Question 5 of 6: Problem 4 — current-source-inverter drive, torque and slip

Question 5: Problem 4 — current-source-inverter drive, torque and slip (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.

Find. (a) the slip that develops 400 N·m; (b) the torque developed at a slip of 0.05.

Approach. Compute the synchronous mechanical speed, group the constants of the supplied torque expression, then clear the fractions to get a quadratic in slip for part (a) and a direct substitution for part (b). The printed expression and the expression that Figure (1) actually yields differ by one factor of s, so both are evaluated and the discrepancy is quantified rather than asserted.

1. Find the synchronous speed and group the constants. For an eight-pole machine on 50 Hz, $$\begin{aligned} n_s&=\frac{120f}{P}=\frac{120\times 50}{8}=750\ \text{rpm}\ \omega_s&=\frac{4\pi f}{P}=\frac{4\pi\times 50}{8}=78.540\ \text{rad/s} \end{aligned}$$ and the two groups that appear throughout are $$X_T=X_m+X_s+X_r=10.42+1.0+1.5=12.92\ \Omega,\qquad 3(X_m I_i)^2=3(10.42\times 12.5)^2=50\,895.19$$
2. Take the printed formula at face value and clear the fractions. Writing the supplied expression with $R_r/s$ substituted and multiplying numerator and denominator by $s^2$, $$T=\frac{3(X_mI_i)^2 R_r}{\omega_s\bigl[(R_s s+R_r)^2+X_T^2 s^2\bigr]}$$ which, remarkably, is a quadratic in s with the slip appearing only in the denominator. Setting $T=400$ N·m, $$400\times 78.540\times\bigl[(0.2s+0.3)^2+166.926\,s^2\bigr]=50\,895.19\times 0.3$$ $$166.9664\,s^2+0.12\,s-0.396013=0$$
3. Solve the quadratic and select the root. The discriminant is $0.0144+4(166.9664)(0.396013)=264.498$, so $$s=\frac{-0.12\pm 16.26340}{333.9328}\quad\Longrightarrow\quad s=+0.048343\ \text{or}\ s=-0.049062$$ The negative root corresponds to super-synchronous running, that is to generating, and is rejected for a motoring problem. Hence $$\boxed{\;s_a=0.04834\;}\qquad n=(1-s)n_s=713.7\ \text{rpm}$$ Note that this family of question is usually two-rooted — the circuit-correct torque expression is quadratic in $R_r/s$ and gives roots straddling the breakdown slip — but the printed formula's extra factor of s makes it quadratic in s itself, with a single positive root, so no flux-based root selection is needed here.
4. Part (b) — substitute the given slip. With $s=0.05$, $R_r/s=6.0$ and $(R_s+R_r/s)^2=38.44$, so with the printed formula $$T=\frac{50\,895.19\times 6.0}{0.05\times 78.540\times\bigl[38.44+166.926\bigr]} =\frac{305\,371.1}{806.47}$$ $$\boxed{\;T_b=378.7\ \text{N}\cdot\text{m}\;}$$ The two parts corroborate each other: 400 N·m at $s=0.04834$ and 378.7 N·m at the neighbouring slip of 0.05 lie on the same falling branch of the printed characteristic, and the ratio of the two torques matches the ratio of the two denominators to five figures.
5. Evaluate what Figure (1) actually yields. Solving the circuit of Figure (1) directly, the injected current divides between $jX_m$ and the series branch, so $I_r=I_i\,jX_m/(R_s+R_r/s+j X_T)$ and the developed torque is $3I_r^2(R_r/s)/\omega_s$: $$T_{circuit}=\frac{3(X_m I_i)^2 (R_r/s)} {\omega_s\bigl[(R_s+R_r/s)^2+X_T^2\bigr]}$$ This is the printed expression without the factor s that precedes $\omega_s$ in the denominator. Evaluated at the two operating points above it gives 19.34 N·m and 18.93 N·m — a factor of about twenty smaller.
6. Quantify the discrepancy rather than assert it. The circuit-correct expression has a maximum over slip at $s_{max}=R_r/\sqrt{R_s^2+X_T^2}=0.3/12.9215=0.023217$, where it develops only $$T_{max,circuit}=24.69\ \text{N}\cdot\text{m}$$ The 400 N·m demanded by part (a) is therefore unattainable at any slip with 12.5 A injected into this circuit — the corresponding quadratic in $R_r/s$ has a negative discriminant and no real root. The printed formula, by contrast, reproduces both sub-parts consistently. The examiner's intent is unambiguous: the paper instructs “use the following torque formula”, and only that formula closes the data.
7. Record the power balance at the chosen operating point. Solving the circuit at $s=0.04834$ gives $|I_r|=9.03$ A, $|I_m|=5.96$ A, a terminal voltage of 62.1 V per phase and an input power factor of 0.673. The balance $3V_{ph}I_i\cos\theta=T_{circuit}\,\omega_s+3I_r^2R_s$ closes at 1567.7 W on both sides to five figures, which confirms the current divider, the parallel combination and the circuit-correct torque. It also shows what the disagreement really is: 12.5 A of injected current can support about 1.5 kW of air-gap power at 78.5 rad/s, and 400 N·m at that speed would be 31.4 kW.

[Figure not reproduced: Torque against slip on the same machine, computed two ways. Left: the formula printed on the exam paper, which passes through both stated operating points. Right: the expression the circuit of Figure (1) actually yields, whose maximum over all slips is 24.69 N.m, so the 400 N.m demanded by part (a) . See the official exam paper.]