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MODULE III
Distribution Feeders
That part of power system which distributes electric power for local use is
known as distribution system. In general, the distribution system is the electrical system between
the sub-station fed by the transmission system and the consumers meters. It generally consists of
feeders, distributors and the service mains. The distribution system is commonly broken down
into three components: distribution substation, distribution primary and secondary. At the
substation level, the voltage is reduced and the power is distributed in smaller amounts to the
customers.
(i) Feeders. A feeder is a conductor which connects the sub-station (or localised generating
station) to the area where power is to be distributed. Generally, no tappings are taken from the
feeder so that current in it remains the same throughout. The main consideration in the design of
a feeder is the current carrying capacity.
(ii) Distributor. A distributor is a conductor from which tappings are taken for supply to the
consumers. In fig , AB, BC, CD and DA are the distributors. The current through a distributor is
not constant because tappings are taken at various places along its length. While designing a
distributor, voltage drop along its length is the main consideration since the statutory limit of
voltage variations is ± 6% of rated value at the consumers’ terminals.
(iii) Service mains. A service mains is generally a small cable which connects the distributor to
the consumers’ terminals
Classification of Distribution Systems
A distribution system may be classified according to ;
(i) Nature of current. According to nature of current, distribution system may be classified as (a)
d.c. distribution system (b) a.c. distribution system. Now-a-days, a.c. system is universally
adopted for distribution of electric power as it is simpler and more economical than direct current
method.
(ii) Type of construction. According to type of construction, distribution system may be
classified as (a) overhead system (b) underground system. The overhead system is generally
employed for distribution as it is 5 to 10 times cheaper than the equivalent underground system.
In general, the underground system is used at places where overhead construction is
impracticable or prohibited by the local laws.
(iii) Scheme of connection. According to scheme of connection, the distribution system may be
classified as (a) radial system (b) ring main system (c) inter-connected system.
A.C. Distribution
The a.c. distribution system is classified into (i) primary distribution
system and (ii) secondary distribution system
(i) Primary distribution system. It is that part of a.c. distribution system which operates
atvvoltages somewhat higher than general utilisation and handles large blocks of electrical
energy than the average low-voltage consumer uses. The voltage used for primary distribu tion
depends upon the amount of power to be conveyed and the distance of the substation required to
be fed. The most commonly used primary distribution voltages are 11 kV, 6·6 kV and 3·3 kV.
Due to economic considerations, primary distribution is carried out by 3-phase, 3-wire system
Fig shows a typical primary distribution system. Electric power from the generating station is
transmitted at high voltage to the substation located in or near the city. At this substation, voltage
is stepped down to 11 kV with the help of step-down transformer. Power is supplied to various
substations for distribution or to big consumers at this voltage. This forms the high voltage
distribution or primary distribution.
(ii) Secondary distribution system. It is that part of a.c. distribution system which includes the
range of voltages at which the ultimate consumer utilises the electrical energy delivered to him.
The secondary distribution employs 400/230 V, 3-phase, 4-wire system.Fig. shows a typical
secondary distribution system. The primary distribution circuit delivers power to various
substations, called distribution substations. The substations are situated near the consumers’
localities and contain stepdown transformers. At each distribution substation, the voltage is
stepped down to 400 V and power is delivered by 3-phase,4-wire a.c. system. The voltage
between any two phases is 400 V and between any phase and neutral is 230 V. The single phase
domestic loads are connected between any one phase and the neutral, whereas 3-phase 400 V
motor loads are connected across 3-phase lines directly
FEEDER LOADING – voltage drop in feeder lines with different loadings
The load connected to feeder can be of any manner
In figure the load points with load currents I1, I2, I3 are at distaces l1, l2,l3 from the substation
feeding point. There are different types of loads.
1)
2)
3)
4)
Load at the end of the line
Load uniformly distributed
Uniformly incresing load
Uniformly decreasing load
D.C Distributors
Concentrated
loads
Fed at both the
end
Ends at Unequal
voltages
Fed at one end
End at Equal
voltages
Distributed
loads
Fed at one end
Fed at both the
end
End at Equal
voltages
Ends at Unequal
voltages
(i) Distributor fed at one end. In this type of feeding, the distributor is connected to the supply
at one end and loads are taken at different points along the length of the distributor.
Fig shows the single line diagram of a d.c. distributor AB fed at the end A (also known as singly
fed distributor) and loads I1, I2 and I3 tapped off at points C, D and E respectively.
The following points are worth noting in a singly fed distributor :
(a) The current in the various sections of the distributor away from feeding point goes on
decreasing. Thus current in section AC is more than the current in section CD and current in
section CD is more than the current in section DE.
(b) The voltage across the loads away from the feeding point goes on decreasing. the minimum
voltage occurs at the load point E.
(c) In case a fault occurs on any section of the distributor, the whole distributor will have to be
disconnected from the supply mains. Therefore, continuity of supply is interrupted.
D.C. Distributor Fed at one End—Concentrated Loading
shows the single line diagram of a 2-wire d.c. distributor AB fed at one end A and having
concentrated loads I1, I2, I3 and I4 tapped off at points C, D, E and F respectively.
Let r1, r2, r3 and r4 be the resistances of both wires (go and return) of the sections AC, CD, DE
and EF of the distributor respectively.
Current fed from point A = I1 + I2 + I3 + I4
Current in section AC = I1 + I2 + I3 + I4
Current in section CD = I2 + I3 + I4
Current in section DE = I3 + I4
Current in section EF = I4
Voltage drop in section AC = r1 (I1 + I2 + I3 + I4)
Voltage drop in section CD = r2 (I2 + I3 + I4)
Voltage drop in section DE = r3 (I3 + I4)
Voltage drop in section EF = r4 I4
∴ Total voltage drop in the distributor
= r1 (I1 + I2 + I3 + I4) + r2 (I2 + I3 + I4) + r3 (I3 + I4) + r4 I4
It is easy to see that the minimum potential will occur at point F which is farthest from the
feeding point A.
Uniformly Loaded Distributor Fed at One End
shows the single line diagram of a 2-wire d.c. distributor AB fed at one end A and loaded
uniformly with i amperes per metre length. It means that at every 1 m length of the distributor,
the load tapped is i amperes. Let l metres be the length of the distributor and r ohm be the
resistance per metre run.
Consider a point C on the distributor at a distance x metres from the feeding point A as shown in
Fig
Then current at point C is
= i l − i x amperes = i (l − x) amperes
Now, consider a small length dx near point C. Its resistance is r dx and the voltage drop over
length dx is
dv = i (l − x) r dx = i r (l − x) dx
Total voltage drop in the distributor upto point C is
The voltage drop upto point B (i.e. over the whole distributor) can be obtained by putting x = l in
the above expression.
∴ Voltage drop over the distributor AB
where i l = I, the total current entering at point A
r l = R, the total resistance of the distributor
Thus, in a uniformly loaded distributor fed at one end, the total voltage drop is equal to that
produced by the whole of the load assumed to be concentrated at the middle point.
Power loss in a uniformly loaded distributor fed at one end
(ii) Distributor fed at both ends. In this type of feeding, the distributor is connected to the
supply mains at both ends and loads are tapped off at different points along the length of the
distributor. The voltage at the feeding points may or may not be equal.
Fig shows a distributor AB fed at the ends A and B and loads of I1, I2 and I3 tapped off at points
C, D and E respectively. Here, the load voltage goes on decreasing as we move away from one
feeding point say A, reaches minimum value and then again starts rising and reaches maximum
value when we reach the other feeding point B. The minimum voltage occurs at some load point
and is never fixed. It is shifted with the variation of load on different sections of the distributor.
Advantages
(a) If a fault occurs on any feeding point of the distributor, the continuity of supply is maintained
from the other feeding point.
(b) In case of fault on any section of the distributor, the continuity of supply is maintained from
the other feeding point.
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 crosssection of the conductor. The two ends of the distributor may be supplied with
(i) equal voltages (ii) unequal voltages.
(i) Two ends fed with equal voltages. Consider a distributor AB 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 shown in Fig.
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.
If these currents are x and y respectively, then,
I3 = x + y
Therefore, we arrive at a very important conclusion that at the point of minimum potential,
current comes from both ends of the distributor.
Point of minimum potential. It is generally desired to locate the point of minimum potential.
There is a simple method for it. Consider a distributor AB having three concentrated loads I1, I2
and I3 at points C, D and E respectively. Suppose that current supplied by feeding end A is IA.
Then current distribution in the various sections of the distributor can be worked out as shown in
Fig.
(i). Thus
IAC = IA ; ICD = IA − I1
IDE = IA − I1 − I2 ; IEB = IA − I1 − I2 − I3
Voltage drop between A and B = Voltage drop over AB
or V −V = IA RAC + (IA − I1) RCD + (IA − I1 − I2) RDE + (IA − I1 − I2 − I3) REB
From this equation, the unknown IA can be calculated as the values of other quantities are
generally given. Suppose actual directions of currents in the various sections of the distributor
are indicated as shown in Fig.(ii). 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.
(ii) Two ends fed with unequal voltages.
Fig.
shows the distributor AB fed with unequal voltages ; end A being fed at V1 volts and end B at V2
volts. The point of minimum potential can be found by following the same procedure as
discussed above. Thus in this case, Voltage drop between A and B = Voltage drop over AB
or V1−V2 = Voltage drop over AB
(iii) Distributor fed at the centre. In this type of feeding, the centre of the distributor is
connected to the supply mains .
It is equivalent to two singly fed distributors, each distributor having a common feeding point
and length equal to half of the total length.
(iv) Ring mains. In this type, the distributor is in the form of a closed ring as shown in Fig
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. A distributor arranged to form a closed loop and fed at one or more points is
called a ring distributor.Such a distributor starts from one point, makes a loop through the area to
be served, and returns to the original point. For the purpose of calculating voltage distribution,
the distributor can be considered as consisting of a series of open distributors fed at both ends.
The principal advantage of ring distributor is that by proper choice in the number of feeding
points, great economy in copper can be affected.
Important points regarding ring main system:
Thevenin Eqlnt
TRANSFORMER APPLICATION FACTOR
TAF= ( Total capacity of network transformers/ Total Secondary network load)
When secondary banking or parallelling of secondary distribution network is used TAF is to be
considered. This is about 2 to 2.2 for two or three feeders and 1.5 for 4 or more feeders. The
contingency and service reliability is desired by this factor.
Design considerations of distribution Feeder
Design Considerations in Distribution System:
Good voltage regulation of a distribution network is probably the most important factor
responsible for delivering good service to the consumers. For this purpose, design of feeders and
distributors requires careful consideration.
(i) Feeders. A feeder is designed from the point of view of its current carrying capacity while
the voltage drop consideration is relatively unimportant. It is because voltage drop in a
feeder can be compensated by means of voltage regulating equipment at the substation.
(ii) Distributors. A distributor is designed from the point of view of the voltage drop in it. It is
because a distributor supplies power to the consumers and there is a statutory limit of voltage
variations at the consumer’s terminals (± 6% of rated value). The size and length of the
distributor should be such that voltage at the consumer’s terminals is within the permissible
limits.
Kelvin’s law
The most economical area of conductor is that for which the total annual cost of transmission
line
is
minimum.
This is called as kelvins law after Lord Kelvin who first stated in 1881.
The transmission line cost forms major part in the annual charges of a power system.
The cost is due to
1. Depreciation
2. Repair and maintenance
3. Loss of energy in the line due to its resistance
4. The cost towards the production of the lost energy is considered
 If we decreases the area of the conductor in order to reduce the capital cost, the line
losses increase.
 Similarly, if we increase the conductor cross-section to save the cost towards copper loss
in the line, the weight of copper increases and hence the capital cost will be more.
Because of the above reasons, it is difficult to find the economical size of the conductor. But it
becomes easy with the help of kelvin’s law.
In this post we will understand about the kelvin’s law and limitations of the kelvin’s law.
Assume
A = Cross section of conductor
C = total initial cost towards conductor
C is directly proportional to A
C∝A
C = PA
where P is a constant.
Let r be the annual rate of interest and depreciation.The annual fixed cost C1 = Cr = PAr
Since line losses are inversely proportional to the area of the conductor
The annual cost on lost energy,
C2 = Q/A where Q is a constant.
Total annual cost
C = C 1 + C2
= PAr + Q/A
For C to be minimum,
C/dA = 0
Pr – Q/A2 =0
Pr = Q/A2
Pr.A2 = Q
A2 = Q/Pr
A = √ (Q/Pr)
The equation shows that
“The economical cross-section of the conductor is that for which the annual
charge on the conductor equals the annual charge for the loss of energy in the conductor”.
This is known as Kelvin’s law.
Limitations
of
Kelvin’s
Law
This law has many problems and limits as we are selecting the cross-section from an economical
point of view.We did not consider the electrical behaviour of the line.
1. It is not easy to estimate the energy loss in the line without actual load curves, which are
not available at the time of estimation.
2. Kelvin’s law did not consider many physical factors like voltage regulation, corona loss,
temperature rise etc.
3. The assumption that annual cost on account of interest and depreciation on the capital
outlay is not 100% true.
4. The conductor size determined by this law may not be always practicable one.
5. The rates of interest and depreciation may vary from time to time.
6. The diameter of the conductor may be so small as to cause high corona loss.
7. The conductor may be too weak to stamp from mechanical point of view.
8. Cost of insulation in cables is assumed to be independent of the cross-section of the
conductor which is only an approx. assumption.
MODULE IV
Voltage drop in DC 2 wire system, DC 3 wire system
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
•
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.
Using 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
•
Advantages of AC
•
Cheaper transformation between voltages
•
Easy to switch off
•
Less equipment needed
•
More economical in general
•
Rotating field
A.C. Distribution Calculations
A.C. distribution calculations differ from those of d.c. distribution in the following
respects :
(i) In case of d.c. system, the voltage drop is due to resistance alone. However, in a.c. system,
the voltage drops are due to the combined effects of resistance, inductance and capacitance.
(ii) In a d.c. system, additions and subtractions of currents or voltages are done arithmetically
but in case of a.c. system, these operations are done vectorially.
(iii) In an a.c. system, power factor (p.f.) has to be taken into account. Loads tapped off form the
distributor are generally at different power factors. There are two ways of referring power
factor viz
(a) It may be referred to supply or receiving end voltage which is regarded as the reference
vector.
(b) It may be referred to the voltage at the load point itself.
There are several ways of solving a.c. distribution problems. However, symbolic notation
method has been found to be most convenient for this purpose. In this method, voltages, currents
and impedances are expressed in complex notation and the calculations are made exactly as in
d.c. distribution.
Methods of Solving A.C. Distribution Problems
In a.c. distribution calculations, power factors of various load currents have to be considered
since currents in different sections of the distributor will be the vector sum of load currents and
not the arithmetic sum. The power factors of load currents may be given (i) w.r.t. receiving or
sending end voltage or (ii) w.r.t. to load voltage itself. Each case shall be discussed separately.
(i) Power factors referred to receiving end voltage.
Consider an a.c. distributor AB with concentrated loads of I1 and I2 tapped off at points C and
B as shown in Fig.
Taking the receiving end voltage VB as the reference vector, let lagging power factors at C and
B be cos φ1 and cos φ2 w.r.t. VB. Let R1, X1 and R2, X2 be the resistance and reactance of
sections AC and CB of the distributor. Impedance of section AC, ZA
Impedance of section AC, ZAC = R1 + j X1
Impedance of section CB, ZCB = R2 + j X2
Load current at point C, I1 = I1 (cos φ1 − j sin φ1)
Load current at point B, I2 = I2 (cos φ2 − j sin φ2)
Current in section CB, ICB = I2 = I2 (cos φ2 − j sin φ2)
Current in section AC, IAC = I1 I2 +
= I1 (cos φ1 − j sin φ1) + I2 (cos φ2 − j sin φ2)
Voltage drop in section CB, VCB = ICB ZCB = I2 (cos φ2 − j sin φ2) (R2 + j X2)
Voltage drop in section AC, VAC = ( ) I AC ZAC = I1 + I2 ZAC
= [I1(cos φ1 − j sin φ1) + I2 (cos φ2 − j sin φ2)] [R1 + jX1]
Sending end voltage, VA = VB+ VCB+ VAC
Sending end current, IA = I I + I 2
Voltage drop in section CB = I2 ZCB = I2 (cos φ2 − j sin φ2) (R2 + j X2)
Voltage at point C = VB + Drop in section CB = VC ∠ α (say)
Now I1 = I1 ∠ − φ1 w.r.t. voltage VC
∴ I1 = I1 ∠ − (φ1 − α) w.r.t. voltage VB
i.e. I1 = I1 [cos (φ1 − α) − j sin (φ1 − α)]
Now IAC = I I + I 2
= I1 [cos (φ1 − α) − j sin (φ1 − α)] + I2 (cos φ2 − j sin φ2)
Voltage drop in section AC = IAC ZAC
∴ Voltage at point A = VB + Drop in CB + Drop in AC
3-Phase Unbalanced Loads
The 3-phase loads that have the same impedance and power factor in each phase are called
balanced loads. The problems on balanced loads can be solved by considering one phase only ;
the conditions in the other two phases being similar. However, we may come across a situation
when loads are unbalanced i.e. each load phase has different impedance and/or power factor. In
that case, current and power in each phase will be different. In practice, we may come across the
following unbalanced loads :
(i)
(ii)
(iii)
Four-wire star-connected unbalanced load
(ii) Unbalanced Δ-connected load
(iii) Unbalanced 3-wire, Y-connected load
The 3-phase, 4-wire system is widely used for distribution of electric power in commercial and
industrial buildings. The single phase load is connected between any line and neutral wire while
a 3-phase load is connected across the three lines. The 3-phase, 4-wire system invariably carries
unbalanced loads. In this chapter, we shall only discuss this type of unbalanced load.
Four-Wire Star-Connected Unbalanced Loads
We can obtain this type of load in two ways. First, we may connect a 3-phase, 4-wire unbalanced
load to a 3-phase, 4-wire supply as shown in Fig . Note that star point N of the supply is
connected to the load star point N′. Secondly, we may connect single phase loads between any
line and the neutral wire as shown in Fig
This will also result in a 3-phase, 4-wire unbalanced load because it is rarely possible that single
phase loads on all the three phases have the same magnitude and power factor. Since the load is
unbalanced, the line currents will be different in magnitude and displaced from one another by
unequal angles. The current in the neutral wire will be the phasor sum of the three line currents
i.e. Current in neutral wire, IN = IR + IY + IB
The following points may be noted carefully :
(i) Since the neutral wire has negligible resistance, supply neutral N and load neutral N′ will be at
the same potential. It means that voltage across each impedance is equal to the phase voltage of
the supply. However, current in each phase (or line) will be different due to unequal impedances.
(ii) The amount of current flowing in the neutral wire will depend upon the magnitudes of line
currents and their phasor relations. In most circuits encountered in practice, the neutral current is
equal to or smaller than one of the line currents. The exceptions are those circuits having severe
unbalance.
Voltage drop with underground cable system
power loss estimation in distribution systems
power factor improvement using capacitors
Special electrical requirement of inductive loads Most loads in modern electrical distribution
systems are inductive. Examples include motors, transformers, gaseous tube lighting ballasts,
and induction furnaces. Inductive loads need a magnetic field to operate. Inductive loads require
two kinds of current: • Working power (kW) to perform the actual work of creating heat, light,
motion, machine output, and so on. • Reactive power (kVAR) to sustain the magnetic field
Working power consumes watts and can be read on a wattmeter. It is measured in kilowatts
(kW). Reactive power doesn’t perform useful “work,” but circulates between the generator and
the load. It places a heavier drain on the power source, as well as on the power source’s
distribution system. Reactive power is measured in kilovolt-amperes-reactive (kVAR). Working
power and reactive power together make up apparent power. Apparent power is measured in
kilovolt-amperes (kVA).
Fundamentals of power factor:
Power factor is the ratio of working power to apparent power. It measures how effectively
electrical power is being used. A high power factor signals efficient utilization of electrical
power, while a low power factor indicates poor utilization of electrical power. To determine
power factor (PF), divide working power (kW) by apparent power (kVA). In a linear or
sinusoidal system, the result is also referred to as the cosine θ.
sub harmonic oscillations and ferro resonance due to capacitor banks
Capacitor banks and transformers can create dangerous resonance conditions when capacitor
banks are installed at the service entrance. Under these conditions, harmonics produced by
nonlinear devices can be amplified manyfold. Problematic amplification of harmonics becomes
more likely as more kVAR is added to a system that contains a significant amount of nonlinear
load. You can estimate the resonant harmonic by using this formula:
optimum power factor for distribution systems