Question 5 of 6: Current-Source-Inverter-Fed Induction Motor

Problem 5: Current-Source-Inverter-Fed Induction Motor (20 marks)

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.

Find. (a) what happens when frequency is reduced at constant voltage; (b) the slip and rotor speed at which the machine develops 180 N·m with 40 A injected; (c) the resulting terminal voltage per phase and the input power factor.

Approach. Answer (a) from the flux relation φ ∝ V/f. For (b), the constant-current torque expression rearranges into a quadratic in the slip, which has two roots; select the one consistent with a sensible air-gap flux. For (c), reduce the equivalent circuit to a single impedance seen by the current source and multiply by I_i.

Part (a) — reducing frequency at constant voltage. The stator e.m.f. obeys V ≈ E = 4.44 f N k_w Φ, so with V held at its rated value the air-gap flux is inversely proportional to frequency: Φ ∝ V/f. Lowering f therefore drives the flux up. Because the magnetic circuit is designed to sit just below the knee of the B–H curve at rated conditions, even a modest reduction in frequency pushes the iron into saturation. The consequences are all unfavourable. The magnetising current rises steeply and non-linearly — far faster than the flux itself once saturation begins — so the stator current increases at constant load and both stator copper loss and, on a current-fed drive, the inverter rating are penalised. Core loss rises as well: hysteresis loss varies roughly as fΦ^1.6–2 and eddy-current loss as (fΦ)², and the flux increase outruns the frequency reduction, so total iron loss grows. The waveform of the magnetising current becomes strongly peaked and rich in triplen harmonics, worsening the supply power factor and adding stray loss. Thermally the machine is doubly disadvantaged, since the shaft-driven cooling fan slows with the rotor while the losses increase, and overheating follows quickly.

On the torque side, breakdown torque rises approximately as (V/f)² and the synchronous speed falls in proportion to f, so the machine appears to gain pull-out capability; but that gain is illusory once saturation limits the useful flux, and the excess current is what actually destroys the machine. This is the reason variable-frequency drives use constant volts-per-hertz control below base speed, reducing the terminal voltage in step with frequency (with a low-speed boost to offset the stator resistance drop) so that flux and hence breakdown torque are held constant.

1. Part (b) — synchronous speed and the constant-current torque expression. For eight poles at 50 Hz, $$n_s = \frac{120f}{P} = \frac{120 \times 50}{8} = 750\ \text{rpm}, \qquad \omega_s = \frac{4\pi f}{P} = 78.540\ \text{rad/s}.$$ From the equivalent circuit of Figure (2) the rotor current is a current-divider fraction of the injected current, $$I_r = \frac{I_i X_m}{\sqrt{(R_s + R_r/s)^{2} + (X_m + X_s + X_r)^{2}}},$$ and the developed torque is the air-gap power divided by synchronous speed, $T = 3I_r^{2}(R_r/s)/\omega_s$, giving $$T = \frac{3\,[X_m I_i]^{2}\,(R_r/s)}{\omega_s\left[(R_s + R_r/s)^{2} + (X_m + X_s + X_r)^{2}\right]} .$$
2. Part (b) — substitute and reduce to a quadratic in s. With $X_m I_i = 12(40) = 480$, $(X_m+X_s+X_r) = 14.7\ \Omega$ so $(X_m+X_s+X_r)^2 = 216.09$, multiply numerator and denominator by $s^{2}$: $$T\,\omega_s\left[(R_s s + R_r)^{2} + 216.09\,s^{2}\right] = 3(480)^{2}R_r\,s .$$ Numerically $T\omega_s = 180(78.540) = 14\,137.2$ and $3(480)^2R_r = 138\,240$, so $$3\,055\,466\,s^{2} - 137\,109\,s + 565.49 = 0 \;\Longrightarrow\; s = 0.04028 \;\text{or}\; s = 0.004595 .$$
3. Part (b) — select the root and state the speed. Both roots satisfy the torque equation exactly; the breakdown slip is $s_{max} = R_r/\sqrt{R_s^{2} + 14.7^{2}} = 0.01360$ at $T_{max} = 295.3\ \text{N}\cdot\text{m}$, so the two solutions lie either side of the peak. The low-slip root drives almost all of the injected current into the magnetising branch — $I_m = 38.0\ \text{A}$ of the 40 A, giving a terminal voltage of 456 V per phase and roughly twice rated air-gap flux, which the iron cannot support. The high-slip root leaves $I_m = 15.0\ \text{A}$ and a sensible flux, so it is the operating point: $$\boxed{s = 0.0403\ (4.03\ \%), \qquad n_r = 750(1 - 0.0403) = 719.8\ \text{rpm}}$$ The rotor therefore runs 30.2 rpm below synchronous speed.
4. Part (c) — reduce the circuit to the impedance seen by the current source. At $s = 0.040279$ the rotor branch resistance is $R_r/s = 0.2/0.040279 = 4.9657\ \Omega$, so the series branch is $$Z_{br} = (R_s + R_r/s) + j(X_s + X_r) = 5.1657 + j2.7\ \Omega,$$ and in parallel with the magnetising branch $jX_m = j12\ \Omega$: $$Z_{in} = \frac{jX_m Z_{br}}{jX_m + Z_{br}} = \frac{(j12)(5.1657 + j2.7)}{5.1657 + j14.7} = \frac{69.95\,\angle 117.60^{\circ}}{15.581\,\angle 70.64^{\circ}} = 4.4889\,\angle 46.96^{\circ}\ \Omega .$$
5. Part (c) — terminal voltage and power factor. The inverter forces $I_i = 40\ \text{A}$ through that impedance, so $$\boxed{V_s = I_i |Z_{in}| = 40 \times 4.4889 = 179.6\ \text{V per phase}}$$ (equivalently $\sqrt{3}(179.6) = 311.0\ \text{V}$ line-to-line), and the phase angle of $Z_{in}$ is the angle between terminal voltage and injected current, so $$\boxed{\text{p.f.} = \cos 46.96^{\circ} = 0.683\ \text{lagging}}$$ Checking the power balance: input power $3V_sI_i\cos\varphi = 3(179.6)(40)(0.6825) = 14\,707\ \text{W}$, against air-gap power $T\omega_s = 14\,137\ \text{W}$ plus stator copper loss $3I_r^{2}R_s = 3(30.81)^{2}(0.2) = 569\ \text{W}$, a total of 14 706 W. The two agree, which confirms both the slip and the impedance reduction.