Design of Electrical Machines Notes

7/28/2019 Design of Electrical Machines Notes

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Anna University

Department of Electrical and Electronics Engineering

Notes on Lessons

Class : VI semester B.E. Sub. EE2355 Design of Electrical Machines

Faculty Name: A.Sangari, Senior Lecturer

UNIT -1

INTRODUCTION

Flux () = (MMF/ Reluctance) , Wb ; Reluctance(S) = (l/a) , A/Wb ;

Permeance = ( 1/S) , Wb/A ; where l = length of the flux path, m ;

A = area of cross section for the flux path, m2

; = permeability = o r ;o = absolutepermeability = 4 x 10

-7 H/m and r = relative permeability.

H = Ampere turns / m = MMF/ l = S / l = S Ba /l = (l/a) Ba/l = B/ .or B = H

Series magnetic circuit : S = S1 + S2+Parallel magnetic circuit: Per= Per1+ Per2+...

Leakage Coeff. = total flux/useful flux; total flux = useful flux + leakage flux

Expressions for reluctance: Sg = lg/ o l ys

1. Various configurations of slotting

i) Smooth iron surface on both sides of the air gap : ysl = ys

ii) slotted armature : ysl

= ys - Ws = Wt (no fringing)

iii) slotted armature : ysl = ys Kcs Ws ; Ws = slot width , Wt = tooth width

Kcs = Carters Coefficient for slots depends on the ratio of slot opening /airgap length orthe empirical relation is 1/ {1+ (5lg/Ws )}

If radial ventilating ducts are provided: Ll= L Kcd nd Wd

where Kcd = Carters Coefficient for ducts,

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nd = No.of ducts, Wd = width of each duct

If ducts are provided on the stator and on the rotor, then Kcd should be basedon half the air gap.

Considering the effect of both slotting and ducts, Kg ( gap contraction factor) = Kgs Kgd

where Kgs ={ ys/ ys l } and Kgd ={L /Ll }

If slots are provided on both sides of the airgap, Kgs = Kgss Kgsr (ss and sr denoting stator

and rotor slots respectively).

MMF for airgap = H Kg lg ={ B/ o }Kg lg = 8,00,000 B Kg lg .

Effect of Saliency : K f=Field form factor = Bave / Bg= = pole arc/pole pitch ;

pole pitch = D/P ; Bave = /( DL/P)_______________________________

2. MMF calculation for teeth

The calculation of MMF for producing flux in the Teeth of the machine is difficultbecause :

i) the teeth are tapered when parallel sided slots are used and this results in variation in

the flux density over the depth of the tooth.

ii) the slots provide another parallel path for the flux flow, the teeth are normally workedin saturation and hence r becomes low.

Following methods are usually employed for the calculation of MMF required for the

tapered teeth:-

i) Graphical method :ATt = Mean ord. x ltMean ord. is the mean ord. of at variation with tooth depth.

ii) Simpsons rule :at mean = (at1 +4 at2 + at3)/6 A/m

iii) Bt1/3 method : ATt = at1/3 x lt ,

where at1/3 = MMF for corresponding to B at 1/3 rd height from the narrow end.

3. Real and Apparent flux densities

Breal = Bapp - 4 x 10-7

at (Ks 1) ;where Ks = Atotal/ Airon = Lys/ Li Wt

Specific Permeance : = Permeance per unit length or depth of the field.

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Parallel sided slot :

=o[(h1/3ws) + (h2/ws )+ {2h3/ws+wo)} + (h4/wo) ]

4. Leakage reactance of transformers

Concentric winding

Total leakage reactance of the transformer referred to the primary =

Xp = 2f oTp2

(Lmt/Lc) {a+ (bp + bs)/3}

The per unit reactance can also be calculated as x = Ip Xp /Vp ; where Ip , and Vp arerated phase current and voltage respectively.

Sandwich winding:

Xp = f o(Tp2/n)(Lmt/w) {a+ (bp + bs)/6}

5. Temperature rise calculations

Q = Power loss(heat produced ), J/s or WG = weight of the active material of the Machine, kg

h = specific heat, J/kg-

CS = cooling surface area, m2

= specific heat dissipation, W/ m2 -Cc = 1/ = cooling coefficient, m2 -C / Wm = final steady temperature rise,

C

The temperature of the machine rises when it is supplying load. As the temperature

rises, the heat is dissipated partly by conduction, partly by radiation and in most cases

largely by air cooling. The temperature rise curve is exponential in nature. Assuming thetheory of heating of homogeneous bodies ,

Heat developed = heat stored + heat dissipated

Q dt = Gh d + S dt ; solving = m ( 1- et/Th ) + i e t/Th

If the machine starts from cold, i = 0 = m ( 1- et/Th

)

Th=heating time constant (time taken by the machine to attain 0.632 times m) = Gh/S

Tc = cooling time constant (time taken by the machine to fall to 0.368 times i)

Cooling curve is an exponentially falling curve.

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6. Rating of machines

IS: 4722-1968: specification for Rotating Electrical machinery:1. continuous duty2. short time duty (T Th )3. intermittent periodic duty4. intermittent periodic duty with starting5. intermittent periodic duty with starting and braking6. continuous duty with intermittent periodic loading7. continuous duty with starting and braking8. continuous duty with periodic speed changes

7. Determination of motor rating

From the point of calculation of motor rating, the various duty cycles listed earlier, can

be broadly classified asi) Continuous duty ii) Fluctuating duty and iii) Short time and intermittent duty.

Continuously duty motors workThese motors work with the same load through out the duty cycle.

Fluctuating loadsThe motor is switched on for a period T1 and kept off for a period T2 .

To calculate the power rating of motors to be used with fluctuating loads,

the commonly used methods are :

i) Method of average losses ii) Equivalent current methodiii) Equivalent torque method.

Short time duty

The motor carries a load much higher than the rated continuous load for a short time.The time for which the motor may be allowed to carry the short time higher load is th =

Th log {ph/(ph 1) }

where ph = ml

/m ; ml= final steady temperature that would be attained if the machine

is allowed to run indefinitely at its short time rating.

8. Cooling of rotating electrical machines

In most cases, the cooling electrical machines is carried out by air flow and this cooling

is called ventilation. In high speed machines such as turbo alternators, hydrogen is usedfor cooling.

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Ad f h d li


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Advantages of hydrogen cooling:

Compared with air, hydrogen has the following properties:-

i) (1/14) th density thereby the windage losses and noise reduced ii) 14 times specific

heat and 1.5 times heat transfer leading to improved cooling iii) 7 times thermal

conductivity resulting in reduced temperature gradient iv) reduced corona effect v)will not support combustion so long as the hyd /air mixture exceeds 3/1.

In operation, the fans mounted on the rotor circulate hydrogen through the ventilating

ducts and internally mounted gas coolers. The required gas pressure is maintained by aregulator. The precaution to be observed is the stator frame must be gas tight and

explosion proof and oil film gas seals at the rotor shaft ends are essential.

Induced and Forced ventilation: In induced ventilation, the fan produces decreased air

pressure inside of the machine, causing air to be sucked into

the machine under the external atmospheric pressure ; and in the forced ventilation, theair is forced into the fan by the fans mounted internally or externally.

The ventilation can also be classified as i) Radial, ii) axial and iii) combined radial and

axial.

9. Quantity of the cooling medium employed

Volume of air required = Va = (Q/cp ) 103

x (760/H) x {( i + 273)/273} V, m3

/s

cp = 995 and V = 0.775 m3

assumed; then

Va = 0.78 (Q/) x (760/H) x {( i + 273)/273} , m3

The capacity of the fan required = Pfan = P Va/fan, W

Similar calculations can be made for the volume of hydrogen or air or oil used for

cooling the machine.

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UNIT II

DC MACHINES

D = stator bore or armature diameter, m

L = stator core length, mp = number of poles

Z = Total numbers of armature conductors

Iz = current in each conductor(Ia/A) , A

E = induced EMF, VP = machine rating (power output),kW

Pa = power developed by the armature, kWQ = kVA rating of the machine = flux per pole, Wb = pole pitch ( D/p), m

Total Electric loading = p ; Total Magnetic loading = Iz Z

Specific Electric loading = ac = Iz Z

(amp.conductor/m) D

Specific magnetic loading = Bave = p(Wb/m

2) DL

1. Output equation

Pa = E Ia x 10-3 = (p/A)(ZN/60) Iz A 10

-3

= {2 Bave ac x10-3

}D

2L n = Co D

2L n where Co = output coefficient

___________________________________

Generator: Pa = (P/) (FW & Iron losses) ; Motor : Pa = P + (FW & Iron losses)

For large machines : FW & Iron losses are neglected i.e.,

Pa = P/ (Generator)= P (Motor)

For small machines : FW & Iron losses can be taken as 1/3rd

of the total losses. So,

Pa = (P/) (1/3) P (1-)/ = P(2+ )/(3 ) generator

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= P + (1/3) P (1 )/ = P(1+2)/(3 ) motor


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= P + (1/3) P (1-)/ = P(1+2)/(3 ) motor

2. Choice of specific magnetic loading (Bave)

i) B max in the iron part of the magnetic circuit:

B max maximum allowable densityi.e., Bt = Bave Ys/wt (non-salient pole machines); Ys = slot-pitch and wt = tooth width.

&Bt = (Bave/)Ys/wt (salient pole machines)where = pole arc /pole-pitch ratio.If Ys =2 wt and = 0.667 , thenBt = 3 Bave : For example,if Bt is to be limited to 2.2 Wb/m

2,

Bave should excced 2.2/3 = 0.73 Wb/m2.

ii) Magnetizing current:

large Bave high magnetizing current large core loss

3. Choice of specific electric loading (ac)

i)Temperature rise()

depends on Q (losses),which in turn depends on ac.Allowable depends on insulating material used.

ii) Cooling coefficient (C)

is also proportional to the cooling coefficient;a machine with a better ventilation has a lower C and then higher ac can be used.

iii)Operating voltage (V)In high voltage machines, the slot space factor, Sfis less and so only smaller ac can

be used. It also depends on the shape of the conductors, circular or rectangular in

cross section.

iv) Current Density ()choice of depends on cooling; higher C higher value in the choice of ac.

4. Constraints in the design of DC machines

i) Peripheral speed,v 45 m/sii) Frequency of flux reversal, f 50 Hziii) Current per brush arm 400 Aiv) Armature MMF per pole 7000 A

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The MMF required for the airgap = 50% of the armature MMF and gap


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The MMF required for the airgap = 50% of the armature MMF and gap

contraction factor = 1.1.

The current per brush arm (Ib) = 2Ia/p , A

For square poles: L = D/p

In the design process, choose p based on f& Ib and then calculate D and L

5. Armature Design

Considerations in choice of number of armature slots :

i) mechanical difficulties ii) cooling of armature

iii) pulsation of flux iv) cost v)commutation (Slots/pole 9 )vi) slot pitch (ys =25 to 35 mm) vii) slot loading (Iz Zs 1500 Amp-cond)viii) suitability of the winding doublelayer Lap or wave )

Slot dimensions:

i) the slot area should accommodate the armature conductors and the required

insulation depending on the operating voltage ii) Bt1/3 2.1 Wb/m2

iii) deep slots

cause eddy current losses iv) slot opening should be narrow to reduce the fluxpulsation and hence to reduce eddy current losses.

Armature voltage drop = Ia ra ;

ra = (Z/2) Lmt /(a2 az ) ;

where Lmt = length of mean turn, m = 2L + 2.3 +5ds ; a = no.of parallel paths and az =area of each conductor, m

2.

6. Design of the Field system

Area of each pole (Ap) = Flux in the pole body / Flux density = cl / Bpwhere cl = leakage coefficient

Width of the pole = Ap /Li ; where Li = net iron length = 0.9 L

Height of the pole(hf) chosen based on the MMF to be provided by the pole at full-load.

(ATfat full-load) /(ATArm. at full-load) =1.0 to 1.25

(to overcome armature reaction)

7. Tentative design of Field winding

C l i th fi ld i di I2

R

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= (faf)2 {Lmt Tf}/af = f

2 Lmt Tf afid thi l


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(faf) {Lmt Tf}/af f Lmt Tfaf

= f2 Lmt Sfhfdf; .(1)

where df= depth of the pole winding. Tf= no.of turns in each field coil.

Permissible loss = S qf= 2Lmt hfqf..(2)_______

Equating (1) and (2) ; f=104 qf/ Sfdf

MMF per metre of the field winding = ATf/hf = IfTf/hf = f(afTf)/ hf

= fSfhfdf/hf= fSfdf= Sqrt {2 qf Sfdf/}

8. Design of commutator and brushes

i)The number of commutator segments is equal to the number of coils.ii)The commutator diameter Dc = 60 70 % of the armature diameter (D)

iii)Peripheral speed of commutator = DcN/60 20 m/s

iv) Pitch of the commutator segment = c = Dc/C 4.0 mm

(3.2 mm for the conducting portion and 0.8 mm for mica separator)

v) Distance between brush spindles = Dc/p (25 30 cm)

vi) Length of the commutator = Lc = nb (wb + cb) + c1+ c2

where nb = number of brushes per brush arm, wb = width of the brush, cb =

clearance between brushes , c1 = clearance for staggering (1030 mm), c2 = clearance

for end ply (10-25 mm).

vii) Current carried by each brush spindle

= Ib = 2Ia/p= Abb ; where Ab = nbwbtb

= total brush contact area ; b = current density in the brushes (0.1 A/mm2).

nb is selected such that each brush does not carry more than about 70 A .viii) The thickness of brush = (2-3) c

ix) Com. surface area = Sc= DcLc m2

x) Commutator surface losses:1.brush contact loss = Wbc = VbIa ; Vb = brush contact drop 1.0 V/brush

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2.brush friction loss =Wbf = bpAbVcvidyarthiplus com


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bf bp b c = coefficient of friction (0.1 - 0.3) ; b = brush pressure(10 15 kN/ m

2)

3. Cooling coefficient = c =k/(1+0.1Vc) ; k = 0.015 0.025

4. Temperature rise c = Qc c /Sc

9. Losses and efficiency in DC machines1.I2R losses : copper loss in i) armature ii) Field iii) Inter pole winding2.Rotational losses :i) Friction &windage

ii)Iron loss a) Hysterises =Kh Bm1.6 f b) Eddy current loss =Ke Bm

2 f2t2

For the calculation of copper losses , the total length and area cross section of each

of the windings should be first calculated.

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UNIT III

TRANSFORMERS

1. Single-Phase Transformers

Voltage per turn = E/T = 4.44fm = Et

The window in a single-phase transformer contains one primary and one secondary

winding. The total copper area in the window:

Ac = Tpap + Tsas = ( TpIp + TsIs )/ since ap = Ip/ and as = Is/

= 2AT/ ; since TpIp = TsIs neglecting magnetizing current

Aw = total window area ; Kw = window space factor = Ac/ Aw

Ac = Kw Aw = 2AT/ AT = Kw Aw /2

Rating in kVA = Q = Vp Ip x 10-3 = Ep Ip x 10

-3 = Et ( Tp Ip )x 10-3

= 4.44 fm (Kw Aw /2 ) 10-3

; where m = Bm Ai

= 2.22 fm (Kw Aw ) 10-3

2. Three-phase transformers

Each window contains two primary and two secondary windings

Ac =2 (Tpap + Tsas ) = 4 AT / AT = Kw Aw /4

Rating in kVA = Q = 3Vp Ip x 10-3

= 3Ep Ip x 10-3

=3 Et ( Tp Ip )x 10-3

= 3 x 4.44 fm (Kw Aw /4 ) 10-3 where m = Bm Ai

= 3.33 fm (Kw Aw ) 10-3

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Using the output equation it can also be shown thatvidyarthiplus com


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_____ __________

E t = K kVA where K = 4.44 f r 103

; r = m / AT

r is a constant for transformer of a given type ,service and method of connection, since

m determines the core section and AT fixes the total copper area.

3. Ratio of Iron loss to copper loss

Copper loss/m3 = (l/a) (current)2 /(la) = (l/a) (a)2/la

= 2 W/m3 = 2/density W/Kg

Taking =0.021 x 10-6 and density =8.9 x 103 Kg/m3

Copper loss/Kg = specific copper loss = pc= 2.36 x 1012 2 W/Kg . Then total copper loss =

Wc = pcGc where Gc weight of copper , Kg

In addition to the above , we must add stray losses which may be 5 to 25 % of copperloss.

The total iron loss / Kg = specific iron loss (p i) can be found from the iron loss curves.

Then the total iron loss = Wi = pi GiWhere Gi = weight of iron . Ratio of Iron loss / copper loss = pi Gi / pcGc

4. Design of the core

The core section of the core type transformer may be rectangular , square or stepped.

Shell type transformers use cores with rectangular cross section. For the rectangularcore the ratio of depth to width 1.4 to 2.0.

Square or stepped coreWhen circular coils are required for high voltage transformers ,

square and stepped cores are used . Circular coils are preferred because of their superiormechanical characteristics.

As the size of the transformer increases, it becomes wasteful to use rectangular cores.Square cores are used and the surrounding circle, representing the inner surface of the

tubular form carrying the windings, is called the circumscribing circle. Even now a lot of

useful space is wasted and the length of the mean turn increases causing higher I2R

losses.

With larger transformers, cruciform cores, with better utilization of the space, are used. It

should be born in mind that two different types of laminations are used in cruciform

Wi h ill l f f h i i d d ili h

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even more effectively. However, larger step sizes larger number of lamination sizes vidyarthiplus com


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higher labor cost.

Assuming a stacking factor for iron = 0.9 , Net core area / area of circumscribing circle

= 0.637 for square core and = 0.710 for stepped core with a= 0.851d , b= 0.526 d ;

where d = diameter of the circumscribing circle.

5. Choice Flux density and Current density( Bm and )

Bm determines the core area.

Higher Bm smaller area smaller Lmtsaving in the cost of iron and copper. Buthigher Bm increases the iron loss and temp rise. For Distribution transformer Bm = 1.1 to

1.35 Wb/m2

. For Power transformer Bm = 1.25 to 1.45 Wb/m2

.

The area of conductors for the primary and secondary windings determined afterchoosing a suitable value for which depends on the method of cooling.

6. Types of Windings

i) Cylindrical winding with circular conductorsii) Crossover winding with circular or rectangular conductors

iii) Continuous disc type winding with rectangular conductorsiv) Helical winding

7. Design of insulation

i) Electrical insulation: depends on the operating voltageii) Eddy current loss in the conductors and tank walls

iii) Mechanical considerations: high mechanical forces during fault

iv) Thermal considerations: depends on cooling

Major insulation : between windings and core (grounded)

Minor insulation ; between turns and layers

Materials : cotton thread, cotton tape, leatheriod paper, millinax paper etc.

8. Window dimensions

a = width of the largest samping ; d = dia of the circumscribing circleD = distance between centres of adjacent limbs

Ww , Hw = width and height of the window ( length of the window)

Hy = height of the yoke

For core type: D = d + Ww ; W = D+a ; H = Hw + 2Hy

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For core type, the yoke section may be either rectangular or stepped. In rectangular yoke

ti d th f th k d th f th h t d dvidyarthiplus com


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sections, depth of the yoke = depth of the core; when stepped cores are used

the core depth = width of the largest stamping; area rectangular yokes = Ay= Dy Hy = a Hy

For three phase transformers : W = 2d + a

Window space factor, Kw = Ac/Aw = 10 /(30+kV)Area of the window = Aw = Hw Ww ;

Aw = 2apTp/ Kw (single-phase) = 2apTp/ Kw ( three-phase)

For shell type : Dy = b ; Hy = a ; W = 2Ww+4a ; H = Hw + 2a

9. No-load current calculations

The phasor sum of the magnetizing current (Im) and the loss component of current (Il) ;

Im is calculated using the MMF/m required for the core and yoke and their respectivelength of flux path. Il is determined using the iron loss curve of the material used for the

core and yoke and the flux density employed and their weight.

10. Temperature rise of transformers

Losses dissipated in transformers in the core and windings get converted into thermal

energy and cause heating of the corresponding transformer parts.The heat dissipation

occurs as follows: i) from the internal heated parts to the outer surface in contact with oilby conduction ii) from oil to the tank walls by convection and iii) from the walls of the

tank to the atmosphere by radiation and convection.

11. Transformer oil as a cooling medium

The specific heat dissipation due to convection of oil

= conv = 40.3 ( /H)

W/m2-oC ; = temp difference of the surface relative to the oil and

H = height of the dissipating surface.

Experimentally found that a plain tank surface dissipates 6.0 W/m2-oC by radiation and

6.5 W/m2-oc by convection (.for a temp rise of 40

oC above an ambient temp of 20

oc).

Thus a total of 12.5 W/m

2

-

o

C is taken.The temp rise = total loss/ ( St) = (Pi + Pc) / ( St)Where St = heat dissipating surface area of the tank.

For small transformers , plain walled tank is enough to dissipate the losses. As the rating

of the transformer increases, the volume increases as he cube of the linier dimensions but

the heat dissipating surface area increases only as square of the linier dimensions. So

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above certain rating, plain tank becomes inadequate to dissipate losses and the area is

increased by providing tubes For larger ratings forced air cooling is usedvidyarthiplus.com


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increased by providing tubes. For larger ratings forced air cooling is used.

If tubing is provided, the oil circulation is improved due to the head of the oil, and this

causes an additional dissipation by convection of about 35 % .

Let x St be the area of the cooling tubes. Then

Loss dissipated by the tank surface 12.5 St W/0C

Loss dissipated by the tubes (1.35 x 6.5) x St W/0 C = 8.8 x St W/

0 C

Total loss dissipated by the tank and oil tubes

= (12.5 St + 8.8 x St ) W/0C

Hence = ( Pi + Pc)/ (12.5 St + 8.8 x St ) Total tube area x St

= (1/8.8) [ {(Pi + Pc)/} 12.5 St ]The number of tubes = nt = Total tube area /( dt lt )

The arrangement of the tubes on tank side walls should be made uniformly with a

spacing of usually 75 mm. Examples of calculation of n t and the arrangement of the tubesshould be studied.

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UNIT IV

INDUCTION MOTORS

1. Output Equation

Q(kVA) = m Eph Iph 10-3

= m (4.44fm Tph Kw ) {(ac D)/( m 2Tph )}

= 4.44 (pNs/120) (Bave DL/p) Kw (ac D/ 2) x 10-3

= 1.11 2 Bave ac Kw x 10-3

D2L ns

= 11 Bave ac Kw x 10-3

D2L ns = Co D

2L ns

Co = output coefficient , Q is calculated as ( hp x 0.746 )/( cos)

2. Choice of Bave

i) Low Bavelarge size machine for a given hpii) high Bave large magnetizing current low power factoriii) high Bave high iron lossiv) high Bave high m less Tph low leakage reactance

larger diameter for the circle diagram larger over load capacityFor 50 Hz motors Bave : 0.3 to 0.6 Wb/m

2

3. Choice ac (ampere conductor /m)

Low ac large size machine for a given hp

High ac higher copper loss and temp rise

High ac large Tph large leakage reactance

y p

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For 50 Hz motors ac : 10,000 to 45,000 amp.cond/mThe value ac chosen depends on the ventilation and cooling

It should be remembered that the Power factor (PF) and efficiency() of the motor at fullload increases with the rating of the machine. Again and Pf are higher for high speed

motors compared to low speed motors.

4. Separation of D and L and air gap length

i) L/ can be assumedii) for best power factor = Sqrt( 0.18 L) or D= 0.135p Liii) D can be chosen based on peripheral speed = DN/60

The stator winding can be connected in star or delta. The motors meant for starting withstar- delta starter should be designed with delta connected stator winding.

Stator slot pitch = yss = D/Ss where Ss =number of stator slotsStator outer diameter Do = D + 2 dss + 2dcsWhere dss = stator slot depth and dcs = stator core depth

Air gap length = 0.2 + 2 sqrt(DL) where Dand L are in metre.

After designing the main dimensions, the following calculations can be made:

i) The flux per pole, m = Bave DL/pii) Turns in series per phase, Tph = V/ (4.44fm Kw )iii) Number of slots per pole per phase can be suitably assumediv) Slot pitch should not exceed about 24 mmv) Number of conductors per slot should be rounded offvi) Tph recalculatedvii) Iph = [Q x 103/(3 400)] for star

and [Q x 103/(3 400)] /3 for delta connection

viii) Assuming suitable current density (s) , conductor area required iscalculated (as = Iph/s)

y p

ix) Total copper area in the slot = (conductors /slot) asvidyarthiplus.com


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x) Total area of the slot = copper area /slot space factorSlot space factor is to account for the space for insulation

xi) A suitable tooth flux density is assumed and tooth width is calculated

xii) slot width = slot pitch- tooth width (at different diameters)

xiii) The core flux(m/2) , core area and core depth are calculatedassuming a core flux density. Core depth = core area/ Li

xiv) The rotor bar and end rings are designed (sq.cage rotor)

xv) The rotor winding is designed similar to stator winding (wound rotor)

5. Losses in the Induction motor

i) stator copper loss ii) rotor copper loss iii) iron loss in the stator teeth and

core iv) friction and windage loss (1- 1.5 % of output)

The rotor resistance in stator terms can be obtained as

rotor copper loss/ I2 ; where I2 = 0.85 I1

6. No-load current

Iron loss component = Il =(Iron loss/phase) /Vph

Magnetizing component = Im =(0.427 p ATtotal )/( Kws Tph )

ATtotal = sum of the ampere turns for airgap ,stator tooth and core and rotor tooth andcore.

No-load current = In = { Il2

+ Im2

}1/2

The leakage reactance calculations are made to find x1 , x2 and x01 .

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UNIT V

SYNCHRONOUS MACHINES

1. Construction

Stationaryarmature, rotating field type of construction is preferred.High speed alternators have non-salient pole rotor (Turbo alternators) and they have

either 2-pole or 4-pole. Slow speed alternators have salient pole rotor (water

wheel alternators) and they have more than 4 poles.

2. Output Equation

Q = m Eph Iph 10-3

= 11 Bave ac Kw x 10-3

D2L ns = Co D

2L ns

Where Co = output coefficient

(see Unit -4 for derivation)

D and L are separated using L/ ratio or maximum specified peripheralspeed of the rotor.

Damper winding is used for starting and damping rotor oscillations that occur duringsudden load changes.

3. Choice of specific magnetic loading (Bave)

i) High Bave high flux density in the teeth and core high iron loss higher temperature rise.

ii) high Bave low Tph low leakage reactance (Xl ) high short circuit current

iii) In high voltage machines slot width required is more to accommodatethicker insulation smaller tooth width small allowable Bave

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iv) stability : Pmax =VE/Xs . Since high Bave gives low Tph and hence low XlPmax increases and improves stability.

v) Parallel operation : Ps = (VE sin)/Xs ; where is the torque angle. Solow X

sgives higher value for the synchronizing power leading stable parallel

operation of synchronous generators.

Guide lines : Non-salient pole alternator : 0.54 0.65 Wb/m2

Salient pole alternator : 0.52 0.65 Wb/m2

4. Choice of specific Electric loading (ac)

i) Copper loss and temperature rise: High value of ac higher copper lossleading high temperature rise. So choice of depends on the cooling method used.

ii) Operating voltage : High voltage machines require large insulation and so the slot

space available for conductors is reduced. So a lower value for ac has to be chosen.

iii) Synchronous reactance (Xs) : High value of ac results in high value of

Xs , and this leads to a) poor voltage regulation b) low steady state stability limit.

iv) Stray load losses increase with increase in ac.

Guide lines : Non-salient pole alternators : 50, 000 75,000 A/mSalient pole alternators : 20,000 40,000 A/m

5. Short Circuit Ratio (SCR)

SCR = Field current required to produce rated voltage on opencircuit

Field current required to produce rated current on short circuit

= 1/ direct axis synchronous reactance = 1/Xd

Thus SCR is the reciprocal of Xd

, if Xd

is defined in p.u.value for rated voltage andrated current. But Xd for a given load is affected by saturation conditions that then

exists, while SCR is specific and univalued for a given machine.

Non-salient pole alternators : 1- 1.5 ; Salient pole alternators : 0.5 0.7

Effect of SCR on machine performance

i) Voltage regulation : A low SCR high Xd large voltage drop

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poor voltage regulation..

ii) Parallel operation : A low SCR high Xd low synchronizingpower parallel operation becomes difficult.

iii) Short circuit current : A low SCR high Xdlow short circuit

current. But short circuit current can be limited by other means not

necessarily by keeping a low value of SCR.

iv) self excitation : Alternators feeding long transmission lines should

not be designed with small SCR as this would lead to large terminalvoltage on open circuit due to large capacitance currents.

Summarizing ,high value of SCR leads to i) high stability limit ii) low voltageregulation iii) high short circuit current iv)large air gap

The present trend is to design machines with low value of SCR, this is due to therecent development in fast acting control and excitation systems.

6. Length of airgap

The length of air gap very much influences the performance of a synchronous

machine. A large airgap offers a large reluctance to the path of the flux produced by

the armature MMF and thus reduces the effct of armature reaction. Thus a machinewith large airgap has a small Xd and so has

i)small regulation ii) high stability limit iii) high synchronizing power which makes

the machine less sensitive to load variations iv) better cooling at the gap surface v)

low magnetic noise and smaller unbalanced

magnetic pull.

But as the airgap length increases, a large value of Field MMF is required resulting inincreased cost of the machine.

7. Number of stator slots

Factors to be considered in the selection of number of slots :

1. Balanced 3-phase winding to be obtained2. With large number of slots i)large number of coils increased

labor cost ii) cooling is improved iii) tooth ripples are less iv) Flux

density in the iron increases due to decreased tooth width.

Guide lines : Slot pitch (ys ) 25 mm for low voltage machines;

40 mm for machines upto 6 kV ; 60 mm for machines upto 15 kV.

8 h d f li i i i

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8. Methods of Eliminating Harmonics

By using i)distributed windings ii) fractional coil pitch

iii) fractional slot windings iv) skewing v) large airgap

Further calculations needed after determining D and L :

i) Flux per pole = = Bave ( DL/p )ii) Tph is calculated from the EMF equation taking Eph = Vphiii) Iph = (Qx 10

3) / 3 VL

iv) Armature MMF/pole = Ata = 2.7 Iph Tph Kw/p

v) Effective area per pole = 0.6 0.65 times actual area

9. Field Design (Salient poles)Data needed for the design of the Field winding :

i) Flux density in the pole core ii)Winding depth (df)iii) Leakage factor (pole flux/gapflux) iv) Field winding space factor (Sf) v) Power dissipation (qf) in W/m

2 v) The

ratio of field MMF to armature MMF vi) Allow about 30 mm for insulation , flanges

and height of the pole shoe.MMF per unit height of the winding = 104 Sqrt (Sfdfqf)

Computer Aided Design of Electrical Machines

The process of design any electrical may be broadly divided into three major aspects: i)

Electrical design ii) Mechanical design iii) Thermal design. Even though, theseproblems can be solved separately, there are many inter- related features.

The advantages of computer aided design are : i) The computer can handle large

volume of data to make a number of trial designs. And speed and accuracy ofcalculations are very high. iii) It can be programmed to satisfying take logical decisions

iv) An optimized design with least cost and the required performance can be easily

obtained.

Generally any design method can be

i) analysis method ii) Synthesis method iii) Hybrid method

In the analysis method of design , a preliminary design is made by the designer

regarding the machine dimensions, materials and other constructional features and

these are given as input data to the computer and the performance quantities arecalculated. The designer examines the performance and accordingly alters the input

data and then feed them to the computer again. The computer calculates the newperformance with the revised data. This process is repeated till the required

f i hi d

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performance is achieved.

In the synthesis method, the required performance values are also given to the

computer as input. The computer through an iterative process alters the dimensions tillthe required performance is obtained.

In the hybrid method, by some human intervention, a combination of analysis and

synthesis methods are adopted.

The method of design optimization using computers :

i) Choice of independent variablesii) Variable transformationiii) Forming the constraint functions for the performanceiv) Forming the objective function (OBJ)v) Applying the minimization technique till the OBJ becomes with in the chosentolerance.Example of Design of optimization of Induction Motors

The independent variables which has a significant effect on the performance are

stator core diameter, stator core length , stator core depth, stator slot depth, stator slot

width, rotor slot depth, rotor slot width, end ring depth, end ring width, . airgap lengthand airgap flux density.

The other variables in the design are either taken as constants dased on the voltage and

power rating of the machine or they are in some way related to the above 11 variables.

During the course of optimization when the variables undergo incrementing or

decrementing, they should also be constrained to be with in practical ranges. This is

obtained by variable transformation. For example for airgap Xact = X tran + Lg min ;where they respectively denote actual and transformed values and Lg min = minimum

airgap required.

Performance Specifications:

1. Starting torque 2. maximum torque 3. Full- load power factor 4. full -load efficiency5. full load slip 6.tooth and core flux densities 7. starting current 8. temperature rise 9.

cost of the machine

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cost of the machine.

Objective function

The objective function is formed by comparing the specified and calculated values of

the performance quantities at each iteration. Objective function minimization can be

carried out either using conventional methods such as Powels algorithm or

Rosenbrock method or the recent techniques such as Genetic algorithm.

It should be noted that the independent variables or the performance specifications vary

with the type of machine and its application.

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Design of Electrical Machines Notes

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