Question 6 of 8: DC Test and Slip Behaviour of an Induction Motor

Question 6: DC Test and Slip Behaviour of an Induction Motor (equal value)

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) per-phase stator resistance $r_1$; (b) synchronous speed; (c) rotor speed; (d) rotor electrical frequency; (e) rotor speed at double load; and, in Part II, the graphical motor–pump operating point.

Part I

Approach. Reduce the delta network seen by the dc source to get $r_1$, then apply the standard slip relations, which need only $f_e$, $P$ and $s$.

1. (a) Unfold the delta. Between two terminals of a delta winding, one phase of resistance $r_1$ is in parallel with the other two in series: $$R_{dc}=\frac{V_{dc}}{I_{dc}}=\frac{3.32}{3.1}=1.071\ \Omega,\qquad R_{dc}=\frac{r_1\cdot2r_1}{r_1+2r_1}=\frac{2}{3}r_1 \;\Longrightarrow\;r_1=\tfrac{3}{2}R_{dc}=\boxed{1.606\ \Omega\text{ per phase}}.$$
2. (b) Synchronous speed. $$n_{sync}=\frac{120f_e}{P}=\frac{120\times60}{6}=\boxed{1200\ \text{r/min}}.$$
3. (c) Rotor speed at 3.5% slip. By definition $s=(n_{sync}-n_m)/n_{sync}$, so $$n_m=(1-s)n_{sync}=(1-0.035)(1200)=\boxed{1158\ \text{r/min}}.$$
4. (d) Rotor electrical frequency. $$f_r=sf_e=0.035\times60=\boxed{2.1\ \text{Hz}}.$$
5. (e) Rotor speed at double load. Near synchronous speed the torque–slip curve is essentially linear ($T\propto s$, since $sX_2\ll R_2$ there), so doubling the load torque doubles the slip: $$s'=2\times0.035=0.07,\qquad n_m'=(1-0.07)(1200)=\boxed{1116\ \text{r/min}}$$ — only a 42 r/min (3.5%) drop for a doubling of load, the near-constant-speed behaviour that makes the induction motor an industrial workhorse.

Part II — graphical operating point

This part asks for a method, since neither the motor curve nor $K$ is given numerically ($K_p$ is symbolic). Steady operation occurs where the torque the motor develops equals the torque the pump demands, at a common shaft speed.

1. Plot both characteristics. Torque on the vertical axis, speed $n$ (rev/s) on the horizontal axis; draw the wound-rotor motor curve $T_m(n)$ from the supplied graph, converting its speed axis to rev/s if it is given in r/min, which rises from locked-rotor torque to a breakdown peak and then falls steeply toward zero at synchronous speed.
2. Superimpose the load law. Plot $T_L=K_pn^{2}$, a parabola through the origin (a centrifugal-type load takes negligible torque at standstill, rising with the square of speed).
3. Read the intersection. $$T_m(n)=K_pn^{2}\;\Longrightarrow\;\boxed{n=n_{op}}$$ gives the operating speed (in r/min, $60\,n_{op}$); the shaft power follows as $P=2\pi n_{op}T_{op}$.
4. Check stability. The intersection must lie where $dT_m/dn \lt dT_L/dn$ (the steep, high-speed side of the motor curve), so that a small speed excursion is self-correcting.