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

[VinnyCee]

Homework Statement

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

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

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

c) For speed 1.2\cdot\omega_1, voltage the same, 500\,\sqrt{3}\,V, 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.

T\,=\,3\,\frac{p}{2}\,L_M\,I_S\,I_F

V_S\,=\,\omega_S\,L_M\,I_F

\omega_S\,=\,\frac{p}{2}\,\cdot\,Speed\,in\,RPM\,\cdot\,\frac{2\,\pi}{60}

The Attempt at a Solution

a) Using the first equation above...

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

I_S\,=\,10.8\,A

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

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

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

Does that look right for part (a)? Also, how do I proceed for parts (b) and (c)?

 

[VinnyCee]

Trying part (b)...I_M\,=\,\left|\bar{I_S}\,+\,\bar{I_F}\right|\,=\,10.8\,A\,+\,27\,A\,=\,37.8\,A

\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}

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

Does part (b) look right? I've no idea how to do part (c). Probably something to do with finding an angle between I_S and I_F I think...

 

[VinnyCee]

Part (a) was done wrong...

V_S\,=\,\omega_S\,L_M\,I_M and not V_S\,=\,\omega_S\,L_M\,I_S

I_M\,=\,\sqrt{I_S^2\,+\,I_F^2}

I_S\,=\,10.8\,A as above, and I_F is given.

So...

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

Now...

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\,VV_S\,=\,487.3\,V and I_S\,=\,10.8\,APart (b) is apparently correct (not sure though - is it really?) So I'll try part (c).

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

So...

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

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

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

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

Using the first to get the angle \beta between V_S and I_F...

\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

\beta\,=\,106.6^{\circ}

\theta\,=\,\beta\,-\,90^{\circ}\,=\,16.6^{\circ}

Now I use the second torque equation to get the new I_S...

\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)

I_S\,=\,8.052\,\angle\,16.6^{\circ}

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