The point on the excitation curve where a 10% increase in secondary exciting voltage results in a 50% increase in secondary exciting current is the British Standard definition of knee-point voltage.
The IEEE 45% and 30% definitions apply to Class C cts and refer to the point on the excitation...
Consider the probably high X/R ratio of the generator and the large fault current that may be available. It may not be possible (or practical) to size the cts large enough to avoid saturation when accounting for remanent flux and dc offset. Then the unequal performance of the mismatched cts...
So a pissing match it shall be. I have never disputed your theory. I poorly worded my response on Dec. 13 in saying that the breakdown torque relationship didn't apply with a VFD. The VFD can't change the basic motor parameters, so the motor will operate the same on VFD as off it. That...
Ron,
I'm too young to have grown up in the days of the right foot shift, and my Triumph is one of the new breed from the Hinckley works, not the old Meriden factory. However, a couple of years ago I took a spin on a right foot shift BSA Rocket 3. It took some getting used to, but I managed to...
jomega - if you have read the thorough posts Mark has made in the past you will realize you didn't tell him anything he didn't already know. I'm still trying to figure out where in Mark's post he mentioned constant torque loads in a way that would cause you to take a run at him in such a...
Greta - please don't cross-post in the forums. It makes the threads very hard to follow as separate discussions arise.
I have red-flagged this post as a duplicate and hopefully everyone will respond to your post in the "electric power engineering" forum.
"Multiply (Hz @ base speed / Hz @ field weakend speed) x motor rated torque at base speed... to obtain the available rated torque at the field weakened speed."
My apologies - the above formula from jomega can be used. It yields the same answer for rated torque that I gave in my...
vpxxx - you cannot use the formula given by jomega with a VFD. The breakdown torque point will never be reached. The VFD operates the motor only in the linear portion of the speed-torque curve, from 0% to perhaps 5% slip depending on the motor. In this range torque will be nearly proportional...
It has been my experience that a motor operated on VFD will exhibit a substantially constant HP characteristic to about 150% overspeed, which is roughly the range vpxxx mentioned in his post. Above that speed the horsepower will decrease as stated by jomega.
Also, a motor on VFD will drive the...
I'm reading these posts too fast. Overspeeding from 50 Hz (not 60 Hz as I thought) to 80 Hz is a 60% overspeed, so continuous torque will reduce to 62.5% of rated. Peak torque will only be in the neighborhood of 90-95% of rated.
Small clarification to my previous post:
P = torque x rpm/5252 where P is in horsepower and torque is in foot-pounds. Not sure about SI, but I think power (W) = torque (N) x angular velocity (rad/s).
Also, motor operates in constant power mode above rated speed because the VFD is flux-limited...
I agree with RDK there is a role for engineers in certain specific instances of public welfare. The Winnipeg flood dikes is one, the North Battleford water inquiry where APEGS had standing is another.
However, this is a far cry from the form of activism on global issues that PM has suggested.
