FIELD WEAKENING
FIELD WEAKENING
(OP)
I have an extruder motor that has rating of 700HP 500vdcARM 300v/150vdc.field 8.9a/17.8 amp field @ 1750RPM. Currently under load we have 800amps on the armature @1750RPM. We are using 300vdc.field EXC Shunt.The manufacture say the motor is capable of 2400RPM. If we purchase a field weaking board. My first question is what will happen to arm.current at the same load go up or down and why?My 2nd question is where can I get some mathematical formulae on dcmotors because I measured the field resistance before installation and got 26ohms on the field.Using ohms law does not apply 300v/26ohms does not equal 8.9amps as rated on nameplate.





RE: FIELD WEAKENING
My first question is what will happen to arm.current at the same load go up or down and why?
The equation for predicting effects on armature current is
Ia=(Vt-Eg)/Ra = (Vt-NKPhi)/Ra
where
Ia = armature current
Vt = Terminal voltage
Eg = generated voltage = NKPhi
N = speed
K= constant
Phi = flux.
Perhaps it depends what you mean by constant load.
If speed is kept constant (in spite of change in motor torque-speed characteristic with increasing field current), then the equation predicts that armature current will go down.
If horsepower is kept constant... let me think about that one (anyone else have answer?)
My 2nd question is where can I get some mathematical formulae on dcmotors because I measured the field resistance before installation and got 26ohms on the field.Using ohms law does not apply 300v/26ohms does not equal 8.9amps as rated on nameplate.
If you have measured the resistance at room temperature, then that is probably the cause of the difference.
Copper resistance increases approx 0.4% per increases in temperature in degrees C. If you increase the temp on the order of 60C, you get approx 23% increase in resistance, which I believe would account for the difference.
RE: FIELD WEAKENING
The DC Variable Speed Drive will not just put out the specified voltage, but will actually act as a current source to supply the desired field current (the current is the important value here). Make sure and put this parameter (field current) in correctly; I have changed DC motors in the past and failed to make a presumably small change in the field current parameter for the new motor(less than 2 amps) and caused a motor to fail in 24 hours (nothing like learning the hard way).
Also note that if your drive is currently set up for "Voltage Feedback" instead of tach feedback, you will be unable to activate the field weakening feature in your drive; in field-weakened state, the armature voltage is no longer an accurate measure of the speed of the motor. You must have tach (or resolver) feedback to use this function. I'm assuming from your question that the motor is rated for this higher speed.
I believe I would also increase the frequency of brush/bearing maintenance on this motor at this increased speed.
Good Luck!
RE: FIELD WEAKENING
http://www.ktech-usa.com
RE: FIELD WEAKENING
Mars
RE: FIELD WEAKENING
Mars
RE: FIELD WEAKENING
RE: FIELD WEAKENING
I use the following equations to help me understand the characteristic of a DC motor.
Terminal Voltage Equation:
Vt = Vcemf + Ia*Ra + BD
Where Vt = terminal volts (what you can measure)
Vcemf = volts counter EMF (used by motor)
Ia = armature current
Ra = armature resistance
Ia*Ra = internal motor voltage loss
BD = brush drop (consider 2volts and ignored)
Speed equation
n = Vcemf/(N*If)*K1
Where n = speed
N = Number of turns in field winding
If = field current
N*If = field flux
K1 = constant
Torque Equation
T = Ia * (N*If) * K2
Where T = Torque
K2 = Constant
Horsepower Equation
HP = (T * n)/746
Where HP = horsepower
746 = constant KW to HP conversion
From the equations you can see that speed is directly proportional to Vcemf and inversly proportional to If. Double the armature voltage and you double the speed. Cut the field current in half and you also double the speed. However you notice that the Torque equation is not related to speed or voltage. As you weaken the field in a constant torque application the speed will go up but the armature current will have to also increase to keep the torque at the same value. So you can only weaken the field until you either reach top speed or the armature current reaches 100% then your out of Torque and HP.
RE: FIELD WEAKENING
RE: FIELD WEAKENING
1. Gordon R. Slemon "Magnetoelectric Devices, Transducers, Transformers and Machines," John Wiley and Sons, Inc., 1966
Section 4.6.2 Armature and Field Resistance Control
Figure 4.38 Shunt-connected motor
Please, notice that a variable resistor Re in the shunt branch sets the flux of the machine, which in turn controls machine. The armature starting current can be controlled by a resistor Rd in series in the armature circuit.
The DC shunt motor equations are often found in DC motor textbooks and on web sites. Visit
http://www.ee.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.h
for more info