Different Horsepower, Same Load

[emoltzan]

Scenario: We have 2 motors, different horsepower, driving a vacuum pump that is loaded the same.
-Motor #1 is a 3phase, 460V, 60hp motor
-Motor #2 is a 3phase, 460V, 75hp motor

Motor #1 runs at about 66 amps and then climbs to 80 amps and eventually to 100 amps then the overloads trip out, as expected.
Motor #2 runs at about 66 amps and then climbs to 80-85 amps and does not trip out, as expected.

My main question is; with 2 different sized hp motors on equal loads, should we expect to see similar or different amp draw and why?

-Thanks

 

[TugboatEng]

The amp draw is based primarily on the load. There may be slight differences due to efficiency between a correctly sized and oversized motor.

 

[waross]

That opens a lot of questions;
Are the motors similar types? eg: ODP vs TEFC?
Do the motors drive the same pump or similar pumps?
Do the motors work independently or in tandem?
As the load on a motor increases, the reactive current increases well as the real current. This is due to the torque distorting the magnetic field in the air gap and lengthening the effective air gap. This will require more Amp-turns and thus more current.
As the motor heats up under load, the winding resistance and thus the I[sup]2[/sup]R losses will increase.
Thus as the motor heats up, the real current to supply I[sup]2[/sup]R losses as well as the reactive current to supply magnetizing current both rise. On an overloaded motor, both the motor temperature and the motor current will stabilize at higher than nameplate values.
Motor #2 stabilizes at around 80 to 85 Amps.
Motor #1, being smaller, can be expected to stabilize at higher current and temperature values than motor #2.
Motor #1 trips on overload before the current and temperature stabilize.
Disclaimer: This is a theory based on assumptions derived from motor basics. This theory may be in error.
Comments welcome.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[Compositepro]

The amp draw of a motor is not a very good indication of load due to changes in power factor. As load increases to 100% of motor rating the power factor increases. So I expect the larger motor will be operating at a lower power factor and thus slightly higher amps.

 

[emoltzan]

Both TEFC
Drive the same type of pump
Motors are independent of each other

 

[waross]

Beware of using percentages of changing values. It is often misleading.
As a motor is loaded, the VARs increase, but not as fast as the kW. Thus both the real and reactive currents increase while the ratio between real and apparent power decreases.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[emoltzan]

waross, I see you put up a ton of great comments/material. Are you open to direct PM? I would love to talk with you in the future as I am young in my career. Thanks

 

[waross]

emoltzan; Thank you for the respect. I don't mind e-mails, but is suggest that it may be more valuable to you to converse on Eng-Tips.
My strength is field work and seat of the pants engineering. There are several gurus here that have a much stronger academic background then myself. If I make a silly mistake or a typo in an IM, you will have been led astray. On the fora, one of my friends will invariably add a correction.
I may be found on-line at the motors and generators forum, the Circuit Design forum, at the Electric power and Distribution forum, at the Electrical Engineering General Discussion forum, and the Hobbies forum.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

Dear Mr. emoltzan (Electrical)
"... We have 2 motors, different horsepower, driving a vacuum pump that is loaded the same....My main question is; with 2 different sized hp motors on equal loads, should we expect to see similar or different amp draw and why?..... Both TEFC. Drive the same type of pump. Motors are independent of each other..."
Che. 1. As a guide, a 440-480V 60Hz 4-pole cage motor 60hp rated 77A; 75hp rated 96A, approx.
2. Check the thermal over-load current settings. It is likely that 60hp set at around 77A; while that of 75hp at 96A, or there about.
3. Motor #1 runs up eventually to 100A. Trips out as expected. This shows that the load#1 on (Motor+pump)#1 is >>60hp; also >75hp.
4. Motor #2 runs at about 66 amps and then climbs to 80-85 amps and does not trip out, as expected. This shows that the load#2 on (Motor+pump) #2 is >>60hp; but <75hp.
5. Attention:
a) (same TYPE of pump) but with slight [difference in the impeller size or the total distance or head]; would result to [different wattage (hp)/current (A)].
b) thermal over-load is sensitive/calibrated based on the current(A) only, irrespective of the wattage (hp) or the pf of the load. It is correct to say higher wattage needs higher current and the pf improves with higher percentage loading.
Che Kuan Yau (Singapore)

 

[waross]

Mr Che:

Beware of using percentages of changing values. It is often misleading.
As a motor is loaded, the VARs increase, but not as fast as the kW. Thus both the real and reactive currents increase while the ratio between real and apparent power decreases.

Increased power factor under load DOES NOT imply reduced reactive current.
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[edison123]

@OP

Same type of pump is not the same as same pump in the same location.

If you want one to one comparison, use both the motors in the same pump in the same location.

As others have said, the motor load current is dictated by the load, not the motor.


Bill

"As a motor is loaded, the VARs increase"

Could you please clarify that? Since the flux density is almost constant from open shaft to full load, why would the VAR's increase under load?

Muthu

http://www.edison.co.in

 

[waross]

My understanding is that the magnetic field distorts or stretches as the load increases, extending the effective air gap. A significant part of the magnetizing current goes towards magnetizing the air gap, and a little additional effective length of the air gap makes a noticeable increase in the reactive current.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

Are these liquid ring pumps? Losing water will increase the displacement and increase the motor load. Are you adding enough makeup water?
It may be a pump/operation issue rather than a motor issue.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[TugboatEng]

Waross, your last comment just enlightened me some, I had never thougjt of the increased load.as water is lost. My experience with liquid ring pumps is for condensers. Despite the increased volume of pump flow, the mass flow rate.from the condenser won't change based on pump displacement so the power shouldn't change, either.

 

[Compositepro]

Waross, my reasoning is that if the load is equal then theoretically the kW draw of both motors would be the same. The power factor of any motor decreases with decreasing load (as a percentage of rated load). So the larger motor will be operating at a lower power factor, and thus higher current. Nothing complicated. So I do not see your point about vars.

An often useful technique in analyzing problems is to take things to extremes. If you have a 10 hp load driven by a ten hp motor vs. a 100 hp motor, which will draw more current?

 

[waross]

So the larger motor will be operating at a lower power factor, and thus higher current.

Fun with numbers.
A lightly loaded motor draws 10 Amps reactive current and 10 Amps real current. The PF is 50%
The load is increased and the motor draws 15 Amps reactive current and 100 Amps real current. The PF has increased to 99% despite the reactive current increasing by 50%.

Beware of using percentages of changing values. It is often misleading.
As a motor is loaded, the VARs increase, but not as fast as the kW. Thus both the real and reactive currents increase while the ratio between real and apparent power decreases.

Further, the sweet spot for motor efficiency is often around 75% of full load.
I would expect a 75 HP motor to be more efficient than a 60 HP motor at anything above around 56 or 57 HP load.
As the applied load increases above 60 HP, and considering the increase in I[sup]2[/sup]R loses, I would expect the larger motor to become progressively more efficient.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[Compositepro]

In these motors the load is the same, so the real current will be the same.

 

[waross]

The real current includes the losses which will be greater with the smaller motor.
The motor overloads will see the total current, not just the real component of the voltage.
Heating losses are related to the square of the total current, not just the real component of the current.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

Dear all,
I am of the opinion that the difference in the [current value] between the 60hp and 75hp motor with same load :
a) the term "same load and type of pump" should be looked into. Presumably "same" load but actually Load#1, Load #2 may? differ slightly in length, head, valve opening, flow rate etc. The "same" type pumps (#1 and #2) may? actually of slight difference in the impeller dimension/characteristics etc. A better way is to test with [a pump and a load]; change only the motor(hp) size.
b) unlikely due to [pf] as in both cases the load are >90%, where the pf increment with increase in loading is diminishing.
c)unlikely due to efficiency as in both cases the load are >90%, where the increment in efficiency is diminishing.
Che Kuan Yau (Singapore)

 

[Gr8blu]

Emoltzan: The basic relationship is that the process is regulated on either volume (aka “flow”) or on pressure. Both of these are a direct result of developed torque, which in turn is directly proportional to “real” (active) current. If the command is to create a desired process condition, you are effectively demanding a motor line current level to maintain the process state.

If the process demands more current than the machine rating, the machine is “undersized” in that it does not have enough capability at the desired speed /power point. This can be a result of one or more design decisions: power factor, efficiency, material cost, labor cost, and loss distribution (e.g. how to apportion losses across core, copper, friction, windage, magnetization, and eddy/stray).

A very small change in driven load will result in noticeable current demand on a given machine, particularly where the nameplate “power” is low (i.e. a 3 hp swing is significant on a 100 hp design but not on a 2000 hp design). These changes can be a result of impeller design/wear, clearance to housing, turbulence factors related to housing smoothness, slight differences in amount of fluid in the system, the routing (and therefore pressure drop) through the piping, guide bearing losses, the density of the fluid being moved, differing restrictions at inlet or outlet, and so on. That’s all external to the differences inherent to the two drive motors being tested! (Hence the reason that apples-to-apples requires mounting the two machines in exactly the same fashion to the same driven load.)

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