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EE 504 .74 1
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.
EE 504 .74 2
• In the previous class, we discussed
• A.C. distribution calculations , p.f of various load currents
• Solved problem on A.C. distribution
RECAP
EE 504 .74 3
OBJECTIVES
Upon the completion of this lesson you would
be able to
• Solve some more problems on A.C. radial distributor.
EE 504 .74 4
PROBLEM
• A 2-wire feeder ABC has a load of 120A at C and of
60A at both at p.f 0.8 lagging .the impedance AB is
(0.04+j0.08) Ώand that of BC is (0.08+j0.12) Ώ.if the
voltage at the far end C is to be maintained at
400V,determine the voltage (i) at A and (ii) at B.
EE 504 .74 5
(0.04+j0.08)Ώ (0.08+j0.12)Ώ
I1=60A,0.8 Lag I2=120A,0.8 Lag
Fig.1
EE 504 .74 6
SOLUTION
• Impedance of section AB,
Z1=(0.04+j0.08) Ώ
• Impedance of section BC,
Z2 =(0.08+j0.12) Ώ
EE 504 .74 7
• Taking voltages at far end C, Vc as reference vector
• Assuming power factors of currents at load points
Band Care given referred to the receiving end voltage
at C, Vc
Vc=400+j0
EE 504 .74 8
• current through section BC=I2=120(0.8-J0.6)
=(96-j72)A
• current through section AB=I1+I2
I=60(0.8-j0.6)+120(0.8-j0.6)
=(144-J108)A
EE 504 .74 9
•Voltage drop in section BC
=I2Z2=(96-j72)(0.08+j0.12)
=(16.32+j5.76)volts
EE 504 .74 10
•Voltage drop in section AB
=(I1+I2)Z1
=( 144-j108)(0.04+j0.08)
=(14.4+j7.2)volts
EE 504 .74 11
• Voltage at B,VB =VC+I2Z2
=(400+J0)+(16.32+J5.763)
=416.32+J5.76
=416.36 0.7390 V

EE 504 .74 12
•Voltage at supply end A,VA=VB+(I1+I2)Z1
=(416.32+J5.76)+(14.4+J7.2)
=(430.72+j12.96)volts
=430.915 1.720Volts

EE 504 .74 13
SUMMARY
We have discussed in this session about
• the problem solving on A.C. radial distributor
EE 504 .74 14
QUIZ
1) For calculation of voltage drop in A.C. distributor, the
data required is
a) Resistance of the distributor
b) Reactance of the distributor
c) Load currents and power factors
d) All of the above
EE 504 .74 15
QUESTIONS
1) Explain the method of calculating voltage drop in
A.C. distributor ?
2) Explain the differences in voltage drop calculation
methods of D.C. and A.C. distributions?
EE 504 .74 16

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EE504.74.ppt

  • 1. EE 504 .74 1 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.
  • 2. EE 504 .74 2 • In the previous class, we discussed • A.C. distribution calculations , p.f of various load currents • Solved problem on A.C. distribution RECAP
  • 3. EE 504 .74 3 OBJECTIVES Upon the completion of this lesson you would be able to • Solve some more problems on A.C. radial distributor.
  • 4. EE 504 .74 4 PROBLEM • A 2-wire feeder ABC has a load of 120A at C and of 60A at both at p.f 0.8 lagging .the impedance AB is (0.04+j0.08) Ώand that of BC is (0.08+j0.12) Ώ.if the voltage at the far end C is to be maintained at 400V,determine the voltage (i) at A and (ii) at B.
  • 5. EE 504 .74 5 (0.04+j0.08)Ώ (0.08+j0.12)Ώ I1=60A,0.8 Lag I2=120A,0.8 Lag Fig.1
  • 6. EE 504 .74 6 SOLUTION • Impedance of section AB, Z1=(0.04+j0.08) Ώ • Impedance of section BC, Z2 =(0.08+j0.12) Ώ
  • 7. EE 504 .74 7 • Taking voltages at far end C, Vc as reference vector • Assuming power factors of currents at load points Band Care given referred to the receiving end voltage at C, Vc Vc=400+j0
  • 8. EE 504 .74 8 • current through section BC=I2=120(0.8-J0.6) =(96-j72)A • current through section AB=I1+I2 I=60(0.8-j0.6)+120(0.8-j0.6) =(144-J108)A
  • 9. EE 504 .74 9 •Voltage drop in section BC =I2Z2=(96-j72)(0.08+j0.12) =(16.32+j5.76)volts
  • 10. EE 504 .74 10 •Voltage drop in section AB =(I1+I2)Z1 =( 144-j108)(0.04+j0.08) =(14.4+j7.2)volts
  • 11. EE 504 .74 11 • Voltage at B,VB =VC+I2Z2 =(400+J0)+(16.32+J5.763) =416.32+J5.76 =416.36 0.7390 V 
  • 12. EE 504 .74 12 •Voltage at supply end A,VA=VB+(I1+I2)Z1 =(416.32+J5.76)+(14.4+J7.2) =(430.72+j12.96)volts =430.915 1.720Volts 
  • 13. EE 504 .74 13 SUMMARY We have discussed in this session about • the problem solving on A.C. radial distributor
  • 14. EE 504 .74 14 QUIZ 1) For calculation of voltage drop in A.C. distributor, the data required is a) Resistance of the distributor b) Reactance of the distributor c) Load currents and power factors d) All of the above
  • 15. EE 504 .74 15 QUESTIONS 1) Explain the method of calculating voltage drop in A.C. distributor ? 2) Explain the differences in voltage drop calculation methods of D.C. and A.C. distributions?