Re: what is the max bus voltage, 300V?

[Lee Hart <leeahart@earthlink.net>]

Galen Kehler wrote:
> 
> * LP8.2: HTML/Attachments detected, removed from message  *

You need to set your email software for plain text, not HTML, Microsoft
Word or other formats fancy formats. Otherwise, all most people see is
the above message.

> Sorry, but now I'm confused. I thought the whole point of going with
> a higher voltage system, was that the current decreased, and lowered
> the resistive (I^2*R) losses. I thought that, for example, a 25kW
> power output would give, ideally:
>
> 347A @ 72V
> 173A @ 144V
> 71A @ 350V

Yes; that is true. But, let's keep thinking about it.

Let's build a battery pack for these voltages. Suppose we use a 12v 50
amphour battery as our basic building block (roughly an Optima "Yellow
Top"). For each pack, let's say we want 7.2kwh of total energy storage.
Each battery is 12v x 50ah = 600wh, so we need 12 batteries.

For 72v:

We'd wire the 12 batteries in two series strings of six (12v x 6 = 72v),
and then wire these two 72v strings in parallel. The 12 batteries will
behave like a 72v 100amphour pack.

Each battery has an internal resistance of about 3.5 milliohms. Each
battery also has a jumper wire with a terminal at each end, to conect it
to the adjacent battery; they add about 1.5 milliohms, so our total
resistance per battery is 5 milliohms. Six of these in series is 6 x 5 =
30 milliohms. And we have two such strings in parallel; so our total
pack resistance is 15 milliohms.

If we draw 347 amps from this pack, this resistance makes the voltage
sag 347a x 0.015ohm = 5.2 volts; our 72v pack sags to 66.8 volts. Power
lost in this resistance is 5.2v x 347a = 1800 watts.

Since there are two parallel strings, each battery delivers half the
load current, or 347a / 2 = 173a.

For 144v:

We'd wire the 12 batteries all in series for 12 x 12v = 144v. They will
behave like a 144v 50amphour pack.

Each battery internal resistance is 3.5 milliohms, plus 1.5 milliohm for
the jumper and terminals, so our total resistance per battery is still 5
milliohms. Twelve of these in series is 12 x 5 = 60 milliohms. Note that
doubling pack voltage quadrupled the resistance.

If we draw 173 amps from this pack, this resistance makes the voltage
sag 173a x 0.06ohm = 10.4 volts; our 144v pack sags to 133.6 volts.
Power lost in this resistance is 10.4v x 173a = 1800 watts. Exactly the
same as the 72v case!

All batteries are in series, so they all deliver 173 amps. This is also
exactly the same as the 72v case. The higher voltage is no benefit as
far as the batteries are concerned.

For 350v:

To get more voltage, we have to use a smaller battery, or use more of
the same battery. If we change to a smaller battery so the total pack
weight is the same, we'd find we were right back where we started, with
the same pack losses.

If we use more of the same battery, pack size and weight go up. But we
DO get lower pack losses!

The point is that if you keep the same weight pack, then the voltage you
wire it for has no effect on pack losses, percent voltage sag, Peukert
effects, etc.

It's total pack weight that matters. Increasing the size and weight of
the battery pack lowers losses, cuts percent voltage sag, reduces
Peukert effects, etc.

> The power companies raise the voltage up to multi kilovolt levels to
> transmit power over long distances, so there must be an advantage to
> higher voltage.

There is; cost! Instead of 1 foot of wire between batteries, they have 1
mile of wire between customers. Therefore, their resistive losses in the
connecting wires are very large. Higher voltages cut the wire size they
need, which is a huge cost savings when you're using it by the mile.

A secondary effect is that all the supporting towers and insulators can
be lighter, too. More cost savings.

> I thought that if you doubled the pack voltage, the current would drop
> by half, assuming the same power level and series string configuration.

It does. But the resistance quadruples, putting you back where you
started.

> If the current per battery is the same, won't the high voltage string
> be putting out much more power than the lower voltage string, so how
> can you compare? That's like saying that the Optima YT 12V stores
> about the same energy as the 6V, both having ~50Ah.

You forgot that when you double the voltage, you have to cut the amphour
capacity of each battery in half; otherwise the pack doubles in size,
weight, cost, etc.

> And the resistive losses can vary greatly, depending on how you wire
> the string. If you use 16 gauge wire, the resistance will be higher
> than 00. Although I am no expert, I think that how you connect the
> batteries together makes a difference in the resistive losses, not
> to mention safety, especially when the current rises. If this were
> not the case, we could wire our EV's with 22 gauge, to save cost. :-)

If the current per battery is the same, then the wiring to each battery
should be the same size.

You *can* deliberately wire a car with undersized wire. The automakers
do it all the time! They will tolerate a 20% voltage drop in 3 feet of
wire, just to save cost and weight. All that happens is that the wire
gets hot. If you can handle the heat, and the efficiency loss doesn't
bother you, it's OK. Drag racers trying to eke out every bit of
performance will pick a smaller wire size to save weight; it won't
overheat in the few seconds it takes for a race.
-- 
Lee A. Hart                Ring the bells that still can ring
814 8th Ave. N.            Forget your perfect offering
Sartell, MN 56377 USA      There is a crack in everything
leeahart_at_earthlink.net  That's how the light gets in - Leonard Cohen