I am trying to visualize what happens when a motor drops a phase. Let's imagine a 3-phase motor that runs 100 A per phase. If one of the phases gets dropped, is it going to now run at 66.7 A or 57.7 A on the remaining two phases? I know an open delta transformer configuration only delivers 57.7% of the power of a closed delta configuration, and it seems to me like the motor, basically a moving transformer, should behave the same way.
Dropped Phase
- Thread starter bcorbin
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hardworkingstiff
Senior Member
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- Wilmington, NC
I believe the amperage will behave differently depending on if the motor windings are connected in a Delta or Wye configuration.
If it is in a Delta, then if you drop a phase, there is still a path for current flow from the two active phases through the two windings connected to the dropped phase (the two windings will be connected in series between the two active phases).
If it is connected in a Wye configuration, then no current will flow through the winding connected to the drop phase.
I wouldn't know how to calculate what the amperages would be, but I am curious as to how to do that.
If it is in a Delta, then if you drop a phase, there is still a path for current flow from the two active phases through the two windings connected to the dropped phase (the two windings will be connected in series between the two active phases).
If it is connected in a Wye configuration, then no current will flow through the winding connected to the drop phase.
I wouldn't know how to calculate what the amperages would be, but I am curious as to how to do that.
davidr43229
Senior Member
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- Columbus, Oh
On a 3 phase motor, the motor will change pitch and each of the 2 legs will go up approx 170%-193%, ultimately kicking out the overloads. If your overloads are oversized, damage will occur!@
Just my $.02
Just my $.02
winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
What happens to the motor circuit loading during phase loss will depend on the way the motor is connected and the mechanical characteristics of the motor load.
If the motor continues to drive the load, and if we presume that the motor remains near synchronous speed, then you will have essentially the same mechanical kW going to the load, and thus would expect at least as much electrical power to be supplied to the motor (efficiency certainly won't go up if you drop a phase under load). So if the motor continues to drive the load and you lose a phase, you would expect the remaining leads to supply the total power, and the current drawn on the remaining phases to increase.
It is possible that the motor will stall under the mechanical load, in which case the current drawn on the remaining phases will increase substantially, since a stalled motor draws far more than its normal full load current.
On the other hand, if there is no mechanical load on the motor, then it will probably continue to spin with no difficulty at all. In fact, if you run an unloaded three phase induction motor via two terminals from a single phase supply, the third terminal will _develop_ the third phase. This is the basis of the rotary phase converter, used to derive three phase power from single phase power.
-Jon
If the motor continues to drive the load, and if we presume that the motor remains near synchronous speed, then you will have essentially the same mechanical kW going to the load, and thus would expect at least as much electrical power to be supplied to the motor (efficiency certainly won't go up if you drop a phase under load). So if the motor continues to drive the load and you lose a phase, you would expect the remaining leads to supply the total power, and the current drawn on the remaining phases to increase.
It is possible that the motor will stall under the mechanical load, in which case the current drawn on the remaining phases will increase substantially, since a stalled motor draws far more than its normal full load current.
On the other hand, if there is no mechanical load on the motor, then it will probably continue to spin with no difficulty at all. In fact, if you run an unloaded three phase induction motor via two terminals from a single phase supply, the third terminal will _develop_ the third phase. This is the basis of the rotary phase converter, used to derive three phase power from single phase power.
-Jon
davidr43229
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Jon,
Thank you I learned from your last post. I do have a question for you though. What kinds of applications for 3 phase motors have you seen that have no mechnical or pump loads? Maybe fractional motors?
I am curious.
Dave
winnie
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- Springfield, MA, USA
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- Electric motor research
davidr43229 said:
Sometimes motors are simply very oversized for the attached load, due to bad engineering.
Another situation where a motor is essentially not loaded is a 'rotary phase converter' (mentioned in my post above). These can be had as induction motors built without any sort of output shaft; the rotor simply turns in a housing, perhaps with an internal cooling fan. There is no mechanical load; the load is purely the electrical load of developing the third service leg.
-Jon
hardworkingstiff
Senior Member
- Location
- Wilmington, NC
bob said:
Bob, can you help us understand the math on how one gets to these numbers? (I'm not questioning your answer, just trying to understand the math).
davidr43229
Senior Member
- Location
- Columbus, Oh
The explaination is on page 3-5
http://www.progress-energy.com/cust...ntations/MotorProtection_VoltageUnbalance.pdf
Just my $.02
http://www.progress-energy.com/cust...ntations/MotorProtection_VoltageUnbalance.pdf
Just my $.02
eric stromberg
Senior Member
- Location
- Texas
Quote: "Sometimes motors are simply very oversized for the attached load, due to bad engineering."
True. However...
Sometimes, whereas pumps need to be a certain size to run, they must be a much larger size to get the process going. (Start-up in a chemical plant that requires larger volumes than when the plant is running). In this case, you might have a 30Hp motor that typically sees a load of around 7.5 to 10 horsepower. Efficiency goes out the window, power factor goes out the window. Lots of solutions. Install two pumps, one sized for running conditions, a helper pump for start-up conditions. VFD's. Or, as is often the case, just put in a big motor and leave it at that.
True. However...
Sometimes, whereas pumps need to be a certain size to run, they must be a much larger size to get the process going. (Start-up in a chemical plant that requires larger volumes than when the plant is running). In this case, you might have a 30Hp motor that typically sees a load of around 7.5 to 10 horsepower. Efficiency goes out the window, power factor goes out the window. Lots of solutions. Install two pumps, one sized for running conditions, a helper pump for start-up conditions. VFD's. Or, as is often the case, just put in a big motor and leave it at that.
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