1
\$\begingroup\$

I was trying to understand load angle in electrical machines, generally denoted by delta, and I found out that while delta needs to be less than 90° in synchronous machine for steady state stability, it is not so in induction machines as delta is always larger than 90° in an induction motor.

The source of this knowledge is Stephen Chapman's Electrical Machine Fundamentals. Screenshot below:

enter image description here

One thing which boggles my mind is the angle by which Br (rotor magnetic field) is shown to lag behind rotor current Ir (that is 90°.) How is this possible? Isn't the rotor MMF supposed to be in phase with the rotor current? Am I missing something or is the book wrong?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ The text says \$I_M\$ and \$E_l\$. But the figure shows \$I_R\$ and \$E_R\$. Please add the descriptions of those variables also into the question. Google books excerpt \$\endgroup\$
    – AJN
    Aug 4, 2021 at 12:28
  • 1
    \$\begingroup\$ That's the -di/dt in Lenz 's law. \$\endgroup\$
    – user16324
    Aug 4, 2021 at 13:22

2 Answers 2

1
\$\begingroup\$

One thing which boggles my mind is the angle by which Br (rotor magnetic field) is shown to lag behind rotor current Ir (that is 90°.)

Couldn't find your version of the textbook on Google books. I found this instead. The figure in this book is slightly better.

google books excerpt showing a marginally better version of the current and magnetic field figure given in the question.

The image caption says

... the rotor current produces a rotor magnetic field BR lagging 90deg behind itself ...

The word lag in your question as well as the image caption and the arrows in the figure may lead one to think that the arrows represented are phasors. If that interpretation were true, then the magnetic field phasor would be expected to be aligned to the current.

(IMO) But the arrows are not phasors. They are physical directions of the magnetic fields. The dotted line is not the direction of the current. In the version of the book I posted above, the the arrow representing \$I_R\$ is clearly shown perpendicular to the plane of the page. Magnetic field \$B_R\$ would then be on the plane of the paper and perpendicular to the loop of the current as given by the right hand rule Wikipedia Figure.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you! The way it was presented I could not think of anything else except phasor diagram. \$\endgroup\$ Aug 4, 2021 at 14:50
1
\$\begingroup\$

In your model, figure 6-15(a) shows the motor running with little slip; that is, with the rotor almost keeping up with the stator. If the rotor were running at true synchronous speed, the rotor bars would not experience any change of magnetic field, and there would be no voltage induced in the rotor.

Since the rotor is running a little slower than the stator, there is a change in magnetic field. This change in the field induces a voltage in the rotor bars, which is E(R) in your diagram. I(R) lags E(R) some because the rotor is an inductor.

Now consider where the magnetic fields come from. The effect of B(S), the stator magnetic field, on the rotor is pretty intuitive, but B(R) is a little more complicated. The angle between I(R) and B(S) changes as the motor is loaded, and slip increases; the same is true for the angle between E(R) and I(R). B(R) is always 90 degrees from I(R) because of the right-hand rule; in your diagram, if the induced current I(R)'s direction is into the page for a bar at the top of the rotor, it must be coming out of the page at the bottom and the resulting contribution to the magnetic field from these two bars must be horizontal. The resulting of angle between the field vectors yields a counter-clockwise torque as shown in the picture.

I think the confusion often comes from the reference point. From the reference of the stator, all of the fields are rotating at the same speed, although the rotor is turning more slowly. However, from the rotor's point of view, the magnetic field is rotating at only a few Hz and this low frequency is what generates B(R).

\$\endgroup\$
1
  • \$\begingroup\$ Thanks a lot! The problem with me was to presume that it was a phasor diagram of rotor current and rotor mmf when it was actually a matter of right hand rule. \$\endgroup\$ Aug 4, 2021 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.