Andrew

• Principles of Power System
• Introduction


  1. In the beginning of the electrical age, electricity was generated as a direct current and voltages were low

  2. The resistance losses in the lines made it impracticable to transmit and distribute power for more than a few localities of the city

  3. With the development of the transformer, a.c. has taken over the load formerly supplied by d.c.

  4. Now-a-days, electrical energy is generated, transmitted and distributed in the form of a.c. as an economical proposition

  5. The transformer permits the transmission and distribution of a.c. power at high voltages

  6. This has greatly reduced the current in the conductors (and hence their sizes) and the resulting line losses
• For certain applications, d.c. supply is absolutely necessary
• Examples of applications requiring d.c. supply

  • Variable speed machinery (e.g. d.c. motors)
  • Electrochemical work
  • Electric traction
• Converting a.c. to d.c.

  A.c. power is converted into d.c. power at the sub-station by using converting machinery e.g. mercury are rectifiers, rotary converters and motor-generator sets
• Types of D.C. Distributors

  • Distributor fed at one end
  • Distributor fed at both ends
  • Distributor fed at the centre
  • Ring distributor
• Distributor fed at one end

  • The current in the various sections of the distributor away from feeding point goes on decreasing
  • The voltage across the loads away from the feeding point goes on decreasing
  • If a fault occurs on any section of the distributor, the whole distributor will have to be disconnected from the supply mains
• Distributor fed at both ends

  • The load voltage goes on decreasing as we move away from one feeding point, reaches minimum value and then again starts rising and reaches maximum value when we reach the other feeding point
  • The minimum voltage occurs at some load point and is never fixed
  • If a fault occurs on any feeding point of the distributor, the continuity of supply is maintained from the other feeding point
  • In case of fault on any section of the distributor, the continuity of supply is maintained from the other feeding point
  • The area of X-section required for a doubly fed distributor is much less than that of a singly fed distributor
• Distributor fed at the centre

  • It is equivalent to two singly fed distributors, each distributor having a common feeding point and length equal to half of the total length
• Ring mains


  • It is equivalent to a straight distributor fed at both ends with equal voltages, the two ends being brought together to form a closed ring
  • The distributor ring may be fed at one or more than one point
• Concentrated loads


  Loads which act on particular points of the distributor
• Distributed loads


  Loads which act uniformly on all points of the distributor
• The distributor is so designed that the minimum potential on it is not less than 6% of rated voltage at the consumer's terminals
• D.C. Distributor Fed at one End - Concentrated Loading

  1. Current fed from point A = I1 + I2 + I3 + I4

  2. Current in section AC = I1 + I2 + I3 + I4

  3. Current in section CD = I2 + I3 + I4

  4. Current in section DE = I3 + I4

  5. Current in section EF = I4

  6. Voltage drop in section AC = r1 (I1 + I2 + I3 + I4)

  7. Voltage drop in section CD = r2 (I2 + I3 + I4)

  8. Voltage drop in section DE = r3 (I3 + I4)

  9. Voltage drop in section EF = r4 I4

  10. Total voltage drop in the distributor = r1 (I1 + I2 + I3 + I4) + r2 (I2 + I3 + I4) + r3 (I3 + I4) + r4 I4
• Maximum permissible drop in the distributor

  10 V
• Voltage at the feeding point

  To maintain 250 V at the far end of the distributor
• [260 V]
• Distributor fed at both ends

  • 13.5
  • 13.5
  • 13.5
  • 13.5
  • 13.5
• Distributor Fed at Both Ends
• Distributor Fed at Both Ends — Concentrated Loading
• Whenever possible, it is desirable that a long distributor should be fed at both ends instead of at one end only, since total voltage drop can be considerably reduced without increasing the cross-section of the conductor
• Two ends of the distributor may be supplied with

  • (i) equal voltages
  • (ii) unequal voltages
• Distributor A B fed at both ends with equal voltages V volts and having concentrated loads I1, I2, I3, I4 and I5 at points C, D, E, F and G respectively
• As we move away from one of the feeding points, say A, p.d. goes on decreasing till it reaches the minimum value at some load point, say E, and then again starts rising and becomes V volts as we reach the other feeding point B
• All the currents tapped off between points A and E (minimum p.d. point) will be supplied from the feeding point A while those tapped off between B and E will be supplied from the feeding point B
• The current tapped off at point E itself will be partly supplied from A and partly from B
• At the point of minimum potential, current comes from both ends of the distributor
• Distributor A B having three concentrated loads I1, I2 and I3 at points C, D and E respectively
• Voltage drop between A and B = Voltage drop over A B
• The load point where the currents are coming from both sides of the distributor is the point of minimum potential i.e. point E in this case
• Distributor A B fed with unequal voltages; end A being fed at V 1 volts and end B at V 2 volts
• Equation 1


  90 − (300 − IA) × 0·04 (6 − x)
• Equation 2


  600 − 0·04 x IA = 590 − 0·04 (1800 − 300 x − 6 IA + IA × x)
• Equation 3


  600 − 0·04 x IA = 590 − 72 + 12 x + 0·24 IA − 0·04 xIA
• Equation 4


  0·24 IA = 82 − 12 x
• Equation 5


  IA = 341·7 − 50 x
• Equation 6


  VM = V A − (341·7 − 50 x) × 0·04 x
• Equation 7


  VM = 600 − 13·7 x + 2x2
• To find minimum VM

  1. Differentiate equation 7 w.r.t. x and set to 0

  2. 0 − 13·7 + 4x = 0

  3. x = 13·7/4 = 3·425 km
• Minimum potential occurs at a distance of 3·425 km from the sub-station A