Does 10kw @48v equal 20kw @ 96v?

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gregted
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Does 10kw @48v equal 20kw @ 96v?

Post by gregted » Tue, 31 Jul 2012, 15:31

Quick question.

I have a 10kw 48 volt ex fork lift motor that I want to use in my future ev.

I am looking at using a Mightyboy or Capri for the donor roller.

The stock motor in the mighty boy has the following power according to wiki.

the single-carb F5A motor delivered 28 PS (21 kW) JIS at 6,000 rpm.

From what I have learned about electric motors compared to ice is that electric delivers max kw at zero rpm up to the max rpm and the controller is basicly a pwm that delivers pulses of power but not full power for a sustained period so it is hard to compare.

What I am wondering is how does the kw of an electric compare in real time on the road to an ice and how is the kw affected when you increase the voltage say from the standard 48 volts to 96 volts.

I understand that increasing the voltage increases speed and increasing the amount of batteries increases the range but how does that effect power at the wheels eg... kw.
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Does 10kw @48v equal 20kw @ 96v?

Post by antiscab » Tue, 31 Jul 2012, 16:15

doubling the voltage means the same torque at twice the rpm for twice the power

in reality you can push it a bit harder at higher rpm for more than double the power as the shaft mounted fan is moving more cooling air

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Does 10kw @48v equal 20kw @ 96v?

Post by woody » Tue, 31 Jul 2012, 17:08

I'm not really a DC guy, but I thought at the same speed:
double voltage => double current => quadruple power (with quadruple losses => heating).
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Does 10kw @48v equal 20kw @ 96v?

Post by gregted » Tue, 31 Jul 2012, 18:58

antiscab wrote: doubling the voltage means the same torque at twice the rpm for twice the power

in reality you can push it a bit harder at higher rpm for more than double the power as the shaft mounted fan is moving more cooling air

Matt


So for example, with a supply voltage of 48 volts, a 10 kw motor at maximum 3000 rpm may give me a top speed of 40kph but a supply voltage of 96 volts for the same 10kw motor at 6000 rpm will give me a top speed of 80kph for the same duration.

Ok. That makes sense. So what I am trying to find out is how will my 10kw, 48 volt motor compare to the original motor of twice the torque.

I assume it will have much more at the bottom end as others have said they only drive in 3rd or 4th gear which with the original 560cc motor in the Mighty boy, would have been impossible.

My other direction for a roller may be a Capri with a clapped motor but going on the specs of the 1.6L motor of 61kw, I may be up for a much bigger motor to maintain anywhere near the original power or at least something better than a snails pace.
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Does 10kw @48v equal 20kw @ 96v?

Post by coulomb » Wed, 01 Aug 2012, 00:19

woody wrote: I'm not really a DC guy, but I thought at the same speed:
double voltage => double current => quadruple power (with quadruple losses => heating).

Err, if you double the voltage and keep the speed the same you'll get about 10x or 100x the current. The current is basically the applied voltage less the back EMF, divided by the armature resistance, which is quite small. [ Edit: and of course, the back EMF is proportional to speed, so same speed implies the same back EMF. ] More or less the same thing happens with AC, but it's the quadrature voltage (hope I got that right; one of the two components) minus the back emf, divided by the reactance (which has some leakage inductance as well as stator resistance). You can adjust the V/Hz of an AC motor, but that only applies to the direct (again, hope I got that right) part of the applied voltage; that effectively "powers the field".

So back to a DC motor: with double the voltage available, you'll be able to get to double the speed with the same torque, so double the power, as has been stated above. If you can't increase the motor speed because of other limitations, then the extra voltage does you no good, as far as power output goes; the controller will never find a situation where it is needed.
Last edited by coulomb on Tue, 31 Jul 2012, 14:21, edited 1 time in total.
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Does 10kw @48v equal 20kw @ 96v?

Post by antiscab » Wed, 01 Aug 2012, 00:49

coulomb wrote:
woody wrote: I'm not really a DC guy, but I thought at the same speed:
double voltage => double current => quadruple power (with quadruple losses => heating).

if you double the voltage and keep the speed the same you'll get about 10x or 100x the current. The current is basically the applied voltage less the back EMF, divided by the armature resistance, which is quite small.


with series wound motors, increasing the current also increases the field strength, which increases the back emf

so doubling the voltage at the same speed quadruples the power if running below ssaturation, and somewhat more than that above saturation (less bemf increase means lots more current)

but yes, I was considering the case where you are running with a motor controller that is limiting current

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Does 10kw @48v equal 20kw @ 96v?

Post by coulomb » Wed, 01 Aug 2012, 03:41

antiscab wrote: with series wound motors, increasing the current also increases the field strength, which increases the back emf

Arrgh! I forgot about that. I guess I was thinking permanent magnet DC or something.

Sorry to have doubted you, Woody. I guess I'm not much of a DC motor person, either Image
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Does 10kw @48v equal 20kw @ 96v?

Post by woody » Wed, 01 Aug 2012, 04:01

No probs, I wasn't sure, I was just repeating something from my early days on this or the zeva forum. Probably Richo :-)
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