The document discusses load curves and selection of generating units for a power station. It explains that the load varies over time so a single generating unit would have poor efficiency. Therefore, multiple units of different sizes are installed to meet the varying load. The number and sizes of units are selected based on the annual load curve to allow the units to operate near maximum efficiency. The total capacity is divided among units to fit the load curve. Operating schedules are determined to run the appropriate units according to demand. Selecting proper units allows them to operate efficiently and reduce energy production costs.

1. Load Curves and Selection of
Generating Units
The load on a power station is seldom constant; it
varies from time to time. Obviously, a single
generating unit (i.e., alternator) will not be an
economical proposition to meet this varying load. It
is because a single unit will have very poor
efficiency during the periods of light loads on the
power station. Therefore, in actual practice, a
number of generating units of different sizes are
installed in a power station. The selection of the
number and sizes of the units is decided from the
annual load curve of the station.

2. The number and size of the units are selected in
such a way that they correctly fit the station
load curve. So it becomes possible to operate
the generating units at or near the point of
maximum efficiency.
In the Figure (i), the annual load curve of the
station is shown. It is clear form the curve that
load on the station has wide variations; the
minimum load being somewhat near 50 kW and
maximum load reaching the value of 500 kW. It
hardly needs any mention that use of a single
unit to meet this varying load will be highly
uneconomical.

4. The total plant capacity is divided into several
generating units of different sizes to fit the load
curve. This is illustrated in Figure (ii) where the
plant capacity is divided into three units
numbered as 1, 2 and 3. The cyan colour outline
shows the units capacity being used. The three
units employed have different capacities and are
used according to the demand on the station. In
this case, the operating schedule can be as
following :

5. Time Units in operation
From 12 midnight to 7
A.M.
Only unit no.1 is put in operation.
From 7 A.M. to 12.00
noon
Unit no. 2 is also started so that both
units 1 and 2 are in operation.
From 12.00 noon to 2
P.M.
Unit no. 2 is stopped and only unit
1operates.
From 2 P.M. to 5 P.M. Unit no. 2 is again started. Now units
1 and 2 are in
operation.
From 5 P.M. to 10.30
P.M.
Units 1, 2 and 3 are put in operation.
From 10. 30 P.M. to
12.00 midnight
Units 1 and 2 are put in operation.

6. Thus by selecting the proper number and sizes of
units, the generating units can be made to
operate near maximum efficiency. This results in
the overall reduction in the cost of production of
electrical energy.

7. Important Points in the Selection
of Units
 The number and sizes of the units should be so selected that they
approximately fit the annual load curve of the station.
 The units should be preferably of different capacities to meet the
load requirements. Although use of identical units ensures saving
in cost, they often do not meet the load requirement.
 The capacity of the plant should be made 15% to 20% more than
the maximum demand to meet the future load requirements.
 There should be a spare generating unit so that repairs and
overhauling of the working units can be carried out.
 The tendency to select a large number of units of smaller capacity
in order to fit the load curve very accurately should be avoided. It
is because the investment cost per kW of capacity increases as
the size of the units decreases.

8. Base Load and Peak Load on Power
Station
The changing load on the
power station makes its
load curve of variable
nature. The Figure shows
the typical load curve of a
power station. It is clear
that load on the power
station varies from time to
time.
However, a close look at the load curve reveals that load
on the power station can be considered in two parts,
namely;

9. (i) Base load. The unvarying load which occurs
almost the whole day on the station is known as
base load. It is clear that 20 MW of load has to
be supplied by the station at all times of day
and night i.e. throughout 24 hours. Therefore,
20 MW is the base load of the station. As base
load on the station is almost of constant nature,
therefore, it can be suitably supplied without
facing the problems of variable load.
(ii) Peak load. The various peak demands of load
over and above the base load of the station is
known as peak load. It is clear that there are
peak demands of load excluding base load.
These peak demands of the station generally
form a small part of the total load and may
occur throughout the day.

10. Base load Plant
It is a type of plant which caters to a constant load demand. Such
plants run 100% of the time. Nuclear and Coal fired plants are
suitable for this
Peak Load Plant
These plants helps tide over short term (15%) demand peak. Gas
turbine, hydro plant can be used.

11. Method of Meeting the Load
The total load on a power station consists of two parts,
base load and peak load. In order to achieve overall
economy, the best method to meet load is to
interconnect two different power stations. The more
efficient plant is used to supply the base load and is
known as base load power station. The less efficient
plant is used to supply the peak loads and is known as
peak load power station. There is no hard and fast rule
for selection of base load and peak load stations as it
would depend upon the particular situation. For
example, both hydro-electric and steam power stations
are quite efficient and can be used as base load as well
as peak load station to meet a particular load
requirement.

12. The interconnection of steam and hydro plants is a
beautiful illustration to meet the load. When water is
available in sufficient quantity as in summer and rainy
season, the hydroelectric plant is used to carry the base
load and the steam plant supplies the peak load as
shown in Figure (i).

13. However, when the water is not available in
sufficient quantity as in winter, the steam plant
carries the base load, whereas the hydro-electric
plant carries the peak load as shown in Figure
(ii).

14. The plant is continuously operated in base load mode, which
results in excess electricity production during off peak periods.
Electrical storage is then need to hold this excess electricity for
use during peak hours.
Storage = 5 MW × 6 hrs = 30 MWh.
Supply = 10 MW × 3 hrs = 30 MWh.
Storage for longer time  supply for short time.

15. schemes of energy management
There are 4 schemes for energy management for
maximum capacity utilization:-
1. Supply power peaks by interconnecting power
network that might have different power
demands on them.
2. Use newer and more efficient power plants for
base load generation and use older, less efficient
plants for peak power generation.
3. Construct smaller, low capital cost through not so
efficient power plants as peaking unit, i.e. gas
turbine, small hydroelectric power plant.
4. Add energy storage system to meet the peak
demand.

16. Example 17: A proposed station has the following daily load cycle :
Time in hours 6—8 8—11 11—16 16—19 19—22 22—24 24—6
Load in MW 20 40 50 35 70 40 20
Draw the load curve and select suitable generator units from the
10,000, 20,000, 25,000, 30,000 kw. Prepare the operation schedule for
the machines selected and determine the load factor from the curve.
The load curve of the power
station can be drawn to some
suitable scale as shown in the
Figure.

17. Selection of number and sizes of units : Assuming power factor of
the machines to be 0·8, the output of the generating units
available will be 8, 16, 20 and 24 MW. There can be several
possibilities. However, while selecting the size and number of
units, it has to be borne in mind that (i) one set of highest capacity
should be kept as standby unit (ii) the units should meet the
maximum demand (70 MW in this case) on the station (iii) there
should be overall economy.

18. Keeping in view the above facts, 3 sets of 20 MW and 2 sets of 16
MW each may be chosen. Four sets will meet the maximum demand
of 70 MW and one unit will serve as a standby unit.
Operational schedule. Referring to the load curve shown in the
Figure, the operational schedule will be as following:
 Set No. 1 of 20 MW will run for 24 hours.
 Set No. 2 of 20 MW will run from 8.00 hours to 24.00 hours.
 Set No. 3 of 16 MW will run from 11.00 hours to 16 hours and
again from 19 hours to 22 hours.
Set No. 4 of 16 MW will run from 19 hours to 22 hours.
Set No. 5 of 20 MW will be stand by.

19. Example 18: A generating station is to supply four regions of load
whose peak loads are 10 MW, 5 MW, 8 MW and 7 MW. The diversity
factor at the station is 1·5 and the average annual load factor is 60%.
Calculate :
(i) the maximum demand on the station.
(ii) annual energy supplied by the station.
(iii) Suggest the installed capacity and the number of units.

20. Example 19: The load duration curve for a typical heavy load being
served by a combined hydro-steam system may be approximated by a
straight line; maximum and minimum loads being 60,000 kW and
20,000 kW respectively. The hydro power available at the time of
minimum regulated flow is just sufficient to take a peak load of 50,000
kWh per day. It is observed that it will be economical to pump water
from tail race to the reservoir by utilising the steam power plant
during the off-peak periods and thus running the station at 100% load
factor. Determine the maximum capacity of each type of plant.
Assume the efficiency of steam conversion to be 60%.
OCBA represents the load duration curve for the combined system as
shown in the Figure. The total maximum demand (i.e., 60,000 kW) is
represented by OC, whereas the minimum demand (i.e., 20,000 kW)
is represented by OD.
Let OE = Capacity of steam plant, EC = Capacity of hydro plant
Area CHI = The energy available from hydro plant in the low flow
period.

22. Area FGB = The off-peak period energy available from steam plant
Obviously, the energy of hydro plant represented by area HEFI and
available from reservoir has been supplied by steam power plant
represented by area FGB. As steam electric conversion is 60%,

24. Problems:
1. A generating station has a connected load of 40 MW and a
maximum demand of 20 MW : the units generated being 60 × 106.
Calculate (i) the demand factor (ii) the load factor. [(i) 0·5 (ii) 34·25%]
2. A 100 MW powers stations delivers 100 MW for 2 hours, 50 MW for
8 hours and is shut down for the rest of each day. It is also shut down
for maintenance for 60 days each year. Calculate its annual load factor.
[21%]
3. A power station is to supply four regions of loads whose peak
values are 10,000 kW, 5000 kW, 8000 kW and 7000 kW. The diversity
factor of the load at the station is 1.5 and the average annual load
factor is 60%. Calculate the maximum demand on the station and
annual energy supplied from the station.
[20,000 kW ; 105·12 × 106 kWh]

25. 4. A generating station supplies the following loads : 15000 kW, 12000
kW, 8500 kW, 6000 kW and 450 kW. The station has a maximum
demand of 22000 kW. The annual load factor of the station is 48%.
Calculate (i) the number of units supplied annually (ii) the diversity
factor and (iii) the demand factor. [(i) 925 × 105 kWh (ii) 52·4% (iii) 1·9]
5. A generating station has a maximum demand of 20 MW, a load
factor of 60%, a plant capacity factor of 48% and a plant use factor of
80% . Find :
(i) the daily energy produced
(ii) the reserve capacity of the plant
(iii) the maximum energy that could be produced daily if the plant was
running all the time
(iv) the maximum energy that could be produced daily if the plant was
running fully loaded and operating
as per schedule. [(i) 288 × 103 kWh (ii) 0 (iii) 4·80 × 103 kWh (iv) 600 ×
103 kWh]

26. 6. A generating station has the following daily load cycle :
Time (hours) 0—6 6—10 10—12 12—16 16—20 20—24
Load (MW) 20 25 30 25 35 20
Draw the load curve and find
(i) maximum demand,
(ii) units generated per day,
(iii) average load,
(iv) load factor, [(i) 35 MW (ii) 560 × 103 kWh (iii) 23333 kW (iv)
66·67%]
7. A power station has to meet the following load demand :
Load A 50 kW between 10 A.M. and 6 P.M.
Load B 30 kW between 6 P.M. and 10 P.M.
Load C 20 kW between 4 P.M. and 10 A.M.
Plot the daily load curve and determine (i) diversity factor (ii) units
generated per day (iii) load factor. [(i) 1·43 (ii) 880 kWh (iii) 52·38%]

27. 8. The yearly load duration curve of a certain power station can be
approximated as a straight line ; the maximum and minimum loads
being 80 MW and 40 MW respectively. To meet this load, three
turbine generator units, two rated at 20 MW each and one at 10
MW are installed. Determine (i) installed capacity (ii) plant factor
(iii) kWh output per year (iv) load factor. [(i) 50MW (ii) 48% (iii) 210
× 106 (iv) 60%]
9. The annual load duration curve for a typical heavy load being
served by a steam station, a run-of-river station and a reservoir
hydro-electric station is as shown in the Figure below. The ratio of
number of units supplied by these stations is as follows :
Steam : Run-of-river : Reservoir : : 7 : 4 : 1
The run-of-river station is capable of generating power
continuously and works as a base load station. The reservoir station
works as a peak load station. Determine (i) the maximum demand
of each station and (ii) load factor of each station.