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#### Jason is Crazy

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Does this make sense?
it is a for a non car project, I just need a 45hp motor running at 600rpm. I found a 15hp at 1800. A quick gear change and bam!! Or??

#### kennybobby

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No the gear box is a speed reducer and a torque multiplier, and the output power will always be less than the input power.

#### Jason is Crazy

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I sort of understand. So your saying HP will be the same (less the efficiency loss) but the torque will be 3fold (less efficiency loss)??

this is a snowblower build, I was planning a 40hp Kohler with a 26gpm at 2500psi pump to power the hydraulic snowblower.
I want to make this set up Electric with a generator.

#### kennybobby

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what rpm does the pump require? Do you have or know the pump impeller power curve? This would give you an idea of what size electric motor could do the task--likely sized less than a 40hp ICE.

An electric motor makes high torque right off the line, whereas an ICE needs high rpm to reach its torque peak.

#### brian_

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this is a snowblower build, I was planning a 40hp Kohler with a 26gpm at 2500psi pump to power the hydraulic snowblower.
I want to make this set up Electric with a generator.
When power is transmitted by a shaft, the power can be calculated as the rotational speed multiplied by the torque.
The electrical equivalent is that power is voltage multiplied by current.
Similarly, the power transmitted by a flow of incompressible fluid is the pressure multiplied by the volumetric flow rate.
In each case, consistent base units must be used, or a unit conversion constant must be multiplied in.

If the pump in this setup actually pumps 26 GPM at 2500 RPM, it is producing 38 HP (which is 28 kW) so the 40 HP engine must be running at full rated output. If that's really needed, then the generator must be able to produce at least 28 kW at whatever speed the engine runs (1800 RPM?) or a different speed if a gearbox or chain or belt drive is between them, the motor must be able to use that at the speed the blower needs to turn (or some other speed if a gearbox or chain or belt drive is used), and the motor and generator need to run at the same voltage.

#### Jason is Crazy

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The blower wants to be driven at 600 rpm. The 1800rpm electric 15 hp motor...would be 3:1 speed reduction.

I would really like to hear “yes Jason, your 15hp motor reduced 3 fold is about 45hp, and a 8kw generator should be sufficient for your needs”...that’s what I want someone to say lol. Because that means it will work. Lol

#### kennybobby

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Are you building this from scratch or is this a modification to an already existing device?

do you have the blower make and model number? Maybe there is a datasheet available with performance parameters to determine what is the minimum power needed to blow some snow?

Until there are some output power values established then it will be difficult or impossible to size the input or driving power device(s). First figure out the output power or Load of the system, then back it thru the gearbox, then thru the motor, then to the power supply (batteries or generator).

Your design should start at the desired output and then work back to what is needed to drive it. Don't start with a motor and then try to cluge parts together to make it work unless you have a fat wallet.

#### Jason is Crazy

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Yes I am building this from scratch. I don’t have a blower yet...but I am thinking of the WoodMax SB-85 (85” width). It is PTO driven and 3 point hitch carried. I am installing this unit on the back of a 2020 Ram 5500.
The blower has a pto hp rating of 25hp-65hp lol. So there is plenty of room, but the more the marrier. I was going to power the blower with hydraulics (essentially buy a hydraulic snowblower of the same width), but I would much rather do electric

ps: thank you for taking the time to help me

#### Jason is Crazy

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Here is all the info I get

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#### brian_

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The blower wants to be driven at 600 rpm. The 1800rpm electric 15 hp motor...would be 3:1 speed reduction.

I would really like to hear “yes Jason, your 15hp motor reduced 3 fold is about 45hp, and a 8kw generator should be sufficient for your needs”...that’s what I want someone to say lol. Because that means it will work. Lol
But if you do follow what we're saying, you know it doesn't work that way. If you put the 15 HP output of the electric motor through any kind of gearbox, belt drive, chain drive, hydrostatic transmission, or whatever, you only get (less than) 15 hp out.
15 hp at 1800 rpm means there must be 44 lb-ft of torque on the shaft;
the same 15 hp but at 600 rpm would mean there must be three times as much torque on the shaft, or 131 lb-ft.

If you need 45 hp on the shaft into the blower at 600 rpm, then you need 394 lb-ft of torque at that speed.
(use the appropriate conversion factor for mismatched units, or a handy online torque/speed/power calculator)
If your motor runs at 1800 rpm and the blower runs at 600 rpm, so you use a 3:1 reduction drive, and you need 45 hp to the blower, then you only need 131 lb-ft of torque from the motor (at 1800 rpm) to produce three times that much torque to the blower... and it's still 45 hp.

Having said that, electric motor ratings are very dependent on conditions. You might be able to force a motor rated at 15 hp to produce 45 hp by supplying it with more voltage so it can run faster while using the same current... but it might also overheat or fail in some other way.

#### brian_

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Since the blower is designed to be operated by a tractor PTO, 600 RPM is a strange speed - I thought that the industry standard was 540 RPM... and sure enough, the 24" diameter fan run at a tip speed of 3,394 ft/min, which means that it must be turning at 3394/2π =540 revolutions per minute, and the sheet includes the 540 PRM PTO speed specification near the bottom. It looks like the PTO shaft drives the fan directly, and the auger is driven through a reduction box; since a right-angle turn is also needed, this reduction is commonly done on small blowers with a worm gear in the middle of the auger shaft, but in this case there's a 1:1 bevel gear box, a shaft to one end of the machine, and a reduction chain drive to the auger.

You could drive the whole thing faster or more slowly, although faster risks mechanical disaster and slower won't throw the snow effectively.

The PTO power is specified as a range because the actual power used at any moment depends on the load. It is intended to always turn at 540 RPM, so the tractor (or electric motor in this case) must provide enough to torque to keep it moving. If the PTO can provide 65 HP, then it must be able to provide 633 lb-ft of torque (and it will really be able to rip into that snow); if the PTO can only provide 25 HP, then it must be able to provide only 243 lb-ft of torque (and its performance will be only barely be acceptable). The heavier the snow and the harder you're forcing the blower into the snow, the more torque is required.

Connect the motor to the blower through a gearbox and the required motor torque is divided by the same ratio as the speed is increased, but the power requirement doesn't change. So for instance, with a 3.33:1 drive (so a motor turning at 1800 RPM turns the blower input shaft at 540 RPM), the same 25 HP to 65 HP range corresponds to 73 ft-lb to 190 ft-lb.

#### brian_

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The blower is the WoodMaxx SB-84 (usually used on the back of an agricultural tractor). The hydraulic equivalent would presumably be the SS-84 (usually used on a skid-steer loader). The SS-84 is listed as requiring a range of 16-30 GPM (at a fixed pressure), just as the SB-84 is listed as requiring a range of power (which corresponds to a range of torque, at a fixed speed). It doesn't matter how the power is transmitted, the power requirement stays the same.

#### Jason is Crazy

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Okay, I am following you.
I guess I can simplify this now.

I was going to buy the Woodmax hydraulic snowblower and mount it to the truck. But them at required me to get a 40hp motor to power a 35gpm pump at 2500psi. Then buy a hydraulic reservoir, a hydraulic cooler, lines, spools, and the deal with maintenance.
I was hoping I could simplify things by buying a PTO drive blower and putting an electric motor on it, and powering it with an appropriate generator.

but the way it is sounding is...the motor will require such a large generator that it will off set the costs of the hydraulic power unit.

Unless I am missing something....the electric route will be unfeasible relative to hydraulic.

#### Jason is Crazy

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And yes 540rpm on the pto is correct. I said 600rpm for simple math (1800rpm with 3:1 gearing = 600 lol)

#### brian_

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And yes 540rpm on the pto is correct. I said 600rpm for simple math (1800rpm with 3:1 gearing = 600 lol)
That's what I thought might have happened.

#### brian_

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...
I was hoping I could simplify things by buying a PTO drive blower and putting an electric motor on it, and powering it with an appropriate generator.

but the way it is sounding is...the motor will require such a large generator that it will off set the costs of the hydraulic power unit.

Unless I am missing something....the electric route will be unfeasible relative to hydraulic.
Right - the electric configuration is reasonably straightforward, and requires the same size of engine as the hydrostatic configuration, but a big generator is expensive (unless you get parts salvaged from scrapped EVs). Hydrostatic drive components are cheap and compact compared to electric, for the same power, in a moderate power range.

Do you have a transmission PTO which you can use instead of a separate auxiliary engine? A direct shaft PTO drive isn't practical because the truck's PTO won't run at the required constant speed (because, unlike a tractor, the truck's engine isn't governed to run at nearly constant speed while working), but the Ram 5500 (when equipped with PTO) does have a Mobile Mode, which allows the PTO to operate while the vehicle is driven. The PTO output speed will change with truck speed, so it can't run a blower directly (by shaft) or even hydraulically (it looks like there would be no power when the truck is stopped), but it could run a generator if there is enough of a battery that the blower drive motor can run as required and the battery can charge when generator output is available.

Since the average power demand of the blower is much lower than the peak, you might even be able to run a second dedicated alternator on the engine (presumably with at least 48 V running voltage) to charge a battery to run a suitably matched motor driving the blower. The Ram eTorque mild hybrid system might even be an adequate power source.

#### Jason is Crazy

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Hey Brian, thanks for your help. FWIW: I really want to do an electric snowblower....I drive a P90D Model S lol.

question: say I use a warp 9 (or similar) motor, and appropriate controller, and for simplicity sake, I make a 48v traction pack with lead acid batteries (this 5500 ram needs weight, lots of it), then I manage to get a 48v alternator on either the engine or the pto drive...
Assuming I run 24 lead acid batteries....is this viable?

I do residential snow removal, I run the blower for no more than 1 minute at a time- every 4 mins. I rarely blow heavy snow, and never hard pack (I pull the snow onto street and blow it off).
I am having a hard time imagining just what power is inside 24 car batteries. Lol

thoughts?

#### brian_

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I think that could work, if you can get enough power out of the motor at only 48 volts. And since it's just an alternator that you'll be driving, it can just go on the engine, so you don't have to worry about the PTO.

#### remy_martian

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Yes I am building this from scratch. I don’t have a blower yet...but I am thinking of the WoodMax SB-85 (85” width). It is PTO driven and 3 point hitch carried. I am installing this unit on the back of a 2020 Ram 5500.
The blower has a pto hp rating of 25hp-65hp lol. So there is plenty of room, but the more the marrier. I was going to power the blower with hydraulics (essentially buy a hydraulic snowblower of the same width), but I would much rather do electric
Why when you can bolt a hydraulic pump directly to the PTO of your transmission?

What is the benefit of going electric - it's not like your regenning from the falling snow

#### Jason is Crazy

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Well, the Pto on the ram, requires a \$3500 driveshaft through the transfer case to support a hydraulic pump large enough to run a hydraulic snow blower. And the Ram Pto doesn’t turn when the truck is not moving but is still in gear.
The f550 has a live drive pto off the motor, which requires the same relocation driveshaft (due to pump size) but is an engine driven Pto. How ever, ford has rated the pto power I. Mobile mode to only 25hp, 50 when in park.

so running the blower off the pto in any event seems not possible.

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