Question on motor speed/torque for press.

Salem Straub

Salem Straub

Joined
Oct 20, 2008
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5,547
I'm building a hydraulic press, and right now I'm assembling parts for the build. I found a guy locally that will sell me a 1phase 5hp motor to use for the pump. My calcs say I'll need 5hp, but also the pump (a splitter-type two stage) needs 3650 rpm to put out it's rated GPM. The motor I've found locally is 1725 rpm.

I'd like to buy this motor off the guy and use it, because it's a 5hp Baldor industrial motor, almost like new, for $100. I'd much rather do that than get a wimpy 5hp compressor motor for $200+ new.

I know that if I doubled the motor speed with pulleys or a gearbox, I'd lose torque. Anyone know if there is a way of doing it without losing any or much Hp? Is it just an immutable law of physics that in this case doubling speed halves horsepower?

Also, the pump would still work fine at 1725 rpm, but would be half as fast. Maybe I could live with that, using an 11 gpm splitter pump. I don't know.

Any advice is appreciated.
 
I remember read that some guys where using 1750 motors rather then 3450.I can't rember all the reasone's but one was the the high picth noise that the pumps with 3450 motors put out is very loud this was not the main reason but a benifit. I think it might have been Wheeler how made post about motor speed seem like a year or two back.
 
I would expect a Leeson compressor motor to be pretty high quality, what I'm referring to more is some relatively no-name brands I would suspect of being low quality/not rated at true hp. It's kind of a shady world, cheap motors. But that's all I can afford right now.

Thanks for the replies, guys. I'll try to find that thread, Brett.
 
What size cylinder are you planning to run? The dual stage pump will help you a good bit. Try it and see. Worst case, you have a great motor for your next grinder :)

-d
 
It's a 5" cylinder with a 1.75" ram. Haldex 3, 11 gpm 2 stage pump.

I did find a thread where Don Hanson said he'd changed out his pump motor from 3450 to 1725, resulting in a quieter press and not terribly reduced speed. I think he was using a 2 stage, 16 gpm pump. I wonder if I went to a faster pump it would help. I do already have the 11 gpm one, though.
 
Perhaps I'm stating the obvious, but:

Why not get a 22GPM pump and run it at 1750RPM ????????
 
I think I can answer your question.

Horsepower = (torque X rpm)/5252

So the torque of your motor (assuming the 5hp rating is at 1750) is ~15ftlbs.
With your approximate doubling of speed, your torque would be cut in half to ~7.61ftlbs.

So the horsepower of that setup would be = (7.61 X 3450)/5252 = ~5hp. In other words, the amount of horsepower does not change.
 
Speedemon, thanks for the info. That's interesting.

Stacy, no doubt you are right on that. I'm exploring my options with the 11gpm pump currently, as I have one I can use. It does seem more prudent at this point to consider increasing the pump speed. I'm going to go have a look at surplus center for 16 or 22 gpm pumps. Thanks.
 
the motor HP is the same, since its speed is the same
it's shaft torque is the same also ~15 lb ft
but the shaft torque of the smaller pulley (hydraulic pump) is 1/2 (or whatever ratio you choose)

P = T x w (w = rotational speed = 2Pi rev/sec)...1/2 the torque and double the speed, not net change
5252 is a conversion factor to get from rev/min (rpm), HP and incorporating the 2 Pi
= (550 ft lb/sec (or 1 HP) x 60 sec/min) / 2Pi ~ 5252

but the torque (or force) is what you are concerned about, it makes the pressure...get the 3450 motor
 
operated at 650 psi (factory setting)
it will make 6 ton press w/your ram
size to use 4.5 HP ~ 90% loading

sheeve the pullies to run the hydraulic motor at ~3200, 10 gpm
the required torque at 10 gpm at 650 psi ~ 6.2 lb ft
the available motor torque ~ 5252/1760 x 5 x 1760/3200 ~ 8.2 lb ft or 75% loading
 
Thanks for the good info, a lot of stuff that's new to me. Although, the factory setting of 650 only applies to the pressure ceiling for the lo pressure stage. With the system pressure at 2500 psi, the cylinder I have would yield approx. 49,087 lbs. force, or nearly 25 tons. I'd be capable with the 1725 5 hp motor of reaching this pressure, but the speed will suffer too much with the 11 gpm pump.

So I'm scrapping the 11 gpm pump idea. I ran all the calcs with 16 and 22 GPM 2 stage pumps. (Just crunched the numbers through the calculator at Surplus Center.)

Here are the numbers I ended up with for a 16 gpm. pump. I cut the speed in half for my 1725 motor, the pump's listed speeds are for 3600 rpm. I checked to see if this was valid by calcing the pump displacement/motor speed/desired flow. It seemed to check out. I ran the numbers with a 22 gpm pump as well, also halving the speeds for my motor. The cylinder speed improved a fair amount, 2.2 ips ext. speed for 11 gpm. lo-press stage. A 5hp motor still cut it for the 22 gpm pump.

If anyone cares to comment or correct, here's what I got:

Pump
Displacement:
0.256 cu. in. / rev. HI press. stage
1.049 cu. in. / rev. LO press. stage
Pressure: 3000 PSI max.
Speed: 3600 RPM max.
Flow:
4 GPM HI press. stage
16 GPM LO press. Stage
(I'll halve the speeds, since motor is 1725 rpm.)

Cylinder:
Throw: 20”
Bore: 5”
Rod Diameter: 1.75”

Motor:
5 hp.
1725 r.p.m.

(Checking here to see if halving rpm indeed halves flow)
Pump Displacement (Lo Pressure)
Desired Flow: 8 GPM
Operating Speed: 1725 RPM
Displacement Required: 1.1 cubic inches

Pump Displacement (Hi Pressure)
Desired Flow: 2 GPM
Operating Speed: 1725 RPM
Displacement Required: 0.3 cubic inches

Cylinder Speed
Bore Size: 5 inches
Rod Diameter: 1.75 inches
Pump Flow: 16 GPM (8GPM @1725 rpm)
Extension Speed: 1.6 inches per second (@8 GPM)
Retraction Speed: 1.8 inches per second

Cylinder Push / Pull Force
Bore Size: 5 inches
Rod Diameter: 1.75 inches
Pump Pressure: 2500 PSI
Push Force: 49,087 lbs.
Pull Force: 43,074 lbs.

(With the 22 gpm pump at 1/2 speed HP reqs. below)
Horsepower to Drive Pump (Lo Pressure)
Desired Flow: 8 GPM
System Pressure: 650 PSI
Horsepower: 4 HP
(11 gpm @ 500 psi= 4hp)

Horsepower to Drive Pump (Hi Pressure)
Desired Flow: 2 GPM
System Pressure: 2500 PSI
Horsepower: 3 HP
(3.15 gpm @ 2500 psi= 5hp)

I'll probably go with the 22 gpm pump, I'll just have to scrimp and save a bit longer.
 
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I really do appreciate the help, as I'm not well versed in hydraulics. But...
I don't get it. I mean, I've never heard of anyone with a forging press at 20 tons needing even an 8hp motor, let alone 17.

Maybe you did not notice that the high volume stage is at 500 psi max? So, while the pump is cruising along at 11 gpm, it only does so up to 500 psi- then kicks into the 3.15 gpm stage up to 2500 psi.

I'm using the below constant to calculate hp requirement.
Hp to drive hydraulic pump = Psi x GPM / 1714

So, with the 22/6.3 gpm 2 stage pump, running at 1725 rpm, the gpm output of the respective stages is halved.
Thus: Hi pressure stage, 3.15 gpm @ 2500 psi: 2500x3.15=7875, 7875/1714= 4.59hp.
Lo pressure stage, 11 gpm @ 500 psi: 500x11=5500, 5500/1714= 3.2088hp.

If the equation is done with 11 gpm @ 2500 psi,
2500x11=27500, 27500/1714=16.044hp.
Is that what you mean? I'd weep if I needed that size motor!

On the pump I got these specs off of, it does say "minimum of 12 hp required if using gas engine." Usually you'd at least double the required horsepower from electric to gas...

Please let me know if I'm wrong again, just trying to figure it out here.
 
Leeson 5 hp, 1750 rpm motor.

16 gpm 2 stage pump

5" cylinder

The press has plenty of power and is plenty fast.

My first 3600 rmp motor made the pump squeal, just about drove my crazy.
 
Plenty of power and plenty fast is great for me. The 16 gpm pump is quite a bit cheaper than the 22.

I think you all have helped me find out what I needed to know. Thanks a bunch.
 
Psi x GPM / 1714 is hydraulic HP, does not factor in pump or motor efficiency/power factor
Hi pressure stage, 3.15 gpm @ 2500 psi: 2500x3.15=7875, 7875/1714= 4.59hp.
Lo pressure stage, 11 gpm @ 500 psi: 500x11=5500, 5500/1714= 3.2088hp.

derived from (gpm x 8.34) x FT HD/(60 sec/min x 1 HP/ (550 lb ft/sec)) = gpm x ft hd/3957
to convert to psi 1 psi = 2.31 ft hd = (gpm x psi)/(3957/2.31) = psi x gpm/1713
it assumes the density of the hydraulic fluid is the same as water 8.34/gal

pump efficiency (positive displacment) typically 0.85
so 4.6 pump (hydraulic) HP, the work/time the pump does is 4.6
but the motor needs to supply ~ 4.6/(0.85) ~ 5.4 HP

if you look at the curve for the nominal 11 gpm pump at 2500 psi it is ~5 HP
but that is the hydraulic power, the motor will need to be larger

what tonnage do you want? 5? 10?
then work back
 
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