Numerical
Problems:
Ex. 1 Design
the stator frame of a 500 kVA, 6.6 kV, 50 Hz, 3 phase, 12 pole, star connected
salient pole alternator, giving the following informations.
(i)
Internal diameter and gross length of the frame
(ii)
Number of stator conductors
(iii)
Number of stator slots and conductors per slot
Specific
magnetic and electric loadings may be assumed as 0.56 Tesla and 26000 Ac/m
respectively. Peripheral speed must be less than 40 m/s and slot must be less
than 1200.
Soln:
(i)
Diameter and gross length of stator: Assuming the winding to be full pitched Kw
= 0.955 Output coefficient Co = 11 x Bav q Kw
x 10-3
= 11 x 0.56
x 26000 x 0.955 x 10-3
= 153
Speed in rps ns = 2f/p = 2 x 50/12 =
8.33 rps
Output Q
= C0 D2Lns =
D2L = Q / C0 ns =
500/( 153 x 8.33) = 0.392 m3
Using
round poles for the salient pole alternator and assuming ratio of pole arc to
pole pitch as 0.65 and pole arc equal to core length
Pole arc/
pole pitch = core length/ pole pitch = 0.65
L = πD/p = πD/12
L = 0.17D
Substituting
this relation in D2L product and solving for D and L
D = 1.32
m and L = 0.225 m.
Peripheral speed =
πDns m/s
= π x 1.32 x 8.33
= 34.6
m/s (with in limitations)
(ii)
Number of stator conductors
Eph
= 6600/√3 = 3810 volts
Air gap
flux per pole = Bav x πDL/p
= 0.56 x π x 1.32 x 0.225/12
= 0.0436 wb
We have Eph
= 4.44f Φ Tph Kw
Hence Tph = 3810/ ( 4.44 x 50 x 0.955
x 0.0436)
= 412
Total
number of stator conductors/phase = 412 x 2 =824 conductors
Total
number of conductors = 412 x 6 = 2472
(iii) Number of stator slots and conductors per slot
Considering
the guide lines for selection of number of slots
Selecting
the number of slots/pole/phase = 3
Total
number of slots = 3 x 12 x 3 =108
Slot
pitch = πD/S
= π x 132/ 108
= 2.84 cm
(quite satisfactory)
Number of conductors per slot = 2472/108 ≈ 24
Hence
total number of conductors = 24 x 108 = 2592
Turns per
phase = 2592/6 = 432
Slot
loading:
Full load current = 500 x 103 / (√3 x 6600) = 43.7 amps
Slot
loading = current per conductor x number of conductors/ slot
= 43.7 x 24
= 1048.8 (
satisfactory)
Ex. 2. A 3 phase 1800 kVA, 3.3 kV, 50
Hz, 250 rpm, salient pole alternator has the following design data.
Stator
bore diameter = 230 cm Gross length of stator bore = 38 cm Number of stator
slots = 216 Number of conductors per slot = 4
Sectional
area of stator conductor = 86 mm2 Using the above data, calculate
(i)
Flux per pole
(ii)
Flux density in the air gap
(iii)
Current density
(iv)
Size of stator slot
Soln:
(i)
Flux per pole
Eph
= 3300/√3 = 1905 volts
Number of
slots per phase 216/3 = 72
Number of
conductors per slot = 4
Total
number of conductors per phase = 72 x 4 = 288
Number of
turns per phase Tph = 288/2 =144
We have
from emf equation Eph =
4.44f Φ Tph Kw
Assuming
Kw =0.955
Flux per
pole Φ = Eph/ (4.44f Tph
Kw)
= 1905/(
4.44 x 50 x 144 x 0.955)
= 0.0624 wb
(ii) Flux
density in the air gap
Air gap
flux per pole = Bav x πDL/p
D = 230
cm,
L = 38
cm,
Ns = 250
rpm
P = 24
Bav
= Φ / πDL/p
= 0.0624 x
24 / (π x 2.3 x 0.38)
= 0.55
Tesla
(iii) Current
density
Sectional
area of the conductor = 86 mm2
Full load current of the machine = 1800 x 103
/ (√3x 3300) = 314.9 amps
Hence
Current density = 314.9/86
= 3.7
amp/mm2
(iv) Size of
the stator slot
Before
fixing up the width of the slot flux density in the middle section of the tooth
has to be assumed as 1.7 Tesla. Based on this flux density width of the slot at
the middle section can be found.
Flux per
pole = 0.0624 wb Gross length of the core = 38 cm Assume
Number of
ventilating duct = 4 Width of the ventilating duct = 1cm Iron space factor
=0.92
Net iron
length of the core li
=( L – nd x wd)ki
= ( 38 – 4
x 1) 0.92
= 31.28 cm
Pole
pitch = πD/p
= π x 230/24
= 30.12 cm
Pole arc/
pole pitch = 0.65 ( Assumed)
Pole arc
= 0.65 x pole pitch
= 0.65 x
30.12
= 19.6 cm
Number of
stator teeth = 216
Slot
pitch =
πD/s
= π x 230/216
= 3.35 cm
Number of
teeth per pole arc = pole arc/ slot pitch
= 19.6/3.35
= 6
Flux density in stator teeth = flux per pole /( bt
x li x number
of teeth per pole arc) bt = 0.0624/( 1.7 x 0.3128 x 6)
= 1.95 cm
Thus the width of the slot should not exceed = 3.35
– 1.95 = 1.4 cm
Slot insulation width wise:
(i) Conductor insulation 2 x 0.5 =
1.0 mm
(ii) Micanite slot liner 2 x 1.5 =
3.0 mm
(iii) Binding tape 2
x 0.4 = 0.8 mm
(iv) tolerence =
1.2 mm
Total =
6.0 mm
Maximum space available for the conductor width
wise = width of the slot – insulation width wise = 1.4 – 0.6
= 0.8 cm
Area of
cross section of the conductor = 86 mm2 Hence thickness of the
conductor = 86/8 = 10.75 mm
Hence the
dimension of the standard conductor selected = 7.8 mm x 11.0 mm Hence the width
of the conductor = 7.8 + 6.0 = 13.8 mm =1.38 cm
Arrangement
of the conductor:
All the
four conductors are arranged depth wise
Depth of
the slot:
(i) Space
occupied by the conductor 4 x 11 = 44.0 mm
(ii)
Conductor insulation 4 x 2 x 0.5 = 4.0 mm
(iii)
Micanite slot liner 2 x 1.5 = 3.0 mm
(iv)Bituminous
insulation between the
insulated
conductors (4-1) x 0.2 = 0.6 mm
(v)Binding
tape on the conductors 2 x 0.4 = 0.8 mm
(vi) Lip = 1.5 mm
(vii)
Wedge = 3.5 mm
(viii)
Tolerance = 1.6 mm
Total 59
mm
Size of
the slot = 1.38 cm x 5.9 cm
Ex.3. A water
wheel generator with power output of 4750 kVA, 13.8 kV, 50 Hz, 1000 rpm, working
at a pf of 0.8 has a stator bore and gross core length of 112 cm and 98 cm
respectively. Determine the loading constants for this machine.
Using the
design constants obtained from the above machine determine the main dimensions
of the water wheel generator with 6250 kVA, 13.8 kV, 50 Hz, 750 rpm operating
at a power factor of 0.85. Also determine (i) Details of stator winding (ii)
Size of the stator slot, (iii) Copper losses in the stator winding.
For 4750
kVA Generator:
D = 112
cm
L = 98 cm
Ns
= 1000 rpm
Ns
= 1000/60 = 16.67 rps
kVA out put Q = C0 D2Lns
C0 = Q / D2Lns
= 4750 /
[(1.12)2 x 0.98 x 16.67]
= 232
Output
coefficient Co = 11 x Bav q Kw x 10-3
Hence Bav
x q = Co / (11 x Kw x 10-3)
= 232 / (11
x 0.955 x 10-3)
= 22200
Assuming
the flux density of 0.6 Tesla
Hence q =
22200/0.6 = 37000 Ac/m
Main
Dimensions of the second machine:
kVA out
put Q = C0 D2Lns
C0
= 232
Q = 6250
kVA
Ns
= 750 rpm
Ns
= 750/60 = 12.5 rps
D2L
= Q / C0 ns
= 6250 /
232 x 12.5
= 2.16 m3
For the
first machine pole pitch τp = πD/p
= π x 112/6
= 58.6 cm
Core
length / pole pitch = gross length/ pole pitch
= 98/58.6
= 1.67
No. of
poles for the second machine p = 120f/Ns= 120 x 50 / 750 = 8
Assuming
the same ratio of gross length to pole pitch for the second machine as that of
first machine
L / πD/p = 1.67
L = 1.67 x πD/8
= 0.655 D
We have D2L
= 2.16 m3
Substituting
the value of L in D2L and solving for D & L
D = 149
cm and L = 97.5 cm
Peripheral
speed for machine 1: πDNs
/60 = π x 1.12 x 1000/60 = 58.5 m/s
Peripheral
speed for machine 2: πDNs
/60 = π x 1.49 x 750/60 = 58.5 m/s
As the
peripheral speed is same for both the machines the diameter and length of the
machine are satisfactory.
Stator
winding details:
Assuming
star connection emf per phase Eph =
13.8/√3 = 7960 volts
We have from emf equation Eph = 4.44f Φ Tph
Kw
Assuming
Kw =0.955, f = 50 Hz
Air gap
flux per pole = Bav x πDL/p
Assuming
the air gap flux density of machine 2 same as that of machine 1 Bav
= 0.6 Tesla
Hence = Bav x πDL/p = 0.6 x π x 1.49 x
0.975/ 8 = 0.342 wb
Hence Tph
= Eph/4.44f Φ Kw
= 7960/
(4.44 x 50 x 0.342 x 0.955)
= 110
Total
number of Conductors =110 x 6 = 660
Full load current per phase Iph = 6250 x
103 / √3 x 13.8
x 103 = 262 amps
Assuming number of slots per pole per phase = 4 1/2
Total number of slots = 4.5 x 8 x 3 = 108
Slot
pitch = πD/s = π x 149/108 = 4.35 cm ( quite
satisfactory)
Number of
conductors per slot = 660/108 ≈ 6
Total
number of conductors revised = 108 x 6 = 648 Number of turns/phase = 108
Total
slot loading = Iph x Cond/slot
= 262 x 6 =
1572 amp cond (quite satisfactory)
Dimension
of the stator slot:
Full load current per phase Iph = 6250 x
103 / √3 x 13.8
x 103 = 262 amps
Assuming
a current density of 4.2 amps/mm2
Area of
cross section of the conductor = 262/4.2 = 62.4 mm2
Based on
the allowable flux density, width of the stator tooth can be calculated and
then the width of the slot can be estimated.
Flux
density in stator tooth Bt = /
(Number of teeth/pole arc x width of the teeth x Iron length)
In a
large salient pole alternator the flux density in the tooth along the depth of
the tooth does not vary appreciably. Thus the flux density at the top of the
tooth may be assumed as 1.7 Tesla and the width of the tooth is calculated at
the top section.
Hence
number of teeth per pole arc = pole arc/ slot pitch
Assuming pole arc/ pole pitch = 0.65
Pole arc
= 0.65 x 58.6 = 38.1 cm
Thus the
number of teeth per pole arc = 38.1/4.35 = 9
Net Iron
length = (L – ndwd) ki
Assuming
10 ventilating ducts of each 1 cm width and an iron space factor of 0.92 Li
= (97.5
-10 x 1)0.92 = 80.5 cm = 0.805 m
Bt = / (Number of teeth/pole arc x x Li) = 0.342/
( 9 x bt x 0.805)
Assuming
the flux density Bt as 1.7 Tesla
Hence
width of the teeth = 2.78 cm
We have
the slot pitch = 4.35 cm
Thus the
slot pitch = 4.35 – 2.78 = 1.55 cm
Slot
insulation width wise:
Slot
insulation width wise:
(i)
Conductor insulation 2 x 0.5 = 1.0 mm
(ii)
Micanite slot liner 2 x 1.5 = 3.0 mm
(iii)
Binding tape 2 x 0.25 = 0.5 mm
(iv)
tolerence = 1.0 mm
Total =
5.5 mm
Insulation
depth wise:
(i)
Conductor insulation 6 x 2 x 0.5 = 6.0 mm
(ii)
Micanite slot liner 2 x 1.5 = 3.0 mm
(iii)Bituminous
insulation between the
insulated
conductors (6-1) x 0.3 = 1.5 mm
(iv) coil
separator between layers =
0.4 mm
(iv)Binding
tape on the conductors 6 x 2 x 0.25 = 3.0 mm
(v) Lip = 1.0 mm
(vi)
Wedge = 3.0 mm
(vii)
Tolerance = 1.6 mm
Total 19.5 mm
Maximum space available for the conductor width
wise = width of the slot – insulation width wise = 1.55 – 0.55
= 1.0 cm
The area
of cross section of the conductor = 62.4 mm2 Approximate depth of
the conductor = 62.4/ 10 = 6.2 mm Selecting the standard conductor of size 9 mm
x 7 mm Thus the area of the conductor = 63 mm2
Six
conductors are arranged as 3 conductors depth wise in two layers.
Hence
width of the slot = 9 mm + 5.5 mm = 14.5 mm = 1.45 cm
Depth of
the slot = 6 x 7 + 19.5 mm = 61.5 mm =6.15 cm
Copper
loss in stator winding
Approximate
length of the mean turn = ( 2L + 2.5 τp + 5 x kV
+ 15)
= ( 2 x
97.5 + 2.5 x 58.6 + 5 x 13.8 + 15)
= 426 cm
= 4.26 m
Resistance
of the stator winding = ζ x lmt
x Tph /a
= 0.021 x
4.26 x 108 / 63
= 0.153 ohm
Total
Copper losses = 3 I2R
= 3 x (262)2
x 0.153
= 31500
watts
Ex.
4. Two
preliminary designs are made for a 3 phase alternator, the two designs
differing only in number and size of the slots
and the dimensions of the stator conductors. The first design uses two slots
per pole per phase with 9 conductors per slot, each slot being 75 mm deep and
19 mm wide, the mean width of the stator tooth is 25 mm. The thickness of slot insulation
is 2 mm, all other insulation may be neglected. The second design is to have 3
slots per pole per phase. Retaining the same flux density in the teeth and
current density in the stator conductors as in the first design, calculate the
dimensions of the stator slot for the second design. Total height of lip and
wedge may be assumed as 5 mm.
Slon.
First
Design:
Slot per
pole per phase q = 2
Total
height of the conductor = 75 – 5 – 2 x 2 = 66 mm
Height of
each conductor = 66/9 = 7.33 mm
Width of
each conductor = 19 -2 x 2 = 15 mm
Area of
each conductor = 7.33 xx 15 = 110 mm2
Slot
pitch at mean diameter = slot width + tooth width = 19 + 25 = 44 mm
Second
Design:
Slots per
pole per phase = 3
Hence,
the number of stator slots in this design are 3/2 times that in the first
design. Retaining the same flux density in the teeth and current density in the
stator conductors The number of conductors per slot in this design is 2/3 times
that in the first design. Number of conductors per slot = 2/3 x 9 = 6
Slot
pitch at mean diameter = 2/3 x 44 = 29.3 mm
Tooth
width at the same flux density = 2/3 x 25 = 16.7 mm Hence slot width = 29.3
-16.7 = 12.6 mm
Width of
each conductor = 12.6 – 2 x 2 = 8.6 mm Height of each conductor = 110/8.6 =
12.8 mm Total height of the conductor = 6 x 12.8 = 76.8 mm Conductor dimensions
12.8 x 8.6 mm2
Depth of
the slot = 76.8 + 5 + 2 x 2 = 85.8 mm Slot dimensions = 85.8 x 12.6 mm2
Ex. 5. A 1000 kVA, 3300 volts, 50 Hz,
300 rpm, 3 phase alternator has 180 slots with 5 conductors per slot.
Single layer winding with full pitched coils is used. The winding is star
connected with one circuit per phase. Determine the specific electric and
magnetic loading if the stator bore is 2 m and core length is 0.4 m. Using the
same specific loadings determine the design details for a 1250 kVA, 3300 volts,
50 Hz, 250 rpm, 3 phase star connected alternator having 2 circuits per phase.
The machines have 600 phase spread.
Slon: Total stator conductors = 180 x 5
=900
Turns per
phase = 900 / 6 = 150 Synchronous speed =300/60 =5 rps
Number of
poles = 120f / Ns = 120 x 50/ 300 = 20 Slots per pole per phase =
180/(20 x 3) = 3 Distribution factor = (sin 60/2) / (3 sin 60/6) = 0.96 For
Full pitched winding, pitch factor kp = 1 Winding factor = kp
x kd = 0.96
Eph
= 3300/√3 =1910 volts
Flux per
pole = 1910 / ( 4.44 x 50 x 150 x 0.96) = 59.8 mwb Pole pitch = πD/p = π x 2 / 20 = 0.314 m
Area of
one pole pitch Ap = pole pitch x core length = 0.314 x 0.4 = 125.6 x
10-3 m2 Specific magnetic loading = / Ap = 59.8
x 10-3 / 125.6 x 10-3 = 0.476 Tesla Current per phase Iph
= 1000 x 103 / ( 3 x 1910) = 175 amps
As there
is one circuit per phase current per conductor = 175 amps
Specific
electric loading = 3 Iph zph/ πD = 6 Iph Tph/
πD = 6 x 175 x 150/ (π x 2) = 25000 Ac/m Peripheral
speed = πDNs/60 = π x 2 x 300/60 = 31.4 m/s
1250 kVA
generator
Synchronous
speed = 250/60 = 4.167 rps Number of poles = 120f / Ns = 120 x 50/
250 = 24 Winding factor = 0.96
Output
coefficient C0 = 11 Bav q Kw x 10-3 = 11 x 0.476 x 0.96 x 25000 x 10-3 = 126
D2L
= Q/ C0 ns = 1250 / (126 x 4.167) = 2.39 m3
Keeping
the peripheral speed same as that of the first machine πDNs/60 = π x D x 250/60 = 31.4 m/s
Hence D =
2.4 m and L 0.414 m
Pole
pitch = πD/p = π x 2.4 / 24 = 0.314 m
Flux per
pole = Bav x πDL/p =
0.476 x 0.314 x 0.414 = 0.062 wb
When
there are more than one circuit per phase ( number of parallel paths = a)
Voltage per phase Eph = 4.44f Φ Tph
Kw / a; a = 2
Hence Tph
= (2 x 1910) / ( 4.44 x 50 x 0.062 x 0.96) = 289 Total number of conductors = 6
Tph = 6 x 289 = 1734 Total number of slots = 3 x 24 x 3 = 216
Number of
conductors per slot = 1734/216 ≈ 8
Revised number of conductors = 8 x 216 = 1728 Revised number of turns per phase
= 1728/ 6 = 288
Ex. 6. Determine the main dimensions of
a 75 MVA, 13.8 kV, 50 Hz, 62.5 rpm, 3 phase star connected alternator. Also find
the number of stator slots, conductors per slot, conductor area and work out
the winding details. The peripheral speed should be less than 40 m/s. Assume
average gap density as 0.65 wb/m2, Specific electric loading as
40,000 AC/m and current density as 4 amp/ mm2.
Slon:
Synchronous
speed = 62.5/60 = 1.0417 rps Number of poles = 120f / Ns = 120 x 50/
62.5 = 96 Winding factor = 0.955
Output
coefficient C0 = 11 Bav q Kw x 10-3
= 11 x 0.65 x 0.955 x 40000 x 10-3 = 273 D2L = Q/ C0
ns = 75000 / (273 x 1.0417) = 264 m3
Taking
the peripheral speed as 40 m/s Peripheral speed = πDNs/60
Hence D =
40 x 60/ π x 300 =
12.2 m and L = 1.77 m Pole pitch = πD/p = π x 12.2 / 96 = 0.4 m
Flux per
pole = Bav x πDL/p =
0.65 x 0. 4 x 1.77 = 0.46 wb Eph = 13800/√3 = 7960 volts
Assuming
one circuit per phase
Turns per
phase Tph = Eph / 4.44f Φ Kw
= 7960 / ( 4.44 x 50 x 0.46 x 0.955) ≈ 82
As the
terminal voltage of the machine is 13.8 kV slot pitch of about 5.5 cm should be
used. Hence the number of slots = πD / τs = π x 12.2 x 100/5.5 = 696
Number of
slots per pole per phase = S/ 3p = 696 / (3 x 96) = 2.42
The above
fractional value of slots per pole per phase indicates that fractional slot
winding is used. Number of slots and turns per phase must be finalized such
that they should not differ significantly from the earlier calculated values.
It is
also to be noted that for fractional slot winding double layer winding is a
must and hence conductors per slot must be an even.
Assuming
number of slots per pole per phase as 2.5 Total number of slots = 2.25 x 3 x 96
= 648
Total
number of conductors = 2 x 3 x Tph = 6 x 82 = 492 Hence number of
conductors per slot = 492/648 = fraction
Hence a
double layer winding is not possible with one circuit per phase. Hence the
number of circuits is to be selected in such a way that number of conductors
per slot is even and the winding becomes symmetrical.
Taking
the number parallel circuits as a = 8
Turns per
phase Tph = a x Eph / 4.44f Φ Kw = 8 x 7960 / (
4.44 x 50 x 0.46 x 0.955) ≈ 654
Hence total number of conductors = 2 x 3 x Tph = 6 x 654 = 3924
Number of
conductors per slot = 3924/ 648 ≈ 6 Hence
the number of conductors = 6 x 648 = 3888 Hence turns per phase Tph
= 3888/6 = 648
Current
per phase = (75000 x 103) / ( 3 x 7960) = 3140 amps
Current
in each conductor = Current per parallel path = 3140/8 = 392.5 amps Area of
cross section of each conductor = 392.5/4 = 98.125 mm2
Area of
cross section of conductor being very large conductors are stranded and used.
Ex.7.
Calculate the stator dimensions for 5000 kVA, 3 phase, 50 Hz, 2 pole
alternator. Take mean gap density of 0.5 wb/m2,
specific electric loading of 25,000 ac/m, peripheral velocity must not exceed
100 m/s. Air gap may be taken as 2.5 cm.
Soln:
Output Q = Co D2Lns kVA Co = 11 Bav q Kw x 10-3 Assuming
Kw = 0.955
Co = 11 x 0.5 x 25000 x 0.955 x 10-3 =
130
Ns =
120f/p = 120 x 50/ 2 = 3000 ns = 3000/60 = 50 rps
D2L = Q/
Cons
= 5000/(130
x 50)
= 0.766 m3
Peripheral velocity = πDrNs/60 = 100 m/s
Dr = 100/(50 x π) = 63.5
cm
D = Dr +
2lg
= 63.5 + 2
x 2.5
= 68.5 cm
L = 163
cm
Numerical
Problems: Turbo alternators
Ex.1.
Calculate the stator dimensions for 5000 kVA,
3 phase,
50 Hz, 2 pole alternator. Take mean gap density of 0.5 wb/m2, specific electric
loading of 25,000 ac/m, peripheral velocity must not exceed 100 m/s. Air gap
may be taken as 2.5 cm.
Soln:
Output Q = Co D2Lns kVA
Co = 11
Bav q Kw x 10-3 Assuming Kw = 0.955
Co = 11 x 0.5 x 25000 x 0.955 x 10-3 = 130
Ns =
120f/p = 120 x 50/ 2 = 3000 rpm ns = 3000/60 = 50 rps
D2L = Q/
Cons
= 5000/(130
x 50)
= 0.766 m3
Peripheral velocity = πDrNs/60 = 100 m/s
Dr = 100/(50 x π) = 63.5
cm
D = Dr +
2lg
= 63.5 + 2
x 2.5
= 68.5 cm
L = 163
cm
Ex.2. A
3000 rpm, 3 phase, 50 Hz, turbo alternator Has a core length of 0.94 m. The
average gap density
is 0.45
Tesla and the ampere conductors per m are 25000. The peripheral speed of the
rotor is 100 m/s and the length of the air gap is 20mm. Find the kVA output of
the machine when the coils are (i) full pitched (ii) short chorded by 1/3rd
pole pitch. The winding is infinitely distributed with a phase spread of 600.
Soln:
Synchronous speed Ns = 3000 rpm ns= 3000/60 = 50
rps
Peripheral speed np = πDrNs/60 = 100 m/s
Hence diameter of the rotor Dr = 100 x 60 / (π x 3000) = 0.637 m
Hence
inner diameter of stator D = Dr + 2lg
= 0.637 + 2
0.02
= 0.677 m
(i) With
infinite distribution and 600 phase spread the distribution factor may be given
by where α is the
phase spread
Kd = sin σ/2 / σ/2 = sin π/6 / π/6 = 0.955
With full
pitched coils Kp = 1
Winding
factor = Kp x Kd = 0.955
Output of
the machine Q = C0 D2Lns
= 11 Bav q
Kw x D2Lns x 10-3
= 11 x 0.45
x 25000 x 0.955 x 0.6672 x 0.94 x 50 x 10-3
= 2480 kVA
(ii) With
chording of 1/3rd of pole pitch: chording angle α = 180/3
=600
Pitch
factor = cos α /2 =
0.866
Winding
factor = Kp x Kd = 0.955 x 0.866 = 0.827
Output of
the machine Q = C0 D2Lns
= 11 Bav q
Kw x D2Lns x 10-3
= 11 x 0.45
x 25000 x 0.827 x 0.6672 x 0.94 x 50 x 10-3
= 2147 kVA
Ex. 3.
Estimate the stator dimensions, size and number of conductors and number of
slots of 15 MVA 11kV,
3 phase,
50 Hz, 2 pole turbo alternator with 600
phase spread.
Assume
Specific electric loading = 36000 AC/m, specific magnetic loading = 0.55 Tesla,
Current density = 5 Amp/mm2 , peripheral speed = 160 m/s. The winding must be
designed to eliminate 5th harmonic.
Soln: Synchronous speed Ns = 120f/p = 120 x 50/ 2 =
3000 rpm ns= 3000/60 = 50 rps
Peripheral speed np = πDrNs/60 = 160 m/s
Hence
diameter of the rotor Dr ≈D
= 160 x 60
/ (π x 3000)
= 1 m
With a
phase spread of 600 distribution factor
Kd = sin σ/2 / σ/2 = sin π/6 / π/6 = 0.955
In order
to eliminate 5th harmonic chording angle α = 180/5=
360
Pitch
factor Kp = cos α /2 =
0.951
Winding
factor = Kp x Kd = 0.955 x 0.951 = 0.908
Output
coefficient C0 = 11 Bav q Kw x 10-3
= 11 x 0.55
x 36000x 0.908 x10-3
= 198
D2L = Q / C0 ns
= 15 000/
(198 x 50) = 1.51 m3 We have D = 1 m and D2L =1.51 m3 Solving for L, L= 1.51 m
Flux per
pole = Bav x πDL/p
= 0.55 x π x 1 x 1.51 / 2
= 1.3 wb
|
Eph |
= 1100/√3 =6360 volts |
|
Hence |
Tph =
Eph/4.44f Φ Kw |
= 6360 / (
4.44 x 50 x 1.3 x 10-3 x 0.908)
= 24
Total number of conductors = 6 x 24 =144
For the
turbo alternator selecting slots/pole/phase = 5
Total
number of stator slots = 5 x 2 x 3 = 30
Conductors/slot
= 144 /30 = 5
can not
use double layer winding, using two circuits per phase conductors/slot = 10
Total
conductors 10 x 30 = 300.
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