Induction Generator 5 - Heyland Circle
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| Number of Parts | 6 | |
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| License | CC Attribution - ShareAlike 4.0 International: You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this | |
| Identifiers | 10.5446/64556 (DOI) | |
| Publisher | 014nnvj65 (ROR) | |
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00:00
DiagramComputer animation
00:26
Lecture/Conference
06:45
Lecture/Conference
11:11
Lecture/ConferenceComputer animation
Transcript: English(auto-generated)
00:13
Hello and welcome. In this video, now we want to transfer what we have learned about the equivalent circuit,
00:20
how we graphically can display those behavior of induction generators. And therefore, we have first to look to this impedance Z21, that is equivalent is R21 divided by S plus Jx. And that we display in the complex area where the imaginary part is on the horizontal and the real part is on the vertical.
00:51
And when we have a look and have the circumstances here to all the slips, then we have the imaginary part, the x that is always the same.
01:03
That's the reason why we have always this distance x from the real axis. And depending when the S is zero, this term is getting infinite. And when the S is infinite, the term for the real part is getting zero and therefore is on the imaginary line.
01:24
But in the highland circle, we do not display the impedance, we are looking for the admittance dy, that means the one divided by the Z21. And when we do so, we result in such a circle.
01:41
That means when we have S equal to zero, that we do the opposite one, we result just in the origin of this table. And where here is S zero, and for the S equivalent to the infinite,
02:00
the real part disappears and only the imaginary part will be there. And everything in between is on that circle. And it's on the circle above the imaginary line for the motor case and below the imaginary line for the generator case.
02:22
But then we also have this vertical branch here. And there we add the active current representing the iron losses. It's then just moving the circle a bit upwards. And when we add also the magnetization losses, it's moving this circle towards the right.
02:44
And then we have always starting with this I zero that is pointing here to the circle. And then we have for the motor case, starting from the no load condition where N is equal to ND, the current is increasing, the phi is decreasing and therefore the cosine phi is increasing too.
03:08
And the reactive power demand is decreasing. For the generator case, it is then the opposite. The I21 is pointing downwards resulting in this I1.
03:22
And again here we have the angle phi and we have cosine phi. And the angle phi is getting at its minimum when we have this tangential behavior where the tangent is just touching the circle. Then also the cosine phi is maximum. That means the generator requires the least inductive power in this position.
03:49
When we have a look to the short circuit point or the startup point where the slip S is one, then we will see that the I1 where we have a very large startup current that is typical for electric motors.
04:08
We always can see in the highland circle the part with the reactive current and we also can see the active current. Or when we later talk about powers, we see always the active power on the vertical lines.
04:25
We see here the active power that is taken from the grid between the imaginary line and the operation point on the highland circle. And that is the reason why this imaginary line also is called the grid power line.
04:43
Then we have a second line also horizontally going through this vector I0 and there we can read the iron losses in the stator between those two lines. Then we can plot an additional line between this I0 and the short circuit point pk.
05:08
And when we read then here the difference between the second horizontal line and this inclined line, that is then the losses we have in the rotor.
05:20
And finally we can read then here the delivered mechanical power including the friction losses here. And that is also the reason why this line here is called the power line. So what else can we see? What we see here displayed is the power that is exchanged between the stator and the rotor.
05:46
And by this relation that the pd is equivalent to m times 2pnd, we have the distance also representing the machine torque.
06:01
And that is the reason why this second horizontal line is also called the machine torque line. What else do we have? We know that the mechanical power is equivalent to the pd times in brackets 1 minus the slip or the slip is equivalent to 1 minus mechanical power divided by the pd.
06:28
And we have the speed n is equivalent to nd times in brackets 1 minus s. And that we also can see then here in the highland circle the mechanical power
06:42
and once more the pd. But what is when we not have given the slip for a given working point but the working point for a given slip is sought? Then we can solve that problem graphically.
07:01
So first we draw a line between s the slip equivalent 1 and equivalent the infinite. That is this line. As a second we draw a perpendicular line to the machine torque line crossing this line 1. And we do that best when it is quite high in the picture.
07:25
And then we are marking s equal to 1 and s equal to 0. And in between those two points we add an equidistant scale. And then we can read here for the slip from s equal to 0, s equal to 10%, 20% up to s equal to 1.
07:47
And when we for example then are looking for the operation point of a slip of 10% we are marking a line between s equal to infinite and the mark of the slip for this 10%.
08:04
And where then this new line is crossing the highland circle we have the operation point for s equal to 0.1 or 10%. And then we again can read all the powers, all the moments we are interested for this operation point.
08:25
So once more what can we see in the highland circle? In the point we have here that is where we have no mechanic moments, we have no power provision,
08:40
we have no power loss in the rotor, the only thing we have in the no-load condition is iron losses. Of interest is for sure when we start up a motor, the startup, then we see here the operation point and again we see there is no power provision.
09:01
But we have a mechanical torque to start and a quite high mechanical torque to start. If we know from electric vehicles that they start up very fast and we have high power loss also in the rotor. And then we start up the motor and then we see that here in these operational points
09:20
we also see that the mechanical power is increasing and increasing but at a certain moment it is decreasing once more. When we have for a given electrical machine the highland circle and we are interested to read out the figures of current, of power, of moment
09:45
then we need to have the correct scales. And that way when we start for example with the currents and there we have that on our sketch
10:01
we have a certain distance is then x ampere per centimeter, that is our starting point. And then we can look for the powers for example when we have delta connection then the power is equivalent to three times the string power is equivalent to three times the nominal voltage times the string current.
10:26
And then we then replace the string current by the current scale, we end up in three times the nominal voltage times this x ampere per centimeter and result then in a figure of y kilowatt per centimeter.
10:43
And then we just can add the moment scale, that is moment is equivalent to the power divided by 2PND and that is then the y divided by 2PND and then we multiply by 60 seconds per minute
11:01
in order to result in the unit Newton meter or finally in our scale z Newton meter per centimeter. But finally we are talking about wind turbines and not interested really in the motor operation but more in the generator case and that we easily can transfer now to the generator case
11:24
that we once more have this current going to the lower part of the circle and then we have here what we can read is the mechanical power supplied by the rotor blades to the generator we have here the losses in the rotor and we have the losses in the stator
11:44
and then we have had here the active power from the grid in the motor case and in the same way now we can read the active power we are now feeding to the grid in this distance here. And then we have had here the power exchange between stator and rotor
12:04
and that is equivalent in our generator case we can read here this power that is transmitted between rotor and stator. And finally we have learned now a lot of induction machines
12:21
and we will end this series to explain why the double fed induction generator is so well suited to operate wind turbines. Thank you very much.
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