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Field weakening in dc motors 9
- Thread starter edison123
- Start date
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: Reference:
Donald G. Fink, H. Wayne Beaty "Standard Handbook for Electrical Engineers," 14th Ed., McGraw-Hill, 2000.
Rearrange Equation 8-15 on page 8-21, namely,
D**2 x L = (watts x 6.8 x 10**8)/[(r/min) x Bs x FI x q]
where
q = Z x Ic / (pi x D)
D = Armature Diameter
L = Armature Gross Core Length
Bs = Main Pole Air Gap Density in Maxwells
FI = Ratio of pole arc to pole pitch
Ic = Current per Armature Conductor in Amps
The Watts versus Bs relationship is directly proportional
Donald G. Fink, H. Wayne Beaty "Standard Handbook for Electrical Engineers," 14th Ed., McGraw-Hill, 2000.
Rearrange Equation 8-15 on page 8-21, namely,
D**2 x L = (watts x 6.8 x 10**8)/[(r/min) x Bs x FI x q]
where
q = Z x Ic / (pi x D)
D = Armature Diameter
L = Armature Gross Core Length
Bs = Main Pole Air Gap Density in Maxwells
FI = Ratio of pole arc to pole pitch
Ic = Current per Armature Conductor in Amps
The Watts versus Bs relationship is directly proportional
-
3
- #4
The answer is yes. In field weakening, you are in the "constant HP range".
While torque reduces by 2, the speed increases by 2. And so, the HP remains the same....
HP = T x N /5252
at 900 rpm torque is 584.55 ft-lbs
at 1800 rpm torque is 291.77
plug these numbers into the equation for HP above and you see that HP is the same at 1800 RPM as it is at 900 rpm ..
While torque reduces by 2, the speed increases by 2. And so, the HP remains the same....
HP = T x N /5252
at 900 rpm torque is 584.55 ft-lbs
at 1800 rpm torque is 291.77
plug these numbers into the equation for HP above and you see that HP is the same at 1800 RPM as it is at 900 rpm ..
- Thread starter
- #5
jb & jO,
I see now that I should have rephrased the question.
Yes, under field weakening, the HP remanins the same due to the equations given by you. I wanted to know that if the connected load requires more HP at the higher speed, will the motor deliver it or simply stall or spark viciously or overheat? Have you had experience with such an operation ?
Thx for yr responses.
I see now that I should have rephrased the question.
Yes, under field weakening, the HP remanins the same due to the equations given by you. I wanted to know that if the connected load requires more HP at the higher speed, will the motor deliver it or simply stall or spark viciously or overheat? Have you had experience with such an operation ?
Thx for yr responses.
electricuwe
Electrical
You will not be able to get higher power than 100hp at higher speed continously because you are operating at the limit of armature current.
Field weakening might only be helpful if your load doesn't need the full power at nominal speed
Field weakening might only be helpful if your load doesn't need the full power at nominal speed
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion to the original posting:
Reference:
Donald G. Fink, H. Wayne Beaty "Standard Handbook for Electrical Engineers," 14th Ed., McGraw-Hill, 2000.
Equation 8-15 on page 8-21, namely,
Equation 1:
D**2 x L = (watts x 6.8 x 10**8)/[(r/min) x Bs x FI x q]
where
q = Z x Ic / (pi x D)
D = Armature Diameter
L = Armature Gross Core Length
Bs = Main Pole Air Gap Density in Maxwells
FI = Ratio of pole arc to pole pitch
Ic = Current per Armature Conductor in Amps
By rearranging Equation 1:
Watts=D**2 x L x (r/min) x Bs x FI x q /(6.8 x 10**8)
it is seen that for Bs=0, Watts=0
I.e. if the Bs was isolated from the rotor or removed, then the Watts will be zero since r/min cannot increase to infinity.
Therefore, it will be possible to run the motor at 100HP at 1800RPM if this is within the motor "constant HP range" only as indicated in JOmega posting, i.e. (Bs900)/900RPM=(Bs1800)/1800RPM=constant…………….Equation 2
If Equation 2 does not hold, then the HP will be lowered somewhat.
Reference:
Donald G. Fink, H. Wayne Beaty "Standard Handbook for Electrical Engineers," 14th Ed., McGraw-Hill, 2000.
Equation 8-15 on page 8-21, namely,
Equation 1:
D**2 x L = (watts x 6.8 x 10**8)/[(r/min) x Bs x FI x q]
where
q = Z x Ic / (pi x D)
D = Armature Diameter
L = Armature Gross Core Length
Bs = Main Pole Air Gap Density in Maxwells
FI = Ratio of pole arc to pole pitch
Ic = Current per Armature Conductor in Amps
By rearranging Equation 1:
Watts=D**2 x L x (r/min) x Bs x FI x q /(6.8 x 10**8)
it is seen that for Bs=0, Watts=0
I.e. if the Bs was isolated from the rotor or removed, then the Watts will be zero since r/min cannot increase to infinity.
Therefore, it will be possible to run the motor at 100HP at 1800RPM if this is within the motor "constant HP range" only as indicated in JOmega posting, i.e. (Bs900)/900RPM=(Bs1800)/1800RPM=constant…………….Equation 2
If Equation 2 does not hold, then the HP will be lowered somewhat.
edison - isn't doubling the design speed of the motor risky?
Electrically, the commutation is set up (i.e. design of the interpoles and compensation windings if fitted) to cover the range of operating speed and currents. You could be operating beyond these limits, although at least this can be clearly judged by viewing the commutation.
Mechanically, the commutator and the banding of the end-windings are designed for a particular speed. If the machine is self-ventilated with an armature mounted fan, this may not be adequately rated to withstand centrifugal forces. Unfortunately operating beyond the acceptable limit isn't obvious until it is too late. Incidentally, a good guide for the maximum peripheral speed of a commuatator for a machine of that size is 10,000ft/min - for a well designed commutator (a dc traction motor say).
Ultimately it is for the design office of the motor manufacturer to say whether this is acceptable. Having said all that, 900rpm doesn't sound that fast!
Electrically, the commutation is set up (i.e. design of the interpoles and compensation windings if fitted) to cover the range of operating speed and currents. You could be operating beyond these limits, although at least this can be clearly judged by viewing the commutation.
Mechanically, the commutator and the banding of the end-windings are designed for a particular speed. If the machine is self-ventilated with an armature mounted fan, this may not be adequately rated to withstand centrifugal forces. Unfortunately operating beyond the acceptable limit isn't obvious until it is too late. Incidentally, a good guide for the maximum peripheral speed of a commuatator for a machine of that size is 10,000ft/min - for a well designed commutator (a dc traction motor say).
Ultimately it is for the design office of the motor manufacturer to say whether this is acceptable. Having said all that, 900rpm doesn't sound that fast!
- Thread starter
- #9
ukpete,
The dc motors meant for variable speed in the ratio of 1 to 3 are designed mechanically to take care of the stresses. My case is a variable speed 100 HP motor (with a base speed of 900 RPM) designed to run at 1800 RPM by field weakening. As discussed in the baove threads, the output, as per theory, is the same 100 HP at 1800 RPM. My doubt was whether the motor can deliver more than 100 HP at 1800 RPM (in case it is required). Or will it stall or result in vicious sparking ? or simply overheat (which I can remove by improved cooling which is also enhanced by the higher speed).
The dc motors meant for variable speed in the ratio of 1 to 3 are designed mechanically to take care of the stresses. My case is a variable speed 100 HP motor (with a base speed of 900 RPM) designed to run at 1800 RPM by field weakening. As discussed in the baove threads, the output, as per theory, is the same 100 HP at 1800 RPM. My doubt was whether the motor can deliver more than 100 HP at 1800 RPM (in case it is required). Or will it stall or result in vicious sparking ? or simply overheat (which I can remove by improved cooling which is also enhanced by the higher speed).
Edison123
I've designed systems with motors having up to 6:1 field weakening (winders/unwinders/coilers/uncoilers, etc.)
Trying to get more than design HP from a motor in the weakend field range, will move you towards a problem in commutation.
There are many factors to consider which prevents a definitive yes or no answer from being given; not the least of which is the design of the motor.
Some other considerations of overloading the motor is the higher current density at the brushes over time. (I assume you are not talking about short-term overloading but rather, continuous overload.)
The machine manufacturer is the best source to learn of the motor's ability to operate at 2:1 field weakend and overloaded by some defined but unstated percentage.
If you want to find out empiracally, take it up in speed and increase the load while observing the action at the brushes.... you'll soon see whether or not you'll have problems. (Might want to wear welders goggles to prevent the flashing and arcing from damaging your vision![[dazed]](/data/assets/smilies/dazed.gif "[dazed] [dazed]")
I've designed systems with motors having up to 6:1 field weakening (winders/unwinders/coilers/uncoilers, etc.)
Trying to get more than design HP from a motor in the weakend field range, will move you towards a problem in commutation.
There are many factors to consider which prevents a definitive yes or no answer from being given; not the least of which is the design of the motor.
Some other considerations of overloading the motor is the higher current density at the brushes over time. (I assume you are not talking about short-term overloading but rather, continuous overload.)
The machine manufacturer is the best source to learn of the motor's ability to operate at 2:1 field weakend and overloaded by some defined but unstated percentage.
If you want to find out empiracally, take it up in speed and increase the load while observing the action at the brushes.... you'll soon see whether or not you'll have problems. (Might want to wear welders goggles to prevent the flashing and arcing from damaging your vision
- Thread starter
- #11
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion to edison123 (Electrical) Oct 31, 2003 marked ///\\jb & jO,
I see now that I should have rephrased the question.
///Any time.\\Yes, under field weakening, the HP remanins the same due to the equations given by you. I wanted to know that if the connected load requires more HP at the higher speed,
///It depends. Namely, the load torque-speed characteristic usually increases with increases with increased speed, however, not always. It may stay constant or decrease, which are rare applications.\\
will the motor deliver it or simply stall or spark viciously or overheat?
///It depends on the design margin the motor manufacturer imbeds into the motor. Normally, sparking will increase first, and later on, either protective device trip or motor stalls with improper protective device.\\ Have you had experience with such an operation ?
///Usually, one experiences the aftermath, unless one is in some university laboratory and watches the motor being overloaded.\\\
I see now that I should have rephrased the question.
///Any time.\\Yes, under field weakening, the HP remanins the same due to the equations given by you. I wanted to know that if the connected load requires more HP at the higher speed,
///It depends. Namely, the load torque-speed characteristic usually increases with increases with increased speed, however, not always. It may stay constant or decrease, which are rare applications.\\
will the motor deliver it or simply stall or spark viciously or overheat?
///It depends on the design margin the motor manufacturer imbeds into the motor. Normally, sparking will increase first, and later on, either protective device trip or motor stalls with improper protective device.\\ Have you had experience with such an operation ?
///Usually, one experiences the aftermath, unless one is in some university laboratory and watches the motor being overloaded.\\\
- Thread starter
- #17
- Thread starter
- #18
jb,
"Namely, the load torque-speed characteristic usually increases with increases with increased speed, however, not always. It may stay constant or decrease, which are rare applications".
Doesn't the speed drop with increasing load torque ? After all that is called the stability. or am I misunderstanding your post ?
"Namely, the load torque-speed characteristic usually increases with increases with increased speed, however, not always. It may stay constant or decrease, which are rare applications".
Doesn't the speed drop with increasing load torque ? After all that is called the stability. or am I misunderstanding your post ?
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion to the previous posting: A distinction shall be made between the motor torque Tm curve and load torque Tl curve. I was referring to the Tl not Tm. It is obvious that the fan will draw more current (higher electricity bill) when switch is turned fan to the higher speed. The same vacuum cleaner, etc., which are more ubiquitous motor-load applications. Visit
etc. for more info (many torque curves)
etc. for more info (many torque curves)
- Thread starter
- #20
To continue this thread, pls consider the following:
A 100 HP/900 RPM motor is loaded only to 75 HP at the base speed with constant armature voltage. Now, I increase the speed to 1200 RPM and the connected load requires more HP (but within 100 HP) at this higher speed.
At this field weakened speed and with the constant armature voltage (dictated by the Drive), I see an increased armature current to meet the new load demand. Due to the increased current and reduced field flux, the torque remains the same (as before at the base speed). With such a constant torque and an increased speed, applying the above HP-Torque-Speed equation, the output HP increases to meet the new load requirement at 1200 RPM.
Is the above scenario right or am I being totally way off the base ?
A 100 HP/900 RPM motor is loaded only to 75 HP at the base speed with constant armature voltage. Now, I increase the speed to 1200 RPM and the connected load requires more HP (but within 100 HP) at this higher speed.
At this field weakened speed and with the constant armature voltage (dictated by the Drive), I see an increased armature current to meet the new load demand. Due to the increased current and reduced field flux, the torque remains the same (as before at the base speed). With such a constant torque and an increased speed, applying the above HP-Torque-Speed equation, the output HP increases to meet the new load requirement at 1200 RPM.
Is the above scenario right or am I being totally way off the base ?
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