Three phase induction motor design

The output equation relates the kVA input rating $Q$ of a 3-phase induction motor to its main dimensions ($D$ = stator bore diameter, $L$ = core length), its specific loadings ($B_{av}$, $ac$) and its synchronous speed $n_s$.

Symbols

• $P$ = number of poles, $f$ = frequency, $n_s$ = synchronous speed (rps)
• $\Phi$ = flux per pole (Wb), $T_{ph}$ = turns per phase, $K_w$ = winding factor
• $E_{ph}$, $I_{ph}$ = phase voltage and phase current
• $B_{av}$ = specific magnetic loading (Wb/m²), $ac$ = specific electric loading (A/m)

Step 1: kVA input

Neglecting the small drop, $V_{ph} \approx E_{ph}$:

$$Q = 3E_{ph}I_{ph}\times 10^{-3}\ \text{kVA}$$

Step 2: Induced emf

$$E_{ph} = 4.44\, f\, \Phi\, T_{ph} K_w, \qquad f = \frac{P n_s}{2}$$

Step 3: Magnetic loading

Total flux around the gap is $P\Phi$ over area $\pi D L$:

$$B_{av} = \frac{P\Phi}{\pi D L} \;\Rightarrow\; \Phi = \frac{B_{av}\pi D L}{P}$$

Step 4: Electric loading

Total stator conductors $= 3 \times 2T_{ph} = 6T_{ph}$, each carrying $I_{ph}$:

$$ac = \frac{6 I_{ph} T_{ph}}{\pi D} \;\Rightarrow\; I_{ph}T_{ph} = \frac{ac\,\pi D}{6}$$

Step 5: Substitute

$$\begin{aligned} Q &= 3 \times 4.44 \times \frac{P n_s}{2} \times \frac{B_{av}\pi D L}{P} \times K_w \times \frac{ac\,\pi D}{6} \times 10^{-3} \\ &= 1.11\,\pi^2 B_{av}\, ac\, K_w\, D^2 L\, n_s \times 10^{-3} \\ &= 11\, B_{av}\, ac\, K_w \times 10^{-3}\, D^2 L\, n_s \end{aligned}$$

So

$$Q = C_0 D^2 L n_s, \qquad C_0 = 11 B_{av}\, ac\, K_w \times 10^{-3}$$

$C_0$ is called the output coefficient. The kVA input is found from the rated shaft output:

$$Q = \frac{\text{kW output}}{\eta \cos\phi}$$

Meaning of the result

• $D^2L = \dfrac{Q}{C_0 n_s}$, so the volume of the active part is directly proportional to rating and inversely proportional to speed and specific loadings.
• Higher $B_{av}$ and $ac$ give a smaller, cheaper machine, but power factor, temperature rise and overload capacity limit them.
• A high-speed motor is smaller than a low-speed motor of the same rating.
• Typical ranges: $B_{av} = 0.3$–$0.6$ Wb/m², $ac = 5000$–$45000$ A/m, $K_w \approx 0.955$ for a full-pitch, 60° phase-spread winding.

The product $D^2L$ is then split into $D$ and $L$ using a chosen ratio $L/\tau$ (for example $L/\tau = 1$ for good overall design) and a check on peripheral speed ($\pi D n_s \le 30$ m/s for normal designs).

The air gap of an induction motor is kept as small as mechanically possible, because the gap needs most of the magnetising mmf. Its final length is a compromise between the following factors.

1. Power factor (magnetising current)

The gap needs the largest share of magnetising ampere-turns. A larger gap increases the magnetising current and lowers the power factor. This is the main reason to keep the gap small.

2. Overload capacity

A larger gap reduces the zigzag and other leakage reactance. Lower leakage reactance gives a larger maximum (breakdown) torque, so the overload capacity improves with a larger gap.

3. Pulsation (tooth) losses and noise

Slot openings cause flux pulsations in the tooth tips. With a larger gap these pulsations are smoothed, so pulsation losses, magnetic noise and vibration are reduced.

4. Cooling

A larger gap allows more air to pass and gives better cooling of the rotor and stator surfaces.

5. Unbalanced magnetic pull

Any eccentricity of the rotor produces a one-sided pull. Its effect is relatively smaller with a larger gap because the percentage change in gap is smaller.

6. Mechanical considerations

The gap must allow for shaft deflection, bearing wear, and manufacturing tolerances of stator and rotor. Machines with long cores or large diameter need larger gaps.

7. Harmonics

A larger gap reduces the effect of slot harmonics, so harmful harmonic torques (cogging, crawling) are reduced.

Empirical formula

For small and medium machines (Sawhney):

$$l_g = 0.2 + 2\sqrt{DL}\ \text{mm}$$

with $D$ and $L$ in metres. Typical values are 0.25 mm to 2 mm; for example, a motor with $D = 0.16$ m and $L = 0.085$ m has $l_g = 0.2 + 2\sqrt{0.0136} = 0.43$ mm.

The number of stator slots $S_s$ is fixed after the main dimensions are known. A large number of slots and a small number of slots each have advantages, so the choice is a compromise.

Advantages of a large number of slots

• Lower leakage reactance (conductors spread over more slots), so higher overload capacity.
• Lower tooth pulsation losses and less magnetic noise.
• Better heat dissipation, because the copper is spread over more slot surface.
• Smaller slots, so fewer conductors per slot and lower slot insulation stress.

Disadvantages of a large number of slots

• More slots means more insulation and more labour, so higher cost.
• Teeth become narrow and mechanically weak, and the tooth flux density may exceed the limit.
• A larger fraction of the slot area is taken by insulation, so the slot space factor is poor.
• Magnetising current increases because the teeth may saturate.

Practical rules for selecting $S_s$

1. Slot pitch: $y_s = \pi D/S_s$ should normally be between 15 mm and 25 mm (smaller for small motors, about 10–15 mm).
2. Integral slot winding: $S_s = 3 \times P \times q$ where $q$ (slots per pole per phase) is a whole number, preferably 2 or more ($q \ge 3$ for good emf waveform).
3. Tooth flux density: the minimum tooth width must keep $B_t \le 1.7$ Wb/m².
4. Conductors per slot: $6T_s/S_s$ must be a whole number (even for a double-layer winding).
5. Slot combination: $S_s$ must suit the rotor slot number $S_r$; $S_s \ne S_r$ and certain differences are avoided to prevent cogging, crawling, noise and vibration.

Example

For a 4-pole motor with $D = 0.18$ m, try $q = 3$: $S_s = 3 \times 4 \times 3 = 36$, slot pitch $= \pi \times 180/36 = 15.7$ mm, which lies in the normal range, so 36 slots are chosen.

Specific electric loading ($ac$) is the total stator ampere-conductors per metre of gap periphery:

$$ac = \frac{6 I_{ph} T_{ph}}{\pi D}$$

Typical values for induction motors are 5000–45000 A/m. Its choice depends on:

1. Copper loss and temperature rise: a higher $ac$ means more copper per unit surface, so more $I^2R$ loss and a higher temperature rise. Good ventilation allows a higher $ac$.

2. Operating voltage: high-voltage machines need thicker insulation, so less slot space is left for copper and a lower $ac$ must be used.

3. Size and cost: from $Q = C_0D^2Ln_s$, a higher $ac$ gives a smaller and cheaper machine.

4. Overload capacity: a higher $ac$ means more turns and higher leakage reactance, which lowers the maximum torque and overload capacity.

5. Size of machine: larger machines (larger $D$) have more space for copper and can use a higher $ac$.

6. Space factor and slot design: a high $ac$ needs more copper in the slots; deeper slots raise slot leakage, wider slots weaken the teeth.

Example: a motor with 1800 conductors carrying 11.16 A on a 0.276 m bore has $ac = 1800 \times 11.16/(\pi \times 0.276) \approx 23000$ A/m, a normal value for a medium motor.

In short, a high value saves material but raises temperature rise and leakage reactance; the designer chooses a value consistent with the required overload capacity and cooling.

Method: find the specific loadings (as the output coefficient $C_0$) from the 15 kW motor, then use the same $C_0$ for the 11 kW motor.

Output equation: $Q = C_0 D^2 L n_s$, $C_0 = 11 B_{av}\, ac\, K_w \times 10^{-3}$.

Specific loadings from the 15 kW motor

1480 rpm at 50 Hz means 4 poles, $n_s = 1500/60 = 25$ rps.

$$\begin{aligned} Q_1 &= \frac{15}{0.84 \times 0.85} = 21.01\ \text{kVA} \\ C_0 &= \frac{Q_1}{D^2 L n_s} = \frac{21.01}{0.25^2 \times 0.16 \times 25} = 84.03 \\ B_{av} &= \frac{C_0}{11\, ac\, K_w \times 10^{-3}} = \frac{84.03}{11 \times 23000 \times 0.955 \times 10^{-3}} = 0.348\ \text{Wb/m}^2 \end{aligned}$$

(i) Main dimensions of the 11 kW motor

6 poles: $n_s = 1000/60 = 16.67$ rps.

$$\begin{aligned} Q &= \frac{11}{0.84 \times 0.85} = 15.41\ \text{kVA} \\ D^2L &= \frac{Q}{C_0 n_s} = \frac{15.41}{84.03 \times 16.67} = 0.0110\ \text{m}^3 \end{aligned}$$

Assume $L = \tau$ (good overall design), $\tau = \pi D/6$:

$$D^3 = \frac{0.0110 \times 6}{\pi} = 0.02101 \Rightarrow D = 0.276\ \text{m}, \quad L = \tau = 0.1445\ \text{m}$$

(ii) Number of stator slots

Slot pitch 15–25 mm. With $q = 2$: $S_s = 3 \times 6 \times 2 = 36$, slot pitch $= \pi \times 275.9/36 = 24.1$ mm (within range). Take 36 slots.

(iii) Stator conductors and conductor area

$$\begin{aligned} \Phi &= B_{av}\tau L = 0.348 \times 0.1445 \times 0.1445 = 7.26 \times 10^{-3}\ \text{Wb} \\ T_s &= \frac{E_{ph}}{4.44 f \Phi K_w} = \frac{460}{4.44 \times 50 \times 7.26\times10^{-3} \times 0.955} = 298.9 \end{aligned}$$

(Delta connection, so $E_{ph} = 460$ V.) Conductors $= 6T_s = 1793$; per slot $= 1793/36 = 49.8$, take 50. Total conductors $Z_s = 36 \times 50 = 1800$, $T_s = 300$.

$$\begin{aligned} I_{ph} &= \frac{15406}{3 \times 460} = 11.16\ \text{A} \\ a_s &= \frac{11.16}{4} = 2.79\ \text{mm}^2 \ (\text{dia} \approx 1.89\ \text{mm}) \end{aligned}$$

Slot area:

$$A_{slot} = \frac{50 \times 2.79}{0.4} = 348.9\ \text{mm}^2$$

(iv) Slot dimensions

$d = 3w$, so $3w^2 = 348.9$: $w = 10.8$ mm, $d = 32.4$ mm.

(v) Minimum width of stator teeth

Maximum tooth density 1.7 Wb/m², net iron length $L_i = 0.9L = 0.130$ m, teeth per pole $= 36/6 = 6$:

$$W_{ts(min)} = \frac{\Phi}{1.7 \times \frac{S_s}{P} \times L_i} = \frac{7.26\times10^{-3}}{1.7 \times 6 \times 0.130} = 5.47\ \text{mm}$$

(Stator core depth with $B_c = 1.2$ Wb/m²: $d_{cs} = \Phi/(2B_cL_i) = 23.3$ mm.)

Answer: $D = 27.6$ cm, $L = 14.45$ cm; 36 slots; 1800 conductors (50 per slot, $T_s = 300$), $a_s = 2.79$ mm², slot area 349 mm²; slot $10.8 \times 32.4$ mm; $W_{ts(min)} = 5.47$ mm.

i) Stator slots should never equal rotor slots

• If $S_s = S_r$, every rotor tooth lines up with a stator tooth at the same time. The reluctance of the magnetic path is then minimum at these positions and the rotor tends to stay there.
• This strong alignment torque is called cogging or magnetic locking. The motor may fail to start, especially at reduced voltage, even though full voltage is applied.
• Equal slot numbers also give strong slot-harmonic torques, which cause noise and vibration.
• Therefore $S_r$ is always made different from $S_s$ (and differences like $S_s - S_r = \pm P$, $\pm 2P$, $\pm 5P$ are also avoided).

ii) Higher-speed machine is smaller for the same output

From the output equation:

$$Q = C_0 D^2 L n_s \;\Rightarrow\; D^2 L = \frac{Q}{C_0 n_s}$$

• For the same output $Q$ and the same specific loadings ($C_0$ fixed), the active volume $D^2L$ is inversely proportional to the synchronous speed $n_s$.
• A higher speed means fewer poles. The same power is produced with less torque ($T = P/\omega$), and torque decides the size of the machine.
• Example: a 4-pole motor ($n_s = 25$ rps) needs only half the $D^2L$ of an 8-pole motor ($n_s = 12.5$ rps) of the same rating.
• Hence a high-speed machine is smaller, lighter and cheaper.

Data: 2.2 kW, 440 V, 50 Hz, 1430 rpm (4 poles, $n_s = 1500$ rpm $= 25$ rps), $\eta = 0.8$, $\cos\phi = 0.85$, $B_{av} = 0.44$ Wb/m², $ac = 21000$ A/m. Assumptions: $K_w = 0.955$; delta-connected stator ($E_{ph} = 440$ V); tooth flux density limit 1.7 Wb/m²; iron stacking factor 0.9.

(i) Main dimensions

$$\begin{aligned} Q &= \frac{2.2}{0.8 \times 0.85} = 3.235\ \text{kVA} \\ C_0 &= 11 B_{av}\, ac\, K_w \times 10^{-3} = 11 \times 0.44 \times 21000 \times 0.955 \times 10^{-3} = 97.07 \\ D^2L &= \frac{Q}{C_0 n_s} = \frac{3.235}{97.07 \times 25} = 1.333 \times 10^{-3}\ \text{m}^3 \end{aligned}$$

Best power factor: $\tau = \sqrt{0.18L}$, so $L = \tau^2/0.18 = (\pi D/4)^2/0.18$:

$$\begin{aligned} D^2 \cdot \frac{\pi^2 D^2}{16 \times 0.18} &= 1.333\times10^{-3} \\ D^4 &= 3.889 \times 10^{-4} \Rightarrow D = 0.1404\ \text{m} \\ L &= \frac{1.333\times10^{-3}}{0.1404^2} = 0.0676\ \text{m} \end{aligned}$$

Check: $\tau = \pi \times 0.1404/4 = 0.1103$ m $= \sqrt{0.18 \times 0.0676}$.

(ii) Stator turns per phase

$$\begin{aligned} \Phi &= B_{av}\tau L = 0.44 \times 0.1103 \times 0.0676 = 3.281 \times 10^{-3}\ \text{Wb} \\ T_s &= \frac{440}{4.44 \times 50 \times 3.281\times10^{-3} \times 0.955} = 632.6 \end{aligned}$$

(iv) Number of stator slots

Slot pitch 12–15 mm. With $q = 3$: $S_s = 3 \times 4 \times 3 = 36$, slot pitch $= \pi \times 140.4/36 = 12.26$ mm (OK). 36 slots. Conductors per slot $= 6 \times 632.6/36 = 105.4$, take 106, so $Z_s = 3816$ and $T_s = 636$.

(iii) Area of stator conductor

$$I_{ph} = \frac{3235.3}{3 \times 440} = 2.451\ \text{A}, \qquad a_s = \frac{2.451}{4} = 0.613\ \text{mm}^2$$

(Bare diameter $\approx 0.88$ mm.)

(v) Area and dimensions of each slot

$$A_{slot} = \frac{106 \times 0.613}{0.4} = 162.4\ \text{mm}^2$$

$d/w = 4$: $4w^2 = 162.4$, $w = 6.37$ mm, $d = 25.5$ mm.

(vi) Minimum width of stator teeth

$L_i = 0.9 \times 0.0676 = 0.0608$ m, teeth per pole $= 36/4 = 9$:

$$W_{ts(min)} = \frac{\Phi}{1.7 \times 9 \times L_i} = \frac{3.281\times10^{-3}}{1.7 \times 9 \times 0.0608} = 3.53\ \text{mm}$$

(vii) Stator slot loading

$$I_{ph} \times \text{cond/slot} = 2.451 \times 106 = 259.8\ \text{A-conductors}$$

(Well below the usual limit of about 1000–1200.)

Answer: $D = 14.04$ cm, $L = 6.76$ cm; $T_s = 636$; $a_s = 0.613$ mm²; 36 slots; slot 162.4 mm² ($6.37 \times 25.5$ mm); $W_{ts(min)} = 3.53$ mm; slot loading 260 A-conductors.

Data: 120 kW, 2200 V, 50 Hz, $N_s = 750$ rpm, star connected. $P = 120f/N_s = 8$ poles, $n_s = 12.5$ rps.

(i) and (ii) Diameter and length

$$\begin{aligned} Q &= \frac{120}{0.92 \times 0.88} = 148.22\ \text{kVA} \\ C_0 &= 11 \times 0.48 \times 26000 \times 0.955 \times 10^{-3} = 131.10 \\ D^2L &= \frac{148.22}{131.10 \times 12.5} = 0.09045\ \text{m}^3 \end{aligned}$$

$L = 1.25\tau = 1.25 \times \pi D/8 = 0.4909D$:

$$D^3 = \frac{0.09045}{0.4909} = 0.1843 \Rightarrow D = 0.569\ \text{m}, \quad L = 0.279\ \text{m}$$

(iii) Turns per phase

$$\begin{aligned} \tau &= \frac{\pi \times 0.569}{8} = 0.2235\ \text{m} \\ \Phi &= B_{av}\tau L = 0.48 \times 0.2235 \times 0.2793 = 0.02996\ \text{Wb} \\ E_{ph} &= \frac{2200}{\sqrt3} = 1270.2\ \text{V} \\ T_s &= \frac{1270.2}{4.44 \times 50 \times 0.02996 \times 0.955} = 199.97 \approx 200 \end{aligned}$$

(iv) Full-load current and conductor section

$$\begin{aligned} I_s &= \frac{Q \times 10^3}{\sqrt3 \times 2200} = \frac{148221}{3810.5} = 38.90\ \text{A} \\ a_s &= \frac{38.90}{5} = 7.78\ \text{mm}^2 \end{aligned}$$

(v) Total stator $I^2R$ loss

Conductors per phase $= 2T_s = 400$, each 0.75 m long, so length per phase $= 400 \times 0.75 = 300$ m.

$$\begin{aligned} R_s &= \frac{\rho\, l}{a} = \frac{0.021 \times 300}{7.78} = 0.810\ \Omega \\ \text{Loss} &= 3 I_s^2 R_s = 3 \times 38.90^2 \times 0.810 = 3676\ \text{W} \end{aligned}$$

($\rho = 0.021$ Ω per m per mm² is copper at about 75 °C.)

Answer: $D = 56.9$ cm, $L = 27.9$ cm, $T_s = 200$, $I_s = 38.9$ A, $a_s = 7.78$ mm², stator copper loss $\approx 3.68$ kW.

Data: 15 kW, 400 V, 50 Hz, 2810 rpm. Nearest synchronous speed is 3000 rpm, so $P = 2$ and $n_s = 50$ rps. Assume $K_w = 0.955$.

Output coefficient and $D^2L$

$$\begin{aligned} Q &= \frac{15}{0.88 \times 0.9} = 18.94\ \text{kVA} \\ C_0 &= 11 B_{av}\, ac\, K_w \times 10^{-3} = 11 \times 0.5 \times 25000 \times 0.955 \times 10^{-3} = 131.31 \\ D^2L &= \frac{Q}{C_0 n_s} = \frac{18.94}{131.31 \times 50} = 2.885 \times 10^{-3}\ \text{m}^3 \end{aligned}$$

Diameter from peripheral speed

$$v_a = \pi D n_s = 20\ \text{m/s} \Rightarrow D = \frac{20}{\pi \times 50} = 0.1273\ \text{m}$$

Core length

$$L = \frac{2.885\times10^{-3}}{0.1273^2} = 0.178\ \text{m}$$

Check

Pole pitch $\tau = \pi D/2 = 0.200$ m, so $L/\tau = 0.89$. This lies in the usual range (about 0.6 to 2), so the proportions are acceptable.

Answer: $D \approx 12.7$ cm, $L \approx 17.8$ cm.

Output equation

Let $D$ = bore diameter, $L$ = core length, $P$ = poles, $n_s$ = synchronous speed (rps), $T_{ph}$ = turns per phase.

kVA input:

$$Q = 3E_{ph}I_{ph}\times10^{-3}, \qquad E_{ph} = 4.44 f\Phi T_{ph}K_w, \qquad f = \frac{Pn_s}{2}$$

Specific magnetic loading:

$$B_{av} = \frac{P\Phi}{\pi DL} \Rightarrow \Phi = \frac{B_{av}\pi DL}{P}$$

Specific electric loading (6$T_{ph}$ conductors):

$$ac = \frac{6I_{ph}T_{ph}}{\pi D} \Rightarrow I_{ph}T_{ph} = \frac{ac\,\pi D}{6}$$

Substituting:

$$\begin{aligned} Q &= 3 \times 4.44 \times \frac{Pn_s}{2} \times \frac{B_{av}\pi DL}{P} \times K_w \times \frac{ac\,\pi D}{6} \times 10^{-3} \\ &= 11 B_{av}\, ac\, K_w \times 10^{-3}\, D^2 L n_s = C_0 D^2 L n_s \end{aligned}$$

where $C_0 = 11B_{av}\,ac\,K_w\times10^{-3}$ is the output coefficient and $Q = \text{kW}/(\eta\cos\phi)$.

Separation of $D$ and $L$

The output equation gives only the product $D^2L$. To get $D$ and $L$ separately, a ratio $L/\tau$ (with $\tau = \pi D/P$) is chosen according to the design aim:

With $L = k\tau = k\pi D/P$:

$$D^3 = \frac{D^2L \cdot P}{k\pi}, \qquad L = \frac{D^2L}{D^2}$$

Checks after separation

• Peripheral speed: $v_a = \pi D n_s$ must not exceed about 30 m/s for normal rotors (higher only with special construction).
• Ventilating ducts: if $L$ exceeds about 100–125 mm, radial ducts (about 10 mm wide) are provided.
• Practical proportions: very long, thin machines are hard to cool and the shaft deflects; very short, large-diameter machines have a high end-winding share.

Example

For $D^2L = 0.0110$ m³, 6 poles, $L = \tau$: $D^3 = 0.0110 \times 6/\pi = 0.0210$, so $D = 0.276$ m and $L = 0.1445$ m, with $v_a = \pi \times 0.276 \times 16.67 = 14.4$ m/s (acceptable).

Why stator slots should never equal rotor slots

If $S_s = S_r$, all rotor teeth face stator teeth at the same time. At these positions the reluctance is minimum, and a strong alignment force holds the rotor there. This cogging (magnetic locking) can stop the motor from starting, and it also produces noise and vibration. So $S_r \ne S_s$ is always used.

Output equation in terms of main dimensions

Let $D$ = stator bore, $L$ = core length, $P$ = poles, $n_s$ = synchronous speed (rps).

1. kVA input: $Q = 3E_{ph}I_{ph}\times10^{-3}$
2. Emf: $E_{ph} = 4.44f\Phi T_{ph}K_w$, with $f = Pn_s/2$
3. Specific magnetic loading:

$$B_{av} = \frac{P\Phi}{\pi DL} \Rightarrow \Phi = \frac{\pi DL B_{av}}{P}$$

4. Specific electric loading (total conductors $6T_{ph}$):

$$ac = \frac{6I_{ph}T_{ph}}{\pi D} \Rightarrow I_{ph}T_{ph} = \frac{\pi D\,ac}{6}$$

5. Substitute into $Q$:

$$\begin{aligned} Q &= 3 \times 4.44 \times \frac{Pn_s}{2} \times \frac{\pi DLB_{av}}{P} \times K_w \times \frac{\pi D\,ac}{6} \times 10^{-3} \\ &= 1.11\pi^2 B_{av}\,ac\,K_w\times10^{-3}\, D^2Ln_s \\ &= 11B_{av}\,ac\,K_w\times10^{-3}\, D^2Ln_s \end{aligned}$$

Hence

$$Q = C_0D^2Ln_s, \qquad C_0 = 11B_{av}\,ac\,K_w\times10^{-3}$$

This shows the output (kVA) is proportional to:

• the main dimensions through $D^2L$ (active volume),
• the specific loadings $B_{av}$ and $ac$ through $C_0$,
• the speed $n_s$.

The kW rating is related by $Q = \text{kW}/(\eta\cos\phi)$. For example, with $B_{av} = 0.44$, $ac = 21000$, $K_w = 0.955$: $C_0 = 97.07$, so a 3.235 kVA, 4-pole motor needs $D^2L = 3.235/(97.07 \times 25) = 1.33 \times 10^{-3}$ m³.

The air gap is the main reluctance in the magnetic circuit of an induction motor, so its length strongly affects performance.

Effects of a too-wide air gap

• High magnetising current: the gap mmf $= 800000\,B_g K_g l_g$ rises in proportion to $l_g$, so the no-load current rises sharply.
• Low power factor: the larger magnetising (reactive) current lowers the power factor; this is the most serious effect.
• Higher copper loss and lower efficiency at light loads because of the large no-load current.
• On the good side: lower leakage reactance (higher overload capacity), less noise, better cooling and less unbalanced magnetic pull.

So a wide gap is used only when overload capacity or mechanical reasons are more important than power factor.

Factors in selecting the air gap length

1. Power factor: small gap for low magnetising current and high power factor.
2. Overload capacity: larger gap lowers zigzag leakage reactance and raises the maximum torque.
3. Pulsation loss and noise: larger gap smooths the tooth-ripple flux, reducing pulsation losses, noise and vibration.
4. Cooling: larger gap allows more cooling air.
5. Unbalanced magnetic pull: larger gap reduces the effect of rotor eccentricity.
6. Mechanical factors: shaft deflection, bearing wear, length of core and manufacturing tolerances set a minimum gap.
7. Harmonics: larger gap reduces slot harmonic fields and their parasitic torques (cogging, crawling).

Empirical formula

$$l_g = 0.2 + 2\sqrt{DL}\ \text{mm} \quad (D, L \text{ in m})$$

Example: $D = 0.25$ m, $L = 0.16$ m gives $l_g = 0.2 + 2\sqrt{0.04} = 0.6$ mm.

Specific magnetic loading ($B_{av}$) is the average flux density over the air-gap surface:

$$B_{av} = \frac{P\Phi}{\pi DL}$$

For induction motors it usually lies between 0.3 and 0.6 Wb/m². The following factors decide its value.

1. Power factor

The magnetising current depends mostly on the gap flux density. A higher $B_{av}$ needs more magnetising mmf, so the magnetising current rises and the power factor falls. Since an induction motor takes all its magnetising current from the supply, this is the main limit on $B_{av}$.

2. Iron loss and efficiency

Iron loss rises roughly as $B^2$ (and with frequency). A high $B_{av}$ raises core loss, lowers efficiency and raises temperature rise. High-frequency machines use lower $B_{av}$.

3. Overload capacity

A higher $B_{av}$ means fewer turns for the same voltage ($T \propto 1/\Phi$), so leakage reactance is lower and maximum torque is higher. A high value therefore improves overload capacity.

4. Flux density in teeth

The teeth carry the gap flux. Tooth flux density must not exceed about 1.7 Wb/m² (to avoid saturation and high magnetising current). With narrow teeth, $B_{av}$ must be lower.

5. Size and cost

From $Q = C_0D^2Ln_s$ with $C_0 \propto B_{av}$, a higher $B_{av}$ gives a smaller machine with less material and lower cost.

6. Quality of core material

Better (low-loss, cold-rolled) steel allows a higher $B_{av}$ without excessive loss.

Summary

So the designer chooses a value (about 0.4–0.5 Wb/m² for 50 Hz motors) that balances power factor and efficiency against size, cost and overload capacity.

The minimum width of a stator tooth is the narrowest tooth that can carry the flux without its flux density exceeding about 1.7 Wb/m² (above this the teeth saturate, the magnetising current and iron loss rise sharply).

Derivation

• Flux per pole: $\Phi = B_{av}\tau L$.
• This flux passes through the teeth under one pole, number of teeth $= S_s/P$.
• Iron area of these teeth $= \dfrac{S_s}{P} \times W_{ts} \times L_i$, where $L_i = k_i(L - n_dw_d)$ is the net iron length ($k_i \approx 0.9$).
• Setting $B_t \le 1.7$ Wb/m²:

$$\frac{\Phi}{\frac{S_s}{P}W_{ts}L_i} \le 1.7 \;\Rightarrow\; W_{ts(min)} = \frac{\Phi}{1.7 \times \frac{S_s}{P} \times L_i}$$

Example

$\Phi = 4.51 \times 10^{-3}$ Wb, $S_s = 24$, $P = 4$, $L = 0.124$ m, $L_i = 0.9 \times 0.124 = 0.1116$ m:

$$W_{ts(min)} = \frac{4.51\times10^{-3}}{1.7 \times 6 \times 0.1116} = 3.96\ \text{mm}$$

The actual tooth width (slot pitch minus slot width, $13.78 - 6.75 = 7.0$ mm) is larger, so the design is safe.

Data: 2.2 kW, 400 V, 50 Hz, 1420 rpm, so 4 poles and $n_s = 25$ rps; $\eta = 0.8$, $\cos\phi = 0.825$. Assumptions: $K_w = 0.955$; star-delta starting, so the stator runs in delta ($E_{ph} = 400$ V); tooth flux density limit 1.7 Wb/m²; iron factor 0.9.

i) Main dimensions

$$\begin{aligned} Q &= \frac{2.2}{0.8 \times 0.825} = 3.333\ \text{kVA} \\ C_0 &= 11 \times 0.44 \times 21000 \times 0.955 \times 10^{-3} = 97.07 \\ D^2L &= \frac{3.333}{97.07 \times 25} = 1.374 \times 10^{-3}\ \text{m}^3 \end{aligned}$$

$L = 1.5\tau = 1.5 \times \pi D/4 = 1.178D$:

$$D^3 = \frac{1.374\times10^{-3}}{1.178} = 1.166\times10^{-3} \Rightarrow D = 0.1053\ \text{m}, \quad L = 0.124\ \text{m}$$

ii) Stator turns per phase

$$\begin{aligned} \tau &= \pi \times 0.1053/4 = 0.0827\ \text{m} \\ \Phi &= 0.44 \times 0.0827 \times 0.124 = 4.51 \times 10^{-3}\ \text{Wb} \\ T_s &= \frac{400}{4.44 \times 50 \times 4.51\times10^{-3} \times 0.955} = 418.3 \end{aligned}$$

iv) Number of stator slots

Slot pitch 12–15 mm. $q = 2$: $S_s = 3 \times 4 \times 2 = 24$; slot pitch $= \pi \times 105.3/24 = 13.78$ mm (OK). 24 slots. Conductors per slot $= 6 \times 418.3/24 = 104.6$, take 105: $Z_s = 2520$, $T_s = 420$.

iii) Size of stator conductor

$$I_{ph} = \frac{3333.3}{3 \times 400} = 2.778\ \text{A}, \qquad a_s = \frac{2.778}{4} = 0.694\ \text{mm}^2$$

Bare diameter $= \sqrt{4 \times 0.694/\pi} = 0.94$ mm.

v) Area and dimensions of each slot

$$A_{slot} = \frac{105 \times 0.694}{0.4} = 182.3\ \text{mm}^2$$

$d = 4w$: $4w^2 = 182.3$, $w = 6.75$ mm, $d = 27.0$ mm.

vi) Minimum width of stator teeth

$L_i = 0.9 \times 0.124 = 0.1116$ m; teeth per pole $= 24/4 = 6$:

$$W_{ts(min)} = \frac{4.51\times10^{-3}}{1.7 \times 6 \times 0.1116} = 3.96\ \text{mm}$$

Actual tooth width at the gap $= 13.78 - 6.75 = 7.0$ mm, so tooth density is safe.

Answer: $D = 10.53$ cm, $L = 12.4$ cm; $T_s = 420$; $a_s = 0.694$ mm² (0.94 mm dia); 24 slots; slot 182.3 mm² ($6.75 \times 27.0$ mm); $W_{ts(min)} = 3.96$ mm.

Data: 2.2 kW, 440 V, 50 Hz, 1430 rpm, so 4 poles and $n_s = 1500/60 = 25$ rps; $\eta = 0.8$, $\cos\phi = 0.85$; $B_{av} = 0.44$ Wb/m², $ac = 21000$ A/m, $L/\tau = 1.5$. Assumptions: winding factor $K_w = 0.955$; star-delta starter, so the motor runs in delta ($E_{ph} = 440$ V); maximum tooth flux density 1.7 Wb/m²; stacking factor 0.9.

i) Main dimensions

$$\begin{aligned} Q &= \frac{\text{kW}}{\eta\cos\phi} = \frac{2.2}{0.8 \times 0.85} = 3.235\ \text{kVA} \\ C_0 &= 11B_{av}\,ac\,K_w\times10^{-3} = 11 \times 0.44 \times 21000 \times 0.955 \times 10^{-3} = 97.07 \\ D^2L &= \frac{Q}{C_0n_s} = \frac{3.235}{97.07 \times 25} = 1.333\times10^{-3}\ \text{m}^3 \end{aligned}$$

With $L = 1.5\tau = 1.5\pi D/4 = 1.178D$:

$$\begin{aligned} D^3 &= \frac{1.333\times10^{-3}}{1.178} = 1.131\times10^{-3} \\ D &= 0.1042\ \text{m}, \qquad L = 1.178 \times 0.1042 = 0.1228\ \text{m} \end{aligned}$$

Peripheral speed $= \pi \times 0.1042 \times 25 = 8.2$ m/s (well within limit).

ii) Stator turns per phase

$$\begin{aligned} \tau &= \frac{\pi \times 0.1042}{4} = 0.0819\ \text{m} \\ \Phi &= B_{av}\tau L = 0.44 \times 0.0819 \times 0.1228 = 4.421\times10^{-3}\ \text{Wb} \\ T_s &= \frac{E_{ph}}{4.44f\Phi K_w} = \frac{440}{4.44 \times 50 \times 4.421\times10^{-3} \times 0.955} = 469.4 \end{aligned}$$

iv) Number of stator slots (needed before rounding turns)

Slot pitch 12–15 mm:

• $q = 2$: $S_s = 24$, slot pitch $= \pi \times 104.2/24 = 13.64$ mm (OK)
• $q = 3$: $S_s = 36$, slot pitch $= 9.09$ mm (too small)

Take 24 slots. Conductors per slot $= 6 \times 469.4/24 = 117.4$, take 118. Total conductors $Z_s = 24 \times 118 = 2832$, so $T_s = 472$.

iii) Size of stator conductor

$$\begin{aligned} I_{ph} &= \frac{Q\times10^3}{3E_{ph}} = \frac{3235.3}{3 \times 440} = 2.451\ \text{A} \\ a_s &= \frac{I_{ph}}{\delta_s} = \frac{2.451}{4} = 0.613\ \text{mm}^2 \end{aligned}$$

Bare diameter $= \sqrt{4 \times 0.613/\pi} = 0.88$ mm (use the nearest standard SWG wire).

v) Area and dimensions of each slot

$$A_{slot} = \frac{\text{cond/slot} \times a_s}{\text{space factor}} = \frac{118 \times 0.613}{0.4} = 180.8\ \text{mm}^2$$

With $d = 4w$: $4w^2 = 180.8$, so $w = 6.72$ mm and $d = 26.9$ mm.

vi) Minimum width of stator teeth

Net iron length $L_i = 0.9 \times 0.1228 = 0.1105$ m; teeth per pole $= 24/4 = 6$.

$$W_{ts(min)} = \frac{\Phi}{1.7 \times \frac{S_s}{P} \times L_i} = \frac{4.421\times10^{-3}}{1.7 \times 6 \times 0.1105} = 3.92\ \text{mm}$$

Tooth width at the gap $= 13.64 - 6.72 = 6.9$ mm $> 3.92$ mm, so tooth density is within limits.

Results

Data: 2.2 kW, 400 V, 50 Hz, 1420 rpm (4 poles, $n_s = 25$ rps), $\eta = 0.8$, $\cos\phi = 0.825$, $K_w = 0.955$, $L/\tau = 1.5$, $d/w = 2$. Assumptions: delta-connected stator ($E_{ph} = 400$ V); maximum tooth flux density 1.7 Wb/m²; stacking factor 0.9.

i) Main dimensions

$$\begin{aligned} Q &= \frac{2.2}{0.8 \times 0.825} = 3.333\ \text{kVA} \\ C_0 &= 11 \times 0.44 \times 21000 \times 0.955 \times 10^{-3} = 97.07 \\ D^2L &= \frac{3.333}{97.07 \times 25} = 1.374\times10^{-3}\ \text{m}^3 \end{aligned}$$

$L = 1.5 \times \pi D/4 = 1.178D$:

$$D^3 = 1.166\times10^{-3} \Rightarrow D = 0.1053\ \text{m}, \quad L = 0.124\ \text{m}$$

ii) Stator conductors and area of conductor

$$\begin{aligned} \tau &= \pi \times 0.1053/4 = 0.0827\ \text{m} \\ \Phi &= 0.44 \times 0.0827 \times 0.124 = 4.51\times10^{-3}\ \text{Wb} \\ T_s &= \frac{400}{4.44 \times 50 \times 4.51\times10^{-3} \times 0.955} = 418.3 \end{aligned}$$

Stator conductors $= 6T_s \approx 2510$; after choosing slots (below), $Z_s = 24 \times 105 = 2520$ ($T_s = 420$).

$$I_{ph} = \frac{3333.3}{3 \times 400} = 2.778\ \text{A}, \qquad a_s = \frac{2.778}{4} = 0.694\ \text{mm}^2$$

iii) Number of stator slots

Slot pitch 12–15 mm. $q = 2$ gives $S_s = 24$ and slot pitch $= \pi \times 105.3/24 = 13.78$ mm. 24 slots, $2510/24 = 104.6$, so 105 conductors per slot.

iv) Area and dimensions of each slot

$$A_{slot} = \frac{105 \times 0.694}{0.4} = 182.3\ \text{mm}^2$$

$d = 2w$: $2w^2 = 182.3$, $w = 9.55$ mm, $d = 19.1$ mm.

v) Minimum width of stator teeth

$L_i = 0.9 \times 0.124 = 0.1116$ m; teeth per pole $= 6$:

$$W_{ts(min)} = \frac{4.51\times10^{-3}}{1.7 \times 6 \times 0.1116} = 3.96\ \text{mm}$$

Tooth width at the gap $= 13.78 - 9.55 = 4.2$ mm $> 3.96$ mm, so it is just safe.

(The given core density 1.2 Wb/m² fixes the stator core depth: $d_{cs} = \Phi/(2 \times 1.2 \times L_i) = 16.8$ mm.)

Answer: $D = 10.53$ cm, $L = 12.4$ cm; 2520 conductors ($T_s = 420$), $a_s = 0.694$ mm²; 24 slots; slot 182.3 mm² ($9.55 \times 19.1$ mm); $W_{ts(min)} = 3.96$ mm.