Assignment - 12 : Solution - NPTEL...Assignment - 12 : Solution Q1.Solution Armaturecurrent, I a =...
Transcript of Assignment - 12 : Solution - NPTEL...Assignment - 12 : Solution Q1.Solution Armaturecurrent, I a =...
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 = KfI2a1
T2 = KfI2a2
Ia2 = Ia1 ×√T2T1
= 100×√
2 = 141.4A
E1 = KeIa1N1
E2 = KeIa2N2
E2 = E1 ×N2
N1× Ia2Ia1
= (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 aresame.
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 = 0.08Ωa220V
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
8Ia2
Since, load torque remains the same, T1 = T2
KtI2a1 = KtIf2Ia2
452 = If2Ia2
452 =5
8I2a2
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 ona 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 ∝ Iaφ2φ1
=Ia2Ia1
Ia2 = 10A =⇒ φ2φ1
=1
3
Speed (N) ∝ Eb
φ
=⇒ N2
N1=Eb2φ1Eb1φ2
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 ohmand 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=Eb2φ1Eb1φ2
=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 ∝ I2aAlso 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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Lecture 38 :Speed Controlof DC SeriesMotors
Lecture 39 :Testing of DCSeries Motors
Lecture 40 :Characteristicsof CompoundDC SeriesMotors
Quiz : Week 12:Assignment
AdditionalQuestions onSpeed Controlof DC Motor
Due on 2017-10-18, 23:59 IST.
10 points1)
10 points2)
3)
Week 12: AssignmentThe 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 243Ia, 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. Whatis the maximum speed the motor can achieve when the armature current is 70A and the breakingresistance 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 seriesmotor are 0.05Ω each. The motor is operated under dynamic breaking at twice the rated torque and at1000 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 otherone 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
14/12/2017 Electrical Machines - I - - Unit 13 - Week 12
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Week 12 :AssignmentSolution
10 points
4)
10 points
10 points5)
6)
10 points
10 points7)
Voltage drop across the field winding = 42 V Resistance of armature of each machine = 0.2Ω Calculate the efficiency of the motor in percentage? [Enter only the numerical value. Do not enter any symbols]
No, the answer is incorrect. Score: 0
Accepted Answers:(Type: Range) 71,75
For the data given in problem-5, find the efficiency of the generator in percentage? [Enter only the numerical value. Do not enter any symbols]
No, the answer is incorrect. Score: 0
Accepted Answers:(Type: Range) 66.5,68.5
Swinburne’s test cannot be performed on which of thefollowing motors?
Shunt DC motor
Series DC motor
Separately excited DC motor
Swinburne’s test can be conducted for all types of DC motors.No, the answer is incorrect. Score: 0
Accepted Answers:Series DC motor
A 240V series motor runs at 1600rpm and draws a current of 30A.The sum of armature resistance and series field resistance is 0.2ohm.What should be the value of external resistance that has to beconnected so that the motor runs at 3200rpm at 10A? Note:Please enter only the numeric value. No units
No, the answer is incorrect. Score: 0
Accepted Answers:(Type: Range) 8-8.5
Speed torque characteristics of various DC motors are showby curves A,B and C. A, B and C respectively represents motors:
14/12/2017 Electrical Machines - I - - Unit 13 - Week 12
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10 points8)
Compound motor, Series motor and Shunt motor
Series motor, Shunt motor and Compound motor
Shunt motor, Compound motor and Series motor
Shunt motor, Series motor and Compound motorNo, the answer is incorrect. Score: 0
Accepted Answers:Shunt motor, Compound motor and Series motor
Which of the following DC motors can run on zero speedregulation even at loaded condition?
Shunt motor
Differential compound motor
Series motor
Cumulative compound motorNo, the answer is incorrect. Score: 0
Accepted Answers:Differential compound motor
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