Machines Formulas

8/20/2019 Machines Formulas

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Institute of Engineering Studies (IES,Bangalore) Electrical Machines Formula Sheet

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DC MACHINES : -

Lap Winding Wave Winding

(1) Coil Span :

=

=

(2) Back Pitch = = (3) Commutator Pitch = 1for Progressive winding = -1for Retrogressive winding

= for Progressive winding = - for Restrogressive winding

(Must be integer)(4) Front Pitch = +2

for Progressive winding

=

-2

for Retrogressive winding

= -

(5) Parallel Paths A = P A = 2

(6) Conductor Current = = (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 = C = No of coils on the rotor A = No of armature parallel paths

= Armature current Distribution factor ( ) = = = Pitch factor ( ) = *100%

Armature mmf/Pole (Peak) , A = AT (Compensating Winding) = *


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AT(Inter pole) = A + Where = Flux density in inter pole airgap

= length of inter pole airgap ,

No of turns in each interpole , = The no of compensating conductor per pole, /pole = ( ) The Mechanical power that is converted is given by =

Where T = Induced torque

= Angular speed of the machines rotor

The resulting electric power produced = The power balance equation of the DC Machine is = The induced emf in the armature is = Torque developed in Dc machine , =

Where = Flux\pole , Z = No of armature conductors , P = No of poles , N = Speed in rpm ,A = No of armature parallel paths,

rmature current

The terminal voltage of the DC generator is given by = - The terminal voltage of the DC motor is given by = + Speed regulation of dc machine is given by ,SR =

* 100 % = * 100 % Voltage regulation , VR =

* 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


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The armature induced voltage Ea and torque T with magnetic flux at angularspeed are:Ea = k f = km T = k f Ia = 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 = T = EaIa = km 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 at angular

speed are:

Ea = k f Ia = km Ia

T = k f Ia2 = kmIa

2

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 = T = EaIa = km Ia

2

Cumulatively compounded DC generator : - ( long shunt)

(a) = + (b) = - ( + )(c) = = shunt field current(d) The equivalent effective shunt field current for this machine is given by


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= + - ( )Where

= No of series field turns

= No of shunt field turns Differentially compounded DC generator : - ( long shunt)(a) = + (b) = - ( + )(c) = = shunt field current(d) The equivalent effective shunt field current for this machine is given by

= - - ( )Where = No of series field turns = No of shunt field turns

Shunt Motor:

For a shunt motor with armature induced voltage Ea, armature current Ia andarmature 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 at angularspeed are:Ea = k f = km T = k f Ia = 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 = T = EaIa = km Ia The speed of the shunt motor , = - T


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Where K =

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 at angular

speed are:

Ea = k f Ia = km Ia

T = k f Ia2 = kmIa

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 Ea

The airgap power Pe for a series motor is:Pe = T = EaIa = km Ia

2

Losses:

constant losses (P k) = Pw f + Pi o

Where, = No of load core loss = Windage & friction loss Variable losses () = + +

where = Copper losses = = Stray load loss = α

= Brush Contact drop = , Where = Brush voltage drop The total machine losses , = + + Efficiency


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The per-unit efficiency of an electrical machine with input power Pin, outputpower Pout and power loss Ploss is:

= Pout / Pin = Pout / (Pout + Ploss) = (Pin - Ploss) / Pin

Rearranging the efficiency equations:

Pin = Pout + Ploss = Pout / = Ploss / (1 - )

Pout = Pin - Ploss = Pin = Ploss / (1 - )

Ploss = Pin - Pout = (1 - )Pin = (1 - )Pout /

Temperature Rise:

The resistance of copper and aluminium windings increases with temperature,and the relationship is quite linear over the normal range of operatingtemperatures. For a linear relationship, if the winding resistance is R1 at

temperature 1 and R2 at temperature 2, then:

R1 / (1 - 0) = R2 / (2 - 0) = (R2 - R1) / (2 - 1)where 0 is the extrapolated temperature for zero resistance.

The ratio of resistances R2 and R1 is:

R2 / R1 = (2 - 0) / (1 - 0)

The average temperature rise of a winding under load may be estimated frommeasured values of the cold winding resistance R1 at temperature 1 (usuallyambient temperature) and the hot winding resistance R2 at temperature 2, using: = 2 - 1 = (1 - 0) (R2 - R1) / R1

Rearranging for per-unit change in resistance Rpu relative to R1:Rpu = (R2 - R1) / R1 = (2 - 1) / (1 - 0) = / (1 - 0)

.Copper Windings:

The value of 0 for copper is - 234.5 °C, so that: = 2 - 1 = (1 + 234.5) (R2 - R1) / R1


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If 1 is 20 °C and is 1 degC:Rpu = (R2 - R1) / R1 = / (1 - 0) = 1 / 254.5 = 0.00393

The temperature coefficient of resistance of copper at 20 °C is 0.00393 perdegC.

Aluminium Windings:

The value of 0 for aluminium is - 228 °C, so that: = 2 - 1 = (1 + 228) (R2 - R1) / R1

If 1 is 20 °C and is 1 degC:Rpu = (R2 - R1) / R1 = / (1 - 0) = 1 / 248 = 0.00403

The temperature coefficient of resistance of aluminium at 20 °C is 0.00403 perdegC.

Dielectric Dissipation Factor:

If an alternating voltage V of frequency f is applied across an insulation systemcomprising capacitance C and equivalent series loss resistance RS, then thevoltage VR across RS and the voltage VC across C due to the resultingcurrent I are:VR = IRS VC = IXC

V = (VR2

+ VC2

)

½

The dielectric dissipation factor of the insulation system is the tangent of thedielectric loss angle between VC and V:tan = VR / VC = RS / XC = 2fCRS RS = XCtan = tan / 2fC

The dielectric power loss P is related to the capacitive reactive power QC by:P = I

2RS = I

2XCtan = QCtan

The power factor of the insulation system is the cosine of the phase

angle between VR and V:cos = VR / Vso that and are related by: + = 90°


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tan and cos are related by:tan = 1 / tan = cos / sin = cos / (1 - cos2)½ so that when cos is close to zero, tan cos

TRANSFORMERS:

Gross cross sectional area = Area occupied by magnetic material + Insulationmaterial.

Net cross sectional area = Area occupied by only magnetic material excluding areaof insulation material.

Hence for all calculations, net cross sectional area is taken since (flux) majorlyflows in magnetic material.

Specific weight of t/f =

Stacking/iron factor :- ( ) = is always less than 1 Gross c.s Area =

= length × breadth

Net c.s Area = = Utilization factor of transformer core = U.F of cruciform core = 0.8 to0.85

Flux = =

According to faradays second law

Transformer emf equations :- = 4.44 (1) = 4.44 (2)

Instantaneous value

of emf in primary


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Emf per turn in = = 4.44 Emf per turn in

= 4.44

⟹ Emf per turn on both sides of the transformer is same ⟹ Transformation ratio = K =

Turns ratio =

For an ideal two-winding transformer with primary voltage V1 applied

across N1 primary turns and secondary voltage V2 appearing across N2 secondaryturns: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 appliedacross (N1 + N2) primary turns and secondary voltage V2 appearingacross 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 primarycurrent I1, rated secondary voltage V2 and rated secondary current I2, the voltampererating S is:S = V1I1 = V2I2

For a balanced m-phase transformer with rated primary phase voltage V1, ratedprimary 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 idealtransformer with N1 primary turns and N2 secondary turns is:Z12 = Z1(N2 / N1)

2


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During operation of transformer :-

⟹ = constantEquivalent ckt of t/f under N.L condition :-

No load current = No load power = Iron losses.

⟹

Transferring from to :-

=

∴

From to :-

No load /shunt branch.


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Total resistance ref to primary = / Total resistance ref to secondary = Total Cu loss =

Or Per unit resistance drops :-

P.U primary resistance drop =

P.U secondary resistance drop = Total P.U resistance drop ref to = Total P.U resistance drop ref to = The P.U resistance drops on both sides of the t/f is same Losses present in transformer :-

1. Cu losses in t/f:

Total Cu loss = =

= Rated current on Similarly current on =

Cu losses or . Hence there are called as variable losses.

1. Copper losses

2. Iron losses

3. Stray load losses

4. Dielectric losses

major losses

minor losses

t/f windings

t/f corecu parts

Iron parts

insulating materials.


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P.U Full load Cu loss =

=

If VA rating of t/f is taken as base then P.U Cu loss

as remaining terms are constant.

P.U Cu loss at x of FL = FL Cu loss

=

∴ P.U Resistance drop = P.U FL cu loss% FL Cu loss = % R = % Resistance dr op.

Iron (or) Core losses in t/f :-

1. Hysteresis loss :

Steinmetz formula :-

Where = stienmetz coefficient = 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 ,( As area decreases in laminated core resistance increases as a result conductivity decreases.

Constant

Supply freq

thickness of laminations.

it is a function of

Area under one hysteresis loop.

. f . v


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During operation of transformer :- Case (i) :-

= constant,

= const.

Case (ii) :-

constant,

const.

P.U iron loss :-

P.U iron loss = 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 :-

= Losses in t/f under no load condition= Iron losses + Dielectric loss + no load primary loss ( )

Constant losses =

Where , = LV winding resistance.To find out variable losses :- = Loss in t/f under S.C condition

Const.

∴ When = const.


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= F.L Cu loss + stray load losses (Cu and Iron) + Iron losses in both wdgs

Variable losses = Iron losses corresponding to O.C test :-

rated S.C test :-

∴ Variable losses =

Under the assumption that small amount of iron losses corresponds to and stray loadlosses are neglected the wattmeter reading in S.C test can be approximately taken as F.LCu losses in the transformer .

F.L Cu loss Efficiency :-

= =

O.C testS.C test


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Transformer efficiency =

Efficiency = =

Total losses in transformer =

output Condition for maximum effieciency is, Cu losses = Iron losses Total losses at = 2 %load at which maximum efficiency occurs % x = *100 %= *100 % KVA corresponding to = F.L KVA Voltage drop in t/f at a Specific load p.f = % Voltage regulation =

100=

P.U resistance P.U reactance

% Regulation = × 100Condition for max. regulation :-

% regulation = (% R) cos = 0Tan lagging

At maximum regulation


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= Value of maximum regulation :-

% Regulation = (% R) cos At max. regulation cos

Sin max. % regulation = (% R)

=

=

max. % regn = % Z

= % of rated voltage required to produce rated short ckt current

.

Condition for zero regulation :-

If the voltage regulation in the t/f is zero, the t/f voltages are maintained at their nominalvalues even under load condition

% Regn = (% R) cos For zero regulation occurs at leading p.f’s

(% R) cos sin = 0Tan

leading. At zero regulation condition : Regulation at x of FL = x [% R cos X sin ]

= x × F.L regn

Regulation at U.P.F:-Regulation at UPF = % R

= % F.L Cu loss


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Scott Connection:

= 0.866 = 0.577 = 0.866 0.577 = 0.289 = 0.577 0.289 = 2 : 1

If a neutral pt is located on 3 side, such that, voltage between any terminal to that neutralpoint is 0.577 then such neutral point divides the primary of teaser transformer in the ratioof 2 : 1

Location of neutral point from top = 0.866 Location of neutral point from bottom = 0.866 Operation of Scott Connection with 2 balanced load at UPF :-Teaser t/f :-

Let

Main t/f

86.6%

0.289

2

:

1

M

N

0.866 0.577 A

B

C


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Let

Capacity of Scott Connection :- ↙ ↓ Vol. rating of 1 – t/f Current rating of 1 – t/f Utilization factor =

= = 0.866 Utilization factor of Scott connection with 2 identical 1 – t/f’s is 86.6%

AUTO TRANSFORMER:

Primary applied voltage, = Secondary voltage referred to primary + primary leakage impedancedrop + secondary leakage impedance drop ref. to primary.

K of auto transformer = I/P KVA =

= 1 –

= 1 – K

∴ (KVA) induction = (1 – K) i/p KVA

(KVA) conduction = I/p KVA – I/p KVA


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Wt. of conductor in section AB of auto t/f Wt of conductor in section BC of auto t/f

∴ Total wt. of conductor in auto t/f is Total wt. of conductor in 2 wdg transformer

= 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.

.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 () :- It is the distance between two sides of the coil. It is expressed in terms ofdegrees, 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.

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 =


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Pitch factor or coil span factor or chording factor :- ( )

=

=

Pitch factor for harmonic i.e, chording angle to eliminate harmonics (α)= coil spam to eliminate harmonics ,() = 180 Distribution factor | spread factor | belt factor | breadth factor(kd) :-

= = Kd =

γ

γ The distribution factor for uniformly distributed winding is

For harmonic, = γ γ To eliminate harmonics ,phase spread (mγ) =

Generally, KVA rating, power output kd and (induce emf) .∴ = = 1.15

= cos /2 =

=


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= 1.06

= 1.5 1.414 Speed of space harmonics of order (6k ± 1) is

where = synchronous speed = The order of slot harmonics is

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 = = γ (slot angle) = 2 harmonic pole pitches

= 1 slot pitch.

Distribution factor for slot harmonics, Is

γ γ i.e., same that of fundamental Pith factor for slot harmonics, =

The synchronous speed Ns and synchronous angular speed s of a machine with p polepairs running on a supply of frequency f s are:

s = 2f s / p

Slip S = Where = synchronous speed

The magnitude of voltage induced in a given stator phase is =


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Where K = constant

The output power Pm for a load torque Tm is:

Pm = sTm

The rated load torque TM for a rated output power PM is:

TM = PM / s = PM p/ 2f s = 120PM / 2Ns

Synchronous Generator :

For a synchronous generator with stator induced voltage Es, stator current Is andsynchronous 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 = + lag p.f leading p.f.

Synchronous Motor:

For a synchronous motor with stator induced voltage Es, stator current Is and synchronousimpedance 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 :

% regulation = 100E – V = ∴ % regulation =


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= 100∴ regulation

∴ As

increases, voltages regulation increases.

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

SCR Voltage regulation Armature reaction∴

SCR

∴ Small value of SCR represent poor r egulation. But reluctance Air gap∴

Armature reaction ∴ SCR Airgap length

∴ machine size

SCR.

Cost

SCR

Power = P SCR

Air gap length SCR


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∴ Large value of SCR represent more power output. Synchronizing power coefficient or stability factor

is given as

= =

is a measure of stability∴ stability But SCR

Stability

SCR

Air gap length

When the stator mmf is aligned with the d – axis of field poles then flux perpole is set upand the effective reactance offered by the alternator is . = Direct axis reactance

When the stator mmf is aligned with the q – axis of field poles then flux

per pole is set up

and the effective reactance offered by the alternator is . = = Quadrature axis reactance Cylindrical rotor Synchronous machine ,

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

Salient pole synchronous machine ,

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

P = δ δ

Power SCR

∴ Stability SCR

∴ Stability Air gap length


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Condition for max. power:-

For cylindrical rotor machine :-

At constant

and

, the condition for max. power is obtained by putting

= 0

∴ = 0Cos δ = 0

δ = 90

Hence maximum power occurs at δ = 90

For salient – pole synchronous machine :- = 0

= 0

Cos δ = The value of load angle is seed to be less than 90°.∴ max. power occurs at δ < 90 Synchronizing power = . ∆ δ.

= .

Synchronizing torque =

.

Power flow in Alternator :-

Complex power = S = P + jQ = V Where Active power flow (P) =

;Reactive power flow (Q) =

; Condition for max. power output :-

P =

= 0 for max power condition

ie – δ = 0

θ =


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If = 0; = δ = 90 ; then max power is given by

SYNCHRONOUS MOTORS:

Speed regulation =

= = 0%

Slip S = = 0%

The speed can be controlled by varying the f requency

ratio control is preferred for rated torque operationPower flow in synchronous motor is given by

complex power i/p s = p + jQ = V where P = Resl power flow , Q = Reactive power flow

:

:

If = 0 ; Condition for max power :-

= 0 0 +

Sin ( + δ) = 0 = sin 180

=

cos θ

= Q =

8 0

P =

Q =


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Expression for mechanical power developed :-

Mechanical power developed = = active component +

Condition for max. mechanical power developed :- δ δ = 0Sin ( – δ) = 0 = sin 0

δ =

This is the expression for the mechanical power developed interms of load angle and the

internal machine angle , for constant voltage and constant E i.e., excitation Gross Torque =

=

synchronous speed in r.p.m

∴ Tg =

Condition for excitation when motor develops :-For max power developed is = 0

Condition for excitation when motor develops is The corresponding value of max. power is

= Power flow in synchronous motors :-

Tg =


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For leading p.f

tan δ =

The mechanical power developed per phase is given by,s δ δ

INDUCTION MACHINES:

The power flow diagram of 3 – induction motor is

The slip of induction machine is (S) =

Mechanical power

developed, Rotor i/p power= airgap powerPower 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

R

loss

Stator

core

loss

Stator

R

loss

(input)

Stator

copper loss

3 Mechanical power

developed in armature = 3 . cos( ± )+ ve lead–ve for lag

= 2 π Ns

60. output

Friction

and Iron

losses

= 2π Ns

60


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=

Where is synchronous speed in rpm

is synchronous speed in rps

= S ∴ Rotor frequency, For an induction machine with rotor resistance Rr and locked rotor leakage reactance Xr , therotor 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 s running at angular speed m andslip 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 = sTm = Pr / s = Pm / (1 - s)Pr = sPt = sPm / (1 - s)

Pm = mTm = (1 - s)Pt The power ratios are:Pt : Pr : Pm = 1 : s : (1 - s)

The gross motor efficiency m (neglecting stator and mechanical losses) is:m = Pm / Pt = 1 - s

Rotor emf, Current Power :- At stand still, the relative speed between rotating magnetic field and rotor conductors is

synchronous speed ; under this condition let the per phase generated emf in rotor circuitbe .∴ = 4.44 = 4.44 = Rotor winding factor But during running conditions the frequency of the rotor becomes, running with speed

S

∴ ∴ Emf under running conditions isE =

= S


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Rotor leakage reactance = 2 (Rotor frequency) (Rotor leakage Inductance)

∴ Rotor leakage reactance at stand still = 2 = Rotor leakage reactance at any slips = 2 Rotor leakage impedance at stand still

= At any slip s, rotor

Per phase rotor current at stand still

=

Per phase rotor current at any slip s is given by

The rotor current lags the rotor voltage by rotor power factor angle given by

Per phase power input to rotor is

=

= s

=


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∴ `

=

= is the power transferred from stator to rotor across the air gap. There f ore is called air

gap power =

= (Rotor ohmic loss) + Internal mechanical power developed in rotor (

)

= S

∴ = Rotor ohmic loss = = S

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

Electromagnetic torque can also be expressed as ∴

Power available at the shaft can be obtained from as follows.Output or shaft power, Mechanical losses Mechanical losses implies frication and windage losses


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Rotor ohmic loss – Friction and windage losses= Net mechanical power output or net power output

Output or shaft torque

If the stator input is known. Then air gap power is given by stator power input – stator loss – stator core loss. Ratio of Rotor input power, rotor copper losses and gross mechanical output is : :

∴ Rotor copper losses = S × Rotor inputGross mechanical output =(1 – S) × Rotor input.

Rotor copper losses = (Gross Mechanical output) × Efficiency of the rotor is approximatelyEqual to

=

= 1 – S

= 1 =

Total torque is m is the number of stator phases.

Torque equation can be written as

rotor input per phase.

Thus the slip at which maximum torque occurs is given by Substituting the value of maximum slip in the torque equation, gives maximum torque

1 : S : (1 – S)


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If stator parameters are neglected then applying maximum transfer theorem to

then

= Slip corresponding to maximum torque is

(Breakdown slip)

is the stalling speed at the maximum torque

Starting torque:- At starting, slip S = 1.00, starting torque is given by

Test =

Motor torque in terms of : The torque expression of an induction motor can also be expressed in terms of maximum

torque and dimension less ratio . In order to get a simple and approximateexpression, stator resistance , or the stator equivalent resistance , is neglected.∴

Since r 1 or Re is neglected The slip at which maximum torque occurs is ∴

∴


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Power slip characteristics :-

The total internal mechanical power developed is Maximum power transfer theorem is invoked again to obtain maximum value of internal

mechanical power developed. Since per phase is the power delivered to , internalmechanical power developed is maximum, when

In order to get maximum power ,substitute , in place of in power equation In order to get maximum power output from an induction generator, the rotor must be deiven

at a speed given by

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 Given

is constant. As is constant

∴ and

Case (ii) : If the ratio of voltage to frequency is not constant and flux is also not constant

≠ const


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∴ Case (iii) : If frequency is constant and voltage is variable then

=

Short circuit current with normal voltage applied to stator is

I = short circuit current with normal voltage

= short circuit current with voltage . Power factor on short circuit is found from

As is approximately equal to full load copper losses

The blocked rotor impedance is

∴ Blocked rotor reactance = Efficiency of Induction machines :-

Generally efficiency = ∴ Efficiency of Induction motor =

∴ Efficiency of Induction generator = Squirrel cage rotor:

Stator Cu loss = 3

I =

Cos =


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∴ Rotor Cu loss = ∴ Wound rotor Direct – on line (across the line) starting :-

The relation between starting torque and full load torque is ∴ =

The above equation valids of rotor resistance remains constant.

Where

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

Where

Here shunt branch parameters of equivalent circuit are neglected. Therefore, for direct switching,

∴ .Stator resistor (or reactor) starting :-

Since per phase voltage is reduced to xv, the per phase starting current is given by

As be fore =

In an induction motor, torque


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∴

Auto transformer starting :-

Per phase starting current from the supply mains is

Star – delta method of starting :

=

∴

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 = = 0.58 i.e with 58% tapping.

= =

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