Department of Electrical and Electronics Engineering s2
Department of Electrical and Electronics Engineering
EE2355 Design of Electrical Machines
UNIT -1
INTRODUCTION
Flux (Φ) = (MMF/ Reluctance) , Wb ; Reluctance(S) = (l/aµ) , 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 ; µ = permeability = µo µr ;
µo = absolute permeability = 4π x 10 -7 H/m and µr = relative permeability.
H = Ampere turns / m = MMF/ l = S Φ/ l = S Ba /l = (l/aµ) Ba/l = B/µ .
or B = µH
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 / µo l ys
1. Various configurations of slotting
i) Smooth iron surface on both sides of the air gap : ys l = ys
ii) slotted armature : ys l = ys - Ws = Wt (no fringing)
iii) slotted armature : ys l = ys – Kcs Ws ; Ws = slot width , Wt = tooth width
Kcs = Carter’s Coefficient for slots depends on the ratio of slot opening /airgap length or the empirical relation is 1/ {1+ (5lg /Ws )}
If radial ventilating ducts are provided: Ll = L – Kcd nd Wd
where Kcd = Carter’s Coefficient for ducts,
nd = No.of ducts, Wd = width of each duct
If ducts are provided on the stator and on the rotor, then Kcd should be based
on 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/ µo }Kg lg = 8,00,000 B Kg lg .
Effect of Saliency : K f = Field form factor = Bave / Bg
= ψ = pole arc/pole pitch ;
pole pitch = πD/P ; Bave = Φ /( πDL/P)
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2. MMF calculation for teeth
The calculation of MMF for producing flux in the Teeth of the machine is difficult because :
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 worked
in saturation and hence µr becomes low.
Following methods are usually employed for the calculation of MMF required for the
tapered teeth:-
i) Graphical method :ATt = Mean ord. x lt
Mean ord. is the mean ord. of “at” variation with tooth depth.
ii) Simpson’s 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 - 4π x 10 -7 at (Ks –1) ;where Ks = Atotal / Airon = Lys / Li Wt
Specific Permeance : λ = Permeance per unit length or depth of the field.
Parallel sided slot :
λ=µo[(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 = 2πf µoTp2 (Lmt/Lc) {a+ (bp + bs)/3}
The per unit reactance can also be calculated as εx = Ip Xp /Vp ; where Ip , and Vp are rated phase current and voltage respectively.
Sandwich winding: Xp = πf µo(Tp2 /n)(Lmt/w) {a+ (bp + bs)/6}
5. Temperature rise calculations
Q = Power loss(heat produced ), J/s or W
G = weight of the active material of the Machine, kg
h = specific heat, J/kg-◦C
S = cooling surface area, m2
λ = specific heat dissipation, W/ m2 -◦C
c = 1/ λ = cooling coefficient, m2 -◦C / W
θm = 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 the theory of heating of homogeneous bodies ,
Heat developed = heat stored + heat dissipated
Q dt = Gh dθ + Sλθ dt ; solving → θ = θm ( 1- e –t/Th ) + θi e –t/Th
If the machine starts from cold, θi = 0 →θ = θm ( 1- e –t/Th )
Th=heating time constant (time taken by the machine to attain 0.632 times θm) = Gh/Sλ
Tc = cooling time constant (time taken by the machine to fall to 0.368 times θi)
Cooling curve is an exponentially falling curve.
6. Rating of machines
IS: 4722-1968: specification for Rotating Electrical machinery:
1. continuous duty
2. short time duty (T‹‹ Th )
3. intermittent periodic duty
4. intermittent periodic duty with starting
5. intermittent periodic duty with starting and braking
6. continuous duty with intermittent periodic loading
7. continuous duty with starting and braking
8. 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 as
i) Continuous duty ii) Fluctuating duty and iii) Short time and intermittent duty.
Continuously duty motors work
These motors work with the same load through out the duty cycle.
Fluctuating loads
The 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 method
iii) 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 = θml / θm ; θml = 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 used for cooling.
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 a regulator. 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, the air 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 θ) 103 x (760/H) x {( θi + 273)/273} V, m3 /s
cp = 995 and V = 0.775 m3 assumed; then
Va = 0.78 (Q/ θ) x (760/H) x {( θi + 273)/273} , m3
The capacity of the fan required = Pfan = P Va /ηfan, W
Similar calculations can be made for the volume of hydrogen or air or oil used for cooling the machine.
UNIT II
DC MACHINES
D = stator bore or armature diameter, m
L = stator core length, m
p = number of poles
Z = Total numbers of armature conductors
Iz = current in each conductor(Ia/A) , A
E = induced EMF, V
P = machine rating (power output),kW
Pa = power developed by the armature, kW
Q = kVA rating of the machine
Φ = flux per pole, Wb
τ = pole pitch (π D/p), m
Total Electric loading = pΦ ; Total Magnetic loading = Iz Z
Specific Electric loading = ac = Iz Z
(amp.conductor/m) π D
Specific magnetic loading = Bave = pΦ
(Wb/m2 ) π DL
1. Output equation
Pa = E Ia x 10 -3 = (p/A)(ΦZN/60) Iz A 10 -3
= {π2 Bave ac x10-3 } D2 L n = Co D2 L n where Co = output coefficient
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Generator: Pa = (P/η) – (FW & Iron losses) ; Motor : Pa = P + (FW & Iron losses)
For large machines : FW & Iron losses are neglected i.e.,
Pa = P/ η (Generator)
= P (Motor)
For small machines : FW & Iron losses can be taken as 1/3 rd of the total losses. So,
Pa = (P/η) – (1/3) P (1-η)/ η = P(2+ η)/(3 η) …generator
= P + (1/3) P (1-η)/ η = P(1+2η)/(3 η) …motor
2. Choice of specific magnetic loading (Bave)
i) B max in the iron part of the magnetic circuit:
B max ≤ maximum allowable density
i.e., Bt = Bave Ys /wt (non-salient pole machines); Ys = slot-pitch and wt = tooth width.
&Bt = (Bave /ψ)Ys /wt (salient pole machines)
where ψ = pole arc /pole-pitch ratio.
If Ys =2 wt and ψ = 0.667 , then
Bt = 3 Bave : For example,if Bt is to be limited to 2.2 Wb/m2 ,
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(θ)
θ depends on Q (losses),which in turn depends on ac.
Allowable θ depends on insulating material used.
ii) Cooling coefficient (C)
θ 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, Sf is 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 (δ)
choice of δ 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/s
ii) Frequency of flux reversal, f ≤ 50 Hz
iii) Current per brush arm ≤ 400 A
iv) Armature MMF per pole ≤ 7000 A
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 = ψ π 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 flux
pulsation and hence to reduce eddy current losses.
Armature voltage drop = Ia ra ;
ra = (Z/2) ρLmt /(a2 az ) ;
where Lmt = length of mean turn, m = 2L + 2.3τ +5ds ; a = no.of parallel paths and az = area of each conductor, m2 .
6. Design of the Field system
Area of each pole (Ap) = Flux in the pole body / Flux density = cl φ / Bp
where 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.
( ATf at full-load) /(ATArm. at full-load) =1.0 to 1.25
(to overcome armature reaction)
7. Tentative design of Field winding
Copper loss in the field winding = If2 Rf
= (δfaf)2 {ρLmt Tf}/af = δf2 ρLmt Tf af
= δf2 ρLmt Sf hfdf ; ………………….(1)
where df = depth of the pole winding. Tf = no.of turns in each field coil.
Permissible loss = S qf = 2Lmt hf qf ..(2)
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Equating (1) and (2) ; δf =104 √ qf / Sf df
MMF per metre of the field winding = ATf/ hf = If Tf /hf = δf (af Tf )/ hf
= δf Sf hfdf/hf = δf Sf df = Sqrt {2 qf Sf df /ρ}
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 = π DcN/60 ≤ 20 m/s
iv) Pitch of the commutator segment = βc = π Dc/C ≥ 4.0 mm
(3.2 mm for the conducting portion and 0.8 mm for mica separator)
v) Distance between brush spindles = π 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 (10–30 mm), c2 = clearance for end ply (10-25 mm).
vii) Current carried by each brush spindle
= Ib = 2Ia/p = Ab δb ; where Ab = nbwbtb