Question 5 of 6: Problem 4 — current-source-inverter-fed induction motor

Question 5: Problem 4 — current-source-inverter-fed induction motor (15 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An eight-pole cage machine is fed at constant input current by a current-source inverter at rated frequency, and the developed torque at that operating point is stated; the per-phase parameters of the approximate equivalent circuit are supplied.

Find. (a) the slip and the rotor speed at which the machine develops 180 N·m on 40 A; (b) the resulting terminal voltage per phase and the input power factor.

[Figure not reproduced: Figure (1) redrawn from the paper: the inverter behaves as a current source, so the rotor branch is fed through a current divider formed by the magnetising reactance and the series branch. See the official exam paper.]

Approach. Because the input current rather than the terminal voltage is held constant, the torque expression is a rational function of $R_{r}/s$ with a maximum; setting it to the demanded 180 N·m gives a quadratic with two roots, and the operating point is chosen on air-gap flux.

1. Compute the synchronous quantities. For eight poles at 50 Hz, $$n_{s}=\frac{120f}{P}=\frac{120\times50}{8}=750\ \text{rpm}\qquad \omega_{s}=\frac{4\pi f}{P}=78.540\ \text{rad/s}$$ and the total reactance is $X_{T}=X_{m}+X_{s}+X_{r}=13.017\ \Omega$.
2. Write the constant-current torque from the circuit of Fig. (1). The rotor branch current follows from the current divider, $I_{r}=I_{i}\,jX_{m}/(Z_{br}+jX_{m})$, and substituting it into $T=3I_{r}^{2}(R_{r}/s)/\omega_{s}$ gives $$T=\frac{3\,(X_{m}I_{i})^{2}(R_{r}/s)}{\omega_{s}\left[(R_{s}+R_{r}/s)^{2}+X_{T}^{2}\right]}$$ The numerator constant is $3(10.417\times40)^{2}/78.540=6631.9$.
3. Locate the breakdown point before solving. Differentiating with respect to $u=R_{r}/s$ puts the maximum at $u=\sqrt{R_{s}^{2}+X_{T}^{2}}=13.019$, i.e. $$s_{max}=\frac{R_{r}}{13.019}=0.01690\qquad T_{max}=250.9\ \text{N}\cdot\text{m}$$ The demanded 180 N·m is below this, so a solution exists — and there will be two of them, one either side of $s_{max}$.
4. Solve the quadratic in $u$. Rearranging $T=180$ gives $$180u^{2}-6559.88\,u+30506.81=0\ \Rightarrow\ u=30.972\ \text{or}\ 5.4722$$ so $$s=\frac{R_{r}}{u}=0.00710\quad\text{or}\quad 0.04020$$

Choosing between the roots is the substance of part (a), and the deciding test is air-gap flux, not static stability. The magnetising current is what the injected 40 A does not send into the rotor branch, and the terminal voltage follows from it directly as $V_{ph}=I_{m}X_{m}$.

5. Test each root on flux. With $Z_{br}=(R_{s}+R_{r}/s)+j(X_{s}+X_{r})$ and $I_{m}=I_{i}\,Z_{br}/(Z_{br}+jX_{m})$: at $s=0.00710$, $I_{m}=37.04$ A of the 40 A injected, giving $V_{ph}=385.8$ V; at $s=0.04020$, $I_{m}=17.58$ A giving $V_{ph}=183.1$ V. The low-slip root starves the rotor branch and demands roughly twice the flux, which the iron cannot support, so it is rejected: $$\boxed{s=0.0402}$$
6. Convert to rotor speed. $$n_{r}=n_{s}(1-s)=750\times(1-0.040203)=\boxed{719.8\ \text{rpm}}$$
7. Take the terminal voltage from the parallel combination. $Z_{eq}=jX_{m}\parallel Z_{br}=3.0529+j3.4110\ \Omega$, so $$V_{ph}=I_{i}|Z_{eq}|=40\times4.5777=\boxed{183.1\ \text{V per phase}}$$
8. Read the power factor off the same impedance. $$\cos\varphi=\frac{\Re\{Z_{eq}\}}{|Z_{eq}|}=\frac{3.0529}{4.5777}=\boxed{0.667\ \text{lagging}}$$

One power balance closes the whole problem. The input to the machine must equal the converted power plus the stator copper loss: $3V_{ph}I_{i}\cos\varphi=3\times183.11\times40\times0.66691=14\,654$ W, while $T\omega_{s}+3I_{r}^{2}R_{s}=180\times78.540+3\times29.345^{2}\times0.2=14\,654$ W. Agreement to five figures confirms the root, the current divider and the parallel combination all at once.