# [EVDL] Real HP in an EV ?

659 7
I ran some number to find what the real hp would be in a typical EV,
just to get an idea of the kind of performance to expect from a
flooded conversion.

For this I used some basic equations from zeva.com.au (great website
by the way!)

So I tallied up the internal resistance (or ohms) in all the batteries
plus the wiring losses. I left out the losses in the contactors and
whatnot since I'm assuming it would be minuscule and I don't know
where to find that info.

So, as I'm sure many of you know, to find power loss you multiply
resistance by the square of current.

So I took the internal resistance, or ohms, of each battery and
multiplied it by how many batteries there would be. In this case, I
took 12 12 volt J-150 batteries, which have an internal resistance of
0.0036, as per evconvert.com. I assumed a use of about 20 feet of
copper, which came up to 0.00154 ohms. I know that basically the max.
amount of current you can safely pull out of flooded lead-acid is 500
amps, so I used that as my current. Here's the math!

12 * 0.0036 + 0.00154 = 0.04474 ohms.

P=500*500*0.044174

So the power loss is around 11 kilowatts. The amount of power
available, without resistance, is 72 kilowatts (144votls*500amps) So
the real amount of power is 61 kilowatts. Then you have to multiply
that by the controller and motor. So it comes down to: 61*.95*.90=

I have come to the conclusion that you can either have cheap flooded
batteries with so-so performance, pricey AGMs with great performance,
or the lithium based batteries that cost a fortune but in the long run
are the better solution.

So anyway, were my numbers right?

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I'm a newbie, so I'm only guessing, but I think pulling 500 amps from 12V
floodies will trash the batteries pretty quickly. I think you can only do
that from 6V floodies, and then not often. Might want to model a pack of 6V
floodies.

Now if we could model a dozen 12V floodies in parallel with a dozen 12V
AGM's...

Marty

----- Original Message -----
From: "Joseph T. " <[email protected]>
To: <[email protected]>
Sent: Monday, August 13, 2007 1:35 AM
Subject: [EVDL] Real HP in an EV ?

>I ran some number to find what the real hp would be in a typical EV,
> just to get an idea of the kind of performance to expect from a
> flooded conversion.
>
> For this I used some basic equations from zeva.com.au (great website
> by the way!)
>
> So I tallied up the internal resistance (or ohms) in all the batteries
> plus the wiring losses. I left out the losses in the contactors and
> whatnot since I'm assuming it would be minuscule and I don't know
> where to find that info.
>
> So, as I'm sure many of you know, to find power loss you multiply
> resistance by the square of current.
>
> So I took the internal resistance, or ohms, of each battery and
> multiplied it by how many batteries there would be. In this case, I
> took 12 12 volt J-150 batteries, which have an internal resistance of
> 0.0036, as per evconvert.com. I assumed a use of about 20 feet of
> copper, which came up to 0.00154 ohms. I know that basically the max.
> amount of current you can safely pull out of flooded lead-acid is 500
> amps, so I used that as my current. Here's the math!
>
> 12 * 0.0036 + 0.00154 = 0.04474 ohms.
>
> P=500*500*0.044174
>
> So the power loss is around 11 kilowatts. The amount of power
> available, without resistance, is 72 kilowatts (144votls*500amps) So
> the real amount of power is 61 kilowatts. Then you have to multiply
> that by the controller and motor. So it comes down to: 61*.95*.90=
>
> I have come to the conclusion that you can either have cheap flooded
> batteries with so-so performance, pricey AGMs with great performance,
> or the lithium based batteries that cost a fortune but in the long run
> are the better solution.
>
> So anyway, were my numbers right?
>
> _______________________________________________
> For subscription options, see
> http://lists.sjsu.edu/mailman/listinfo/ev
>
>

_______________________________________________
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Marty Hewes <[email protected]> wrote:
>> I think pulling 500 amps from 12V floodies will trash the batteries
>> pretty quickly. I think you can only do that from 6V floodies, and
>> then not often. Might want to model a pack of 6V floodies.

Joseph T. wrote:
> I have the impression that 12 volt flooded batteries can put out 500
> amps.

I agree with Marty. You won't do this for long!

You can draw 500 amps or so for a matter of seconds, to start an ICE or
accelerate an EV. But if you try to do it for more than 10 seconds or
so, you will find that their life will be very short.

Even the big flooded golf cart batteries can't supply a sustained
current of 500 amps for more than 30 seconds or so.

--
Ring the bells that still can ring
Forget the perfect offering
There is a crack in everything
That's how the light gets in -- Leonard Cohen
--
Lee A. Hart, 814 8th Ave N, Sartell MN 56377, leeahart_at_earthlink.net

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You missed a couple things. At least one minor consideration and one
major one.
Minor, battery resistance spec is for a fully charged battery and goes up
aas it discharges.

Major point, you forgot to account for battery sag. At 500 amps that 12v
battery will sag by at least 2 volts

> I ran some number to find what the real hp would be in a typical EV,
> just to get an idea of the kind of performance to expect from a
> flooded conversion.
>
> For this I used some basic equations from zeva.com.au (great website
> by the way!)
>
> So I tallied up the internal resistance (or ohms) in all the batteries
> plus the wiring losses. I left out the losses in the contactors and
> whatnot since I'm assuming it would be minuscule and I don't know
> where to find that info.
>
> So, as I'm sure many of you know, to find power loss you multiply
> resistance by the square of current.
>
> So I took the internal resistance, or ohms, of each battery and
> multiplied it by how many batteries there would be. In this case, I
> took 12 12 volt J-150 batteries, which have an internal resistance of
> 0.0036, as per evconvert.com. I assumed a use of about 20 feet of
> copper, which came up to 0.00154 ohms. I know that basically the max.
> amount of current you can safely pull out of flooded lead-acid is 500
> amps, so I used that as my current. Here's the math!
>
> 12 * 0.0036 + 0.00154 = 0.04474 ohms.
>
> P=500*500*0.044174
>
> So the power loss is around 11 kilowatts. The amount of power
> available, without resistance, is 72 kilowatts (144votls*500amps) So
> the real amount of power is 61 kilowatts. Then you have to multiply
> that by the controller and motor. So it comes down to: 61*.95*.90=
>
> I have come to the conclusion that you can either have cheap flooded
> batteries with so-so performance, pricey AGMs with great performance,
> or the lithium based batteries that cost a fortune but in the long run
> are the better solution.
>
> So anyway, were my numbers right?
>
> _______________________________________________
> For subscription options, see
> http://lists.sjsu.edu/mailman/listinfo/ev
>

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Isn't the voltage sag another way of stating the loss across the internal
resistance, which was a factor he took into effect?

----- Original Message -----
From: "Peter VanDerWal" <[email protected]>
To: "Electric Vehicle Discussion List" <[email protected]>
Sent: Tuesday, August 14, 2007 9:35 AM
Subject: Re: [EVDL] Real HP in an EV ?

> You missed a couple things. At least one minor consideration and one
> major one.
> Minor, battery resistance spec is for a fully charged battery and goes up
> aas it discharges.
>
> Major point, you forgot to account for battery sag. At 500 amps that 12v
> battery will sag by at least 2 volts
>
>
>> I ran some number to find what the real hp would be in a typical EV,
>> just to get an idea of the kind of performance to expect from a
>> flooded conversion.
>>
>> For this I used some basic equations from zeva.com.au (great website
>> by the way!)
>>
>> So I tallied up the internal resistance (or ohms) in all the batteries
>> plus the wiring losses. I left out the losses in the contactors and
>> whatnot since I'm assuming it would be minuscule and I don't know
>> where to find that info.
>>
>> So, as I'm sure many of you know, to find power loss you multiply
>> resistance by the square of current.
>>
>> So I took the internal resistance, or ohms, of each battery and
>> multiplied it by how many batteries there would be. In this case, I
>> took 12 12 volt J-150 batteries, which have an internal resistance of
>> 0.0036, as per evconvert.com. I assumed a use of about 20 feet of
>> copper, which came up to 0.00154 ohms. I know that basically the max.
>> amount of current you can safely pull out of flooded lead-acid is 500
>> amps, so I used that as my current. Here's the math!
>>
>> 12 * 0.0036 + 0.00154 = 0.04474 ohms.
>>
>> P=500*500*0.044174
>>
>> So the power loss is around 11 kilowatts. The amount of power
>> available, without resistance, is 72 kilowatts (144votls*500amps) So
>> the real amount of power is 61 kilowatts. Then you have to multiply
>> that by the controller and motor. So it comes down to: 61*.95*.90=
>>
>> I have come to the conclusion that you can either have cheap flooded
>> batteries with so-so performance, pricey AGMs with great performance,
>> or the lithium based batteries that cost a fortune but in the long run
>> are the better solution.
>>
>> So anyway, were my numbers right?
>>
>> _______________________________________________
>> For subscription options, see
>> http://lists.sjsu.edu/mailman/listinfo/ev
>>
>
>
> --
> If you send email to me, or the EVDL, that has > 4 lines of legalistic
> junk at the end; then you are specifically authorizing me to do whatever I
> wish with the message. By posting the message you agree that your long
> legalistic signature is void.
>
> _______________________________________________
> For subscription options, see
> http://lists.sjsu.edu/mailman/listinfo/ev
>
>

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