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=
> 52kw or about 65-70hp.
>
> 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
>
>
_______________________________________________
For subscription options, see
http://lists.sjsu.edu/mailman/listinfo/ev
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=
> 52kw or about 65-70hp.
>
> 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
>
>
_______________________________________________
For subscription options, see
http://lists.sjsu.edu/mailman/listinfo/ev