ppt

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DEPARTMENT OF TECHNICAL EDUCATION
ANDHRA PRADESH.
Name : P.V.KRISHNA MURTHY
Designation : Senior Lecturer
Branch : Electrical & Electronic Engg.
Institute : G.M.R. Polytechnic., Srisailam
Year/semester : v Semester
Subject : Power systems-II
Subject code : EE 504
Topic : Distribution
Sub topic : Problems on AC distribution
Duration : 50 min
Teaching Aids : ppt, diagrams.

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We have learnt in the last session about
• Solving some problems on A.C radial distributor
RECAP

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OBJECTIVES
On the completion of this lesson the you would
be able to
• Solve problems on A.C. radial distributor.

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PROBLEM:
1)The loading on a distributor is shown in fig.1 the distributor
is a two- core cable for which the resistance and
reactance are 0.25Ώ and 0.125 Ώper 1000 meters of
cable run respectively. what should be the voltage at the
point A to maintain 400Vat the point D?

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C
I1=50A
0.8 lag
I3=40 A
u. p.f
Fig.1
A B D
I2=20A
0.5 lag
100 m 150 m 150 m

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SOLUTION:
• The impedance of cable
=(0.25+j0.125)Ώper meters

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• Impedance of section AB,
Z1=(0.25+j0.125)×100/1000
=90.025+j0.0125) Ώ

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• Impedance of section BC,
Z2=(0.25+j0.125) ×150/1000
=(0.0375+j0.01875) Ώ

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• Impedance of section CD,
Z3 =(0.0375+j0.01875)Ώ

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• Current flowing from B,
I1 =50(0.8-j0.6)
=(40-j30)A

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• Current flowing from C,
I2 =20(0.5-j0.866)
=(10-j17.32)A

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Current flowing from D,
I3 =40(1.0-j0)
=(40-j0)A

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• Assuming all power factors referred to far end voltage
VD
voltage at D,
VD=400+j0

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• taking VD as reference vector
voltage at C, VC=VD+I3Z3
=(400+j0)+(40-j0)(0.0375+j0.01875)
=(401.5+j0.75)V

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• voltage at B, VB=Vc+(I2+I3)Z2
=(401.5+j0.75)[(10-j17.32)+(40-j0)](0.01875)
=(401.5+j0.75)+(50-j17.32)(0.0375+j0.01875)
=(403.7+j1.0380)volts

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• Voltage at A,VA=VB+(I1+I2+I3)Z1
=(403.7+j1.038)+[(40-j30)+(10-j17.32)
+(40-](0.025+j0.0125)
=(403.7+j0.038)+(90-j47.32)(0.025+j0.0125)
=(406.542+j0.980)
=406.542<0.14° volts

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SUMMARY
• Solved problem on A.C. radial distribution
We have discussed in this session about

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QUIZ
1) In case of radial distribution , the section nearest feeding
point is
a) Over loaded always
b) Under loaded always
c) Over loaded or under loaded depends upon load p.f
d) None

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1. A two wire AC feeder is loaded as shown in fig-1.
the power factors are lagging and are referred to the
voltage at the respective load points.
The section impedance FA=0.03+j0.05 and
AB=0.05+j0.08 ohm. If the voltage at the far and is to
be maintained at 230 volt. calculate the voltage at the
supply end. (MARCH/APRIL-2005)
Frequently Asked In Examination

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F A B
100 A
0.707
p.f
200 A
0.8 p.f
Fig.2

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2. A two wire distributor 1200 meters long is loaded as
shown in fig-2. B-is the mid point .The power factors at
the two load points refer to the voltage at ’C’. The
impedances of each line is (0.15+j0.2)Ώfor 1200m.
Calculate the sending end voltage and current. the
voltage at point C is 220volts
(MARCH/APRIL-2006)

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A B C
600m 600m
100 Amp
0.8 p.f. lag
60 Amp
0.9 p.f lag
Fig.3

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