DC Motor Operations

Hello Everyone:

Maybe this is not related to PLC but I hope anybody can help me regarding DC motor operations:

Here is the scenario:

I have the ff. motor:
DC motor
Armature current: 8A
Armature Voltage: 220VDC
Field current: 0.8A
Field voltage: 180VDC

My objective is to run this motor during brown-out with our standy Emergency battery system of 220VDC. What should I do? Im planning to directly supply the armature and field circuits directly to 220VDC. Am i on the right track? I'm afraid it will damaged my DC motor if I will directly supply the 220 VDC. Do I need additonal resistors to correctly tune-up the DC motor? Please help if somebody knows this.

Thanks,
Alex

This type of motor is driven by a 90/180 VDC controller. For 220 VAC, the line voltage is feed to the field via a full wave bridge. What it's getting is a DC voltage with plenty of ripple.
The armature is driven by SCR's. I don't think you could run the controller directly off of DC, by it might work. I think the CR's need a pulsing supply to unlatch.
The field is a bit forgiving, but direct 220VDC would be high. Check the motor specs. You may need to use a resistor in series. 50 ohms, 50 watts. The motor will run at full speed with 220VDC on the armature. Do you want to run at full speed? If not, find a controller that will run on DC. You could also use a resistor here to adjust speed, but that will be big and expensive.
There's nothing complicated here. Just use simple Ohms law for DC circuits to make calculations. If you're voltages are too high, the first indication will be motor temperature. Monitor it for a while. If it's running at the proper temp, you're fine.

DC Motors have been driven off batteries or from DC generators for eons.

The motor will love being supplied from a battery no ripple etc etc.

Now for the hurdles
1. Do you have to have speed control
2. How are you starting motor now
3. Are you using a DC controller fed from AC
AND how are you going to switchover to battery?
4. What does this motor drive
5. When on battery alone how long will you need to drive motor ie what is total Amp hour (better yet Kw Hr) of battery needed to support this?
6. What is the driven load - what is so vital about this you must have battery backup - ie is this a bilge pump on a ship etc.

Dan Bentler

leitmotif said:

DC Motors have been driven off batteries or from DC generators for eons.

The motor will love being supplied from a battery no ripple etc etc.

Now for the hurdles
1. Do you have to have speed control
No speed control is needed here
2. How are you starting motor now
I will start the motor by just activating a DC contactor in line with the DC Emergency power supply
3. Are you using a DC controller fed from AC AND how are you going to switchover to battery?
I am using Mentor MP DC Drive for normal power operations, but during AC power loss, it will automatically switch on to the emergency power supply just to return the load to its home position. It will just take 20-30 minutes for the motor to return to its home position
4. What does this motor drive
The motor is driving an Anode furnace in a copper smelting plant. During normal power operation, a Mentor MP DC drive is supplying the DC Motor. But for cases where commercial AC power is loss, the Anode furnace needs to return to its home position for safety and process purposes by using the Emergency battery system (220VDC).
5. When on battery alone how long will you need to drive motor ie what is total Amp hour (better yet Kw Hr) of battery needed to support this?
Now, the system is running but I need to revised the design as part of the upgrading. The motor will run 20-30 minutes just to return the anode furnace to its home position.
6. What is the driven load - what is so vital about this you must have battery backup - ie is this a bilge pump on a ship etc.
Its driving an Anode Furnace in a copper smelting plant. It must return to its home position during brown out for process and safety purposes.

Dan Bentler

Click to expand...

Thanks for asking those questions. Please see above answers.

OK now I have a fair grasp of what is going on.
1. Melting operation using carbon electrode furnace
2. electrodes I assume are supplied by 3 phase
3. On AC power loss you want to withdraw electrodes.
4. Max time to pull electrodes to full raise postion 30 minute.
5. rounding off motor - draws 9 A on a 220 VDC line so that is 1.98 Kw for one hour call it 1 Kwhr battery capacity to meet need of one half hour.
6. Is motor just shunt or compound?

1. On loss of AC I would have a dropout relay to parallel motor to the battery and disconnect from the normal drive (both armature and field). Coil on relay fed by AC.
2. When dropout activates and on return of AC it will not switch back over - switchover must be done manually.
3. I assume you have limit switch at top of travel for the normal controller you may need to add another or a relay to act as a top travel stop switch for both normal control and for emergency control.
4. With electrodes anywhere below top position a relay fed by 220 DC will connect motor control to battery
AND will activate the motor circuit in the up direction
5. I would recommend a dropping resistor in series with motor and a drop resistor bypass contact to short dropping resistor after a second or two to prevent motor burnout when starting at full line voltage. A second volt drop resistor is a good idea.
6. Shunt field voltage can be supplied direct from battery with a dropping resistor (reostat if you wish).

Gonna be a few bucks in design and relays but shoud be cheaper than digging electrodes out of solidified copper. I worked in a steel mill with a 125 ton electric arc furnace.

Dan Bentler

Just for emphasis, a leitmotif states, you cannot put the 220V directly on the 180V field. You can figure the resistance with the voltage drop of 40V divided by whatever the field current is. The resistor wattage would be the 40V times the field current times a safety factor of about 1.5. That keeps the resistor from running too hot under DC power conditions.

leitmotif said:

OK now I have a fair grasp of what is going on.
1. Melting operation using carbon electrode furnace
*Its not carbon electrode furnace, its a furnace fired with heavy oil fuel combined with LPG & air. Molten blister copper is melted on this furnace. Once the melting point is ok, the furnace will be tilted to pour the mixture into a channel for casting into solid copper. The DC motor here is the one controlling the tilting speed of the furnace. During AC power failure and the furnace is at the middle of the operation. It must be return to its home position to avoid possible spillage of the melted copper.
2. electrodes I assume are supplied by 3 phase
3. On AC power loss you want to withdraw electrodes.
4. Max time to pull electrodes to full raise postion 30 minute.
5. rounding off motor - draws 9 A on a 220 VDC line so that is 1.98 Kw for one hour call it 1 Kwhr battery capacity to meet need of one half hour.
6. Is motor just shunt or compound?
*Shunt wound motor. I found 4 wires out from the DC motor (YASKAWA)

1. On loss of AC I would have a dropout relay to parallel motor to the battery and disconnect from the normal drive (both armature and field). Coil on relay fed by AC.
2. When dropout activates and on return of AC it will not switch back over - switchover must be done manually.
3. I assume you have limit switch at top of travel for the normal controller you may need to add another or a relay to act as a top travel stop switch for both normal control and for emergency control.
4. With electrodes anywhere below top position a relay fed by 220 DC will connect motor control to battery
AND will activate the motor circuit in the up direction
5. I would recommend a dropping resistor in series with motor and a drop resistor bypass contact to short dropping resistor after a second or two to prevent motor burnout when starting at full line voltage. A second volt drop resistor is a good idea.
6. Shunt field voltage can be supplied direct from battery with a dropping resistor (reostat if you wish).

Gonna be a few bucks in design and relays but shoud be cheaper than digging electrodes out of solidified copper. I worked in a steel mill with a 125 ton electric arc furnace.

Dan Bentler

Click to expand...

Please see above for the answer of your one question and some explanations. Thank you very much for this information. As of now I have no problem with the control interlocking since i am using SIEMENS CPU315-2DP as my main controller. My Mentor MP is running on Profibus DP via ET200M modules. My tricky part is only on the part during AC power failures.

Don't start the motor "Across The Line"

You cannot start this motor by applying 220 VDC directly onto the armature. The starting current will be VERY high, probably several hundred Amps. You must use a starting resistor in series with the armature to limit the current. Limiting the current to about 200% of full load is a good place to start. You would need a 220/8x2 = 13.75 ohm resistor for the initial resistance. Depending on the load on the motor you may have to decrease the resistance in steps as the motor accelerates. You want to keep the current below 200% at all times. You may need 2 or 3 steps to do this. If the motor doesn't need to run at full speed, some resistance may be left in continuously, as long as the resistor is sized for the continuous current.

Vic said:

You cannot start this motor by applying 220 VDC directly onto the armature. The starting current will be VERY high, probably several hundred Amps. You must use a starting resistor in series with the armature to limit the current. Limiting the current to about 200% of full load is a good place to start. You would need a 220/8x2 = 13.75 ohm resistor for the initial resistance. Depending on the load on the motor you may have to decrease the resistance in steps as the motor accelerates. You want to keep the current below 200% at all times. You may need 2 or 3 steps to do this. If the motor doesn't need to run at full speed, some resistance may be left in continuously, as long as the resistor is sized for the continuous current.

Click to expand...

Thank you very much for this inputs.