4WD and regen

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markrmarkr
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4WD and regen

Post by markrmarkr »

7circle wrote: Have checked out the Curtis website.
They have a section for the 1238-75. Maybe the model name variation makes it hard to find.
Manual 1238-75
And Data Sheet 1238-75

But they only refer to volatge range of 72-96V. and 2min RMS Power Rating 62.3 kVA.

Being such a compact controller requiring heatsinking, it would be useful to know what size/rating DC capacitor bus is used.
If the controller was operating close to peak voltage supply extra DC bus capacitance may help protect against battery & cable circuit inductance overshoots.

Under regen the bus voltage may also be driven high and beyond Max causing damage to controller.



I was impressed that thunderstuck site provides a spreadsheet of an analysed system. Power Curve AC-50 (?? 1238-7501)


Is this the info you need?


Thanks for that 7circle. That was very helpful - and yet it raises more questions.

The "System Voltage" is 72 - 96,
The 2 min RMS Current rating is 550 Amps
The 2 min RMS Power rating is 62.3kVA


I'm not quite clear what "System Voltage" is - possibly the voltage seen by the motor rather than the battery voltage.   Is that right? and the KVA multiplied by the PF gives the actual KW of power is this right?   I apologise if I'm a bit slow here but Uni was a long time ago and I'm not familiar with some of the Jargon.

I understand that if you want to push the voltage/power limits of this controller you definitely want to do some heat sinking along with fans or water cooling.

Looking at the full manual for the other 1238 controllers - they seem to have lots of protection built in for things like the maximum regen voltage, so I think this should be manageable.

I've heard that the bateries used for the Thunderstruck test were some old lead acids, hence the sag at max power.   I think Thunderstruck are into performance EV's. They must have a dyno availabe to have done the test.

Thanks again for your help 7circle.
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antiscab
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4WD and regen

Post by antiscab »

markrmakr wrote:I'm not quite clear what "System Voltage" is - possibly the voltage seen by the motor rather than the battery voltage.   Is that right? and the KVA multiplied by the PF gives the actual KW of power is this right?   I apologise if I'm a bit slow here but Uni was a long time ago and I'm not familiar with some of the Jargon.
system voltage is the nominal voltage of the battery pack.
so that controller is intended for between 36 and 48 lead acid cells.

so minimum voltage to run is at least ~63v, and maximum voltage is at least 115v (from other sources we know it is of course higher).

kva*PF = kw, yes
but thats only kw into the motor.
getting the PF of the motor will be difficult at that loading condition, but can be considered to be between 0.8 and 0.5 (rated torque, 3x torque respectively).

66.3kva@550A suggests 113v at the batteries.
they would have to be some very stiff batteries indeed to do that when the max voltage is 120v.

Matt

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markrmarkr
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4WD and regen

Post by markrmarkr »

antiscab wrote:
system voltage is the nominal voltage of the battery pack.
so that controller is intended for between 36 and 48 lead acid cells.

so minimum voltage to run is at least ~63v, and maximum voltage is at least 115v
..
66.3kva@550A suggests 113v at the batteries.
they would have to be some very stiff batteries indeed to do that when the max voltage is 120v.

Mattt

OK That clears up a lot.

36 and 48 cells at 2V per cell giving 72V to 96V. OK so far.

and ~63v is 36 cells which are discharged or 1.75V per cell.
and 115v is 48 cell which are fully charged or 2.39v per cell. I'm still OK.

2 min RMS power rating is 66.3kVA

because it's kVA it's not PF corrected and actual power would be say

0.8 x 66.3k = 53kW (2 min rating) at rated torque
or
0.5 x 66.3K = 33kW (2 min rating) at 3 x rated torque
               (3 x rated torque has a LOT of slip and P ~ T x RPM )

I'm still OK.

BUT

66.3KVA / 550A = 113V       - problem equation

This looks like an apples and oranges equation to me. They quote 96V as the max system voltage and 550A is 2 min RMS current and no mention of PF so presumably the problem equation should be:

550A x 96V = 52.8kW (2 min power at rated torque) - Note there are some assumptions in this but it should be close to the mark. Also it agrees with the value derived using guestimated PF of 0.8 above which gave 53kW.

Also the 550 Amps is RMS current (what the motor sees) so presumably the battery current could be more (losses in the controller).

Notice that max battery voltage of 96V is assumed.

This also gells with the

Thunderstruck data which shows even though battery voltage sagged to 86V at peak power, battery current was 556A!!   Which is more than the controller current 550A (what the motor see).   Also notice the controller current hovered at about the same value for the whole test - implying it was conducted with Wide open throttle, as you'd want.
    the controller current actually hovered at about the same value up to about 4000rpm (edit my correction)

The problem I have is I'd like to use 120V on this thing but I can't find any proof that it'll do it. As I've said previously I know some people are going to try it at 120, but doing so appears to clearly violate Curtis's specs for the controller. It boils down to how over engineered is the controller. Maybe Curtis knows but they haven't said anything. I guess they can't because of Warranty issues.

Thanks for your help.
Last edited by markrmarkr on Fri, 28 May 2010, 06:48, edited 1 time in total.
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Tritium_James
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4WD and regen

Post by Tritium_James »

At full current and full speed, the battery current will be (sqrt(3) / sqrt(2)) times the motor phase rms current. So if they're quoting 550Arms, you can expect to see 673A from the batteries when the motor is at full speed.

antiscab
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Post by antiscab »

some of the motor controllers are rated as if they were DC controllers, for the less "engineer" markets (kelly do this)

so I was expecting the 550A RMS motor current to be "between" the phases so to speak. so 550A is what you measure battery side when the motor is at full speed.

but yes, if they are rated in true phase RMS current, then James is right.

that would mean the 66.3kva rating would be at battery voltage of ~98v.
66.3kva / 550A* sqrt(3) = 69v RMS phase-phase
69v * sqrt(2) = 98.5v at battery

which actually looks more correct (thanks James).

Matt
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4WD and regen

Post by 7circle »

Hi Guys just want to keep it accurate. The 1238-75 spec quotes
2 min RMS power rating is 62.3kVA not 66.3

Anitiscab mstyped kVA as 66.3 not 62.3
markrmarkr wrote: ... 2 min RMS power rating is 66.3kVA
...
Also the 550 Amps is RMS current (what the motor sees) ...
...
Thunderstruck data which shows even though battery voltage sagged to 86V at peak power, battery current was 556A!!   Which is more than the controller current 550A (what the motor see).   

To produce the test results this could have taken 10seconds. So 556A has limited evidence to its 2min capability. But 556 is very close to (edit 550) so the controller battery drain of 47.6kW would be okay for 2min.

The spec is defining the Controller capability and power factor is detiremined by the motor.

From the Thunderstruck spreadsheet (found it at other sites too)
Motor   O/P 3245 RPM, 113 Nm Gives 38kW
Battery I/P 560 Arms, 85 V DC rms Gives 47.6kW
Their readings show @ 3245 RPM
Controller Eff 94.65%
Motor Eff 84.75%

So what would happen with a very "Stiff" battery that keeps the volts at 100VDC? (Edit and pushing the 2min 62.3kVA Controller O/P)
Edit 550 not 560. Image
Last edited by 7circle on Fri, 28 May 2010, 12:08, edited 1 time in total.

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4WD and regen

Post by 7circle »

So calcs expanded -
P = 3 x Vphase x Iphase x pf
Vphase = Vline/√3 and Iphase = Iline
Vline = P/(√3xIline x pf)
      = 63.2k/(√3x550 x 100%)
      = 65.4V AC rms
Vbat = FetIRloss + √2 x Vline
      = 0.01*550 + 92.5V (Assuming 2 x FET Res of 5mOhm)
      = 98 V DC

PowerOut/PowerIn = Eff% (Assuming efficiency is the same as Vbat 85V)
        PowerIn = Vbat*Ibat = PowerOut/94.65%
            Ibat = (62300/94.65%)/98 (Edit <A )
                 = 65821/98
            Ibat = 671 A DC for pf 1.0 (Edit)

So with some p.f of 0.85 for 50kW Induction motor (ie AC-50)
     Vline = 65.4V AC (Edit Same)
     Vbat = 98V DC (edit Same)

PowerOut/PowerIn = Eff% (Assuming efficiency is the same as Vbat 85V)
        PowerIn = Vbat*Ibat = VA.pf/94.65%
            Ibat = 62300*0.85/(94.65% * 98) <B-Edit
                 = 55948.2/98
            Ibat = 571 A DC for pf 0.85

So with Vbat at 98V DC system looks like rating of 62.3kVA is feasable with p.f, eff and Fet-Res assuumptions. P into Motor 53kW * Eff-M = 45kW
Thats with "rock brittle" Battery Image (Edit not so Rock Brittlenow)

So at 100V DC on Bat Side pf 0.85 (and spreasdsheet)
63656.11 VA
54107.7     P(w)
Motor Eff 84.75%     
45856.27 W Motor Power

So Vbat@100V Pm=45.8kW
   Vbat@85V Pm=38kW
But I'm sure the induction motor demand at higher RPM the pf will change.
So I'd like to know the AC50 Motor per phase parameters.
I'm hoping the guys here haven't already covered this in the forum.
Well I'm learning. Image

edit Checking calcs, driv'n me nuts, looks better Image
Last edited by 7circle on Fri, 28 May 2010, 22:39, edited 1 time in total.

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Post by coulomb »

7circle wrote: P = 3 x Vphase x Iphase x pf
Vphase = Vline/√3 and Iphase = Iline

A minor pedantic point: the immediately above assumes star connecting of the motor windings. Very low voltage motors like this are more likely to be delta connected, so then
Vphase = Vline and Iphase = Iline/√3

It works out to the same power in the end, of course. That's why the power equation is usually written as
P = √3 * Vline * Iline * PF

The 3 on the top line is divided by a √3 coming from either the voltage (star) or current (delta), resulting in a factor of √3 on top, provided that you use line voltage and current.

For completeness, PF = Power Factor = real power / total VA. (VA = Volts times Amps, ignoring their phase relationship). Since it's usually the current that is thought of as being split into real and imaginary (magnetising) components, this can also be written as
PF = Ireal/Itotal (into the motor).

Edit: total power -> VA. "Power" is generally held to be "real power", or power averaged over a cycle.
Last edited by coulomb on Sat, 29 May 2010, 05:48, edited 1 time in total.
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