more HNC crap assignment question confusion: induction motors. - Page 1
Forum section: electronics / beginners

author: Simon
author_rank: Unknown
timestamp: December 18, 2018, 09:23:33 pm
content: So having suffered the mind bogling experience of trying to understand the one weirdo's on the planet idea of a topic on induction motors i now have to answer the even weirder put questions. So I have stumbled at this one, the first question that attempted to be clever (badly).Code:A four-pole, star-connected, squirrel-cage induction motor operates froma variable voltage 50 Hz three-phase supply. The following results wereobtained as the supply voltage was gradually reduced with the motorrunning on no-load.Stator line voltage 220 164 112 88 42Stator line current (amperes) 6.8 5.4 3.9 3.8 3.7Stator power (watts) 470 360 278 244 232(a) By plotting a suitable graph from these results, determine the total ofwindage and friction losses, the no load magnetising current I0(assume no load current I0 is magnetising losses) and the equivalentcircuit magnetising circuit parameters R0 and X0.(b) With locked rotor and 50 volts applied to the stator, the current was23 amperes and power input 985 watts. The resistance between twolines is 0.5 ohms.If the rotor turns per phase equals the stator turns per phase, draw theequivalent circuit of the motor and determine the values of thevarious components. Assume the stator phase reactance is the samevalue as the rotor phase reactance.The equivalent circuit is shown in FIGURE 1.Well I'm not sure how a graph helps. but I made a table. So fram what i can tell here they are giving power into the motor (line volts X line amps) and the power specifically into the stator allowing by simple subtraction (remember this is the clever question) to calculate the rest of the motor losses which are mechanical. So my table reads as:V Is Ps "mechanical loss"220 6.8 470 1026164 5.4 360 525.6112 3.9 278 158.888 3.8 244 90.442 3.7 232 -76.6Right so just as I thought they had not tied themselves in too many knots oh look i have a perpetual motion machine......-76.6WSo are my assumptions correct so far and this is an oddity? i could answer these stupid questions by using other books but the questions are always styled as per the course material. Obviously they managed to not mention at what voltage they want friction and windage loss and they have stated the windage loss anyway.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 18, 2018, 09:55:09 pm
content: I've not performed this experiment, but based on my understanding:The RPM should be fairly stable in the decreasing-voltage sequence, right?Then, the mechanical losses will be fairly stable across that sequence.We know that* electrical losses correspond to winding voltage and current, so the electrical losses should be dropping to nearly zero at the same time. The zero intercept would seem to reflect mechanical losses.This makes sense, as, supposing we spin the motor from an external (mechanical) source, and short out the windings, now the electrical loss is zero (assuming the rotor doesn't have any residual field, which may not actually be the case with real silicon steel -- if so, then stator current will be nonzero, dissipating some power in DCR; or if we open-circuit the windings, then there will be nonzero core loss), and all the mechanical losses must be supplied from external torque * RPM = power.*From where? -- It might not be covered very well in class. As it happens, DCR (series loss due to current flow) and core loss (eddy currents and hysteresis, manifests as an equivalent parallel resistance) are present, and so we can separate them with several tests.We certainly do not simply multiply the numbers. If those are actual measurements, then for example, 220V * 6.8A = 1496 VA, but measured P was 470W so the power factor is small (470/1496 ~= 0.3 PF), presumably inductive, as motors tend to be. And that's the sum of electrical and mechanical losses. But as we go down to, say, 42V * 3.7A = 155W, we get... well, a lot less than 232W, so maybe those aren't measurements, I don't know what's up with that.Setting aside the exact voltage and current numbers for a moment: assuming the mechanical losses are constant, then the zero intercept gives pure mechanical losses. Which looks to be around 200W. Which means the remainder at high power (~270W) is electrical, which puts core loss for example at about 200 ohms equivalent.But I don't know what they mean by "plotting a suitable graph". I can only assume that was covered in lecture or the textbook somewhere.Tim

author: Simon
author_rank: Unknown
timestamp: December 19, 2018, 07:29:37 am
content: Thanks Tim. You are the second knowledgeable person to look at this question and be a bit confused. The material is a bit wooly spending 2 A4 pages on resistive losses that are explained in 4 simple mathematical statements coming out instead with a A:B:C statement that I have never seen in any math or text book.

author: Simon
author_rank: Unknown
timestamp: December 19, 2018, 07:30:11 am
content: And no they never said that the motor will stop dropping in speed at lower voltages

author: Simon
author_rank: Unknown
timestamp: December 22, 2018, 08:37:10 pm
content: Right I have gone back through the course material (see attached page 8 ). So they say that if i draw a graph of the values using the square of the voltage and the power I can see where the graph will cross the power axis at 0V by continuing the line. This will give the losses asked for. So now of course the obvious question, are we talking the VxI power or the power given. i take it that the "line" volts and amps are the apparent power and the power given is the real part of the power used?I am still trying to work out what I0 is in their eyes but need to revise further.