How much log splitter can a 5hp electric motor drive?

[user 140828]

Okay, so I was hoping to build a (honest) 20-ton log splitter powered by an (honest) 5hp electric motor.

So I played around with the oft-recommended calculators on surpluscenter.com, and according to these calculators ("Calculated for electric motors. Double this figure for gas engines.") a 5hp electric motor can produce 3gpm at 2550 psi without exceeding its rated output.

Happily, the calculators tell me that at that pressure a 4.5" cylinder will produce just over 20 tons of force.

Looking at the two-stage log splitter pumps offered on surpluscenter.com, I found that their 9-12446 MTE 13gpm pump has a high pressure stage flow of 3gpm.

So I could use a good 5hp electric motor, that 13gpm MTE pump, and a 4.5" cylinder to get reasonably fast log splitting up to 20 tons. Voila, right?

Then I called product support at surpluscenter.com to confirm my thinking, and happened to mention I intended to use a 5hp electric motor to run the 13gpm pump. The conversation went something like this:

Support Guy: "5hp isn't enough. You need at least 6.5hp*."

Me: "But I used your calculators."

Support Guy: "They're just estimates."

*(To my knowledge, a 6.5hp electric motor, if it exists, would be exceedingly rare. So, 7.5hp.)

So, hmm, is it kinda like Pirates of the Caribbean, "They're more like guidelines?"

Don't get me wrong, he may be absolutely right and it may be that I can't get 20 tons out of that setup. The problem is trying to find out for sure. A good motor will be my largest single expense in building a log splitter so I surely don't want to get it wrong.

I've spent hours searching forums trying to find out. While there is some discussion of electric splitters here and there over the years, it is rarely (in what I could find, anyway) accompanied by complete descriptions of working setups. Someone will say they or their cousin has been happily using an electric for years and it does everything they need and they may mention some details, but not their whole setup--actual motor type, pump size and hi/lo flow numbers, cylinder size, etc. Or they do provide some numbers which disagree with both the calculators and the support guy, and even I am skeptical.

Not sure where to go from here. I would appreciate thoughts and suggestions but would particularly welcome hard facts based on experience.

Which is what I thought I was getting from the calculators.

 

[Streblerm]

No experience here other than using a 22ton box store splitter that is probably nowhere near 22 tons in real life. I can say that I have never encountered anything that it wouldn’t split. I do have a few thoughts.

First I wouldn’t get too wrapped up with 20 tons. There’s nothing magical about 20. 15, 18, 19.4 will all get the job done.

I suspect the setup you are talking about would be fine. Usually when you are looking at those type of ratings they are calculated at continuous duty which you won’t have with a log splitter. The log splitter only has to reach that pressure for a few seconds. That same pump/cyl combo driven by a 6.5 HP gas engine would be rated 22-25 tons on a box store splitter. If you’re going by the roughly double for gas engine guidelines...Worst case scenario you could run at a slightly lower pressure if it stalls the motor. That would only happen if you stalled the cylinder.

I would consider a 4” cylinder 13gpm 2 stage pump and 5hp electric motor. That would give you a little boost in speed and require a little less power. My gut tells me that what you have planned would work just fine though.

 

[ChoppyChoppy]

GPM X Pressure..

Divide that by 1714 x effeciency (85% is a good # to use)

That will equal the horsepower.

So 3gpm at 2500psi is about 5hp

 

[ChoppyChoppy]

Here's a chart for you, that might be easier.

https://www.womackmachine.com/engin...electric-motor-size-for-hydraulic-pump-drive/

 

[deadtrees]

Are you looking at a 3400 RPM motor to match 3600 RPM in a gas engine? Many elec motors are 1750 RPM.

 

[user 140828]

I picked 20 tons because I've been using a bottle jack splitter I made last summer as a first try, with a Harbor Freight 20 ton bottle jack. I figured why take a step back. Of course, I have no way of knowing whether that bottle jack is honestly rated.

From my impressions reading a lot of forum posts and doing the calcs it did seem like it should work. Then the guy at surpluscenter said no right away and basically lost interest in the conversation, I felt. So I wondered whether he knew something I and the creators of their own calculators didn't.

The motor I'm looking is a Leeson 131641, single phase, 3400-3500 rpm, capacitor start capacitor run, continuous duty, service factor 1.15. Not yer typical compressor duty motor.

 

[user 140828]

I recalculated using the formula GPM x Pressure divided by 1714 x .85 and the result is that the 5hp motor can produce 3gpm up to 2400 psi before overloading (not 2550 as with the calculator on surpluscenter.com). This produces more like 19 tons but that's still right in the ballpark for what I want to do. The control valve relief could be set to 2400psi so the motor doesn't get overloaded. Further, the MTE pump I'm looking at unloads to high pressure only at 500 psi, which by the same formula requires a little under 4.5 horsepower from the motor.

On negative side, the characteristics of this 13gpm pump, surpluscenter number 9-12446, appears to match those of item 21404 in this chart on MTE's site:

https://mtehydraulics.com/wp-content/uploads/2016/02/MTE-Logsplitter-Hi-Lo-Specs.pdf

and the chart says the pump requires 7 horsepower.

So the formula says 5hp is okay, surpluscenter's calculator says 5hp is okay, surpluscenter's support person says at least 6.5hp is required (electric), MTE's chart says 7hp is required (not specifying gas or electric that I could find). Maybe there is another factor being taken into account (pump internal losses)? According to the chart, even their 11gpm pump requires at least 6hp.

Starting to look like I may need a 7.5hp motor unless I want to gamble. Curiously, on ebay they don't appear to cost that much more than 5hp motors.

(Edit correction: the relief adjustment is on the system control valve, not the pump.)

 

[Haywire Haywood]

Personally, I would not want to build a machine where the motor is pushed to max duty every time I used it. Would you redline your car between every gear on every trip to the grocery store? Give it a little headroom and step up to the next bigger motor if you can. Now I'm no hydraulic expert, I have just always operated on the premise that if I need X, use a piece of equipment that is capable of X +10%. I can't stand tools that just barely meet the minimum requirement to do the job.

 

[muddstopper]

A couple of things to consider. Most of those 2 stage pumps are rated at a certain flow at a certain rpm. If that flow number is 3gpm@3600rpms and your motor only turns 3400rpms, then you are not pumping 3 gpm. Another thing, max hp is only required at max flow and max pressure. If the wood will split with 2000psi, but the pump is capable of produceing 3000psi, the motor will never reach max hp requirement and the 3000psi number will never be seen. Then you have the whole efficiency thing, 3gpm at 85%, is only 2.55gpm. Will your motor pull a pump capable of pumping 2.55gpm at your 2400psi numbers you are looking at. Now is your motor also turning the 3600rpms required for max flow or are the rmps lower by a few rpms. If less than 3600rps, you are not even pumping the full 2.55gpm of your 3gpm rated pump..

Check the cuin size of the pump you intend to use. That is the amount of fluid the pump produces with each rpm the shaft turn. It takes 231cuin to equal a gallon. Muliply the cuin of pump by the rpms your motor will turn and divide by 231 and that will give you the actual gpm the pump will produce. Multiply that by the rated pump efficienecy, usually 85%. and then use that number to select the proper size motor to use for the flow produced and the pressure you are shooting for. Two stage pumps have two different size pump gearsets with the larger gearset combined with the smaller gear set for max flow and low pressure. When the pu,p kicks down into low flow/high pressure, it is only using the smaller of the gearsets to produce pressure.

 

[user 140828]

Two very good points, on redlining and on adjusting for lower rpms on the electric motor.

The small additional cost to jump up to 7.5hp makes it a no-brainer. The 5hp Leeson 131641 is selling on ebay for around $430, and the 7.5hp Leeson 132044 for $460-$480. I had never looked, assuming the larger one would be way more. I'll also ask around for used at motor repair shops, though my experience buying used here in the LA area hasn't been that great.

 

[kevin j]

1.15 service factor means you can run that all day long at about 5.7 hp then there is the intermittent load cycle calculations for heat that you can probably spike through the high-pressure small displacement setting and typically a 5 hp could spike to six or seven quite easily typically up to 10 under max amperage conditions for brief periods.
look for the root mean square load cycle calculation for an electric motor I don’t have any articles at my fingertips

if the price is that close I’m go with the 7.5 although it’s also quite possible that their frame sizing is very similar and they’re not really 50% apart on actual horsepower.
Will this be hardwired or do you have some sort of extension cord power to get to the power source?

 

[user 140828]

One is 184T and the other is 184JM (different shaft) so yes same frame. One weighs 95 lbs and the other is 110 lbs. I'll power it through the 8 gauge 50 amp extension cord I use for my arc welder, with a heavy duty on-off switch in there somewhere.

(Edited to add motor weights)

 

[Mustang71]

I was thinking that a gas engine would probably make a lot more rpm than an electric motor of the same size. You make hp at a specific rpm vs amount of current needed to make the hp. Also what kind of power does a true 5hp electric motor need? I know a 1hp furnace blower motor uses about 10amps at 120 volts. It's a cool idea that I have thought about too but I think it's more about turning the correct size pump fast and not about the actual hp needed to turn the pump I'm general. Just my 2 cents I could be completely wrong.

 

[user 140828]

No I think you're absolutely right, all other things being equal a gas engine is the way to go. I'm in a suburb of LA and my next door neighbor's home is maybe 10-12 feet from where my splitter would be along the side of mine. They are kind enough not to complain when I chainsaw logs to length for an hour now and then but I would not impose on them to listen to an afternoon of gas powered log splitting. It could be managed but it would be a delicate scheduling dance.

The 5hp motor I'm looking at draws 24-22 amps at full load, the 7hp one draws 34-31. At 208-230 volts. And as muddstopper pointed out they do turn a bit slower than the 3600 rpm the pumps are rated for and which I believe the gas motors are happy to run at, so you have to take that into consideration when looking at pump specs.

(Edited to add voltage. I seem to think of things after I post. )

 

[muddstopper]

I was thinking that a gas engine would probably make a lot more rpm than an electric motor of the same size. You make hp at a specific rpm vs amount of current needed to make the hp. Also what kind of power does a true 5hp electric motor need? I know a 1hp furnace blower motor uses about 10amps at 120 volts. It's a cool idea that I have thought about too but I think it's more about turning the correct size pump fast and not about the actual hp needed to turn the pump I'm general. Just my 2 cents I could be completely wrong.

Not sure what you are saying. You can turn a pump as fast as you want with minimal power. When the pump starts to build pressure, your going to need that hp if you wish to build any appreciable amount of pressure.

 

[kevin j]

Motor synchronous speed on 60 Hz power will be either 3600 or 1800 (excluding 1200). Actual speed depends on load. With almost no load, it still won’t exceed 3600/1800. As torque increases, speed drops. Rated full load speeds are usually 3450 and 1725 or 1750, so they are close to synch speed.

Pumps have a rated maximum speed due to mechanical limits and inlet suction conditions, and a minimum recommended speed due to leakage decreasing the effiency drastically at low rpm, but that is typically below 750 or 500 rpm or so. In between, without doing the leakage and efficiency calculations, it is close enough to just ratio the speed and gpm. So a ‘nominal 13 gpm’ pump at 3600 is probably actual about 12-12.5 at 3600, 11.7 at 3450, 10.2 at 3000 rpm, 8.5 at 2500 rpm, etc. These are the total flow with both gear sets, before the large gear set unloads at some pressure. Then do the same ratio process with the small gear set for flow at various rpm.

I think it is good politics to do everything you can to make neighbors happy and be a considerate neighbor.

 

[Mustang71]

Not sure what you are saying. You can turn a pump as fast as you want with minimal power. When the pump starts to build pressure, your going to need that hp if you wish to build any appreciable amount of pressure.

I was saying that a gas engine should make more rpm at its rated hp than an electric motor. I get that you need the hp but I'd think you need the speed too to make it all work. Wouldn't a lower rpm pump require more hp to get the same flow and power out of it? Kind of like the hydraulics on my diesel tractor.

 

[muddstopper]

I was saying that a gas engine should make more rpm at its rated hp than an electric motor. I get that you need the hp but I'd think you need the speed too to make it all work. Wouldn't a lower rpm pump require more hp to get the same flow and power out of it? Kind of like the hydraulics on my diesel tractor.

I understand what you mean, I think. The question would also have to include torque. I have never fooled using electric to power pumps so havent done any torque hp calculations. I suspect which power source can maintain the best torque is going to be the winner. The torque and hp curves are going to cross at 5250rpms with either type of power source regardless. Neither the gas or electric motor are going to reach those rpms, (in this application), so I think which ever motor has the best torque curve at the recommended rpm range would work the best.

As for a lower rpm pump and power requirements, I havent studied this aspect but a pump produces a certain amount of flow per revolution of the shaft. That doesnt mean you can turn more or less rpms and produce the same amount of flow. There are limits, a pump can turn super slow and still produce flow or super fast and still produce more flow, but the limits are internal slippage and fluid bypass. A pump will need a certain rpm speed to overcome the internal slippage before it can produce flow and overspeeding the pump can cause cavatation burning up the pump. Not to even mention bushings, bearings and seals.. A one cuin pump should produce one cuin of flow with each revolution of the shaft, double the input speed and you double the flow,(gpm), but the whole efficiency thing hasnt been addressed. The pump could have enough clearance between the gear set that at one rpm it doesnt produce anything and all the oil just bypasses around the gear set and case housing as they turn. At high speed the gear set could cause air bubbles to appear in the oil which creates friction and will quickly erode the pump housing. At least thats the way I understand it and welcome other opinions.

Pumps have a rated maximum speed due to mechanical limits and inlet suction conditions, and a minimum recommended speed due to leakage decreasing the effiency drastically at low rpm, but that is typically below 750 or 500 rpm or so. In between, without doing the leakage and efficiency calculations, it is close enough to just ratio the speed and gpm. So a ‘nominal 13 gpm’ pump at 3600 is probably actual about 12-12.5 at 3600, 11.7 at 3450, 10.2 at 3000 rpm, 8.5 at 2500 rpm, etc. These are the total flow with both gear sets, before the large gear set unloads at some pressure. Then do the same ratio process with the small gear set for flow at various rpm.

 

[muddstopper]

Just found this, http://www.epi-eng.com/piston_engine_technology/power_and_torque.htm
Need to read the whole article but here is a cut and paste of the pump parts.
Power to Drive a Pump
In the course of working with lots of different engine projects, we often hear the suggestion that engine power can be increased by the use of a "better" oil pump. Implicit in that suggestion is the belief that a "better" oil pump has higher pumping efficiency, and can, therefore, deliver the required flow at the required pressure while consuming less power from the crankshaft to do so. While that is technically true, the magnitude of the improvement number is surprisingly small.

How much power does it take to drive a pump delivering a known flow at a known pressure? We have already shown that power is work per unit time, and we will stick with good old American units for the time being (foot-pounds per minute and inch-pounds per minute). And we know that flow times pressure equals POWER, as shown by:

Flow (cubic inches / minute) multiplied by pressure (pounds / square inch) = POWER (inch-pounds / minute)

From there it is simply a matter of multiplying by the appropriate constants to produce an equation which calculates HP from pressure times flow. Since flow is more freqently given in gallons per minute, and since it is well known that there are 231 cubic inches in a gallon, then:

Flow (GPM) x 231(cubic inches / gal) = Flow (cubic inches per minute).

Since, as explained above, 1 HP is 33,000 foot-pounds of work per minute, multiplying that number by 12 produces the number of inch-pounds of work per minute in one HP (396,000). Dividing 396,000 by 231 gives the units-conversion factor of 1714.3. Therefore, the simple equation is:

Pump HP = flow (GPM) x pressure (PSI) / 1714.

That equation represents the power consumed by a pump having 100% efficiency. When the equation is modified to include pump efficiency, it becomes:

Pump HP = (flow {GPM} x pressure {PSI} / (1714 x efficiency)

Common gear-type pumps typically operate at between 75 and 80% efficiency. So suppose your all-aluminum V8 engine requires 10 GPM at 50 psi. The oil pump will have been sized to maintain some preferred level of oil pressure at idle when the engine and oil are hot, so the pump will have far more capacity than is required to maintain the 10 GPM at 50 psi at operating speed. (That's what the "relief" valve does: bypasses the excess flow capacity back to the inlet of the pump, which, as an added benefit, also dramatically reduces the prospect cavitation in the pump inlet line.)

So suppose your 75%-efficient pump is maintaining 50 psi at operating speed, and is providing the 10 GPM needed by the engine. It is actually pumping roughly 50 GPM ( 10 of which goes through the engine, and the remaining 40 goes through the relief valve ) at 50 psi. The power to drive that pressure pump stage is:

HP = ( 50 gpm x 50 psi ) / ( 1714 x 0.75 efficiency ) = 1.95 HP

Suppose you succumb to the hype and shuck out some really big bucks for an allegedly 90% efficient pump. That pump (at the same flow and pressure) will consume:

HP = ( 50 gpm x 50 psi ) / ( 1714 x 0.90 efficiency ) = 1.62 HP.

WOW. A net gain of a full 1/3 of a HP. Can YOUR dyno even measure a 1-HP difference accurately and repeatably?

 

[Mustang71]

The power thing still gets me if you are out there all day splitting wood then your electric bill is going to be a killer. The only electric splitters I have seen are the screw type that splits the wood by screwing into it and those ones where you turn it on and it moves the wedge into the wood. I would think if an electric hydraulic splitter was that amazing someone would be selling one at a box store if they were comparable to a no name gas 22ton splitter.