continuation of machne design

Item 24 – 25: Ventilating Ducts

Not more than three(3) ducts should be necessary in an armature

9 to 11 inches long with each ducts 3/8 inch wide.

Item 26: Net length of the armature

ln = 0.92(la – lv)

where: la = axial length of the armature core

lv = total width of all vent ducts

ln = 0.92( 8.52 – 3 x 3/8) ; ln = 6.8 inches

Item 27: Net cross section of teeth under pole

Net cross section of teeth under pole = ln x Atw x ( slots / pole) x r

Where: ln = net length of armature

Atw = average tooth width

r = ratio of pole arc to pole pitch

Net cross section of teeth under pole = 6.8 x 0.150 x ( 201 / 6 ) x

0.64

Preliminary Net cross section of teeth under pole = 21.87 sq.

inches

Final Net cross section of teeth under pole = 6.84 x 0.150 x ( 201/

6 ) x 0.64

Final Net cross section of teeth under pole = 22 sq. inches

Item 28: Flux density in teeth

Flux density in teeth = Ф / Net cross section of teeth under pole

=2,712,700 / 21.87

Flux density in teeth = 124,037.49 lines/ sq.inch

By formula:

laD2 = (P/N)(6.06 x 108 / (B”g x q x r))

= (160,000 / 1200)( 6.06 x 108 / (51,100 x 781.12 x 0.64))

laD2 = 2933.34

laD2 = 3014.85 / D2 = 3014.85 / 18.552 = 8.76

B”g = 2,712,700 / (6.21 x 8.76) = 49866.18 lines/ sq.inch

lar τ = 6.21 x 8.76 = 54.40 sq.inch

ln = ( 8.76 – 1.125 ) x 0.92

ln = 7.02 inches

B”t = ln old / ln new x Фteeth

B”t = ( 6.34 / 7.02 ) x 124,037.49

B”t = 112,022.46 line/ sq.inch

Item 29: Length per turn of armature core

sin θ = (1.15 x s) / λ

sin θ = (1.15 x 0.1160) / 0.2899

θ = 27.4

By formula:

lc = ( 2τ / cos θ + 4d + 3)

lc = ( 2 x 9.71)/ cos 27.4 + (4 x 1.53) + 3

lc = 31.54 inches

total length per turn = lc + 2la

= 31.53 +( 2 x 8.76)

total length per turn = 49.05 inches

Item 30: Resistance of one turn, ohm at 60 o C

R = total length per turn / (2 x Ls x dc x (4/π)x106 )

Where: Ls = space for each conductor

Dc = depth of each conductor

R = 49.05 / (2 x 0.0665 x 0.6015 x (4/π) x106 )

R = 481.55 μΩ

Item 31: Resistance of armature

With a total of 415 / 2 = 208 turns divided into six parallel paths, the

resistance per ckt. will be 481.55 x ( 811 / 2(6)) = 0.0325 and the total

armature resistance will be 1/6 of 0.0325 / 6 = 5.42 ohms.

R = (481.55 μΩ)(2 / 2P)

= 481.55 (811 / 2(6))

= 0.0325 ohms

Ra = 0.0323 / 6

= 5.42 μΩ

Item 32: IR drop

V = IR

Where: V = is the voltage drop across the winding

I = current at full load

R = resistance per ckt.

V = (0.0325)(56.13) = 1.82 V

Item 33: Wattage loss in armature winding

W = I2R = 56.13 x 1.82 x 6 = 612.94 watts

Item 34: Full load flux

In addition to the specified increase in terminal voltage from 220 to

240 volts, it is necessary to develop enough voltage to overcome

internal resistances. Assuming a brush contact drop of about ½ that of

the voltage drop in the armature winding, the total generated voltage

at the full load must be:

480 + 2 + 1.82 + (0.94 / 2) = 484.29 volts

The full load flux therefore is:

2,712,700 x ( 484.29/ 440) = 2,985,757.92 maxwells

Item 35 – 36: Flux density in armature core and Internal

Diameter

Assume a density of 73000 from the table of upper limits of flux

density in dynamo armatures is suitable.

By formula:

Rd x ln x 73,000 = (Ф/2)

Where: Rd = radial depth

ln = net length of the armature

Ф = full load flux

Rd = 2,985,757.92 / ( 2 x 7.02 x 73,000 )

Rd = 2.91 inches

The internal diameter(di ) of the core stamping is therefore:

di = D – ( 2 x Rd ) – ( 2 x d)

where: D = armature diameter

d = slot depth

di = 18.55 – (2 x 2.91) – ( 2 x 1.53)

di = 9.67 inches

Item 37: Weight of Iron in Core

The weight of iron in cubic inch is 0.28, then the total weight of iron

in core below the teeth will be:

=0.28 x ln x (π/4)( (D – 2xd)2 - di 2)

= 0.28 x 7.02 x (π/4)[ ( 18.55 – 2 x 1.53)2 – (9.67)2

= 230 lbs.

Item 38: Wight of iron in teeth

0.28 x d x Atw x s x ln

= 0.28 x 1.53 x 201 x 0.15 x 7.02

= 90.67 lbs.

Item 39: Total weight of the armature stamping

Total weight of the armature stamping = Wic + Wit

Total weight of the armature stamping = 230 + 90.67

Total weight of the armature stamping = 320.67 lbs.

B.)Design of commutator and brushes

Item 40: Diameter of the commutator

Diameter of the commutator = (armature diameter / 2) + 4

Diameter of the commutator = (18.55 / 2) + 4

Diameter of the commutator = 13.28 inches

Item 41- 43 : Average voltage per turn of armature winding,

Number of turns of armature winding, total number of

commutator bars.

As shown in the table of Volts between commutator segment, a

480volts machine can operate from 4 to 12 volts. Therefore:

E/ ( Z/ 2P) = 480 / ( 811 / 2(6)) = 7.1 V

C = 811 / 2

Number of turns of armature winding = 1

Total number of commutator bars = 405.5 commutator segments

Item 44 – 45: Bar pitch and Bar width

Bar pitch = ( π x 13.28) /405.5 = 0.103 inch

With mica 0.03 thick in inch, the bar width is 0.103 – 0.03 = 0.073

inch

Item 46: Radial depth of segment (h)

h = (Diameter of the commutator +15) / 15 x (4700/4500)2

h = ( 13.28 + 15) / 15 x (4700/4500)2

h = 2.05 (assume)

Item 47 - 51: Dimensions of Brushes

Unless a very soft quality of carbon is used, the current density over

the brush contact surface is about 30 to 50 ampere/ inch.

Use 40 amperes/ inch as a preliminary value.

Contact area per brush set = ( 6 x 56.13) / (3 x 40) = 2.81 sq.inch

Assume brush arc (Circumferential width) = 1 inch

Axial length per set = Contact area per brush set / brush arc

Axial length per set = 2.81 sq.inch / 1 inch

Axial length per set = 2.81 inches say 3 inches

The current density will then be increase to:

( 40 x 281 ) / 3 = 37.47 amperes/ inch

Assume the number of brushes as indicated up to eight(8) – 1¼ by 1

inch

In addition to the 8.8 inches which must be provided for eight(8) – 1¼

by 1 inch carbon brushes, the axial length of the commutator face

must allow for the following:

a. brush holder clearance = 8 x 5/16 = 2.5 inches

b. stage ring of positive(+) and negative(-) brushes = 5/8 inch

c. end clearance for brushes = 1inch

d. end play = 3/8 inch

total axial length = 3 +2.5 + 5/8 + 1 + 3/8

total axial length = 7.5 or 8 inches

Item 53: Brush contact drop

Referring to the figure of Potential drop due to surface resistance of

carbon brushes, the brush contact drop for hard carbon at about 40

amperes/ sq.inch is 2.08. allowing 10% for roughness, chipping and

irregularities, this drop will be about:

2.08 + 2.08 x 10% = 2.29 volts

Item 54 – 56: Brush Losses

Brush contact loss = 2.29 x 6 x 56.13 = 771.23 watts

Brush friction loss(Wf)

= cPAND x π x 746 / ( 12 x 33,000)

Where: c = coefficient of friction with carbon brushes

= 0.23 for hard carbon

P = pressure of the brush in the commutator, use 2

lbs/sq.inch of contact

pressure.

N = speed

D = commutator diameter

A = total area of the brush surface

Brush friction loss = ( 0.25 x 2 x 2.81 x 6 x 1,200 x 13.28 x π x

746) / (12 x 33,000)

Brush friction loss = 795.06 watts

Total brush losses = 771.23 watts + 795.06 watts

Total brush losses = 1566.29 watts

Item 57: Diameter of the shaft supporting the armature

Diameter = ( output power / speed)1/3

Diameter =0.84 x ( 160,000 watts / 1200 rpm) 1/3

Diameter = 4.29 inches

C) Design of Pole cores, Frames and Windings

Item 1: Number of Poles

Number of Poles = 6 poles

Item 2: Number of commutating poles

Number of Commutating poles = 6 poles

Item 3: Diameter of armature core

Diameter of armature core = 19 inches

Item 4: Axial length of Armature core, gross

Axial length of Armature core = 8.56 inches

Item 5: Ratio of pole arc to pole pitch

r = 0.64

Item 6: Air Gap under center of pole(δ)

δ = ( Z x Ic) / ( P x B”g )

where: Z = Total number of face conductor

Ic = load current

P = Number of poles

B”g = total apparent air gap density

δ = ( 415 x 112.31 ) / ( 6 x 48,620)

δ = 0.16 inch

By using formula:

δt = 0.04 x ( D + 3)1/2

δt = 0.04 x ( 19 + 3 )1/2

δt = 0.19 inch

Item 7: Leakage factor

In the table of leakage factor, by interpolation:

( 200 – 100) / ( 200 – 160) = ( 1.22 – 1.11) / (1.22 – lf)

Lf = 1.18

Item 8: Maximum Flux

The flux estimated in item 34 of this design is 2,983,219 maxwells.

From the B-H curve it appears that 95,000 lines/sq.inch will not

require as excessive number of ampere-turns.

Item 9: Cross section of the pole core

Cross section of the pole core = (Фe x lf) / Фm

where: Фe = estimated flux

lf = leakage factor

Фm = maximum flux

Cross section of the pole core = (2,983,219 x 1.18) / 95000

Cross section of the pole core = 37.05 sq.inch

Item 10: Length of winding space

Lcore = la – 1.5δ

where: la = axial length

Lcore = length of the core

δ = total air gap

Lcore = 8.56 – 1.5(0.19)

Lcore = 8.28 inches

The width of the pole core should therefore be:

= 37.05 / 8.28 = 4.475 inches

The dimension will tentatively be made 6 ¼.

Item 11: Flux Density in Frame

Select 90,000 lines/sq.inch

Item 12: Cross-Sectional Area of the Yoke

Cross-Sectional Area of the Yoke = (Фe x lf ) / ( 2 x 90,000)

Cross-Sectional Area of the Yoke = ( 2,983,219 x 1.18 ) / ( 2 x

90,000 )

Cross-Sectional Area of the Yoke = 20 sq. inch

Assumed yoke overhang of 2 inches per end,

axial length of the yoke = 8.56 + ( 2x2) = 12.56 inches

Width of the yoke = 19.56 / 12.56

Width of the yoke = 1.56 inches say 2 inches

Item 13: Outside diameter of Frame

AB = ( rτ – pole width)

AB = ( 6.37 – 4.29) inch

AB = 2.08 inch

Flux passing at portion of X:

Фx = ( lf -1) x Фfl / 4 + ABx Фfl / rτ

Фx = (( 1.18 -1) x 2,983,219 / 4) + (2.08x 2,983,219 / 6.37)

Фx = 1,108,357,182 maxwells

x = 1,108,357.182 / ( 8.28 x 90000) = 1.49 inches

Item 14 - 17: Ampere turns per pole for Total Magnetic Circuit.

Plotting the Open – circuit.

saturation curve.

δe = equivalent air gap

= 0.5685 / ( 0.1685 / 0.19 + 2ln ( 0.5 x 0.4/0.19 + 1))

δe = 0.2445 inch

de = equivalent slot depth

de = 0.905 + 0.19 – 0.2445

de = 0.4705

From Item 23 of Design 1

t = 0.1685 inch

tr = 0.1143 inch

tave = 0.1414 inch

From formula 40:

Bg = Bt (de + δe) + de λℓa - 1

µ tmln

de λℓa + δe

tmln

Bg = Bt [1.3965 + 0.3965/µ]

Table 1:

Bt at tm

assumed

H,

magnetizing

force

µ

Bt/H

Bg

Formula 40

12, 000 8 1, 500 16,755

13, 000 12

1,

083.33 18,150

14, 000 21 667 19,543

16, 000 58 276 22,321

Table 2:

Bg At the

middle

At

root

At top H (TI)t

Bt Hm Bt Hr Bt He

22,321 16,

000

60 19,79

4

130 15,87

7

15 172.

5

164.0

5

19,543 14,

000

23 17,31

9

34 13,89

2

9 22.5 21.4

18,150 13,

000

15 16,08

2

22 12,90

0

7 14.8

3

14.1

16,755 12,

000

9 14,84

5

14 11,90

7

6 9.33

3

8.88

te = de(t - tr)/d + tr

= [ 0.4705( 0.1685 - 0.1143)/0.905 ] + 0.1143te =

0.1425 inches

For Column 1: Value of Bg in Table 1

For Column 2: Bt at middle corresponding to Bg(Base on Curve No. 1)

For Column 3: Base on the graph (Fig 47 & 48) corresponding to Bg at middle

For Column 4: Bt at root = Bt at middle x tm/tr

Bt1 = 16, 000 x ( 0.1414 / 0.1143 ) = 19,784 gauss

Bt2 = 14, 000 x ( 0.1414 / 0.1143 ) = 17,319gauss

Bt3 = 13, 000 x ( 0.1414 / 0.1143 ) = 16,082 gauss

Bt4 = 12, 000 x ( 0.1414 / 0.1143 ) = 14,845 gauss

For Column 5: Corresponding Hr for Bt at root (Base on graph Fig 47 & 48)

For Column 6: Bt at top = Bt at middle x tm/te

Bt1 = 16, 000 x ( 0.1414/0.1425) = 15,877 gauss

Bt2 = 14, 000 x ( 0.1414/0.1425) = 13,892 gauss

Bt3 = 13, 000 x ( 0.1414/0.1425) = 12,900 gauss

Bt4 = 12, 000 x ( 0.1414/0.1425) = 11,907gauss

For Column 7: He for Bt at top (base on the graph Fig 47 & 48)

For Column 8: Average H

H = 2/3Hm + (Hr + He)/6

H1 = (2/3)(60) + (130 + 15)/6 = 172.5 Gilberts/cm

H2 = (2/3)(23) + (34 + 9)/6 = 22.5 Gilberts/cm

H3 = (2/3)(15) + (22 + 7)/6 = 14.83 Gilberts/cm

H4 = (2/3)(9) + (14 + 6)/6 = 9.333 Gilberts/cm

For Column 9: Ampere-turns at teeth, (TI)t

(TI)t = ( Hde x 2.54)/0.4π

280

260

TI1 = (172.5 x 1.224 x 2.54)/ 0.4π = 164.05 A-t

TI2 = (22.5 x 1.224 x 2.54)/ 0.4π = 21.4 A-t

TI3 = (14.83 x 1.369 x 2.54)/ 0.4π = 14.1 A-t

TI4 = (9.333 x 1.369 x 2.54)/ 0.4π = 8.88 A-t

Ampere turn required for air gap:

TIg = ( Bg x δe x 2.54)/ 0.4π

TIg = ( Bg x 0.2445 x 2.54)/ 0.4π = 0.4942 Bg

ITEM 14. TI per pole for total magnetic circuit (no load)

Intersection at 220 V and DC curve = 2490.37 A-t

ITEM 15. TI per pole on shunt field(full load)

Intersection at 240 V and resistance line = 3231.12 A-

ITEM 16. TI per pole total at full-load

Intersection of 247.57V and DC curve = 4378.5 A-t

ITEM 17 . TI per pole in series winding

4378.5 A-t - 3231.12 A-t = 1147.38 A-t

Curve

Bg (gauss) TI g

22,321 11,031

19,543 9,658

18,150 8,970

16,755 8,280

No

Load

Voltage

200

240

220

1 2 3 4 5 6

Table 3No load voltageFlux entering armature per pole, maxwells

210

3,500,000

220

2,651,000

240

2,983,219

260

4,330,000

Flux density: lines/in2

Armature core 59,509 45,100 50,735 73,700

Pole core 76,200 57,700 64,920 94,200

Yoke ring 71,900 54,440 61,300 89,000

Air gap 42,700 32,410 36,500 53000

Ampere turn per inch

Armature 10 11 13 16

Pole core 17 19.5 25 38

Yoke ring 25 28.5 35 52

Ampere turn

Armature 38 36.52 43.12 61

Pole core 119 136.5 175 266

Yoke ring 250 285 350 520

Air gap and teeth 3,200 2032.35 2,663 4,250

Total 3,607 2,490.37 3,231.12 5,097

For 210V

Flux entering armature per pole

Ø = 3,500,000

Flux density

At armature = Ø = 3,500,000 = 59,509 lines/in2

2 x At 2 x 29.4

At pole core = [1.125] Ø/ 51.7 = 76,200 lines/in2

Shunt field-open circuit

Shunt field-full load

Shunt & series full load

Series field

Ampere-turn (x1000)

At yoke ring = 1.15(3,500,000) = 71,900lines/in2

2 x 28.06

At air gap = 3,500,000

81.8

= 42,700lines/ in2


Armature: Annealed = 10

Pole core: Annealed = 17

Yoke ring: Magnet = 25

At armature = 10 A-t x 3.8 in = 38 A-t inches

At pole core = 17 A-t x 7 in = 119 A-t inches

At yoke ring = 275A-t x 10 in = 250 A-t inches

Air gap and teeth = 3,200 A-t inches

Total = 3,607A-t inches

For 220 V


Ø = 2,651,000

Flux density

At armature = Ø/2x At = 45,100 lines/in2


At yoke ring = 1.15(Ø)/2x28 = 54,440 lines/in2

At air gap = 2,651,000/81.8 = 32,410 / 6.44 = 5,032.35



Pole core: Annealed = 19.5

Yoke ring: Magnet = 28.5

Amp-turns

At armature = 11 A-t x 3.32 in = 39.6 A-t inches

At pole core = 19.5 A-t x 7 in = 149.5 A-t inches

At yoke ring = 28.5 A-t x 10 in = 275.9 A-t inches

Air gap and teeth = 2,032.35 A-t inches

Total = 2,490.37 A-t inches

For 240 V


Ø = 2,983,219

Flux density

At armature = Ø/2 x At = 2,983,219/2 x 29.4 = 50,735

lines/in2


At yoke ring = 1.15(2,983,219)/2 x 28 = 61,300 lines/in2

At air gap = 2,983,219/81.8 = 36,500/6.44 = 5663 gauss





At armature = 13 A-t x 3.32 in = 43.12 A-t inches

At pole core = 25 A-t x 7 in = 175 A-t inches

At yoke ring = 35 A-t x 10 in = 350 A-t inches

Air gap and teeth = 2663 A-t inches

Total = 3231.12 A-t inches

For 260 V


Ø = 4,330,000

Flux density

At armature = Ø/2 x At = 4,330,000/ 2 x 29.4= 73,700 lines/in2


At yoke ring = 1.15(4,330,000)/2 x 28.06 = 89,000

lines/in2

At air gap = 4,330,000/81.8 = 53000 lines/ in2





At armature = 16 A-t x 3.6 in = 61 A-t inches

At pole core = 38 A-t x 6.5 in = 266 A-t inches

At yoke ring = 52 A-t x 8.9 in = 520 A-t inches

Air gap and teeth = 4,250 A-t inches

Total = 5,097 A-t inches

ITEM 18-22. Shunt field winding

Assuming about ½ inch for insulation at the top and bottom and

for space between shunt and series coils, the total length of the

winding space will be:

6 ¼ - ½ = 5 ¾ = 5.75 inches

Dividing this in the proportion to the shunt and series ampere turns

equals:

(3231.12 x 5.75) / 4378.5 = 4.24 inches

Series coil winding apace:

5.75 – 4.24 = 1.51 inch

Width of shunt field coil = 4.24 + ¼ = 4.49 inches

Length of the shunt field coil = 8.28 + 0.25 = 8.53 inches

Assuming a total thickness of the winding(w) = 1 ¼ inches, the

mean length per turn(m) will therefore be:

m = 2(lsf + wsf) + (w x π)

where: lsf = length of the shunt field coil

wsf = width of the shunt field coil

w = total thickness of the winding

m = 2 x ( 8.53 + 4.59 ) + (1 ¼ x π )

m = 30.17 inches

It will be therefore supposed that the shunt field rheostat

adsorbs 15% of the voltage on the open circuit. By formula,

(m) = ( m x TInl x P ) / ( E x 0.85 )

where: m = mean length per turn

TInl = ampere turns per pole at no load

P = number of poles

E = voltage at no load

( m ) = ( 30.07 x 2,490.37 x 6 ) / ( 220 x 0.85 )

( m ) = 2410.73 circular mils

Refer to wire table II at the appendix of the reference book

Gage # CM Nearest turns/in2 R in ohms per 1000ft @

75oC

# 14 4545.0 182 2.78

Having a resistance of 2.78 ohms per 1000ft., the resistance at

60oC will be:

(R60 / R75) = (T + 60) / (T + 75)

Where T = 234.5 oC for a annealed copper conductor is

R60 = (2.78)( 234.5 + 60) / (234.5 + 75)

R60 = 2.65 ohms per 1000ft.

Since there is 182 wires per sq. inch, it means about 13 ½

wires per inch will be used.

Wires per layer = 13 ½ x width of the shunt field coil

= 13 ½ x 4.49

Wires per layer = 61 wires per layer

Number of layers per coil = 1 ¼ x 13 ½ = 16.88 say 17 layers per coil

Making allowance for manufacturing and paper insulation layers,

it will be assumed there will be 61 x 17 = 1037 turns per coil

Resistance per coil = (m x TC x R60 ) / 12, 000

Where: m = mean length per turn

TC = turns per coil

R60 = resistance at 60 oC

Resistance per coil = ( 30 x 1037 x 2.65 ) / 12000

Resistance per coil = 68.7 or 69 ohms

and the current will be:

At no load:

(220 X 0.85) / ( 6 x 68.7 ) = 0.454 amperes

At full load:

( 240 / 220 )( 0.454) = 0.545 amperes

Checking on the ampere turns:

At no load:

1037 turns per coil x 0.454 amperes = 470.8 ampere-turns /

coil

At full load:

1120 turns per coil x 0.545amperes = 565.17 ampere–turns /

coil

The current at full load is:

= 0.545 / ( 4545 / (4/π x 106 )) = 152.68 or 163 amperes / sq.

inch

Item 23 to 26: Series Field Coils

Available winding space = 0.5 inch

Number of turns = ( NI per pole / Ia )

= 1147.38 / ( 666.67 + 0.545)

Number of turns = 1.72 or 2 turns

Assume a current density = 1600 ampere/ sq. inch

Cross section of copper = (Ia / ∆) = ( 666.67 + 0.545) / 1600

Cross section of copper = 0.417 sq. inch

Thickness of copper = cross section of copper / available winding

space

Thickness of copper = 0.417 sq. inch / 0.5 inch

Thickness of copper = 0.834 inches

Assuming length connections equal to 50 inches, therefore,

Total length of copper = ( 6 x 2 x 30) + 50

Total length of copper = 410 inches

Total resistance of series field = total length of copper / ( As x (4/π x

106))

Total resistance of series field = 410 / ( 0.417 x (4/π x 106))

Total resistance of series field = 7.72 x 10-4 ohms

Total copper loss in the series field will be:

I2R = ( 666.67 + 0.545)2 x (7.72 x 10-4)

I2R = 343.68 watts

Item 27: Temperature Rise of Field Coils

Total Cooling Surface = 2( total winding space + w )(m)

Where: w = total thickness of winding

m = mean length per turn

Total cooling surface = 2( 4.49 + 1 1/4 ) ( 30)

Total cooling surface = 344.44 sq. inches

Watts lost in Series and shunt field winding per pole:

= (Total loss in the series field / P) + ( If2 ) x Resistance per coil of

shunt field

= ( 344.44 / 6 ) + ( 0.545)2( 69) = 77.9 or 78 watts

For V = 5970 fpm (from Design 1)

The coefficient of cooling, c = 0.0087 W/in2/ºC

Temperature rise, t = ω/c x s = 78 watts/(0.0087 x 344.44 in2 ) =

26.03ºC

Tables:

Approximate Values of Apparent Air gap Density

Output, KW B”g Output, KW B”g

5 37000 750 61000

10 42000 1000 62000

20 45000 1500 62500

30 47000 2000 63000

40 48000 2500 63500

50 50000 3000 64000

100 53000 4000 65000

200 56500 5000 65500

300 57500 7500 66500

400 58500 10000 67000

500 59000 Larger 67500

Approximate values of q for interpole dynamos

Output, KW q Output, KW q5 400 750 950

10 450 1000 1000

20 500 1500 1050

30 550 2000 1100

40 600 2500 1150

50 625 3000 1200

100 700 4000 1225

200 800 5000 1250

300 850 7500 1275

400 875 10000 1300

500 900 Larger 1300

Output, KW Number of poles

Speed, rpm

2 or less 2 Over 1250

2 to 75 4 900 to 1750

75 to 200 6 up to 1200

200 to 500 6 or 8 up to 1200

500 to 1500 8 to 12 up to 900

1500 to 2500 12 or 14 up to 500

2500 to 5000 14 to 24 up to 375

Number of poles and usual speed limits of dynamos

Upper limits of Flux density in Dynamos

Armature

Frequency Density in

teeth

Density in core

10 150,000

95,000

20 142,000

90,000

30 135,000

85,000

40 130,000

80,000

50 126,000

76,000

60 123,500

73,000

DESIGN SHEET FOR ARMATURE OF D-C GENERATOR

ITEM

NO.

SPECIFICATIONS: 160KW; 220/240 VOLTS; 1160RPM

SYMBOLPRELIMINAR

Y ORASSUMEDVALUES

FINALVALUES

1

ARMATURE CORE AND WINDING

NUMBER OF POLES

FREQUENCY

p

f6…..

6

602 RATIO OF POLE ARC TO POLE

PITCHr 0.64 0.64

3 SPECIFIC LOADING Q 760 7804 APPARENT AIR-GAP FLUX

DENSITY (OPEN CIRCUIT)Bg” 55,100 48,620

5 LINE CURRENT, Ampere I ….. 666.676 TYPE OF WINDING ….. ….. Lap7 ARMATURE CURRENT PER

CIRCUITIc 112.31 112.31

8 OUTPUT FACTOR ….. 3,014.85 3,0909 ARMATURE DIAMETER, INCHES D 19

10 PERIPHERAL VELOCITY, rpm v 5970 597011 TOTAL NUMBER OF FACES

CONDUCTORSZ 404 415

12 NUMBER OF SLOTS S 101 10513 NUMBER OF CONDUCTORS

PER SLOTS….. …… 4

14 AXIAL LENGTH OF ARMATURE CORE; GROSS, INCHES

la 8.144 8.56

15 FLUX PER POLE, MAXWELL Φ 2,651,00016 POLE PITCH, INCHES r 9.9517 POLE ARC, INCHES rr 6.3718 AREA COVERED BY POLE

FACE, INCHES²….. 51.88

19 DIMENSION OF ARMATURE CONDUCTORS

….. ….. 2(0.0705x1/3)

20 SLOT PITCH, INCHES Λ ….. 0.568521 SLOT WIDTH, INCHES S …. 0.422 SLOT DEPTH, INCHES D …. 0.90523 TOOTH WITDH

AT TOP, INCHES AT ROOT, INCHES AVERAGE, INCHES

t …..….

….….….

0.16850.11430.1414

24 NUMBER OF RADIAL VENTILATING DUCTS

n ….. 3

25 WIDTH OF RADIAL DUCTS ….. ….. 0.37526 NET LENGTH OF ARMATURE

COREln 6.46 6.84

27 NET TOOTH SECTION UNDER POLE, AT CENTER, SQ. IN.

….. 10.23 10.833

28 APPARENT DENSITY IN TEETH UNDER POLE, AT CENTER, LINES PER SQUARE INCHES

BT 259,000 245,000

29 LENGTH PER TURN OF THE ARMATURE COIL, in.

….. ….. 57.6

30 RESISTANCE OF ONE TURN, OHMS AT 60°C

….. …. 0.000963

31 RESISTANCE OF THE ARMATURE, OHMS

….. ….. 0.00555

32 IR DROP IN THE ARMATURE, VOLTS

….. ….. 3.71

33 I²R LOSS IN ARMATURE WINDIN, WATTS

…. …. 2,500

34 ESTIMATED FULL-LOA FLUX PER POLE, MAXWELL

….. ….. 2,983,219

35 FLUX DENSITY IN THE ARMATURE CORE BELOW TEETH

….. 73,000

36 INTERNAL DIAMETER OF THE CORE STAMPINGS, INCHES

….. ….. 11.19

37 WEIGHT OF IRON IN CORE (WITHOUT TEETH), lb

….. ….. 260

38 WEIGHT OF THE IRON IN TEETH, lb

….. ….. 26

39 TOTAL WEIGHT OF ARMATURE STAMPINGS, lb

….. ….. 286

40 DIAMETER OF COMMUTATOR,INCHES

Dc ….. 13.5

41 AVERAGE VOLTS PER TURN OF ….. ….. 6.94

ARMATURE WINDING, VOLTS42 NUMBER OF TURNS BETWEEN

BARS….. ….. 1

43 TOTAL NUMBER OF COMMUTATOR BARS

….. ….. 208

44 BAR PITCH, INCHES ….. ….. 0.204

45 WIDTH OF COPPER BAR (ON SURFACE), INCHES

….. ….. 0.174

46 RADIAL DEPTH OF BAR, INCHES

….. ….. 2.073

47 CURRENT DENSITY AT BRUSH CONTACT SURFACE, AMPERE PER SQUARE INCHES

∆b 40 43

48 CONTACT AREA FOR BRUSH SET, SQUARE INCHES

….. 5.62 6

49 BRUSH ARC (CIRCUMFERENCE WIDTH), INCHES

….. ….. 1.0

50 AXIAL BRUSH LENGTH (TOTAL) PER SET, INCHES

….. ….. 6

51 NUMBER OF BRUSHES PER SET

….. ….. 8

52 AXIAL LENGTH OF COMMUTATOR, INCHES

Lc ….. 10.5

53 BRUSH-CONTACT DROP, VOLTS

….. ….. 2.29

54 BRUSH-CONTACT LOSS, WATTS

….. ….. 1,543.14

55 BRUSH-FRICTION LOSS, WATTS

….. ….. 1,487.15

56 TOTAL BRUSH LOSS, WATTS ….. ….. 3,030.29

57 DRAWING TO SCALE GIVING LEADING DIMENSION OF ARMATUREAND COMMUTATOR

….. ….. …..

DESIGN SHEET FOR POLE CORES, FRAME, AND YIELD WINDINGS OF D-C GENERATOR

ITEM

SPECIFICATION: 160KW; 220/240 VOLTS; 1160 RPM; ALLOWABLE TEMPERATURE RISE OF FIELD COILS = 26ºC

SYMBOL

PRELIMINARY

ASSUMED

VALUES

FINALVALUES

GIVEN OR ASSUMED DATA1 NUMBER OF MAIN POLES p ….. 62 NUMBER OF COMMUTATING

POLES….. ….. 6

3 DIAMETER OF ARMATURE CORE

D ….. 20.6

4 AXIAL LENGTH OF ARMATURE la ….. 9.89

CORE; GROSS5 RATIO OF POLE ARC TO POLE

PITCHr ….. 0.65

6 AIRGAP UNDER CENTER OF POLE

….. 0.19

7 LEAKAGE FACTOR lf 1.18 1.18

CALCULATIONS8 MAXIMUM FLUX DENSITY IN

POLE CORE ( FULL LOAD )….. 95,000 95,000

9 CROSS SECTION OF POLE CORE

Ac ….. 37.05

10 LENGTH (RADIAL) OF WINDING SPACE ON POLE

…..8.28 8.28

11 FLUX DENSITY IN FRAME ….. 90,000 …..12 CROSS SECTION OF YOKE

RING….. 20

13 OUTISIDE DIAMETER OF FRAME

….. ….. 2.08

14 TI PER POLE FOR TOTAL MAGNETIC CIRCUIT ( NO LOAD )

….. ….. 2,490.37

15 TI PER POLE ON SHUNT FIELD ( FULL LOAD )

….. ….. 3,231.12

16 TI PER POLE TOTAL AT FULL LOAD, VOLTS

….. ….. 4,378.5

17 TI PER POLE IN SERIES WINDING, VOLTS

….. ….. 1,147.38

18 LENGTH OF WINDING SPACE FOR SHUNT COILS

…. ….. 4.49

19 WIDTH OF WINDING SPACE FOR SHUNT COILS

w ….. 1.51

20 SIZE OF SHUNT-FIELD WIRE ( B&S GAGE)

….. ….. 14

21 NUMBER OF TURNS IN WIRE PER SHUNT-FIELD COIL

….. ….. 1120

22 SHUNT FIELD CURRENT (FULL LOAD ) Amperes

….. ….. 0.545

23 NUMBER OF TURNS IN SERIES WINDING PER POLE

….. ….. 2

24 CROSS SECTION OF SERIES WINDING

….. ….. 0.417

25 RESISTANCE (HOT) OF SERIES WINDING ( OHMS )

….. ….. 7.72x10-4

26 I²R LOSS IN SERIES FIELD COILS ( WATTS) watts

….. ….. 343.68

27 SUREFACE TEMPERATURE RISE ON FIELD COILS (DEGRES C)

….. ….. 26.03

TECHNOLOGICAL INSTITUTE OF THE PHILIPPINESDEPARTMENT OF ELECTRICAL ENGINEERING

EE583D1 EE52FC1FIRST SEMESTER SY 2011 – 2012

DESIGN NO. 1

DIRECT CURRENT GENERATOR

SPECIFICATIONS:1. POWER OUTPUT 240kw2. SPEED 1600 rpm3. TERMINAL VOLTAGE

A. OPEN CIRCUIT 440VB. FULL – LOAD 480 V

SUBMITTED BY:SIBAG, JOHN GLEN MARI S.BSEE 2011

APPROVED BY:ENGR EDGARDO O. DOMINGO, PEE

INSTRUCTOR

continuation of machne design

Documents

rd x ln x

d x atw x s x ln

bg x q x r

ls x dc x 4x106

ln x atw x slots pole

inch item

armature ln

ln old lnnew x teeth