Field weakening in dc motors 9

[edison123]

If a run 100 HP/900 RPM dc motor at 1800 RPM by field weakening, can I still get 100 HP output at 1800 RPM ?
I know the torque at 1800 RPM will be 50% of 900 RPM torque but what about the motor output ?

 

[ScottyUK]

Hi Edison,

I think you are on the right track. But it early here, and my brain is only in second gear!

 

[jbartos]

Suggestion: A proper way to go about it is to follow the motor characteristics to avoid the motor damage. Try to obtain the HP versus speed and torque versus speed motor characteristics (including 1200RPM) from the manufacturer.

 

[TomG33]

There is no simple answer to this issue. There are a few different things to consider.

1 The type of load
2 The type of Field COntrol
3 The type of Armature current limit
4 THe type of speed characteristic required
Each of these will have an impact on how the control and the motor will respond to different situations.
I will write a little about each of these.

1 Types of Loads
There are a few different types of load.

Consider a fan , the faster it goes the more torque required to drive it. thus as the speed goes up, the power, which is a function of both speed and torque, will go up in some exponent, proportional to speed.

you could have a Conveyor type drive, where basically the torque required to drive it remains constant throughout the operating speed range, in this case, with increasing speed, the power required to drive it goes up linearly.

Another type of load could be in the form of a cyclic load ( mining excavators or batch processing) where the load depends on how much product is collected in each pass. It could vary either up or down for each cycle.

The motor and drive system will respond to each of these types of loads in different ways.

2 Type of Field control.

There are a few different ways to control the motor when it is operated above it's Base Speed.

The first is fixed field. Here the field is set below rated until you achieve the desired top motor speed at the top motor volts. You can have either a voltage or a speed regulator. In either case the motor volts will go up fairly linearly with speed. the problem with this is that the weakened field will result in lower motor torque for any given amount of armature current. (You need more armature current to produce the necessary torque)

The second is to have a "Crossover Regulator" Here you use a tachometer to measure motor speed, thus these are predominantly speed regulators. Consider starting an unloaded motor slowly and then bringing it up to top speed. When the motor is initially started, the field is at maximum nameplate value. AS the speed reference is increased, the voltage will increase with the speed until you reach nameplate voltge. at this point, the motor will be operating at "Base Speed". After this the regulator changes in some way, so that any further increase in speed is achieved by weakening the field. The motor nameplate (or manufacturers data sheet)will specify the level of field to give top speed of the motor. Sometimes the maximum safe speed is also specified. The point at which the control "crosses over" to field control is usually set as a voltage.

3 Types of Armature current control.

You must keep in mind that motor will only provide power that a load can absorb, So the power delivered is controlled by the load and not the motor itself. Overloading a motor will simply mean that more current is drawn to match the load. THe control should be set up to limit the amount of current to the motor. The most simple form is to have a hard currnet limit at some multiple of motor nameplat current (150% is a good starting point)

Another method of Current limiting is to use a " Tapered Current Limit" Here there is a hard current limit at some acceptable value for a motor at stand still, AS the speed or voltage increases, the limit is "Tapered away to a lower value.


Most motors can handle some form of overload (cyclic or occasional) but typically the RMS current of the motor, over say a 10 minute period, should not exceed the nameplate value. So if you are running a continuous process, You would not operate it above the motor rated current, whereas if you had some sort of batch process, of a cyclic nature, Large overloads in current are permissable, as long as there is a cooling time in the cycle to compensate the overloads.

Normally it is the responsibility of the equipment manufacturer to control his process so it does not overload the motor. If you wish to maintain set speed, the only way to reduce the motor current is to remove some of the mechanical load.

4 THe type of speed characteristic required

In continuous processes it is usually desirable to keep the process at uniform speed independant of load. In these situations a fixed hard current limit is the way to go because this allows the motor to operate in the overload area for a short time and still maintain speed. Some sort of Timed overload functions are included in the control to prevent a process failre from overloading the motor to destruction.

In, say, A mining application, Where stalling a motor to a standstill is a normal operating condition, The tapered current limit is the best method. here, Constant speed is not critical, but the fastest possible speed for any given load is the requirement.

Keeping all of these things in mind, lets choose a particular type of load, Field Control and armature current Control to answer your question about increasing load in the field weakened area.

Lets choose a crossover regulator, in a converyor type application with a fixed current limit at 150% of rated.

Initially you are at 75% load, so at full volts and rated full field the current will be pretty close to 75% of rated current at the motor base speed of 900 as per your example.

Now you wish to increase your speed to 1200. To achieve this , the field is weakened. Because the conveyor is a constant torque type load, the Torque will be almost constant. To provide the torque at the higher speed, more armature current is necessary.

so, as the field is weakened, the speed increases, and the armature current increases. At some point, you will reach rated motor current, and your motor will then be operating at rated power (100 HP). This may or may not be at your 1200 RPM top speed. Assuming it is not at 1200 rpm, and you still wish to go there, further field weakening will result in armature current beyond the rated current. This is normally ok for a short while, but if you wish to have your motor last a long time, You are going to have to reduce the load somehow. For a conveyor you would need to reduce the amount of product going onto the belt at it's feed end.

So you see, in order to get the motor up to your top speed and still keep the motor operating within it's design limits, You will have to reduce the load on it.

Consider this from the motor point of view.
The volts are constant, and the armature current is increasing, so power, ( Volts times Current) will increase with the increasing speed. This makes sense from the Load's point of view as well. You are initially moving the same amount product, but moving it faster , so The torque is constant, but the speed is increasing. so Power, (Torque times speed) will go up.

This, of course, is only true until you reach the rated motor current. If the belt feed is reduced to keep the current at rated, The load torque will go down, this coupled with the increasing speed, will result in a constant hP as long as the load is adjusted to keep the motor Armature current at motor rated.

Typically we do not set the current limit at 100% but allow it to go over for a short time, thus in a transient situation, the power may be momentarily over the motor rated.

If the armature current limit was set hard at 100%, your attempts to reach your desired speed of 1200, would be cut short at the speed where the current first reached the 100% value. Less load would allow you to faster and more load would slow it down.

This is very draw out and I appologize for the lenght of this but as you see, there are lots of variables at play here.

 

[edison123]

Dear Tom,

That was an excellent explanation which cleared my doubts quite emphatically. Not only was your explanation lucid, it was grounded solidly in basics. Thx for taking the time to post this. I look forward to reading your posts in this excellent forum.

Regards,
Kumar

BTW, you get a star.

 

[jbartos]

Eng-tip: Visit
file:///C:/doc/Paper/Motor/DC%20Motor%20Control_files/DC%20Motor%20Control.htm
for:
Field weakening:
a. Shunt DC motors
b. Series DC motors
The original posting does not differentiate between Shunt and Series motors. Is that posting for educational purposes?

 

[rhatcher]

A star for TomG33's excellent post.

 

[jbartos]

Correction: My previous posting link should be since the posted one has become unavailable on the web:

http://www.powerqualityanddrives.com/torque_constant_horsepower/

for:
A replacement DC motor could be run in armature control from zero to base speed of 1750 RPM and above base speed by field weakening.
When exceeding rated base speed in the field weakening range of a DC shunt wound motor, the torque developed will be inversely proportional to the speed, which is a true constant horsepower. Torque is therefore proportional to 1/RPM.
The torque available from the DC motor in the field weakening range will be proportional to 1 / RPM. See Figure 2.

 

[TomG33]

Jbartos,
I am not sure of the point you are trying to make here but You must remember that these curves and equasions you quote are all calculated for given motor designs and tell what Torque and speed the motor will be at for any given combination of field strength, Voltage, load torque and speed.
They do not imply that the motor will limit the power delivered, to follow any curve,

The power delivered by an unregulated motor is purely a function of the power demanded by the load.

If you are running your motor at maximum power at top weakened field speed, and then hit it with a step increase in Load, It will slow slightly, The COunter EMF will fall and thus the armature current will increase, (torque will increase as well) and thus the delivered power (Speed * Torque) will go up.

As I said previously, Most motors can handle quite a bit of overloading on a short time basis. On the long term we must intervene in some way to get acceptable life out of the motor.

The problem is two-fold.

First the increased current will put thermal stress on the winding insulation (shorter motor life) and

Secondly, the extra current at the higher speed will push the commutating ability of the commutator to it's limit, resulting in extra sparking initially and if sustained, there is the possibility of a commutator "Flash over", which is a glorified way of saying that the current in each of the armature winding segments is not successfully reversed while the two commutator bars are under the brush. The sparking (caused by the reversal without the shorting brush) can be carried around by the fast turning commutator and if allowed to continue it can join up with tne next Brush arm, resulting in a dead short between the two brush arms ( one + and the other -)

Both the Driving generator or Static bridge and the turning motor armature with it's load inertia can feed energy into this short and the results can be Quite Catastropic.

This outcome is not really desirable so we put controls on the motor drive to limit it's current in the overload condition. Usually we do calculations for most load conditions we would normally expect to see and our aim is to Keep the RMS current to within rated current over a 10 minute period, and to keep the motor within it's commuataion limits (which are published by the factory) in the high speed motoring and regeneration quadrants.

This is where we use the Equasions and the calculations you quote, or reference to, To make sure the motor works correctly within it's design limits and that the load can be successfully driven by it.

No Motor will follow the constant horsepower line on it's own accord unless some sort of control keeps it there.

Torque is always a function of the load unless some type of current limit is brought into play.

Tom

 

[jbartos]

Comment on the previous posting:
Jbartos,
I am not sure of the point you are trying to make here
///Just providing eng-tips to the original posting, if you have not noticed. I was not addressed to you.\\ but You must remember that these curves and equations you quote are all calculated for given motor designs and tell what Torque and speed the motor will be at for any given combination of field strength, Voltage, load torque and speed.
///Self-evident.\\They do not imply that the motor will limit the power delivered, to follow any curve,
///Not quite. There are curves or characteristics representing the system integration intended performance. In some instances, if the curves are exceeded the equipment protective devices trip or limit the device performance.\\The power delivered by an unregulated motor is purely a function of the power demanded by the load.
///Yes, until the motor burns in the worst case.\\If you are running your motor at maximum power at top weakened field speed, and then hit it with a step increase in Load, It will slow slightly, The COunter EMF will fall and thus the armature current will increase, (torque will increase as well) and thus the delivered power (Speed * Torque) will go up.
///The increased armature current may increase heating and commutator sparking.\\As I said previously, Most motors can handle quite a bit of overloading on a short time basis.
///Yes, when new. When old who knows?\\ On the long term we must intervene in some way to get acceptable life out of the motor.
///Agree, that is why I presented link with figures.\\The problem is two-fold.

First the increased current will put thermal stress on the winding insulation (shorter motor life) and
///Yes, that is why I presented the link with Figures.\\Secondly, the extra current at the higher speed will push the commutating ability of the commutator to it's limit, resulting in extra sparking initially and if sustained, there is the possibility of a commutator "Flash over", which is a glorified way of saying that the current in each of the armature winding segments is not successfully reversed while the two commutator bars are under the brush. The sparking (caused by the reversal without the shorting brush) can be carried around by the fast turning commutator and if allowed to continue it can join up with tne next Brush arm, resulting in a dead short between the two brush arms ( one + and the other -)
///Agree, nothing new.\\Both the Driving generator or Static bridge and the turning motor armature with it's load inertia can feed energy into this short and the results can be Quite Catastropic.
///Agree, nothing new. Nothing posted to the contrary.\\This outcome is not really desirable so we put controls on the motor drive to limit it's current in the overload condition. Usually we do calculations for most load conditions we would normally expect to see and our aim is to Keep the RMS current to within rated current over a 10 minute period, and to keep the motor within it's commuataion limits (which are published by the factory) in the high speed motoring and regeneration quadrants.

This is where we use the Equasions and the calculations you quote, or reference to, To make sure the motor works correctly within it's design limits and that the load can be successfully driven by it.
///Agree. The original posting includes data, apparently with the intent to have some analytical eng-tips presented.\\No Motor will follow the constant horsepower line on it's own accord unless some sort of control keeps it there.
///Yes, self-evident.\\Torque is always a function of the load unless some type of current limit is brought into play.
///Yes, self-evident. This is the Eng-Tips Forum.\\Tom

 

[Warpspeed]

Great read Tom, well worth another star.

 

[jbartos]

Suggestion: Visit

http://www.trainweb.org/railwaytechnical/tract-01.html

for more info on DC motor field weakening. There are probably similarities between the railroad locomotives and mining locomotives. Nothing has been addressed about this from mining expert Tom yet.

 

[edison123]

jb,

Tom has answered my original and subsequent posts quite clearly (which is also appreciated apparently by many, who are really interested in learning something useful). I don't see any point in your nitpicking over his post other than to confuse the issue. If you can't appreciate a good post, just leave well enough alone.

 

[jbartos]

Comment on the previous posting: Thank you for letting me know. I feel that I have posted a legitimate posting without any "nitpicking." In fact, my posting was not understood by Tom; therefore, I provided answers. What is wrong with that? The original posting should have been more specific by providing guidelines to volunteers what to post.

 

[Modula2]

edison123, I want to hear everyone's opinion, not just those that are approved by you. Then, I'll decide for me. Especially, I appreciate the input of jBartos, who may be the overall most knowledgeable person herein this NG.