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Numerical Problems on Field System Design of Salient pole machines

Design of Electrical Machines - Synchronous Machines - Solved Numerical Problems on Field System Design of Salient pole machines

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 ATf0 =  SCR x ATa    and  ATa=1. 35 Iph Tph Kw /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 103/ 3 x 6600 = 43.7 amps

ATa=1. 35 Iph Tph Kw /p = 1.35 x 43.7 x 424 /6 = 4169 AT

Assuming a short circuit ratio of 1.1 ATf0 = SCR x ATa = 1.1 x 4169 = 4586 AT Assuming AT required for the air gap as 70 % of the no load field ampere turns per pole ATg = 0.7 x ATfo = 0.7 x 4586 = 3210 AT

 

Assuming Carter’s coefficient for the air gap kg as 1.15 and field form factor Kf as 0.7 Bg = Bav/Kf = 0.56/0.7 = 0.8 Tesla

We have air gap ampere turns ATg = 796000 Bg kg lg

Hence air gap length lg = 3210 / ( 796000 x 0.8 x 1.15) = 0.0044 m = 4.4 mm

(ii)               Diameter of the rotor Dr = D - 2 lg = 1.2 – 2 x 0.0044 = 1.191m

(iv)                       Peripheral speed = πDrNs / 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 Lp= 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 Φp = 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 m2

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 ATfl = 1.8 x ATa = 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/m2 Height of the filed coil hf = (If Tf) / [ 104 x (sf df qf)]

= 7504 / [104 x (0.04 x 0.7 x 700)]

= 0.17 m

Hence the height of the pole = hf + 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/m2 of total coil surface. The mean length of the turn is 1.2 m. The resistivity of copper is 0.021 /m and mm2.

 

Soln. Area of field conductor      af = ζ x x (If Tf ) / Vc

= 0.021 x 1.2 x 12000/ 5

= 60.4 mm2

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

Revised area of the conductor = 60 mm2

Total heat dissipating surface      S = 2 x lmt  (hf + df )

= 2 x 1.2 (hf + 0.03)

= 2.4 hf  + 0.072 m2

 

 

Hence total loss dissipated Qf = 1200 (2.4 hf + 0.072) watts = 2880 hf + 86.4 watts

 

Field current If = Qf/vc= (2880 hf + 86.4)/ 5 = 5.76 hf + 17.3 And If Tf = (5.76 hf + 17.3) Tf =12000

If Tf = 5.76 hf Tf + 17.3 Tf =12000

 

Height occupied by the conductor including insulation = 2 + 0.15 = 2.15 mm Hence height of the field winding hf = Tf x 2.15 x 10-3

Substituting this value in the expression for If Tf  we get

 

If Tf = 5.76 x Tf x 2.15 x 10-3 Tf + 17.3 Tf =12000 Solving for Tf, Tf = 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 Vc = 110 – 30 / 16 = 5 volts

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

Mean length of the turn = 2( lp + bp) + π (df + 2ti) = 2 ( 0.3 + 0.15) + π ( 0.03 + 2 x 0.01) = 1.05 m Sectional area of the conductor af = ζ x lmt x If Tf / Vc

= (0.021 x 1.05 x 6000)/5 = 28.66 mm2

Standard size of the conductor available = 28.5 mm2 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/ mm2

 

Field current If = af x δf =28.5 x 2.6 = 74 amps Number of turns: Tf = If Tf/ If = 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 Rf = ζ x lmt x Tf / af

= 0.021 x 1.05 x 88/ 28.5

= 0.068

Filed Copper loss:         If2 Rf = 742 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/ cm2. Assume missing data suitably.

 

Soln:          Area of the field coil:

 

Number of field coils or poles = 120f/Ns = 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( lp + bp) + π (df + 2ti)

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

 

Sectional area of the conductor  af = ζ x lmt  x If Tf / Vc

= (0.021 x 0.957 x 6000)/5 = 24.2 mm2 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/cm2 Area of the outer surface of the field coil = ( lmt + π df ) hf = (95.7 + π x 3) hf

 

= 105.1 hf cm2 Hence heat dissipated = 0.35 x 105.1 hf = 36.8 hf = Vc x If

= Vc x If Tf / Tf

 

Hence 36.8 hf  = Vc x If Tf / Tf

= 5 x 6000/ Tf Hence hf Tf = 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 hf = 0.185 x Tf /2 Hence hf Tf = 0.185 x Tf /2 x Tf = 815

Tf  = 94

 

Hence height of the field coil hf = 0.185 x Tf /2 = 0.185 x 94/2 = 8.7 cm Field current If = 6000/94 = 64 amps

Resistance of the field coil  Rf = ζ x lmt  x Tf / af

= 0.021 x 0.957 x 94/ 24.2

= 0.078

Filed Copper loss:     If2 Rf = 642 x 0.078 = 320 watts

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

 

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