Gear speed multiplier for pump 2

[erik3000]

I have an unusual situation where I am trying to run a pump manually with a mechanical device that is very slow (40 RPM max) but high torque (1500 in-lbs), while the pump should be run closer to 1500 RPM and 100 in-lbs torque.

Most gearheads seem to be made to reduce high speed motors, while I'm looking for the opposite (convert manual crank to motor speeds). Is this as simple as turning a gear speed reducer around the other way? Any advice where I can find what I'm looking for?

 

[TheTick]

You need magic. Your power out is greater than your power in.

 

[erik3000]

How could I calculate the maximum possible output from my input?

 

[btrueblood]

Find out how to calculate rotational power (hint: your physics and/or machine design textbook should have it). Theoretically Pin = Pout, but in reality, Pout = Pin - Ploss (where Ploss is lost power due to noise, friction, etc.) Where'd you get your degree, by the way?

 

[Occupant]

I don't know what you are trying to accomplish. The pump demands ~2.5 HP to put out at the rated capacity, but you are driving at very, very slowly and are not getting even remotely close to that.

 

[erik3000]

I rechecked my pump requirements, see attachment. My initial estimates where slightly off. If I want 125 PSI with the 3/8" port looks like I need 900 RPM, 45 in-lbs, requiring 1/3 HP based on the performance chart.

My crank input power, P = 125 (ft-lbs) * 40 (RPM) / 5252 = 0.95 HP

0.95 HP >> 1/3 HP

What am I missing?

 

[kjoiner]

I agree with the others. Your input power is already 2.5 times too low(not including losses) for the required output.

Kyle

 

[ivymike]

latest info [900 rpm * 40 in.lb (output)] < [40 RPM * 1500 in.lb (input)] which doesn't violate any laws
first info [1500 rpm * 100 in.lb (output)] > [ 40 RPM * 1500 in.lb (input)] which is not possible

1500 / 40 is a 37.5x speed increase, which seems a lot to me. A quick internet search didn't turn up anything higher than about 7x speed increaser, and those are way too big for your application.

Perhaps you would have better luck using a small DC generator and DC motor to accomplish the speed change? You could also look at a generator + battery + PWM controller if you need variable output speed/power.

 

[3DDave]

900 RPM * 45IN-LBS out < 40 RPM * 12IN/FT * 125FT-LBS in, so that's a good start over the original request.

The next problem is the 22.5:1 Output to input ratio. That's a pretty large jump in ratio; maybe 3 stage spur gear or two stage planetary. It might be cheaper to get a motor that is matched to the load; likely to be more reliable.

 

[3DDave]

"I am trying to run a pump manually "

What does this mean??

ivymike - you typed faster than I did.

 

[HPost]

Power = Torque x rads/sec

For the given pressure and flow, your pump needs 0.333HP

You want more flow, so the pump has to turn faster. Increasing the speed leads to a lower torque, but the extra flow generates extra pressure and pressure x flow means more power...which needs more torque.

You need to work out how much power you need, as a function of the required pressure and flow.

Add in the efficiency losses.

That will tell you the total power you need.

HPost CEng MIMechE

 

[erik3000]

Obviously I would use a motor if possible, but this must be a purely mechanical system. The rotational input I have described is the only source of power. Essentially I am trying to gauge the feasibility of a manual input hydraulic system. Pump selection is not final, but it needs to be stainless, bi-directional (reversible), and around 125 psi, 1-2 gpm. Struggling to find something that will work with my slow input.

 

[3DDave]

If you intend a human operated crank, recognize that this is close to World class athelete performance levels for short durations.

"A trained cyclist can produce about 400 watts of mechanical power for an hour or more, but adults of good average fitness average between 50 and 150 watts for an hour of vigorous exercise. A healthy well-fed laborer over the course of an 8-hour day can sustain an average output of about 75 watts.[1] The yield of electric power is decreased by the efficiency of the human-powered generator." Wikipedia: Human Power

 

[erik3000]

A hand crank is a good way to think about it, but it's a robot not a human.

 

[3DDave]

Regular bicycle chain drives are about 5:1 max speed increasers handling power in the range you are looking for; put them in series to get appox. 25:1 total and that would be close to what you want.

 

[MintJulep]

but it's a robot not a human

What?

So you intend:
Power source --> Motor of robot --> Mechanical, hydraulic, pneumatic or whatever linkage --> Robot "arm" --> Crank handle --> Speed increasing gear train --> Pump

Don't you think that it would be more efficient to do:
Power source --> Jumper cables from the auto part store --> Motor --> Pump

 

[erik3000]

Yes, but it must be done the first way you mentioned, hence why it is such a challenge. Imagine a Mars rover with a crank arm on it, and you want to drive it around and crank the pumps which are not connected to any power source.

 

[MintJulep]

What is the business case for requiring that something be done in the least efficient way possible?

 

[Occupant]

Rino has small speed increasing gear boxes, but not in the ratio you need. I guess you could, however, stick two of them together - (1) at 1/5 and (1) at 1/6 that gives you an input speed of 30 RPM.

 

[dvd]

low rpm source --> positive displacement pump --> hydraulic motor --> high speed pump --> Nobel Prize in economics