Assignment - 12 : Solution - NPTEL ... Assignment - 12 : Solution Q1.Solution Armaturecurrent, I a =

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  • Assignment - 12 : Solution

    Q1.Solution

    Armature current, Ia = 70A

    From the table, at N = 900 rpm and Ia = 70A, Ea = 188V

    In regenerative breaking Ea = (Ra +Rse + δRB)× Ia

    At maximum speed δ = 0.9. Hence, E ′ a = (0.15 + 0.9× 3)× 70 = 199.5V

    E ′ a

    Ea = Nmax 900

    Nmax = 199.5

    188 × 900 = 955 rpm

    Q2.Solution

    T1 = KfI 2 a1

    T2 = KfI 2 a2

    Ia2 = Ia1 × √ T2 T1

    = 100× √

    2 = 141.4A

    E1 = KeIa1N1

    E2 = KeIa2N2

    E2 = E1 × N2 N1 × Ia2 Ia1

    = (220− (100× 0.1))× 1000 1500

    × 141.4 100

    = 210× 1000 1500

    × 141.4 100

    = 197.96V

    E2 = Ia2 × (RB + 0.1)

    197.96 = 141.4× (RB + 0.1)

    RB = 1.3 Ω

    Q3.Solution

    When the diverter resistance is not connected, the armature current and field current of the machine are same.

    Without field diverter If1 = Ia1 = 45A

    Assignment No : 12 onlinecourses.nptel.ac.in Page 1 / 7

  • I = 45Aa1

    n1 = 1500 rpm

    R = 0.06Ωf

    R = 0.08Ωa 220V

    n2

    R = 0.1Ωd

    If2

    Ia2R = 0.06Ωf

    Figure 1:

    With field diverter If2 = Rd

    Rf +Rd × Ia2

    = 0.1

    0.06 + 0.1 =

    5

    8 Ia2

    Since, load torque remains the same, T1 = T2

    KtI 2 a1 = KtIf2Ia2

    452 = If2Ia2

    452 = 5

    8 I2a2

    I2a2 = 3240

    Ia2 = 56.92A

    If2 = 35.57A

    Back emfs, Eb1 = V − Ia1(Ra +Rf )

    KgIa1n1 = 220− 45(0.08 + 0.06)

    Kg × 45× 1500 = 213.7V

    With diverter, Eb2 = V − Ia2 ( Rf ×Rd Rf +Rd

    +Ra

    ) KgIf2n2 = 220− 56.92

    ( 0.06× 0.1 0.1 + 0.06

    + 0.08

    ) Kg × 35.57× n2 = 213.3119V Kg × 45× 1500 Kg × 37.57× n2

    = 213.7

    213.3119

    n2 = 1793 rpm

    Assignment No : 12 onlinecourses.nptel.ac.in Page 2 / 7

  • Q4.Solution

    Total input Pin = (Vtm + Vfm + Vgm)Iam

    = (600 + 42 + 42)× 57 = 38, 988W

    Generator output Pout = 520× 38 = 19, 760W

    Total losses in the two machines Ploss = 38, 988− 19, 760 = 19, 228W

    Total Cu-loss = (572 × 0.2) + (382 × 0.2) + 57(42 + 42) = 5726.6W

    No-load rotational loss of both the machines = 19, 228− 5726.6 = 13501.4W

    No-load rotational loss of each machine Prot = 13501.4

    2 = 6750.7W

    Motor input, Pin_m = (600 + 42)× 57 = 36, 594W

    Motor losses, Ploss_m = (572 × 0.2) + (57× 42) + 6750.7 = 9794.5W

    Efficiency of Motor ηm = (

    1− 9794 36, 594

    ) × 100 = 73.23%

    Q5.Solution

    Total losses in the generator, Ploss_g = (382 × 0.2) + (57× 42) + 6750.7 = 9433.5W

    Generator input, Pin_g = Pout + Ploss_g

    = 19, 760 + 9433.5 = 29193.5W

    Efficiency of Generator ηg = (

    1− 9433.5 29193.5

    ) × 100 = 67.68%

    Q6.Solution

    DC series motor cannot be run at no load condition. Hence Swinburne’s test cannot be be conducted on a DC series motor.

    Assignment No : 12 onlinecourses.nptel.ac.in Page 3 / 7

  • Q7.Solution

    Back emf Eb1 when running at 1600rpm at 40A = 240− 30× 0.2 = 234V

    φ ∝ Ia φ2 φ1

    = Ia2 Ia1

    Ia2 = 10A =⇒ φ2 φ1

    = 1

    3

    Speed (N) ∝ Eb φ

    =⇒ N2 N1

    = Eb2φ1 Eb1φ2

    3200

    1600 = Eb2 Eb1 × 3

    =⇒ Eb2 = Eb1 × 2

    3

    ∴ Eb2 = 234× 2/3

    = 156V

    ∴ Let the external resistance to be added be Rext

    Rtot = Rext +Ra +Rse

    Where, Ra +Rse = 0.2ohm

    Eb2 = V − Ia2Rtot

    156 = 240− 10×Rtot

    =⇒ Rtot = 240− 156

    10 = 8.4ohm

    ∴ Rext = Rtot −Ra −Rse

    Rext = 8.4− 0.2 = 8.2ohm

    Q8.Solution

    In the given question, data regardingArmature resistance and field winding resistance we not given. Hence the question will be cancelled. But the solution considering Armature resistance=1.1 ohm and resistance of each field winding=0.4 ohm is solved below

    Assignment No : 12 onlinecourses.nptel.ac.in Page 4 / 7

  • Ra1 = 1.1 + 2× 0.4 = 1.9ohm

    Ra2 = 1.1 + 2× 0.4/2 = 1.3ohm

    Eb1 = 250− Ia1Ra1 = 250− 20× 1.9 = 212V

    Eb2 = 250− Ia2Ra2

    = 250− 1.3Ia2 −−−−−−(1)

    Since ouput for both motors are same

    Eb1Ia1 = Eb2Ia2

    Eb2Ia2 = 250× 20−−−−−−− (2)

    Using (2) in (1)

    (250− 1.3Ia2)Ia2 = 4240

    Reaaranging, we get

    1.3I2a2 − 250Ia2 + 4240 = 0

    Ia2 = 18.8A

    =⇒ Eb2 = 4240

    18.8 = 225.5V

    Field current in the second case = 18.8

    2 = 9.4A

    N2 N1

    = Eb2φ1 Eb1φ2

    = 225.5× 20 212× 9.4

    N2 = 2263.467rpm

    Assignment No : 12 onlinecourses.nptel.ac.in Page 5 / 7

  • Q9.Solution

    For any DC motor T ∝ φIa

    For Shunt motor

    ==================

    For a shunt motor flux (φ)is constant and hence torques T rises linearly with Ia

    Speed for a shunt motor is nearly constant with slight reduction in speed due to IaRa drop

    Hence the Torque-speed charecteristics will be a straight line almost parallel to x axis

    For Series motor

    ==================

    In series motor T ∝ I2a Also flux φ ∝ Ia

    Hence as load current increases Speed decreases

    Hence we get an inverse relationship between Torque and speed

    For Compound motor

    ==================

    A compound motor combines the charecteristics of a shunt and series motor.

    Hence we expect the characteristics of it to lie between that of shunt and series motor

    Ans = Shunt motor, Compound motor and Series motor

    Assignment No : 12 onlinecourses.nptel.ac.in Page 6 / 7

  • Q10.Solution

    The load-speed charecteristics of a Differential compund motor is shown in figure

    It can be seen that at some point in the characteristics crosses the no load speed represented by dashed lines

    At this point the speed regulation is zero

    Assignment No : 12 onlinecourses.nptel.ac.in Page 7 / 7

  • 14/12/2017 Electrical Machines - I - - Unit 13 - Week 12

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    Courses » Electrical Machines - I

    Unit 13 - Week 12

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    Lecture 38 : Speed Control of DC Series Motors

    Lecture 39 : Testing of DC Series Motors

    Lecture 40 : Characteristics of Compound DC Series Motors

    Quiz : Week 12: Assignment

    Additional Questions on Speed Control of DC Motor

    Due on 2017-10-18, 23:59 IST.

    10 points1)

    10 points2)

    3)

    Week 12: Assignment The due date for submitting this assignment has passed.

    Submitted assignment

    A 240V,70A dc series motor has a combined armature circuit resistance Ra+Rse = 0.15Ω. The magnetization curve expressed in terms of Ea vesus Ia at 900 rpm is given by the following table.

    EA,V 95 150 188 212 229 243 Ia, A 30 52 70 78 85 90

    A chopper whose duty ratio can be changed from 0.1 to 0.9 is used to brake the motor dynamically. What is the maximum speed the motor can achieve when the armature current is 70A and the breaking resistance is 3Ω?

    850 rpm

    1045 rpm

    894 rpm

    955 rpm

    No, the answer is incorrect. Score: 0

    Accepted Answers: 955 rpm

    The armature winding and field winding resistance of a 220V, 100A, 1500 rpm dc series motor are 0.05Ω each. The motor is operated under dynamic breaking at twice the rated torque and at 1000 rpm. Calculate the value of braking resistance. Assume linear magnetic circuit.

    5.1Ω

    2.6Ω

    1.3Ω

    No, the answer is incorrect. Score: 0

    Accepted Answers: 1.3Ω

    Field test is performed on two identical DC series machines, one acting as a motor and the other one as a generator. The readings obtained are given below: For Motor: Armature current = 57 A Armature voltage = 600 V Voltage drop across field winding = 42 V For Generator Armature current = 38 A Armature voltage = 520 V

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  • 14/12/2017 Electrical Machines - I - - Unit 13 - Week 12

    https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=93&assessment=98 2/4

    Week 12 : Assignment Solution

    10 points

    4)

    10 points

    10 points5)

    6)

    10 points

    10 points7)

    Voltage drop