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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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
4Ω
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
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Week 12 : Assignment Solution
10 points
4)
10 points
10 points5)
6)
10 points
10 points7)
Voltage drop