Machines Ty s

download Machines Ty s

of 14

  • date post

    12-Jan-2016
  • Category

    Documents

  • view

    215
  • download

    0

Embed Size (px)

description

read

Transcript of Machines Ty s

  • Electrical EngineeringElectrical Machines

    WORKBOOKWORKBOOKWORKBOOKWORKBOOKWORKBOOK

    2016

    Detailed Explanations ofTry Yourself Questions

  • Copyrightwww.madeeasypublications.org

    T1 : Solution

    (c)(c)(c)(c)(c)As shown in figure below,

    A A

    0.3

    0.2

    0.2

    0.3

    Ia1 Ia2

    If1 If2

    300 V300 V

    500 AA

    B

    300 0.2 I = 300 0.3

    I

    I (0.2) =

    I (0.3)

    II

    =

    =I I

    +I I = 500

    +I I = 500

    I =

    = 200

    I = 1.5

    I = 1.5 200 = 300 A

    andI (0.3) =

    I (0.2)

    II

    =

    I = 1.5

    I

    I +

    I = 500 A

    I + 1.5

    I = 500

    I =

    = 200

    DC Machines1

  • Copyright www.madeeasypublications.org

    3Try yourself

    T2 : Solution

    (34)(34)(34)(34)(34)

    = 500 0.5 50 = 475 ...(1)

    Let the current when joined in series = I2 AThe resistance = 0.5 2 = 1

    = 500 I2 1 = 500 I2 ...(2)

    =

    = = ...(3)

    where, k =

    From (1), (2) and (3)

    I=

    500 I2 =

    ...(4)

    =

    = =

    I II

    ( T2 = 2T1)

    100 = kI2

    k =

    I...(5)

    From (5) and (4),

    500 I2 =

    =

    I I I2 = 34 A

    T3 : Solution

    (0.85)(0.85)(0.85)(0.85)(0.85)

    Input power =

    =

    Input line current =

    =

    Shunt field current =

    =

    Back emf,

    = 500 0.4 85.7 = 466.5 V

    Since the load and the constant losses are constant

    I =

    I = 466.5 83.7 = 39046

    Total resistance after the series field is added= 0.4 + 0.1 = 0.5

  • 4 Electrical Engineering Electrical Machines

    Copyrightwww.madeeasypublications.org

    (500 0.5 I2)I2 = 39046

    I 1000 I2 + 78092 = 0

    This gives, I2 = 85 AThe input current = 85 + 2 = 87 A

    The efficiency =

    =

    = 0.85

    T4 : Solution

    (403)(403)(403)(403)(403)

    T =

    I

    209 =

    I = 10.34 Ia

    Ia = the line current =

    Eb =

    =

    =

    440 =

    or, N =

    =

    T5 : Solution

    (1989)(1989)(1989)(1989)(1989)

    Eg =

    200 =

    or, = ! !

    The flux density, B = "#

    =

    =

    $

    The ampere-turns required for 0.8 cm gap

    =

    l =

    = 1989

  • Copyright www.madeeasypublications.org

    T1 : Solution

    (d)(d)(d)(d)(d)Maximum allowable flux, max = Bmax core area

    = 1.2 0.025= 0.03 Wb

    Voltage on primary (high voltage) side,V1 = 6000 V

    Number of turns of (low voltage) side of main transformer

    N2 =

    = = 60

    Number of turns on low voltage side of teaser transformer= 0.866 N2 = 0.866 60 = 52

    Point to be tapped =

    =

    T2 : Solution

    (c)(c)(c)(c)(c)

    pu resistance =

    =

    pu reactance =

    =

    Voltage regulation will be maximum, when phase angle,

    =

    $

    $

    II

    = "#"#

    "%%"#

    =

    =

    Power factor, cos = cos 63.435 = 0.447 lagging (for maximum regulation)Maximum regulation = pu resistance cos + pu reactance sin

    = 0.025 0.447 + 0.05 0.8944

    = 0.0559 or 5.59%

    Transformers2

  • 6 Electrical Engineering Electrical Machines

    Copyrightwww.madeeasypublications.org

    T3 : Solution

    (b)(b)(b)(b)(b)At 3/4 Full load and unity power factor output

    =

    =

    Maximum efficiency, max = 97%

    Input =

    = =

    Total loss = Input output = 386.6 375 = 11.6 kW

    This loss is equally divided between iron and copper loss, so

    = Pi = =or, Pc =

    =

    Percentage resistance =

    I

    =

    II

    = &

    =

    Percentage impedance = 10%

    Percentage reactance = = 9.786%

    Percentage regulation =

    #% %

    + I I

    = 2.06 0.8 + 9.786 0.6 = 7.52%

    T4 : Solution

    (c)(c)(c)(c)(c)Rated output, S = 25 kVASecondary voltage, V2 = 220 VFull-load (or rated) secondary current,

    I2 =

    = = 113.64 A

    Total resistance referred to secondary,

    = K2R1 + R2 =

    +

    = 0.02

    Full-load copper loss, Pc = I = (113.64)2 0.02

  • Copyright www.madeeasypublications.org

    7Try yourself

    = 258 W or 0.258 kWIron loss, Pi = 80% of full load copper loss

    =

    = 206.4 W or 0.2064 kW

    Full-load output at 0.8 p.f. = 25 cos = 25 0.8= 20 kW

    Full-load efficiency at 0.8 p.f., = '

    + i

    =

    + +

    = 97.93%

  • Copyrightwww.madeeasypublications.org

    T1 : Solution

    (a)(a)(a)(a)(a)Since the stator resistance is neglected

    Stator out-put = Rotor out-put

    =( !

    =

    =

    s

    if E2 = rotor stand-still voltage per phase

    rotor current at slips =

    if s1 and s2 are the slips with two given stator out-puts we have

    = 9 ...(i)

    and

    = 4 ...(ii)

    From (i) and (ii)

    = 2.25

    since, s1 =

    )

    =

    so,

    = 2.25 =

    2.25 (ns n2) = ns n1 2.25 ns ns = 2.25 n2 n1 1.25 750 = 2.25 n2 n1 ...(iii)

    as

    = 1.043 (given)

    from equation (iii)1.25 750 = 2.25 n2 1.043 n2

    n2 = 776.719 777 rpmn1 = 1.043 776.719 810 rpm

    Induction Machine3

  • Copyright www.madeeasypublications.org

    9Try yourself

    T2 : Solution

    (c)(c)(c)(c)(c)

    Torque at any slip, T =

    +...(1)

    This will be maximum,when R = sXSince, maximum torque occurs at s = 0.2

    R = 0.2 X X = 5 R ...(2)At starting, s = 1

    Tst =

    +

    with R increased to R + 0.5,

    =

    +

    + +( X = 5R) ...(3)

    putting, S =

    in (1)

    Tmax =

    $

    = = ( X = 5R) ...(4)

    Tst is 75% of Tmax, from (3) and (4)

    +

    + +=

    =

    38 R2 17R + 0.75 = 0 R = 0.395 or 0.05 and X = (5 0.395) or (5 0.05)or X = 1.97 or 0.25

    T3 : Solution

    (b)(b)(b)(b)(b)

    Synchronous speed, Ns =

    =

    Slip, s =

    =

    Since the torque in maximum (stalling)

    = s = 0.16

    X 2 =

    = = 0.44

  • 10 Electrical Engineering Electrical Machines

    Copyrightwww.madeeasypublications.org

    If the total rotor resistance = (in the second case)

    = s at starting = 1

    = X 2 = 0.44

    The resistance to be added = 0.44 0.07 = 0.37

    T4 : Solution

    (b)(b)(b)(b)(b)

    Ns =

    =

    N1 = 1425 rpm S1 =

    =

    N2 = 1275 rpm S2 =

    =

    Slip = # %%

    =

    (

    I

    slip R2

    =

    + +=

    R =

    = 0.5 /ph

  • Copyright www.madeeasypublications.org

    T1 : Solution

    (a)(a)(a)(a)(a)Given data,Given data,Given data,Given data,Given data, z s = (0.5 + j4) = 4.03182.8Vt = 400 V, Ef = 500 V, 2 = 90 2 = (90 82.8) = 7.12

    shaft power output = maximum output power friction windage and iron losses

    maximum output power =

    =

    = 41.922 kWshaft power output = (3 42.922 kW) 0.9 kW = 124.867 kW

    T2 : Solution

    (0.98)(0.98)(0.98)(0.98)(0.98)Taking 11 kV, 1600 KVA as base

    The pu synchronous reactance =

    = 0.4

    if V1 = 10 = terminal voltage and I = 1Then the generated voltage

    = + I

    or, = + = + + = 1 + 0.4(sin + jcos) = 1 0.04 sin + j0.4cos

    For, zero regulation, magnitude of this is 1 (1 0.4 sin )2 + (0.4cos)2 = 1 1 + 0.42 0.8 sin = 1

    sin =

    = 11.5 power factor, cos = cos11.5 = 0.98

    Synchronous Machine4

  • 12 Electrical Engineering Electrical Machines

    Copyrightwww.madeeasypublications.org

    T3 : Solution

    (b)(b)(b)(b)(b)

    Vt =

    = 265.58 V

    Ef =

    = 317.54 V

    zs =

    + = + = 1.67585.44

    Gross Pm(output)maximum =

    +

    =

    + = 45.53 kW

    net Pm (output) maximum = 45.53 3 1.4 = 135.19 kWT4 : Solution

    (c)(c)(c)(c)(c)

    P =

    % %

    +

    =

    + =

    +

    = 0 for P = Pmax

    0 =

    +

    1.1 cos + cos 2 = 0 2cos2 1 + 1.1 cos = 0

    cos =

    +=

    = 61.12

  • Copyright www.madeeasypublications.org

    13Try yourself

    T5 : Solution

    (b)(b)(b)(b)(b)

    Phase voltage =

    *

    =

    I =

    =

    I

    as reference,

    drop = I

    = 131(0.4 + j6) = 52 + j786The terminal voltage = 3810(0.8 + j 0.6)

    = 3048 + j 2286 V The generated voltage = 3100 + j3072 V

    = 4.36444.7 kV ...(i )with I as reference, if x kV is the load phase voltage

    Terminal voltage = 0.8 x j 0.6 x kV

    I

    = 0.052 + j 0.786 kV Generated voltage = (0.8 x + 0.052) + j (0.786 0.6 x) ...(ii )From equation (i) and (ii), (0.8 x + 0.052)2 + (0.786 0.6x)2 = 4.3642

    x 2 0.86 x 18.424 = 0x = 4.744

    line voltage = 4.744 1000 = 8220 V

  • Copyrightwww.madeeasypublications.org

    Single-Phase Motor& Special Machine,

    Energy Conversion Principles5T1 : Solution

    (5.22)(5.22)(5.22)(5.22)(5.22)

    S =

    = 0.06

    The circuit model is given below,

    Im Im 1.8 7.8

    1.8 7.8

    Im

    j48

    j48

    220 V

    3.4 s = 5.67

    3.4(2 s) = 1.75

    = j48 (1.8 + 56.7 + j7.8) = j48 (58.5 + j7.8)= 3554 = 20.6 + j28.3

    = j48 (1.8 + 1.75 + j7.8) = j48 (3.55 + j7.8)= 7.3669.1 = 2.63 + j6.88

    = (20.6 + j 28.3) + (2.63 + j6.88) = 42.1656.6

    I

    =

    =

    IL = IM = 5.22 A