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Chapter: Design of Electrical Machines - Synchronous Machines

Design of the field System: Non Salient pole Alternator

In case of turbo alternators, the rotor windings or the field windings are distributed in the rotor slots. The rotor construction of the turbo alternator is as shown in fig. below.

Design of the field System: NonSalient pole Alternator:

In case of turbo alternators, the rotor windings or the field windings are distributed in the rotor slots. The rotor construction of the turbo alternator is as shown in fig. below.


Normally 70% of the rotor is slotted and remaining portion is unslotted in order to form the pole. The design of the field can be explained as follows.

 

(i)                Selection of rotor slots: Total number of rotor slots may be assumed as 50 – 70 % of stator slots pitches. However the so found rotor slots must satisfy the following conditions in

 

order to avoid the undesirable effects of harmonics in the flux density wave forms.

 

(a)  There should be no common factor between the number of rotor slot pitches and number of stator slot pitches.

 

(b) Number of wound rotor slots should be divisible by 4 for a 2 pole synchronous machine. That means the number of rotor slots must be multiple of 4.

(c)  Width of the rotor slot is limited by the stresses developed at the rotor teeth and end rings.

(ii)             Design of rotor winding

 

(a) Full load field mmf can be taken as twice the armature mmf. ATfl = 2 x ATa = 2 x 1.35 x Iph x Tph x kw /p

 

(b) Standard exciter voltage of 110 - 220 volts may be taken. With 15-20 % of this may be reserved for field control. Hence voltage across each field coil Vf = (0.8 to 0.85) V/p

 

(c)  Length of the mean turn lmt = 2L + 1.8 τp + 0.25 m

(d) Sectional area of each conductor  af  = ζ x lmt x (If x Tf) / vf

 

(e) Assume suitable value of current density in the rotor winding. 2.5 – 3.0 amp/mm2 for conventionally cooled machines and 8 – 12 amp/mm2 for large and special cooled machines.

 

(f)   Find area of all the rotor conductors per pole = 2 x (If x Tf) /δf

(g) Find the number of rotor conductors per pole = 2 x (If x Tf) / (δf x af)

(h) Number of field conductors per slot = 2 x (If x Tf) / (δf x af x sr), where sr is the number of rotor slots.

 

(i)    Resistance of each field coil Rf = ζ x lmt x Tf / af

 

(j)    Calculate the current in the field coil If = vf/ Rf

 

Based on the above data dimensions may be fixed. The ratio of slot depth to slot width may be taken between 4 and 5. Enough insulation has to be provided such that it with stands large amount of mechanical stress and the forces coming on the rotor.

 

The following insulation may be provided for the field coil.

(i)                All field conductors are provided with mica tape insulation.

 

(ii)             Various turns in the slots are separated from each other by 0.3 mm mica separators.

(iii)           0.5 mm hard mica cell is provided on all the field coil.

(iv)           Over the above insulation, 1.5 mm flexible mica insulation is provided.

(v)             Lastly a steel cell of o.6 mm is provided on the whole field coil.

 

Ex. 1. Design the rotor of a 3 phase 20 MVA, 11 kV, 3000 rpm, 50 Hz, turbo alternator with the following design data. Diameter at the air gap = 0.8 m, Gross length = 2.4 m, stator turns per phase = 18, Number of stator slots = 36, Exciter voltage = 220 volts, Estimate (i) Number of rotor slots, (ii) area of the field conductor (iii) Turns in the filed coil and (iv) Field current

 

Soln: (i) Number of rotor slots : Selection of rotor slots: Total number of rotor slots may be assumed as 50 – 70 % of stator slots. Normally 70% of the rotor is slotted and remaining portion is unslotted.

 

Number of stator slots = 36

 

Hence number of slots pitches must be between 18 to 26 Satisfying the conditions number of rotor slot pitches = 23 Number of wound slots = 16

 

(ii)Area of the field conductor

 

Assuming 40 volts in the field regulator voltage across filed coil = 220 – 40 /2 = 90 volts Armature ampere turns /pole ATa=1. 35 Iph Tph Kw /p

 

= 1.35 x 1050 x 18 x 0.955/ 1 = 24300 AT Assuming full load field ampere turns/pole = 2 x ATa = 2 x 24300 = 48600 AT Mean length of the turn is given by lmt = 2L + 1.8 τp + 0.25 m

=   2 x 2.4 + 1.8 x 1.256 + 0.25

= 7.31 m

Area of the field conductor  af  = ζ x lmt x (If x Tf) / vf

= 0.021 x 7.31 x 48600/90

= 83.22 mm2

 

(iii)           Number of field turns : Full load field ampere turns/pole = 48600 AT Full load field ampere conductors/pole = 2 x 48600 AT

Assuming a current density of 2.6 amp/mm2

Area of all the rotor conductors = 2 x 48600 / 2.6 = 37400 mm2

Number of rotor conductors/pole = 37400/84 = 445

 

Number of wound slots per pole = 16/2 = 8

 

Number of conductors per slot = 445/8 = 56

Modified value of conductors per pole = 56 x 8 = 448

Number of field turns per pole Tf = 448/2 = 224

Number of coils per pole = 8/2 = 4

(iv)        Field current: Resistance of the field coil Rf  = ζ x lmt x Tf / af

= 0.021 x 7.31 x 224/ 84

= 0.41

Current in the field winding If = Vc/ Rf  = 90/0.41 = 219 Amps.


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Design of Electrical Machines - Synchronous Machines : Design of the field System: Non Salient pole Alternator |


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