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Distribution systems

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Distribution systems

  1. 1. DISTRIBUTION SYSTEMS
  2. 2. ELECTRICAL POWER SYSTEM • The distribution system is a part of the power system, existing between distribution sub-stations and the consumers.
  3. 3. DISTRIBUTION SYSTEMS
  4. 4. INTRODUCTION(DISTRIBUTION SYSTEMS) • Distribution systems To distribute the electric power among the consumer. Below a certain voltage General distribution scheme
  5. 5. Requirements of good distribution systems • Continuity in the power supply must be ensured. • Voltage must not vary more than the prescribed limits.(∓5%). • Efficiency of line must be high as possible. • Safe from consumer point of view. • Layout should not effect the appearance of locality. • Line should not be overloaded.
  6. 6. Distribution system is further classified on the basis of voltage 1.primary distribution systems 2.secondary distribution systems • Primary Distribution:-The part of the electrical-supply system existing between the distribution substations and the distribution transformers is called the primary system. • Secondary Distribution:-The secondary distribution system receives power from the secondary side of distribution transformers at low voltage and supplies power to various connected loads via service lines.
  7. 7. Distribution system DC Distribution system D.C Three wire system 1.General Distribution system AC Distribution system Ring main Distribution system Radial Distribution system
  8. 8. DC Distribution system 1.General Distribution system 2. D.C Three wire system1.General Distribution system • Feeder are used to feed the electrical power from the generating station to the substation. • Distributors are used to distribute the supply further from the substation. • Service mains are connected to the distributors so as to make the supply available at the consumers.(simplest two wire distribution system)
  9. 9. 2.D.C Three wire system • Voltage level can not be increased readily like a.c. • Method:-two generators are connected in series -each is generating a voltage of V volts -common point is neutral from where neutral wire is run. (too expensive , use to double the transmission voltage) • Demand :-consumers demanding higher voltage are connected to the two lines. -consumers demanding less voltage are connected between any one line and neutral.
  10. 10. D.C Three wire system
  11. 11. • Balanced:-One line carries current I1 while the other current I2.when the load is balanced(loads connected on either sides of the neutral wire are equal) .neutral current is zero. • Out of balance current:-I1 is greater than I2 then neutral wire carries current equal to I1-I2 -I2 is greater than I1 then neutral wire carries current equal to I2- I1. (Direction).neutral potential will not remain half of that between the 2 lines.
  12. 12. Single generator having twice the line • Two small d.c machines are connected across the line in series which are mechanically coupled to a common shaft . These are called balancers. • load is balanced:-machines work as the d.c motors. • Out of balance:-machine connected to lightly loaded side acts as motor , heavily loaded side acts as generator. Energy is transferred from lightly loaded side to heavily loaded side as machine as motor drives the machine as generator.
  13. 13. AC Distribution system 1.Radial Distribution system 2.Ring main Distribution system 1.Radial Distribution system • only one/single path is connected between each Distributor and substation is called radial Distribution system. • Fault occurs either on feeder or a distributor, all the consumers connected to that distributor will get affected. • In India, 99% of distribution of power is by radial distribution system only.
  14. 14. Advantages: • Its initial cost is minimum. • Simple in planning, design and operation. • Useful when the generation is at low voltage.. • Station is located at the center of the load Disadvantages: • Distributor nearer to the feeding end is heavily loaded. • The consumers at the far end of the feeder would be subjected to series voltage fluctuations with the variations in load.
  15. 15. 2.Ring main Distribution system • Feeder covers the whole area of supply in the ring fashion and finally terminates at the substation from where it is started. • Closed loop form and looks like a ring.
  16. 16. Advantages: • Less conductor material is required as each part of the ring carries less current than in the radial system. • Less voltage fluctuations. Disadvantage: • It is difficult to design when compared to the designing of a radial system.
  17. 17. Types of D.C Distributors D.C Distributors Concentrated loads Fed at both the end Ends at Unequal voltages End at Equal voltages Fed at one end Distributed loads Fed at one end Fed at both the end End at Equal voltages Ends at Unequal voltages
  18. 18. • Concentrated loads:-load which are acting at particular points of the distributor are called concentrated loads. • Distributed loads:-load which spread over the particular distance of the distributor are called distributed load.(no load condition practical) D.C Distributor with Concentrated loads • Classified 1. Fed at one end 2.Fed at both the ends
  19. 19. 1.Fed at one end(Concentrated loads)
  20. 20. • Fed at one end A-A’. • Applying KCL at various points we get, i1=I1+I2+I3,i2=I2+I3 and i3=I3 • the wire A’B’ is the return wire of the distributor r’=resistance per unit length of conductor in Ω • Voltage drop tabulated as, section Drop section Drop Aa i1l1r’ A’a’ i1l1r’ ab i2(l2-l1)r’ a’b’ i2(l2-l1)r’ bc i3(l3-l2)r’ b’c’ i3(l3-l2)r’
  21. 21. • In practice , the resistance of go and return conductor per unit length is assumed to be r=2r’. • r1=2r1’, r2=2r2’, r3=2r3’ • The total drop in the distributor is =r1i1l1+r2i2(l2-l1)+r3i3(l3-l2)
  22. 22. Current loading and voltage drop diagram
  23. 23. 2.Fed at both the ends(Concentrated loads) 1.End at Equal voltages • A and B maintained at equal voltage • ‘b’ be the point of minimum potential(the load point where the current are coming from both the side of distributor is the point of minimum potential. Let x be supplied by point A while y be supplied by point B, y=I2-x
  24. 24. • As both the point A and B are maintained at same voltage, drop in section Aa must be equal to drop in section Bb. i1r1+i2r2=i3r3+i4r4 (I1+x)r1+xr2=(I2-x)r3+(I2+I3-x)r4 • All current known , x and voltage drop can be calculated Current loading and voltage drop diagram
  25. 25. 2.Ends at Unequal voltages • A and B maintained at different voltage • ‘b’ be the point of minimum potential. • Let x be supplied by point A while y be supplied by point B, y=I2-x • Voltage drop between A and B = Voltage drop over AB • If voltage of A is V1 and is greater than voltage of B which is V2 then, V1-V2=drop in all the section of AB • The same equation can be written as, V1-drops over Ab= V2-drops over Bb
  26. 26. V1-i1r1-i2r2=V2-i3r3-i4r4 V1-(I1+x)r1-xr2=V2-(I2-x)r3-(I2+I3-x)r4 Here V1 and V2 are known ,obtain x Current loading and voltage drop diagram
  27. 27. D.C Distributor with Uniformly Distributed load • Classified 1. Fed at one end 2.Fed at both the ends 1.Fed at one end (Distributed load ) • Uniformly distributed load on 2 wire distributor , fed at one end I amperes per meter Total voltage drop is to be obtained by considering a point C(distance x),feeding end A • Current tapped at point C is =total current – current up to point ‘C’=i× 𝑙 − 𝑖 × 𝑥=i(l× 𝑥)
  28. 28. • dx near point C , its resistance rdx dV=i(l-x)rdx Total voltage drop upto point C Upto B, x=l 𝑉𝐴𝐶 = 0 𝑥 𝑖 𝑙 − 𝑥 𝑟 𝑑𝑥 = 𝑖𝑟 0 𝑥 𝑙 − 𝑥 𝑑𝑥 = 𝑖𝑟(𝑙𝑥 − 𝑥2 2 )0 𝑥 𝑉𝐴𝐶=𝑖𝑟(𝑙𝑥 − 𝑥2 2 ) volts equation of parabola 𝑉𝐴𝐵=𝑖𝑟(𝑙 × 𝑙 − 𝑙×𝑙 2 )=ir 𝑙2 2 = 1 2 (il)(rl) 𝑉𝐴𝐵 = 1 2 𝐼𝑅
  29. 29. • Power loss=I2R,elementary length dx • At point C X=l 𝑑𝑃 = [𝑖 𝑙 − 𝑥 ]2 𝑟 𝑑𝑥 𝑃 = 0 𝑙 𝑖2 𝑙2 − 2𝑙𝑥 + 𝑥2 𝑟 𝑑𝑥 = 𝑟 𝑖2(𝑙2 𝑥 − 2𝑙𝑥2 2 + 𝑥3 3 ) 𝑃 = 𝑖2 𝑟𝑙3 3
  30. 30. 2.Fed at both the ends (Uniformly Distributed load) 1.End at Equal voltages fed at point A and B are maintained at equal voltage The total current to be supplied is il amperes. As two end voltage are equal ,each end will supply half the required current i.e 𝑖𝑙 2 . Midpoint distance l/2,point C at a distance x , current feeding is il/2 (A) Current at C= 𝑖𝑙 2 − 𝑖𝑥 = 𝑖 𝑙 2 − 𝑥 Voltage drop over length dx is, 𝑑𝑣 = 𝑖 𝑙 2 − 𝑥 𝑟 𝑑𝑥
  31. 31. • Upto point C is, 𝑉𝐴𝐶 = 0 𝑥 𝑖 𝑙 2 − 𝑥 𝑟 𝑑𝑥 = 𝑖𝑟 𝑙𝑥 2 − 𝑥2 2 = 𝑖𝑟 2 [𝑙𝑥 − 𝑥2] Maximum voltage drop at midpoint x=l/2 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑟𝑜𝑝 = ir 𝑙2 4 − 𝑙2 8 = 𝑖𝑟𝑙2 8 = 1 8 𝑖𝑙 𝑟𝑙 = 𝐼𝑅 8 ¼ drop of fed at one end Power loss ,point c 𝑑𝑃 = [𝑖 𝑙 2 − 𝑥 ]2 𝑟 𝑑𝑥 𝑃 = 𝑖2 𝑟 0 𝑙 𝑙2 4 − 𝑙𝑥 + 𝑥2 𝑑𝑥 𝑃 = 𝑖2 𝑟𝑙3 12
  32. 32. • 2.Ends at Unequal voltages • Let point C be the point of minimum potential which at a distance x from feeding point A • The current supplied by the feeding point A is ix • The current supplied by the feeding point B is i(l-x) • V1-drops over AC= V2-drops over BC • In case of distributed load the drop is given by 𝑖𝑟𝑙2 2 for a length of l 𝑉𝐴𝐶 = 𝑖𝑟𝑥2 2 𝑣𝑜𝑙𝑡𝑠 𝑉𝐵𝐶 = 𝑖𝑟(𝑙 − 𝑥)2 2 𝑣𝑜𝑙𝑡𝑠 𝑉1 − 𝑖𝑟𝑥2 2 = 𝑉2 − 𝑖𝑟(𝑙 − 𝑥)2 2 X?
  33. 33. Ring main distributor with interconnection • Cable is arranged in the Loop fashion,fed at only one point • Use for large area hence voltage drop across the various section become larger(excessive voltage drop). • Solution:-distant point of ring distributor are joined together by a conductor this is called interconnection. • Thevenin’s theorem 𝐼 = 𝐸 𝑜 𝑅 𝑇𝐻 + 𝑅 𝐷𝐺
  34. 34. AC Distribution • Advantages of AC • Cheaper transformation between voltages • Easy to switch off • Less equipment needed • More economical in general • Rotating field
  35. 35. Methods of solving A.C Distribution problem • 1.power factor referred to receiving end voltage • Resistance R , reactance X • Impedance of section PR is given by, 𝑍 𝑃𝑅 = 𝑅1 + 𝑗𝑋1. • Impedance of section RQ is given by, 𝑍 𝑅𝑄 = 𝑅2 + 𝑗𝑋2. • The load current at point R is 𝐼1, 𝐼1 = 𝐼1(cos ∅1 − 𝑗 sin ∅1) • The load current at point Q is 𝐼2, 𝐼2 = 𝐼2(cos ∅2 − 𝑗 sin ∅2) • Current in section RQ is nothing but 𝐼 𝑅𝑄 = 𝐼2 = 𝐼2(cos ∅2 − 𝑗 sin ∅2) • Current in section PR is 𝐼 𝑃𝑅 = 𝐼1 + 𝐼2=𝐼1(cos ∅1 − 𝑗 sin ∅1)+𝐼2(cos ∅2 − 𝑗 sin ∅2)
  36. 36. 1.power factor referred to receiving end voltage
  37. 37. • Voltage drop in section RQ, 𝑉𝑅𝑄 = 𝐼 𝑅𝑄 𝑍 𝑅𝑄 𝑉𝑅𝑄=[𝐼2(cos ∅2 − 𝑗 sin ∅2)].[ 𝑅2 + 𝑗𝑋2] Voltage drop in section PR , 𝑉𝑃𝑅 = 𝐼 𝑃𝑅 𝑍 𝑃𝑅 =[𝐼1(cos ∅1 − 𝑗 sin ∅1) + 𝐼2(cos ∅2 − 𝑗 sin ∅2)][𝑅1 + 𝑗𝑋1]. Sending end voltage 𝑉𝑃 = 𝑉𝑄 + 𝑉𝑅𝑄 + 𝑉𝑃𝑄 Sending end current 𝐼 𝑃 = 𝐼1 + 𝐼2
  38. 38. 2. power factor referred to respective load voltages
  39. 39. • 2. power factor referred to respective load voltages Voltage drop in section RQ is given, 𝑉𝑅𝑄 = 𝐼2 𝑍 𝑅𝑄 𝑉𝑅𝑄=[𝐼2(cos ∅2 − 𝑗 sin ∅2)].[ 𝑅2 + 𝑗𝑋2] 𝑉𝑅 = 𝑉𝑄 +drop of voltage in section RQ=𝑉𝑅 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 α. 𝐼1 = 𝐼1(cos ∅1 − 𝑗 sin ∅1) w.r.t voltage 𝑉𝑅 𝐼1 = 𝐼1(cos(∅1−α) − 𝑗 sin(∅1 − α)) w.r.t voltage 𝑉𝑄 𝐼 𝑃𝑅 = 𝐼1 + 𝐼2 =𝐼1(cos(∅1−α) − 𝑗 sin(∅1 − α)) +𝐼2(cos ∅2 − 𝑗 sin ∅2) 𝑉𝑅𝑄=[𝐼1(cos(∅1−α) − 𝑗 sin(∅1 − α)) +𝐼2(cos ∅2 − 𝑗 sin ∅2)].[𝑅1 + 𝑗𝑋1] Sending end voltage 𝑉𝑃, 𝑉𝑃 = 𝑉𝑄 + 𝑉𝑅𝑄 + 𝑉𝑃𝑅
  40. 40. THANK YOU

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