Slip Ring Starter |
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Slip Ring Starter |
May 18 2002, 09:58 PM
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#1
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Posting Freak ![]() ![]() ![]() ![]() ![]() ![]() Group: Moderator Posts: 2,099 Joined: 24-April 02 From: Christchurch, New Zealand Member No.: 1 |
How does the starting torque increase and the starting
current decrease by adding resistance? Thank You K.B. -------------------- Mark Empson administrator
Skype Contact = markempson | phone +64 274 363 067 LMPForum | LMP Tech Warehouse | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Mark Empson Website |
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May 18 2002, 09:59 PM
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#2
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Posting Freak ![]() ![]() ![]() ![]() ![]() ![]() Group: Moderator Posts: 2,099 Joined: 24-April 02 From: Christchurch, New Zealand Member No.: 1 |
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. -------------------- Mark Empson administrator
Skype Contact = markempson | phone +64 274 363 067 LMPForum | LMP Tech Warehouse | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Mark Empson Website |
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Jan 23 2003, 11:26 PM
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#3
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Junior Member ![]() ![]() Group: Members Posts: 2 Joined: 20-January 03 From: Cleveland Member No.: 248 |
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 |
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Jan 24 2003, 07:05 PM
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#4
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Posting Freak ![]() ![]() ![]() ![]() ![]() ![]() Group: Moderator Posts: 2,099 Joined: 24-April 02 From: Christchurch, New Zealand Member No.: 1 |
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, -------------------- Mark Empson administrator
Skype Contact = markempson | phone +64 274 363 067 LMPForum | LMP Tech Warehouse | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Mark Empson Website |
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