6 Relationship Between Load Factor (LF) and Loss Factor (LLF)

12 Electrical Power Systems



LF =



P2 ´ t + P1 ´ (T - t)

P2 ´ T



LF =



P1

t

+

×

P2

T



or



FG T - t IJ

HTK



...(1.25)



The loss factor is

LLF =



Lavg

Lmax



=



Lavg

L2



...(1.26)



where

Lmax = maximum power loss = L2

From Fig. 1.6, we obtain



Lavg = average power loss.



L2 ´ t + L1 ´ (T - t)

T

From eqns. (1.26) and (1.27), we get

Lavg =



LLF =



L2 ´ t + L1 ´ (T - t)

L2 ´ T



...(1.27)



...(1.28)



where

t = peak load duration

(T – t) = off-peak load duration.

The copper losses are the function of associated loads. Therefore, the loss at off-peak and

peak load can be expressed as:

L1 = K × P12



...(1.29)



L2 = K × P22



...(1.30)



From eqns. (1.28), (1.29) and (1.30), we get,



t

+

LLF =

T



FG P IJ FG T - t IJ

HP K H T K

1



2



2



...(1.31)



By using eqns. (1.25) and (1.31), the load factor can be related to loss factor for three

different cases:

Case-1: Off-peak load is zero.

Here, P1 = 0 and L1 = 0, therefore, from eqns. (1.25) and (1.31), we have



t

T

That is load factor is equal to loss factor and they are equal to t/T constant.

LF = LLF =



....(1.32)



Structure of Power Systems and Few Other Aspects 13



Case-2: Very short lasting peak.

Here, t ® 0

Hence in eqns.(1.25) and (1.31),



Therefore,



FG T - t IJ ® 1.0

HT K



LLF ® (LF )2

...(1.33)

Case-3: Load is steady

Here, t ® T

That is the difference between peak load and off-peak load is negligible. Thus from eqns.

(1.25) and (1.31),

LLF ® LF

...(1.34)

Therefore, in general, the value of the loss factor is

(LF)2 < LLF < LF

...(1.35)

Therefore, the loss factor cannot be determined directly from the load factor. However, an

approximate formula to relate the loss factor to the load factor as:

LLF = 0.3LF + 0.7(LF)2

...(1.36)



1.7 LOAD GROWTH

Load growth is the most important factor influencing the expansion of distribution system.

Forcasting of load increases is essential to the planning process. If the load growth rate is

known, the load at the end of the m-th year is given by

Pm = P0(1 + g)m

...(1.37)

where

Pm = load at the end of the m-th year.

P0 = initial load (load at the base year)

g = annual load growth rate

m = number of years.



1.8 MULTIPHASE SYSTEMS

Three-phase system is universally used. However, attention has been given in recent years to

the use of more than three phases for power transmission purposes. In particular, six and

twelve phase systems have been studied. Advantages of six and twelve phase systems relative

to three phase systems are as follows:

1. Thermal loading capacity of lines is more.

2. Corona effects is less because for a given conductor size and tower configuration the stress

on the conductor surface decreases with the number of phases.

3. The higher the number of phases, the smaller the line-to-line voltage becomes relative to

the phase voltage, resulting in increased utilization of rights of way because of less phaseto-phase insulation requirement.



14 Electrical Power Systems



4. Existing double-circuit lines (two three-phase circuits on each tower) could be converted to

single circuit six-phase lines. It is always advantageous to describe multiphase systems in

terms of the phase voltage rather than line-to-line, as in the case for three-phase systems.

The transmission efficiency is higher.

A six-phase supply can be obtained by suitable arrangement of the secondary windings of

a three phase transformer. Figure 1.7 shows the transformer connections and phasor diagram.

The windings on the three limbs of the transformer are centre-tapped with the taps mutually

connected.



(a)



(b)



Fig.1.7: Six-phase system (=) transformer connection (>) phasor diagram.



From Fig. 1.7(b)

Vaf = Van – Vfn = Vph

...(1.38)

Therefore for six-phase system line-to-line voltage is equal to the phase voltage in magnitude.

Example 1.4: A sub-station supplies power to four feeders. Feeder-A supplies six consumers

whose individual daily maximum demands are 70 KW, 90 KW, 20 KW, 50 KW, 10 KW and

20 KW, while maximum demand on the feeder is 200 KW. Feeder-B supplies four consumers

whose daily maximum demands are 60 KW, 40 KW, 70 KW and 30 KW while maximum demand

on feeder-B is 160 KW. Feeders C and D have a daily maximum demand of 150 KW and 200 KW

respectively, while the maximum demand on the station is 600 KW.

Determine the diversity factor for consumers of feeder-A and B and for the four feeders.

Solution:

From eqn.(1.6), diversity factor is,

FD =



Sum of individual maximum demands

Coincident maximum demands



For feeder-A, Coincident maximum demand = 200 KW



\

For feeder-B



FDA =



(70 + 90 + 20 + 50 + 10 + 20)

= 1.3

200



(60 + 40 + 70 + 30 )

= 1.25

160

Diversity factor for the four feeders,

FDB =



FD =



(200 + 160 + 150 + 200 )

= 1.183

600



Structure of Power Systems and Few Other Aspects 15



1.9 DISADVANTAGES OF LOW POWER FACTOR

For a three-phase balanced system, if load is PL, terminal voltage is V and power factor is cos f,

then load current is given by

IL =



PL

3V cos f



... (1.38)



If PL and V are constant, the load current IL is inversely proportional to the power factor,

i.e., if cos f is low, IL is large. The poor power factor of the system has following disadvantages:

1. Rating of generators and transformers are inversely proportional to the power factor.

Thus, generators and transformers are required to deliver same load (real power) at low

power factor. Hence, system kVA or MVA supply will increase.

2. At low power factor, the transmission lines, feeders or cable have to carry more current for

the same power to be transmitted. Thus, conductor size will increase, if current density in

the line is to be kept constant. Therefore, more copper is required for transmission line,

feeders and cables to deliver the same load but at low power factor.

3. Power loss is proportional to the square of the current and hence inversely proportional to

the square of the power factor. More power losses incur at low power factor and hence poor

efficiency.

4. Low lagging power factor results in large voltage drop which results in poor voltage

regulation. Hence, additional regulating equipment is required to keep the voltage drop

within permissible limits.

Electric utilities insist the industrial consumers to maintain a power factor 0.80 or above.

The power tariffs are devised to penalize the consumers with low lagging power factor and force

them to install power factor correction devices for example shunt capacitors.



1.10 VARIOUS CAUSES OF LOW POWER FACTOR

1. Most of the induction motors operate at lagging power factor. The power factor of these

motor falls with the decrease of load.

2. Occurrence of increased supply mains voltage during low load periods, the magnetizing

current of inductive reactances increase and power factor of the electrical plant as a whole

comes down.

3. Very low lagging power factor of agriculture motor pump-set.

4. Arc lamps, electric discharge lamps and some other electric equipments operate at low

power factor.

5. Arc and induction furnaces operate on very low lagging power factor.

The average power factors of some of the electrical equipments are given in Section-1.4.

Example 1.5: Peak demand of a generating station is 90 MW and load factor is 0.60. The plant

capacity factor and plant use factor are 0.50 and 0.80 respectively. Determine (a) daily energy

produced (b) installed capacity of plant (c) reserve capacity of plant (d) utilization factor.

Solution:

(a)



Maximum demand = 90 MW

Load factor = 0.60



16 Electrical Power Systems



Average demand = (Maximum demand) × (Load factor)



\



Average demand = 90 × 0.60 = 54 MW.

Daily energy produced = (Average demand) × 24

= 54 × 24 = 1296 MWhr.



(b) From eqn. (1.3),

Plant factor =

\



\

\



Actual energy produced

Maximum plant rating ´ T



Plant factor = 0.50

Actual energy produced = 1296 MWhr.

Maximum plant rating =



1296

= 108 MW

0.50 ´ 24



Installed capacity = 108 MW.



(c)



Reserve capacity = (Installed capacity) – (Peak demand)



\



Reserve capacity = (108 – 90) = 18 MW.



(d) From eqn.(1.2), utilization factor is,

UF =



\



Maximum demand of the system

Rated system capacity



UF =



90

= 0.833.

108



EXERCISE–1

1.1 Load duration data of a system are given below:

Load (MW)

2.0

4.0

6.0

8.0

10.0

12.0



Duration (hours)

8760

7000

4380

2628

1752

876



15.0

87

Plot the load duration curve and determine the load factor.

Ans: 0.39

1.2 A power plant has a peak demand of 15 MW, load factor is 0.70, plant capacity factor is 0.525 and

a plant use factor of 0.85. Determine (a) daily energy produced (b) reserve capacity of the plant

(c) the maximum energy that could be produced daily if the plant operating schedule is fully loaded

when in operation.

Ans: (a) 252 MWhr (b) 5 MW (c) 296.47 MWhr.



Structure of Power Systems and Few Other Aspects 17

1.3 A generating station has peak demand of 120 MW and its connected load is 200 MW. The energy

produced annually is 4 × 105 MWhr. Determine (a) load factor (b) demand factor

Ans. (a) 0.38 (b) 0.60.

1.4 A power plant has to meet the following load demand:

Load A: 100 MW from 8 AM – 6 PM

Load B: 150 MW from 6 AM – 10 AM

Load C: 50 MW from 6 AM – 10 AM

Load D: 20 MW from 10 AM – 6 AM

Plot the daily load curve and determine (a) diversity factor (b) load factor (c) daily energy produced.

Ans: (a) 1.067 (b) 0.439 (c) 3160 MWhr.

1.5 From a load duration curve, following data are available:

Maximum demand on the system is 40 MW. The load supplied by the two generating units is 28

MW and 20 MW. Unit no. 1 is the base unit and works for all the time. Unit no. 2 is peak load unit

and works only for 40% of the time. The energy produced annually by unit 1 is 2 × 10 8 units and

that by unit 2 is 15 × 166 units. Find the (a) load factor (b) plant capacity factor (c) plant use factor

of both the units. Also (d) determine the load factor of the total plant.

Ans: (a) 0.815, 0.356 (b) 0.815, 0.856 (c) 0.815, 0.214 (d) 0.613



18 Electrical Power Systems



2

Resistance and Inductance of

Transmission Lines

2.1 INTRODUCTION

Power system engineering is that branch of Electrical Engineering which concerns itself with

the technology of generation, transmission and distribution of electrical energy. The power

system growing into a vast and complex system represents one of the most vital systems in

every modern nation. The basic purpose of a transmission network is to transfer electric energy

from generating units at various locations to the distribution system which ultimately supplies

the load. Transmission line also interconnect neighbouring power utilities which allows not

only economic dispatch of electrical power within regions during normal conditions, but also

transfer of power between regions during emergencies.

An overhead transmission line consists of a group of conductors running parallel to each

other and carried on supports which provide insulation between the different conductors and

between each conductor and earth. A transmission line has four parameters—resistance,

inductance, capacitance and shunt conductance. The shunt conductance accounts for leakage

currents flowing across insulators and ionized pathways in the air. The leakage currents are

negligible as compared to the current flowing in the transmission lines. The series resistance

causes a real power loss in the conductor. The resistance of the conductor is very important in

transmission efficiency evaluation and economic studies. The power transmission capacity of

the transmission line is mainly governed by the series inductance. The shunt capacitance

causes a charging current to flow in the line and assumes importance for medium and long

transmission lines. These parameters are uniformly distributed throughout but can be lumped

for the purpose of analysis on approximate basis.



2.2 LINE RESISTANCE

The dc resistance of a solid round conductor is given by

Rdc =

where



r×l

A



r = resistivity of conductor

l = length of conductor

A = cross sectional area of conductor



...(2.1)



Resistance and Inductance of Transmission Lines 19



The conductor resistance is affected by three factors: frequency, spiraling, and temperature.

The dc resistance of a stranded conductor is greater than the value given by eqn. (1.1)

because spiralling of the strands makes them longer than the conductor itself. The increase in

resistance due to spiralling is around 1% for three strand conductors and about 2% for

concentrically stranded conductors.

When an alternating current flows through a conductor, the current distribution is not

uniform over the conductor cross-sectional area and the degree of non-uniformity increases

with increase in frequency. The current density is greatest at the surface of the conductor. This

causes the ac resistance to be somewhat higher than the dc resistance. This effect is known as

skin effect. The ac resistance is usually referred as the effective resistance of the conductor.

The conductor resistance increases with the increase of temperature. Since the value of r

is given at a specific temperature and the line operates at higher temperature, the actual

resistance is higher than the value found in eqn. (2.1). For small changes in temperature, the

resistance increases linearly as temperature increases and the resistance at a temperature T is

given by

RT = Ro(1 + ao · T)

where



...(2.2)



RT = resistance at T°C

Ro = resistance at 0°C

a0 = temperature coefficient of resistance at 0°C.



By using eqn. (2.2), the resistance R2 at a temperature T2°C can be found if the resistance

R1 at a temperature T1°C is known, i.e.



R2

T + 1 ao

= 2

R1

T1 + 1 a o

For aluminium



...(2.3)



1/ao » 228



2.3 INDUCTANCE—BASIC CONCEPTS

From our basic understanding of electromagnetic field theory, we know that a conductor

carrying current has a magnetic field around it. The magnetic lines of force are concentric

circles having their centres at the centre of the conductor and are arranged in planes

perpendicular to the conductor.

The voltage induced in a conductor is given by



dy

volt

...(2.4)

dt

where y represents the flux linkages of the conductor in Wb-Turns. Eqn. (2.4) can be written in

the form

d y di

di

×

= L×

E=

...(2.5)

di dt

dt

E=



dy

is the inductance in Henrys.

di

In a linear magnetic circuit, flux linkages vary linearly with the current such that the

inductance remains constant and is given by



when L =