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
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
