DISTRIBUTION SYSTEMS

ELECTRICAL POWER SYSTEM

• The distribution system is a part of the power system, existing between distribution

sub-stations and the consumers.

DISTRIBUTION SYSTEMS

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 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)

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.

D.C Three wire system

• 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

Current loading and voltage drop diagram

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

Current loading and voltage drop diagram

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)0𝑥

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

2𝐼𝑅

• 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.e𝑖𝑙

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 = 𝑖𝑙

2− 𝑥 𝑟 𝑑𝑥

• Upto point C is,

𝑉𝐴𝐶 = 0

𝑥

𝑖𝑙

2− 𝑥 𝑟 𝑑𝑥 = 𝑖𝑟

𝑙𝑥

2−𝑥2

2=𝑖𝑟

2[𝑙𝑥 − 𝑥2]

Maximum voltage drop at midpoint x=l/2

𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑟𝑜𝑝 = ir𝑙2

4−𝑙2

8=𝑖𝑟𝑙2

8=1

8𝑖𝑙 𝑟𝑙 =

𝐼𝑅

8

¼ drop of fed at one end

Power loss ,point c

𝑑𝑃 = [𝑖𝑙

2− 𝑥 ]2𝑟 𝑑𝑥

𝑃 = 𝑖2𝑟 0

𝑙 𝑙2

4− 𝑙𝑥 + 𝑥2 𝑑𝑥

𝑃 =𝑖2𝑟𝑙3

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

2for a length of l

𝑉𝐴𝐶 =𝑖𝑟𝑥2

2𝑣𝑜𝑙𝑡𝑠

𝑉𝐵𝐶 =𝑖𝑟(𝑙 − 𝑥)2

2𝑣𝑜𝑙𝑡𝑠

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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