# I need help with a Baldor dc motor, 15kw 240 volt ,150 volt field

• odbob
In summary, lowering the field voltage on a shunt wound motor will result in a decreased torque and a decreased RPM.
odbob
I need help with a Baldor dc motor, 15kw 240 volt armature and 150 volt field, I don't need the 15kw and so want to feed it with 150 volts to both field and armature, does anyone know the effect this will have on the motor, can I cause damage, what would be the approximate power output, I know will be greatly reduced as I am aware of the usual effect of reducing voltage to a normal DC motor but am simply not familiar with this type, thank you

Lower armature voltage, will lower the RPM and lower the torque.
Power will be about 15 kW * ( 150 / 240 )² = 5.86 kW.

odbob
the lower output is fine, but will it harm the motor with the field at same voltage, I don't really understand the relationship, ie what happens if I raise or lower the field voltage ? what happens to the current/ output etc

Everything is connected.

Lower field voltage results in a lower field current, so a lower magnetic field.
Motor torque is proportional to the product of the field by armature current.
Back EMF is proportional to the field and RPM.

The difference between the back EMF and the applied armature voltage decides the armature current.
The unloaded motor RPM occurs when the applied voltage is equal to the back EMF due to rotation.

Will it harm the motor? Specify the load.
If you stall the motor, the armature current will be at a maximum.

thanks but I am non the wiser, this motor is intended to power a small cabin cruiser and so should never be in a position where it would stall and I can always guard against that happening, from what you have said so far, I feel comfortable with connecting this motor to a 150 volt DC supply, and in terms of speed control will simply vary this voltage downwards in say, three steps 150, 100, 50 volts, if I did need more power, I can always add a further 50 volts, the voltage will be applied to both the field and armature simultaneously, thank you for your help

I would be careful about lowering the field voltage. Shunt wound motors are designed for good speed regulation with changing load. Lowering the field voltage is likely to decrease performance in that area. Think about it. Is there ever a problem lowering the voltage to a permanent magnet motor? Also, care should be taken to prevent the loss of field current in a shunt wound motor. Very bad things happen when field current is lost.

odbob
odbob said:
this motor is intended to power a small cabin cruiser
Like a boat? With only a few horsepower?

https://www.divein.com/boating/cabin-cruiser/

I have a cabin cruiser that is about 19 feet long. I put a 10 HP Johnson outboard on it for just putting around if someone has a fishing line out. So the motor in question here is probably not out of line for size.

berkeman, odbob and russ_watters
Averagesupernova said:
I have a cabin cruiser that is about 19 feet long. I put a 10 HP Johnson outboard on it for just putting around if someone has a fishing line out. So the motor in question here is probably not out of line for size.
this is a 26 feet long cruiser but only intended for the canals, 4 miles per hour

you can only really damage a motor by overheating it, (not counting some weird cases like overvoltage etc)
So one very known way to overheat a motor is during a stall, there is no back EMF produced and the stator/rotor windings pass through their maximum available current which in AC is only limited by the winding self inductance and in DC case only limited by the resistance of the actual wires, so a stalled DC motor would pass through a current that would be only limited by the wire resistance assuming a constant source that doesn't sag under load.

I believe one would have to do the math for each individual case but a lower B field strength would result in a lower back EMF produced which would allow more current to pass through , so I think you need to calculate how much current passes trough that motor at rated voltage operation and then calculate how much would pass through at the lower voltage operation and then compare.It's very counterintuitive but it is a known fact that you can actually damage a car starter (the classical series universal/dc motor) by using it with a "half dead" battery.
The reason is that a battery that cannot support the current required by the motor will sag in voltage, this will result in a situation where the starter RPM decreases so the back EMF will also decrease , if the RPM decreases by a large factor the back EMF becomes so small that the starter almost becomes a simple resistor put across the battery, this dissipates alot of heat in it's windings and can cause damage.

So I don't think the answer is straight forward, I think you have to calculate for each case for each individual motor based on it's parameters.

artis said:
you can only really damage a motor by overheating it
Motors can be over sped. Depending upon whether the motor in question here can be disconnected from the prop, this is a real possibility here with the loss of field current.

artis and berkeman
Averagesupernova said:
Motors can be over sped. Depending upon whether the motor in question here can be disconnected from the prop, this is a real possibility here with the loss of field current.
I liked your comment first and then thought about it, well series wound universal motors can be oversped or so i recall reading, without a load attached, but what has that to do with losing field current?
How can such a motor turn at all without the stator provided field in which the rotor coils turn against?

Averagesupernova said:
I would be careful about lowering the field voltage. Shunt wound motors are designed for good speed regulation with changing load. Lowering the field voltage is likely to decrease performance in that area. Think about it. Is there ever a problem lowering the voltage to a permanent magnet motor? Also, care should be taken to prevent the loss of field current in a shunt wound motor. Very bad things happen when field current is lost.

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