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**Numerical
Problems on Field System Design of Salient pole machines:**

**Ex.1. **The following information has
been obtained during the preliminary design of a 3 phase 500** **kVA, 6.6
kV, 12 pole, 500 rpm, star connected salient pole alternator.

Stator
diameter = 1.3 m, gross length of stator = 0.21m, air gap flux per pole =
0.0404 wb Based on the above information, design the field system of the
alternator giving the following details.

(i)
Length of the air gap

(ii)
Diameter of the rotor at the air gap surface

(iii)
Dimension of the pole

**Soln:**

(i) Length of
the air gap : Air gap flux per pole = Bav x πDL/p

= (12 x
0.0404)/( π x1.3 x
0.21)

= 0.56
Tesla

We have AT_{f0}
= SCR x AT_{a} and
AT_{a}=1. 35 I_{ph} T_{ph} K_{w} /p

We have Eph = 4.44 f Tph kw and

Hence Tph x Kw = Eph/(4.44f ) = 6600/√3/ ( 4.44 x 50 x 0.0404) = 424
Full load current = 500 x 10^{3}/ √3 x 6600
= 43.7 amps

AT_{a}=1.
35 I_{ph} T_{ph} K_{w} /p = 1.35 x 43.7 x 424 /6 = 4169
AT

Assuming
a short circuit ratio of 1.1 AT_{f0} = SCR x AT_{a} = 1.1 x
4169 = 4586 AT Assuming AT required for the air gap as 70 % of the no load
field ampere turns per pole AT_{g} = 0.7 x AT_{fo} = 0.7 x 4586
= 3210 AT

Assuming
Carter’s coefficient for the air gap k_{g} as 1.15 and field form
factor K_{f} as 0.7 B_{g} = Bav/K_{f} = 0.56/0.7 = 0.8
Tesla

We have
air gap ampere turns AT_{g} = 796000 B_{g} k_{g} *l*_{g}

Hence air
gap length *l*_{g} = 3210 / ( 796000 x 0.8 x 1.15)
= 0.0044 m = 4.4 mm

(ii)
Diameter of the rotor D_{r} = D - 2 *l*_{g} = 1.2 –
2 x 0.0044 = 1.191m

(iv)
Peripheral speed = πD_{r}N_{s}
/ 60 = π x 1.191 x 500/60 =31.2 m/s

(v)
Dimensions of the pole : Assuming the axial length
as 1 cm less than that of the gross length of the stator

(a) Axial
length of the pole L_{p}= 0.21 – 0.01 = 0.2 m

(b) Width of
the pole: Assuming the leakage factor for the pole as 1.15

Flux in
the pole body Φ_{p} = 1.15 x
0.0404 = 0.0465 wb Assuming flux density in the pole body as 1.5 Tesla Area of
the pole = 0.0465/1.5 = 0.031 m^{2}

Assuming
a stacking factor of 0.95

Width of
the pole = area of the pole / stacking factor x Lp = 0.031/ (0.95 x 0.2) = 0.16
m Height of the pole: Assuming AT_{fl} = 1.8 x AT_{a} = 1.8 x
4169 = 7504 AT

Assuming : Depth of the field coil = 4 cm Space
factor for the filed coil = 0.7

Permissible loss per unit area = 700 w/m^{2}
Height of the filed coil h_{f} = (I_{f} T_{f}) / [ 10^{4}
x √(s_{f} d_{f} q_{f})]

= 7504 /
[10^{4} x √(0.04 x
0.7 x 700)]

= 0.17 m

Hence the
height of the pole = h_{f} + height of the pole shoe + height taken by
insulation Assuming height of the pole shoe + height taken by insulation as
0.04 m

Height of
the pole = 0.17 + 0.04 = 0.21 m

**Ex.2. **The field
coils of a salient pole alternator are wound with a single layer winding of
bare copper** **strip 30 mm deep, with a separating
insulation of 0.15 mm thick. Determine a suitable winding length, number of
turns and thickness of the conductor to develop an mmf of 12000 AT with a
potential difference of 5 volts per coil and with a loss of 1200 w/m^{2}
of total coil surface. The mean length of the turn is 1.2 m. The resistivity of
copper is 0.021 /m and mm^{2}.

**Soln. **Area of field conductor a_{f} = ζ x x (I_{f} T_{f}
) / V_{c}

= 0.021 x
1.2 x 12000/ 5

= 60.4 mm^{2}

Hence
height of the conductor = 60.4/30 = 2 mm

Revised
area of the conductor = 60 mm^{2}

Total heat dissipating surface S = 2 x *l _{mt}*
(h

= 2 x 1.2
(h_{f} + 0.03)

= 2.4 h_{f} + 0.072 m^{2}

Hence total loss dissipated Q_{f} = 1200
(2.4 h_{f} + 0.072) watts = 2880 h_{f} + 86.4 watts

Field
current I_{f} = Q_{f}/v_{c}= (2880 h_{f} +
86.4)/ 5 = 5.76 h_{f} + 17.3 And I_{f} T_{f} = (5.76 h_{f}
+ 17.3) T_{f} =12000

I_{f}
T_{f} = 5.76 h_{f} T_{f} + 17.3 T_{f} =12000

Height
occupied by the conductor including insulation = 2 + 0.15 = 2.15 mm Hence
height of the field winding hf = T_{f} x 2.15 x 10^{-3}

Substituting
this value in the expression for I_{f} T_{f} we get

I_{f}
T_{f} = 5.76 x T_{f} x 2.15 x 10^{-3} T_{f} +
17.3 T_{f} =12000 Solving for T_{f}, T_{f} = 91

Hence
height of the field winding = 2.15 x 91 = 196 mm

**Ex. 3 **Design the field coil of a 3
phase, 16 pole, 50 Hz, salient pole alternator, based on the following** **design
information. Diameter of the stator = 1.0 m, gross length of the stator = 0.3
m, section of the pole body = 0.15 m x 0.3 m, height of the pole = 0.15 m,
Ampere turns per pole =6500, exciter voltage = 110 volts, Assume missing data
suitably.

**Slon**.
Sectional area of the conductor:

Assuming 30 volts as reserve in field regulator V_{c}
= 110 – 30 / 16 = 5 volts

Assuming
depth of the field coil = 3 cm, thickness of insulation = 1 cm

Mean length of the turn = 2( l_{p} + b_{p})
+ π (d_{f} + 2t_{i})
= 2 ( 0.3 + 0.15) + π ( 0.03 +
2 x 0.01) = 1.05 m Sectional area of the conductor a_{f} = ζ x *l _{mt}* x I

= (0.021
x 1.05 x 6000)/5 = 28.66 mm^{2}

Standard
size of the conductor available = 28.5 mm^{2} with the size 16 mm x 1.8
mm Assuming an insulation thickness of 0.5 mm over the conductor

size of
the conductor = 16.5 mm x 2.3 mm Assuming an insulation of 2mm between the
layers

Actual
depth of the field winding = 16.5 + 2 + 16.5 = 35 mm or 3.5 cm Field current:
Assuming a current density of 2.6 amps/ mm^{2}

Field
current I_{f} = a_{f} x δ_{f} =28.5 x
2.6 = 74 amps Number of turns: T_{f} = I_{f} T_{f}/ I_{f}
= 6000/74 = 88 turns

Arrangement of turns: As decided above 88 turns are
arranged in two layers with 44 turns in each layer. Height of each field turn =
2.3 mm

Hence
height of the field coil = 44 x 2.3 = 10.1 cm

As height
of the pole is 15 cm, height of the field coil is satisfactory. Resistance of
the field coil R_{f} = ζ x *l _{mt}* x T

= 0.021 x
1.05 x 88/ 28.5

= 0.068

Filed
Copper loss: I_{f}^{2}
R_{f} = 74^{2} x 0.068 = 372 watts

Total
field cu loss = 16 x 372 = 5.95 kW.

**Ex.4. **Design the field coil of a 500
rpm, 3 phase, 50 Hz alternator having the following design data.** **Diameter
of stator = 95 cm, Core length = 30 cm, Pole body = 10 cm x 30 cm, Field ampere
turns = 6000, Excitation voltage = 80 volts. Heat dissipation from the outer
surface = 0.35 watts/ cm^{2}. Assume missing data suitably.

**Soln:** Area of the field coil:

Number of
field coils or poles = 120f/N_{s} = 120 x 50 /500 =12 Assuming 20 volts
in the field regulator

Voltage
per coil = 80 -20/ 12 = 5 volts Ampere turns /pole =6000

Pole body
= 10 cm x 30 cm,

Assuming
depth of the field coil = 3 cm, Thickness of insulation = 1 cm

Mean
length of the turn = 2( l_{p} + b_{p}) + π (d_{f} + 2t_{i})

= 2 ( 0.3
+ 0.1) + π ( 0.03 +
2 x 0.01) = 0.957 m

Sectional
area of the conductor a_{f} = ζ x *l _{mt}*
x I

= (0.021 x 0.957 x 6000)/5 = 24.2 mm^{2}
Standard size of the conductor available 14.2 mm x 1.7 mm

Assuming
an insulation thickness of 0.5 mm over the conductor Assuming an insulation of
1.6 mm between the layers

Actual
depth of the field winding = 14.2 + 1.6 + 14.2 = 3.0 cm

Number of
turns: Heat dissipation from the outer surface = 0.35 watts/cm^{2} Area
of the outer surface of the field coil = ( *l _{mt}* + π d

= 105.1 h_{f} cm^{2} Hence heat
dissipated = 0.35 x 105.1 h_{f} = 36.8 h_{f} = V_{c} x
I_{f}

= V_{c}
x I_{f} T_{f} / T_{f}

Hence
36.8 h_{f} = V_{c} x I_{f}
T_{f} / T_{f}

= 5 x 6000/ T_{f} Hence h_{f} T_{f}
= 5 x 6000/ 36.8 = 815

Assuming
an insulation thickness of 0.15 mm between the conductors Height of each
conductor = Height of conductor + insulation

= 1.7 + 0.15 = 1.85 mm =0.185 cm Assuming that the
turns are arranged in two layers

Height of
turns / layer h_{f} = 0.185 x T_{f} /2 Hence h_{f} T_{f}
= 0.185 x T_{f} /2 x T_{f} = 815

T_{f} = 94

Hence
height of the field coil h_{f} = 0.185 x T_{f} /2 = 0.185 x
94/2 = 8.7 cm Field current I_{f} = 6000/94 = 64 amps

Resistance
of the field coil R_{f} = ζ x *l _{mt}*
x T

= 0.021 x
0.957 x 94/ 24.2

= 0.078

Filed Copper loss: I_{f}^{2}
R_{f} = 64^{2} x 0.078 = 320 watts

Total
field cu loss = 12 x 320 = 3.84 kW.

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