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ELECTRICAL POWER SYSTEM
• The distribution system is a part of the power system, existing between distribution
sub-stations and the consumers.

INTRODUCTION(DISTRIBUTION SYSTEMS)
• Distribution systems
To distribute the electric
power among the consumer.
Below a certain voltage
General distribution scheme

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.

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.

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 system
1.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)

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.

• 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.

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.

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.

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.

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.

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.

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

• 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

1.Fed at one end(Concentrated loads)

• 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’

• 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)

Current loading and voltage drop diagram

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

• 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

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

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

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)

• 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

• 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(𝑙2𝑥 −
2𝑙𝑥2
2
+
𝑥3
3
)
𝑃 =
𝑖2𝑟𝑙3
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.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

• 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

• 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?

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 =
𝐸𝑜
𝑅𝑇𝐻 + 𝑅𝐷𝐺

AC Distribution
• Advantages of AC
• Cheaper transformation between voltages
• Easy to switch off
• Less equipment needed
• More economical in general
• Rotating field

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, 𝑍𝑃𝑅 = 𝑅1 + 𝑗𝑋1.
• Impedance of section RQ is given by, 𝑍𝑅𝑄 = 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 𝐼𝑅𝑄 = 𝐼2 = 𝐼2(cos ∅2 − 𝑗 sin ∅2)
• Current in section PR is 𝐼𝑃𝑅 = 𝐼1 + 𝐼2=𝐼1(cos ∅1 − 𝑗 sin ∅1)+𝐼2(cos ∅2 − 𝑗 sin ∅2)

1.power factor referred to receiving end voltage

• Voltage drop in section RQ, 𝑉𝑅𝑄 = 𝐼𝑅𝑄 𝑍𝑅𝑄
𝑉𝑅𝑄=[𝐼2(cos ∅2 − 𝑗 sin ∅2)].[ 𝑅2 + 𝑗𝑋2]
Voltage drop in section PR , 𝑉𝑃𝑅 = 𝐼𝑃𝑅 𝑍𝑃𝑅
=[𝐼1(cos ∅1 − 𝑗 sin ∅1) + 𝐼2(cos ∅2 − 𝑗 sin ∅2)][𝑅1 + 𝑗𝑋1].
Sending end voltage 𝑉𝑃 = 𝑉𝑄 + 𝑉𝑅𝑄 + 𝑉𝑃𝑄
Sending end current 𝐼𝑃 = 𝐼1 + 𝐼2

2. power factor referred to respective load
voltages

• 2. power factor referred to respective load voltages
Voltage drop in section RQ is given, 𝑉𝑅𝑄 = 𝐼2 𝑍𝑅𝑄
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
𝐼𝑃𝑅 = 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

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