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Chapter: Design of Electrical Machines - Synchronous Machines

Solved Numerical Problems - Design of Synchronous Machines

Design of Electrical Machines - Synchronous Machines - Solved Numerical Problems - Design of Synchronous Machines

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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