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Transcript of gradeup ELECRIAL MACHINES (Formula Notes) gradeup ELECRIAL MACHINES (Formula Notes) gradeup 1...

  • 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→ Specific weight of t/f =

    K

    We

    VA

    i

    g

    ra

    ht

    t

    i

    of

    n fg of 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

    0.85 → Flux = mmF

    Reluctance = = ϕm sin ωt

    → According to faradays second law e1 = −N1 dϕ dt

    = −N1 d dt

    �ϕm sin ωt�

    → Transformer emf equations :- E1 = 4.44 N1 Bmax Anf  (1) E2 = 4.44 N2 Bmax Anf  (2)

    N1 → Emf per turn in Iry = E1 = 4.44 BmaxAnf

    N2 → Emf per turn in IIry = E2 = 4.44 Bmax An f

    ⟹ Emf per turn on both sides of the transformer is same E1 N1 N2

    = E2

    ⟹ E1 E2

    = N1 N2

    = 1 k

    Transformation ratio = K = E2 E1

    =

    Instantaneous value of emf in primary

    e1 = N1ϕm ω sin�ωt − π�2�

    N2 N1

    Turns ratio = 1 K

    = N1 ∶ N2

    ELECRIAL MACHINES (Formula Notes)gradeup

    gradeup 1

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  • → 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 f

    ∝ V1

    Bmax = constant ⟹ V1 f

    = constant

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

    No load /shunt branch.

    N1 N2

    E2 E1 V1

    I0

    Iw R0 X0

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  • → No load current = I0 = Iµ + Iw = I0 �−ϕ0

    Iw = I0 cos ϕ0 Iµ = I0 sin ϕ0 → No load power = v1I0 cos ϕ0 = v1Iw = Iron losses.

    R0 = v1

    Iw1 ; X0 =

    v1 Iµ

    ⟹ Iw = No load power

    V1

    Transferring from 𝐈𝐈𝐈𝐈𝐫𝐫𝐫𝐫 to 𝐈𝐈𝐫𝐫𝐫𝐫:-

    I 2 2 R2 = I 12 R 21

    R 21 = R2 � I2�

    2

    K2

    I1

    = R2

    ∴ R 21 = K R2

    2

    From 𝐈𝐈𝐫𝐫𝐫𝐫 to 𝐈𝐈𝐈𝐈𝐫𝐫𝐫𝐫 :- I 1 2R1 = I 2 2. R 11

    R 11 = I 1 2

    I 2 2 . R1

    R 1 1 = R1. K2

    → Total resistance ref to primary = R1 + R 21 R01 = R1 + R2/k2

    → Total resistance ref to secondary = R2 + R 1 1 R02 = R2 + k2R1

    → Total Cu loss = I 12R01 Or I 2 2 R02

    R R211

    Per unit resistance drops :-

    E1 → P.U primary resistance drop = I1R1

    E2 → P.U secondary resistance drop = I2 R2

    E1 → Total P.U resistance drop ref to Iry = I1R01

    E2 → Total P.U resistance drop ref to IIry = I2 R02

    → The P.U resistance drops on both sides of the t/f is same I1 R01 = I2R02

    E1 E2

    Losses present in transformer :-

    1. Copper losses

    2. Iron losses

    3. Stray load losses

    4. Dielectric losses

    major losses

    Iron parts minor losses

    t/f windings

    t/f core cu parts

    insulating materials.

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  • 1. Cu losses in t/f:

    Total Cu loss = I 1 2R1 + I 2 2R1 = I 12R01

    E1

    = I 2 2 R02 → Rated current on Iry = VA rating of t/f

    E2 Similarly current on IIry = VA rating of t/f

    VA rating of t/f

    → Cu losses ∝ I 1 2 or I 2 2. Hence there are called as variable losses. → P.U Full load Cu loss = FL Cu loss in watts

    = I 1 2R01 E1I1

    → 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 I ry

    or P. U resistance ref to Iry

    � = I1R01 E1 I1

    × I1

    = I 1 2R01 E1I1

    ∴ 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 ma2 x f 2. t2

    Constant Supply freq

    thickness of laminations.

    Area under one hysteresis loop. xη B max . f . v Wh =

    (it is a function of σ )

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  • During operation of transformer :- Bm ∝

    V1 f

    f Case (i) :- V1 = constant, Bmax = const.

    we ∝ f 2

    we = B f 2

    Const.

    ∴ wi = wh + we

    wi = Af + Bf 2

    When Bmax = const.

    Case (ii) :- V1 f

    ≠ constant, Bm ≠ const.

    we ∝ � V1 f

    � 2

    . f 2

    we ∝ V1 2

    wi = wh + we

    wi = A V

    0

    . 1

    6

    1.6

    f + BV1 2

    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 :-  W0 = Losses in t/f under no load condition

    = Iron losses + Dielectric loss + no load primary loss (I 0 2 R1)

     Constant losses = W0 − I 0 2R1 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 VSC

    � 2

    (Wi)S.C = Wi × � VSC

    V1 rated �

    2

    ∴ Variable losses = WSC − (Wi)SC � VSC

    V1 rated �

    2

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  • → 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

    SC

    2 SC

    Efficiency :-

     Efficiency of transformer is given by η = output power input power

    = output power output power+losses

    = E2 I2 cos ϕ2 E2 I2 cos ϕ2+ F.L cu losses+Iron losses

    ηF.L = E2 I2 cos ϕ2

    2E2I2 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 V R

    ′ 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�