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Motor Current-Horsepower(kW) Relation

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controlnovice

Electrical
Jul 28, 2004
975
Can anyone direct me to a graph showing the typical relationship between the current drawn by a motor and the load of the motor?

I'm trying to convince the company that we should be looking at the kW of the motor, not the amps because up to 25-30% of motor load the amps will be ~flat.
 
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I know. Most people have a hard time understanding that current doesn't reflect load. And it doesn't help saying it to them. They just do not believe you.

Allen Bradley has a good page on motors. See there are some diagrams showing load, current, power factor, efficiency, spped and so on.

There are also so-called Ossanna diagrams that show how current and torque are related in an induction motor. The diagram shows clearly that the relationship is far from linear - in fact, increasing load can sometimes decrease current (due to better power factor). See and figure 2 for a typical diagram,

Gunnar Englund
 
To get the power provided to the load (HPout) of a 3-phase induction motor, input kW (kWin) and motor efficiency (EFF) are required.

HPout = kWin * EFF / 0.746

The kWin can be electrically measured on the supply line; however, the efficiency (EFF) changes with the load.
The motor manufacturer can provide typical performance curves including the efficiency-load curve. These curves are different for each specific motor.
Attempts to use the equivalent circuit or circular diagrams are not practical unless you have a good knowledge of electrical machines.

The current does not change in lineal proportion to the load, but around the full load range (+/- 10% of Ifl ) it performs close to lineal proportion of load changes, as far as other parameters like voltage and frequency are constant.
 
Hello controlnovice

It is not too difficult to generate your own approximate curves.
Essentially, there are two major components of the current, the magnetising current and the load current. There are other elements also but the contribution of them to the total current under reasonable loading conditions is not large.
The magnetising current is essentially constant and does not change with load provided that the voltage is constant. The load current does follow the shaft load. The magnetising current (inductive) is at 90 degrees to the load current (resistive). There is additional resistive current which is due to losses in the motor and aditional inductive current due to the leakage reactance of the motor. If you are interested in real shaft loads which are above the open shaft load, you can consider the influence of these additional elements to be small.

For a range of shaft loads, you can draw the magnetising vector and the load vector and establish the resultant vector this is the current drawn by the motor.
The magnetising current is typically in the order of 20 - 25% for large high speed motors and 40-60% for small low speed motors. - assume that open shaft current is primarily magnetising current for a reference.
As you can see, when the load current is equal to the magnetising current, the line current increases from the magnetisng (open shaft current) by 40%. Double the load current and the line current will increase to rt5 2.236 times the magnetixsing current etc.

Best regards,

Mark Empson
 
Controlnovice;

Just a practical observation, before we all jump in technical definitions. The term "load" refers more often
to the current than the kW among the facilties personnel in plants.

You did not meniton the purpose of watching the "load" on the motor. If it is to monitor energy consumption than you are right, kW is more useful.

On the other hand if someone wants to keep an eye on the motor to see that it is not overloaded, its the acutal current (amps) that matters. The thermal capacity of the the motor winding is limited by how much current it draws and not really on kW. If there are power factor correcting caps are involved, things get little more complicated. But that is a separate discussion.

Also for a given set up and within certain range, the current will vary approximately in proportion to change in the mechanical load it is driving. So it will be helpful if you post the purpose of monitoring the 'load'.

Plus it is much easier and cheaper to monitor the current only, if it serves the purpose, than kW.



 
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