Question 6 of 8: Induction Motor — DC Test, Slip Relations and Load Matching

Question 6: Induction Motor — DC Test, Slip Relations and Load Matching (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. A 208 V, six-pole, 60 Hz induction motor whose stator is delta connected, subjected to a dc test at 3.32 V and 3.1 A, then run at 3.5 % slip and finally at double load. Part II supplies the motor's speed–torque characteristic and a pump torque that varies as the square of speed.

Find. The per-phase stator resistance, the synchronous and rotor speeds, the rotor-current frequency, the rotor speed when the load is doubled, and a method for locating the motor–pump operating point.

Approach. The dc test reduces the machine to a resistor network, so the whole of part [a] is deciding what that network is for a delta connection. The remaining parts follow from the synchronous-speed formula and the definition of slip, with the doubled-load case resting on the near-linear torque–slip relation in the normal operating region.

1. Find the resistance actually measured. $$R_{meas} = \frac{V_{DC}}{I_{DC}} = \frac{3.32}{3.1} = 1.071\ \Omega.$$ This is the resistance between two of the three motor terminals, not the resistance of one phase.
2. Convert the terminal measurement to a per-phase value. In a delta connection, a measurement between two terminals sees one phase winding in parallel with the other two in series: $$R_{meas} = \frac{r_1 \cdot 2r_1}{r_1 + 2r_1} = \frac{2}{3}\,r_1, \qquad\text{so}\qquad r_1 = \frac{3}{2}R_{meas}.$$ Hence $$r_1 = 1.5 \times 1.071 = \boxed{1.61\ \Omega\ \text{per phase}}$$ Identifying the connection first is essential: for a wye stator the same measurement would give $r_1 = R_{meas}/2 = 0.536\ \Omega$, a factor of three away.
3. Compute the synchronous speed (part [b]). The rotating field turns at $$n_s = \frac{120 f}{P} = \frac{120 \times 60}{6} = \boxed{1200\ \text{rev/min}}$$
4. Compute the rotor speed (part [c]). By the definition of slip, $s = (n_s-n)/n_s$, so the rotor runs at $$n = n_s(1-s) = 1200\,(1-0.035) = \boxed{1158\ \text{rev/min}}$$ The rotor must always lag the field: at synchronous speed there would be no relative motion, no induced rotor current and therefore no torque.
5. Compute the rotor-current frequency (part [d]). The rotor conductors are cut by the field at the slip speed, so the induced currents alternate at $$f_r = s f = 0.035 \times 60 = \boxed{2.1\ \text{Hz}}$$ The very low rotor frequency at normal slip is why rotor iron losses are negligible in steady running.
6. Find the speed at double load (part [e]). Over the normal operating region — that is, between no load and full load, well below breakdown — the torque–slip characteristic is very nearly a straight line through the origin, so $T \propto s$. Doubling the load torque therefore doubles the slip: $$s' = 2 \times 0.035 = 0.07, \qquad n' = 1200\,(1-0.07) = \boxed{1116\ \text{rev/min}}$$ The speed falls by only 42 rev/min for a doubling of load, which is the characteristic stiffness that makes the induction motor behave as a near-constant-speed drive.

Part II — locating the operating point. The motor and the pump are rigidly coupled, so in the steady state they must share a single speed and the torque the motor develops must equal the torque the pump demands. The procedure is therefore graphical:

1. Plot both characteristics on one pair of axes. Draw the motor's measured speed–torque curve, and on the same axes draw the pump's parabola $T = K_p n^{2}$, using the constant $K_p$ obtained from any one known duty point of the pump (for instance its rated torque at rated speed).
2. Read the intersection. The steady-state operating point is where the two curves cross, since only there is the net accelerating torque zero: $$T_{motor}(n) = K_p\,n^{2}.$$ The abscissa of that crossing is the operating speed of the system and the ordinate is the common shaft torque.
3. Confirm the intersection is stable. Below the crossing the motor torque exceeds the pump torque and the set accelerates; above it the pump demands more than the motor can supply and the set decelerates. Because the pump curve rises more steeply than the motor curve falls at that point, any disturbance is self-correcting and the operating point is stable. Only intersections satisfying this slope condition are usable.
4. Use the same construction for speed control. For a wound-rotor machine, adding external rotor resistance lowers the speed at which the motor develops any given torque, sliding the motor curve to the left; the intersection with the fixed pump parabola moves down and the pump runs slower. Repeating the construction for several rotor resistances yields the achievable speed range directly.