Q9Computer Aided Design of Electrical Machines

Question

Q.5. (a) Deduce the output equation of a 3-phase induction motor in terms of its specific loadings. Why the length of air gap in induction motor is kept as minimum as possible? [7]

(b) Determine the main dimensions, turns per phase of a 250 hp, 3-phase, 50 Hz, 400 V, 1410 rpm, slip ring induction motor. Assume: Average flux density in air gap = 0.5 Wb/m^2, specific elect. loading = 30000 Ac/m, efficiency = 0.9, power factor = 0.9, winding = 0.955, ratio of core length to pole pitch = 1.2. The machine is delta connected. [9]

Answer 1

(a) Output Equation of 3-Phase Induction Motor

The output equation of a 3-phase induction motor is derived using an approach directly analogous to that used for the alternator output coefficient (discussed in the corresponding earlier answer), relating the motor's rated input kVA to its main dimensions (stator bore diameter D and core length L) and its synchronous speed.

Derivation: the per-phase induced EMF in the stator winding (approximately equal to the applied per-phase voltage, neglecting stator impedance drop for this design-level approximation) is given, exactly as for the alternator, by Eph=4.44KwfphiTph, with flux per pole phi=BavtauL. Using f=PNs(rps), tau=piD/P, and combining with the specific electric loading relation ac=(6TphIph)/(pi*D) (using the same standard 3-phase winding ampere-conductor relation as in the alternator derivation), the induction motor's per-phase input kVA (input rather than output, since the induction motor's electrical input rating, before subtracting losses/slip effects, is the natural quantity directly related to these winding parameters) combines to give:

Q (input kVA) = 11 K_{w} B_{a v} a c \times 1 0^{- 3} \times D^{2} L N_{s} = C_{o} D^{2} L N_{s}

where Ns is the synchronous speed expressed in revolutions per second (rps), and Co=11KwBavac10^-3 is the output coefficient, identical in form to the alternator's output coefficient derived in the corresponding earlier answer, reflecting the shared underlying electromagnetic design principle common to all AC rotating machines. This output equation is the fundamental relationship used to determine an induction motor's main dimensions D and L for a specified input kVA rating, synchronous speed, and chosen specific loadings, exactly as applied in the accompanying numerical design problem.

Why the Air Gap Length is Kept Minimum in Induction Motors

The air gap of an induction motor is deliberately made as small as practically achievable (consistent with mechanical manufacturing tolerances, bearing clearances, and rotor mechanical deflection/vibration safety margins) for several important electromagnetic performance reasons: a smaller air gap reduces the magnetizing (exciting) current required to establish the working air-gap flux, since the air gap represents the dominant reluctance component of the motor's magnetic circuit (exactly as discussed in the corresponding air-gap MMF answer elsewhere in this paper) — since this magnetizing current is a purely reactive (lagging) component that does not contribute to useful torque production, minimizing it directly improves the motor's overall power factor, a particularly important performance characteristic for induction motors, which inherently operate at a lagging power factor and for which magnetizing-current-related reactive power represents a substantial fraction of total reactive power drawn from the supply, especially at light load. A smaller air gap also improves the motor's efficiency (by reducing magnetizing-current-related I^2R losses in the stator winding) and can improve certain aspects of the torque-producing mutual coupling between stator and rotor. In practice, however, the air gap cannot be made arbitrarily small, since it must remain large enough to reliably accommodate manufacturing tolerances in bore/shaft concentricity, bearing wear over the motor's service life, and rotor mechanical deflection under load, without risking a rotor-stator rub (mechanical contact) that would cause serious damage — the actual minimum practical air gap therefore represents a careful engineering trade-off between these competing electromagnetic performance and mechanical reliability considerations, with typical induction motor air gaps ranging from a fraction of a millimetre for small motors up to several millimetres for very large machines.

Slip and Rotor Frequency

The magnetizing-current and air-gap discussion above is closely linked to the concept of slip, defined as s=(Ns-N)/Ns, where Ns is synchronous speed and N is the actual rotor speed; slip is the fundamental quantity that permits induction motor operation in the first place, since it is only the relative (slip) speed between the rotating stator field and the rotor conductors that induces the rotor EMF, rotor current, and hence torque, exactly as noted for the 6% full-load slip found in the numerical solution below. The frequency of the currents actually induced in the rotor circuit, the rotor (slip) frequency fr=sf, is correspondingly only a small fraction of the supply frequency at normal running speeds (for example, fr=0.0650=3Hz at the 6% slip found here), which is why rotor-circuit reactance is comparatively small at running speed even though it can be substantial at standstill (s=1, fr=f). The air-gap flux and the associated magnetizing current calculated from the output equation in part (a) are, notably, essentially independent of slip (being determined by the stator applied voltage and frequency, exactly as for a transformer's exciting current), which is precisely why minimizing the air gap to control magnetizing current and power factor is a stator-and-air-gap-geometry design question rather than one resolved by the rotor's operating slip.

Squirrel-Cage vs Slip-Ring Rotor Choice

The motor in this numerical problem is specified as slip-ring (wound-rotor) type, a choice with direct bearing on both starting performance and air-gap/magnetizing-current considerations for a machine of this rating (250hp). A slip-ring rotor, with its three-phase winding brought out to external slip rings and connected to an external rotor (starting) resistance, allows the starting current to be limited and the starting torque to be substantially boosted (up to close to maximum/pull-out torque) by inserting external resistance at standstill and progressively cutting it out as the motor accelerates, making it well suited to applications requiring high starting torque with limited starting current, or where the connected load (such as a high-inertia fan, crane, or compressor) demands smooth, controllable acceleration — a particularly relevant consideration for a motor in the 250hp range, where direct-on-line starting of an equivalent squirrel-cage motor would draw a very large starting current from the supply. A squirrel-cage rotor, by contrast, is mechanically simpler, more rugged, and requires no slip rings or brushes (eliminating a maintenance item and a potential source of sparking), but offers comparatively little external control over its starting characteristic (limited to stator-side methods such as star-delta or autotransformer starting, or, in modern practice, solid-state soft starters and variable-frequency drives) and inherently provides lower starting torque per ampere of starting current than an externally-resistance-augmented slip-ring rotor. The choice between the two rotor types for a given rating is therefore governed primarily by the starting-torque and starting-current requirements imposed by the driven load, rather than by the air-gap and magnetizing-current considerations discussed above, which apply essentially identically to both rotor types since they concern the stator-side magnetic circuit common to both constructions.

(b) Numerical: Main Dimensions of 250hp Induction Motor

Given: rated output=250 hp, 3-phase, f=50Hz, V=400V (delta-connected), N=1410rpm, Bav=0.5 Wb/m^2, ac=30000 A/m, efficiency=0.9, power factor=0.9, Kw=0.955, L/tau ratio=1.2.

Step 1: Input kVA

P_{o u tp u t} = 250 \times 746 = 186, 500 W

Input kVA = \frac{P _{o u tp u t}}{η \times cos ϕ \times 1000} = \frac{186 , 500}{0.9 \times 0.9 \times 1000} \approx 230.2 kVA

Step 2: Number of Poles and Synchronous Speed

Since the rated (full-load) speed is 1410 rpm, the nearest standard synchronous speed above this value (for a positive, physically reasonable slip) at 50 Hz is 1500 rpm, corresponding to P=4 poles (giving a slip of (1500-1410)/1500=6%, a typical, physically reasonable full-load slip value for an induction motor of this size):

N_{s} = 1500 rpm = 25 rps

Step 3: Output Coefficient and D-Squared-L Product

C_{o} = 11 \times 0.955 \times 0.5 \times 30000 \times 1 0^{- 3} \approx 157.58

D^{2} L = \frac{230.2}{157.58 \times 25} \approx 0.05845 m^{3}

Step 4: Splitting D and L Using L/tau=1.2

Since tau=piD/P and L=1.2tau=1.2piD/4:

D^{2} \times (1.2 \times \frac{π D}{4}) = 0.05845 \Rightarrow D^{3} = \frac{0.05845 \times 4}{1.2 π} \approx 0.06204 \Rightarrow D \approx 0.3958 m

τ = \frac{π D}{4} \approx 0.3109 m, L = 1.2 τ \approx 0.3731 m

Step 5: Turns per Phase

For a delta-connected stator winding, phase voltage equals line voltage, Vph=400V. Flux per pole:

ϕ = B_{a v} \times τ \times L = 0.5 \times 0.3109 \times 0.3731 \approx 0.05799 Wb

T_{p h} = \frac{V _{p h}}{4.44 K _{w} f ϕ} = \frac{400}{4.44 \times 0.955 \times 50 \times 0.05799} \approx 32.5 turns

Result: the calculated main dimensions are D approximately 0.396m (stator bore diameter) and L approximately 0.373m (core length), with turns per phase approximately 33 (rounding to the nearest practical integer value, since a physical winding must have a whole number of turns) — these dimensions represent physically reasonable proportions for a 250 hp, 4-pole induction motor of this rating, consistent with typical industrial induction motor design practice.