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 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 10³/ √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_rN_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 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 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/m² Height of the filed coil h_f = (I_f T_f) / [ 10⁴ x √(s_f d_f q_f)]

= 7504 / [10⁴ 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/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      a_f = ζ x x (I_f T_f ) / V_c

= 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 l_mt  (h_f + d_f )

= 2 x 1.2 (h_f + 0.03)

= 2.4 h_f  + 0.072 m²

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_f T_f / V_c

= (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 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_f / a_f

= 0.021 x 1.05 x 88/ 28.5

= 0.068

Filed Copper loss:         I_f² R_f = 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/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_f T_f / V_c

= (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 = ( l_mt + π d_f ) h_f = (95.7 + π x 3) h_f

= 105.1 h_f cm² 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_f / a_f

= 0.021 x 0.957 x 94/ 24.2

= 0.078

Filed Copper loss:     I_f² R_f = 64² x 0.078 = 320 watts

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