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

Question 6: DC Test and Slip Behaviour of an 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] 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 method for the 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 nothing but $f_e$, $P$ and $s$.

1. [a] Resistance seen by the dc source. From Figure 6 the dc supply is applied between two of the three line terminals: $$R_{DC} = \frac{V_{DC}}{I_{DC}} = \frac{3.32}{3.1} = 1.0710\ \Omega .$$ Because the test uses dc there is no induced voltage, no rotor current and no reactance — the reading is pure stator resistance.
2. Unfold the delta. Between two terminals of a delta winding, one phase of resistance $r_1$ appears in parallel with the other two in series: $$R_{DC} = \frac{r_1 \cdot 2r_1}{r_1 + 2r_1} = \frac{2}{3}\,r_1 \quad\Longrightarrow\quad r_1 = \tfrac{3}{2} R_{DC},$$ $$r_1 = 1.5 \times 1.0710 = \boxed{\;r_1 = 1.606\ \Omega\ \text{per phase}\;}$$ (Had the machine been wye connected the two phases would be in series and $r_1 = R_{DC}/2 = 0.536\ \Omega$ — a factor-of-three difference, which is why the connection must be read from the nameplate.)
3. [b] Synchronous speed. The stator field rotates at $$n_{\text{sync}} = \frac{120 f_e}{P} = \frac{120 \times 60}{6} = \boxed{\;1200\ \text{r/min}\;}$$
4. [c] Rotor speed at 3.5 % slip. By definition $s = (n_{\text{sync}} - n_m)/n_{\text{sync}}$, so $$n_m = (1-s)\,n_{\text{sync}} = (1 - 0.035)(1200) = \boxed{\;1158\ \text{r/min}\;}$$
5. [d] Rotor electrical frequency. The rotor conductors are cut by the field at the slip speed, so $$f_r = s\,f_e = 0.035 \times 60 = \boxed{\;2.1\ \text{Hz}\;}$$ Checking through the slip speed: $P(n_{\text{sync}} - n_m)/120 = 6(42)/120 = 2.1$ Hz.
6. [e] Rotor speed at double load. In the normal low-slip operating region the torque–slip curve is essentially linear, $T \propto s$, because $s X_2 \ll R_2$ and the rotor branch is resistance dominated. Doubling the load torque therefore doubles the slip: $$s' = 2 \times 0.035 = 0.07, \qquad n_m' = (1 - 0.07)(1200) = \boxed{\;1116\ \text{r/min}\;}$$ The speed falls by only 42 r/min — 3.5 % — for a doubling of load, which is the near-constant-speed behaviour that makes the induction motor an industrial workhorse.

Part II — graphical determination of the operating point

This part asks for a method, not a number, since neither the motor curve nor $K_P$ is given numerically. The system is in steady state when the torque the motor develops equals the torque the pump demands, at a common shaft speed.

1. Plot both characteristics on one set of axes. Take torque on the vertical axis and speed $n$ (rev/s, to match the pump law) on the horizontal axis, and draw the supplied wound-rotor motor characteristic $T_m(n)$: it starts at the locked-rotor torque at $n = 0$, rises to breakdown torque, then falls steeply to zero at synchronous speed $n_s$.
2. Superimpose the load law. On the same axes plot the pump demand $T_L = K_P n^{2}$ — a parabola through the origin, since a centrifugal pump takes essentially no torque at standstill and its torque rises with the square of speed.
3. Read the intersection. Steady operation requires zero net accelerating torque: $$T_m(n) = K_P n^{2} \quad\Longrightarrow\quad \boxed{\;n = n_{\text{op}}\;}$$ The abscissa of the intersection is the operating speed; its ordinate is the operating torque, and the shaft power follows as $P = 2\pi n_{\text{op}} T_{\text{op}}$.
4. Check that the intersection is stable. The intersection must lie on the steep, high-speed side of the motor curve, where $\mathrm{d}T_m/\mathrm{d}n < \mathrm{d}T_L/\mathrm{d}n$. Then a small speed rise makes the load torque exceed the motor torque and the machine decelerates back to $n_{\text{op}}$; a small speed drop does the reverse. An intersection on the rising side of the motor curve, below breakdown, would be unstable and the drive would either stall or run away to the stable point.
5. Confirm starting is possible. Finally verify that the motor curve lies above the parabola over the whole range from standstill to $n_{\text{op}}$; otherwise the set cannot accelerate from rest. With a wound-rotor machine this is straightforward — external rotor resistance shifts the peak torque toward standstill for starting and is then shorted out for running, which also lets the operating speed be adjusted deliberately by re-inserting resistance.