STATEMENT OF THE PROBLEMDESIGN #1:DESIGN OF AN ARMATURE OF A DC GENERATORPROBLEM: Design an armature of a DC generator having the following specifications:

1) Full load output 1550 kW2) Open circuit / No load Voltage 750 V3) Full load Voltage 660 V4) Ratio of pole arc to pole pitch (r) 0.645) Speed (N) up to 500 rpm6) Winding Simplex Lap Winding7) Tolerance ± 10 %

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DESIGN SHEET FOR ARMATURE OF D-C GENERATOR (PART – 1)ITEM NO.

SPECIFICATION:1550 kW ; 660/750 Volts ; 500 rpm SYMBOLPreliminary or Assumed Values Final Values

ARMATURE CORE AND WINDING 1 Number of Poles - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - P 12 12 Frequency, Hz - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ƒ - - - - - - - - - 502 Ratio of pole arc to pole pitch - - - - - - - - - - - - - - - - - - - - - - - - - - - - r 0.64 0.643 Specific Loading - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - q 1,055 1,0554 Apparent Air-gap flux density (open circuit) - - - - - - - - - - - - - - - - Bg

' ' 62,550 70,706.506 5 Line Current (full load) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - I - - - - - - - - - 2,348.485 6 Type of Winding - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - lap 7 Armature current per circuit - - - - - - - - - - - - - - - - - - - - - - - - - - - - Ic 392.59 8 Output factor (laD2) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 44,481.004 50,682.602 9 Armature diameter, in - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - D 57.34 57.34 10 Peripheral velocity, fpm - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ν 7,509.72 7,509.7211 Total number of face conductors - - - - - - - - - - - - - - - - - - - - - - - - - Z 484 484 12 Number of slots - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - S - - - - - - - - - 20 13 Number of conductors per slot - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 24 14 Axial length of armature core; gross, in - - - - - - - - - - - - - - - - - - - - la 13.53 15.415 15 Flux per pole (open circuit) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ϕ 16.363 x106 16 Pole pitch, in - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - τ 15.012 17 Pole arc, in - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - rτ 9.61 18 Area covered by pole face (rτla), sq in - - - - - - - - - - - - - - - - - - - - - - - - - - - 130.023 148.13819 Dimensions of armature conductors, in. units - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (0.185x0.345) 20 Slot pitch, in - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - λ - - - - - - - - - 9.007 21 Slot width, in - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - s - - - - - - - - - 0.8 22 Slot depth, in - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - d - - - - - - - - - 1.10423 Tooth width, in. At top - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - t - - - - - - - - - 8.207 At root - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 7.86 Average - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 8.034

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24 Number of radial ventilating ducts - - - - - -- - - - - - - - - - - - - - - - - - n - - - - - - - - - 1 25 Width of radial ducts, in - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.375 26 Net length of armature core, in - - - - - - - - - - - - - - - - - - - - - - - - - - - ln 12.103 13.837 27 Net tooth section under pole, at center, sq in - - - - - - - - - - - - - - - - - - - - - - 103.718 118.577 28 Apparent density in teeth under pole, at center, sq in - - - - - - - - - - Bt

' ' 157,770.458 91,999.51 29 Length per turn of armature coil, in - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 68.602 30 Resistance of one turn, ohms at 60°C - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.00042231 Resistance of armature, ohms - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.0085 32 IR drop in armature, volts - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 6.674 33 I2R loss in armature winding, watts - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 31,441.889 34 Estimated full-load flux per pole - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 16.661x106 35 Flux density in armature core below teeth - - - - - - - - - - - - - - - - - - - - - - - - 76,000 76,002.571 36 Internal diameter of core stampings, in - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 39.572 37 Weight of iron in core (without teeth), lb - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 4,484.022 38 Weight of iron core in teeth, lb - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 665.487 39 Total weight of armature stampings, lb - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 5,149.509

DESIGN OF ARMATURE CORE AND WINDINGItem1: Number of poles (P) and frequency (ƒ)Referring to Art. 3, for an output of 1550 kW, a speed of up to 500 rpm and 12 poles would be suitable for this trial design, therefore,

ƒ = P N120

=12(500)120

= 50 HzItem 2: Ratio of Pole Arc to Pole PitchRefer to Art. 2. Since the machine will be provided with commutating poles, a suitable value for this ratio is; r = 0.64Item 3: Specific Loading (q)

2,000−1,5001,100−1,050

=1,550−1,500q – 1,050

q = 1,055 amp .−conductor¿

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Item 4: Apparent Air-gap Flux Density, open circuit (Bg’’)63,000−62,5002,000−1,500

= 1,550−1,500Bg’ ’ –62,500

Bg’’ = 62,550 lines¿2

Item 5: Line Current , Full Load ( IL )IL = full load output

full load voltage = 1,550,000

660 = 2,348.485 Amp

Item 6: Type of Winding Simplex Lap WindingRefer to Art. 6. Since the current per path should not exceed 250 to 300 amp, lap winding with two parallel paths could be used for this trial design. Item 7: Armature Current per Circuit ( Ic )

Ic = I LP1

(1+% excitation) From Art. 47, % of excitation for 1000 and larger is 0.3 to 0.4, thus, 0.3 may be used.Then,

Ic = 2,348.4856

(1+0.003) = 392.59 AmpItem 8: Output factor ( IaD2 )

IaD2 = WN

¿ = 1,550,000500

[ 6.06 x108

(62,550 ) (1,055 )(0.64)] = 44,481.003 in3

Item 9: Armature Diameter (D)Using a rectangular pole face, where k = 0.71la = πr

Pk(D) = π (0.64)

12(0.71)(D) = 0.236 D

Hence, D = 3√44,481.003 /0.236 = 57.34 in.Item 10: Peripheral Velocity (fpm)

4

ν=πDN12

= π (57.34 )(500)12

= 7,509.72 fpm Item 11: Total number of Face Conductors (Z)

Z=πDqI c

= π (57.34 )(1,055)392.59

= 484 conductorsItem 12: Number of Slots (S) Assuming that there are 24 conductors per slot

S= Zconductors per slot

= 48424

= 20 slotsAnd the Slot per Pole ( S

P) would be,SP

=2012

=1.67 slots per pole

SP2

=206

=3.33 slots per pair pole

Item 13: Number of Conductors per Slot ( CS ) C s=

ZS=48420

=24conductors per slot

Item 14: Axial length of Armature Core ( la )Solving for la ,la=

laD2

D2 = 44,481.003(57.34)2

=13.53∈¿

Item 15: Flux per Pole, Open Circuit ( ϕ)From Voltage Equation for dynamo;E= ϕPNZ

P160 x108

Solving for ϕ,ϕ=

E P160x 108

PNZ=660(12)(60 x108)12(500)(484 )

=16,363,636.36maxwells

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Item 16: Pole Pitch ( τ) τ=π D

P=π (57.34)12

=15.012∈¿

Item 17: Pole Arc ( rτ ) rτ=0.64 x 15.012=9.61∈¿

Item 18: Area covered by Pole Face (rτ la) rτ la=9.61 (13.53 )=130.023¿2

Item 19: Dimensions of the Armature Conductors Assuming k = 700,000∆ = k

q + ν

4 = 700,000

1,055+ 7,509.72

4 = 2,540.94 amp .

¿2Then,Area of Cross Section = I c

∆ = 392.592,540.94

= 0.155 ¿2Referring to Appendix I (Wire Table, bare and DCC Copper Ribbon), an insulated copper conductor having a cross sectional area of 0.184 ¿2 has a thickness of (t=0.185) and width of (w=0.345) respectively.

Item 20: Slot Pitch (λ ) λ=π D

S=π (57.34)20

=9.007∈¿

Item 21: Slot Width (s ) Insulation thickness = 27 + E

30 = 27 + 660

30 = 49 mils or 0.049 in

s=3t+5(insulation thickness)

s=3 (0.185 )+5 (0.049 )=0.8∈¿

Item 22: Slot Depth (d ) d=2w+3 ( insulation thickness )+ 1

3s

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d=2(0.345)+3 (0.049 )+ 13(0.8)

d=1.104∈¿

Fig. I Arrangement of Conductors in Slot (Scale: NTS)Item 23: Tooth Width (t ) At Top,

tTop=λ−s=9.007−0.8=8.207∈¿At Root,tRoot=

π (D−2d )S

−s=π [57.34−2(1.104 )]

20−0.8=7.86∈¿

Therefore the Average Tooth Width is,t Ave=

tTop+ tRoot2

=8.207+7.862

=8.034∈¿

Item 24: Number of Radial Ventilating Ducts 7

Refer to Art. 16. Not more than three ducts should be necessary in an armature 9 to 11 in long, with each duct 38

in. wide.Therefore, Use one (1) radial ventilating duct.

Item 25: Width of Radial Duct ( lv) lv=

38

in or 0.375 inItem 26: Net length of Armature Core ( ln )Refer to Art. 16.

ln=0.92 (la−lv )=0.92 [13.53−1(0.375)]=12.103∈¿

Item 27: Net Cross Section of Teeth Under Pole ln (t ave)( SP ) (r )=12.103 (8.034 )( 2012 ) (0.64 )=103.718¿2

Item 28: Apparent Density in Teeth Under Pole (Bt' ')

Bt' '=( ϕ

Net Cross SectionoTeeth )=16,363,636.36103.718=157,770.458 lines

¿2

But referring to the table on number 17, a 50 cycles have a Bt' 'of 126,000 lines

¿2

and applying the 10 % tolerance, a 50 cycles should be below 138,600 lines¿2

.Since the preliminary value of Bt

' ' is above the upper limit, therefore, it will be necessary to correct (increase) the length of the armature core to bring the Flux Density down to a reasonable figure.Assuming Bt

' ' = 138,000 lines¿2

Net Cross Section of Teeth = 16,363,636.36

138,000 = 118.577 ¿2

Net Length Armature is.8

ln=Net Cross Sectionof Teeth

t ave xSPx r

= 118.577

8.034( 2012 )(0.64)=13.837∈¿

To solve for the new and final value of la ,la=

ln0.92

+lv=13.8370.92

+0.375=15.415∈¿

Then,laD

2=(15.415 )(57.34 )2=50,682.602¿3

rτla= (9.61 )(15.415)=148.138¿2

Bg' '=16,363,636.36

148.138=110,462.112 lines

¿2

Bt' '=105,180.305( 12.10313.837 )=91,999.51 lines¿2

Now both Bg' '∧Bt

' ' have an acceptable values wherein both values are within ± 10 % tolerance.Item 29: Number of Radial Ventilating Ducts By referring to Art. 16,

sinα=1.15 sλ

=1.15(0.8)9.007

=0.102

Then,α=sin−10.102=5.854 ° ; cos α=cos38.16=0.995

I c=2 τcosα

+4d+3=2(15.012)0.995

+4 (1.104 )+3=37.772∈¿ Therefore,

Total Length per Turn = Ic + 2 Ia = 37.772+2 (15.415 )=68.602∈¿

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Item 30: Resistance of one turn, Ohms at 60R=Total Length per Turn

2 (t x w ) x 4πx106

= 68.602

2 [ (0.185 )(0.345) ] x 4πx106

=0.000422Ω

Item 31: Resistance of Armature, (R¿¿ c)¿

Rc=R (Total Number of TurnsP1 )=0.000422( 48412 )=0.017Ω

Finally, the Total Resistance in the ArmatureTotal Armature Resistance=1

2(0.017 )=0.0085Ω

Item 32: IR Drop in ArmatureRc x I c=0.017 x392.59=6.674Volts

Item 33: I2R Loss in Armature WindingI 2R Loss=I c

2 x RcP=392.592 (0.017 ) (12 )=31,441.889watts

Item 34: Full load FluxAssuming a brush contact drop of 2 volts, and a series field and commutating-field drop of about one-half that in the armature winding, the total generated voltage at full load must be,

TotalGenerated Voltage at Full Load=V FL+Brushdrop+ IR drop+IR drop2

¿660+2+6.674+ 6.6742= 672.011Volts

10

Therefore, the Full Load Flux must be,Full Load Flux=ϕ(Total GeneratedVoltage at Full loadV FL

) ¿16,363,636.36( 672.011660 ) ¿16,661,429.75maxwells

Item 35: Flux density in Armature Core below TeethFrom Art. 17, the suitable density for a frequency of 50 Hz is 76,000. Bearing in mind that the air-gap is into two equal parts below the teeth, the armature flux is one-half of the total flux. Therefore, density in core is 76,000 lines

¿2 .

Rd=ϕ

2x ln x Density∈Core= 16,363,636.362 (13.837 )(76,000)

=7.78∈¿

And the final value of Flux Density in Armature Core below Teeth is,FluxDensity∈ArmatureCore= ϕ

2 x ln x Rd

= 16,363,636.362 (13.837 )(7.78)

=76,002.571 lines¿2

Item 36: Internal Diameter of Core StampingsDi=D−2 Rd−2d=57.34−2 (7.78 )−2 (1.104 )=39.572∈¿

Item 37: Weight of iron CoreThe weight of a cubic inch of Iron is 0.28 lb, and the total weight of iron Core below the Teeth will be,

Weight of IronCore=0.28x ln xπ4

¿

¿0.28(13.837)( π4 ) [57.34−2 (1.104)]2−(39.572)2

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¿4,484.022 lbItem 38: Weight of iron Teeth

Weight of IronTeeth=0.28 x d x ( tave x S ) x ln ¿0.28(1.069)(8.034 x20)(13.837) ¿665.487 lbItem 39: Total Weight of Armature Stampings

TotalWeight=Weight of IronCore+Weight of IronTeeth ¿4,484.022+665.487 ¿5,149.509 lb

DESIGN SHEET FOR ARMATURE OF D-C GENERATOR (PART – 2)ITEM NO.

SPECIFICATION:1550 kW ; 660/750 Volts ; 500 rpm SYMBOLPreliminary or Assumed Values Final Values

COMMUTATOR AND BRUSHES 40 Diameter of Commutator, in Dc - - - - - - - - - 36.45 41 Average volts per turn of armature winding - - - - - - - - - - - - - - - - 37.19 42 Number of turn between bars - - - - - - - - - - - - - - - - 1 43 Total number of Commutator bars - - - - - - - - - - - - - - - - 23944 Bar pitch, in - - - - - - - - - - - - - - - - 0.479 45 Width of copper bar (on surface), in - - - - - - - - - - - - - - - - 0.449 46 Radial depth of bar, in - - - - - - - - - - - - - - - - 3.43 47 Current density at brush-contact surface, amp per sq in ∆b

40 40.061 48 Contact area per brush set, sq in - - - - - - - 19.63 19.6 49 Brush arc (circumferential width), in - - - - - - - - - - - - - - - - 1.0 50 Axial brush length (total) per set, in - - - - - - - - - - - - - - - 19.6 51 Number of brushes per set - - - - - - - - - - - - - - - - 31.36 52 Axial length of Commutator, in Lc - - - - - - - - - 22.85

12

53 Brush-contact drop, volts - - - - - - - - - - - - - - - - 2.3 54 Brush-contact loss, watts - - - - - - - - - - - - - - - - 9,096.684 55 Brush-friction loss, watts - - - - - - - - - - - - - - - - 12,684.355 56 Total brush loss, watts - - - - - - - - - - - - - - - - 21,781.039 57 Drawing to scale giving leading dimensional ofArmature and Commutator - - - - - - - Fig. II

DESIGN OF COMMUTATOR AND BRUSHESItem 40: Diameter of Commutator (Dc)From Art. 14, a diameter of Commutator not exceeding 80 % of the Armature Core Diameter is generally found practical, although a reasonably good rule to follow is to make Dc=

D2

+Rd∈¿ . Thus,

Dc=57.342

+7.78=36.45∈¿

This is 63.57 % of the core diameter and makes the peripheral velocity, νc = 0.6357 x 7,509.72 = 4,773.929 fpm

Item 41: Average Volts per Turn of Armature Winding13

EZ2 P1

= 7504842 (12)

=37.19Volts

Item 42: Number of Turns between Bars One (1)Item 43: Total Number of Commutator Bars

C=(Cs2 x S)−1=( 242 x20)−1=239Commutator Bars

Item 44: Bar Pitchπ Dc

C=π (36.45)239

=0.479∈¿

Item 45: Width of Copper Bar (On Surface)Using 0.03 in mica for insulation, the width of copper bar = Bar Pitch ― 0.03, therefore, 0.479― 0.03 = 0.449in at the Commutator surface.Item 46: Radial Depth of BarSince the Peripheral Velocity is less than 4,500 fpm, the Radial Depth of the Commutator Segment should be about,

h=D c+1515

=36.45+1515

=3.43∈¿

Item 47: Current Density at Brush-Contact Surface (∆b)The Current Density over the Brush Contact Surface is about 30 to 50 amp .

¿2 ,

taking 40 amp .¿2

as preliminary value.Item 48: Contact Area per Brush Set

P I c

( P2 )(∆b)=12(392.59)6(40)

=19.63¿2

Item 49: Brush Arc (Circumferential Width)14

1.0inchItem 50: Axial Brush Length per Set

Contact AreaBruh Arc

=19.631.0

=19.63∈¿

Item 51: Number of Brushes per SetThe actual area per Brush Set is,AxialBrush Length x Brush Arc=19.63 x 1.0=19.6¿2

From Art. 34, the usual width of Brush is something between the limits 58

in and 114

in., and as a further check on the desirable dimensions, the width should not exceed one-tenth of the pole pitch referred to the Commutator surface. Thus, use a Brush of 58

in. width.Then,

No .of Brushes per Set=19.658

= 19.60.625

=31.36 Brushes

The Current Density will be increased slightly to;40 ( 19.6319.6 )=40.061 amp .¿2

Item 52: Axial Length of Commutator (Lc)In addition to the 19.6 in which must be provided for the four 5

8 in Carbon Brushes, the Axial Length of the Commutator Face must allow for the following:

a. Brush holder and clearances = 4 516

=54∈¿

b. Staggering of (+) and (-) Brushes = 58∈¿

c. End clearance for Brushes = 1 ind. End Play = 3

8∈¿

Total Length will be,15

Lc=19.6+54+ 58+1+ 3

8=22.85∈¿

Item 53: Brush Contact DropReferring to Fig. 38, the Brush Contact Drop for Hard Carbon at about 40. amp .

¿2 is 2.085 volts. Allowing the 10 % of roughness, chipping and irregularities, thi drop will be about 2.3 volts.

Item 54: Brush Contact LossBrush Contact Drop x P x Ic = 2.3 (12 x 329.59) = 9,096.684 wattsItem 55: Brush Friction LossFor a peripheral speed greater than 4,000 fpm, use c = 0.25 for hard carbon; P = 2 lb

¿2

W f=cPAN Dc π x746

12 x33,000=0.25 (2 )(19.6 x12)(500)(36.45)π x746

12 x33,000=12,684.355watts

Item 56: Total Brush LossTotal Brush Loss=BrushContact Loss+Brush Friction Loss ¿9,096.684 + 12,684.355 ¿21,781.039watts

Item 57: drawing to Scale giving leading Dimensions of Armature and CommutatorShaft Diameter=0.84 3√ Output∈wattsN

¿0.84 3√ 1,550.000500 ¿12.25∈¿

16

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