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TRANSFORMERS:
→ Gross cross sectional area = Area occupied by magnetic material + Insulation
material.
→ Net cross sectional area = Area occupied by only magnetic material excluding area
of insulation material.
→ Hence for all calculations, net cross sectional area is taken since ϕ (flux) majorly
flows in magnetic material.
ϕ = BAn
t
→ Specific weight of t/f = f
K
We
VA
i g
ra
ht
ti
of
n g off
t
→ Stacking/iron factor :- (ks) =
Net Cross Sectional area
Gross Cross Sectional area
Total C.S Area
→ ks is always less than 1
→ Gross c.s Area = AG = length × breadth
→ Net c.s Area = An = ks × AG
→ Utilization factor of transformer core =
Effective C.S.Area
U.F of cruciform core = 0.8 to
→ Flux =
mmF
Reluctance
= = ϕm sin ωt
→ According to faradays second law e 1 = −N 1
dϕ
dt
= −N 1
d
dt
�ϕm sin ωt�
→ Transformer emf equations :-
E 1 = 4.44 N 1 Bmax Anf (1)
E 2 = 4.44 N 2 Bmax Anf (2)
N 1
→ Emf per turn in Iry^ =
E 1
= 4.44 BmaxAnf
N 2
→ Emf per turn in IIry^ =
E 2
= 4.44 Bmax An f
⟹ Emf per turn on both sides of the transformer is same
E 1
N 1 N 2
E 2
E 1
E 2
N 1
N 2
k
Transformation ratio = K =
E 2
E 1
Instantaneous value
of emf in primary
e 1 = N 1 ϕm ω sin�ωt − π� 2 �
N 2
N 1
Turns ratio =
K
= N 1 ∶ N 2
gradeup ELECRIAL MACHINES (Formula Notes)
→ For an ideal two-winding transformer with primary voltage V 1 applied
across N 1 primary turns and secondary voltage V 2 appearing across N 2 secondary turns: V 1 / V 2 = N 1 / N 2
→ The primary current I 1 and secondary current I 2 are related by:
I 1 / I 2 = N 2 / N 1 = V 2 / V 1
→ For an ideal step-down auto-transformer with primary voltage V 1 applied
across (N 1 + N 2 ) primary turns and secondary voltage V 2 appearing across N 2 secondary turns: V 1 / V 2 = (N 1 + N 2 ) / N 2
→ The primary (input) current I 1 and secondary (output) current I 2 are related by:
I 1 / I 2 = N 2 / (N 1 + N 2 ) = V 2 / V 1.
→ For a single-phase transformer with rated primary voltage V 1 , rated primary current I 1 , rated secondary voltage V 2 and rated secondary current I 2 , the voltampere rating S is: S = V 1 I 1 = V 2 I 2
→ For a balanced m-phase transformer with rated primary phase voltage V 1 , rated
primary current I 1 , rated secondary phase voltage V 2 and rated secondary current I 2 , the voltampere rating S is: S = mV 1 I 1 = mV 2 I 2
→ The primary circuit impedance Z 1 referred to the secondary circuit for an ideal
transformer with N 1 primary turns and N 2 secondary turns is: Z 12 = Z 1 (N 2 / N 1 )^2
→ During operation of transformer :-
Bm ∝
E 1
f f
∝
V 1
Bmax = constant ⟹
V 1
f
= constant
Equivalent ckt of t/f under N.L condition :-
No load /shunt branch.
N 1 N 2
V 1 E 1 E 2
I 0
Iw
R 0 X 0
Iμ
1. Cu losses in t/f:
Total Cu loss = I 12 R 1 + I 22 R 1
= I 12 R 01
E 1
= I 22 R 02
→ Rated current on Iry^ =
VA rating of t/f
E 2
Similarly current on IIry^ =
VA rating of t/f
VA rating of t/f
→ Cu losses ∝ I 12 or I 22. Hence there are called as variable losses.
→ P.U Full load Cu loss =
FL Cu loss in watts
I 12 R 01
E 1 I 1
→ If VA rating of t/f is taken as base then P.U Cu loss ∝ I 12 as remaining terms are constant.
→ P.U Cu loss at x of FL = x^2 × PU FL Cu loss
P. U resistance drop ref to Iry
or
P. U resistance ref to Iry
I 1 R 01
E 1 ×^ I 1
I 1
I 12 R 01
E 1 I 1
∴ P.U Resistance drop = P.U FL cu loss
% FL Cu loss = % R = % Resistance drop.
Iron (or) Core losses in t/f :-
1. Hysteresis loss :
Steinmetz formula :-
Where
η = stienmetz coefficient
Bmax = max. flux density in transformer core.
f = frequency of magnetic reversal = supply freq.
v = volume of core material
x = Hysteresis coeff (or) stienmetz exponent
= 1.6 (Si or CRGo steel)
2. Eddycurrent loss:
Eddy current loss ,(We) ∝ Rce × I e^2
As area decreases in laminated core resistance increases as a result conductivity decreases.
We = K. Bma^2 x f 2. t^2
Constant
Supply freq
thickness of laminations.
Area under one hysteresis loop.
x
Wh = η Bmax. f. v
(it is a function of σ )
During operation of transformer :-
Bm ∝
V 1
f
f
Case (i) :-
V 1
= constant, Bmax = const.
we ∝ f 2
we = B f 2
Const.
∴ wi = wh + we
wi = Af + Bf 2
When Bmax = const.
Case (ii) :-
V 1
f
≠ constant, Bm ≠ const.
we ∝ �
V 1
f
. f 2
we ∝ V 12
wi = wh + we
wi =
A V
1 6
- 6
f + BV^1
VA rating of t/f
P.U iron loss :-
→ P.U iron loss =
Iron loss in watts
→ As VA rating is choosen as base then the P.U iron loss are also constant at all load conditions.
To find out constant losses :-
W 0 = Losses in t/f under no load condition
= Iron losses + Dielectric loss + no load primary loss (I 02 R 1 )
Constant losses = W 0 − I 02 R 1
Where , R 1 = LV winding resistance.
To find out variable losses :-
Wsc = Loss in t/f under S.C condition
= F.L Cu loss + stray load losses (Cu and Iron) + Iron losses in both wdgs
Variable losses = WSC − Iron losses corresponding to VCC
O.C test :-
V 1 rated → Wi
S.C test :-
VSC → (Wi)S.C
Wi ∝ V 12
W (^) i
(W i)SC^ =^ �
V 1 rated VSC
(Wi)S.C = Wi × �
VSC
V 1 rated�
∴ Variable losses = WSC − (Wi)SC �
VSC
V 1 rated
AUTO TRANSFORMER :
→ Primary applied voltage, Vab = Secondary voltage V 2 referred to primary + primary leakage impedance
drop + secondary leakage impedance drop ref. to primary.
Vab = �
N 1 −N 2
N 2 �^ V^2 + I^1 (r^1 + jx^1 ) + (I^2 −^ I^1 )(r^2 + j x^2 )^ �
N 1 − N 2
N 2 �
→ K of auto transformer =
LV
HV
(KVA)induction = (V 1 − V 2 ) I 1
I/P KVA = V 1 I 1
(KVA)induction
i/p KVA =^ V 1 I 1
(V 1 − V 2 ) I 1
HV
LV
= 1 – K
∴ (KVA) induction = (1 – K) i/p KVA
(KVA) conduction = I/p KVA – (KVA)ind
(KVA)conduction = K × I/p KVA
→ Wt. of conductor in section AB of auto t/f ∝ (N 1 − N 2 ) I 1
→ Wt of conductor in section BC of auto t/f ∝ (I 2 − I 1 )N 2
∴ Total wt. of conductor in auto t/f is
∝ I 1 (N 1 − N 2 ) + (I 2 − I 1 )N 2
∝ 2 (N 1 − N 2 ) I 1
→ Total wt. of conductor in 2 wdg transformer
∝ I 1 N 1 + I 2 N 2
∝ 2 I 1 N 1
wt.of conductor in an auto t/f wt.of conductor in 2 wdg t/f 2N 1 I 1
2 (N 1 −N 2 )I 1
= 1 – N 1
N 2
= 1 – K
Wt. of conductor in auto t/f = (1 – K) (wt. of conductor in 2 wdg t/f)
→ Thus saving of conductor material if auto – t/f is used} = K × {conductor wt in 2 wdg transformer.
→ (% FL losses)Auto t/f = (1 − K)(% FL losses) 2 wdg t/f
→ (% Z)AT = (1 − K) (% Z) 2 wdg t/f
→ (KVA)AT =
1−K
(KVA) 2 wdg t/f.
DC MACHINES :
Lap Winding Wave Winding (1) Coil Span : (^) Ycs = S
P
Ycs =
S
P
(2) Back Pitch Yb = U Ycs Yb = U Ycs (3) Commutator Pitch Yc = 1 for Progressive winding Yc = - for Retrogressive winding
Yc =
2 (c+1)
p
for Progressive winding Yc = -
2 (c+1)
p
for Restrogressive winding (YcMust be integer) (4) Front Pitch Yf =Yb + for Progressive winding Yf =Yb - for Retrogressive winding
Yf =2Yc - Yb
(5) Parallel Paths A = P A = 2 (6) Conductor Current (^) Ic =I^ a
A
Ic =
I a
(7) No of brushes No of brushes = A = P No of brushes = 2
- S = No of commutator segments
- P = No of poles
- U = No of coil sides / No of poles =
2C
S
- C = No of coils on the rotor
- A = No of armature parallel paths
- Ia = Armature current
→ Distribution factor (Kd) =
phasor sum coil emf
arthematic sum of coil emf
=
chord
arc
=
→ Pitch factor ( Kp) =
elecrrical angle of coil
*100%
→ θ (^0) electrical
=
P
θ (^0) mechanical
→ Armature mmf/Pole (Peak) , ATa =
ZI a
2AP
2AP
→ AT (Compensating Winding) =
ZI a
pole arc
pole pitch
→ AT(Inter pole) = ATa +
Bi
lgi
Where Bi= Flux density in inter pole airgap
lgi = length of inter pole airgap , μ 0 = 4π ∗ 10 −
→ No of turns in each interpole , Ninterpole =
AT(Inter pole)
I a
→ The no of compensating conductor per pole, Ncw/pole =
Z
2 A P
(
pole arc
pole pitch
)
→ The Mechanical power that is converted is given by Pconv = Tind ωm Where T = Induced torque (^) ω m = Angular speed of the machines rotor
→ The airgap power P e for a series generator is: P e = ωT = E aI a = kmω I a^2 → Cumulatively compounded DC generator : - ( long shunt) (a) Ia = If + IL (b) Vt = Ea - Ia (Ra + Rs )
(c) Isf =
Vx
R f
= shunt field current
(d) The equivalent effective shunt field current for this machine is given by
Isf =Isf +
N se
I - (
Armature reaction MMF
N f a^ N f
)
Where Ns e = No of series field turns Nf = = No of shunt field turns
→ Differentially compounded DC generator : - ( long shunt) (a) Ia = If + IL (b) Vt = Ea - Ia (Ra + Rs )
(c) Isf =
Vx
R f
= shunt field current
(d) The equivalent effective shunt field current for this machine is given by
Isf =Isf -
N se
N f
Ia - (
Armature reaction MMF
N f
)
Where Ns e = No of series field turns , Nf = = No of shunt field turns
Shunt Motor:
→ For a shunt motor with armature induced voltage E a, armature current I a and armature resistance Ra, the terminal voltage V is: V = E (^) a + I (^) aRa The field current I f for a field resistance R f is: I f = V / R f → The armature induced voltage E a and torque T with magnetic flux Φ at angular speed ω are: E a = k fΦω = kmω T = k fΦI a = km I a where k (^) f and km are design coefficients of the machine.
Note that for a shunt motor:
- induced voltage is proportional to speed,
- torque is proportional to armature current.
→ The airgap power P e for a shunt motor is: P e = ωT = E aI a = kmω I a → The speed of the shunt motor , ω =
V
K∅
R a
(K∅)^2
T (^) Where K =
PZ
2πA
Series Motor :
→ For a series motor with armature induced voltage E a, armature current I a, armature resistance Ra and field resistance R (^) f , the terminal voltage V is:
V = E (^) a + I (^) aRa + I (^) aR (^) f = E (^) a + I (^) a(Ra + R (^) f )
The field current is equal to the armature current. → The armature induced voltage E a and torque T with magnetic flux Φ at angular speed ω are: E (^) a = k (^) fΦω I (^) a = k (^) mω I (^) a T = k (^) fΦI (^) a^2 = km I (^) a^2 where k (^) f and km are design coefficients of the machine.
Note that for a series motor:
- induced voltage is proportional to both speed and armature current,
- torque is proportional to the square of armature current,
- armature current is inversely proportional to speed for a constant E a
→ The airgap power P (^) e for a series motor is: P e = ωT = E aI a = kmω I a^2
Losses:
→ constant losses (P^ k) =^ Pw f +^ P^ i o
Where ,^ Pio = No of load core loss
→ Pwf = Windage & friction loss → Variable losses (Pv) = Pc + Ps t + Pb
where Pc= Copper losses =^ Ia (^2) R
a
Ps t = Stray load loss = α I^2
Pb = Brush Contact drop = VbIa , (^) Where Vb = Brush voltage drop
→ The total machine losses , PL = Pk +VbIa+ Kv Ia^2
Efficiency
→ The per-unit efficiency η of an electrical machine with input power P in, output power P (^) out and power loss P (^) loss is:
η = P out / P in = P out / (Pout + P loss ) = (P in - Ploss ) / P in
→ Rearranging the efficiency equations:
Dielectric Dissipation Factor:
→ If an alternating voltage V of frequency f is applied across an insulation system comprising capacitance C and equivalent series loss resistance RS , then the voltage V (^) R across RS and the voltage V (^) C across C due to the resulting current I are: V (^) R = IRS V C = IXC V = (V R^2 + V C^2 )½
→ The dielectric dissipation factor of the insulation system is the tangent of the dielectric loss angle δ between V C and V: tanδ = V (^) R / V (^) C = RS / XC = 2πfCRS RS = XCtanδ = tanδ / 2πfC
→ The dielectric power loss P is related to the capacitive reactive power Q C by: P = I 2 RS = I 2 XCtanδ = Q (^) Ctanδ
→ The power factor of the insulation system is the cosine of the phase angle φ between V (^) R and V: cosφ = V (^) R / V so that δ and φ are related by: δ + φ = 90°
→ tanδ and cosφ are related by: tanδ = 1 / tanφ = cosφ / sinφ = cosφ / (1 - cos^2 φ) ½ so that when cosφ is close to zero, tanδ ≈ cosφ
SYNCHRONOUS MACHINES:
→ Principle of operation :-
Whenever a conductor cuts the magnetic flux, an emf is induced in that conductor”
Faraday’s law of electromagnetic induction.
→ Coil span (β) :- It is the distance between two sides of the coil. It is expressed in terms of
degrees, pole pitch, no. of slots / pole etc
→ Pole pitch :- It is the distance between two identical points on two adjacent poles.
Pole pitch is always 180° e = slots / pole.
→ θelec =
P
2 θmech
→ Slot pitch or slot angle :- (T)Slot angle is the angle for each slot.
→ For a machine with ‘P’ poles and ‘s’ no. of slots, the slot angle = γ =
P( 180 °)
d
→ Pitch factor or coil span factor or chording factor :- (KP)
KP =
The emf induced | coil in short pitched winding The emf induced |coil in full pitched winding
= Arithmetic sumof induced emf | coil
The vector sum of induced emf | coil
KP =
2E cos∝/ 2 2E
→ chording angle to eliminate nth^ harmonics (α)=
n
→ coil spam to eliminate nth^ harmonics ,(β) = 180 �
n−
n
→ Distribution factor | spread factor | belt factor | breadth factor(kd) :-
Kd =
The emf induced when the winding is distributed The emf induced when the winding is concentrated
Kd =
Vector sum of emf induced Arithmetic sum of emf induced
� �p
s
K p n =
cos n ∝ 2
Kp = cos ∝/
→ Pitch factor for nth harmonic i.e,
m γ
Is kd^2
1 =^
s m
in sinγ 2
i.e., same that of fundamental
→ Pith factor for slot harmonics, kp �
2s
p ± 1�^ =^ kp^1 = cos^
→ The synchronous speed Ns and synchronous angular speed of a machine with p pole
pairs running on a supply of frequency f s are:
ωs = 2πf s / p
N S
→ Slip S =
N S − N
Where NS =
120 f
= synchronous speed
p → The magnitude of voltage induced in a given stator phase is Ea= √ 2 π Nc ∅ f = K∅ω Where K = constant
→ The output power Pm for a load torque Tm is:
Pm = ωs T m
→ The rated load torque T (^) M for a rated output power P (^) M is:
T (^) M = P (^) M / ωs = P (^) M p/ 2πf (^) s = 120P (^) M / 2πNs
Synchronous Generator :
→ For a synchronous generator with stator induced voltage Es, stator current Is and
synchronous impedance Zs, the terminal voltage V is:
V = E - Is Zs = E (^) s - I (^) s (Rs + jXs )
where Rs is the stator resistance and X s is the synchronous reactance
E = �(V cos ϕ + Ia Ra)^2 + (V sin ϕ ± Ia Xs)^2
+ ⇒ lag p.f
− ⇒ leading p.f.
Synchronous Motor:
→ For a synchronous motor with stator induced voltage Es, stator current I s and synchronous
impedance Zs, the terminal voltage V is:
V = E (^) s + I (^) s Zs = Es + Is (Rs + jXs )
where Rs is the stator resistance and Xs is the synchronous reactance
Voltage regulation :
→ % regulation =
|E|− |V|
|V|
× 100
V
E – V = IaZs
∴ % regulation =
E − V
Ia Zs
V ×^100
∴ regulation ∝ Zs
∴ As Zs increases, voltages regulation increases.
→ Condition for zero | min. voltage regulation is, Cos (θ + ϕ) = −
Ia Zs 2V
→ Condition for max. Voltage regulation is, ϕ = θ
→ Short circuit ratio (SCR) =
Ifm
Ifa^ =^
Zs(adjusted)|unit^ =^
Xs(adjusted)|unit
SCR ∝
Xa^ ∝^
Armature reaction
Voltage regulation ∝ Armature reaction
∴ SCR ∝
Voltage regulation
∴ Small value of SCR represent poor regulation.
ϕa =
armature mmf reluctance
But reluctance ∝ Air gap
∴ ϕa =
armature mmf airgap
ϕa ∝
Air gap length
Armature reaction ∝ ϕa ∝
Airgap length
→ Cylindrical rotor Synchronous machine ,
The per phase power delivered to the infinite bus is given by P =
E (^) f Vt
Xs^ sin δ
→ Salient pole synchronous machine ,
The per phase power delivered to the infinite bus is given by
Xd 2 Xq
P = Xd
E (^) f Vt
sin δ +
Vt 2
− � sin 2δ
Condition for max. power: -
→ For cylindrical rotor machine :-
At constant Vt and Ef, the condition for max. power is obtained by putting
dp dδ
dδ
dp
E (^) f Vt Xs
cos δ = 0
Cos δ = 0
Hence maximum power occurs at δ = 90°
→ For salient – pole synchronous machine :-
dp dδ
Vt E (^) f
cos δ + Vt 2 �
Xq−^
Xd Xd�^ cos^ 2δ^ = 0
Cos δ = −
E (^) f Xq
4Vt �Xd − Xq � ±^
�^1
2 +^ �^
E (^) f Xq
4Vt �Xd − Xq ��
∴ max. power occurs at δ < 90°
→ Synchronizing power = Psy. ∆ δ.
EV
Xs^ cos^ 𝛿𝛿^.^ ∆^ 𝛿𝛿^.
→ Synchronizing torque =
Synchronizing power
Power flow in Alternator :-
→ Complex power = S = P + jQ = VI a∗^ Where Active power flow (P) =
EV
Zs^ cos(θ −^ δ)^ −^
V^2
Zs^ cos^ θ^ ;
The value of load angle is seed to be less than 90°.
Reactive power flow (Q) =
EV
Z 2 sin(θ −^ δ
V^2
Zs
sin θ ;
→ Condition for max. power output :-
P =
EV
Zs^ cos(θ −^ δ)^ −^
V^2
Z 2 cos^ θ
dp
dδ = 0 for max power condition
If Ra = 0; θ = δ = 90° ; then max power is given by
ie θ – δ = 0
Pmax =
EV
Zs
V^2
Zs
cos θ
INDUCTION MACHINES:
→ The power flow diagram of 3 – ϕ induction motor is
ns
The slip of induction machine is (S) =
ns − nr
N (^) s − Nr N (^) s
Where Ns is synchronous speed in rpm
ns is synchronous speed in rps
⇒ Nr = Ns(1 − s)
⇒ Ns − Nr = SNs
∴ Rotor frequency, f 2 =
P. SNs 120
= S
PN (^) s 120
= Sf 1
For an induction machine with rotor resistance Rr and locked rotor leakage reactance X r, the
rotor impedance Zr at slip s is:Zr = Rr + jsX r
The stator circuit equivalent impedance Zrf for a rotor / stator frequency ratio s is:
Zrf = Rrs / s + jX rs
Rotor emf, Current Power :-
At stand still, the relative speed between rotating magnetic field and rotor conductors is
synchronous speed Ns; under this condition let the per phase generated emf in rotor circuit
be E 2.
∴ E 2 /ph = 4.44 Nphr ϕ 1 f 1 Kdr Kpr
E 2 /ph = 4.44 Nphr ϕ 1 f 1 Kwr
Mechanical power Pg developed, Pm
Rotor i/p power = airgap power
Power i/p to stator from mains
Power of rotor shaft
Windage loss
Friction loss at bearings and sliprings of (if any)
Rotor core loss (negligible for small slips)
Rotor I^2 R loss
Stator core loss
Stator I^2 R loss
For an induction motor with synchronous angular speed ωs running at angular speed ωm and
slip s, the airgap transfer power Pt, rotor copper loss Pr and gross output power Pm for a
gross output torque T m are related by:
Pt = ωsT m = Pr / s = Pm / (1 - s)
Pr = sPt = sPm / (1 - s)
Pm = ωm T m = (1 - s)Pt
The power ratios are:
Pt : Pr : Pm = 1 : s : (1 - s)
The gross motor efficiency ηm (neglecting stator and mechanical losses) is:
ηm = Pm / Pt = 1 - s
