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 = BA n t f → Specific weight of t/f = K We VA ig ra ht ti of n f g of t → Stacking/iron factor :- (k s ) = Net Cross Sectional area Gross Cross Sectional area Total C.S Area → k s is always less than 1 → Gross c.s Area = A G = length × breadth → Net c.s Area = A n = k s × A G → Utilization factor of transformer core = Effective C.S.Area U.F of cruciform core = 0.8 to 0.85 → Flux = mmF Reluctance = = ϕ m sin ωt → According to faradays second law e 1 = −N 1 dϕ dt = −N 1 d dt �ϕ m sin ωt� → Transformer emf equations :- E 1 = 4.44 N 1 B max A n f (1) E 2 = 4.44 N 2 B max A n f (2) N 1 → Emf per turn in I ry = E 1 = 4.44 B max A n f N 2 → Emf per turn in II ry = E 2 = 4.44 B max A n 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 = 1 k Transformation ratio = K = E 2 E 1 = Instantaneous value of emf in primary e 1 = N 1 ϕ m ω sin�ωt − π � 2 � N 2 N 1 Turns ratio = 1 K = N 1 ∶ N 2 ELECRIAL MACHINES (Formula Notes) gradeup gradeup 1
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gradeup ELECRIAL MACHINES (Formula Notes) · gradeup ELECRIAL MACHINES (Formula Notes) gradeup 1 → For an ideal two-winding transformer with primary voltage V1 applied across N1
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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
tf→ Specific weight of t/f =
K
We
VA
i
g
ra
ht
t
i
of
n fg oft
→ Stacking/iron factor :- (ks) = Net Cross Sectional areaGross 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
→ For an ideal two-winding transformer with primary voltage V1 applied across N1 primary turns and secondary voltage V2 appearing across N2 secondary turns: V1 / V2 = N1 / N2
→ The primary current I1 and secondary current I2 are related by: I1 / I2 = N2 / N1 = V2 / V1
→ For an ideal step-down auto-transformer with primary voltage V1 applied across (N1 + N2) primary turns and secondary voltage V2 appearing across N2 secondary turns: V1 / V2 = (N1 + N2) / N2
→ The primary (input) current I1 and secondary (output) current I2 are related by: I1 / I2 = N2 / (N1 + N2) = V2 / V1.
→ For a single-phase transformer with rated primary voltage V1, rated primary current I1, rated secondary voltage V2 and rated secondary current I2, the voltampere rating S is: S = V1I1 = V2I2
→ For a balanced m-phase transformer with rated primary phase voltage V1, rated primary current I1, rated secondary phase voltage V2 and rated secondary current I2, the voltampere rating S is: S = mV1I1 = mV2I2
→ The primary circuit impedance Z1 referred to the secondary circuit for an ideal transformer with N1 primary turns and N2 secondary turns is: Z12 = Z1(N2 / N1)2
→ If VA rating of t/f is taken as base then P.U Cu loss ∝ I 1 2 as remaining terms are constant.
→ P.U Cu loss at x of FL = x2 × PU FL Cu loss
→ P. U resistance drop ref to Iry
or P. U resistance ref to Iry
� = I1R01E1 I1
× I1
= I 1 2R01E1I1
∴ 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. B ma
2x f 2. t2
Constant Supply freq
thickness of laminations.
Area under one hysteresis loop. xη B max . f . v Wh =
→ Under the assumption that small amount of iron losses corresponds to VSC and stray load losses are neglected the wattmeter reading in S.C test can be approximately taken as F.L Cu losses in the transformer.
→ Wse ≃ F.L Cu loss ≃ I
S C 2 . R01
R01 = I
W
SC2
SC
Efficiency :-
Efficiency of transformer is given by η = output powerinput power
= output poweroutput power+losses
= E2 I2 cos ϕ2E2 I2 cos ϕ2+ F.L cu losses+Iron losses
ηF.L = E2 I2 cos ϕ22E2I2 cos ϕ2+ I2 R02 + Wi
→ Transformer efficiency = KVA × cos ϕ KVA × cos ϕ + wi + Cu losses
ηx of F.L = x (E2 I2) cos ϕ2
x (E2 I2) cos ϕ2 + x2 (I 2 2 R02) + Wi
O.C test S.C test
V 1 ′
→ Voltage drop in t/f at a Specific load p.f = I2R02 cos ϕ2 ± I2X02 sin ϕ2
→ % Voltage regulation = I2 R02 cos ϕ2 ± I2X02 sin ϕ2 ×100
= �I2 VR
′02
1� cos ϕ2 ± �I2 X02
V 1 ′ � sin ϕ2
↓ P.U resistance
↓ P.U reactance
% Regulation = �(P. U R) cos ϕ2 + (P. U X) sin ϕ2� × 100
→ Primary applied voltage, Vab = Secondary voltage V2 referred to primary + primary leakage impedance drop + secondary leakage impedance drop ref. to primary.
→ The resulting electric power produced Pconv = EA IA → The power balance equation of the DC Machine is Tind ωm = EA IA
→ The induced emf in the armature is Ea = ∅ZNP60A
→ Torque developed in Dc machine , Te = PZ2πA
∅ Ia
Where ∅ = Flux\pole , Z = No of armature conductors , P = No of poles , N = Speed in rpm ,
A = No of armature parallel paths, Ia = Armature current → The terminal voltage of the DC generator is given by Vt = Ea - Ia Ra → The terminal voltage of the DC motor is given by Vt = Ea + Ia Ra
ωfl → Speed regulation of dc machine is given by ,SR = ωn
l− ωfl * 100 % =
Nnl− NflNfl
* 100 %
→ Voltage regulation , VR = Vnl− Vfl
Vfl* 100 %
Shunt Generator: → For a shunt generator with armature induced voltage Ea, armature current Ia and
armature resistance Ra, the terminal voltage V is: V = Ea - IaRa
→ The field current I f for a field resistance R f is: I f = V / R f
→ The armature induced voltage Ea and torque T with magnetic flux Φ at angular speed ω are: Ea = k fΦω = kmω T = k fΦIa = kmIa where k f and km are design coefficients of the machine.
Note that for a shunt generator: - induced voltage is proportional to speed, - torque is proportional to armature current.
→ The airgap power Pe for a shunt generator is: Pe = ωT = EaIa = kmω Ia
Series Generator:
→ For a series generator with armature induced voltage Ea, armature current Ia, armature resistance Ra and field resistance R f, the terminal voltage V is: V = Ea - ( IaRa + IaR f )= Ea - Ia(Ra + R f) The field current is equal to the armature current.
→ The armature induced voltage Ea and torque T with magnetic flux Φ at angular speed ω are: Ea = k fΦω Ia = kmω Ia T = k fΦIa2 = kmIa2 where k f and km are design coefficients of the machine.
Note that for a series generator: - 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 Ea
→ The airgap power Pe for a series generator is: Pe = ωT = EaIa = kmω Ia2
→ Cumulatively compounded DC generator : - ( long shunt) (a) Ia = If + IL (b) Vt = Ea - Ia (Ra + Rs )
(c) Isf = VxRf
= shunt field current
(d) The equivalent effective shunt field current for this machine is given by
Isf =Isf + Nse I - (Armature reaction MMFNf
a Nf)
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 = VxRf
= shunt field current
(d) The equivalent effective shunt field current for this machine is given by
Isf =Isf - NseNf
Ia - (Armature reaction MMFNf
)
Where Ns e = No of series field turns , Nf = = No of shunt field turns
Shunt Motor:
→ For a shunt motor with armature induced voltage Ea, armature current Ia and armature resistance Ra, the terminal voltage V is: V = Ea + IaRa The field current I f for a field resistance R f is: I f = V / R f
→ The armature induced voltage Ea and torque T with magnetic flux Φ at angular speed ω are: Ea = k fΦω = kmω T = k fΦIa = kmIa 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 Pe for a shunt motor is: Pe = ωT = EaIa = kmω Ia
→ For a series motor with armature induced voltage Ea, armature current Ia, armature resistance Ra and field resistance R f, the terminal voltage V is:
V = Ea + IaRa + IaR f = Ea + Ia(Ra + R f)
The field current is equal to the armature current. → The armature induced voltage Ea and torque T with magnetic flux Φ at angular
speed ω are: Ea = k fΦω Ia = kmω Ia T = k fΦIa2 = kmIa2 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 Ea
→ The airgap power Pe for a series motor is: Pe = ωT = EaIa = kmω Ia2
Losses:
→ constant losses (P k) = Pw f + Pi o
Where, Pio = No of load core loss
→ Pwf = Windage & friction loss → Variable losses (Pv) = Pc + Ps t + Pb
where Pc= Copper losses = Ia2 Ra
Ps t = Stray load loss = α I2
Pb = Brush Contact drop = VbIa , Where Vb = Brush voltage drop
→ The total machine losses , PL = Pk +VbIa+ Kv Ia2
Efficiency
→ The per-unit efficiency η of an electrical machine with input power Pin, output power Pout and power loss Ploss is:
η = Pout / Pin = Pout / (Pout + Ploss) = (Pin - Ploss) / Pin
→ The resistance of copper and aluminium windings increases with temperature, and the relationship is quite linear over the normal range of operating temperatures. For a linear relationship, if the winding resistance is R1 at temperature θ1 and R2 at temperature θ2, then:
R1 / (θ1 - θ0) = R2 / (θ2 - θ0) = (R2 - R1) / (θ2 - θ1) where θ0 is the extrapolated temperature for zero resistance.
→ The ratio of resistances R2 and R1 is: R2 / R1 = (θ2 - θ0) / (θ1 - θ0)
→ The average temperature rise ∆θ of a winding under load may be estimated from measured values of the cold winding resistance R1 at temperature θ1 (usually ambient temperature) and the hot winding resistance R2 at temperature θ2, using: ∆θ = θ2 - θ1 = (θ1 - θ0) (R2 - R1) / R1
→ 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 VR across RS and the voltage VC across C due to the resulting current I are: VR = IRS VC = IXC V = (VR2 + VC2)½
→ The dielectric dissipation factor of the insulation system is the tangent of the dielectric loss angle δ between VC and V: tanδ = VR / VC = RS / XC = 2πfCRS RS = XCtanδ = tanδ / 2πfC
→ The dielectric power loss P is related to the capacitive reactive power QC by: P = I2RS = I2XCtanδ = QCtanδ
→ The power factor of the insulation system is the cosine of the phase angle φ between VR and V: cosφ = VR / 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 - cos2φ)½ so that when cosφ is close to zero, tanδ ≈ cosφ
∴ Large value of SCR represent more power output. → Synchronizing power coefficient or stability factor Psy is given as
dδd
dδ�EV
XsPsy = dp = sin δ�
= EVXs
cos δ
Psy is a measure of stability ∴ stability ∝ Psy
But Psy ∝ 1Xs
∝ SCR
Stability ∝ SCR ∝ Air gap length
→ When the stator mmf is aligned with the d – axis of field poles then flux ϕd perpole is set up
minimum current √3 Ia(min )
and the effective reactance offered by the alternator is Xd.
Xd = maximum Voltage = (Vt)line (at min. Ia ) = Direct axis reactance
→ When the stator mmf is aligned with the q – axis of field poles then flux ϕq per pole is set up and the effective reactance offered by the alternator is Xq.
Xq = minimum voltge maximum voltage √3 Ia(max )
= Vt line (at maximum Ia ) = Quadrature axis reactance
→ The power flow diagram of 3 – ϕ induction motor is
nsThe slip of induction machine is (S) = ns− nr
= Ns− Nr Ns
Where Ns is synchronous speed in rpm ns is synchronous speed in rps ⇒ Nr = Ns(1 − s)⇒ Ns − Nr = SNs
∴ Rotor frequency, f2 = P . SNs120
= S PNs120
= Sf1
For an induction machine with rotor resistance Rr and locked rotor leakage reactance Xr, the rotor impedance Zr at slip s is:Zr = Rr + jsXr The stator circuit equivalent impedance Zrf for a rotor / stator frequency ratio s is: Zrf = Rrs / s + jXrs
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 E2.
∴ E2/ph = 4.44 Nphr ϕ1 f1 Kdr Kpr E2/ph = 4.44 Nphr ϕ1 f1 Kwr
Mechanical power
developed, Pm Pg 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 I2R loss
Stator core loss
Stator I2R 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 Tm are related by: Pt = ωsTm = Pr / s = Pm / (1 - s) Pr = sPt = sPm / (1 - s) Pm = ωmTm = (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
Case (ii) : If the ratio of voltage to frequency is not constant and flux is also not constant
fv⇒ ≠ const ϕ ≠ const
Ph ∝ f Pe ∝ f 2
Ph = Kh f B m
1.6 2Pe = Ke f 2 Bm
1 2Case (iii) : If frequency is constant and voltage is variable then Ph = Kh f Bm
.6 Pe = Khf 2 Bm
f= Kh f �v�
1.6
∴ Ph ∝ v1.6 f −0.6 Pe ∝ v2
Ph ∝ v1.6 Pe ∝ v 1 2
→ Short circuit current with normal voltage applied to stator is
I = short circuit current with normal voltage Ibr = short circuit current with voltage Vbr.
→ Power factor on short circuit is found from Pbr = √3 Vbr Ibr cos ϕbr
⇒
→ As Pbr is approximately equal to full load copper losses
input power
Electrical power input∴ Efficiency of Induction motor = Net mechanical output ∴ Efficiency of Induction generator = Net electrical output
mechanical power input
VI = Ibr × Vbr
Cos ϕbr = Pbr
√3 Vbr Ibr
Rbr = PbrI
b r 2
The blocked rotor impedance is
Zbr = VbrI
br
Losses and efficiency :- There are three cases in iron losses. Case (i) : If the ratio of voltage to frequency is constant and flux is also constant then Iron loss = Hysteresis loss + eddy current loss
∴ Blocked rotor reactance = Xbr = �Z br
2 − R br
2
Efficiency of Induction machines :- Generally efficiency = output power