1. INDUCTION MOTORS Output Equation Q(kVA) = m Eph Iph 10-3 = m (4.44f Φm Tph Kw ) {(ac πD)/( m 2Tph )} = 4.44 (pNs/120) (Bave π DL/p) Kw (ac πD/ 2) x 10-3 = 1.11 π2 Bave ac Kw x 10-3 D2L ns = 11 Bave ac Kw x 10-3 D2L ns = Co D2L ns Co = output coefficient, Q is calculated as ( hp x 0.746 )/(η cosΦ)

2. Choice of Bave i)Low Bave → large size machine for a given hp ii)high Bave → large magnetizing current → low power factor iii)high Bave → high iron loss iv)high Bave → high Φm→ less Tph→ low leakage reactance→ larger diameter for the circle diagram→ larger over load capacity For 50 Hz motors Bave : 0.3 to 0.6 Wb/m2

3. Choice of ac i)Low ac → large size machine for a given hp ii)High ac → higher copper loss and temp rise iii)High ac → large Tph→ large leakage reactance→ lower diameter for the circle diagram→ lower over load capacity For 50 Hz motors ac : 10,000 to 45,000 amp.cond/m The value ac chosen depends on the ventilation and cooling It should be remembered that the Power factor (PF) and efficiency(η) of the motor at full load increases with the rating of the machine. Again η and Pf are higher for high speed motors compared to low speed motors.

4. Separation of D and L and air gap length i)L/τ can be assumed ii)For best power factor τ = Sqrt( 0.18 L) or D= 0.135p√ L iii)D can be chosen based on peripheral speed = πDN/60 The stator winding can be connected in star or delta. The motors meant for starting with star- delta starter should be designed with delta connected stator winding. Stator slot –pitch = yss = π D/Ss where Ss =number of stator slots Stator outer diameter Do = D + 2dss + 2dcs Where dss = stator slot depth and dcs = stator core depth Air gap length = 0.2 + 2 sqrt(DL) where Dand L are in metre.

5. After designing the main dimensions, the following calculations can be made: 1.The flux per pole, Φm = Bave π DL/p 2.Turns in series per phase, Tph = V/ (4.44f Φm Kw ) 3.Number of slots per pole per phase can be suitably assumed 4.Slot pitch should not exceed about 24 mm 5.Number of conductors per slot should be rounded off 6.Tph recalculated 7.Iph = [Q x 103 /(√3 400)] for star and [Q x 103 /(√3 400)] /√3 for delta connection 8.Assuming suitable current density (δs), conductor area required is calculated (as = Iph / δs )

6. 9.Total copper area in the slot = (conductors /slot) as 10.Total area of the slot = copper area /slot space factor. Slot space factor is to account for the space for insulation 11.A suitable tooth flux density is assumed and tooth width is calculated slot width = slot pitch- tooth width (at different diameters) 12.The core flux(Φm/2), core area and core depth are calculated assuming a core flux density. 13.Core depth = core area/ Li 14.The rotor bar and end rings are designed (sq.cage rotor) 15.The rotor winding is designed similar to stator winding (wound rotor)

7. Losses in the Induction motor i)stator copper loss ii)rotor copper loss iii)iron loss in the stator teeth and Core iv)friction and windage loss (1- 1.5 % of output) The rotor resistance in stator terms can be obtained as rotor copper loss/ I2’ ; where I2’ = 0.85 I1

8. No-load current Iron loss component = Il =(Iron loss/phase) /Vph Magnetizing component = Im =(0.427 p ATtotal )/( Kws Tph ) ATtotal = sum of the ampere turns for airgap,stator tooth and core and rotor tooth and core. No-load current = In = { Il^2 + Im^2 }^1/2 The leakage reactance calculations are made to find x1, x2 and x01.