Running 1.5 HP 3 phase motor on a 12V SLA battery

(...)

proof-of-concept phase is only for a utility

deep cycle battery should give at least one

Were that the case, yes. :)

1 HP at 100% efficiency is 746 W or 62.2 A. Figure 30% efficiency and that number goes up to 207 A.

So figure about *6 minutes* at 1 HP, for a 100 AH battery, not an hour. :) See the data for the '121000' battery here:

Read "the small print" about lead acid batteries here:

--Winston

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I wouldnt even contemplate any of the under 3kVA invertors, and really

3kVA is too small. 1.5hp - 1.12kW. Guesstimate another 20% onto the current for lagging pf -> 1.34kVA. Even 6x run current during startup would be 8kVA. If you use a 3kVA invertor, you'll need a way to cope gracefully with excess demand by reducing Vout to limit i.

NT

hour.=20

I can't see where efficiency can be that low. The motor is about 85%=20 efficient, and there will also be some inefficiencies in the = transmission,=20 differential, and rolling friction of the tires on the ground. But I = can't=20 easily measure those, so I'm just assuming that 1 HP as measured by = motor=20 draw will be sufficient. The VF controller is rated to be about 94%=20 efficient. The conversion from AC to DC is probably about 95% efficient. = And=20 the inverter is supposed to be about 90% efficient. So the overall=20 efficiency (not counting the battery itself) is 0.94*0.95*0.90 =3D 80%. = Thus=20 for 1 HP motor draw, the battery would need to provide 1.24 HP or 928 = watts=20 or 77 amps. The battery you mention below would provide that for 30-45=20 minutes. Or the same based on a constant 155 watts per cell.

Yes, my 1 hour figure was too optimistic, but within a binary order of=20 magnitude. Now, to determine the actual power needed, I have an access = road=20 which is approximately 300 feet long and a rise of about 80 feet. I = would=20 like my tractor to be able to carry its own weight and me this distance = at=20 about 5 MPH, which is about 400 pounds in 36 seconds. The actual work is =

400=20 lb * 80 ft / 36 seconds or 392 lb-ft/sec. 1 HP is 550 lb-ft/sec. So it's = in=20 the right ballpark. I figure that the motor efficiency of 85% and = mechanical=20 efficiency of drive train at 80% will be 68% which means 392/0.68 or 576 = HP.=20 Close enough for "gummint work"!

10920-TechManual-Lo.pdf

There have been many tractor conversions that seem to indicate that a=20 sustained draw of about 3 HP is needed for actual mowing work that was=20 originally done with an 8 to 10 HP gas engine. It is generally accepted = that=20 there is a 3:1 ratio of equivalent power for gas/electric, so that seems =

about right. IIRC the tractor used three 100 A-H batteries and they = lasted=20 for about 1/2 hour of mowing, so again my optimistic estimate is not far = off=20 the mark. But the proof is in the pudding (or the "putting" of a = datalogger=20 on the system and actually trying it).

There is a website dedicated to one person's three-year electric mower=20 project, using a larger tractor which happens to be essentially the same = as=20 my other one, which I plan to electrify with a 5 HP electric motor and a =

7.5=20 HP 480V VF controller, using three or four batteries:

And here are videos of my tractor:

(walk-around) (engine test, removal, electric motor and=20 controller)

And my diesel tractor:

(walk-around) (start and run) (ride)

I'm having some fun with these projects, and I hope to learn a lot about =

electric vehicles in general. Eventually I may want to make my own=20 street-legal electric car, as some guys in Canada did for less than = $1000=20 (including the car!):

Thanks,

Paul=20

(...)

You are right. It'll be lower.

tires on the ground. But I can't easily

be sufficient. The VF controller is rated

efficient. And the inverter is supposed to

itself) is 0.94*0.95*0.90 = 80%.

Battery efficiency ranges from 50% 60% when not fully charged so

0.94*0.95*0.90*0.50 = 40.2%

amps. The battery you mention below would

cell.

Even if you have a bottomless pocketbook for batteries, your estimate is still wildly optimistic, IMHO. Figure about half the energy you put into the battery will be available on discharge.

The terminal (!) voltage at that rate is 1.85 V per cell (or discharge to about

10% of capacity). This'll wreck your battery muy pronto. Industry practice is to limit discharge to 40% of capacity, worst case (1.98 V per cell).

This is assuming that you want more than a couple runs off each battery you buy.

Have fun Paul! :)

--Winston

Yes, exceeding rated discharge by a factor of 10 will severely impair the number of cycles. Also, mass produced sealed LA batts have a discharge rating over a specified time, such as 8 hours or 10 hours, or until some voltage end point such as 1.9 volts per cell. Some designs are tested to determine engineering limits, thermal characteristics, internal resistance, etc. SLA discharge curves for SLAs made by Johnson show curves for 0.1, 0.3, 0.5 on up to 7 times capacity. That design showed the 10 hour rating to 1.6 volts/cell. The 2C curve ended after 10 minutes at 1.2 volts per cell. There is a 7C curve that ends at 1.0 volt/cell which was reached in 1 minute. There are many pages of such testing in Linden's Handbook of Batteries. The OP could increase the capacity to avoid over 50 percent discharge. Just increase the size by a factor of 10, for example using high capacity tractor starting batteries in parallel.

"NT" wrote in message=20 news: snipped-for-privacy@x17g2000yqj.googlegroups.com...=

Actually the current limit of the inverter seems to limit its output = during=20 startup which involves charging the two 3300 uF capacitors. Once = charged,=20 they should be able to handle a motor current surge of 3 HP (2200 watts = or 7=20 amps at 320 VDC) for about 75 mSec. If I used 12,000 uF 200 V capacitors =

instead, they would give about 300 mSec surge. The motor controller = would=20 handle the overload stuff.

I just did a bit of testing on the tractor, and it seems that maybe the=20 automotive type inverters are not the way to go. I could only run the = motor=20 in neutral and at 600 RPM it drew 10.5 to 11.5 amps. At about 850 RPM it =

drew 12.5 amps at a battery voltage of 11.5 volts, and at 1080 RPM it=20 reached a peak current of about 15 amps at which point the undervoltage = trip=20 on the VF drive shut it down. I checked the bus link voltage, and it was =

only about 250 VDC with no load other than the drive, and it dropped to =

203=20 VDC under load. This was with 3.36 amps per phase and 43 volts according = to=20 the drive display. When I released the clutch the battery current rose = to=20 about 15 amps and the tractor started to move but then the drive kicked = out.

This was not really surprising, but the low bus link voltage makes me = think=20 it may not be practical to use the automotive inverters, especially the =

120=20 VAC type in a voltage doubler configuration as I have. The stepped sine = wave=20 is really a rectangular waveform with 180 volt peaks at about 35% duty=20 cycle, which is 106 VRMS. I did an LTspice simulation which showed 325 = volts=20 out into 150 ohms, for 707 watts, but this requires 20 amp peak current=20 pulses from the inverter, and it is probably limited to something like = 9.5=20 amps which is 600 watts (peak) / (180V * 0.35). The actual input power = at=20 shut-down is about 11.5V*15A =3D 172 watts, but the power factor of the = load=20 is probably causing a higher output current draw. The motor is drawing = 43 *=20 3.36 * sqrt(3) =3D 250 VA according to the VFD display.

My simulation shows 8.15 amps RMS at 89.3 VRMS (30% duty cycle and 1 ohm =

source resistance), which is 727 VA. But I have a pure resistive load of =

707=20 watts.

Anyway, maybe I should go back to my original plan to make my own DC-DC=20 converter. And I'm wondering if I might be able to modify the 600 watt = UPS I=20 described in another post. I already have a PIC circuit that will drive = the=20 MOSFETs in push-pull configuration, although I probably need to add gate =

drivers to get 10 volts for good results (they probably are not logic=20 level). If I drive the transformer with 49% duty cycle, and a somewhat=20 higher frequency (probably 180-240 Hz), I should be able to get a solid =

160=20 volt square wave which should drive my doubler to 320 VDC. Probably = worth a=20 try, since otherwise I was going to scrap it.

Stay tuned...

Paul

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D = VoltageDoubler.asc = =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D

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0)=20 (+7m 0) (+1u -180) (+3m -180) (+1u 0) ) endrepeat SYMATTR SpiceLine Rser=3D1 SYMBOL polcap -16 480 R0 WINDOW 3 45 31 Left 2 SYMATTR Value 3300=B5 SYMATTR InstName C2 SYMATTR Description Capacitor SYMATTR Type cap SYMATTR SpiceLine V=3D600 Irms=3D2.9 Rser=3D0.018 Lser=3D0 SYMBOL res -80 272 R0 SYMATTR InstName R2 SYMATTR Value 25k SYMBOL res -80 448 R0 SYMATTR InstName R3 SYMATTR Value 25k TEXT -464 472 Left 2 !.tran 1 startup=20

From the chart supplied on the PowerSonic 100 A-H battery, the actual = A-H=20 capacity to 1.85 volts/cell is 93 A-H at 4.65 A (about 1/20 C), 80 A-H = at 10=20 A (1/10 C), and 44 A-H at 58 A ( 0.6 C ). And in the technical data = sheet it=20 says that a deep cycle battery can be discharged at a 1 Hr rate for an=20 effective capacity of 62% of the rated capacity based on 20 hr = discharge.=20 And at high discharge rates, 1.75 V/cell represents about 50% of = capacity,=20 and the battery "easily recovers" from such a discharge.

Discharging to 50% of capacity should give a total of 400-500 cycles, = and=20 even discharging to 100% will still give about 200 cycles. The 100 A-H=20 battery I was considering costs $85. So if I get 400 cycles from it, = that is=20 not even 25 cents per cycle. I can live with that. The energy from the = grid=20 adds about 10 cents per charge. If I can run this vehicle at 5 MPH for =

1/2=20 hour, that's 2.5 miles at about $0.12/mile. Not bad.

An equivalent gas powered vehicle might get 50 MPG so 1/20 gallon for =

2.5=20 miles, at $5/gallon, is actually about the same. But electric has=20 advantages: quieter, cleaner, less maintenance, and can be obtained from =

renewable energy sources. And, hey, I just like it!

Paul=20

If you hack off the FWB and chopper inside the inverter and rewire it as a full wave doubler, and shove that straight into the VFD (wired for 240VAC input, which should work on 320VDC just fine assuming there isn't some hunk of iron providing control voltages). Voila, now the only power factor you have to worry about is the inverter switching supply's -- which is an important concern with cap-input filters, but they're made pretty brutally so it's not really going to notice the difference.

Tim

-- Deep Friar: a very philosophical monk. Website:

"P E Schoen" wrote in message news:jldirl$l1h$ snipped-for-privacy@dont-email.me... "NT" wrote in message news: snipped-for-privacy@x17g2000yqj.googlegroups.com...

Actually the current limit of the inverter seems to limit its output during startup which involves charging the two 3300 uF capacitors. Once charged, they should be able to handle a motor current surge of 3 HP (2200 watts or 7 amps at 320 VDC) for about 75 mSec. If I used 12,000 uF 200 V capacitors instead, they would give about 300 mSec surge. The motor controller would handle the overload stuff.

I just did a bit of testing on the tractor, and it seems that maybe the automotive type inverters are not the way to go. I could only run the motor in neutral and at 600 RPM it drew 10.5 to 11.5 amps. At about 850 RPM it drew 12.5 amps at a battery voltage of 11.5 volts, and at 1080 RPM it reached a peak current of about 15 amps at which point the undervoltage trip on the VF drive shut it down. I checked the bus link voltage, and it was only about 250 VDC with no load other than the drive, and it dropped to 203 VDC under load. This was with 3.36 amps per phase and 43 volts according to the drive display. When I released the clutch the battery current rose to about 15 amps and the tractor started to move but then the drive kicked out.

This was not really surprising, but the low bus link voltage makes me think it may not be practical to use the automotive inverters, especially the 120 VAC type in a voltage doubler configuration as I have. The stepped sine wave is really a rectangular waveform with 180 volt peaks at about 35% duty cycle, which is 106 VRMS. I did an LTspice simulation which showed 325 volts out into 150 ohms, for 707 watts, but this requires 20 amp peak current pulses from the inverter, and it is probably limited to something like 9.5 amps which is 600 watts (peak) / (180V * 0.35). The actual input power at shut-down is about 11.5V*15A = 172 watts, but the power factor of the load is probably causing a higher output current draw. The motor is drawing 43 *

3.36 * sqrt(3) = 250 VA according to the VFD display.

My simulation shows 8.15 amps RMS at 89.3 VRMS (30% duty cycle and 1 ohm source resistance), which is 727 VA. But I have a pure resistive load of 707 watts.

Anyway, maybe I should go back to my original plan to make my own DC-DC converter. And I'm wondering if I might be able to modify the 600 watt UPS I described in another post. I already have a PIC circuit that will drive the MOSFETs in push-pull configuration, although I probably need to add gate drivers to get 10 volts for good results (they probably are not logic level). If I drive the transformer with 49% duty cycle, and a somewhat higher frequency (probably 180-240 Hz), I should be able to get a solid 160 volt square wave which should drive my doubler to 320 VDC. Probably worth a try, since otherwise I was going to scrap it.

Stay tuned...

Paul

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.

during=20

charged,=20

or 7=20

=20

would=20

Geez Paul, typical induction motor start times are about 5 seconds at = that power level, not milliseconds. Make no mistake the VFD helps a lot, but does not remedy everything.

motor=20

=20

trip=20

=20

203=20

to=20

to=20

out.

think=20

120=20

wave=20

volts=20

9.5=20

at=20

load=20

43 *=20

I agree, direct conversion to 360 V(dc) is a better choice, you can even get some regulation done at that stage. Don't know if the VFD will = accept that input though. Much better efficiency to be had though.

=20

707=20

UPS I=20

the=20

=20

160=20

worth a=20

VoltageDoubler.asc = =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D

0)=20

[generous snip]

though.

I finally built a DC-DC converter that should handle up to about 2 HP. = It=20 uses a toroid transformer made from a 500 VA powerstat, with two = primaries=20 of 8 turns each of #10 AWG, and a secondary of about 100 turns of about = #18=20 AWG. I am driving it push-pull using MOSFETs driven by a 500 Hz square = wave=20 generated by a PIC16F684, and it has some current limiting to throttle = down=20 the PWM over about 50 amps. It draws about 2 amps from the 12 VDC SLA=20 battery (17 A-H), with no load, and produces 320 VDC on the two 3300 uF =

400=20 V capacitors in series, using a doubler circuit as previously described. =

This connects to the VFD DC link and the output drives a 3450 RPM 2 HP=20 induction motor.

This was installed, rather crudely, on a stripped-down riding mower, and = I=20 took it for a ride. I was pleasantly surprised that the battery draw was =

only 15-30 amps, and the VFD input was about 250 VDC and 1-2 amps. These = are=20 analog meters and not very steady or accurate, but this calculates to=20

180-360 watts total battery draw, and 250-500 watts into the VFD. Rough=20 efficiency about 72%, but probably better. Here is a video I made of the =

battery powered vehicle's maiden voyage on battery power:

These power figures correlate pretty well to another battery vehicle = project=20 posted here:

His figures were 100 watts coasting, 250 watts on flat ground, and a = maximum=20 of 1114W on a 36% grade and 700 watts on a 17% grade, towing 200 kg. All = at=20 low speed, of course. It's all about torque, then HP if you want to get = it=20 done quicker.=20

I didn't read the whole thread, but why don't you just get a DC motor a skip all the conversion. Like this,

Not exactly a lawn mower but the details are there. Mikek

Or this:

:)

Well, if all I wanted was a utility vehicle or a go-cart or riding = mower, I=20 would just buy one or take the easiest route by using stock components = that=20 are available for conversion. And then I would have the end product but = not=20 the learning experience and thrill of the design and prototyping = process.=20 Also, I claim that standard three-phase motors and controllers with a = DC-DC=20 converter close to the batteries is potentially more efficient, and = there=20 are other advantages of low maintenance, high reliability, ruggedness, = and=20 low cost. The control is also more precise.

As I go through the design process, I am considering and trying various=20 approaches, and I think I have a concept for something that may be=20 competitive with many other technologies that tend to become "fixated" = on=20 certain parameters. At one time I thought it would be a great idea to = rewind=20 an induction motor for much lower voltage and higher current, so that it =

could be run directly on 24 to 48 VDC batteries, and also overclock it = to=20 get possibly as much as 6 to 8 times the nominal 60 Hz horsepower. It's=20 possible, but not really practical, yet it was a valuable experience to = have=20 taken a 1/2 HP single phase 120 VAC PSC motor, and rewinding it as a 12 = pole=20 three phase motor at 8 VAC nominal at 60 Hz. I verified that it ran at = about=20

600 RPM at 60 Hz, and then I ran it up to about 240 Hz where it ran at = about=20 2400 RPM.

The thing is that the small size and weight of high frequency motors for =

vehicles, particularly tractors, is not a significant advantage. = Tractors=20 often add dead weight just for traction, and lead-acid batteries are = perfect=20 for that. A practical small tractor needs three or four batteries at = about=20

50 pounds each, and a very capable 2 HP motor as I used on my = contraption=20 weighs only about 30 pounds. Even a 5 HP motor, which I plan to use in a =

larger tractor, is only about 100 pounds, and it will need probably 300=20 pounds of batteries.

Part of my "research" is to determine just how much power is really = needed=20 for garden tractors and riding mowers. My old 1967 Simplicity Broadmoor = GT=20 had an 8 HP B&S engine, and it was fully capable of running a 36" mower=20 deck, or a dozer blade. My "toy" riding mower is much smaller and = lighter,=20 yet it had a 10 HP engine. And most riding mowers now are at least 16 = HP.=20 But I think the actual power needed is closer to 2 HP, and it seems that =

lawn mowers as well as cars have incorporated ever more powerful engines =

mostly for bragging rights and micro-penile compensation.

Anyway, that was a long answer to a short question. But mainly I make my =

choices for my own reasons of enjoying the process more than the end = result.=20 Much like life itself.

Paul=20

If you want to see some others designing 3 phase converters, check out the DIY electric car forum. You'll need to look through it to find the guys doing the 3 phase stuff.

Mikek

out=20

Thanks. I joined and posted. There was a thread on DC-DC converters and = I=20 might be able to help with that. It seems like most of those guys are = into=20 high power vehicles, looking for 40-50 kW motors and pushing them to = several=20 times rating for Tesla-like acceleration. And some of them are using=20 hundreds of small batteries to get the voltage and current they need (or =

want).

Paul=20

Shows that sometimes you just have to put something to the test. :)

It will be interesting to take some more accurate measurements and run = at=20 higher power. There seems to be about 24 watts with no load at 12 VDC, = at=20

500 Hz. I had thought there was a problem at higher frequencies, but I = found=20 later that I had a wiring error and I need to try again. I'd rather run = it=20 at about 2 kHz or even 4 kHz, if only to reduce the annoying audible = noise=20 at 500 Hz.

The ultimate test will be at 24 VDC and 1200 watts, which I think will = be=20 more than enough to operate this simple vehicle at useful (and fun) = speeds=20 on moderate slopes. The no-load power will probably go up to about 50 = watts.=20 I estimate the primary winding to be about 8 milliohms, so at 50 amps = that=20 will be about 20 watts. The secondary will draw about 4 amps and the=20 resistance is about 0.6 ohms, so about 10 watts there. The MOSFETs=20 (HUF75645) have about 0.014 ohms each, and now I have two in parallel = but I=20 plan to use four, which will add another 10 watts. So optimistically = I'll=20 have 1200 watts with 90 watts of losses, or 92% efficiency.

I have found some interest in such a DC-DC converter on the forum:=20

set-upi-73181.html

They were discussing the conversion of the main battery pack voltage=20 (144-300 VDC or so) to 12 VDC for accessories, including a cab heater! I =

suggested using ordinary 120 and 240 volt electric fin-strip heaters=20 directly, or a heat pump to extract the waste heat from the motors and = drive=20 electronics, which also gives the option of running it in reverse for = A/C.

I was impressed with some of the motor controllers they were using. = There is=20 a "Soliton 1" which is rated for 300 kW (340 VDC and 1000 amps), for = $3300.=20 They even have a 1.2 MW unit for 10 grand! I question the need for such=20 power in a passenger vehicle, but apparently these are used on electric=20 dragsters and super-high-performance vehicles. And these controllers are = for=20 brush type series wound motors. But it's still awesome to consider a =

1500 HP=20 electric motor, which is equivalent to something like a 5000 HP ICE! = See:=20

Stay tuned...

Paul=20

though.

It=20

primaries=20

#18=20

wave=20

down=20

400=20

=20

I=20

=20

are=20

=20

project=20

maximum=20

at=20

it=20

Hmmm. Works better than i expected.

?-)