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DISTRIBUTION SYSTEMS
ELECTRICAL POWER SYSTEM
โข The distribution system is a part of the power system, existing between distribution
sub-stations and the consumers.
DISTRIBUTION SYSTEMS
INTRODUCTION(DISTRIBUTION SYSTEMS)โข Distribution systems
To distribute the electric
power among the consumer.
Below a certain voltage
General distribution scheme
Requirements of good distribution systems
โข Continuity in the power supply must be ensured.
โข Voltage must not vary more than the prescribed limits.(โ5%).
โข Efficiency of line must be high as possible.
โข Safe from consumer point of view.
โข Layout should not effect the appearance of locality.
โข Line should not be overloaded.
Distribution system is further classified on the basis of voltage 1.primary distribution systems
2.secondary distribution systems โข Primary Distribution:-The part of the electrical-supply system existing between
the distribution substations and the distribution transformers is called the primary
system.
โข 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.
Distribution system
DC Distribution
system
D.C Three wire system
1.General Distribution
system
AC Distribution
system
Ring main Distribution
system
Radial Distribution
system
DC Distribution system 1.General Distribution system
2. D.C Three wire system1.General Distribution system
โข Feeder are used to feed the electrical power from the generating station to the substation.
โข Distributors are used to distribute the supply further from the substation.
โข Service mains are connected to the distributors so as to make the supply available at the consumers.(simplest two wire distribution system)
2.D.C Three wire system
โข Voltage level can not be increased readily like a.c.
โข 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)
โข Demand :-consumers demanding higher voltage are connected to the two lines.
-consumers demanding less voltage are connected between any one line
and neutral.
D.C Three wire system
โข 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.
โข 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.
Single generator having twice the line
โข Two small d.c machines are connected across
the line in series which are mechanically coupled
to a common shaft . These are called
balancers.
โข load is balanced:-machines work as the d.c
motors.
โข 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.
AC Distribution system 1.Radial Distribution system
2.Ring main Distribution system
1.Radial Distribution system
โข only one/single path is connected between each Distributor and substation is called radial Distribution system.
โข Fault occurs either on feeder or a distributor, all the consumers connected to that distributor will get affected.
โข In India, 99% of
distribution of power
is by radial distribution
system only.
Advantages:
โข Its initial cost is minimum.
โข Simple in planning, design and operation.
โข Useful when the generation is at low voltage..
โข Station is located at the center of the load
Disadvantages:
โข Distributor nearer to the feeding end is heavily loaded.
โข The consumers at the far end of the feeder would be subjected to series voltage
fluctuations with the variations in load.
2.Ring main Distribution system
โข Feeder covers the whole area of supply in the ring fashion and finally terminates at
the substation from where it is started.
โข Closed loop form and looks like a ring.
Advantages:
โข Less conductor material is required as each part of the ring carries less current than
in the radial system.
โข Less voltage fluctuations.
Disadvantage:
โข It is difficult to design when compared to the designing of a radial system.
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
โข Concentrated loads:-load which are acting at particular points of the distributor are
called concentrated loads.
โข 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
โข Classified 1. Fed at one end
2.Fed at both the ends
1.Fed at one end(Concentrated loads)
โข Fed at one end A-Aโ.
โข Applying KCL at various points we get,
i1=I1+I2+I3,i2=I2+I3 and i3=I3
โข the wire AโBโ is the return wire of the distributor
rโ=resistance per unit length of conductor in ฮฉ
โข Voltage drop tabulated as,
section Drop section Drop
Aa i1l1rโ Aโaโ i1l1rโ
ab i2(l2-l1)rโ aโbโ i2(l2-l1)rโ
bc i3(l3-l2)rโ bโcโ i3(l3-l2)rโ
โข In practice , the resistance of go and return conductor per unit length is assumed to be r=2rโ.
โข r1=2r1โ, r2=2r2โ, r3=2r3โ
โข The total drop in the distributor is
=r1i1l1+r2i2(l2-l1)+r3i3(l3-l2)
Current loading and voltage drop diagram
2.Fed at both the ends(Concentrated loads)
1.End at Equal voltagesโข A and B maintained at equal voltage
โข โbโ 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
โข 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
โข All current known ,
x and voltage drop
can be calculated
Current loading and voltage drop diagram
2.Ends at Unequal voltages
โข A and B maintained at different voltage
โข โbโ be the point of minimum potential.
โข Let x be supplied by point A while y be supplied by point B,
y=I2-x
โข Voltage drop between A and B = Voltage drop over AB
โข 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
โข The same equation can be written as,
V1-drops over Ab= V2-drops over Bb
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
D.C Distributor with Uniformly Distributed loadโข Classified 1. Fed at one end
2.Fed at both the ends
1.Fed at one end (Distributed load )
โข 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
โข Current tapped at point C is
=total current โ current up to point โCโ=iร ๐ โ ๐ ร ๐ฅ=i(lร ๐ฅ)
โข dx near point C , its resistance rdx
dV=i(l-x)rdx
Total voltage drop upto point C
Upto B, x=l
๐๐ด๐ถ = 0
๐ฅ
๐ ๐ โ ๐ฅ ๐ ๐๐ฅ = ๐๐ 0
๐ฅ
๐ โ ๐ฅ ๐๐ฅ = ๐๐(๐๐ฅ โ๐ฅ2
2)0๐ฅ
๐๐ด๐ถ=๐๐(๐๐ฅ โ๐ฅ2
2) volts equation of parabola
๐๐ด๐ต=๐๐(๐ ร ๐ โ๐ร๐
2)=ir
๐2
2=1
2(il)(rl)
๐๐ด๐ต =1
2๐ผ๐
โข Power loss=I2R,elementary length dx
โข At point C
X=l
๐๐ = [๐ ๐ โ ๐ฅ ]2 ๐ ๐๐ฅ
๐ = 0
๐
๐2 ๐2 โ 2๐๐ฅ + ๐ฅ2 ๐ ๐๐ฅ = ๐ ๐2(๐2๐ฅ โ2๐๐ฅ2
2+๐ฅ3
3)
๐ =๐2๐๐3
3
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.e๐๐
2.
Midpoint distance l/2,point C at a distance x , current feeding is il/2 (A)
Current at C=๐๐
2โ ๐๐ฅ = ๐
๐
2โ ๐ฅ
Voltage drop over length dx is,
๐๐ฃ = ๐๐
2โ ๐ฅ ๐ ๐๐ฅ
โข Upto point C is,
๐๐ด๐ถ = 0
๐ฅ
๐๐
2โ ๐ฅ ๐ ๐๐ฅ = ๐๐
๐๐ฅ
2โ๐ฅ2
2=๐๐
2[๐๐ฅ โ ๐ฅ2]
Maximum voltage drop at midpoint x=l/2
๐๐๐ฅ๐๐๐ข๐ ๐๐๐๐ = ir๐2
4โ๐2
8=๐๐๐2
8=1
8๐๐ ๐๐ =
๐ผ๐
8
ยผ drop of fed at one end
Power loss ,point c
๐๐ = [๐๐
2โ ๐ฅ ]2๐ ๐๐ฅ
๐ = ๐2๐ 0
๐ ๐2
4โ ๐๐ฅ + ๐ฅ2 ๐๐ฅ
๐ =๐2๐๐3
12
โข 2.Ends at Unequal voltages
โข Let point C be the point of minimum potential which at a distance x from feeding point A
โข The current supplied by the feeding point A is ix
โข The current supplied by the feeding point B is i(l-x)
โข V1-drops over AC= V2-drops over BC
โข In case of distributed load the drop is given by ๐๐๐2
2for a length of l
๐๐ด๐ถ =๐๐๐ฅ2
2๐ฃ๐๐๐ก๐
๐๐ต๐ถ =๐๐(๐ โ ๐ฅ)2
2๐ฃ๐๐๐ก๐
๐1 โ๐๐๐ฅ2
2= ๐2 โ
๐๐(๐ โ ๐ฅ)2
2
X?
Ring main distributor with interconnection
โข Cable is arranged in the Loop fashion,fed at only one point
โข Use for large area hence voltage drop across the various section become larger(excessive voltage drop).
โข Solution:-distant point of ring distributor are joined together by a conductor this is called interconnection.
โข Theveninโs theorem
๐ผ =๐ธ๐
๐ ๐๐ป + ๐ ๐ท๐บ
AC Distribution
โข Advantages of AC
โข Cheaper transformation between voltages
โข Easy to switch off
โข Less equipment needed
โข More economical in general
โข Rotating field
Methods of solving A.C Distribution problem
โข 1.power factor referred to receiving end voltage
โข Resistance R , reactance X
โข Impedance of section PR is given by, ๐๐๐ = ๐ 1 + ๐๐1.
โข Impedance of section RQ is given by, ๐๐ ๐ = ๐ 2 + ๐๐2.
โข The load current at point R is ๐ผ1, ๐ผ1 = ๐ผ1(cos โ 1 โ ๐ sinโ 1)
โข The load current at point Q is ๐ผ2, ๐ผ2 = ๐ผ2(cos โ 2 โ ๐ sinโ 2)
โข Current in section RQ is nothing but ๐ผ๐ ๐ = ๐ผ2 = ๐ผ2(cos โ 2 โ ๐ sinโ 2)
โข Current in section PR is ๐ผ๐๐ = ๐ผ1 + ๐ผ2=๐ผ1(cosโ 1 โ ๐ sinโ 1)+๐ผ2(cosโ 2 โ ๐ sinโ 2)
1.power factor referred to receiving end voltage
โข Voltage drop in section RQ, ๐๐ ๐ = ๐ผ๐ ๐ ๐๐ ๐
๐๐ ๐=[๐ผ2(cos โ 2 โ ๐ sin โ 2)].[ ๐ 2 + ๐๐2]
Voltage drop in section PR , ๐๐๐ = ๐ผ๐๐ ๐๐๐
=[๐ผ1(cos โ 1 โ ๐ sin โ 1) + ๐ผ2(cos โ 2 โ ๐ sin โ 2)][๐ 1 + ๐๐1].
Sending end voltage ๐๐ = ๐๐ + ๐๐ ๐ + ๐๐๐
Sending end current ๐ผ๐ = ๐ผ1 + ๐ผ2
2. power factor referred to respective load
voltages
โข 2. power factor referred to respective load voltages
Voltage drop in section RQ is given, ๐๐ ๐ = ๐ผ2 ๐๐ ๐
๐๐ ๐=[๐ผ2(cos โ 2 โ ๐ sin โ 2)].[ ๐ 2 + ๐๐2]
๐๐ = ๐๐ +drop of voltage in section RQ=๐๐ ๐๐๐๐๐ ๐๐ ฮฑ.
๐ผ1 = ๐ผ1(cos โ 1 โ ๐ sin โ 1) w.r.t voltage ๐๐ ๐ผ1 = ๐ผ1(cos(โ 1โฮฑ) โ ๐ sin(โ 1 โ ฮฑ)) w.r.t voltage ๐๐
๐ผ๐๐ = ๐ผ1 + ๐ผ2
=๐ผ1(cos(โ 1โฮฑ) โ ๐ sin(โ 1 โ ฮฑ)) +๐ผ2(cos โ 2 โ ๐ sin โ 2)
๐๐ ๐=[๐ผ1(cos(โ 1โฮฑ) โ ๐ sin(โ 1 โ ฮฑ)) +๐ผ2(cos โ 2 โ ๐ sinโ 2)].[๐ 1 + ๐๐1]
Sending end voltage ๐๐ , ๐๐ = ๐๐ + ๐๐ ๐ + ๐๐๐