Q5Electric Drives

Given: 3-phase induction motor, P=6 poles, f=50Hz supply frequency, slip s=4%=0.04, stator supply V=400V (line, 50Hz), full-load efficiency η=92%=0.92, power factor cosφ=0.85.

(i) Synchronous Speed

The synchronous speed of an induction motor is determined purely by the supply frequency and the number of poles, independent of load or slip:

(ii) Actual Speed at Full Load

The actual rotor speed is related to synchronous speed and slip by N=Ns(1-s):

(iii) Rotor Frequency at Full Load

The rotor circuit frequency (the frequency of the EMF/current actually induced in the rotor windings) is directly proportional to slip, since it represents the relative speed between the rotating stator field and the actual rotor:

(iv) Developed Torque at Full Load — Additional Data Required

Developed torque is fundamentally related to the motor's mechanical power output and its rotational speed by T=P_mech/ωm=P_mech×60/(2πN), or, if computed from air-gap power instead, T=Pag/ωs. To compute a specific numerical torque value, the motor's actual output (or input) power rating must be known — however, the problem as given does not specify the motor's rated power in kW or HP (or any other equivalent power-rating parameter), meaning the developed torque cannot be calculated as a specific absolute numerical value from the data actually provided. If the motor's rated output power (say, Pout in kW) were given, the developed torque could be calculated directly as T=(Pout×1000)/(2πN/60) N·m, using the full-load speed N=960 rpm found in part (ii); alternatively, if the input electrical power were specified, the output power could first be found via Pout=Pin×η, before applying the same torque formula. Without this additional power-rating data point, only the calculation method can be presented, not a specific final numerical torque value.

(v) Current Drawn from the Mains at Full Load — Additional Data Required

Similarly, the full-load line current drawn from the 400V, 50Hz mains supply is related to the motor's input electrical power by the standard three-phase power formula:

Since Pin can be found from the (also not given) output power rating and the stated 92% efficiency (Pin=Pout/η), and Pout itself is not specified in the given problem data, the full-load line current similarly cannot be calculated as a specific numerical value without this missing rated-power information. If, for illustration, the motor's rated output power were (for example) assumed to be 10 kW, then Pin=10/0.92=10.87kW, giving IL=10,870/(√3×400×0.85)=18.47A — but since no such rated power value was actually specified in the given problem statement, this remains only an illustrative example of the calculation method, not the actual answer to this specific problem as posed.

Summary: parts (i), (ii) and (iii) of this problem are fully determined by the given data (synchronous speed, slip, and supply frequency alone), giving Ns=1000 rpm, N=960 rpm, and fr=2 Hz respectively; however, parts (iv) and (v) additionally require the motor's rated power (kW or HP) output, which is not provided in the problem statement as given — a complete numerical solution to parts (iv) and (v) would require this additional data point to be specified, after which the calculation would proceed directly using the standard torque and three-phase power formulas presented above.

Why the Missing Power Rating Cannot Be Worked Around

It is worth being explicit about why the efficiency and power factor values given in the problem (92% and 0.85 respectively) are not, by themselves, sufficient to determine torque or current, even though they look like they should be 'enough' extra data. Efficiency is a dimensionless ratio (η=Pout/Pin) and power factor is likewise dimensionless (cosφ=Pin/(√3·VL·IL) in terms of apparent power) — both relate one power/current quantity to another, but neither one, nor their combination, introduces an actual absolute magnitude (a wattage, kW, or HP figure) into the problem. Mathematically, the equations available are Pin=Pout/η and IL=Pin/(√3·VL·cosφ); both are correctly stated and usable, but both contain Pout (or equivalently Pin) as the one remaining unknown quantity, and no combination of slip, speed, frequency, voltage, efficiency, or power factor alone can generate a wattage value out of thin air — a genuine power magnitude (kW or HP rating, or equivalently a specified line current or torque value from which power could be back-calculated) must be supplied from outside these ratios. This is a basic dimensional-analysis check that is worth applying whenever a problem seems to provide 'enough' parameters: counting the number of independent equations against the number of unknowns shows that, for parts (iv) and (v), there are two unknowns (Pout, and then IL derived from it) but the given data supplies relationships among ratios only, leaving the system under-determined by exactly one independent absolute-magnitude data point.

Why such data-omission errors occur in real exam papers: in practice, numerical induction-motor problems are frequently adapted or abbreviated from larger textbook problem sets, and a rated power/HP value is one of the most common figures to be accidentally dropped during transcription or when a question is shortened to fit a fixed mark allocation — particularly here, where the problem already supplies five other parameters (poles, frequency, slip, voltage, efficiency, power factor) that create the appearance of a fully-specified problem even though the specific combination given cannot support parts (iv) and (v). This is a well-recognized category of exam-paper error, and the professionally correct response, as applied in this answer, is to solve every part that the given data does support fully and rigorously, explicitly identify the specific missing data item preventing further progress, and present the method/formula that would immediately yield a numerical answer once that one missing figure is supplied — rather than silently assuming an arbitrary, unstated power rating merely to force a numerical result, which would misrepresent the actual, verifiable content of the answer.

Physical Significance of the Three Computed Quantities

Beyond their numerical values, Ns, N, and fr each carry distinct physical meaning worth noting. The synchronous speed Ns=1000 rpm represents the speed of the rotating magnetic field produced by the three-phase stator winding — a speed fixed entirely by the electrical supply frequency and the winding's pole configuration, completely independent of the rotor, load, or any mechanical property of the machine; it is the speed the rotor would have to reach (impossible in a normal induction motor) for slip, and hence all rotor-induced torque production, to vanish entirely. The actual rotor speed N=960 rpm, lower than Ns by the slip fraction, reflects the fundamental operating principle of the induction motor: the rotor must always rotate slightly slower than the stator field so that a relative (slip) velocity exists between the two, inducing the rotor EMF and current that are themselves the sole mechanism by which the rotor develops torque — an induction motor operating at exactly zero slip would, by the same mechanism, develop exactly zero torque, so some nonzero slip is an unavoidable structural feature of normal loaded operation, not a design flaw. The rotor frequency fr=2 Hz is the frequency actually seen by the rotor conductors and any rotor-circuit components (in a wound-rotor machine, this is the frequency present at the slip rings), and this low value (compared to the 50Hz stator supply) is precisely why rotor-circuit iron losses in a normally-operating induction motor are small relative to stator iron losses — since core (hysteresis and eddy-current) losses increase with frequency, and the rotor here only experiences a frequency of 2Hz rather than the full 50Hz stator frequency, confirming why rotor core losses are typically neglected or treated as minor in standard induction motor performance calculations (consistent with the problem's own instruction to neglect winding and core losses).