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
Ampere turn per inch
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
Flux entering armature per pole
Ø = 2,651,000
Flux density
At armature = Ø/2x At = 45,100 lines/in2
At pole core = [1.125] Ø/ 51.7 = 57,700 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
Ampere turn per inch
Armature: Annealed = 11
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
Flux entering armature per pole
Ø = 2,983,219
Flux density
At armature = Ø/2 x At = 2,983,219/2 x 29.4 = 50,735
lines/in2
At pole core = [1.125] Ø/ 51.7 = 64,920 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
Ampere turn per inch
Armature: Annealed = 13
Pole core: Annealed = 25
Yoke ring: Magnet = 35
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
Flux entering armature per pole
Ø = 4,330,000
Flux density
At armature = Ø/2 x At = 4,330,000/ 2 x 29.4= 73,700 lines/in2
At pole core = [1.125] Ø/ 51.7 = 94,200 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
Ampere turn per inch
Armature: Annealed = 16
Pole core: Annealed = 38
Yoke ring: Magnet = 52
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
Top Related