
and u=V*I*cos(phi)
You end up with quite a few terms on the numerator divided by the orginal denominator (v) squared. But the terms cancel down to this
dn/di=((c-aI2)/(denominator)2)* v cos(phi)
From this, it is obvious the equation is zero when c=aI2. Further differentiation can prove whether this is a maximum (turning point) or minimum (turning point), but it is a maximum. It would be quite easy to demonstrate this by a small program written in Q BASIC of MS Excel etc.
It is curious that the "bI" term cancels (as it has no meaning) leaving only constant (iron) and I2 (copper) terms, which is why it is dangerous to rely on intuition, rather than sound mathematics to demonstrate an idea is true. It also demostrates conflicting requirements when designing motors and transformers and gives one a good understanding as to why claimed efficiencies are often not relised in the real world.
Lengould,
I asuume you have accepted the max efficiency proof, since you have not disproved it, here is empirical proof of the max power transfer theory. Its written in BASIC, so you will need a copy, easily available from older Windows and late DOS packages or from the web. You will see that max power to the load is when the impedance of the transformer is equal to the load resistance, though not a permitted operating condition in power supply systems it is an important and well known concept. As mentioned above it is used to prove the maximum torque condition in the induction motor
Mathematical proof is derived in a similar way to my last example using differentiation.
CLS
SCREEN 8
t = 0
R = 0
WINDOW (-0, -30)-(1000, 50)
LINE (0, 0)-(1000, 0)
R1 = 100 'transformer winding resistance (referred to secondary)
X = 100 'transformer leakage reactance (ditto)
v = 100 'secondary voltage
FOR R2 = .1 TO 1000 STEP .1 'secondary load resistance
ztotal = (X ^ 2 + (R1 + R2) ^ 2) ^ .5 'secondary impedance including load
z = (X ^ 2 + R1 ^ 2) ^ .5 'secondary impedance excluding load
I = v / ztotal
power = I ^ 2 * R2
dp = power - powerprev 'dp is incremental change in power
powerprev = power
IF (dp / .1) < .00001 AND dp > -.00001 THEN PRINT "load resistance="; R2: PRINT "z="; z
PSET (R2, power), 10
'PRINT dp
PSET (R2, dp * 10)
NEXT R2
What I objected to was your claim that electric drivetrains wouldn't scale because of limitations on rare earths for magnets. True, induction motors have greater losses due to slip; however, they are rugged, cheap and need only iron and conductors. There will be some price point where the savings on RE magnets will buy enough batteries that it yields the same utility.
Fair comment, though I don't recall saying rare earth magnets were rare and would prevent scaling, because many rare earth elements are not infact rare. All I suggested was we would end up with these motors that require position feedback, just as ICE's have been blighted by unecessary complexity to gain ever dwindling emissions improvements.
I am a big fan of the good old Sqirrel Cage induction motor due to its simplicity, but again folk are forever being encouraged to operate them closed loop, which is a sales driven gimmick in many cases, but its happening in the real world. Closed loop is sold for its low speed stability and zero speed maximum torque capability, none of which are required for a train locomotive, but may be saleable feature for a road car, to hold on a hill for example. Complexity is often market driven, engineers simply respond to the market department's demands.
Audi are advertising a 700 mile range for one of their models at the moment, You state this is not required in an earlier post, I would not disagree with you, but its what people will buy that counts.
Does this put us nearer any sort of agreement?
The 700-mile range figure appears to pertain to the VW Blue Sport, and from what I can tell it's about the fuel economy (50 MPG).
I'm not opposed to a car with 700-mile range; I've done well over 700 miles on a tank myself. But if your concern is operating cost, swapping batteries every 100 miles to get 5¢/mile energy cost would do just as well. Plugging in at home so you never need to visit any kind of service station for your typical week of local driving would be a winner too.
I don't think I agree or disagree with you; I think you were just missing the point. ;-)