Electric machines & power electronics - PMAC machine connected in Y, 4 poles

In summary, the stator current must be minimized to achieve minimal losses. The stator current is 8.052 A and the angle is 16.6^{\circ}.
  • #1
VinnyCee
489
0

Homework Statement



A PMAC machine is connected in Y and has 4 poles. [tex]I_F\,=\,27\,A[/tex] and [tex]L_m\,=\,20\,mH[/tex].

a) It is to operate at a torque of 35 Nm and at a speed of 4000 RPM. Calculate the necessary stator current [tex]I_S[/tex] and voltage [tex]V_{S\,line-to-line}[/tex] so that the losses are minimal.

b) If the maximum line-to-line voltage is [tex]500\,\sqrt{3}\,V[/tex], what would the maximum speed the motor would reach without field weakening, [tex]\omega_1[/tex], for the same torque of 35 Nm?

c) For speed [tex]1.2\cdot\omega_1[/tex], voltage the same, [tex]500\,\sqrt{3}\,V[/tex], and torque of 35 NM, what would be the stator current and it's angle?

Homework Equations



For minimum losses, the stator current must be minimized.

[tex]T\,=\,3\,\frac{p}{2}\,L_M\,I_S\,I_F[/tex]

[tex]V_S\,=\,\omega_S\,L_M\,I_F[/tex]

[tex]\omega_S\,=\,\frac{p}{2}\,\cdot\,Speed\,in\,RPM\,\cdot\,\frac{2\,\pi}{60}[/tex]

The Attempt at a Solution



a) Using the first equation above...

[tex](35\,Nm)\,=\,3\,\frac{(4)}{2}\,\left(20\,mH\right)\,I_S\,\left(27\,A\right)[/tex]

[tex]I_S\,=\,10.8\,A[/tex]

To get [tex]V_S[/tex] I need to use the second equation above. And to get that I need [tex]\omega_S[/tex] using the third equation.

[tex]\omega_S\,=\,\frac{(4)}{2}\,\cdot\,\left(4000\,RPM\right)\,\cdot\,\frac{2\,\pi}{60}\,=\,837.8\,\frac{rad}{sec}[/tex]

[tex]V_S\,=\,\left(837.8\,\frac{rad}{sec}\right)\,\left(20\,mH\right)\,\left(27\,A\right)\,=\,452.4\,V[/tex]

Does that look right for part (a)? Also, how do I proceed for parts (b) and (c)?
 
Last edited:
Physics news on Phys.org
  • #2
Trying part (b)...[tex]I_M\,=\,\left|\bar{I_S}\,+\,\bar{I_F}\right|\,=\,10.8\,A\,+\,27\,A\,=\,37.8\,A[/tex]

[tex]\omega_{S,\,max}\,=\,\frac{V_{S,\,max}}{I_M\,\cdot\,L_M}\,=\,\frac{\frac{500\,\sqrt{3}}{\sqrt{3}}}{37.8\,A\,\cdot\,20\,mH}\,=\,661.4\,\frac{rad}{sec}[/tex]

[tex]\omega_{mech,\,max}\,=\,\frac{2}{p}\,\cdot\,\omega_{S,\,max}\,=\,\frac{2}{4}\,\cdot\,661.4\,=\,330.7\,\frac{rad}{sec}[/tex]

Does part (b) look right? I've no idea how to do part (c). Probably something to do with finding an angle between [tex]I_S[/tex] and [tex]I_F[/tex] I think...
 
  • #3
Part (a) was done wrong...

[tex]V_S\,=\,\omega_S\,L_M\,I_M[/tex] and not [tex]V_S\,=\,\omega_S\,L_M\,I_S[/tex]

[tex]I_M\,=\,\sqrt{I_S^2\,+\,I_F^2}[/tex]

[tex]I_S\,=\,10.8\,A[/tex] as above, and [tex]I_F[/tex] is given.

So...

[tex]I_M\,=\,\sqrt{I_S^2\,+\,I_F^2}\,=\,\sqrt{\left(10.8\,A\right)^2\,+\,\left(27\,A\right)^2}\,=\,29.08\,A[/tex]

Now...

[tex]V_S\,=\,\omega_S\,L_M\,I_M\,=\,\left(837.8\,\frac{rad}{sec}\right)\,\left(20\,mH\right)\,\left(29.08\,A\right)\,=\,487.3\,V[/tex][tex]V_S\,=\,487.3\,V[/tex] and [tex]I_S\,=\,10.8\,A[/tex]Part (b) is apparently correct (not sure though - is it really?) So I'll try part (c).

I'll use [tex]\omega_{mech,\,max}[/tex] as [tex]\omega_1[/tex]. Does that seem right?

So...

[tex]1.2\,\cdot\,\omega_{mech,\,max}\,=\,1.2\,\left(330.7\,\frac{rad}{sec}\right)\,=\,396.8\,\frac{rad}{sec}[/tex]

[tex]V_{S,\,(line-to-line)}\,=\,500\,\sqrt{3}[/tex] and [tex]T\,=\,35\,Nm[/tex]

Now I use two variations of the torque equation...

[tex]T\,=\,-3\,\frac{p}{2}\,L_M\,I_M\,I_F\,cos\,\beta[/tex] and [tex]T\,=\,3\,\frac{p}{2}\,L_M\,I_M\,I_S\,cos\,\theta[/tex]

Using the first to get the angle [tex]\beta[/tex] between [tex]V_S[/tex] and [tex]I_F[/tex]...

[tex]\left(35\,Nm\right)\,=\,-3\,\frac{(4)}{2}\,\left(20\,mH\right)\,\left(37.8\,A\right)\,\left(27\,A\right)\,cos\,\beta\,\longrightarrow\,cos\,\beta\,=\,-0.2858[/tex]

[tex]\beta\,=\,106.6^{\circ}[/tex]

[tex]\theta\,=\,\beta\,-\,90^{\circ}\,=\,16.6^{\circ}[/tex]

Now I use the second torque equation to get the new [tex]I_S[/tex]...

[tex]\left(35\,Nm\right)\,=\,3\,\frac{(4)}{2}\,\left(20\,mH\right)\,\left(37.8\,A\right)\,I_S\,cos\,\left(16.6^{\circ}\right)[/tex]

[tex]I_S\,=\,8.052\,\angle\,16.6^{\circ}[/tex]

Seems reasonable, but for some reason I don't think that is correct! What did I do wrong, can someone please explain?
 
Last edited:

1. What is a PMAC machine?

A PMAC (Permanent Magnet Alternating Current) machine is an electric machine that runs on alternating current and uses permanent magnets to generate a rotating magnetic field.

2. What is the significance of connecting a PMAC machine in Y?

Connecting a PMAC machine in Y allows for the machine to be operated in either a star or delta configuration, providing flexibility in its use and control.

3. What does it mean for a PMAC machine to have 4 poles?

The number of poles in a PMAC machine refers to the number of pairs of north and south poles found in the stator. A 4-pole PMAC machine would have 4 pairs of poles, resulting in a total of 8 poles.

4. How does a PMAC machine differ from other types of electric machines?

A PMAC machine differs from other types of electric machines, such as induction machines, in that it uses permanent magnets instead of electromagnets to generate the magnetic field. This results in higher efficiency and power density.

5. What is the role of power electronics in a PMAC machine?

Power electronics play a crucial role in a PMAC machine by controlling the flow of electric power to and from the machine. They are responsible for converting the direct current (DC) power from the permanent magnets to alternating current (AC) power used to drive the machine.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
7
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top