Variable Load on Power Stations


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CHAPTER
!
Variable Load on Power Stations

3.1 Structure of Electric Power System 3.2 Variable Load on Power Station 3.3 Load Curves 3.4 Important Terms and Factors 3.5 Units Generated per Annum 3.6 Load Duration Curve 3.7 Types of Loads 3.8 Typical Demand and Diversity Fac-
tors 3.9 Load Curves and Selection of Gener-
ating Units 3.10 Important Points in the Selection of
Units 3.11 Base Load and Peak Load on Power
Station 3.12 Method of Meeting the Load 3.13 Interconnected Grid System

Introduction
The function of a power station is to deliver power to a large number of consum ers. However, the power demands of different consumers vary in accordance with their activities. The result of this variation in demand is that load on a power station is never constant, rather it varies from time to time. Most of the complexities of modern power plant operation arise from the inherent variability of the load demanded by the users. Unfortunately, electrical power cannot be stored and, therefore, the power station must produce power as and when demanded to meet the requirements of the consumers. On one hand, the power engineer would like that the alternators in the power station should run at their rated capacity for maximum efficiency and on the other hand, the demands of the consumers have wide variations. This makes the design of a power station highly complex. In this chapter, we shall focus our attention on the problems of variable load on power stations.
3.1 Structure of Electric Power System
The function of an electric power system is to connect the power station to the consumers’ loads

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Variable Load on Power Stations

43

(iv) The power demanded by the consumers is supplied by the power station through the transmission and distribution networks. As the consumers’ load demand changes, the power supply by the power station changes accordingly.

3.2 Variable Load on Power Station

The load on a power station varies from time to time due to uncertain demands of the consumers and is known as variable load on the station.

A power station is designed to

meet the load requirements of the con-

sumers. An ideal load on the station,

from stand point of equipment needed

and operating routine, would be one

of constant magnitude and steady du-

ration. However, such a steady load

on the station is never realised in ac-

tual practice. The consumers require

their small or large block of power in accordance with the demands of their

Transmission line

activities. Thus the load demand of one consumer at any time may be different from that of the other

consumer. The result is that load on the power station varies from time to time.

Effects of variable load. The variable load on a power station introduces many perplexities in its operation. Some of the important effects of variable load on a power station are :

(i) Need of additional equipment. The variable load on a power station necessitates to have additional equipment. By way of illustration, consider a steam power station. Air, coal and water are the raw materials for this plant. In order to produce variable power, the supply of these materials will be required to be varied correspondingly. For instance, if the power demand on the plant increases, it must be followed by the increased flow of coal, air and water to the boiler in order to meet the increased demand. Therefore, additional equipment has to be installed to accomplish this job. As a matter of fact, in a modern power plant, there is much equipment devoted entirely to adjust the rates of supply of raw materials in accordance with the power demand made on the plant.

(ii) Increase in production cost. The variable load on the plant increases the cost of the production of electrical energy. An alternator operates at maximum efficiency near its rated capacity. If a single alternator is used, it will have poor efficiency during periods of light loads on the plant. Therefore, in actual practice, a number of alternators of different capacities are installed so that most of the alternators can be operated at nearly full load capacity. However, the use of a number of generating units increases the initial cost per kW of the plant capacity as well as floor area required. This leads to the increase in production cost of energy.

44

Principles of Power System

3.3 Load Curves
The curve showing the variation of load on the power station with respect to (w.r.t) time is known as a load curve.
The load on a power station is never constant; it varies from time to time. These load variations during the whole day (i.e., 24 hours) are recorded half-hourly or hourly and are plotted against time on the graph. The curve thus obtained is known as daily load curve as it shows the variations of load w.r.t. time during the day. Fig. 3.2. shows a typical daily load curve of a power station. It is clear that load on the power station is varying, being maximum at 6 P.M. in this case. It may be seen that load curve indicates at a glance the general character of the load that is being imposed on the plant. Such a clear representation cannot be obtained from tabulated figures.
The monthly load curve can be obtained from the daily load curves of that month. For this purpose, average* values of power over a month at different times of the day are calculated and then plotted on the graph. The monthly load curve is generally used to fix the rates of energy. The yearly load curve is obtained by considering the monthly load curves of that particular year. The yearly load curve is generally used to determine the annual load factor.

Importance. The daily load curves have attained a great importance in generation as they sup-

ply the following information readily :

(i) The daily load curve shows the variations of load on the power station during different hours of the day.

(ii) The area under the daily load curve gives the number of units generated in the day.

Units generated/day = Area (in kWh) under daily load curve.

(iii) The highest point on the daily load curve represents the maximum demand on the station on

that day.
(iv) The area under the daily load curve divided by the total number of hours gives the average load on the station in the day.

Area (in kWh) under daily load curve

Average load =

24 hours

(v) The ratio of the area under the load curve to the total area of rectangle in which it is con-

tained gives the load factor.

Load factor = Average load = Average load × 24 Max. demand Max. demand × 24

=

Area (in kWh) under daily load curve

Total area of rectangle in which the load curve is contained

* For instance, if we consider the load on power station at mid-night during the various days of the month, it may vary slightly. Then the average will give the load at mid-night on the monthly curve.

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(vi) The load curve helps in selecting* the size and number of generating units. (vii) The load curve helps in preparing the operation schedule** of the station.

3.4 Important Terms and Factors

The variable load problem has introduced the following terms and factors in power plant engineering:

(i) Connected load. It is the sum of continuous ratings of all the equipments connected to supply system.

A power station supplies load to thousands of consumers. Each consumer has certain equipment installed in his premises. The sum of the continuous ratings of all the equipments in the consumer’s premises is the “connected load” of the consumer. For instance, if a consumer has connections of five 100-watt lamps and a power point of 500 watts, then connected load of the consumer is 5 × 100 + 500 = 1000 watts. The sum of the connected loads of all the consumers is the connected load to the power station.

(ii) Maximum demand : It is the greatest demand of load on the power station during a given period.

The load on the power station varies from time to time. The maximum of all the demands that have occurred during a given period (say a day) is the maximum demand. Thus referring back to the load curve of Fig. 3.2, the maximum demand on the power station during the day is 6 MW and it occurs at 6 P.M. Maximum demand is generally less than the connected load because all the consumers do not switch on their connected load to the system at a time. The knowledge of maximum demand is very important as it helps in determining the installed capacity of the station. The station must be capable of meeting the maximum demand.

Maximum demand meter

(iii) Demand factor. It is the ratio of maximum demand on the power station to its connected load i.e.,

Maximum demand Demand factor =
Connected load The value of demand factor is usually less than 1. It is expected because maximum demand on the power station is generally less than the connected load. If the maximum demand on the power station is 80 MW and the connected load is 100 MW, then demand factor = 80/100 = 0·8. The knowledge of demand factor is vital in determining the capacity of the plant equipment.

(iv) Average load. The average of loads occurring on the power station in a given period (day or month or year) is known as average load or average demand.

Energy meter

* It will be shown in Art. 3.9 that number and size of the generating units are selected to fit the load curve. This helps in operating the generating units at or near the point of maximum efficiency.

** It is the sequence and time for which the various generating units (i.e., alternators) in the plant will be put in operation.

46

Principles of Power System

Daily average load = No. of units (kWh) generated in a day 24 hours

No. of units (kWh) generated in a month

Monthly average load =

Number of hours in a month

Yearly average load = No. of units (kWh) generated in a year 8760 hours

(v) Load factor. The ratio of average load to the maximum demand during a given period is known as load factor i.e.,
Average load Load factor = Max. demand If the plant is in operation for T hours, Load factor = Average load × T
Max. demand × T Units generated in T hours = Max. demand × T hours The load factor may be daily load factor, monthly load factor or annual load factor if the time period considered is a day or month or year. Load factor is always less than 1 because average load is smaller than the maximum demand. The load factor plays key role in determining the overall cost per unit generated. Higher the load factor of the power station, lesser* will be the cost per unit generated.
(vi) Diversity factor. The ratio of the sum of individual maximum demands to the maximum demand on power station is known as diversity factor i.e.,

Diversity factor = Sum of individual max. demands Max. demand on power station

A power station supplies load to various types of consumers whose maximum demands generally do not occur at the same time. Therefore, the maximum demand on the power station is always less than the sum of individual maximum demands of the consumers. Obviously, diversity† factor will always be greater than 1. The greater the diversity factor, the lesser‡ is the cost of generation of power.

(vii) Plant capacity factor. It is the ratio of actual energy produced to the maximum possible energy that could have been produced during a given period i.e.,

Plant capacity factor =

Actual energy produced

Max. energy that could have been produced

Average demand × T ** = Plant capacity × T = Average demand
Plant capacity

* It is because higher load factor factor means lesser maximum demand. The station capacity is so selected that it must meet the maximum demand. Now, lower maximum demand means lower capacity of the plant which, therefore, reduces the cost of the plant.
† There is diversification in the individual maximum demands i.e., the maximum demand of some consumers may occur at one time while that of others at some other time. Hence, the name diversity factor
‡ Greater diversity factor means lesser maximum demand. This in turn means that lesser plant capcity is required. Thus, the capital investment on the plant is reduced.
** Suppose the period is T hours.

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Thus if the considered period is one year, Annual plant capacity factor = Annual kWh output Plant capacity × 8760
The plant capacity factor is an indication of the reserve capacity of the plant. A power station is so designed that it has some reserve capacity for meeting the increased load demand in future. Therefore, the installed capacity of the plant is always somewhat greater than the maximum demand on the plant.
Reserve capacity = Plant capacity − Max. demand It is interesting to note that difference between load factor and plant capacity factor is an indication of reserve capacity. If the maximum demand on the plant is equal to the plant capacity, then load factor and plant capacity factor will have the same value. In such a case, the plant will have no reserve capacity. (viii) Plant use factor. It is ratio of kWh generated to the product of plant capacity and the number of hours for which the plant was in operation i.e.
Plant use factor = Station output in kWh Plant capacity × Hours of use
Suppose a plant having installed capacity of 20 MW produces annual output of 7·35 × 106 kWh and remains in operation for 2190 hours in a year. Then,
e j Plant use factor = 7⋅ 35 × 106 = 0·167 = 16·7% 20 × 103 × 2190

3.5 Units Generated per Annum

It is often required to find the kWh generated per annum from maximum demand and load factor. The procedure is as follows :
Average load Load factor = Max. demand



Average load = Max. demand × L.F.

Units generated/annum = Average load (in kW) × Hours in a year

= Max. demand (in kW) × L.F. × 8760

3.6 Load Duration Curve

When the load elements of a load curve are arranged in the order of descending magnitudes, the curve thus obtained is called a load duration curve.

48

Principles of Power System

The load duration curve is obtained from the same data as the load curve but the ordinates are arranged in the order of descending magnitudes. In other words, the maximum load is represented to the left and decreasing loads are represented to the right in the descending order. Hence the area under the load duration curve and the area under the load curve are equal. Fig. 3.3 (i) shows the daily load curve. The daily load duration curve can be readily obtained from it. It is clear from daily load curve [See Fig. 3.3. (i)], that load elements in order of descending magnitude are : 20 MW for 8 hours; 15 MW for 4 hours and 5 MW for 12 hours. Plotting these loads in order of descending magnitude, we get the daily load duration curve as shown in Fig. 3.3 (ii).
The following points may be noted about load duration curve : (i) The load duration curve gives the data in a more presentable form. In other words, it readily
shows the number of hours during which the given load has prevailed. (ii) The area under the load duration curve is equal to that of the corresponding load curve.
Obviously, area under daily load duration curve (in kWh) will give the units generated on that day. (iii) The load duration curve can be extended to include any period of time. By laying out the abscissa from 0 hour to 8760 hours, the variation and distribution of demand for an entire year can be summarised in one curve. The curve thus obtained is called the annual load duration curve.
3.7 Types of Loads A device which taps electrical energy from the electric power system is called a load on the system. The load may be resistive (e.g., electric lamp), inductive (e.g., induction motor), capacitive or some combination of them. The various types of loads on the power system are :
(i) Domestic load. Domestic load consists of lights, fans, refrigerators, heaters, television, small motors for pumping water etc. Most of the residential load occurs only for some hours during the day (i.e., 24 hours) e.g., lighting load occurs during night time and domestic appliance load occurs for only a few hours. For this reason, the load factor is low (10% to 12%).
(ii) Commercial load. Commercial load consists of lighting for shops, fans and electric appliances used in restaurants etc. This class of load occurs for more hours during the day as compared to the domestic load. The commercial load has seasonal variations due to the extensive use of airconditioners and space heaters.
(iii) Industrial load. Industrial load consists of load demand by industries. The magnitude of industrial load depends upon the type of industry. Thus small scale industry requires load upto 25 kW, medium scale industry between 25kW and 100 kW and large-scale industry requires load above 500 kW. Industrial loads are generally not weather dependent.
(iv) Municipal load. Municipal load consists of street lighting, power required for water supply and drainage purposes. Street lighting load is practically constant throughout the hours of the night. For water supply, water is pumped to overhead tanks by pumps driven by electric motors. Pumping is carried out during the off-peak period, usually occurring during the night. This helps to improve the load factor of the power system.
(v) Irrigation load. This type of load is the electric power needed for pumps driven by motors to supply water to fields. Generally this type of load is supplied for 12 hours during night.
(vi) Traction load. This type of load includes tram cars, trolley buses, railways etc. This class of load has wide variation. During the morning hour, it reaches peak value because people have to go to their work place. After morning hours, the load starts decreasing and again rises during evening since the people start coming to their homes.
3.8 Typical Demand and Diversity Factors
The demand factor and diversity factor depend on the type of load and its magnitude.

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TYPICAL DEMAND FACTORS

Type of consumer

Demand factor

Residence lighting

R |

1 kW 4

1·00

ST|| 12 kW 0·60

{ Over 1 kW

0·50

Commercial lighting General power service

Restaurants

0·70

R Theatres

0·60

|| Hotels

0·50

S Schools

0·55

| Small industry

0·60

|| Store

0·70

TR 0 –10 H.P.

0·75

| 10 –20 H.P.

0·65

S 20 –100 H.P.

0·55

T|| Over 100 H.P.

0·50

TYPICAL DIVERSITY FACTORS

Residential lighting

Commercial lighting

General power supply

Between consumers

3 – 4

1·5

1·5

Between transformers

1·3

1·3

1·3

Between feeders

1·2

1·2

1·2

Between substations

1·1

1·1

1·1

Illustration. Load and demand factors are always less than 1 while diversity factors are more than unity. High load and diversity factors are the desirable qualities of the power system. Indeed, these factors are used to predict the load. Fig. 3.4 shows a small part of electric power system where a distribution transformer is supplying power to the consumers. For simplicity, only three consumers a, b, and c are shown in the figure. The maximum demand of consumer a is the product of its connected load and the appropriate demand factor. Same is the case for consumers b and c. The maximum demand on the transformer is the sum of a, b and c’s maximum demands divided by the diversity factors between the consumers. Similarly, the maximum demand on the feeder is the sum of maximum demands on the distribution transformers connected to it divided by the diversity factor between transformers. Likewise diversification between feeders is recognised when obtaining substation maximum demands and substation diversification when predicting maximum load on the power station. Note that diversity factor is the sum of the individual maximum demands of the subdivisions of a system taken as they may occur during the daily cycle divided by the maximum simultaneous demand of the system. The “system” may be a group of consumers served by a certain transformer, a group of transformers served by a feeder etc. Since individual variations have diminishing effect as one goes

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Principles of Power System

farther from the ultimate consumer in making measurements, one should expect decreasing numerical values of diversity factor as the power plant end of the system is approached. This is clear from the above table showing diversity factors between different elements of the power system.

Example 3.1. The maximum demand on a power station is 100 MW. If the annual load factor is 40% , calculate the total energy generated in a year.

Solution.

Energy generated/year = Max. demand × L.F. × Hours in a year = (100 × 103) × (0·4) × (24 × 365) kWh = 3504 × 105 kWh

Example 3.2. A generating station has a connected load of 43MW and a maximum demand of 20 MW; the units generated being 61·5 × 106 per annum. Calculate (i) the demand factor and
(ii) load factor.

Solution.

(i)

Demand factor = Max. demand = 20 = 0·465

Connected load 43

(ii)

Average demand = Units generated / annum = 61⋅ 5 × 106 = 7020 kW

Hours in a year

8760

∴ Load factor = AMvearxa.gededmemanadnd = 2070×21003 = 0·351 or 35·1%

Example 3.3. A 100 MW power station delivers 100 MW for 2 hours, 50 MW for 6 hours and is shut down for the rest of each day. It is also shut down for maintenance for 45 days each year. Calculate its annual load factor.

Solution.

Energy supplied for each working day

= (100 × 2) + (50 × 6) = 500 MWh

Station operates for = 365 − 45 = 320 days in a year



Energy supplied/year = 500 × 320 = 160,000 MWh

Annual load factor =

MWh supplied per annum

× 100

Max. demand in MW × Working hours

=

160,000
a100f × b320 ×

g24

×

100

=

20·8%

Example 3.4. A generating station has a maximum demand of 25MW, a load factor of 60%, a plant capacity factor of 50% and a plant use factor of 72%. Find (i) the reserve capacity of the plant (ii) the daily energy produced and (iii) maximum energy that could be produced daily if the plant while running as per schedule, were fully loaded.

Solution.

(i) Load factor = Average demand Maximum demand

or 0·60 = Average demand 25



Average demand = 25 × 0·60 = 15 MW

Plant capacity factor = Average demand Plant capacity



Plant capacity = Average demand = 15 = 30 MW

Plant capacity factor 0 ⋅ 5

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∴ Reserve capacity of plant = Plant capacity − maximum demand

= 30 − 25 = 5 MW

(ii) Daily energy produced = Average demand × 24

= 15 × 24 = 360 MWh

(iii) Maximum energy that could be produced

Actual energy produced in a day

=

Plant use factor

= 360 = 500 MWh/day 0 ⋅ 72
Example 3.5. A diesel station supplies the following loads to various consumers :

Industrial consumer = 1500 kW ; Commercial establishment = 750 kW

Domestic power = 100 kW; Domestic light = 450 kW

If the maximum demand on the station is 2500 kW and the number of kWh generated per year is 45 × 105, determine (i) the diversity factor and (ii) annual load factor.
Solution.

(i) Diversity factor = 1500 + 750 + 100 + 450 = 1·12 2500

(ii)

Average demand = kWh generated / annum = 45 × 105/8760 = 513·7 kW

Hours in a year



Load factor = Average load = 513⋅ 7 = 0·205 = 20·5%

Max. demand 2500

Example 3.6. A power station has a maximum demand of 15000 kW. The annual load factor is

50% and plant capacity factor is 40%. Determine the reserve capacity of the plant.

Solution.

Energy generated/annum = Max. demand × L.F. × Hours in a year

= (15000) × (0·5) × (8760) kWh = 65·7 × 106 kWh

Units generated / annum Plant capacity factor = Plant capacity × Hours in a year



Plant capacity = 65 ⋅ 7 × 106 = 18,750 kW

0 ⋅ 4 × 8760

Reserve capacity = Plant capacity − Max. demand

= 18,750 − 15000 = 3750 kW

Example 3.7. A power supply is having the following loads :

Type of load

Max. demand (k W)

Diversity of group

Demand factor

Domestic

1500

1·2

0·8

Commercial

2000

1·1

0·9

Industrial

10,000

1·25

1

If the overall system diversity factor is 1·35, determine (i) the maximum demand and (ii) connected load of each type.

Solution.

(i) The sum of maximum demands of three types of loads is = 1500 + 2000 + 10,000 = 13,500 kW. As the system diversity factor is 1·35,

∴ Max. demand on supply system = 13,500/1·35 = 10,000 kW

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Variable Load on Power Stations