Question 6 of 8: Induction Motor — DC Test, Slip, and Graphical Operating Point

Question 6: Induction Motor — DC Test, Slip, and Graphical Operating Point (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] the per-phase stator resistance $r_1$; [b] the synchronous speed in rpm; [c] the rotor speed in rpm; [d] the rotor electrical frequency; [e] the rotor speed when the load torque is doubled; and, in Part II, the method for locating the operating speed of the motor-pump combination.

Approach. Reduce the measured terminal resistance to a per-phase value using the delta network's parallel-series geometry, then apply the standard synchronous-speed, slip and rotor-frequency relations. Part II is answered by superimposing the load law on the machine characteristic and finding their intersection.

Part I

1. [a] Reduce the dc measurement to one phase. The dc source is applied between two of the three motor terminals, so the measured resistance is $$R_{\text{meas}}=\frac{V_{DC}}{I_{DC}}=\frac{3.32}{3.1}=1.0710\ \Omega$$ In a delta winding, that measurement sees one phase winding in parallel with the other two in series: $$R_{\text{meas}}=\frac{r_1\cdot 2r_1}{r_1+2r_1}=\frac{2r_1^{2}}{3r_1}=\frac{2}{3}r_1 \quad\Longrightarrow\quad r_1=1.5\,R_{\text{meas}}$$ $$r_1=1.5\times1.0710=1.6065\ \Omega$$ $$\boxed{\;r_1=1.61\ \Omega\ \text{per phase}\;}$$ Rebuilding the delta from this value returns $\left(1.6065\times 2\times 1.6065\right)/\left(3\times 1.6065\right)=1.0710\ \Omega$, confirming the reduction.
2. [b] Speed of the rotating magnetic field. The synchronous speed depends only on the supply frequency and the pole count: $$n_{\text{sync}}=\frac{120f}{P}=\frac{120\times60}{6}=1200\ \text{rpm}$$ $$\boxed{\;n_{\text{sync}}=1200\ \text{rpm}\;}$$
3. [c] Rotor speed at 3.5 % slip. Slip is the fractional shortfall of the rotor behind the field, $s=(n_{\text{sync}}-n_m)/n_{\text{sync}}$, so $$n_m=(1-s)\,n_{\text{sync}}=(1-0.035)(1200)=0.965\times1200=1158\ \text{rpm}$$ $$\boxed{\;n_m=1158\ \text{rpm}\;}$$
4. [d] Electrical frequency of the rotor current. The rotor conductors are cut by the field at the slip speed, so the rotor frequency is the supply frequency scaled by the slip: $$f_r=s\,f=0.035\times60=2.1\ \text{Hz}$$ $$\boxed{\;f_r=2.1\ \text{Hz}\;}$$ This very low frequency is the reason a running induction motor's rotor iron losses are negligible compared with the stator's.
5. [e] Rotor speed with the load doubled. In normal running the machine operates on the steep, nearly linear part of its speed-torque curve, where developed torque is proportional to slip. Doubling the load torque therefore doubles the slip: $$s_{\text{new}}=2\times0.035=0.070 \quad\Longrightarrow\quad n_m=(1-0.070)(1200)=0.930\times1200=1116\ \text{rpm}$$ $$\boxed{\;n_m=1116\ \text{rpm at double load}\;}$$ Equivalently, the slip speed doubles from 42 rpm to 84 rpm, which is the same statement and a quick way to check the arithmetic.

Part II — locating the operating point graphically

The pump imposes $T_{\text{load}} = K_p n^{2}$, a parabola through the origin; the wound-rotor motor supplies the torque given by its speed-torque characteristic. Because motor and pump are on one shaft they must share a single speed, and in steady state the two torques must be equal. The operating point is therefore the intersection of the two curves, and the method is as follows.

1. Put both curves on one set of axes. Plot the supplied motor characteristic with torque on the vertical axis and speed on the horizontal axis, using consistent units. If the motor curve is drawn against slip, convert with $n=n_{\text{sync}}(1-s)$ so that both curves share the same abscissa.
2. Superimpose the load law. Evaluate $T=K_p n^{2}$ at several speeds from zero to synchronous and draw the parabola on the same axes. It starts at the origin, which is why an induction motor can always start a centrifugal pump: at standstill the load demands no torque, so the entire locked-rotor torque is available to accelerate the inertia.
3. Read the intersection. The speed at which the two curves cross satisfies $T_{\text{motor}}(n)=K_p n^{2}$, which is the steady-state condition. Drop a vertical line from the crossing to the speed axis; that abscissa is the operating speed, and the ordinate is the operating torque.
4. Check that the intersection is stable. A crossing is a stable operating point only if the motor torque falls more steeply with speed than the load torque rises — that is, if it lies on the high-speed side of the breakdown point. Then a chance rise in speed makes the load demand exceed the motor supply and the system decelerates back; a chance fall does the reverse. An intersection on the low-speed side of breakdown is an unstable equilibrium and the machine will stall or run away from it.
5. Worked illustration. Taking a machine with $n_{\text{sync}}=20$ rev/s, breakdown torque 120 N·m at 30 % slip, driving a pump with $K_p=0.38$ N·m per (rev/s)$^2$, solving $T_{\text{motor}}(n)=0.38\,n^{2}$ numerically gives $$\boxed{\;n_{\text{op}}=16.54\ \text{rev/s}=992\ \text{rpm},\qquad T_{\text{op}}=103.9\ \text{N}\cdot\text{m}\;}$$ which lies above the breakdown speed of 14 rev/s and is therefore a stable operating point.

A wound-rotor machine adds a useful degree of freedom to this picture. Inserting external resistance in the rotor circuit moves the whole characteristic to the left without changing the breakdown torque, so the intersection with the fixed pump parabola slides down to a lower speed. That is the classical method of speed control for pump and fan drives, and the graphical construction above is exactly how the required rotor resistance would be selected: draw the pump curve, mark the desired operating speed, and choose the rotor resistance whose characteristic passes through that point.