Analyzing car efficiency at different power outputs (accelerations)

Analyzing car efficiency at different power outputs (accelerations)

Physics dictates that the amount of energy required to accelerate a given mass to a specific velocity is independent of the acceleration. (Final kinetic energy = 1/2mV^2 => how fast you got there doesn't change the amount of energy required). This is of course under ideal conditions (in a vacuum, rolling resistance is constant, etc.) but it should still be usable as a guideline in real world situations.
Based on some very crude experiments with my Model S, I've got the idea that heavy acceleration to a given speed is less efficient (uses more energy) than light acceleration. If true, this must be an artifact of some other component of the system: the motor, inverter, battery, etc.

I don't know much about efficiency of 3-phase inverters at varying outputs, so I'll just ignore that. :)
I think we do know that the efficiency of the motor is fairly constant at the low end (say, 0-60mph in the Telsa: real world use).
So that leaves the battery, and I think I have a model that explains why the battery causes the car to "waste" energy during heavy acceleration (or even deceleration?)
Batteries are often modeled as an ideal voltage source and a "internal resistance". This isn't perfect, but works pretty well. Here's an article discussing the internal resistance and other factors in 3 types of EV batteries:

In this model, the "waste energy" from the battery is I^2 * Ri, where I is the current, and Ri is the internal resistance of the battery. I think this waste energy essentially becomes heat. Here is some information about these Lithium 18650 batteries Tesla uses: It says a "healthy battery" has an internal resistance of ~0.1ohm, and a depleted battery more like 0.250ohms. At 0.400ohms internal resistance, the battery is toast.
The power meter on the Model S goes from -60kW (regen, pedal up) to +320kW (pedal down), so lets run throw some numbers on how much "waste energy" is used at 2 different accelerations. Lets accelerate at 40kW and 320kW.
I'll do a lot of rounding here, because I don't know exact numbers--like how many cells there are, or how many are in use at the same time--and hoping that it kind of averages out to the right answer.
I believe there are in the range of 7500 lithium batteries in the 85kWh models. Lets assume they are all in use all the time. These cells deliver ~4V.
So how much current is each cell delivering when they are delivering 40kW of power?

P = I * V or I = P/V
40000W / 4V = 10000Amps
10000A/7500 cells = 1.3Amps per cell
"Waste energy" per cell is I^2 * Ri = 1.3^2 * 0.1ohm == 0.169W
0.169 W * 7500 cells = 1267W
1.267kW / (40+1.267kW) = 3% wasted energy (battery loss) at 40kW

I would say 3% "waste" is small enough that there may be other causes of loss that are much bigger, so it might even be "negligible".

Now lets try it with the pedal floored: 320kW of useful power.

320000W/4v = 80000Amps
80000A/7500 cells = 10.7Amps per cell
"Waste energy" per cell is I^2 * Ri = 10.7^2 * 0.1ohm = 11.4W
11.4 W * 7500 cells = 86kW
86kW / (320+86)kW = 21% wasted energy (battery loss) at 320kW

At the point that you are producing 86kW of heat, presumably you need to start spending some more energy to cool the batteries as well...
My conclusion is that heavy acceleration uses more energy. I think it's also likely that heavy regen is less efficient than light regen. This could explain another post I read that claimed using the "reduced regen" mode was giving better mileage.

The analysis of how the internal resistance changes as the battery degrades over time also has repercussions on how much power the cells can deliver as they age.
If we go forward in time to the point when the batteries have an internal resistance of 0.2ohms (5 years? 8 years?), then that roughly doubles the waste energy (P = I^2 * Ri and Ri doubled)

So if the car could still output 320kW of useful energy, now the waste energy would be:

172kW / (320+172kW) == 35% wasted energy (battery loss)

But note it's still fairly efficient at low power

2.5 / (40 + 2.5) == 6%

Heavy power consumption comes at a price.
As the batteries age, this cost increases.
It seems likely that top acceleration will degrade as the batteries age.

Getting Amped Again | April 9, 2013

Very interesting!

The only thing I'd like to add is that for me, driving about 1000 miles per month, the difference in cost for using 300 Wh/mi and 360 Wh/mi is about $7 (using a spreadsheet posted by a forum member which includes Vampire losses). After 1900 miles I'm averaging 330 Wh/mi, and I like to "get it" at least a few times every day, and I always accelerate faster than my previous ICE.

So it may take more energy to accelerate faster, but it's piddly more (at least at my $0.10/kW-hr), so my recommendation is "Enjoy what ya got!"

Great post!

Brian H | April 9, 2013

Oil kerrect, AFAIK.

Leofingal | April 9, 2013

Good post, and it makes sense based on what I see (hard accel hurts efficiency). But on the other hand, hard accel sells Model S'! I think my neighbor is buying one after I let him drive it this past weekend!

c.bussert67 | April 10, 2013

Very valid point andy! I would've never gone thru all the trouble you just did to explain that detail. But yes, the closer you come to transferring no power, the more you lower parasitic losses. So my little game is to keep the needle as close to zero as possible. My only little gripe with the cruise control is that it seems a little aggressive in maintaining a speed. It could be smoother with lower spikes in power to keep the speed.

lolachampcar | April 10, 2013

someone needs to include those expensive tires in their cost analysis (when using the right foot)

smorgasbord | April 14, 2013

Are you saying:
1) that the battery puts out 320kW, but with 86kW of losses only 234kW makes it to the motor?
2) that the battery puts out 406kW, but with 86kW of losses only 320kW makes it to the motor?

Remember that Tesla specs the HP of the Perf model at 416HP or 310kW. To which do you think that refers?

Mark K | April 15, 2013

smorgasbord - it's the latter, #2.

Motor power figures are quoted net to the motor.

Because of internal battery impedance losses, the battery must use up more of its charge to actually yield the required kilowatts to the motor.

andycrews - your assumptions are good. There are more dimensions to the equation though, and there are nonlinearities. It is not only SOC and degradation that affects impedance, the instantaneous current demand has an effect as well. The battery modeled as an ideal voltage source in series with a linear resistance is a useful simplification, but it is not accurate under acute demand. The battery is a nonlinear animal.

In the future, both fast acceleration and regen will get more efficient with hybrids of supercaps and batteries. The supercap impedance is nearly zero, which makes it the ideal buffer for high rate draw or charge.

That combo is a ways off for cars, but supercaps are already in use on trains for regen.

10kWh of supercap, with 100+kWh of battery will be a sweet combo.

Brian H | April 16, 2013

Slip a few kWh in to replace the dummy cells in a 60S, and you'd have a hot beast. They're lightweight, but bulky (volume), so you might get 5kWh in there.

Winnie796 | April 16, 2013

I always thought it was the motor inverters that determine what power is taken out of the batteries and not the batteries "putting out power".
So I don't see how that will affect acceleration in the future except to say that as the battery degrades you won't get to accelerate as often on one charge but the acceleration itself will still be as intense.

As for efficiency I would estimate 4% losses through the inverter, 3% losses through the motor and other heat losses (although variable) <3%. So, overall about 10% losses.
That is bloody marvelous compared to ICE petrol heat losses which are about >80% and ICE diesel heat losses which are >50%. Basically, nearly all the energy you pour into your ICE car goes out of the exhaust as heat.

Regen is not included, which will only increase the efficiency of the electric car.

Pasadena-S | April 16, 2013

Something doesn't seem right here.

All these cells are not in parallel, but in some combination of parallel and series, right? If the cells are 4v and the inverter takes in 400v (guessing), then there are 100 cells in series. That means that each series has 100 times the current shown in your analysis. But if that were the case, your I^2 calculation would blow up by 10,000.

We're missing something.

larryh | April 16, 2013

Series and parallel are irrelevant to the calculation. Each cell runs at about 4 Volts so, given a specific power output, you can calculate the current.

P= V*A

ghillair | April 16, 2013

Is it possible that under heavy acceleration some of the extra energy goes to stretching face muscle into that famous GRIN?

Mark K | April 16, 2013

Pasadena-S - yes, something is amiss in your statement.

TM has not published its cell array architecture, but if indeed there are clusters of 100 cells in series, it would pencil out as follows:

At 1C, each cell can provide about 3 amps at 3.6V (loaded conditions).

If you stack 100 in series, you get the sum of their voltages at that same current (see Kirchoff's law).

So at 1C discharge, each cluster would provide about 360V at 3A = 1.08 kW

With 80 clusters in parallel comprising 8000 cells, you'd get about 80kW output, net, after internal impedance losses. (That is manifest in the 3.6V loaded number instead of the 4.2V open circuit voltage).

At 4C discharge, you'd get 320kW.

The fact that 4C discharge is already demonstrated is, in turn, why many of us believe we are in for a pleasant surprise on improvements to SuperCharger times.

andycrews | April 17, 2013

This statement
It seems likely that top acceleration will degrade as the batteries age.
was not at all clear, sorry about that. It is based on my assumption that the total power output of the batteries will remain constant (or lessen) over time. Thus if more of the power is wasted as heat as the batteries degrade, there is less useful power available. I'm not sure if that's true, but it seems reasonable to me.

DavidN | April 18, 2013

Battery issues aside, there's another reason that heavy acceleration appears to be less efficient: you're not really comparing apples to apples.

You can't just compare the energy it takes to go from 0-60 in 6 seconds to the energy it takes to go 0-60 in 12 seconds. Obviously, the quicker car is going to be farther down the road at any given point in time, and will have a higher average speed at that point. Obviously, to go farther faster takes a lot more energy, regardless of the rate of acceleration.

To make a fair comparison, you'd have to pick acceleration profiles that, at some point, put the cars at the same place and the same time, going the same speed. Will the faster-accelerating car have used more energy? Interesting question.

If you graph speed vs. time, the area under the curve is distance traveled. You'd need to pick two acceleration profiles that, at some point in time, have equal area under the curves.

Take an example: Car A accelerates at a steady rate of 10 mph/sec. It hits 60 mph in six seconds, then holds a steady 60 mph.

Car B accelerates at 5 mph/sec, and takes 12 seconds to hit 60 mph. At this point, obviously, it is well behind Car A. But if car B continues to accelerate at 5 mph/sec to 90 mph, it will begin to catch up. If car B then begins decelerating at 5 mph/second it will slow to 60 mph just as it catches up to car A, which is also still going 60. Thus we have a fair apples-to-apples comparison, in which rate of acceleration is the only variable.

So which car will have used less energy at this point? Beats me,

With rockets in space, which are totally friction-free, the situation is a lot simpler. Final speed (and therefore energy) achieved is directly proportional to what rocket scientists call the impulse, which is just the thrust times the time the thrust is applied. A million pounds of thrust for one second, or one pound of thrust for a million seconds--the final speed and energy are the same.

With the rolling resistance and aerodynamic drag of cars, it's a lot more complicated.

EVTripPlanner | April 18, 2013

OK, let me try this:

(1) resistive losses are proportional to the square of the current [ri^2]
(2) torque in electrical motors is proportional to current
(3) acceleration is proportional to torque (and thus current)
(4) distance covered to achieve a certain speed is proportional to acceleration][.5*a*t^2)

From (1)-(3), losses are proportional to the square of acceleration. So, when you double acceleration, you have 4x the losses but from (4) you've covered only 2x the distance, so you have 2x the losses per distance, thus reducing your efficiency.

So, for example, if you have 10% electrical losses (90% efficiency) when doing 0-60 in 8.4 seconds, you will have 20% electrical losses (80% efficiency) when doing 0-60 in 4.2 seconds.

The higher dynamic loads (and resulting increased friction, etc) probably don't count for much.

So, I for one, will NOT be holding back!!!

I don't have this in the spreadsheets yet, but you can see other useful calculations at