Chapter 1. Structure of Power Systems and Few Other Aspects

2 Electrical Power Systems



Figure 1.2 shows part of a typical power system.



Fig. 1.1: Schematic diagram of a power supply system.



Fig. 1.2: Part of a power system.



Structure of Power Systems and Few Other Aspects 3



Fig. 1.3: Typical distribution system.



1.2 REASONS FOR INTERCONNECTION

Generating stations and distribution systems are connected through transmission lines. The

transmission system of a particular area (e.g., state) is known as a grid. Different grids are

interconnected through tie-lines to form a regional grid (also called power pools). Different

regional grids are further connected to form a national grid. Cooperative assistance is one of the

planned benefits of interconnected operation. Interconnected operation is always economical

and reliable. Generating stations having large MW capacity are available to provide base or

intermediate load. These generating stations must be interconnected so that they feed into the

general system but not into a particular load. Economic advantage of interconnection is to

reduce the reserve generation capacity in each area. If there is sudden increase of load or loss

of generation in one area, it is possible to borrow power from adjoining interconnected areas. To

meet sudden increases in load, a certain amount of generating capacity (in each area) known as

the “spinning reserve” is required. This consists of generators running at normal speed and

ready to supply power instantaneously.

It is always better to keep gas turbines and hydro generators as “spinning reserve”. Gas

turbines can be started and loaded in 3 minutes or less. Hydro units can be even quicker. It is

more economical to have certain generating stations serving only this function than to have

each station carrying its own spinning reserve. Interconnected operation also gives the flexibility

to meet unexpected emergency loads.



1.3 LOAD CHARACTERISTICS

Total load demand of an area depends upon its population and the living standards of people.

General nature of load is characterized by the load factor, demand factor, diversity factor,

power factor and utilization factor. In general, the types of load can be divided into the following

categories: (1) Domestic (2) Commercial (3) Industrial (4) Agriculture.



4 Electrical Power Systems



Domestic Load: Domestic load mainly consists of lights, fans, refrigerators, airconditioners,

mixer, grinders, heaters, ovens, small pumping motors etc.

Commercial Load: Commercial load mainly consists of lighting for shops, offices,

advertisements etc., fans, heating, airconditioning and many other electrical appliances used in

commercial establishments such as market places, restaurants etc.

Industrial Loads: Industrial loads consists of small-scale industries, medium-scale

industries, large-scale industries, heavy industries and cottage industries.

Agriculture Loads: This type of load is mainly motor pump-sets load for irrigation purposes.

Load factor for this load is very small, e.g., 0.15–0.20.



1.4 POWER FACTOR OF VARIOUS EQUIPMENTS

Total kVA (or MVA) demand depends on the power factor of various equipments and lagging

power factor of some of the equipments are tabulated below:

Induction motors

:

0.6 –0.85

Fractional HP motors

:

0.5–0.80

Fluorescent lamps

:

0.55–0.90

Neon signs

:

0.40 –0.50

Fans

:

0.55–0.85

Induction furnaces

:

0.70 –0.85

Arc welders

:

0.35 –0.55



1.5 BASIC DEFINITIONS OF COMMONLY USED TERMS

Connected Load: Each electrical device has its rated capacity. The sum of the continuous

ratings of all the electrical devices connected to the supply system is known as connected load.

Demand: The demand of an installation or system is the load at the receiving terminals

averaged over a specified interval of time. Here, the load may be given in kW, kVA, kiloamperes,

or amperes.

Demand Interval: It is the time period over which the average load is computed. The time

period may be 30 minute, 60 minute or even longer.

Maximum Demand: The maximum demand of an installation or system is the greatest of

all demands which have occurred during the specified period of time. Maximum demand

statement must express the demand interval used to measure it. For example, the specific

demand might be the maximum of all demands such as daily, weekly, monthly or annual.

Coincident Demand (or Diversified Demand): It is the demand of composite group, as

a whole, of somewhat unrelated loads over a specified period of time. It is the maximum sum of

the contributions of the individual demands to the diversified demand over a specific time

interval.

Noncoincident Demand: It is the sum of the demands of a group of loads with no

restrictions on the interval to which each demand is applicable.

Demand Factor: It is the ratio of the maximum demand of a system to the total connected

load of the system. Thus, the demand factor (DF ) is given as:

DF =



Maximum demand

Total connected load



... (1.1)



Structure of Power Systems and Few Other Aspects 5



The demand factor is usually less than 1.0. Demand factor gives an indication of the simultaneous

operation of the total connected load. Demand factor can also be found for a part of the system,

for example, an industrial or commercial or domestic consumer, instead or the whole system.

Utilization Factor: It is the ratio of the maximum demand of a system to the rated

capacity of the system. Thus, the utilization factor (UF) is

UF =



Maximum demand of the system

Rated system capacity



...(1.2)



The rated capacity of the system may be selected to be the smaller of thermal-or voltage drop

capacity. The utilization factor can also be obtained for a part of the system.

Plant Factor: Also known as capacity factor or use factor. It is the ratio of the total actual

energy produced over a specified period of time to the energy that would have been produced

if the plant (or generating units) had operated continuously at maximum rating. Therefore, the

plant factor is,

Plant Factor =



Actual energy produced

Maximum plant rating ´ T



...(1.3)



Plant factor is mostly used in generation studies. It is also given as,

Annual Plant Factor =



or



Actual energy generation

Maximum plant rating



...(1.4)



Annual Plant Factor =



Actual annual energy generation

Maximum plant rating ´ 8760



...(1.5)



Diversity Factor: It is the ratio of the sum of the individual maximum demands of the

various subdivisions or groups or consumers to the maximum demand of the whole system.

Therefore, the diversity factor (FD) is given as

FD =



Sum of individual maximum demand

Coincident maximum demand



...(1.6)



n



or



FD =



å Pi

i=1



Pc



...(1.7)



where

Pi = maximum demand of load i

Pc = coincident maximum demand of group of n loads.

The diversity factor can be equal or greater than unity. From eqn. (1.1), the demand

factor is

DF =



\



Maximum demand

Total connected load



Maximum demand = Total connected load × DF



...(1.8)



6 Electrical Power Systems



For i-th consumer, let us assume, total connected load = TCPi and demand factor = DFi.

Therefore, eqn.(1.8) can be written as:

Pi = TCPi × DFi

...(1.9)

From eqns. (1.7) and (1.9), we get

n



FD =



å TCPi DFi

i=1



Pc



...(1.10)



Coincidence Factor: It is the ratio of the maximum coincident total demand of a group of

consumers to the sum of the maximum power demands of individual consumers comprising the

group both taken at the same point of supply for the same time. Therefore, coincidence factor

(CF) is

CF =



or



CF =



Coincident maximum demand

Sum of individual maximum demands

Pc



...(1.11)



...(1.12)



n



å Pi

i=1



From eqns. (1.12) and (1.7), we get

CF =



1

FD



...(1.13)



Thus, the coincidence factor is the reciprocal of the diversity factor.

Load Diversity: It is the difference between the sum of the peaks of two or more individual

loads and the peak of the combined load. Therefore load diversity (LD) is defined as



FG å P IJ – P

H K

n



LD =



i=1



i



c



...(1.14)



Contribution Factor: It is given in per unit of the individual maximum demand of the

i-th load. If Ci is the contribution factor of the i-th load to the group of maximum demand,

Then,

Pc = C1 × P1 + C2 × P2 + ... + Cn × Pn

n



\



Pc =



å C i Pi

i=1



...(1.15)



From eqns. (1.12) and (1.15), we get,

n



CF =



å Ci Pi

i=1

n



å Pi

i=1



...(1.16)



Structure of Power Systems and Few Other Aspects 7



Case-1:

If

Then



P1 = P2 = P3 = ... = Pn = P

n



CF =



P ´ å Ci

i=1



P ´n



n



=



å Ci

i=1



n



...(1.17)



That is, the coincident factor is equal to the average contribution factor.

Case-2:

If

C1 = C2 = C3 = ... = Cn = C,

Then

n



CF =



C ´ å Pi

n



i=1



å Pi



=C



...(1.18)



i=1



That is, coincidence factor is equal to the contribution factor.

Load Factor: It is the ratio of the average load over a designated period of time to the peak

load occurring on that period.

Therefore, the load factor (LF ) is defined as:

LF =



Average load

Peak load



LF =



Average load ´ T

Peak load ´ T



... (1.19)



or



Energy served

...(1.20)

Peak load ´ T

where T = time, in days, weeks, months or years. If T is large, LF is small. The reason for this

is that for the same maximum demand, the energy consumption covers a larger time period and

results in a smaller average load. Load factor is less than or equal to unity. Annual load factor

is defined as:

\



LF =



Annual Load Factor =



Total annual energy

Annual peak load ´ 8760



...(1.21)



Loss Factor: It is the ratio of the average power loss to the peak-load power loss during a

specified period of time. Therefore, the loss factor (LLF) is defined as:

LLF =



Average power loss

Power loss at peak load



...(1.22)



Equation (1.22) is applicable for the copper losses of the system but not for iron losses.



8 Electrical Power Systems



Example 1.1: A power station supplies the load as tabulated below:

Time

(hours)

6 AM – 8 AM

8 AM – 9 AM

9 AM – 12 Noon

12 Noon – 2 PM

2 PM – 6 PM

6 PM – 8 PM

8 PM – 9 PM

9 PM – 11 PM

11 PM – 5 AM

5 AM – 6 AM

(a)

(b)

(c)

(d)



Load

(MW)

1.2

2.0

3.0

1.50

2.50

1.80

2.0

1.0

0.50

0.80



Plot the load curve and find out the load factor.

Determine the proper number and size of generating units to supply this load.

Find the reserve capacity of the plant and plant factor.

Find out the operating schedule of the generating units selected.



Solution:

(a) Figure 1.4 show the plot of load curve



Fig. 1.4: Load curve of Ex–1.1.



Units generated during 24 hours

= (2 × 1.2 + 1 × 2 + 3 × 3 + 2 × 1.5 + 4 × 2.5 + 2 × 1.8 + 1 × 2

+ 2 × 1 + 6 × 0.5 + 1 × 0.8) MWhr.

= 37.80 MWhr

Average load =



\



Units generated

Time in hours



Average load =



37.80

= 1.575 MW.

24



Structure of Power Systems and Few Other Aspects 9



Load factor,

LF =



Average load

Maximum load



Maximum load = 3 MW

\



LF =



1575

.

= 0.525

3



(b) Maximum demand = 3 MW. Therefore, 4 generating units of rating 1.0 MW each may be

selected. During the period of maximum demand 3 units will operate and 1 unit will remain as

stand by.

(c) Plant capacity = 4 × 1.0 = 4.0 MW

Reserve capacity = 4 – 3 = 1 MW

From eqn. (1.3),

Plant Factor =



Actual energy produced

Maximum plant rating ´ T



Actual energy produced = 37.80 MWhr

Maximum plant rating = 4 MW

Time duration T = 24 hours

\



Plant Factor =



37.80

= 0.39375.

4 ´ 24



(d) Operating schedule will be as follows:

One generating unit of 1 MW:— 24 hours

Second generating unit of 1 MW:— 6 AM – 9 PM (15 hours)

Third generating unit of 1 MW:— 9 AM – 12 Noon

2 PM – 6 PM

(7 hours)

Example 1.2: A generating station has a maximum demand of 80 MW and a connected load of

150 MW. If MWhr generated in a year are 400 × 103, calculate (a) load factor (b) demand factor.

Solution:



Maximum demand

Connected load

Units generated in one year

Total number of hours in a year T



\



Average load =



= 80 MW

= 150 MW

= 400 ×103 MWhr

= 8760



400 ´ 103

= 45.662 MW

8760



Load factor,



LF =



Average load

Maximum load



\



LF =



45.662

= 0.57

80



10 Electrical Power Systems



Demand factor,

DF =



\



DF =



Maximum demand

Connected load

80

= 0.533

150



Example 1.3: A sample distribution system is shown in Fig. 1.5. One of the feeders supplies an

industrial load with a peak of 2 MW and the other supplies residential loads with a peak of 2

MW. Combined peak demand is 3 MW. Determine (a) the diversity factor of the load connected

to transformer (b) the load diversity of the load connected to transformer. (c) the coincidence

factor of the load connected to transformer.



Fig. 1.5: Sample distribution system of Ex–1.3.



Solution:

(a) From eqn.(1.7), diversity factor is

n



FD =



å



i=1



n=2



Pi



=



Pc



å Pi

i=1



Pc



=



( P1 + P2 )

Pc



P1 = 2 MW, P2 = 2 MW and Pc = 3 MW



\



FD =



(2 + 2)

= 1.33

3



(b) From eqn. (1.14), load diversity is,



F I

LD = G å P J – P

H K

n



i=1



i



c



n = 2, P1 = P2 = 2 MW, Pc = 3 MW

\



LD = (P1+ P2) – Pc = (2 + 2) – 3



Structure of Power Systems and Few Other Aspects 11



\

LD = 1 MW

(c) From eqn.(1.13), coincidence factor is,

1

1

=

FD

133

.

CF = 0.75.



CF =

\



1.6 RELATIONSHIP BETWEEN LOAD FACTOR (LF) AND

LOSS FACTOR (LLF)

In general, loss factor can not be determined from load factor. However, the limiting values of

the relationship can be established. Fig. 1.6 shows an arbitrary and idealized load curve and it

does not represent a daily load curve.



Fig.1.6: Idealized load curve.



Assume that at peak load P2, loss is L2 and at off-peak load P1, loss is L1.

The load factor is,

LF =



Pavg

Pmax



=



Pavg

P2



... (1.23)



From Fig.1.6,

Pavg =



P2 ´ t + P1 ´ (T - t)

T



From eqns. (1.23) and (1.24), we obtain



...(1.24)



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)