This thread has been challenging. As previously stated, I have replied, as always, from experience and that experience is that in an industrial setting a rectifier's output is considered to be DC and is measured with instruments on the DC setting. I cannot, from experience, vouch for the...
Edison123,
LPS, my first thought, based on the size of the motor, was exactly the same thing. However, the little 'port' that is indicated by the OP does look like an oil lube port. However, my doubts about this, as alluded to by you, are why I suggested that he/she contact the manufacturer.
Pete,
What is posted is an excerpt from a manufacturer's manual that, per the OP's question, applies to large PM motors. In my opinion it is a good, non-specific, procedure that should apply to most large PM motors.
You should consult the manufacturer. Based on the size and configuration of the port, my guess is oil but it is just that, a guess. Beyond the question of what lubricant to use, there is also the question of how much lubricant to use and how often to apply it. Unless there is someone with direct...
Pete,
It sounds like you have thoroughly covered all of the possibilities based on the information at hand. It will be informative to see the results of the second rotor TIR with the journals on the rollers. Please share this when you get it.
Pete,
The rotor will be constrained from touching the stator by the ODE, lower, bearing until it is lifted clear from the ODE bearing housing. At that point, the magnetic forces will keep it centered unless the lift is off centered. Having said this, I don't know what degree of 'off center...
The conductors from the disconnect will connect to an individual motor starter or VFD, not to a 'bus'. As mentioned by previous posts, the conductor sizing should be based on local code but in general they will be rated for no less than 125% of the full motor load current. The motor overload...
I have read through the entire post again and I figured out 'what am I missing'. The prevailing analysis in this thread is based on AC RMS voltage. My understanding, education, and experience is that the output of a rectifier is considered to be DC. My equations from 29 Jun 20:00 were correct...
The responses are correct, I left out a step.
- Remove the bearing cap bolts from both ends of the motor.
- Place the motor in a vertical position with the drive end up.
- Remove the DE bracket.
- Lift the rotor straight up.
It sounds like you have a dual voltage field and all of the field leads should be connected. If this is the case, the motor nameplate should indicate it by showing two voltages in the field data such as 90V/180V.
If you consider the source leads to have an S prefix and the motor leads to have...
...not correctly answered. It seems like there is agreement that a half wave rectifier only provides 1/2 voltage and it should be apparent that if only 1/2 of the voltage is present, only 1/2 of the current is drawn. Per Ohm's Law; 1/2V * 1/2R = 1/4P and, coincidentally, P = V^2/R. What am I...
Hi Petronila,
The procedure for a vertical machine having rolling element bearings is to remove the lower bearing cap bolts and, with the motor in the vertical orientation, lift the rotor straight up. The magnetic forces should balance to keep the rotor in the center of the stator bore without...
This size motor with sleeve bearings should have received an alignment before startup, there should be an alignment report that would have indicated a bent shaft before start-up had the motor been damaged during transport or installation. Having said that, it's hard to imagine the shaft being...
Hey Jeff,
I am familiar with the acronym 'MCB' being used for 'molded case breaker' but I am not familiar with the common use of the acronym 'MOP'. The only thing that I can think of in this context is 'motor overload protection', otherwise known as the 'heaters' (very old school) or the...
...the SQRT(2) from the denominator of my voltage equation and it matches yours. I assert that the current equation takes the same form as voltage and that power equals voltage times current.
P = V * I
For a half wave rectifier;
Vrms = Vp/2
Irms = Ip/2
Prms = Vrms * Irms = Vp/2 * Ip/2 = (Vp *...
I gave Zlatkodo the second LPS yesterday and I stand by it.
Sine Wave AC or Full Wave Rectified DC
Vrms= Vp/SQRT(2)
Irms = Ip/SQRT(2)
Prms = (Vp*Ip)/2
Half Wave Rectified DC
Vrms = Vp/(2*SQRT(2))
Irms = Ip/(2*SQRT(2))
Prms = (Vp*Ip)/8
P(half wave) = P(full wave)/4
Keith, add the SQRT(2)...
Wow, I accepted Jeff’s written explanation that the power is the cube of the speed change as being correct, which it is, without checking his math. My bad again, Muthu is correct.
Compositepro, the motor torque above base speed has nothing to do with the drive programming, it is a function of the motor’s volts/hz ratio. The voltage reaches its maximum value at base speed and as the frequency continues to increase the V/hz ratio decreases and reduces the motor torque...
