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...

$$V_S\,=\,\omega_S\,L_M\,I_M\,=\,\left(837.8\,\frac{rad}{sec}\right)\,\left(20\,mH\right)\,\left(29.08\,A\right)\,=\,487.3\,V$$$$V_S\,=\,487.3\,V$$ and $$I_S\,=\,10.8\,A$$Part (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?