# Theoretical battery pack questions.



## bchaffin72 (Jan 4, 2017)

I'm just wondering if I have this right. I'm just designing it on paper, not for any specific use. It's a learning exercise.

Let's say I want a 144 volt pack with 200 Ah. One way would be 45 pairs of CALB LiFePO4 100Ah 3.2v batteries. The pairs would be in parallel, giving each pair 200Ah and still 3.2 volts. 45 parallel pairs connected in series would give the 144 volts, 200 Ah.

The pack would weigh roughly 675 pounds. If the CALB batteries have a continuous discharge of 3C, that's 600 amps. The power would be 144 volts X 600 amps = 86,400 watts or 86.4 kW.

Am I correct, or off somewhere? Thanks in advance. 

I know this is only one way to do it, and maybe not even the best.


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## dcb (Dec 5, 2009)

Yup. That is how you do it. It is a bit optimistic because of various losses, but that is the basic idea and how batteries are marketed.

And fwiw, it would be a 144v * 200ah = 28.8kwh pack (energy storage)

and optimistically you could make that 86kw for 20 minutes straight at 3C (60 minutes/3)


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## bchaffin72 (Jan 4, 2017)

Thanks.  I figured there might be some small losses not accounted for. For now, since I don't have the means to build yet, I'm just learning to work things out on paper. It doesn't cost much and knowing the math in advance make things a lot easier if I do get a chance later on to build.


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## dcb (Dec 5, 2009)

the difference between marketing stats and real world usages are not small, just fyi, lots of variables there.

The two big ones are 1. temperature, and 2. most EVs do not use %100 depth of charge or discharge as that shortens the lifespan the harder you push it to the extremes.


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## bchaffin72 (Jan 4, 2017)

I know temperature affects performance. And, in a real world use, I'd have to account for the battery type's depth of discharge and Peukert effect, in order to maintain the proper power and avoid wrecking the pack.


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## nucleus (May 18, 2012)

CALB 100Ah cells usually have about 117Ah of capacity, so even if you only use 80%, that 93Ah usable energy leaving 20% in reserve.


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## bchaffin72 (Jan 4, 2017)

So, basically, if I wanted to use my pack as is, I'd have to account for where 80% discharge is, thus not getting the full potential. 

So using 93 Ah, the parallel cells would give 186 Ah. At 144v and 186Ah, I'd have a usable potential of 26.78 kwh and 80.35 kW.

And if I wanted my original numbers to be the _usable_ numbers, I'd have to overbuild the pack to account for DoD.


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## Sunking (Aug 10, 2009)

In theory you are correct, but it is only theory and not real. You are not taking into account Peukert Effect, battery internal resistance, and wiring losses. 

At 3C or 600 amps the battery internal resistance losses drops the voltage from 144 volts to 112 volts. Which means you are burning off (144 - 112) x 600 amps = 19,200 watts 19.2 Kw as waste heat on your batteries. Then Mr Peukert will take his cut of about 5% more capacity away at 3C. Then there is the wiring losses. At 3C you are discharge in in about 12 to 14 minutes.

At cruise you only want to be at around C/2 to C/3 discharge rate. Commercial EV's like Tesla use roughly C/6 discharge rate which gives them about 5 hours run time.


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## bchaffin72 (Jan 4, 2017)

Thanks.  I know about Mr. Peukert, but didn't account for him yet. I generally see it given as .98 for Lithium batteries, vs .55 for lead acid. Does it increase as rate of discharge increases? And I didn't account for internal resistance as I don't know how to yet, so guilty as charged.............

Going back to my ideal numbers for a moment, at 1C = 200 amps, at cruise I wouldn't want to be drawing more than 66.6(C/3) to 100(C/2) amps?


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## Sunking (Aug 10, 2009)

bchaffin72 said:


> Thanks.  I know about Mr. Peukert, but didn't account for it yet. I generally see it given as .98 for Lithium batteries, vs .55 for lead acid. Does it increase as rate of discharge increases?


Absolutely. Peukert Law change in capacity of rechargeable batteries at different rates of discharge. As the rate increases, the battery's available capacity decreases. 

LFP capacity is rated at 1C, and Pb depends on the battery. Consumer grade Pb batteries are rated at C/20, and Industrial Traction Batteries like those for Fork Lifts can be rated at 4 to 8 hour rate. What every way they want to spin the numbers. Lithium are not as bad as Pb but still affected. 

Internal Resistance is easy peasy. CALB list the 100 AH cells at .0012 Ohms at room temps, but goes up significantly as temps drop to almost unusable at freezing. Voltage = Current x Amps. Power = Current x Current x Resistance. So at 600 amps you lose 600 x .0012 = .72 volts and 600 x 600 x .0012 = 432 watts per cell. If you have a 144 volt or 45S battery is 45 x .72 volts = 32.4 volts, and 45 x 432 watts = 19,400 watts. So with a 6C on a 45S battery the batteries are being drained at 86,400 watts, but 19,400 is generating heat and sending 67,000 watts out or loosing 22% power.

The above losses are just at the battery terminals. There are more losses in the wiring, motor controller, and motor.


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## bchaffin72 (Jan 4, 2017)

Is that part of why I see battery packs at a higher voltage than the motor, to account for losses. One common spec I keep seeing for AC systems is a 300 volt battery pack with a 240 volt motor. And the Tesla pack, give or take depending on the model, is around 400 volts.


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## dcb (Dec 5, 2009)

a 240v motor typically means 240vrms, as a phase to phase measurement, it has a peak voltage of sqrt(2) * 240, so it needs a battery of 340v or so to spin it properly.

And you are getting into AC motor ratings and stuff now, and it is kinda complicated at first, but basically a typical 240v 4 pole motor can only be pushed to ~1800 rpm at rated load (assuming 60hz) with a 340v battery, but there is a lot more power available with higher rpm (which requires even higher voltage, i.e. 6000 rpm). So I don't know what you mean by a tesla 240v motor. Suffice it to say the motor and windings and battery and controller were carefully matched to get the performance.


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## bchaffin72 (Jan 4, 2017)

I suppose what I'm getting at(and it's more than likely I'm misunderstanding a few things at the moment)is that it seems that, AC or DC, if your battery pack and motor have the same voltage(say 144), by the time you account for losses, especially under increasing draws, the motor will never see it's full voltage potential, since the battery is losing power along the way. And the fact that the battery doesn't just power the motor, but everything else as well.


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## dcb (Dec 5, 2009)

yup, gonna be some voltage drop in the controller, cabling, etc, plus a current dependent voltage drop (sag), plus batteries will lose some voltage as they discharge. AC or DC.


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## bchaffin72 (Jan 4, 2017)

dcb said:


> So I don't know what you mean by a tesla 240v motor. Suffice it to say the motor and windings and battery and controller were carefully matched to get the performance.


Well, let me go back to this and clarify. They were two different examples. Perhaps I should have separated the sentences for clarity.

The first is just an example I've seen used in a few places of an AC system using a 300 volt pack with a 240 volt motor, where the controller takes in 300vdc and converts it to a maximum of up to 240vac for the motor.

The second is my observation from research of Tesla battery packs being(give or take some volts, depending on the size of the pack) around 400 volts. I have no idea what the rated voltage of Tesla's motors are. I was speaking only of the packs.

Didn't mean to be confusing there..............


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## Sunking (Aug 10, 2009)

One thing that might be tripping you up is relating battery voltage to motor voltage. In reality and practice there is no need to have a relationship between the two. I can generate 400 volts AC power with a 12 volt battery, 24, 100, or 1000 volts DC. Although it would be extremely foolish to do any drive train at lower voltages. But it is and can be done. 

DC motors are much easier to control than AC motors. Just about every DC motor controller out there is nothing more than a PWM voltage regulator. PWM has two states, either on or off just like a switch. The controller connects the battery to the motor and disconnects the battery.

AC motors on the other hand are much different and far more complex. They take DC and turn it to 3-phase Variable frequency and Voltage. They can step the voltage up or down. 

Commercial EV's use much higher voltages than DIY EV's because higher voltage is much more efficient and can deliver more more power with less current. Current is what is expensive and causes most of the losses. In fact you will see commercial EV manufacturers using higher voltages than they do now. Porsche is going to compete with Tesla and is releasing a 600 volt EV. Will not be long until you see 1000 volts. 

Think of this a 200 HP motor uses roughly 160 Kwh of power. At 144 volts requires 1100 amps. A 1000 volt system only 160 amps. 

Having said all that life is much easier, less complicated and expensive if the battery voltage is higher than the motor operating voltage.


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## major (Apr 4, 2008)

Sunking said:


> ...
> 
> *AC motors* on the other hand are much different and far more complex. They take DC and turn it to 3-phase Variable frequency and Voltage. They can step the voltage up or down.
> 
> ...


Hi Sun,

I think you mean to say "AC motor controllers". The AC motor itself can be fairly simple. The AC controllers are inverters, DC to AC 3-phase, VVVF. As such they cannot step-up voltage from the battery to the motor unless equipped with a boost converter front end or are of some other topology not commonly seen on EVs. 

Regards,

major


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## bchaffin72 (Jan 4, 2017)

Sunking said:


> One thing that might be tripping you up is relating battery voltage to motor voltage. In reality and practice there is no need to have a relationship between the two. I can generate 400 volts AC power with a 12 volt battery, 24, 100, or 1000 volts DC. Although it would be extremely foolish to do any drive train at lower voltages. But it is and can be done.
> 
> DC motors are much easier to control than AC motors. Just about every DC motor controller out there is nothing more than a PWM voltage regulator. PWM has two states, either on or off just like a switch. The controller connects the battery to the motor and disconnects the battery.
> 
> ...


Maybe I was looking at it from the wrong angle, but that's what I was getting at, whether or not it's advantageous to have a higher pack voltage than what the demands are. So it seems the answer is yes.

So say we're driving two cars cars at a certain state of draw, say 55 mph down the highway, everything else that needs power powered up and we're listening to the radio. X amount of power needed.

One has a 300 volt pack and one has a 450 volt pack. Is it fair to say the higher voltage pack will have an easier time supplying the required current and will do it more efficiently and for a longer period of time, with less waste and heat generated?

So, although both packs are capable of doing the job, the 300 volt pack has to work harder to do so and is less efficient.


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## dcb (Dec 5, 2009)

bchaffin72 said:


> So, although both packs are capable of doing the job, the 300 volt pack has to work harder to do so and is less efficient.


kinda. Yes pack voltage should be as high or higher than the expected motor voltage (a function of the KV rating of the motor and anticipated RPM, which is a function of how it is wound/connected). Stick with that.

but if the motor/controller were suited for 150V instead and twice the current, the same battery pack reconfigured with 2 in parallel at 150v would see the same stress, and the motor would make the same power (assuming you used thicker cables).

power is volts times amps, electrically. (rpm times torque mechanically)


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## Sunking (Aug 10, 2009)

major said:


> Hi Sun,
> 
> I think you mean to say "AC motor controllers".


Yep my bad.


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## Sunking (Aug 10, 2009)

dcb said:


> power is volts times amps, electrically. (rpm times torque mechanically)


 Mechanical HP = T x RPM / 5252. You left out the constant 5252.


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## Sunking (Aug 10, 2009)

bchaffin72 said:


> So, although both packs are capable of doing the job, the 300 volt pack has to work harder to do so and is less efficient.


Pretty much. 

Technically 1 HP mechanical = 745 Watts. At least on paper but not in practice. Once you take the motor efficiency can be as low as 800 watts up to 1000 watts. AC motors are the most efficient. 

So jus tplay with a couple of formulas and you will see what is going on. In school you run the equations a few hundred times and you finally get it and understand Ohm's Law. 

Current = Power / Voltage. 

So say you have a 100 HP motor with an eff = 85%. to find the Power = 745 / .85 x 100 = 87,647 watts. Make it easy and say 88 Kw. So how much current are we talking about?


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## dcb (Dec 5, 2009)

Sunking said:


> Mechanical HP = T x RPM / 5252. You left out the constant 5252.


??? I said power, not horsepower. power is a function of torque and rpm.


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## Sunking (Aug 10, 2009)

dcb said:


> ??? I said power, not horsepower. power is a function of torque and rpm.


Huh? You lost me. Power does not equal T x RPM.

100/ft-pounds x 5252 RPM / 5252 = 100 HP mechanical. To convert that to power or watts requires you to know the motor efficiency, say 80%. So 100 HP x 745 watts / .8 = 93.125 Kw.

100/ft-lbs x 5252 RPM = 525200 what? It is not electrical power or mechanical HP.


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## dcb (Dec 5, 2009)

horsepower is a specific unit of mechanical power, requiring specific units of input and a constant scaling factor.
i.e.
HP=Torque (ft/lbs) x Speed (RPM) / 5252
HP=Torque (lb.in) x Speed (RPM) / 63,025

Or it can be expressed in watts (or kilowatts)
i.e.
KW=Torque (N.m) x Speed (RPM) / 9.5488

but in all practical cases, it is a function of torque multiplied by rpm (for rotating machines). Pretty much like electrical watts are simply amps times volts.
edit: and indeed, torque is generally proportional to motor amps, and speed to motor volts.

Watts is more convenient of course since if the motor was %100 efficient then power in = power out.


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## bchaffin72 (Jan 4, 2017)

Sunking said:


> Pretty much.
> 
> Technically 1 HP mechanical = 745 Watts. At least on paper but not in practice. Once you take the motor efficiency can be as low as 800 watts up to 1000 watts. AC motors are the most efficient.
> 
> ...



Well, I looked at specs for a DC motor, the D&D ES-15-6, a 48-72 volt series wound.

It's rated 9 continuous horsepower at 72 volts.

So, assuming 85% efficiency:

745 / .85 x 9 = 7,888 watts

So current draw is 7,888 watts / 72 volts aka 109 amps?

Its peak hp is 30, which would take 365 amps at 72 volts


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## dcb (Dec 5, 2009)

getting farther from battery here, but necessary eventually 

Always google for a graph for a given motor, gives a much clearer picture. Found this on elmoto (click to expand). Key here is motors are generators. The faster it spins the more back emf it makes, till it approaches pack voltage and you cannot push as much current through it anymore. Note how linear the torque/amp relationship is. The controller will buck the battery voltage and current into a lower voltage and higher current (till it hits %100 duty cycle, trying to deliver the requested torque at higher rpms).


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## bchaffin72 (Jan 4, 2017)

That makes sense. So the controller will draw whatever power it needs from the battery and then do whatever is necessary with that power(and within its limits)to meet the varying demands placed on the motor.


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