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#1 marke

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Posted 18 May 2002 - 09:58 PM

How does the starting torque increase and the starting
current decrease by adding resistance?
Thank You
K.B.

#2 marke

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Posted 18 May 2002 - 09:59 PM

First of all, it is important to understand the equivalent circuit of the
induction motor. The induction motor behaves like a transformer with a very
low impedance in it's secondary circuit. The current is limited by that
impedance. Increasing the impedance will reduce the current flow. Adding
rotor resistance, increases the secondary impedance and therefore reduces
the current.

Secondly, the maximum torque at any rotor speed (slip) occurs when the rotor
resistive impedance equals the rotor reactive impedance at that speed. As
the speed of the rotor reduces, the slip increases and therefore the
frequency of the rotor current increases. The rotor reactance is constant
but the rotor reactance is jwL and so it increases with increasing slip.

If at any speed, the rotor resistance is increased beyond the rotor
reactance, the rotor current will reduce and so will the torque. If you look
at the torque curve for an induction machine, the maximum torque occurs
close to full speed. Increasing the rotor resistance moves that maximum
torque along the speed axis in the direction of increasing slip.

If you are using a slip ring motor and require maximum torque over the whole
speed range, you would use a number of steps and each step would have the
maximum torque occurring at a different value of slip. For example, you may
use a 5 step starter with the first step providing a maximum torque at say
5% speed, the second at 25% speed, the third at 50% speed the fourth at 75%
speed and the fifth at maximum speed (no resistance)

So you don't actually increase the torque capacity of the motor, you just
shift the speed torque curve so that the maximum torque occurs at a useful
speed.

#3 castera

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Posted 23 January 2003 - 11:26 PM

How do you determine the starting current required to move high inertia loads. I have a crane which will not hoist the full SWL of 375 Ton in first notch or second notch hoist. The crane will hoist in third notch. The motors are two 280Kw 440v 470A 310Rv 555RA motors. The resistors are set to give 580RA at standstill, and i've worked out that third notch is delivering 1050RA at standstill, hence it hoists.
This is obviously at the expense of premature motor failures due catastrophic rotor failure. The rotor resistors appear to be correct for the name plate motor details - I've compared them to simillar hoist schemes which work, the rotor resistance tapers are calculated on percentage's of the rotor resistance.
I could work out roughly the hoisting power required - but lets say for instance it came to 325Kw per motor how, do i work that in to a new figure for rotor current at that output and hence new resistor values to give me that power output.
I am sure someone got the motor selection incorrect at the design stage ( 1970) and know that new motors are required - though the price makes this not an option ( 170,000 ea motor plus 60,000 resistors). what i am hoping is that by getting the resistors re-rated so that i can get the load moving at the lowest possible current then have four more resistance steps to accelerate the load after the motor are hoisting i will get better motor life.
I hope this makes sense

#4 marke

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Posted 24 January 2003 - 07:05 PM

Hello castera

The torque produced by the motor is a function of the rotor speed and the rotor resistance. The speed is a function to the motor torque and the load torque.
If at any speed, you select the resistor to be equal to the rotor reactance, you will get maximum torque at that speed. If the resistor is higher or lower than that value, the torque will reduce. The current drawn at maximum torque will typically be 200 - 300 % full load current.
For any value of resistor, there is a speed of maximum torque. Above and below that speed, the torque falls. The greater the number of steps, the higher the average torque across the full speed range. The last stage should always be designed to provide maximum torque at close to full speed to preven a severe step in current as you short the rotor.
Our ElectrcialCalculations software has a section on calculating rotor resistors.
Best regards,

#5 hhpp

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Posted 22 January 2011 - 02:05 AM

QUOTE (marke @ Jan 24 2003, 03:05 PM) <{POST_SNAPBACK}>
Hello castera

The torque produced by the motor is a function of the rotor speed and the rotor resistance. The speed is a function to the motor torque and the load torque.
If at any speed, you select the resistor to be equal to the rotor reactance, you will get maximum torque at that speed. If the resistor is higher or lower than that value, the torque will reduce. The current drawn at maximum torque will typically be 200 - 300 % full load current.
For any value of resistor, there is a speed of maximum torque. Above and below that speed, the torque falls. The greater the number of steps, the higher the average torque across the full speed range. The last stage should always be designed to provide maximum torque at close to full speed to preven a severe step in current as you short the rotor.
Our ElectrcialCalculations software has a section on calculating rotor resistors.
Best regards,


Hello everybody,
I am new here, and i want to talk about the same topic.

I have a 3,3 KV motor for a mill, with a LRS starter, it was working fine for almot 2 years.

Last time during operation, people from operation needed to stop the mill. When they wanted to start up again, we had a big problem with the slip rings, one of them was broken across the ring, the multiloin 269 registered 469 amps stator (FLA 82), i am not sure if this problem was becAUSE OF THE lrs OR A MILL extremely chatrged, so I need some help in this to understand what really happened and work out a solution.

best regards
hp

#6 hhpp

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Posted 25 January 2011 - 03:39 PM

Well I think it is really extarnge what happened to me, I would like to exactly how the LRS works, I mean an analysis of what happen electrically into the LRS, may be if i know exactly the equations, forms, and so on I would work out what is the problem,

thanks.
hp.




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