Formula Notes Electrical Machines Final

ELECRIAL MACHINES (Formula Notes)

gradeup 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 f

Weight of

→ Specific weight of t/f =

t f

KVA rating of

Net Cross Sectional area

→ Stacking/iron factor :- (k s ) = Gross Cross Sectional area

→ k s is always less than 1 → Gross c.s Area = AG = length × breadth → Net c.s Area = An = k s × AG → Utilization factor of transformer core = 0.85

Effective C.S.Area

mmF

→ Flux = Reluctance = = ϕm sin ωt

Total C.S Area dϕ

U.F of cruciform core = 0.8 to

d

→ According to faradays second law e1 = −N1 dt = −N1 dt �ϕm sin ωt� e1 = N1 ϕm ω sin�ωt − π�2�

Instantaneous value of emf in primary

→ Transformer emf equations :E1 = 4.44 N1 Bmax An f  (1) E2 = 4.44 N2 Bmax An f  (2) E

→ Emf per turn in Iry = N1 = 4.44 Bmax An f → Emf per turn in II

ry

=

1

E2

N2

= 4.44 Bmax An f

⟹ Emf per turn on both sides of the transformer is same ⟹

E1

=

N1 E1 E2

E2

N2 N1

=

N2

=

1

k

Transformation ratio = K =

E2 E1

=

N2 N1

1

Turns ratio = K = N1 ∶ N2

gradeup 1

gradeup → 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 → During operation of transformer :Bm ∝

E1 f

∝

V1 f

Bmax = constant ⟹

V1 f

= constant

Equivalent ckt of t/f under N.L condition :-

V1

Iw

R0

I0

N1 Iµ

X0

E1

N2 E2

No load /shunt branch.

gradeup 2

gradeup → No load current = I0 = Iµ + Iw = I0 �−ϕ0 Iw = I0 cos ϕ0 Iµ = I0 sin ϕ0

→ No load power = v1 I0 cos ϕ0 = v1 Iw = Iron losses. R0 =

v1 Iw1

v1 Iµ 𝐫𝐫𝐫𝐫

; X0 =

Transferring from 𝐈𝐈𝐈𝐈

R 21

R1

to 𝐈𝐈 𝐫𝐫𝐫𝐫:-

⟹ Iw =

No load power V1

I22 R 2 = I12 R 21 I

R2 K2 R R 21 = K22

=

∴

From 𝐈𝐈 𝐫𝐫𝐫𝐫 to 𝐈𝐈𝐈𝐈 𝐫𝐫𝐫𝐫 :I12 R1 = I22 . R11 R11 =

2

R 21 = R 2 � 2 � I 1

I12 . R1 I22

R11 = R1 . K 2

→ Total resistance ref to primary = R1 + R 21 R 01 = R1 + R 2 /k 2 → Total resistance ref to secondary = R 2 + R11 R 02 = R 2 + k 2 R1 → Total Cu loss = I12 R 01 Or 2 I2 R 02 Per unit resistance drops :I1 R1 E1 I R drop = 2E 2 2

→ P.U primary resistance drop = → P.U secondary resistance

I1 R01 E1 ry I2 R02 II = E 2

→ Total P.U resistance drop ref to I ry = → Total P.U resistance drop ref to

→ The P.U resistance drops on both sides of the t/f is same I1 R01 E1

=

I2 R02 E2

Losses present in transformer :1. Copper losses 2. Iron losses 3. Stray load losses 4. Dielectric losses

t/f windings major losses t/f core cu parts Iron parts minor losses insulating materials.

gradeup 3

gradeup 1. Cu losses in t/f: Total Cu loss = I12 R1 + I22 R1 = I12 R 01 = I22 R 02

VA rating of t/f E1 VA rating of t/f current on II ry = E2 ∝ I12 or I22. Hence there are called FL Cu loss in watts

→ Rated current on I ry = Similarly

→ Cu losses

→ P.U Full load Cu loss =

=

as variable losses.

VA rating of t/f

I12 R01 E1 I1

→ If VA rating of t/f is taken as base then P.U Cu loss ∝ I12 as remaining terms are constant. →

P.U Cu loss at x of FL = x 2 × PU FL Cu loss

ry → P. U resistance drop ref to I � = or P. U resistance ref to Iry

=

I1 R01 E1

I12 R01 E1 I1

×

I1 I1

∴ 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 :x

Wh = η Bmax . f . v

Area under one hysteresis loop.

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 ) ∝ R ce × Ie2 As area decreases in laminated core resistance increases as a result conductivity decreases. 2 2 2 We = K. Bma x f .t

Constant

thickness of laminations.

Supply freq

(it is a function of σ )

gradeup 4

gradeup During operation of transformer :-

Case (i) :-

V1 f

Bm ∝

V1 f

= constant, Bmax = const. we ∝ f 2

we = B f 2

Const.

∴ wi = wh + we 2

wi = Af + Bf Case (ii) :-

V1 f

When Bmax = const.

≠ constant, Bm ≠ const. 2

V

we ∝ � f1 � . f 2 we ∝ V1 2

wi = wh + we wi =

P.U iron loss :-

→ P.U iron loss =

A V11.6 f0.6

+ BV1 2

Iron loss in watts VA rating of t/f

→ 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 : W0 = Losses in t/f under no load condition = Iron losses + Dielectric loss + no load primary loss (I02 R1)

 Constant losses = W0 − I02 R1 Where , R1 = 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 :V1 rated → Wi S.C test :-

VSC → (Wi )S.C Wi ∝ V1 2

Wi (Wi )SC

V1 rated 2 � VSC

= �

(Wi )S.C = Wi × � V

VSC

1 rated

2

�

∴ Variable losses = WSC − (Wi )SC �V

VSC

1 rated

2

�

gradeup 5

gradeup → 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 ≃ ISC2 . R 01 R 01 =

WSC 2

ISC

Efficiency :output power input power output power output power+losses E2 I2 cos ϕ2

 Efficiency of transformer is given by η =

=

=E

ηF.L =

ηx

of F.L

=

2 I2 cos ϕ2 + F.L cu losses+Iron losses

E2 I2 cos ϕ2

E2 I2 cos ϕ2 + I22 R02 + Wi

x (E2 I2 ) cos ϕ2 x (E2 I2 ) cos ϕ2 + x 2 (I2 2 R 02 ) + Wi

O.C test

S.C test

→ Transformer efficiency =

KVA × cos ϕ KVA × cos ϕ + wi + Cu losses

→ Voltage drop in t/f at a Specific load p.f = I2 R 02 cos ϕ2 ± I2 X02 sin ϕ2 → % Voltage regulation =

I2 R02 cos ϕ2 ± I2 X02 sin ϕ2

I R02

=�2

V1′

V1′

I X02

� cos ϕ2 ± � 2

V1′

×100

� sin ϕ2

↓ ↓ P.U resistance P.U reactance % Regulation = �(P. U R) cos ϕ2 + (P. U X) sin ϕ2 � × 100

gradeup 6

gradeup AUTO TRANSFORMER: → Primary applied voltage, Vab = Secondary voltage V2 referred to primary + primary leakage impedance drop + secondary leakage impedance drop ref. to primary. N1 −N2 � N2

Vab = �

N1 − N2 � N2

V2 + I1 (r1 + jx1 ) + (I2 − I1 )(r2 + j x2 ) � LV

→ K of auto transformer = HV

(KVA)induction = (V1 − V2 ) I1 I/P KVA = V1 I1

(KVA)induction i/p KVA

∴

=

(V1 − V2 ) I1

=1–

V1 I1 LV HV

=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 ∝ (N1 − N2 ) I1

→ Wt of conductor in section BC of auto t/f ∝ (I2 − I1 )N2 ∴ Total wt. of conductor in auto t/f is ∝ I1 (N1 − N2 ) + (I2 − I1 )N2 ∝ 2 (N1 − N2 ) I1 → Total wt. of conductor in 2 wdg transformer ∝ I1 N1 + I2 N2 ∝ 2 I1 N1

→

wt.of conductor in an auto t/f wt.of conductor in 2 wdg t/f

=

=

2(N1 −N2 )I1 2N1 I1 N 1 – N2 1

=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 (KVA)2 wdg t/f . 1−K

gradeup 7

gradeup DC MACHINES : Lap Winding

Wave Winding

Ycs = P

Ycs = P

S

(1) Coil Span : (2) Back Pitch (3) Commutator Pitch

Yb = U Ycs Yc = 1 for Progressive winding Yc = -1 for Retrogressive winding

(4) Front Pitch

Yf =Yb +2 for Progressive winding Yf =Yb -2 for Retrogressive winding

(5) Parallel Paths (6) Conductor Current

A=P I Ic = Aa

(7) No of brushes

S = No of commutator segments P = No of poles

•

U = No of coil sides / No of poles =

• • •

C = No of coils on the rotor A = No of armature parallel paths Ia = Armature current

→ Distribution factor (K d ) = P

Yb = U Ycs Yc =

phasor sum coil emf

→ θ0electrical = 2 θ0mechanical

p

for Restrogressive winding (Yc Must be integer) Yf =2Yc - Yb

A=2 I Ic = 2a

No of brushes = 2

2C

arthematic sum of coil emf elecrrical angle of coil

S

=

*100%

1800

2 (c+1)

for Progressive winding 2 (c+1) Yc = - p

No of brushes = A = P

• •

→ Pitch factor ( K p ) =

S

chord arc

2

=π

ZI

a → Armature mmf/Pole (Peak) , ATa = 2AP

ZI

pole arc

a → AT (Compensating Winding) = 2AP * pole pitch

B

→ AT(Inter pole) = ATa + µ i lgi 0

Where Bi = Flux density in inter pole airgap

lgi = length of inter pole airgap , µ0 = 4π ∗ 10−7

→ No of turns in each interpole , Ninterpole =

AT(Inter pole) Ia

Z

pole arc

→ The no of compensating conductor per pole, Ncw /pole = 2 A P (pole pitch )

→ The Mechanical power that is converted is given by Pconv = Tind ωm Where T = Induced torque ω = Angular speed of the machines rotor m

gradeup 8

gradeup → 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 =

∅ZNP

60A PZ

→ Torque developed in Dc machine , Te = 2πA ∅ 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 R a → The terminal voltage of the DC motor is given by Vt = Ea + Ia R a → Speed regulation of dc machine is given by ,SR =

→ Voltage regulation , VR =

Shunt Generator:

Vnl− Vfl Vfl

* 100 %

ωnl− ωfl ωfl

* 100 % =

Nnl− Nfl Nfl

* 100 %

→ 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: If = V / Rf

→ 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

gradeup 9

gradeup → 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 (R a + R s ) V

(c) Isf = Rx = shunt field current f

(d) The equivalent effective shunt field current for this machine is given by Isf =Isf +

Nse Nf

Armature reaction MMF

Ia - (

)

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 (R a + R s ) (c)

(d)

Isf =

Vx Rf

= shunt field current

The equivalent effective shunt field current for this machine is given by Isf =Isf -

Shunt Motor:

Nse Nf

Armature reaction MMF

Ia - (

Nf

)

Where Ns e = No of series field turns , Nf = = No of shunt field turns

→ 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: If = V / Rf → 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 V Ra → The speed of the shunt motor , ω = K∅ - (K∅) 2T →

Where K =

PZ

2πA

gradeup 10

gradeup Series Motor : → 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 2 where Pc = Copper losses = Ia R a

Ps t = Stray load loss = α I2

Pb = Brush Contact drop = Vb Ia , Where Vb = Brush voltage drop

→ The total machine losses , PL = Pk +Vb Ia + K v Ia 2 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 → Rearranging the efficiency equations:

gradeup 11

gradeup Pin = Pout + Ploss = Pout / η = Ploss / (1 - η) Pout = Pin - Ploss = ηPin = ηPloss / (1 - η) Ploss = Pin - Pout = (1 - η)Pin = (1 - η)Pout / η

Temperature Rise: → 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 → Rearranging for per-unit change in resistance ∆Rpu relative to R1: ∆Rpu = (R2 - R1) / R1 = (θ2 - θ1) / (θ1 - θ0) = ∆θ / (θ1 - θ0) Copper Windings: → The value of θ0 for copper is - 234.5 °C, so that: ∆θ = θ2 - θ1 = (θ1 + 234.5) (R2 - R1) / R1 → If θ1 is 20 °C and ∆θ is 1 degC: ∆Rpu = (R2 - R1) / R1 = ∆θ / (θ1 - θ0) = 1 / 254.5 = 0.00393 → The temperature coefficient of resistance of copper at 20 °C is 0.00393 per degC. Aluminium Windings: → The value of θ0 for aluminium is - 228 °C, so that: ∆θ = θ2 - θ1 = (θ1 + 228) (R2 - R1) / R1 → If θ1 is 20 °C and ∆θ is 1 degC: ∆Rpu = (R2 - R1) / R1 = ∆θ / (θ1 - θ0) = 1 / 248 = 0.00403 → The temperature coefficient of resistance of aluminium at 20 °C is 0.00403 per degC.

gradeup 12

gradeup 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 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

gradeup 13

gradeup 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 = γ=

180° s p

� �

P(180°) d

→ Pitch factor or coil span factor or chording factor :- (K P)

KP = =

KP =

The emf induced | coil in short pitched winding The emf induced |coil in full pitched winding

The vector sum of induced emf | coil Arithmetic sumof induced emf | coil 2E cos∝/2 2E

Kp = cos ∝/2 cos n ∝ → Pitch factor for nth harmonic i.e, K pn = 2 180° n n−1 180 � � n

→ chording angle to eliminate nth harmonics (α)=

→ coil spam to eliminate nth harmonics ,(β) =

→ Distribution factor | spread factor | belt factor | breadth factor(kd) :The emf induced when the winding is distributed

K d = The emf induced when the winding is concentrated Kd =

Vector sum of emf induced Arithmetic sum of emf induced

gradeup 14

gradeup

Kd =

mγ 2 γ m sin 2

sin

→ The distribution factor for uniformly distributed winding is mr 2 π × 2 180

sim

kd4 = mr For nth harmonic, kdn =

mnγ 2 nγ m sin 2

sin

→ To eliminate nth harmonics ,phase spread (mγ) =

360° n

→ Generally, KVA rating, power output ∝ kd and Eph (induce emf) ∝ k d . Tph . ∴

KVA60 (3− ϕ) KVA120 (3− ϕ)

KVA60 (3ϕ) KVA90 (2ϕ)

Pout60 (3ϕ) Pout90° (2ϕ)

=

KVA60 (3ϕ) KVA180 (1ϕ)

=

KVA90 (2− ϕ) KVA180 (1− ϕ)

=

Pout60 (3 ϕ) Pout120 (3ϕ)

=

kd60 kd90

Pout60 (3ϕ) Pout180 (1− ϕ)

=

Pout90 Pout180

=

=

=

kd60 kd120

=

60 2 90 sin 2

sin

kd60 kd180

kd90 kd180

=

=

60 2 120 sin 2

sin

×

90 60

60 2 180 sin 2

sin

sin90⁄2 sin180⁄2

=

×

m120 m60

=

sin 30° sin 60°

×

120 60

= 1.15

= 1.06

×

×

180 60

180 90

= 1.5

= 1.414

1 . Ns ±1)

→ Speed of space harmonics of order (6k ± 1) is (6k where Ns = synchronous speed = 2S P

The order of slot harmonics is �

120 f p

± 1�

where S = no. of slots , P = no. of poles

→ Slot harmonics can be eliminated by skewing the armature slots and fractional slot winding. The angle of skew = θs = γ (slot angle)

= 2 harmonic pole pitches = 1 slot pitch. 2s p

→ Distribution factor for slot harmonics, k d � ± 1�

gradeup 15

gradeup

Is kd1 =

mγ 2 γ m sin 2

sin

i.e., same that of fundamental 2s p

→ Pith factor for slot harmonics, k p � ± 1� = k p1 = cos ∝�2 → The synchronous speed Ns and synchronous angular speed of a machine with p pole pairs running on a supply of frequency fs are:

ωs = 2πfs / p → Slip S =

NS − N NS

Where NS =

120 f p

= synchronous speed

→ 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 = ωsTm

→ The rated load torque TM for a rated output power PM is: TM = PM / ωs = PM p/ 2πfs = 120PM / 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 - IsZs = Es - Is(Rs + jXs) where Rs is the stator resistance and Xs is the synchronous reactance E = �(V cos ϕ + Ia R a )2 + (V sin ϕ ± Ia Xs )2 + ⇒ lag p.f − ⇒ leading p.f.

gradeup 16

gradeup Synchronous Motor: → For a synchronous motor with stator induced voltage Es, stator current Is and synchronous impedance Zs, the terminal voltage V is:

V = Es + IsZs = Es + Is(Rs + jXs) where Rs is the stator resistance and Xs is the synchronous reactance Voltage regulation : |E|− |V|

→ % regulation = E – V = Ia Zs

|V|

E−V V Ia Zs × V

∴ % regulation = =

×100 100

∴ regulation ∝ Zs ∴ As Zs increases, voltages regulation increases.

→ Condition for zero | min. voltage regulation is, Cos (θ + ϕ) = − → Condition for max. Voltage regulation is, ϕ = θ → Short circuit ratio (SCR) = SCR ∝

1 Xa

∝

Ifm Ifa

1 Armature reaction

=

1 Zs (adjusted)|unit

=

Ia Zs 2V

1 Xs (adjusted)|unit

Voltage regulation ∝ Armature reaction ∴ SCR ∝

1 Voltage regulation

∴ Small value of SCR represent poor regulation. ϕa =

armature mmf reluctance

But reluctance ∝ Air gap

∴ ϕa = ϕa ∝

armature mmf airgap

1 Air gap length

Armature reaction ∝ ϕa ∝

1 Airgap length

gradeup 17

gradeup

∴ SCR ∝

1 Armature reaction

∝ Airgap length

Air gap length ∝ SCR

∴ machine size ∝ SCR. Cost ∝ SCR Power =

EV sin δ Xs

⇒P∝

1 Xs

∝ SCR

Power ∝ SCR

∴ Large value of SCR represent more power output. → Synchronizing power coefficient or stability factor Psy is given as Psy =

dp dδ

=

=

d EV � sin δ� dδ Xs EV cos δ Xs

Psy is a measure of stability ∴ stability ∝ Psy 1

But Psy ∝ X ∝ SCR s

∴ Stability ∝ SCR

Stability ∝ SCR ∝ Air gap length ∴ Stability ∝ Air gap length

→ When the stator mmf is aligned with the d – axis of field poles then flux ϕd perpole is set up and the effective reactance offered by the alternator is X d . Xd =

maximum Voltage minimum current

Xq =

minimum voltge maximum voltage

=

(Vt )line (at min. Ia ) √3 Ia (min )

= 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 X q. =

Vt line (at maximum Ia ) √3 Ia (max )

= Quadrature axis reactance

gradeup 18

gradeup → Cylindrical rotor Synchronous machine , The per phase power delivered to the infinite bus is given by P = → Salient pole synchronous machine ,

Ef Vt Xs

sin δ

The per phase power delivered to the infinite bus is given by P=

Ef Vt Vt2 sin δ + Xd 2

1

�X −

Condition for max. power:-

q

1 � sin 2δ Xd

→ For cylindrical rotor machine :At constant Vt and Ef , the condition for max. power is obtained by putting dp dδ

∴

=

Ef Vt cos δ Xs

=0

dp dδ

=0

Cos δ = 0 δ = 90° Hence maximum power occurs at δ = 90° → For salient – pole synchronous machine :⇒

Vt Ef cos δ + Xd

Cos δ = − 4V

t

Vt 2 �

Ef Xq

1 Xq

�Xd − Xq

−

1 � cos 2δ Xd

=0

E Xq

1 f ± � + �4V 2 � �X t

d − Xq

dp dδ

=0

2

� �

The value of load angle is seed to be less than 90°.

∴ max. power occurs at δ < 90° → Synchronizing power = Psy. ∆ δ.

EV cos 𝛿𝛿 . ∆ 𝛿𝛿 . Xs Synchronizing power . torque = ω

=

→ Synchronizing

Power flow in Alternator :→ Complex power = S = P + jQ = VIa∗

Where Active power flow (P) = EV sin(θ Z2

Reactive power flow (Q) =

→ Condition for max. power output :P= dp dδ

EV cos(θ Zs

− δ) −

V2 cos θ Z2

− δ) −

V2 Zs

EV cos(θ − Zs

δ) −

V2 cos θ Zs

;

sin θ ;

= 0 for max power condition ie θ – δ = 0 θ=δ Pmax =

EV Zs

−

V2 Zs

If R a = 0; θ = δ = 90° ; then max power is given by

cos θ

gradeup 19

gradeup INDUCTION MACHINES: → The power flow diagram of 3 – ϕ induction motor is Power i/p to stator Rotor i/p power from mains = airgap power

Stator I2R loss

Mechanical power Pg

Power of rotor shaft

Rotor Rotor core loss Friction loss Windage I2R (negligible for at bearings loss loss small slips) and sliprings of (if any)

Stator core loss

The slip of induction machine is (S) = = Where Ns is synchronous speed in rpm ns is synchronous speed in rps ⇒ Nr = Ns (1 − s) ⇒ Ns − Nr = SNs ∴ Rotor frequency, f2 =

developed, Pm

P . SNs 120

=S

ns − nr ns Ns − Nr Ns

PNs 120

= 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 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

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 K dr K pr E2 /ph = 4.44 Nphr ϕ1 f1 K wr

gradeup 20

gradeup

K wr = Rotor winding factor

→ But during running conditions the frequency of the rotor becomes, running with speed Nr P(Ns − Nr ) 120

=

P SNs 120

= Sf1

∴ fr = Sf1 ∴ Emf under running conditions is E = √2 π fr Kw2 Nphr ϕ1 = SE2 → Rotor leakage reactance = 2π (Rotor frequency) (Rotor leakage Inductance) ∴ Rotor leakage reactance at stand still = 2π f1 𝑙𝑙2 = x2 Ω → Rotor leakage reactance at any slips = 2π f2 𝑙𝑙2 = sx2 Ω

→ Rotor leakage impedance at stand still = �r22 + x22

→ At any slip s, rotor

= �r22 + (sx2 )2

→ Per phase rotor current at stand still =

E

�r22 + x22

→ Per phase rotor current at any slip s is given by SE2 E2 = I2 = �r22 + (sx2 )2 �(r1 /s)2 + x22

→ The rotor current I2 lags the rotor voltage E2 by rotor power factor angle θ2 given by sx θ2 = tan−1 2 r 2

gradeup 21

gradeup

→ Perr phase power input to rotor is Pg = F2 I2 cos θ2

Per phase rotor resistance Per phase rotor impedance

cos θ2 =

=

r2 /s

�(r2 /s)2 + (x2 )2

r2 /s

∴ Pg = E2 I2 ×

�(r2 /s)2 + (x2 )2

`

E2

=

�(r2 /s)2 + (x2 )2

= I22

r2 s

× I2

r2 s

→ Pg is the power transferred from stator to rotor across the air gap. There fore Pg is called air gap power r Pg = I22 2 =

s 2 I2 r2

+ I22 r2 �

1−S � S

Pg = (Rotor ohmic loss) + Internal mechanical power developed in rotor (Pm ) = S Pg + (1 − S)Pg ∴ Pm = (1 − S) Pg = I22 r2 �

Rotor ohmic loss = �

S � 1−S

1−S � S

Pm = SPg

→ Internal (or gross) torque developed per phase is given by Te =

Internal mechanical power developed inrotor Rotor speed in mechanical radian per sec

Te =

Pm ωr

(1−S)Pg

=

(1−S)ωs

=

Pg

ωs

→ Electromagnetic torque Te can also be expressed as Te =

Pg

ωs

=

1 ωs

I22 r2 Rotor ohmic loss = (ωs ) slip S Rotor ohmic loss (ωs ) slip

×

∴ Te =

gradeup 22

gradeup → Power available at the shaft can be obtained from Pg as follows. Output or shaft power, Psh = Pm − Mechanical losses → Mechanical losses implies frication and windage losses

Psh = Pg − Rotor ohmic loss – Friction and windage losses = Net mechanical power output or net power output P

sh Output or shaft torque Tsh = = Rotor speed

Psh (1−s) ωs

→ If the stator input is known. Then air gap power Pg is given by Pg = stator power input – stator I 2 R loss – stator core loss.

→ Ratio of Rotor input power, rotor copper losses and gross mechanical output is 1

Ir2 R 2 /s : Ir2 R 2 : Ir2 R 2 � − 1� s

⇒ 1 : S : (1 – S) ∴ Rotor copper losses = S × Rotor input Gross mechanical output =(1 – S) × Rotor input. Rotor copper losses = (Gross Mechanical output) × Efficiency of the rotor is approximately Gross mechanical power output Rotor input (1−S) Rotor input Rotor input

Equal to ηrotor = =

S 1−S

=1–S =1− =

N Ns

ηrotor ≃

NS − N NS

N Ns

Total torque is Te =

m ωs

×

Ve2 r2 2 �Re + � + (x2 + Xe )2 s

×

m is the number of stator phases. Torque equation can be written as m r Te = × I22 × s2 Te =

ωs m ωs

r2 s

Nm

× rotor input per phase.

gradeup 23

gradeup Thus the slip SmT at which maximum torque occurs is given by r2 SmT = �Re2 + X2

Substituting the value of maximum slip in the torque equation, gives maximum torque Tem =

m ωs

×

Ve2

2�Re + �Re2 + X22 �

If stator parameters are neglected then applying maximum transfer theorem to r2 /s then

=x2

r2 s

Slip corresponding to maximum torque is Sm =

r2 x2

(Breakdown slip)

Nm = Ns (1 − Sm ) ⇒ Nm = Ns (1 − R 2 /x2 )

Nm is the stalling speed at the maximum torque

Starting torque:-

At starting, slip S = 1.00, starting torque is given by Test =

m Ve2 ωs

×

r2 (Re + r2 )2 + X2

Motor torque in terms of 𝐓𝐓𝐞𝐞𝐞𝐞 :

→ The torque expression of an induction motor can also be expressed in terms of maximum torque Tem and dimension less ratio

S . Sm,T

In order to get a simple and approximate

expression, stator resistance r1 , or the stator equivalent resistance R e , is neglected. T

∴Te = em

2�Re + �Re2 + X2 � r 2 �Re + 2� + X2 s

→ Since r1 or Re is neglected Te Tem

=

2X

r 2 � 2� + X2 s

×

×

r2 s

r2 s

→ The slip at which maximum torque occurs is r SmT = 2 ∴ r2 = SmT X ∴

Te Tem T

=

X 2X

⇒Te = em

S X 2 � mT � + X2 s SmT S

Te =

2

+

S SmT

×

SmT X s

2Tem SmT S + S SmT

gradeup 24

gradeup 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 Ph = K h + Bm1.6 Given

V f

Pe = K e f 2 Bm2

is constant. As Bm ∝

⇒ Bm is constant

∴

V f

Ph ∝ f

and

Pe ∝ f 2

Case (ii) : If the ratio of voltage to frequency is not constant and flux is also not constant v ⇒ ≠ const ϕ ≠ const f

2 Pe = K e f 2 Bm

Ph = K h f Bm1.6

∴

Ph ∝ v1.6 f −0.6

Pe ∝ v 2

Case (iii) : If frequency is constant and voltage is variable then 1.6 2 Ph = K h f Bm Pe = K h f 2 Bm v 1.6 f

= Kh f � �

Ph ∝ v1.6

Pe ∝ v12

→ Short circuit current with normal voltage applied to stator is I = Ibr ×

V Vbr

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 ⇒

Cos ϕbr =

Pbr √3 Vbr Ibr

→ As Pbr is approximately equal to full load copper losses R br =

Pbr Ibr2

The blocked rotor impedance is Zbr =

Vbr Ibr

∴ Blocked rotor reactance = X br = �Zbr 2 − R br2

Efficiency of Induction machines :Generally efficiency =

∴ Efficiency of Induction motor =

Net mechanical output Electrical power input

output power input power

∴ Efficiency of Induction generator =

Net electrical output mechanical power input

gradeup 25

gradeup ∴ Wound rotor GD GF

=

I22 r2 I12 r1

GD GF

=

=

r2 r1

Psc − 3 Isc2 r1 3 Isc2 r1

I

2

�I2 � 1

Direct – on line (across the line) starting :→ The relation between starting torque and full load torque is

∴

Te.st Te.fl

1 ωs

Te =

r I22.st 2

=

1 r2 Sfl

I22.ft

× I22

r2 s

2 I I2 fl

= � 2 st � × Sfl

The above equation valids of rotor resistance remains constant. Te.st Te.fl

Where

Ist Ifl

I

2

= � Ist� × Sfl fl

(Effective rotor to stator turns ratio) I2 st (Effective rotor to stator turns ratio) I2 fl

=

→ Per phase short – circuit current at stand still (or at starting) is, Isc =

V1 Zsc

Where Zsc = (r1 + r2 ) + j(x1 + x2 ) Here shunt branch parameters of equivalent circuit are neglected. → Therefore, for direct switching,

∴

Test Tesf

Ist = Isc = I

2

= � Isc � Sfl . fl

V1 Zsc

Stator resistor (or reactor) starting :Since per phase voltage is reduced to xv, the per phase starting current Ist is given by xv Ist = 1 = xIsc Z As be fore

Te.st Te.fl

sc

I

2

= � Ist� × Sfl

=

fl

xI 2 � I sc � fl

× Sfl

→ In an induction motor, torque ∝ (voltage)2 ∴

starting torque with reactor starting starting torque with direct switching

xv

2

= � v 1� = x2 1

gradeup 26

gradeup Auto transformer starting :→ Per phase starting current from the supply mains is Ist = x 2 Isc Te.st Te fl

=

Per phase starting current in motor winding Per−phase motor full load current

Te.st Te.fl

=

Ist Isc Ifl2

× Sfl

× Sfl

Test with an autotransformer Test with direct switching

xv

2

= � v 1� = x2 1

Star – delta method of starting : →

starting torque with star delta starter starting torque with direct switching in delta V � L� √3

= [V

2

L]

2

=

1 3

∴ star delta starter also reduces the starting torque to one – third of that produced by direct switching in delta. → With star – delta starter, a motor behaves as if it were started by an auto transformer starter with x =

→

1 √3

= 0.58 i.e with 58% tapping.

Starting torque with star delta starter Te .st Full load torque with startor winding in delta,Tefld

=

2r 1 � 12 �I ws st .y 1 r (I )2 2 ws fl d Sfl

=

2 1 I � √3 st .d (Ifl .d )2

�

× Sfl =

1 3

I

d 2

�I sc . d� × Sfl fl .

gradeup 27