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14/12/2017 Electrical Machines - I - - Unit 4 - Week 3
https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=15&assessment=35 1/5
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Courses » Electrical Machines - I
Unit 4 - Week 3
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Week1
Week 2
Week 3
Week 4
Week 5
Week 6
Week 7
Week 8
Week 9
Week 10
Lecture 7:VoltageRegulation ofSingle PhaseTransformers
Lecture 8:ParallelOperation ofSingle PhaseTransformers
Lecture 9:Harmonics andSwitchingTransients inSingle PhaseTransformers
Quiz : Week 3:Assignment
Week 3 :AssignmentSolution
Feedback forweek 3
.
1)
10 points
10 points2)
3)
Week 3: Assignment
A 100-kVA, 24/2.4 kV single-phase transformer is connected to a power source through a feeder ofimpedance 24.8 + j132 Ω. The equivalent series impedance of the transformer referred to its low voltageside is 0.20 + j0.40 Ω. The load on the low-voltage side of the transformer is 80 kW at 0.8 pf lagging and2200V. What is the voltage regulation of transformer in percentage?
Accepted Answers:(Type: Range) 0.80,0.86
The equivalent circuit of a 5.5 kVA, 220V/440V, 50 Hz transformer referred to HV side isshown in figure. What should be the applied voltage to the LV side when the transformer delivers ratedcurrent at 0.7 power factor lagging at a terminal voltage of 400V?
422 V
211 V
440 V
220 V
Accepted Answers:211 V
The following data were obtained for a 5 kVA, 50 Hz, 200/1000 V single phase transformer: O.C.Test (L.V. Side) : 200V, 1.2 A, 90W S.C. Test (H.V. Side): 50 V, 5A, 110W. What is the voltage regulation of the transformer in percentage, with full-load on the HV side at ratedvoltage, The load power factor being 0.8 lagging?
14/12/2017 Electrical Machines - I - - Unit 4 - Week 3
https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=15&assessment=35 2/5
Week 11
Week 12
10 points
4)
10 points
10 points5)
10 points6)
10 points7)
Accepted Answers:(Type: Range) 4,4.8
For the transformer given in question-3, find the voltage regulation in percentage, when the loadpower factor is 0.8 leading.
Accepted Answers:(Type: Range) -0.85,-0.78
A single phase transformer of rating 200kVA, 8000/480 V has an equivalent leakagereactance referred to secondary equal to 0.014 ohm. It is connected in parallel with another single phasetransformer of rating 350kVA, 8000/ 440 V having equivalent leakage reactance referred to secondaryequal to 0.026 ohm. What is the circulating current in the secondary windings of the parallel connectedtransformers at no-load, when rated voltage is applied at primary?
500 A
750 A
1000 A
1250 A
Accepted Answers:1000 A
For a transformer
Zero regulation occurs at lagging load and Maximum regulation occurs at leadingload.
Zero regulation occurs at leading load and Maximum regulation occurs at lagging load.
Zero regulation and Maximum regulation occurs at leading load.
Zero regulation and Maximum regulation occurs at lagging load.
Accepted Answers: Zero regulation occurs at leading load and Maximum regulation occurs at lagging load.
Which of the following diagrams approximately represents the inrush current in atransformer?
14/12/2017 Electrical Machines - I - - Unit 4 - Week 3
https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=15&assessment=35 3/5
14/12/2017 Electrical Machines - I - - Unit 4 - Week 3
https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=15&assessment=35 4/5
10 points8)
10 points9)
Accepted Answers:
The harmonic components present in the no load current of a transformer is
Accepted Answers:
Two transformers A and B are connected in parallel. The impedances referred tosecondary are ZA=0.15+j0.5 and ZB=0.1+j0.6. The no load terminal voltage are EA=200<0 and
EB=203<0. If the load impedance is 2+j2, find the power output of each transformer.
1.5kW and 7.6kW
4.7kW and 3.8kW
5.2kW and 6.5kW
10kW and 7kW
I = ∑∞n=2
In
I = ∑∞n=2,4,8
In
I = ∑∞n=3
In
I = ∑∞n=3,5,7
In
I = ∑∞n=3,5,7
In
14/12/2017 Electrical Machines - I - - Unit 4 - Week 3
https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=15&assessment=35 5/5
Accepted Answers:4.7kW and 3.8kW
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Assignment - 3 : Solution
Q1.Solution
The circuit is shown in Fig.1
Figure 1: Circuit Configuration
The secondary current I2 is given by
|I2| =80k
(2200)(0.8)= 45.45A
The power factor 0.8 lagging, corresponds to an impedance angle of 36.87 o.
Hence, the secondary current phasor is given by
I2 = 45.45∠− 36.87 oA
To find the voltage-regulation of the transformer, we have to find the primary side voltage of the transformer
referred to the secondary side
V′source = 2200 + 45.45∠− 36.87(0.2 + j0.40)
V′source = 2218.19∠0.23
Therefore, the voltage regulation of the transformer is
V R =2218.19− 2200
2200× 100 = 0.83%
Q2.Solution
The rated secondary current of the transformer is given by
I2 =5500
440= 12.5A
The primary side voltage, referred to the secondary side voltage, when the transformer delivers rated cur-rent at the secondary at 400V is given by
Assignment No :3 onlinecourses.nptel.ac.in Page 1 / 7
V′1 = 400 + 12.5∠− 45.57× (1.4 + j1.12)
V′1 = 422.26∠− 0.37
Hence, the magnitude of the actual voltage is
V1 =220
440× 422.26 = 211.13V
Q3.Solution
From OC test on LV side;
Equivalent core-loss resistance referrred to LV side is given by
Rc_LV =2002
90= 444.44Ω
The apparent power, under no load, Soc = VocIoc = 200× 1.2 = 240V A
Thus, the reactive power is
Qoc =√
2402 − 902 = 222.48 V AR
The magnetizing reactance, referred to LV side
Xm =2002
222.48= 179.79Ω
The core-loss resistacne, and magnetizing reactance, referred to HV side
Rc_HV =
(1000
200
)2
× 444.44 = 11.11 kΩ
Xm_HV =
(1000
200
)2
× 179.79 = 4.49 kΩ
Rated current at HV side
IHV _rated =5000
1000= 5 A
Since, the short-circuit test is conducted at HV side
The equivalent series resistance referred to HV side is given by
Req_HV =Psc
I2sc=
110
52= 4.4 Ω
The equivalent leakage impedance referred to HV side
Zeq_HV =Vsc
Isc=
50
5= 10 Ω
The equivalent lekage reactance referred to HV side
Xeq_HV =√
102 − 4.42 = 8.97Ω
Assignment No :3 onlinecourses.nptel.ac.in Page 2 / 7
Figure 2: Approximate equivalent circuit referred to HV side
The approximate equivalent circuit referred to the HV side is show in Fig.2.The transformer primary voltage referred to the HV side, when it is delivering full-load at 0.8 pf laggingwith rated voltage at the secondary is given by;
V′L = 1000 + 5∠− 36.86× (4.4 + j8.97)
V′L = 1044.75∠1.24
Hence, voltage regulation is given by
V R =1044.75− 1000
1000× 100 = 4.47%
Q4.Solution
When the load power factor is 0.8 leading, the primary side voltage referred to the HV side is given by;
V′L = 1000 + 5∠36.86× (4.4 + j8.97)
V′L = 991.91∠2.83
Hence, the voltage regulation is given by
V R =991.91− 1000
1000= −0.809%
Assignment No :3 onlinecourses.nptel.ac.in Page 3 / 7
Q5.Solution
Figure 3: Circuit diagram for problem - 5
From the Fig.3, the circulating current can be calculated as follows;
Icir =480− 440
0.014 + 0.026= 1000 A
Assignment No :3 onlinecourses.nptel.ac.in Page 4 / 7
Q7.Solution
Zero regulation occurs at leading load and Maximum regulation occurs at lagging load.
Figure 4: Regulation of transformer
Q8.Solution
Figure 5: Inrush current in a transformer
When a transformer is connected to the supply depending on what point in the supply cycle it is,
the inrush current magnitude may vary from two times rated to no inrush current.
Assignment No :3 onlinecourses.nptel.ac.in Page 5 / 7
Q9.Solution
The no load current in a transformer approximately looks like the figure given above.
Considering the odd nature of the current, only odd harmonics are present in the current
I = I1 +
∞∑n=3,5,7
In
Figure 6: Transformer No Load current
Q10.Solution
Lo
ad
V’ =20011
' V’ =20322
'
VL
Z1
Z2
ZL
Assignment No :3 onlinecourses.nptel.ac.in Page 6 / 7
IL =VL
ZL=
V′11 − VL
Z1+
V′22 − VL
Z2
Solving for VL
VL =V
′11Y1 + V
′22Y2
Y1 + Y2 + YL
Where Y1 = 1/Z1 = 0.55− j1.835
Y2 = 1/Z2 = 0.27− j1.622
and YL = 1/ZL = 0.25− j0.25
Subtituting and solving we get
VL = 185.44∠− 2.78V
∴ I1 =200∠0− VL
0.15 + j0.5
and I2 =203∠0− VL
0.1 + j0.6
I1 = 33.14∠− 42
I2 = 32.75∠− 53.7
∴ Total apparent power delivered by transformer 1 = I1V L = 33.14∠− 42× 185.44∠2.78
= 6145.5∠− 39.2
∴ Total apparent power delivered by transformer 2 = I2V L = 32.75∠− 53.7× 185.44∠2.78
= 6073.16∠− 50.92
∴ Real power delivered by transformer 1 = 6145.5 cos(39.2) = 4762.4 W
Real power delivered by transformer 2 = 6073.16 cos(50.92) = 3828.55 W
Assignment No :3 onlinecourses.nptel.ac.in Page 7 / 7
