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