Max HP from 5 HP motor

[junkhound]

184T frame, 240 V/25.5 A single phase cap run motor, rated 5 HP.

For use as pressure washer drive for a 4 GPM, 4000psi pump. If I add water cooling loops to the case, can I drive the motor at 50A/10HP without burning it up?

 

[petersonra]

184T frame, 240 V/25.5 A single phase cap run motor, rated 5 HP.

For use as pressure washer drive for a 4 GPM, 4000psi pump. If I add water cooling loops to the case, can I drive the motor at 50A/10HP without burning it up?

Maybe. It's not just about cooling it. The wire in the coils are pretty small to be taken 50 amps

 

[Ingenieur]

How do plan on getting 50 A thru it?
aren't those pumps positive displacement?

typical 5 hp 1 ph performance curves
https://www.baldorvip.com/Drawing/ACPerformance?productNumber=L1430T&recordId=10575

 

[drcampbell]

How do you plan to cool the rotor?

 

[JFletcher]

^ Those are positive displacement pumps.

To "drive" the motor at 10HP, you need 10HP of load on it. You'd need to increase the pump flow and/or pressure w/o the other dropping a commensurate amount to do that. or drive 2 pumps...

A 5HP motor is likely going to go poof if you double its amperage. and getting more pressure or flow from the pump would mean replacing it with a bigger/higher pressure unit.

Even if you had a water cooled motor, you can't really affect the heat transfer rate within the windings themselves, which will still be air cooled.

 

[junkhound]

you can't really affect the heat transfer rate within the windings themselves, which will still be air cooled.

Good call JF, came to the same conclusion myself !


Inspired by the air bearing audio noise thread, thought I'd throw this 5 to 10 HP 'project' out there just 'for fun'.
Did this 'conversion' a few years back, had a pump off a MTM 15 HP washer, but only had the 5 HP 184T motor in my 'junk' pile.

Photo is how I ended up cooling it - about 1000 CFM airflow, ducted THRU and around the motor.

I measured 48.7A at 240 Vac to the 184T motor when using a 40 deg pressure washer nozzle, have run it for over 4 hours straight and motor case only reaches 143F on a 70F day. I should have put a thermocouple inside the windings, but only put one on the outside. The fan pulls another few amps, but have never yet tripped the 50A breaker even though in excess of code guidelines. All wires in conduit, 6 AWG feeder.



1000 CFM about 70 # of air per minute, so say 15 BTU per deg F per minute if 100% effective heat transfer, or 900 BTU/hr per deg F

Motor likely only 80% efficient at 100% overload, so around 8000 BTU/hr heat loss.

Typically try to keep airflow differential of 15F or less in forced air cooling setups.
Air leaving at about 85F means 15*900 = 13,500 BTU/hr capability if 15F adhered to
73F rise means air leaving is 8000/900 = only about 9F temp rise in exhaust air, so the setup has cooling margin .

Interesting that the 48+Amps and 80% efficiency guess falls about right on 200% load level of an eyeball extrapolation of the Baldor curves that Ingenier posted.

 

[JFletcher]

Nice! It's obviously not impossible to run a 5HP motor at double its design rating (which may be quite conservative for warranty/reliability reasons), just needs some serious extra cooling. Tho your setup has run 4 hours with no apparent ill effects, I would not run it continuous, day in and out, at temperatures that may well exceed 70*F ambient (like a scrubber recirc pump on an odor control station at a WWTP).

I think a fan ducted as you have it will be much better than a water cooled case and standard air cooling. But if running it in Antarctica, you may well be able to get 3x loading w/o any supplementary cooling. I suppose as long as you dont have localized hot spots that exceed the insulation's breakdown temperature limit, you can load it as much as you want.

What would be the efficiency of a correctly sized motor?

 

[Sahib]

If there is any standard on field emergency/continuous overloading of motors, it will be more helpful as it will provide more complete set of conditions to be observed.

 

[GeorgeB]

^ Those are positive displacement pumps.

To "drive" the motor at 10HP, you need 10HP of load on it. You'd need to increase the pump flow and/or pressure w/o the other dropping a commensurate amount to do that. or drive 2 pumps...

4 gpm at 4000 psi is nominally a 10 HP load.

 

[Ingenieur]

4 gpm at 4000 psi is nominally a 10 HP load.

pump hp = (hd x gpm x 8.34) / (33000 x pump eff)
= (9227 x 4 x 8.34) / (33000 x 0.85) = 11 hp
depending on pump eff may be 90%+ making hp ~ 10

 

[kwired]

If motor has a 1.25 service factor you are a little closer to getting it done, compared to a 1.0 service factor motor.

If you have limited duty cycle you may also have a little better luck in this attempt.

 

[Jraef]

It's still about the cooling however. To double the HP you have to double the torque, which means doubling the current. Adding "water cooling loops" to the (presumably) exterior case is going to cool only the exterior case. The heat in the windings has a transfer rate designed into the stator frame and if the motor was designed for 5HP max, its unlikely that the thermal damage curve of that motor would allow 2x current flow to dissipate out from the core to the case fast enough to accommodate it.

Then on top of that is the rotor. 2x stator current will result in 2x rotor current and in the case of the rotor, heat dissipation is even less forgiving, because that is generally done via the small amount of air that can be pushed through the gap, and what can be transferred via conduction through the shaft into bearings etc., which on a pump is typically only the one end.

I work a lot with a company that makes cryogenic submersible pumps, meaning they are immersed in liquified gases like nitrogen at -320F. The pump rotors are hollow, so the cryo fluid goes straight through the center of the rotors. The motor frames are still about 1/2 of those for equivalent air cooled motors, so you could likely get a 10HP pump in the same frame size as a 5HP in that case. Water cooling is not going to do much for you though.

If you are using a C-face motor, the flange dimmensions are identical for a 215 T frame which goes to 10HP, it's just about 1-1/2" wider diameter and about 3-1/2" longer. You can't live with that?

 

[GeorgeB]

pump hp = (hd x gpm x 8.34) / (33000 x pump eff)
= (9227 x 4 x 8.34) / (33000 x 0.85) = 11 hp
depending on pump eff may be 90%+ making hp ~ 10

I've always used HP=PQ/1714 for theoretical, PQ/1500 for realistic, P in psi, Q in GPM

but then I'm a hydraulic power guy.

 

[Ingenieur]

I've always used HP=PQ/1714 for theoretical, PQ/1500 for realistic, P in psi, Q in GPM

but then I'm a hydraulic power guy.

basically the same
33000 / 2.31 psi/ft hd / 8.34 lb/gal =1714
the 1500 assumes an eff of ~0.88

33000 = 60 sec/min x 550 ft-lb/sec per hp

P = (4000 x 4) /(1500) = 10.7 hp

 

[retirede]

How about just forgetting the whole thing and drink that case of Hennessy I see in the background?

 

[junkhound]

How about just forgetting the whole thing and drink that case of Hennessy I see in the background?

Empty, all gone years ago, along with contents of the other case you see in the background.

 

[GeorgeB]

basically the same
33000 / 2.31 psi/ft hd / 8.34 lb/gal =1714

Now you've piqued my curiosity ... I've never been concerned with s.g. of the fluid, whether typical petroleum at 0.88 or so, or water-glycol at around 1.05 or synthetic esters at around 1.15. But the psi/ft hd would vary linearly with s.g. so that would drop out ...

It's been too long to balance units in my head, some work to do.

THANKS!

 

[drcampbell]

... I've never been concerned with s.g. of the fluid ...

The density of the fluid doesn't matter. Power's just pressure x volume flowrate. (in a consistent set of units)

 

[Ingenieur]

The density of the fluid doesn't matter. Power's just pressure x volume flowrate. (in a consistent set of units)

HP = HD (ft) x Q (Gal/Min) x 8.34 (lb/gal) / (60 (sec/min) x 550 (ft-lb/sec per HP) x eff)

the 8.34 lb/gal is the 'density' of water
so density does have an effect
if a gal of the material weighs 10 lb/gal the power goes up for a given HD and flow
makes sense, since you are lifting more weight up the HD height
???

 

[GoldDigger]

HP = HD (ft) x Q (Gal/Min) x 8.34 (lb/gal) / (60 (sec/min) x 550 (ft-lb/sec per HP) x eff)

the 8.34 lb/gal is the 'density' of water
so density does have an effect
if a gal of the material weighs 10 lb/gal the power goes up for a given HD and flow
makes sense, since you are lifting more weight up the HD height
???

The two formulations come out the same if the only pressure in the system is the static head pressure (due to gravity)
If you are working against a system pressure from other than gravity head you have to use the pressure times volume flow rate form.

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