THEORY: Wattage not equivalent at higher voltages?

[RageNR]

Let me preface this theory with the notion that I DO NOT state these figures as FACTS. This was something that struck my curiosity, and made me ponder on the actual outcome.
Please weigh in if you have theories of your own, or facts to disprove the stated theory. All discussion is welcome.


We were discussing the almost negligible gains of increasing voltage from 64.8v nominal (18s @ 3.6v/cell) to 72v (20s @ 3.6v/cell), which should result in 288w diff if pulling 40amps (2592w vs. 2880w).
The discussion didn't seem to go anywhere, but I am still very curious what the experienced e-bike tinkerers have to say on the matter.
Is it as simple as the voltage x amps = wattage @ any given voltage? Or are other factors at play, like voltage sag?

I apologize if this has been covered in depth before. I did not see anything in direct reference to this idea. Felt it was worthy of its own topic so it can be discussed without being lost amidst the posts in another thread.

I will self-quote a post I made in the "New Cyclone 3000 w mid-drive kit" thread: https://endless-sphere.com/forums/viewtopic.php?f=28&t=69867&start=1150

I was thinking of higher voltage setups. 72-84v range.
In that case a fully charged pack @84v could see a sag to 72v under a 10amp load. Like in my previous statement, this is based on data I could find on the Panasonic 18650 cells from Tesla.

If you take that same theory and apply it to a charged 14s pack (14s x 4.2v = 58.8 ), the sag at 10amps might reach 50.4v. ___ 58.8/50.4 = 1.16666..., so roughly 16.7% drop.
The voltage drop from a 20s pack, 84v to 72v (84/72 = 1.666....) would actually end up being the same? 16.7%
But lets take both those at amps pulled during load. For arguments sake, lets say the motor is pulling 3000w. At 84v that's 35.71amps. At 58.8v that's 51.02amps.
If we consider cells in 8p, 35.71/8 = 4.46a while 51.02/8 = 6.38a per cell. The 18650s in particular like lower discharge rates. The lower the amp draw, the higher the maintained voltage. 6.38/4.46 = 1.43. So a 43% diff in amp draw.
I am stating the packs at fully charged voltage because that is the data I found available to examine. This is based on a discharge chart for the Panasonic 18650s.
Lets say the cells drop to 3.8v @ 4.46a draw, and 3.65v @ 6.38a. Ok, that considered... 35.71a x 3.8v = 135.7w ______ 58.8a x 3.65v = 214.62w. Diff = ~58% (this is the losses, right?)
3.8v to 3.65v is a 4.1% diff. So we have a diff of 4.1% in voltage, but the diff in wattage is 58%.
That is not 100% scientific, but the best napkin math I could muster. Please do correct me where I am wrong.

All these numbers are based on the charts provided here: https://teslamotorsclub.com/tmc/threads ... ata.45063/
The numbers I am using are based on the first chart at the 0.02 capacity mark. I had to estimate voltages based on what I could see since there are no tests at the amp ratings in the figures above.
I know my figures are not spot on, but I tried to get as close as possible.
Again, I am only stating my theory. Looks to make sense to me, but then again, lots of things do. LoL Please add some REAL math in there.

 

[liveforphysics]

Power is power. Watts are watts.

 

[RageNR]

Power is power. Watts are watts.

Oh hey there. Glad you stopped by. Just want to say I like the stuff you have done.

Yes, that is absolutely true. I guess I am trying to understand if there are more losses involved at lower voltages? Of course, the battery chemistry would also have to be considered.
Would you mind explaining why my math above does not hold any merit? I am really trying to understand this on a deeper technical level.

 

[Drunkskunk]

Watt's law:

P = V * I

Or more fashionably with the E crowd,

W = V * A

 

[liveforphysics]

Power is power. Watts are watts.

Oh hey there. Glad you stopped by. Just want to say I like the stuff you have done.

Yes, that is absolutely true. I guess I am trying to understand if there are more losses involved at lower voltages? Of course, the battery chemistry would also have to be considered.
Would you mind explaining why my math above does not hold any merit? I am really trying to understand this on a deeper technical level.

If you are willing to carry say 20 cells if some size, if you put them in a 20p 1s string, or a 20s 1p string , or a 2p 10s string or 4p 5s string or 5p 4s string, you have exactly the same amount of energy available, power available, and for any amount of power demand you have the exact same percentage of pack voltage sag and identical cell heating per amount of power etc.

Same on the motor side , a 1 turn or 10turn or 5turn or whatever all are identically efficient and make identical heating per unit of torque or power output etc provided the %copper fill in the slot is the same for the various winding combos.

From a controller perspective, it gets tricky to keep it small and light more than a few kW as you go under say 12-24v, and just becomes needles inefficiency and MOSFET RdsOn losses to pick a voltage over ~20S (because 100v MOSFETs are currently the power density switching maxima available with today's silicon die technology).

 

[RageNR]

Ok, I perfectly understand the laws of power. I suppose I may have misrepresented my point by wording the description poorly.

I am trying to get to the bottom of real world usage scenarios. There are many factors that could affect final output vs what the raw numbers would suggest. That is the aim of my thoughts.
And I am not implying that I feel I am right. Just trying to clarify this in my own head.

Lets say the cells drop to 3.8v @ 4.46a draw, and 3.65v @ 6.38a. Ok, that considered... 35.71a x 3.8v = 135.7w ______ 58.8a x 3.65v = 214.62w. Diff = ~58% (this is the losses, right?)
3.8v to 3.65v is a 4.1% diff. So we have a diff of 4.1% in voltage, but the diff in wattage is 58%.

Do these figures hold any relevance at all? Of course the wattage does not change, but the effective output seems to be affected by voltage sag as per the example above.
If this has no merit at all, please explain why or why not.

 

[RageNR]

Sorry, I made the reply above before you responded. Somehow it edited the OP instead. Fixed all that mess now. Anyways...

If you are willing to carry say 20 cells if some size, if you put them in a 20p 1s string, or a 20s 1p string , or a 2p 10s string or 4p 5s string or 5p 4s string, you have exactly the same amount of energy available, power available, and for any amount of power demand you have the exact same percentage of pack voltage sag and identical cell heating per amount of power etc.

So based on what you are saying here, the diff would be because of the number of cells in the pack increased. Thus, the amount of available watts would increase.
That is considering 18s8p vs 20s8p (as per the example posted in the OP). 144 vs 160 cells.
Ok, that in mind... where does this come into play when considering the wattage limit induced by the controller?
I would think that the closer you push the cells to their peak capable amp rating, the less performance you would get from them.

 

[ferret]

The discussion didn't seem to go anywhere, but I am still very curious what the experienced e-bike tinkerers have to say on the matter.
Is it as simple as the voltage x amps = wattage @ any given voltage? Or are other factors at play, like voltage sag?


Lets say the cells drop to 3.8v @ 4.46a draw, and 3.65v @ 6.38a. Ok, that considered... 35.71a x 3.8v = 135.7w ______ 58.8a x 3.65v = 214.62w. Diff = ~58% (this is the losses, right?)
3.8v to 3.65v is a 4.1% diff. So we have a diff of 4.1% in voltage, but the diff in wattage is 58%.

It is as simple as the voltage x amps = wattage @ any given voltage.
However, you must measure the actual voltage and amperage at the same time when riding, not use the battery or controller published ratings, or measurements taken at different times. The sag effects the actual voltage you are measuring.

In the example above the power difference is not connected to losses and and is only marginally effected by the differences in voltage.
It is mostly caused by difference in current. The battery supplies about 65% MORE amps (for example because of climbing a slope), it also supplies about 4% LESS volts as you calculated. If you do the exact calculations, combining the power increase from the added current and the power loss from the reduced voltage you will get the 58% net increase in power.

Avner.

 

[dogman dan]

You did mention voltage sag under load.

So on the road, your power will be the working voltage under load x amps. Your actual power is never from a nominal voltage, or voltage not under load. It's always changing, as loads change, or battery voltage drains down. It's always varying according to whatever the voltage is at that moment, under load.

Motor rpm also changes the load, the motor won't draw as much amps at higher rpm as it will at 0 rpm.

So on the road, both amps and volts vary constantly. This is why watching a cycle analyst as you ride is so edifying.


And then, in some cases, you can get into increasing the power beyond what the motor can handle. In this case, the motor just heats itself very fast, and you don't feel more power at the wheel because it's just melting your motor faster. Different windings behave different as well, leading to long arguments about torque of different windings. At one rpm one motor can feel like more torque, when it's really not. It may be running more efficiently at a given rpm leading to that feeling as you ride it.

 

[d8veh]

I think your assumptions about sag are incorrect. The amount of sag depends on the current you take and the number of cells in parallel.

If you were looking at the difference between 18S6P and 20S6P packs using the same controller runnning at say 10 amps, the current would be the same through each cell, so the sag per cell would be the same in both packs, but you have more volts ( more cells) in the 20S6P pack, so the end result will be that at any point in use, the 20S one will have 20/18 times the voltage of the 14S one, which is 11% more regardless of how much sag there is at any point. Therefore, you will get approximately 11% more power.

Running at a higher voltage increases the band at the low speed end where efficiency is low. This means that at low speed your efficiency will be less using the higher voltage. That then means that at low speed, the power taken from the battery will be the same, but you could have substantially lower output power. The higher the speed, the less you get that effect. The cross-over point would be somewhere around 30% of maximum RPM

 

[liveforphysics]

Learn about "C-rates."

Think in terms of total power demand and power per cell.

 

[Ykick]

Do you have a CA (cycle analyst) in use?

And yes, C rate is very useful knowledge for the revolution...

 

[ccmdr]

Hi RageNR

Not sure if I'm deciphering your post completely, but I think your going on about the losses under load at equiv wattage with a 48v 30Ah vs 72v 20Ah battery pack both produce 1440W, however do a quick play on Justins ebike sim will show the low voltage sags more. Voltage = speed , Amps = Acceleration, however you need a balance both over 25ish mph due to increased resistances, which is probably where your perceiving that there isn't much difference?

Also you have to take into account the IR of the pack, generally more cells in parallel = lower IR at a constant max current draw, but joining them with steel wire or having 3 meter long leads going from the battery to controller or controller to motor will also effect the results.

Motors also have a massive effect at any voltage, a 3T motor at 3000W will be much faster than a 5T 3000W, however the 5T will tackle all but the most vertical hills without overheating.

Power as in W is much like BHP, just because it's a massive number you could be talking fast low torque F1 car or a torque munching slow tractor puller .

 

[gogo]

Motors also have a massive effect at any voltage, a 3T motor at 3000W will be much faster than a 5T 3000W, however the 5T will tackle all but the most vertical hills without overheating.

On hills, they will use equal power at equal speeds and with equal heating, but at different voltages.

 

[RageNR]

So let me state what I understand of all this...

The power output at higher voltages COULD be more noticeable due to the batt packs ability to supply more wattage. (larger pack, more cells)
The equivalent total watts would only be affected by the batteries ability to supply more amps while under load.
Even at lower voltages, the sag would only become relevant if the overall available watts is lower.
When considering the losses from other factors, what would the correct term be to describe it? Effective output?


I am looking at building a 20s7p pack from the cells pulled from a 2015 Tesla Model S pack. Those should be able to achieve 10amp sustained and 15amp peak discharge rates without harm to the cells.
That gives me a 70amp to 105amp peak output capability. I had been considering a 18s6p pack before learning about the E-S member selling these Tesla cells.
My powerplant will be the Cyclone 1800-3000w kit, and stock controller. I do plan on making modifications to the controller eventually. Should be able to get 4000w out of it (I hope).

Target speed is as close to 50+mph as possible. Total riding weight should be close to 210lbs. So higher voltage is required and sounds like will be more efficient.
My impression was that the losses at 18s6p (talking 3x 20Ah 10C Lipo cells) would be greater than simply the wattage increase to 20s7p (18650s).
Obviously that pack will be larger, but has lower amps available if called for. Either pack should be more than enough to handle the 3000w Cyclone. Am I headed in the right direction here?


So if the efficiency is better at higher RPMs, then what about when applying higher voltages than the avg kit calls for? Is the efficiency still in the top 20-30% of RPMs @ the applied voltage?
I do not think it will be too much (others have been running 84v charged), but I would like to maintain a modest efficiency if possible. If not... fine. MO POWAH!!

 

[cycborg]

Just leaving this here while I make the first entry in my "foe" list...
http://www.nbcnews.com/id/24943229/#.V2wwQPkrK9I

 

[Ykick]

Just leaving this here while I make the first entry in my "foe" list...
http://www.nbcnews.com/id/24943229/#.V2wwQPkrK9I

I think I'll be joining you...

 

[nutnspecial]

Lol u guyz! What does a mexican car accident have to do with anything?
Or being a 'foe' for that matter. . . Imo: Just do it, or explain your issue

So if the efficiency is better at higher RPMs, then what about when applying higher voltages than the avg kit calls for?

Sure, and if you can use the extra motor rpms you will end up with more watts in total.
Your controller caps will likely be the limit of voltage.

Is the efficiency still in the top 20-30% of RPMs @ the applied voltage?

Yes, the efficiency curve won't change. 60% thru 90-95% is usually a great target. (over gear if you want efficiency at desired top speed, and will be happy with what you are taking away from the bottom end)

I'm not really sure what the rest of the discussion is about. To get all the power available, run the highest balance of voltage and amperage the motor, gearing, and battery work well with together.

I don't know much about the cyclone, and would follow suit with other fellers routes when choosing a battery and controller to juice it.

 

[d8veh]

Motors also have a massive effect at any voltage, a 3T motor at 3000W will be much faster than a 5T 3000W, however the 5T will tackle all but the most vertical hills without overheating.

On hills, they will use equal power at equal speeds

Correct

and with equal heating,

Not correct - assuming running at full throttle and at less than 50% max rpm.

but at different voltages.

??????? What are you saying?

 

[RageNR]

Just leaving this here while I make the first entry in my "foe" list...
http://www.nbcnews.com/id/24943229/#.V2wwQPkrK9I

I think I'll be joining you...

LMFAO!!! i knew someone was bound to try to determine the origin of that picture eventually. That's why I picked it.... Not that what happened was funny.
I was just kickin around and that random image came across. Didn't know the back story till I looked it up later. Pic just looked like it fit my M.O.

If no one got it yet, it was meant to make fun of those that still have to pedal. Super obvious, but some may not get it at first. Anywho...
Thanks for the laugh fellas. Made my day

 

[RageNR]

Do you have a CA (cycle analyst) in use?

And yes, C rate is very useful knowledge for the revolution...

No sir, no CA yet. Its on my list of things to buy right after the battery pack and charging setup. Almost afraid to buy a CA3 in fear that the CA4 with bedazzling color touch screen is just around the corner.