UNIT-I- MAGNETIC CIRCUIT AND COOLING OF ELECTRICAL MACHINES.

UNIT-I-
MAGNETIC CIRCUIT AND COOLING OF ELECTRICAL MACHINES.


1.1.BASIC PRINCIPLES OF MAGNETIC CIRCUITS:
The path of the magnetic flux is called magnetic circuit. Notations used are,φ=flux in Wb;
A=area of magnetic path in m2;l=length of magnetic path in m;H=magnetic field intensity in AT/m;
AT=mmf in ampere turn;μ=μoμr;μ=absolute permeability of magnetic material inH/m;μo=permeability of free space=4πx10-7H/m;μr=relative permeability=1 for air;S=reluctance in Amp/Wb;Δ=permeance.Wb/A.
In Electric circuit,Ohms law gives relation between current,emf and resistance,while in magnetic circuit, a similar relation exists relating flux,mmf and reluctance.This is known as Ohms law of Magnetic circuit and is expressed as
Flux(φ)=mmf(AT)/Reluctance(S).---(1).Also,Reluctance(S)=length/area x permeability →S=l/Axμ—(2)
From,(1),AT= φ x S=φx l/Axμ= φ/Ax l/μ=Bx l/μ;AT/l= B/μ;∴H= B/μ;→ B=μH---(3).,
For the core material of length’l’ and carrying uniform flux,the total mmf is,AT=H x l—(4).
In series magnetic circuit the total reluctance is the sum of reluctance of individual paths.
S=S1+S2+S3+-----+Sn.where,S=Total reluctance;S1,S2—reluctance of individual paths.The total mmf acting around a complete magnetic circuit is mmf=φ x S=φx(S1+S2+S--)=AT1+ AT2+ AT3 – ATn=H1l1+ H2l2-(5).
In parallel magnetic Circuit, the same mmf is applied to each of the parallel paths and the total flux divides between the paths inversely proportional to the reluctance.
Ie φ= φ1+ φ2+ φ3;Dividing by AT, we get,φ/AT= φ1 /AT+ φ2 /AT+ φ3/AT→1/S=1/S1+1/S2→Δ=Δ1 +Δ2—(6).
Note:For Series magnetic circuit;S=S1+S2+S3 →mmf/Reluctance=l1/μoμrA1 + l2/μoμrA2 +---
For parallel magnetic circuit; 1/S=1/S1+1/S2→Δ=Δ1 +Δ2; S=Reluctance;Δ=Permeance(=1/S).
1.2.MAGNETIC CIRCUIT CALCULATIONS:
(a).Calculation of total mmf: The calculation of total mmf required to establish flux in magnetic circuit involves knowledge of dimensions and configurations of magnetic circuits. The magnetic circuit is split into number of paths.The flux density is calculated in every path and mmf per unit length,AT,is found out. The summation of mmf in series gives total mmf.
The various paths for which the calculation of mmf is required are(i).mmf for airgap.(ii).mmf for teeth.
1.3.MMF FOR AIRGAP/RELUCTANCE OF AIRGAP OF SMOOTH/SLOTTED ARMATURE:
l=length of armature in m;D=Diameter of armature in m;lg=length of airgap inmm;ys=slotpitch in mm; ws=slot width in mm;wt=width of the teeth in mm;wo=slot opening in mm;nd=No.of radial ducts;
wd=width of each ducts in mm.
Note:Slot Pitch: Distance between centres of two adjacent slots .





(a).SMOOTHARMATURE:Consider an iron surface on the two sides of the airgap to be smooth as shown in fig(a).The flux is uniformly spread over the entire slot pitch and goes straight across the airgap.
Then the reluctance of the airgap is ,Sg=lg/μoLys---(1)
(b).SLOTTED ARMATURE:In a slotted armature, the effective area of flux path is decreased resulting in an increased reluctance of airgap.Consider the case of a slotted armature with a very small airgap length as shown in fig(b).
The flux in this case is only confined to tooth width.Hence the effective slot pitch is ys’.From fig( b),
ys=ws+wt → wt= ys-ws;As the flux is confined to tooth width,ys’=wt;
Sg=lg/μoLys= lg/μoLys’= lg/μoLwt ∴Sg=lg/μoL (ys-ws)---(2)
Reluctance of AG of smooth armature,Sg=lg/μoLys; Reluctance of AG of slotted arm.,Sg=lg/μoL (ys-ws)
(C).FRINGING:However, some fringing of flux around the edge in a slotted armature.The flux penetrate down the slot as shown in fig(c).Hence the reluctance in this case is more than that in case of a smooth armature but less than in case where the whole flux is assumed to be confined over the tooth width.
Reluctance of AG of slotted armature taking into fringing effect,Sg=lg/μoL (ys-Kcs ws)—(3)
Where Kcs=Cater’s coefficient for slots which depend upon the ratio of slot width to airgap length.
An empirical formula which gives the value of Kcs directly is Kcs=1/1+5lg/ws.
Let Kgs=Gap contraction factor for slots.=Reluctance of AG for slotted arm/Rel.Of AG for smooth arm.
∴ Kgs= lg/μoL (ys-Kcs ws)/ lg/μoLys ∴ Kgs=ys/ (ys-Kcs ws)---(4)








(d).Radial Ventilating Ducts:To imrove the cooling effect or ventilation ,radial ducts are provided in the armature.∴The effective axial length of the machine is reduced due to the presence of ducts and this results in increase in the reluctance of airgap.∴Effective axial lengthL’=L-Kcdndwd.;where Kcd= Cater’s coefficient for ducts.
An empirical formula which gives the value of Kcs directly is Kcd=1/1+5lg/wd
Let Kgd=Gap contraction factor for ducts.=Rel. of AG for arm with ducts/Rel.Of AG without ducts .
∴ Kgd=L/ (L-Kcd nd wd)---(5)
The effect of both slotting and ventilating ducts can be expressed as below.
Reluctance of AG of smooth armature without duct,Sg=lg/μoLys ---(a).
Reluctance of AG of slotted arm.,Sg=lg/μoL’ ys’ ---(b).;(b)/(a)→L/L’xys/ys’→ ys/ (ys-Kcs ws)x L/ (L-Kcd nd wd)
∴Kg=KgsxKgd.---(6).Where Kg=Total gap contraction factor for slots and ducts.
(e)Semienclosed slots:ys=ys-Kcswo;∴Kgs=ys/ ys-Kcswo----(7).
(f)Salient pole machines:
Length of airgap is not uniform.Flux density should be calculated interms of gap density , Bg.
(g).Calculation of MMF:
Flux(φ)=mmf(AT)/Reluctance(S) → mmf = Flux x Reluctance ;
Mmf for airgap = Flux x Reluctance of airgap.→ATg=φ x Sg= φ x lg/μoA=φ/A xlg/4πx10-7=Bx lg/4πx10-7
∴ATg=Blg x800,000.→ATg=800,000 x B lg ---AT orWb/m.---(8)
NOTE:
(i).mmf for airgap for smooth armature, ATg=800,000 x B lg --AT.
(ii). mmf for airgap for slotted armature, ATg=800,000 x B lg Kg --AT
Kg =KgsxKgd; Kgs=ys/ (ys-Kcs ws); Kgd=L/ (L-Kcd nd wd).
(iii). mmf for airgap for semi enclosed slots, ATg=800,000 x B lg Kg ---AT.
Kg =KgsxKgd; Kgs=ys/ (ys-Kcs wo); Kgd=L/ (L-Kcd nd wd).
(iv). mmf for airgap for salient pole machines, ATg=800,000 x Bg lg Kg ---AT.
Bg=Gap density;Bg=φ/Pole arc x Length of armature.;Also,Bg=Bav/Form factor.
ψ=Pole arc/pole pitch;Pole pitch =πD/p;Pole arc=ψxPole pitch.
1.1.Calculate mmf required for the airgap of a machine having length=0.32m,including 4ducts of 10mm each,pole arc =0.19m,slot pitch =65.4mm,slot opening=5mm,airgap length=5mm,flux per pole=52mWb. Given Cater’s coefficient is o.18 for opening/ gap=1 and 0.28for opening / gap=2.width of the slot=5mm.
Data:L=0.32m,nd=4,wd=10mm=10x10-3m,pole arc=0.19m,ys=65.4mm=65.4x10-3m,wo=5mm,
lg =5mm=5x10-3m,φ=52x10-3Wb,Kcs=0.18,Kcd=0.28,ws=5mm=5x10-3m.To find :ATg.
Sol: mmf for airgap for slotted armature, ATg=800,000 x B lg Kg --AT ---(1).
B= φ/Pole arc x Length of armature =52x10-3 /0.19x0.32=0.8552Wb/m2; Kg =KgsxKgd;
Kgs=ys/ (ys-Kcs wo)= 65.4x10-3/65.4x10-3-(0.18x5x10-3)=1.013;Kgd= L/ (L-Kcd nd wd)=1.0362.
Kg =KgsxKgd=1.0504; ATg=800,000 x B lg Kg =3590AT
1.2.A 50 kW,220V,4pole,DC m/c has the following data:Armature diameter=0.25m,Length=0.125m,flux per pole=11.7mWb,length of airgap at pole centre=2.5mm,the ratio of pole arc to pole pitch =0.66. Calculate the mmf required for airgap (i).if the armature is treated as smooth.(ii).if the armature is slotted and the gap contraction factor is 1.18.
Data:Po=50kW,V=220V,p=4,DC m/C,D=0.25m,L=0.125m,φ=11.7x10-3Wb, lg=2.5x10-3m,ψ=0.66,Kg=1.18.
Sol:(i). mmf for airgap for smooth armature, ATg=800,000 x B lg --AT
B= φ/Pole arc x Length of armature;ψ=pole arc/pole pitch →pole arc=0.66xpole pitch;
pole pitch=πD/p;∴pole arc=0.1295m,B=0.7255T;ATg=800,000 x B lg=1,451AT.
(ii). mmf for airgap for slotted armature, ATg=800,000 x B lg Kg =1712.18AT.
1.3.Determine the airgap length of a DC m/C from the following data: Gross length of core=0.12m, No.of ducts=1&is 10mmwide,slot pitch=25mm,slot width=10mm,Cater’s co-efficient for slots&ducts=0.32,flux density at pole centre=0.7Wb/m2,Field mmf/pole=3900AT,mmf required for iron paths of magnetic circuit=800AT.
Data:L=0.32m,nd=1,wd=10x10-3m,ys=25x10-3m,ws=10 x10-3m,Kcs=Kcd=0.32,B=0.7T,ATf=3,900, ATi=800AT.
Sol: mmf for airgap for slotted armature, ATg=800,000 x B lg Kg ; ATf = ATi+ATg →ATg=3100AT
Kg =KgsxKgd; Kgs=ys/ (ys-Kcs ws)=1.15 Kgd= L/ (L-Kcd nd wd) =1.03;Kg=1.18; ATg=800,000 x BlgKg→lg=4.7mm.
1.4.Estimate the effective gap area of a 10pole,slip ring induction motor with the following data:Stator bore=0.65m,bore length=0.25m,No,of stator slots=90,stator slot opening=3mm,rotor slot=120,rotor slot opening=3mm,airgap length=0.95mm,Cater’s co-efficient for ducts=0.68 and for slots =0.46,No.of ventilating ducts=3,each on stator and rotor,width of each ventilating duct=10mm.
Data:p=10,SRIM,D=0.65m,L=0.25m,Ss=90,wo(s)=3 x10-3m,Sr=120,wo(r)= 3 x10-3m,lg=0.953 x10-3m,Kcs=0.46,
Kcd=0.68,nd=3,wd=10 x10-3m,.To find:Ag.
Sol:Effective Airgap area,Ag=Ag’/Kg.;Ag’=πDL/p=0.051m2; Kg =KgsxKgd;Kgs=Kgs(s)Kgs(r); Kgs(s)=ys(s)/ (ys(s)-Kcs wo(s));
ys(s)=πDs/Ss=0.022m;∴ Kgs(s) =1.0169; Kgs(r)=ys(r)/ (ys(r)-Kcs wo(r)); ys(r)=πDr/Sr; Dr=Ds-2lg;∴ ys(r)=0.0169m;
Kgs(r)=1.0889; Kgs=Kgs(s)Kgs(r)=1.16174; Kgd= L/ (L-Kcd nd wd) =1.0888; Kg =KgsxKgd =1.2649; Ag=Ag’/Kg=0.04m2.


1.4.Problems associated with the calculation of mmf for airgap:
(i).One or both of the iron surfaces around the air gap may be slotted, so that the flux tends to concentrate on the teeth rather than distributing itself, uniformly over the airgap.
(ii).There are radial ventilating ducts in the machine for cooling purpose,which affects the calculation.
(iii).In salient pole machine, the gap dimensions are not constant over the whole of the pole pitch.
1.5.Stacking factor: The whole length of the armature is not occupied by iron, some part of the length is taken by irregularity in thickness of laminations. It is usual to define iron space factor called stacking factor as the ratio of actual length of iron in a stack of assembled core plate to the total actual lengths of stack. It is denoted by ‘Ki’. Ki=Net iron length/Length of slot portion of conductor≈0.9.
Stacking factor for iron largely depends upon the thickness of plate and the type of insulating material employed. The expression for net iron length is given by,Li=Ki(L-ndwd).
1.6.MMF FOR TEETH.
(a). Problems associated with the calculation of mmf for teeth:
(i).The teeths are wedge shaped or tapered(width of the slot gradually increases or decreases from top of slot to bottom of slot), when parallel sided slots are used.This means that area present through the path of flux is not constant and this gives different values of flux density over the length of the teeth.
(ii).The slot provides another parallel paths for the flux ,shunting the tooth.As a result some amount of flux goes down the depth of the slots.This causes problem in calculation.
(b).Methods of Determination of mmf for teeth:
(i).Graphical method: Construct a graph showing the variation of AT vs lenth of the tooth.-Total mmf required for the tooth is given by,
att=atmean x lt (or) att=atmean x ds;ds=depth of the slot;lt=height of the tooth.
To determine atmean, it is necessary to determine a graph showing the manner in which the flux density varies with respect to at, throught the length of the tooth.
(ii).Three ordinate method.(Simpson’s rule):Graph drawn between AT vs length of the tooth.Values of ‘at’ are determined at 3 equidistant points,the end of the teeth,its centre, and its beginning.The mean value of ‘at’ is given by atmean=(at1 +4at2 +at3)/6.
(iii).Bt1/3 method: Applicable to teeths of a smaller taper. Values of ‘at’ obtained for flux density at 1/3rd from the narrow end .Let, Bt1/3=flux density at 1/3rd height from narrow end, at1/3=value of mmf/m for
Bt1/3 obtained from B vs at curve.∴Total mmf for teeth, att=at1/3 x lt = at1/3 x ds.









1.7.Calculation of Real and Apparent Flux density:
Slots provides an alternative path for the flux to pass, while most of the flux passes through iron portion.
This means that real flux passing through the teeth is always less than total or apparent flux..
Apparent flux density is defined as Bapp=Total flux in slot pitch/tooth area→Bapp=φs/Ai—(1).
Real flux density is defined as ,Breal =Actual flux in tooth/tooth area→Breal= φi/Ai—(2)
In actual m/c, taking the flux over the pole pitch, there are two parallel paths as below,
(i).Iron paths: Area of iron Ai=tooth width x net iron length→Ai=wtLi—(3)
(ii).Air path:Area of air path,Aa=total area-iron area=[slot pitch x pole length-Ai]=[ysL-wtLi]=Lys-wtLi-(4)
φs=flux over one pole pitch= φi + φa --(5).Divide throughout by Ai→ = φs/Ai = φi/Ai + φa/Ai
→ Bapp= Breal+ φa/Ai → Bapp= Breal+ φa/Aa xAa/Ai→ Bapp= Breal+ Ba xAa/Ai→ Bapp= Breal+ Ba xk.;k=Aa/Ai
But,B=μH=μo μr H→ B=μo H;μr=1;∴ Bapp=Breal+μoHk→ Bapp= Breal+4πx10-7xATreal k=Breal+4πx10-7x ATreal (ks-1)
ks-1=K→ks=1+k;But k=Aa/Ai; k= Lys-wtLi/wtLi= Lys/ wtLi -1;→k= Lys/ wtLi -1 ; ks-1= Lys/ wtLi -1;∴ks=Lys/ wtLi
Summary of apparent and real Flux densities:
Bapp= Breal+4πx10-7x ATreal (ks-1); ks=Lys/ wtLi ; Li=Ki(L-ndwd); ys=ws +wt

1.5.Calculate the apparent flux density at a particular section of a tooth from the following data:Tooth width =12mm,slot width=10mm,gross core length=0.32m,No.of ventilating ducts=4 each 10mmwide, real flux density =2.2Wb/m2,permeability of teeth corresponding to real flux density=31.4x10-6H/m. stacking factor=0.9.
Data:wt=12x10-3m,ws=10 x10-3m,L=0.32m,nd=4,wd=10 x10-3m,Breal=2.2T,μreal=31.4x10-6H/m,ki=0.9.
To find Bapp: Bapp= Breal+4πx10-7x ATreal (ks-1); ys=ws +wt =22 x10-3m,Li=Ki(L-ndwd)=0.252m,
ks=Lys/ wtLi =2.33;Breal=μreal Hreal→ Hreal=70,063.7=ATreal;∴Bapp= Breal+4πx10-7x ATreal (ks-1)=2.316T.
1.6.Determine the apparent flux density in the teeth of a DC m/c when the real flux density is 2.15Tesla.Slot pitch=28mm,slot width=10mm,gross core length=0.35m,No.of ventilating ducts=4,each 10mm wide.The magnetizing force for a flux density 2.15T is 55,000AT/m.The iron stacking factor=0.9.
Data: Breal=2.15T,ys=28 x10-3m,ws=10 x10-3m,L=0.35m,nd=4,wd=10 x10-3m, ,ATreal=55,000AT/m,ki=0.9.
Sol: Bapp= Breal+4πx10-7x ATreal (ks-1); wt=ys –ws =18 x10-3m,Li=Ki(L-ndwd)=0.279m,
ks=Lys/ wtLi =1.951;∴Bapp= Breal+4πx10-7x ATreal (ks-1)=2.215T.
1.7.Calculate apparent flux density at the section of the teeth of an armature of DC M/C from the following data at that section.Slot pitch=24mm,slot width=tooth width=12mm,length of armaturecore including 5 ducts of 10mm each is 0.38m.Iron stacking factor=0.92.True flux density in that section is 2.2Wb/m2 for which the mmf is 70,000AT/m.
Data:ys=24 x10-3m,ws=wt=12 x10-3m,L=0.38m,nd=5,wd=10 x10-3m,Ki=0.92,Breal=2.2T,ATreal=70,000AT/m.
Sol:Li=Ki(L-ndwd)=0.279m,ks=Lys/ wtLi =2.558;∴Bapp= Breal+4πx10-7x ATreal (ks-1)=2.33Wb/m2.
1.8.LEAKAGE REACTANCE CALCULATIONS:
(I).Leakage flux: Flux that do not follow the confined path.-If the leakage flux alternates, it will induce voltage in any winding with which it links. This Voltage is known as reactance voltage.The reactance corresponding to this voltage is known as leakage reactance and plays an important part in the performance of the ac m/cs,such as Voltage regulation,stability of the m/c.Also it affects dc excitation in DC machines and makes commutation difficult.
(ii).Specific permeance:It is defined as permeance per unit length of slot or depth of field.
The leakage flux can be assumed to be consists of flux tube of length ‘y’ and a constant width ‘dx’ along the length’L’ of the field.Permeance of the flux tube,dΔ=μo xarea of the tube/length of the tube.
∴ dΔ=μoLdx/y.; Permeance of the whole field,Δ=∫μoLdx/y=Lμo∫dx/y --(1).
Specific permeance is defined as permeance per unit length,L,∴λ=Δ/L=Lμo∫dx/y/L.→λ=μo∫dx/y-(2)
If the length of the tube is constant over a height’h’,specific permeance,λ=μo ; λ=μoh/y
(iii).Leakage reactance(x):When a leakage flux is associated with a winding carrying a steady state alternating current,a reactive voltage is produced.The reluctance offered to leakage flux is due to the airpath and it is directly proportional to the current giving a constant inductance.
WKT ,Inductane,L=Flux linkage/current=Tφ/I ---(1).;T=No.of turns.Also,Flux=mmf/reluctance
→flux=mmf xpermeance=TI xΔ;∴φ= TI xΔ –(2),Put (2)in (1),L= Tφ/I=T(TI xΔ)/I; ∴L=T2Δ
WKT,X=2πfL=2πf(T2Δ). But λ=Δ/L →Δ=λ L;∴X=2πf(T2 λ L)
1.9.LEAKAGAGE REACTANCE OF CORE TYPE TRANSFORMER:[φ=mmfxΔ;ψ=φT;L=ψ/I;X=2πfL]
The arrangement of winding of a core type transformer is shown in fig.The calculation of’x’ is based on the following assumptions.
(i).The Primary and Secondary winding have equal axial length
(ii).The flux paths are parallel to the windings.
(iii).mmf for iron part is negligible
(iv).Total mmf,AT=IpTp=IsTs.
(v).Half the leakage flux in duct links with each winding.
(vi).The length of mean turn of the windings are equal.
Three leakage flux paths φp,φs φo are shown.
Let,φp,φs φo=Flux leakages in Py&Sy Windings and in duct.
Lo=mean circumference of duct,Lc=axial height of winding,
bp ,bs=radial widths of Primary and Secondary windings,a=width of radial duct.
(a).Flux linkage in conductor portion:Consider an infinitesimal strip of width’dx’a t a distance ‘x’ from the edge of the Primary winding alog its width.Mmf acting across the strip=(IpTp)x/bp.
Permeance of the strip=μoLmtp/Lc dx;∴Flux in the strip=mmf x permeance=(IpTp)x/bp (μoLmtp/Lc dx ).
→φ= (μoLmtp/Lc )(IpTp) (x/bp)dx;This flux lins with (x/bp)Tp turns.∴Flux linage,dψ1=φ x No.of turns
∴ dψ1 =[(μoLmtp/Lc )(IpTp) (x/bp)dx ] (x/bp)Tp→ dψ1=(μoLmtp/Lc ) (IpTp2) (x/bp)2dx.
∴Flux linkage of primary winding due to the flux in strip,ψ1=∫ dψ1=(μoLmtp/Lc ) (IpTp2)
∴ ψ1 =(μoLmtp/Lc ) (IpTp2)(bp/3)----(1)
(ii).Flux linkage in the Duct portion:
Mmf acting across the duct=IpTp; Permeance of the duct=(μoLo/Lc ) a;Flux in the duct,φ=mmfxpermeance
→ φ =( IpTp) (μoLo/Lc ) a ;Half of the duct flux links with each of the two windings.
∴.Duct flux linking with primary winding, φo=1/2 ( IpTp) (μoLo/Lc ) a =1/2 (aμoLo/Lc ) (IpTp).
This flux links with the entire primary winding.∴Flux linkages of primary winding due to duct flux,
→ψo=1/2 (aμoLo/Lc ) (IpTp) (Tp)→ ψo =1/2 (aμoLo/Lc ) (IpTp2)---(2).
∴Total flux linkages of Primary winding=ψp= ψ1+ ψo=(μoLmtp/Lc ) (IpTp2)(bp/3)+ 1/2 (aμoLo/Lc ) (IpTp2)
→ ψp =μo (IpTp2/Lc)[ Lmtp bp/3 +aLo/2];AssumeLmtp=Lmt=Lo;→ ψp = μo IpTp2 Lmt/ Lc[bp/3 +a/2] ----(3)
WKT, Inductance,L=Flux linkage/Current→Lp= ψp/Ip→Lp= μo Tp2 Lmt/ Lc[bp/3 +a/2]—(4); also ,X=2πfL
Leakage reactance of Primary windings,xp=2πf μo Tp2 Lmt/ Lc[bp/3 +a/2] –(5).
Similarly, Leakage reactance of Secondary windings,xs=2πf μo Ts2 Lmt/ Lc[bs/3 +a/2] –(6)
Leakage reactance of secondary winding referred to primary side,xs’=xs/K2;K=N2/N1=Ts/Tp.
∴xs’=xs (Tp/Ts)2→ xs’= 2πf μo Tp2 Lmt/ Lc[bs/3 +a/2] –(7)
Total reactance of transformer referred to PY side,Xp=xp+xs’→ 2πfμoTp2 Lmt/ Lc[ (bp+bs)/3 +a]—(8).
If the windings are divided into 2 parts (usually HV), Xp=πfμoTp2 Lmt/ Lc[ (bp+bs)/6 +a]---(9)
Note: bp= Width of Primary winding,bs= Width of lv winding,Lc= height of coi,Lmt= length of mean turn,
Tp =Primary winding turn, a=width of duct (between hv and lv winding)
1.8.A 300kVA,6600/400V,50Hz,Delat/star,3φ core type transformer has the following data:
Width of hv winding=25mm, Width of lv winding=16mm,height of coil=0.5m,length of mean turn=0.9m, hv winding turn=830,width of duct between hv and lv winding=15mm.(a).Calculate leakage reactance of the transformer referred to the hv side.(b).If the l.v. coil is split into two parts on each side of the h.v.coil
Calculate the leakage reactance referred to the h.v. side.
Data: Vp=6600,Vs=400V,bp=25x10-3m,bs=16 x10-3m,Lc=0.5m,Lmt=0.9m,Tp=830,a=15 x10-3m,,f=50Hz.
(i).Leakage reactance referred to the primary side,Xp=2πfμoTp2 Lmt/ Lc[ (bp+bs)/3 +a]
Xp =2πx50x4πx 10-7x(830)2x0.9/0.5[(25x10-3+16 x10-3)/3 +15 x10-3]=14.03Ω.
(ii).The hv winding is divided into two parts ,then,Xp=πfμoTp2 Lmt/ Lc[ (bp+bs)/6 +a]
Xp=πx50x4πx 10-7x(830)2x0.9/0.5[(25x10-3+16 x10-3)/6 +15 x10-3]=5.34Ω.
1.10.LEAKAGE REACTANCE CALCULATION FOR ROTATING AC MACHINES. (INDUCTION MACHINES).
(a).Parallel sided Slot:(All dimensions are in m).The total height of the slot is divided into different sections h1,h2,h3,h4 and the corresponding specific permeance is λ1, λ2, λ3, λ4.
Let h1=height of conductor portion,h2=height above the conductor but below the wedge,
h3=height of wedge,h4=Limb height,ws=width of the slot,wo=slot opening,
Zs=total No.of conductor per slot.Note: Specific permeance,
λ=μo —Conductor portion; λ=μo — Non-Conductor portion.
λ1= μo ;Zx=x/h1 Zs ,h=h1,y=ws.→ λ1= μo/h1ws → λ1=μoh1/3ws.
λ2=μo ;h=h2,y=ws;λ2=μoh2/ws;λ3=μo ;h=h3,y=(ws+wo)/2;λ3=μo2h3/(ws+wo)
λ4=μoh4/wo;λs= λ1 + λ2+ λ3 + λ4→ λs=μo[h1/3ws+ h2/ws+ 2h3/(ws +wo)+ h4/wo];Xs=8πfTph2L(λs/pq)
(b)Tapered slot: λ1= μo ;Zx=x/h1 Zs ,h=h1,y=(w1+ ws)/2→ λ1=2μoh1/3(ws+w1)
λ2= μo ;h=h2,y=(w1+w2)/2→λ2=μo 2h2/(w1+ w2);λ3=μo ;h=h3,y=(w1+wo)/2;λ3=μo2h3/(w1+wo)
for λ4,h=h4,y=wo.∴λ4=μoh4/wo; λs=μo[2h1/3(ws+w2)+2 h2/(w1+w2)+ 2h3/(w1 +wo)+ h4/wo]
1.9.Determine the specific permeance per metre length of a rectangular semi enclosed slot having the following dimension.Slot width=10mm,slot opening=4.5mm,height of conductor=26mm,height above conductor and below wedge=1mm,height of wedge=3.5mm.Limb height=1.5mm.
Data: ws=10x10-3m,wo=4.5 x10-3,h1=26 x10-3,h2=1 x10-3,h3=3.5 x10-3,h4=1.5 x10-3.To find λ.
Sol: λs=μo[h1/3ws+ h2/ws+ 2h3/(ws +wo)+ h4/wo]=2.23x10-6Wb/AT.
1.11.COOLING OF TURBO-ALTERNATORS:
(a)The various systems of cooling turbo-alternators are,
(i).Axial system(for rotor):This is the conventional method of cooling rotors of turbo-alternators. Narrow sub-slots, just below the main slots, are formed through the rotor core as shown in fig(a).Large quantities of air are forced through these sub-slots.
(ii).Radial system:Here, the stator core has radial ventilating ducts every 5 to 7cm of its length as shown in fig(b).All the air for cooling the stator is forced into the air gap from both ends. This air than flows radially outward through the radial ventilating ducts as shown.
(iii).Radial Axial System: This is also used for stator and is a combination of the first and second systems.
(iv).Multiple Inlet System: This is the modern method of cooling and applicable to any length of machine.In this system, the stator frame and core are divided into a number of compartments as shown in fig(c).In some compartments ,the direction of air is radially outwards and in the others it is radially inwards. Air under pressure is forced into the stator casing from where it flows radially inwards in to the stator ducts.The air passes down and goes into other compartments through the airgap and axial holes from where it flows radially inwards into the stator ducts. The air passes down and goes into other compartments through the airgap and axial holes from where it flows radially outwards through ducts.
The air is drawn from these latter compartments, cooled and recirculated(fig c).
(b).Cooling media:Cooling media nay be either air or hydrogen giving correspondingly air-cooled alternators and hydrogen-cooled alternators.













Air-Cooled alternators:Here air may be circulated through alternator by anyne of the above four methods. In small alternators air may be supplied by fans mounted on their shafts whereas in large machines it is a separate system with motor driven fans.Modern practice has become to spray the air with water so that, in addition to getting cooled, the air loses all dust particles that it may carry.
Hydrogen cooled Alternator: It is applied for alternator of more than 60MW .Advantages of Hydrogen as the medium when compared to air are,
(i).Noise and losses due to windage are less.(density ofH2=1/14 air).(ii).Better cooling medium.(Specific heat of H2=14 times of air-Rapid cooling).(iii).Low-temperature gradient-Thermal conductivity of H2=7 times air.Hot particles of H2distributes its heat to surrounding particles, more rapidly than a particle of air.This property has lower the temperature gradients.(iv).Higher rating can be secured.(v).Power required for circulation is less.(10% of air).
Note:It is estimated that 60MW machine which has a total loss of 1000kW, and requires about 150 tons of air per hour and a fan power of about 100kW.
















SHORT QUESTIONS AND ANSWERS.
1.Write any two similarities and differences between Electric and Magnetic Circuits.
Similarities-
Electric Circuits
1.Emf circulates current in a closed path.
2.Flow of Current is opposed by Resistance
Differences:
1.When current flows,energy is spent continuously
2.Current actually flows in Electric Circuits
3.Ohms Law:emf=Current XResistance
Magnetic Circuits
1.Mmf creates flux in a closed path.
2. Creation of flux is opposed by reluctance.
1.Energy is needed to create the flux,but not to maintain it.
2.Flux does not flow in a magnetic circuit.
3. Ohms Law:mmf=Flux XReluctance
2.What is magnetization curve?
*Curve – Y axis-Magnetic field intensity(H) Vs X axis-Flux density(B).of Magnetic material.
*To estimate mmf required for flux path in the magnetic material.
3.Define gap contraction factor for slots and ducts and gap contraction factor..
Kgs=Gap contraction factor for slots.=Reluctance of AG for slotted arm/Rel.Of AG for smooth arm.
Kgd=GCF for ducts.=Rel. of AG for armature of a machine with ducts/Rel.Of AG without ducts .
K g = Rel. of AG for arm. in m/c with slotted arm& ducts/Rel.Of AG of smooth arm. and without ducts .
K g =Total gap contraction factor for slots and ducts=KgsxKgd.≈1.2
Note:The slots and ducts in the armature (or stator or rotor) of electrical m/c increases the reluctance of air-gap ,which in turn increases the mmf required for air-gap.Kg represents the increase in reluctance as an increase in air-gap length.With the knowledge of Kg,the mmf required for air-gap can be estimated without calculating the increase in reluctance due to slots &ducts.
4.What are ventilating ducts?
The radial ventilating ducts are small gaps of width wd in between the stacks of armature core.They are provided for improving cooling of the core when the length of the core is greater than 0.1m.

5.What is Cater’s coefficient?What is its significance?.
It is a parameter used to estimate the contracted or effective slot pitch in case of armature with open or semi enclosed slots .It is a ratio of wo/lg,where wo=slot opening,lg=length of AG.
It is also used to estimate the effective length of armature when ducts are employed. It is a ratio of wd/lg,where wd=width of duct,lg=length of AG.
6.What is the effect of salient poles on the air-gap mmf?
In salient pole machines, the length of air-gap is not constant over the whole pole pitch.
Hence the effective air-gap is not constant over the whole pole pitch. Hence the effective air-gap length is kglg,where kg is the gap contraction factor. Also for calculating mmf, the maximum gap density Bg,at the centre of the pole is considered instead of average gap density.The field form factor kf, relates the average gap density over the pole pitch to maximum flux density in the air-gap ,given by,
Field form factor,kf=Bav/Bg;Also,kf≈ψ=Pole arc/Pole pitch.
7.What are the problems encountered in estimating mmf for teeth?List the methods of estimating them
Problems:(i).The flux density in different section of a tooth is not uniform.(ii).The slots provides another parallel path for the flux.
Estimation of mmf for teeth: (i).Graphical method.(ii)Three ordinate method.(iii).B1/3 method.
8.Distinguish between real and apparent flux densities in the tooth section of slot.
The real flux density is due to the actual flux through a tooth.The apparent flux density is due to total flux that has to pass through the tooths.Since some of the flux passes through slot,the real flux density is always less than the apparent or total flux density.
Bapp=Total flux in a slot pitch/Tooth area. Breal=Actual flux in a tooth/Tooth area.
9.What is leakage flux and leakage co-efficient?How will you minimize the magnetic leakage?
The leakage flux is the flux passing through unwanted path. It will not help either for transfer or conversion of energy. Leakage co-efficient,Cl=Total flux/Useful flux.The leakage flux affects the excitation demand,regulation,forces on the winding under short circuit conditions, commutation etc.
Leakage flux flows through air-gap of the m/c.If the air-gap of the m/c is kept as low as possible ,LF↓.
10.What is fringing flux?What are the differences between leakage flux and fringing flux?
The bulging of magnetic path at the air-gap is called fringing.The fluxes in the bulged portion are called fringing flux. The leakage flux is not useful for energy transfer or conversion and flows in unwanted path. But the fringing flux is useful flux and flows in the magnetic path.The effect of leakage flux on the m/c performace is accounted by leakage reactance whereas fringing flux increases the slot reactance.
11.List the various types of armature leakage flux.
Slot LF:flux crosses the slot from one tooth to the next and returning through iron.
Tooth topLF:Flux flowing from top of one tooth to the top of another tooth.
Zig-Zag LF:Flux passing form one tooth to another in a Zig-Zag fashion across the air-gap.
Overhang LF:Flux produced by overhang portion of the armature winding.
Harmonic or Differential LF:Flux produced due to difference in stator and rotor harmonic content.
Skew LF:Reduction in mutual flux due to skewing of rotor in induction motor.
Peripheral LF:Flux flowing circumferencially round the air-gap without linking with any of the winding.
12.Define leakage reactance. What is reactance voltage? Is it beneficial?.
The leakage reactance is the reactance offered by leakage flux.The induced voltage due to LF is reactance voltage. It is not beneficial as it affects commutation, and produce sparking in dc machines.
Leakage reactance,Xs=2πfZs2Lλs;L=Length of Arm.λs=Specific slot permeance,Zs=Conductors/slot.
Leakage reactance for poly phase m/c,Xs=8πfTph2Lλs/pq.
13.What is permeance and specific permeance?
Permeance is inverse of reluctance. S=Reluctance=l/μoμrA;Δ=Permeance(=1/S).
Specific permeance: Permeance per unit length of slot.
14.What is stacking factor?
The cores of magnetic circuits are built up with laminated steel plates wherever required. These laminations are insulated from each other by paper, stuck one side of the lamination.Also ventilating ducts are provided along the length of the armature.Hence ,it is clear that the whole length of the armature is not occupied by iron;some part of length is taken up by ventilating ducts and some part by lamination.It is usual to define iron space factor or stacking factor that relates the length of the armature interms of the non-iron portion as Stacking factor=Length of Iron in a stack of assembled core plates to total axial length.
∴Gross iron length,Ls=core length-length of ventilating ducts=L-ndwd;Net iron length,Li=Ki(L-ndwd)
15.On what factors Eddy current and Hysteresis loss depends?
Hysteresis Lossα fBm1.5;Eddy current lossαf2Bm2---wattss
16.What are the various methods of cooling of Turbo-alternator?
(a).Air Cooled Turbo-alternators:
(i).One sided axial ventilation :(upto 3MW).Machine is supplied with air by propeller fan and the air enters the machine from one side and leaves from the other.
(ii).Two sided axial Ventilation:(12MW).Air is forced through the machine from both sides.
(iii).Multiple inlet system:(60MW)Useful for machines having longer core lengths.Outer stator casing is divided into number of compartments, with alternative inlet(air is directed radially inwards) and outlet chambers(air is directed radially outwards).The air is drawn from the outlet chamber and is sent to the coolers where it is cooled and recirculated.
(b). Hydrogen Cooled Turbo-alternators:(more than 60MW).Hydrogen when mixed with air forms an explosive mixture over avery wide range.∴The frame of hydrogen cooled machines has to be made strong enough to withstand possible internal explosion without suffering serious damage.All joints in cooling circuits are made gas tight and oil film shaft seals are used to prevent leakage of hydrogen.
Intially,a pressure of 105kN/m2 was used .For modern conventionally cooled turbo-alternators,the pressure is about 200-300 kN/m2.Fans mounted on the rotor circulate hydrogen through the ventilating ducts and internally arranged coolers.The gas pressure is maintained by an automatic regulating and reducing valve controlling the supply from gas cylinders.
17.Define heating and cooling time constant.
Heating time constant: Time taken by the machine to attain 0.632 of its final steady temperature rise.It is an index of the machine to attain its final steady temperature rise.
θ=θm(1-e-t/Th);If t =Th ;θ=0.632θm;T=Gh/Sλ;λ=specific heat dissipation in W/m2-°C;G=weight of active part of machine inkg,h=specific heat in J/kg-°C,S=Cooling surface in m2.
Cooling time constant:θ=θie-t/Tc; If t =Tc ;θ=0.368θi;∴Cooling time constant is the time taken by the machine for its temperature rise to fall to 0.368 of its initial value.
18.Define rating of electrical machine.
The rating of machines refers to the whole of the numerical values of electrical and mechanical quantities with their duration and sequences assigned to the machines by the manufacturers and stated on the rating plate, the machine complying with the specified conditions.
19.Define Continuous, short time and intermittent short time duty of electrical machine.
(i).Continuous Duty:On this duty,the duration of load is for a sufficiently long time such that all parts of the motor attain thermal equilibrium,ie the motor attains maximum final steady temperature rise.
Continuous rating of a motor is defined as the load that may be carried by the machine for an indefinite time without the temperature rise of any part exceeding maximum permissible value.(eg)Fan,pumps.
(ii).Short time duty:The motor operates at a constant load for some specified time which is then followed by a period of rest.The period for load is so short that the machine cannot reach its thermal equilibrium,ie steady temperature rise while the period for rest is so long that the motor temperature drops to the ambient temperature.(eg).Railway turntable,navigation lock gates.
(iii).Intermittent periodic duty:On intermittent periodic duty,the periods of constant load and rest with machine de-energised alternate.The load periods are too short to allow the motor to reach its final steady state value while periods of rest are also too small to allow the motor to cool down to the ambient temperature.(e.g)Cranes,lifts,metal cutting machine.
20.Distinguish between conventionally cooled and direct cooling of Turbo-alternator.
Conventional cooling:Here machines dissipate their losses to a coolant which is entirely outside the coil.
Direct cooling:Process of dissipating the winding losses to a cooling medium circulating within the winding insulation wall.Here the Coolant is in direct contact with conductor.




PART-A-
1.Write any two similarities between magnetic and electric circuits.
2.Write any two differences between magnetic and electric circuits.
3.Write the Ohm’s law of magnetic circuit.
4.Define gap contraction factor for slots&ducts.
5.Distiguish between continuous rating and short time rating.
6.What are ventilating ducts?
7.What is Carter’s coefficient?
8.Write the expressions for gap contraction factor for slots and ducts.
9.Write the expression for reluctance of air-gap in machines with smooth armature and slotted armature.
10.What is the effect of salient poles on the air gap mmf?
11.Define field form factor.
12.What are the problems encountered in estimating the mmf for teeth?
13.List the methods used for estimating the mmf for teeth.
14.State the relation between apparent flux density and real flux density.
15.What is magnetic leakage and leakage coefficient?
16.What is fringing effect?
17.Define slot leakage reactance.
18.What is reactance voltage?
19.In which way the air gap length influence the design of machines?
20.How to find the total mmf in a series circuit?
21.What is the effect of saliency in a magnetic circuit?
22.What is the use of magnetization curve?
23.What are the major five parts of the magnetic circuit of a D.C.Machine?
24.On what factors the Hysteresis and Eddy current loss depends?
25.What are the principal components of armature leakage flux?
26.What are ventilating ducts?
27.Define permeance and specific permeance .
28.What are the advantages of Hydrogen cooling of Turbo-alternators?
29.What are the components of armature leakage flux?
30.Define rating of a electrical machine.

PART-B-
31.Derive an expression for the leakage reactance of transformer.Calculate the appropriate leakage reactance of the transformer referred to the HV side for a 750kVA, 6.6kV/415V, 50 Hz, phase,delta/star core type transformer with the following data.Width of L.V.winding=3.5cm;width of H.V.winding =2.5cm; width of duct between LV and HV=1.75cm.Height of each of winding is 50 cm. Length of
mean turn=175cm.No. of turns on H.V.side=210. [2.32Ω]
32.Explain the procedure for calculating the ampere-turns of the following parts in the design of d.c.machine. (a) air gap (b) teeth (c) armature core (d) pole(e)yoke
33.A 15 kW, 230V, 4-pole d.c.machine has the following data. armature diameter=0.25m;armature core length=0.125m;length of air gap at pole center=2.5mm; flux per pole=11.7*10-3Wb.
Polearc/polepitch=0.66.Calculate the mmf required for air gap (i) if the armature surface is treated as smooth (ii) if the armature is slotted and the gap Contraction factor is 1.18.[1443A,1700A]
34.What do you understand by fringing and Carter’s coefficient? Explain their Significance in magnetic Circuit calculations.
35.Calculate the mmf required for the air gap of a machine having core length =0.32m including 4 ducts of 10mmeach,polearc=0.19m;slotpitch=65.4mm;slot opening=5mm;airgap length=5mm;flux per pole=52mWb;Given Cartr’scoefficient is 0.18 for opening/gap=1 and is 0.28 for opening/gap=2.[3587A]
36.Estimate the effective air gap area per pole of a 10 pole slip ring induction motor with the following data: Stator bore=0.65m,Core length=0.25m,No.of stator slots=90,stator slot opening=3mm,rotor slots=120,rotor slot opening=3mm,airgap length=0.95mm.Carter’s coefficient for slots=0.46, Carter’s co-efficient for ducts=0.68 number of ventilating ducts=3 each on rotor and stator, width of each
ventilating duct=10mm.[40.52*10-3m2]
37.Determine the air gap length of a d.c. machine from the following data: Gross core length =0.12m;No.of ducts=One of 10mm width; Slot pitch=25mm;slot width=10mm;Carter’s coefficient for slots and ducts=0.32;Gap density at polecenter=0.7T;Field mmf per pole=3900A;mmf required for iron parts of magnetic circuit=800A. [4.7mm]
38.Describe the effect of slot opening on the flux distribution in the air gap of an electrical machine
39.Calculate the AT per pole required for the air gap of a machine having the following data.
Core length =20cm,pole arc=18.5cm,slot pitch=2.7cm.slot opening=1.2cm,core flux per
pole=0.025 Wb, length of the air gap=0.6cm.
40.Calculate the apparent flux density at a section of the teeth of an armature of a d.c.machine from the following data at that section.Slot pitch=24mm;slot width=tooth width=12mm;length of armature core including 5 ducts 10 mm each=0.38m,iron stacking factor=0.92.True flux density at that section is 2.2 Wb/m2 for which mmf is 70,000A/m.[2.332T]
41.Derive an expression for the leakage rectance of (i)Parallel sided semi enclosed slot (ii).Tapered slot.Determine the leakage permeance per metre length of a rectangular semi-enclosed slot having the following dimensions (all in mm) slot width=10;slot opening=4.5;height of conductor portion=26;height above conductor and below wedge =1;wedge height 3.5;lip height=1.5.[1.78*10-6 ]
42.Discuss the various methods of cooling of Turbo-alternator.
43.Write brief notes on the various methods of calculation of mmf for teeth.
44.Calculate the apparent flux density at a particular section of a tooth from the following data: Tooth width=12mm;slot width=10mm;gross core length=0.32;number of ventilating duct=4,each 10mm wide; real flux density=2.2Wb/m2; permeability of teeth corresponding to real flux ensity=31.4*10-6H/m; stacking factor=0.9.[2.371Tesla]
45.Determine the apparent flux density in the teeth of a d.c.machine when the real flux density is 2.15 Wb/m2; slot pitch =28mm;slot width 10mm and the gross core length 0.35m.The number of ventilating duct is 4,each 10mm wide. The magnetizing force for a flux density of 2.15Wb/m2is 55000A/m.The iron stacking factor is 0.9.[2.2156Wb/m2]
46.Write short notes on various types of armature leakage flux.
47.Discuss the various types of ventilation system for the cooling of rotating electrical machines.
48.Define rating of a machine.What do you understand by continuous,short-time and short-time intermittent rating of an electrical machines?
49..Explain the terms ‘continuous duty’, ‘short time duty’,and intermittent periodic duty as applied to electrical machines.Sketch the typical load,electrical losses and temperature rise versus time curves for these ratings to illustrate the answers.
50.Write a short notes on methods of estimation of motor rating for variable load drives?
51.Derive an expression for temperature rise of an Electrical machine.What is heating time constant?what is cooling time constant?
52.The temperature rise of a transformer is 25ºC after one hour and 37.5ºC after two hours of starting from cold conditions.Calculate its final steady temperature rise and the heating time constant.If its temperature falls from the final steady value to 40ºC,in 1.5 Hours when disconnected,calculate its cooling time constant.The ambient temperaure is 30ºC.(Refer Page No.80,EMD,Sawhney)
53.A 400 kVA transformer has its maximum efficiency at 80% of full load.During a short full load heat run the temperature rise after one hour and two hours is observed to be 24ºC and 34ºC, respectively. Find the thermal time constant and final steady temperature rise of the transformer.If,by use of a fan, the cooling is improved so that rate of heat dissipation per unit area per degree rise in temperature is increased by 15 % find the new kVA rating possible(a) for the same final temperature rise as before
(b) if the allowable temperature rise taken as 50ºC.( Refer Page No.81,EMD,Sawhney)


















ASSIGNMENT-I-
A.1.1.Determine the mmf required for the air gap of a d.c.machine having open slots, given the following data:Slot pitch =4.3cm,Gross core length =48cm,Airgap length=0.6cm,slot opening=2.1cm,pole arc=18cm,Flux per pole=0.056Wb.There are 8 ventlating ducts each1.2cm wide.
Slot opening/Gap length 1 2 3 3.5 4
Cater’s Coefficient 0.15 0.28 0.37 0.41 0.43
The above data may be used for ducts also.[Ans.kgs=1.25,kgd=1.059,Bg=0,65T,ATg=4,120AT]
2.Estimate the effective gap area per pole of a 12-pole induction motor with a bore of 90cm,core length 25cm,108stator slots with 3mm opening,144 slots with 2mm opening and gap length 1mm.The cater’s co-efficient is given by ,
Slot opening/Gap length 2 4 6 8 10
Cater’s Coefficient 0.28 0.45 0.55 0.61 0.66
[Ans.kgs(s)=1.044, kgs(r)=1.0294,kg=1.075,Ag’=548cm2]
3.Find the apparent tooth density at a section of the tooth in the following case when the real tooth density at that section is 2.15Wb/m2.Gross armature length 32cms,Number of ventilating ducts=4,each 1cm wide,tooth width at that section =1.2cm,slot width with parallel sides=1cm.Permeability of the teeth corresponding to tooth density=35.8X10-6.[Bapp=2.225Wb/m2]