Voltage Drop on a Battery Due to 3 Motors

[Electro22]

With three motors, 1 - 150 HP VDC motor and 2 - 65 HP VDC motors connected to a 240 VDC Battery, I am trying to find the initial voltage drop on the battery due to the inrush current on the motors.

I have the information off the nameplates of each motor,
150 HP DC Series Motor
545 Amps,230 Volts, 480RPM etc. Would the in-rush current be supplied within the technical data sheets for the motor? Also, I have calculated the voltage drop due to the conductors and have the minimum operateing voltage of the motors would it be possible to calculate the voltage supplied to the 150 HP motor while each motor is experiencing a locked rotor current at the same time?

 

[ScottyUK]

Most motors of that size will normally use a controller of some form to keep the current within acceptable limits. What those limits are depends to some extent on the size of the battery and how tolerant the other loads are of voltage sags. The controller could be an electronic type, or an old-fashioned resistor timestarter where a series of resistors are sequentially removed from the circuit. The starter drops the voltage delivered to the motor in order to limit the current. A reasonable rule-of-thumb value for the maximum current would be 3 - 4 times rated current unless you have an awkward load and a motor designed to cope with it.

A stall condition on a series machine is fairly serious if you have a large battery source and the motor is unable to accelerate - you did remember to include a protection relay didn't you?

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I don't suffer from insanity. I enjoy it...

 

[Electro22]

A Time overcurrent relay has been included. Also the details for the battery were Time to Discharge: 2,200A for 5seconds
1,400A for 10 seconds

Cheers,

 

[waross]

The current at locked rotor is determined by the sum of the resistances in the circuit and the battery open circuit voltage.
The resistance of the rotor.
The resistance of the series field.
The resistance of the supply cables.
The battery internal resistance.
Any other resistance in the circuit.
The brushes add a little, but it is a conservative error to ignore the brush resistance.
Multiply the current times the sum of all resistances except the battery internal resistance to determine the terminal voltage.
Multiply the current squared times the battery internal resistance to determine the heating losses in the battery.
yours

 

[itsmoked]

Electro; It is possible to calculate all this stuff. As waross sez the starting current will be entirely controlled by the complete circuit resistance. This should be included in the motor spec sheet.

If not find a power supply like a multi-amp voltage/current controlled bench unit.

Set the supply's current control low mabe an amp.
Turn up the voltage a little until you hit the current limit.

Now turn up the current and the voltage alternately until you have something like 5-10A running thru the motor circuit. The motor shouldn't be turning(or be allowed to turn).

Now using a good voltmeter and measure the voltage at the point where the supply has been connected to the motor circuit. Not on the power supply leads but on the actual heavy motor copper leads.

Divide this voltage by the current your supply is injecting.(If your supply has a low quality current meter then hook-up in-series a good DMM ammeter so you can get a good milliamp accuracy.)

Divide the injected current by the heavy lead voltage measurement.

This will give you your "starting resistance".

You then must examine the data sheet for your batteries. It will tell you the internal resistance of your batteries. Using this resistance and the motor circuit resistance, you can now calculate the battery pack output voltage at start-up.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[waross]

Good idea Keith.
Direct injection of current is often an accurate way to measure low resistance values. I would try to bypass the brushes. I have never had good luck measuring resistance through brushes. One method is to replace the brushes with copper bars. Use enough pressure to avoid arcing either to the commutator or the brush holders. The error will be conservative.
You may estimate the battery internal resistance by connecting a load and measuring the voltage drop. I would double check with two or more levels of loading. Voltage change divided by current change will give you battery resistance.
respectfully

 

[Electro22]

Just to clarify most DC motors would never have a locked rotor current more than 4 times the rated FLA's?

Generally how long does this current last for before returning to acceptable limits.

DC's motors are rated to operate at 150% FLA's for one minute.

Cheers

 

[ScottyUK]

Hmm, not true.

We have a 15kW DC compound-wound machine driving a pump which draws over 1600A at locked rotor. We discovered this when a problem occurred with the shunt field circuit and it accelerated using the series field only. The starting resistor lit up like a light bulb before exploding, only they don't make 100kW light bulbs as far as I know.

What I said in my earlier post is that a starter will typically limit the current to about 4x LRC. Without some form of controller it will be essentially controlled by the cable and winding resistances. The current could be very high indeed if the cable is short.


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I don't suffer from insanity. I enjoy it...

 

[waross]

What ScottyUK said.
yours

 

[itsmoked]

E; How long is entirely controlled by the load. Which includes the actual motor armature. The current starts dropping from the moment the shaft starts turning.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com