DC permanent Magnet Motor Theory Questions

[PCS74]

Hi guys,

I need some help with my understanding of permanent magnet DC motor operation. I am building a spreadsheet to model motor performance. I'm an ME and my electrical knowledge is poor so I have started with the theoretical side to help develop my understanding. I have a few good books on the way, but would like to start getting feedback here as well so any help you can provide would be greatly appreciated.

The motor I am attempting to model is the AGNI 95R which can be reviewed here:

http://www.agnimotors.com/home/index.php?option=com_content&task=view&id=5&Itemid=60

My spreadsheet is attached.

My first question is with current draw. Based on the information I have been supplied, both measured and derived from manufacturer performance figures, this motor has an armature resistance of 0.0125 ohms. This seems very low to me. Therefore, at 60v terminal voltage and low shaft speeds, hence low counter emf voltages, I see very high current draws. Theoretically speaking, is this correct or am I overlooking something?

If I was correct above, my second question is what occurs when you can't supply the required current? Assume I will only be able to provide this motor with 240 amps continuous. During an unloaded acceleration run, will it draw the full 240 amps from 0 rpm to the rpm point where the terminal voltage and counter emf voltage combine to require less current than 240 amps?

I'm sure I will have more questions, so thanks in advance for your time and help.

Chris Skarzenski

 

[waross]

I always think of the difference between a DC motor and a DC generator as just a few volts or a few RPM. That is for the types of motors that are suitable for use as generators, but I find that the exercise helps to get a feel for other motor applications.
Suppose a generator is running at 1000 RPM and generating 100 Volts. It is now connected to a 100 Volt source. Nothing happens, there is no current flow. Now increase the voltage a few volts. Current will flow in a direction to increase the speed of the machine. Decrease the voltage a few volts. Now the machine will force current through the source.
Likewise, increase the RPM a few Rs and the machine will generate and force current through the source. Drop the RPM a little and the machine will draw current from the source and try to maintain speed.
The next question, How much.
Assume the source is a battery or another generator (large enough to be stable with the changes in load).
At normal speeds things are pretty linear. 1000 RPM / 100 Volts gives 10 RPM/Volt.
If we change the RPM by 10 R, the voltage will change by one volt.
The back EMF 1 volt different from the applied voltage and the effective voltage driving current through the armature will be 1 Volt.

this motor has an armature resistance of 0.0125 ohms

One volt will cause about 80 amps to flow and the motor. How do we change the RPM? We change the load. If we apply enough load to drop 10 RPM the current will increase by about 80 amps. 80 Amps x 100 Volts (Applied voltage) is 8000 Watts. A load of one horsepower will cause the voltage to drop a little less than 0.1 Volt.
This is in a different format than you have presented, but it may give you a feel for motor characteristics.
I hope that you can follow my explanation.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[vannyelec]

waross

Thanks for the excellent information but I have another question for you. We have a 750 VDC, 1000 HP, 1150 rpm motor that is used to test a gear box under a load. However, our customer was curious about running our test at say 25% speed for a sustained period of time without the test gear box being under any load.

I am extremely new to this field and my gut tells me that simply supplying 25% of the voltage to the motor would cause the motor rotation to be 25% of the max speed but I feel like that is too simple and I may be overlooking something.

Any input is appreciated and thanks in advance.

KEV

 

[Skogsgurra]

Your Excel sheet looks OK to me. I really appreciate the use of SI units.

Use that spreadsheet to see what happens when you change DC voltage and load on that 1000 HP motor.

Warning: I haven't checked the numbers in detail. But the principles look right.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[itsmoked]

While the speed will change in a mostly linear fashion you can't really assume 25% voltage will dish up 25% speed.

I suggest you put a shaft speed devise on somewhere. That will allow you to actually know what you're doing. An alternative would be a strobe to "stop" the shaft's apparent motion and thereby determine the speed.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[waross]

Let's consider a motor on a load that will overhaul and tend to overspeed the motor.
There will be a speed at which the back EMF will equal the applied voltage and the current will be zero. Slower and the back EMF will drop and the m,otor will start to draw current and develop positive torque. Faster and the back EMF will increase above the applied voltage and the motor will become a generator and try to export energy.
This relationship is quite linear. You may use this as a rule of thumb but be aware that there may be a few percent error.
Comparing a fully loaded motor at 100% voltage to a lightly loaded motor at 25% voltage will introduce a few percent error. For the majority of applications this error may be ignored. In some of the remaining applications the automatic control systems will correct the error.
If you look at the percent difference between no load speed and full load speed, the errors may be expected to be contained within that envelope.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[vannyelec]

Thank you much gentlemen. Like you said, I expected some error to be present but just wanted to make sure I wasn't looking over something major due to my lack of experience.

KEV

 

[waross]

Another point to be aware of; Small inefficient motors will have more error when estimating speed based on the applied voltage. The motor losses are responsible for much of the error so an inefficient motor with higher percentage losses will have a greater speed error at reduced voltage. When you get up to integral horsepower motors the efficiency is usually pretty good and the error is small.
I would expect that your 100 Hp motor will have good efficiency and the speed error will be small.
The no load speed/voltage linearity on a large motor should be excellent.
If you have a 5% speed drop between no load and full load, you may expect close to the same drop in RPM between no load speed and the speed at rated current at reduced voltage.
If the speed drops from 1200 RPM at no load to 1140 RPM at full load, expect the speed to drop from close to 600 RPM at no load to about 540 RPM at rated current. In both examples the speed from no load to full current has dropped about 60 RPM, but this is 5% at full voltage and 10% at 1/2 rated voltage.
This is still a rule of thumb subject to some assumptions and still ignoring some small effects such as the voltage drop across the brushes, but it will reduce the error inherent in using a straight linear relationship between voltage and speed.
If you need an exact speed, follow itsmoked's advice and monitor the actual speed. Another tip; for small speed adjustments it may be easier to trim the field voltage than the armature voltage. A decrease in field voltage or field strength will cause an increase in motor speed. This technique will work fine at light loading. Be wary of field weakening with a fully loaded motor and particularly a motor that experiences overload conditions.
Another point, when we are controlling the speed of a motor by dropping the voltage, the field should be held at full rated voltage. Drop only the armature voltage.

Bill
--------------------
"Why not the best?"
Jimmy Carter