UNIT-V

ECONOMICS OF POWER
PLANTS
Important Terms and Definitions

1. Connected load:

It is the combined continuous rating of all the receiving apparatus on consumer’s

premises. If a consumer has connections for 3 lamps of 40 watts each, and power point of 500W
for refrigerator and T.V. consuming 60W, then the total connected load of the consumer = 3 x
40 + 500 + 60 = 680W.

2. Demand:

It is the load that is drawn from the source of supply at the receiving terminals averaged
over a suitable and specified internal of time.

3. Maximum demand:

It is the maximum load which is used by a consumer at any time. It is determined by

measurement, according to specifications over a prescribed interval of time. It can be less than
or equal to connected load. But generally, the actual maximum demand is less than the
connected load because all the loads never run in full load at the same time.

4. Demand factor:

It is the ratio of actual maximum demand of the system to the total connected demand
of the syste.

𝐴𝑐𝑡𝑢𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑚𝑎𝑛𝑑

Demand factor =
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑 𝑑𝑒𝑚𝑎𝑛𝑑

5. Load factor:

It is the ratio of average load over a given time interval to the peak load during the
same time interval.

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑 𝑜𝑣𝑒𝑟 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙

Load Factor =
𝑃𝑒𝑎𝑘 𝑙𝑜𝑎𝑑 𝑑𝑢𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙

Load factor is always less than unity. It plays an important part on the cost of power
generation per unit. The higher the load factor, the lesser will be the cost of power generation
per unit for the same maximum demand.

6. Capacity factor or plant capacity factor:

It is the ratio of actual energy produced in kilowatt hours (kWhr) to the maximum
possible energy that could have been produced during the same period.

𝐴𝑐𝑡𝑢𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑘𝑊𝐻𝑟 𝐸

Capacity factor = =
𝑅𝑎𝑡𝑒𝑑 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 the 𝑝𝑙𝑎𝑛𝑡 𝐶𝑥𝑡
 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑
Capacity factor = 𝑅𝑎𝑡𝑒𝑑 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 the 𝑝𝑙𝑎𝑛𝑡
Where, E  Energy produced in kWhr
C  Capacity of the plant in kW
t  Total number of hours in given period.

If the rated capacity of the plant is equal to the peak load, then the load factor and
capacity factor will be nemerically equal.

7. Utilisation factor:

It is the ratio of maximum load to the rated capacity of the plant.

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑙𝑜𝑎𝑑
Utilisation factor = 𝑅𝑎𝑡𝑒𝑑 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 the 𝑝𝑙𝑎𝑛𝑡
8. Reserve factor:

It is the ratio of load factor to capacity factor.

𝐿𝑜𝑎𝑑 𝑓𝑎𝑐𝑡𝑜𝑟
Reserve factor = 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟

9. Diversity factor:

It is defined as the ratio of sum of individual maximum demand to the actual peak load
of the system.

𝑆𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑚𝑎𝑛𝑑

Diversity factor = 𝐴𝑐𝑡𝑢𝑎𝑙 𝑝𝑒𝑎𝑘 𝑙𝑜𝑎𝑑 𝑜𝑓 the 𝑠𝑦𝑠𝑡𝑒𝑚

10. Plant use factor:

It is the ratio of energy produced in a given time to the maximum possible energy that
could have been produced during the same period of operation.

𝐴𝑐𝑡𝑢𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑔𝑖𝑣𝑒𝑛 𝑡𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑

Plant use factor =
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑏𝑦 the 𝑝𝑙𝑎𝑛𝑡

𝐸
= 𝑐 𝑥 𝑡1

Where, 𝑡1  actual number of hours when the plant has been in

operation.
11. Load curve:
It is a graphical representation which shows the power demands for every instant during
a certain time period. It is drawn between load in kW and time in hours. If it is plotted for 1
hour, it is called hourly load curve and if the time is considered is of 24 hours then it is called
daily load curve. When it is plotted for one year (8760 hours), then it is called annual load
curve.
The area under the load curve represents the energy generated in the priod considered.
If we divide the area under the curve by the total number of hours, then it will give the
average load on the power station. The peak load indicated by the load curve represents the
maximum demand of the power station.

Load curve

This curve gives full information about the incoming load and helps to decide the installed
capacity of the power station. It is also useful to decide the economical sizes of various
generating units.

12. Load duration curve:

Load duration curve

This curve represents the re-arrangement of all the load elements of load curve in
A typical daily load curve for a power station is shown in fig. It may be observed that
the maximum load on power station is 40kW from 6 to 9P.M. Similarly, other loads of the load
curve are plotted in decreasing order in the fig (b). This curve is called load duration curve.
 It is observed that the area under both the curves is equal and represents the total energy
delivered by generation station. Load duration curve gives a clear analysis about generation
power economically.
Types of loads:

The various types for loads are described below:

(i) Residential load: This type of load includes domestic lights, power needed for
domestic appliances such as radios, television, electric cookers, water heaters,
refrigerators, grinders etc.
(ii) Commercial load: It includes lighting for shops, advertisements and electric
appliances used in shops, hotels and restaurants etc.
(iii) Industrial load: It consists of load demand of various industries.
(iv) Municipal load: It consists of power required for street, lights, water supply
and drainage purposes.
(v) Irrigation load: It includes electrical power required for pumps to supply water
to fields.
(vi) Traction load: It consists of power required for tram cars, trolley, buses and
railways.

13. Dump power:

This term is used in hydroelectric power plants. It shows the power in excess of the load
requirements.

14. Firm power:

It is the power which should always be available even under emergency conditions,
15. Prime power:
The power which may be mechanical, hydraulic or thermal that are always available
for convertion into electric power.
16. Base load and peak load power plants:

The base load is the load below which the demand never falls and is supplied 100% of
the time. The power plants used to supply base loads are called base load power plants. The
base load power plants are loaded heavily. Operating cost of such plants is very important. A
high capital cost is permissible, if low operating costs can be maintained. Hydro and nuclear
power plants are usually classified as base load power plants.

The peak load is the load which occurs at the top portion of the load curve. The power
plants which are used to supply peak loads are called as peak load power plants. The peaking
load occurs for about 15% of the time. The peak load power plants are of smaller capacity, run
for a shorter period in the year and work at low load factors. Peak load power plants should be
capable of quick starting. Since peaking load plants are used only for a small fraction of time,
the fuel cost is not of major importance. Minimum capital cost should be the criterion. Diesel
power plants are usually classified as peak load power plants.
ACTUAL LOAD CURVES

Load curve is a graphical representation which shows the power demands for every
instant during certain time period. By drawing these load curves, the peak load can be identified
and hence the capacity of power plant can be judged.

Fig. shows two load curves in which the energy consumed by the customer is same but
the way adopted to use is different. The design of any power plant is bases on the way that the
customer adopted to use energy. The customer X and Y consume same amount of energy but
the nature of consumption is different.

Constant and variable demand load curve

In fig, the peak load is far greater than the first. Therefore, the generating capacity of
the plant required to supply the load of Y is greater than the capacity required to supply the
load of X. The plant designed for customer Y is not only bigger in size but it also runes
underload conditions for the majority of the period. Therefore, the cost of energy supplied to
Y may be more than the cost of energy supplied to X even the total energy consumed by both
customers is same.
As explained earlier, the different types of loads for different types of customers are
explained below:

1. Residential load curve:

Fig. shows as typical residential load curve. During the early morning hours (6A.M. to
9A.M.), the energy is required for light, fans, refrigerators, water heaters. After the breakfast
(at 9A.M.), the demand decreases and fairly remains constant till about 4 P.M. After 4 P.M,
the demand is beginning to increase and attains its peak at 6P.M. The peak load remains almost
constant up to 8P.M, after that it decreases rapidly and attains minimum at 12P.M, the foresaid
statements are applicable for summer season.
 Residential load curve

But during winter season, the load remains constant during day time ad will be
minimum. After 5P.M. the load rapidly approaches its peak. The high demand occurs at about
8P.M.
2. Commercial load curves:

Commercial loads of shops, offices and hospitals

Fig. shows a typical load curve for commercial usage like shops, office, restaurants etc. The
lighting in shops and office starts at 6A.M. for cleaning and sweeping and them it reaches peak
at 10A.M. It remains constant more or less during 10A.M. to 4P.M. It increases further during
4 to 7P.M. as more lights are required. Then, the load rapidly falls during 7P.M. to 12P.M. as
the offices remain closed.
3. Industrial load curve:

Fig. shows a typical load curve for industrial community of on shift basis. In early
morning from 5A.M. to 8A.M, the energy demand increases as some for the machinery starts
for warming prior to operation. The entire industry starts running and energy demand remains
constant from 8A.M. to shortly before noon. There is a heavy fall in energy demand during 12
to 1P.M. due to lunch hours. By 2P.M. again, the load attains the same level as at 8A.M, Shortly
before 5P.M, the load starts to drop as the shift of work ends. By 6P.M. most of the machines
are shut down and load gradually decreases to minimum until 10A.M. Then the minimum
demand continues till the start of next working day.

Industrial load curve for one shift

4. Street lighting load curve:

Street lightning load curve

 Fig. shows a typical local curve for street lighting. Street lighting is the only form of
load that does not exhibit peak demands. Normally, all streetlights are simultaneously made
‘On’ at 6P.M. and are turned ‘Off’ at 6A.M. The load demand remains more or less constant
during these hours.

5. Urban traction load curve:

Fig. shows a typical load curve for urban traction. During midnight hours from 12P.M.
to 5A.M, the demand tapers off as the service reaches its minimum level. As the early factory
workers start for their work, the required train services increases rapidly and consequent load
continuously rise as the factory workers are followed by office workers, school children,
college students and early shoppers. The peak load reaches about at 9.30A.M. After 10P.M.,
the load rapidly diminishes as some of the trains return to the yards.

Urban load curve

The minimum load is reached at noon hours and then rises continuously until the evening rush
hours. The load again reaches its peak at 5P.M. when most of the workers go back to their
homes. The load after 6P.M. falls rapidly.
VARIABLE LOAD OPERATINGS

The variable load operation problem affects power plant design and operation as well
as the cost of generation. The necessity of supplying a variable load influences the
characteristics and the method of using the power plant equipment. The generation of power
must be regulated according to the demand. For that purpose governing is necessary to achieve
it. Quick response to varying load is another important requirement of the power plant.

A carful study of the load duration curve helps to decide the capacity of the base load
plant and also of the peak load plant. The base load plant should be run at high load factor. The
peak load plant should be of smaller capacity to reduce the cost of generation. It could be a
gasturbine unit, pumped hydro-system, compressed air energy storage system or a diesel engine
depending on the size and scope of availability.

If a whole load is to be supplied by the same power plant, the generation and prime
mover must be able to take varying load as quick as possible without variation of the voltage
or frequency of the system. When the load on the generator increases, it will show down the
rotor and prime mover and therefore, it reduces the frequency. When the speed of the prime
mover decreases, the governor must act. It is the function of the governor to control the supply
of fuel to prime mover according to the load. It should be enough to bring the speed back to
normal and pick up the load. Frequency stabilizers are used to maintain the frequency constant
which may change due to response of the equipment.

In case of thermal power plant, the raw material used are fuel air, and water to produce
variable power. According to the requirement, the raw materials are supplied correspondingly
with an increase in load on the plant, the governor admits more steam and maintains the turbine
speed up to certain point, the governor responds rapidly with change in load but beyond this
point changes are not so rapid. Because of fluctuating steam demand, it becomes very difficult
to secure good combustion and steady steam pressure. Therefore, the design of thermal plants
for variable loads is always more difficult than diesel or hydroelectric power plants and it is
always desirable to allow the thermal plant to operate as base load plant.
Economic load sharing between base load plant and peak load plant is desirable. Steam
power plant, and nuclear power plant are preferred as base load plants whereas diesel power
plant and hydro power plant can be used as peak load plant. Hydro power plant with larger
water storage can also be used as base load plant.

COST OF ELECTRICAL ENERGY

A power plant should provide a reliable supply of electricity at minimum cost to the
customer. The cost of the electricity is determined by the following costs:

FIXED COST OR CAPITAL COST

It is the cost required for installation of complete power plant. This cost includes the
cost of land, buildings, equipments, transmission and distribution lines, cost of planning and
designing the plant and many others. It also consists of interest, taxes, depreciation, insurance
etc.
1. Land, building and equipment cost:

The cost of land and building does not change much with different types of power plants
but the equipment cost changes considerably. The cost of building can b reduced by eliminating
the superstructure on the boiler house and turbine house. To reduce the cost of equipment, unit
system may be adopted, reduced by simplifying the piping system and elimination of duplicate
system such steam headers and boiler feed heaters.

The cost of the equipment or the plant investment cost is usually expressed on the basis
of kW capacity installed. The per kW capacity may not vary for various thermal power plant
where as for hydro electric power plant, it changes a lot because the cost of hydroelectric power
plant depends on the availability of foundation, types of dam available head and spillways used.

2. Interest:
The money needed for investment in an enterprise may be obtained as loans, through
bonds and shares. Interest is the difference between money borrowed and money returned. The
rate of interest may be simple rate expressed as % per annum or may be compounded. A
suitable rate of interest must be considered on the capital invested.

3. Depreciation cost:
It is the amount to be se aside per year from income to meet the depreciation caused by
the age of service, wear and tear of machinery. It also covers the decrease in value of equipment
due to obsolescence.
The power plant and equipment in the plant will have a certain period of useful life.
After years of use, the equipment loses its efficiency or becomes absolute and needs
replacement. Sometimes, Equipment may have to be changed even when fairly new, if more
efficient equipment has come into the market. Some money is put aside annually to enable this
to be done when necessary. This is known as depreciation fund.
The following methods are generally used to calculate the depreciation cost:
(i) Straight line method.
(ii) Sinking fund method.
(iii) Diminishing value method.

(i) Straight line method:

It is the simplest and commonly used method. It is based on the assumption that
depreciation occurs uniformly for every year according to a straight-line law. They money
saver neglects any interest.

According to this method, annual amount to be set aside is calculated by the following
expressiong.

P−S
A=
n
Where, P = Capital cost of the equipment.
S = Salvage value at the end of plant life.
n = Life of plant in years.
The straight line method is represented by the fig.
(ii) Sinking fund method:
In this method, a sum of money is set aside every year for N years and invested to earn
compound interest. This method is bases on the conception that the annual uniform deduction
from income for depreciation will accumulate to the capital value of the plant at the end of life
of the plant. Let A is the annual deposit and ‘I’ is the interest compounded annually when the
deposit is invested

Then amount set aside at the end of first year = A

Amount at the end of second year = A + interest on A
= A + Ai = A (1 + i)
Amount at the end of third year = A(1 + i) + interest on A (1 + i)
= A(1 + i) + A(1 + i)
= A(1 + i)2

Similarly, at the end of 1th year = A(1 + i)n – 1

Total amount accumulated in N years (say x) = Sum of the amounts accumulated in n

years
X = A +A (1 + i) + A (1 + i)2 + . . . + A (1 + i)n – 1

= A 1 + (1 + i ) + (1 + i ) + ... + (1 + i )
2 n −1
 ---------- (i)
Multiplying the above equation (i) by (1 + i), we get
 
= (1 + i) x = A (1 + i ) + (1 + i) 2 + .... + (1 + i ) n ---------- (ii)
Subtracting equation (i) from (ii), we get

 
xi = A (1 + i) n − 1
  (1 + i ) n − 1
x = A 
 i 
Where x is the sum of amounts saved together with interest earnings
Therefore, x = (P – S)
Where, P = Capital cost of the equipment
S = Salvage value at the end of Nth year
 (1 + i ) n − 1
P − S = A 
 i 
 i 
A= ( P − S )
 (1 + i ) − 1
n

(iii) Diminishing value method:

In this method, the amount set aside per year decreases as the life of the plant increases.
The following example gives clear idea of this method.

Say the equipment cost as Rs. 40000. The amount set aside is 10% of the initial cost at
the beginning of the year and 10% is the remaining cost with every successive year. Therefore,

The amount set aside during first year

10
= 40000 − x 40000 = Rs.36000.Balance
100
The amount set aside during 2nd year
10
= 36000 − x36000 = Rs.32400.Balance
100
The next installment during third year
10
= 32400 − x32400 = Rs.29160.Balance
100

The main disadvantage of this method is that it requires heavy investments in the early
years when the maintenance charges are minimum and it goes on decreasing as the time passes
but the maintenance charges increase.

4. Insurance:
Now-a-days, insurance plays a vital role in the country. It becomes necessary to insure
the costly equipments especially for the \fire or accident risks. A fixed sum is set aside per year
as insurance charges. The annual premium may be 2 to 3% of the equipment cost but the annual
installment is quite heavy when the capital cost of the equipment is high.

5. Management cost:
This cost includes the salary of the management employees working in the plant. This
must be paid whether the plant is working or not. Therefore, this cost is included in the fixed
cost.
OPERATING COSTS

The operational cost includes the cost of fuel, cost of lubricating oil, greases, cooling
water, cost of maintenance and repairs, operating labour cost, supervision cost and taxes. This
cost varies with amount of electrical energy produced.

1. Cost of fuel:

The fuel consumption depends on the amount of electrical energy produced. As load on
the prime movers increases, the fuel consumption will increase and so does the cost. The
efficiency of the prime mover is the highest at the rated load. At lower loads, the efficiency
decreases and so, the fuel consumption will increase. The selection of the fuel and the
maximum economy in its use are, therefore, very important consideration in thermal plant
design. The cost of fuel includes not only its price but also its transportation and handling costs
also. The cost of fuel depends on the calorific value and availability.

2. Lubrication oil, Grease and Water cost:

The cost of these materials is also proportional to the amount of energy generated. This
cost increases with an increase in life of the power plant as the efficiency of the power plant
decreases with the age.

3. Cost of maintenance and repairs:

In order to avoid breakdowns, maintenance is necessary. It includes periodic cleaning,
adjustments and overhauling of equipments. The material used for maintenance is also charged
under this head.

It is necessary to repair when the plant breakdown or steps due to faults in mechanism.
The repairs may be major or minor and are charged to the depreciation fund of the equipment.
This cost is higher for thermal power plants than hydro power plants.

4. Cost of operating labour:

This includes the salary of the operating labour working in the plant. Maximum labours
are needed in a thermal power plant using coal as a fuel. A hydro power plant or a diesel power
plant of same capacity requires a lesser number of labours. In automated power plant, labour
cost is reduced to a great extent.

5. Cost of supervision:
In includes the salary of the supervising staff. A good supervision reduces the
breakdowns and extends the plant life. The supervising staff includes the superintendent, chief
engineer, engineers, store incharges, purchase officers etc.

6. Taxes:
The various taxes are included in this head. These are income tax, sales tax, provisional
tax, commercial tax etc.

The total cost of energy produced is the sum of fixed charges and operating charges.

Total cost = Fixed costs + Operating costs.

ECONOMIC LOAD SHARING BETWEEN POWER PLANTS

Different power plants such as hydro, thermal, nuclear, gas turbine, MHD etc are
operated combinedly to give greater reliability and maximum economic benefits. When the
number of stations works in combination with each other to supply the power to the consumers,
the system is known as interconnected system.
In an interconnected system, the major problem is division of load among the power
plants. The load distribution among the power plants depends upon the operating characteristics
of the power plant. The distribution of load amount the power plant in an interconnected system
is done in such a way that the overall economy is achieved.

Fig.shown as load duration curve. The whole area under the load curve is divided into
three parts as base load, intermediate load and peak load. It is not economical to design a power
plant to load to the maximum peak load as it works in under-load condition for the most of
time. In order to achieve maximum possible is the loading of the most efficient power station
in the order of merit of low fuel cost. It is made possible by establishing central control room
which can control number of power plants simultaneously in the grid system. This also saves
the fuel consumption per kW of power generation. In addition to the saving in fuel consumption,
there is also saving due to reduced spare capacity required and also due to the employment of
large size units.

Base load station takes up the load on the lower region of the load curve. This station
is highly efficient and operates on three shift basis throughout the year. The fixed cost of these
plants is high. Capacity factor is the index of the return on the capital investment on the plant.
Continuous operation of base load plant at high load factor improves the capacity factor and
this makes operation of the base load plant an economic proposition. Hydro and nuclear power
plants are usually classified as base load plants.

Intermediate stations operate on two or single shift basis. The capital cost of such plant
is lower and fuel cost is higher than base load plants. Thermal stations fall under this category.

Peak load plants operate only when required for short times under the upper part of the
load curve. Plant capacity factor is low as it is operating for short duration. Fuel is very high
but total capital cost is less. Diesel and gas turbine plants are classified under this category.
 For a known load duration curve, the economic load sharing between base load and
peak load plants operating in parallel can be found as follows.

Let the operating costs are known and these are given by
C 2 = A2 ( P − Pb ) + B2 ( N − N P )
The cost of the system C is given by

C = C1 + C2 = ( A1 Pb + B1 N p ) + A2 ( P − Pb ) + B2 ( N − N p ) 
The minimum total cost can be obtained when
dc
=0
dPb
dN
 ( A1 − A2 ) + ( B1 − B2 ) b = 0
dPb
dN b  A1 − A2 
 = hrs
dPb  B1 − B2 
Thus for economic load sharing, the area under the load curve is so divided by
horizontal line that its magnitude is given by
A − A2
H= 1 hrs
B2 − B1
The indicates that for economic load sharing, peak load plant should work for H hrs per
years. The value of H should be always higher.

 A1 > A2 and B2 > B1

A1 is higher and B1 is lower for base load plant as compared to peak load plant. Mark the point
 H 
T on the x-axis of load curve for the distance of H hrs in percentage  x100. Draw the
 8760 
vertical line through T which meets the load curve at point P.
Draw the horizontal line PQ as shown in fig. Now, the area Ap above the line PQ gives
the energy generated by peak load plant and area below it gives the energy generated by the
base load plant. The scale taken for drawing load curve is 1cm = x % along time axis, and 1cm
= y in kW along the load axis.
 1cm 2 = ( x%) xy
100% = 8760 hrs
 x 
1cm 2 =  x8760 xy in kWh
 100 
2
If areas Ab and Ap in cm are known, then
 x 
N b = Ab x x8760 xy in kWh ------ for base load plant
 100 

 x 
N p = A p x x8760 xy in kWh ------- for peak load plant
 100 

Thus, the load sharing between the two power plants can be achieved and this results
in overall efficiency of operation.

ENERGY RATES

The rate of energy sold to the consumers depends upon the type of consumers such as
domestic, commercial and Industrial. The rate of energy also depends upon the total energy
consumed and the load factor of the consumer.

The rate of energy is decided in such a way that the following items must cover within
the energy rate:

(i) Recovery of capital cost invested for the generating power plant.
(ii) Recovery of the running costs such as operating cost, maintenance cost,
metering the equipment cost, billing cost etc.
(iii) Minimum profit on the invested capital, since the power plant is considered as
the profitable business for the government.

Determination of cost of each item mentioned here is simple but the allocation of theses
items among various classes of consumers are rather difficult. This repairs considerable
engineering judgment.

The general energy rate or tariff can be given by the following equation:

E = Ax+By+C

Where

E = Total amount of bill for the period considered

A = Rate per kW of maximum demand
x = Maximum demand in kW
B = Energy rate per kW-hr
Y = Energy consumed in kW-hr during the period considered
C = Constant amount charged to the consumer during each bill period. This
charge is independent of demand or total energy.
TYPES OF TARIFFS

The various forms used for changing the consumers as per their energy consumed and
maximum demand are discussed below.

1. Flat demand rate:

In this type of charging, the charging depends only on the connected load and fixed
number of hours of used per month or year. This can be given by the following equation
E = Ax

Flat demand rate

The notations are taken as discussed above. This rate expresses the charge per unit of demand
(kW) of the consumer. Here no metering equipments and manpower are required for charging.
In this system, the consumer can theoretically use any a mount of energy upto that consumer
by all connected loads. The unit energy cost decreases progressively with an increases energy
usage. The variation in total cost and unit cost are shown in fig.

Straight meter rate

This type of charging depends upon the amount of total energy consumed by the consumer.
The bill charge is directly proportional to the energy consumed by the consumer. This can be
represented by the following equation.
E = By
The major drawbacks of this system are:
 (a) In this type of system, the consumer using no energy will not pay any mount
although he has incurred some expenses to the power station.
(b) The rate of energy is fixed, therefore this method of charging does not
encourage the consumer to use more power.

The variation in total cost and unit consumed are shown in fig.

3. Block-meter rate:

In previous straight line meter rate, the unit charge is same for all magnitudes of energy
consumption. The increases consumption spreads the item of fixed charge over a greater
number of units of energy.

Block meter rate

Therefore, the price of energy should decrease with an increase in consumption. The block
meter rate is used to overcome this difficulty. This method of charging is represented by the
equation.
E = B1 y1 + B2 y2 + B2 y3 + ........
Where
B3 < B2 < B1 and
y1 + y2 + y3 + ....... = y (total energy consumption
The level of y1 , y 2 , y 3 ,...... is decided by the government to recover the capital cost. In
is system, the rate of unit charge decrease with increase in consumption of energy as shown in
fig.
4. Hopkinson demand rate of Two-part tariff:
This method of charging depends upon the maximum demand and energy consumption.
This method is proposed by Dr. John Hopknson in 1982.

This method of charging is represented by the equation

E = A+By

Hopkinson method of charging

In this method, two meters are required to record the maximum demand and the energy
consumption of the consumer. This method is generally used for the industrial consumers. The
variation in total cost with respect to the total energy consumption taking x as parameter is
shown in fig.

5. Doherty rate or three part tariff:

This method is proposed by Henry L. Doherty. In this method of charging, the consumer
has to pay some fixed amount in addition to the charges for maximum demand and energy
consumed. The fixed amount to be charged depends upon the occasional increase in prices and
wage charges of the workers etc.

This method of charging is expressed by the equation.

x = Ax+By+C

Doherty method of charging

This method of charging is most commonly used in Tamilnadu and all over India. In this
method, the customers are discouraged to use more power when the generating capacity is less
than the actual demand. For example, for the first 50kW-hr units the charging rate is fexed as,
says Rs. 2.5/kW-hr and if it exceeds than this charge is rapidly increased as Rs. 3.5/kW-hr for
next 100 kW-hr unit (i.e from 51kW-hr to 150kW-hr). This method is unfair to the customer,
but very common in India and many developing nations.

ECONOMICS OF LOAD SHARING BETWEEN GENERATORS

During design of power plants, prime importance is given to the economics of load
sharing. Engineers are designing the power plant components such as boilers, heat turbines,
heat exchangers, condensers, and generators etc for getting the highest thermal efficiency of
the plant, various methods have been developed for economic operation of the power plant
under varying load conditions. Transmission loss is also minimized by introducing the
successful design of transmission lines. The main problem for the electrical power engineers is
the economic load sharing of the output of the generators. The proper sharing of load between
two generators tog give maximum overall efficiency is the major problem in load distribution
amount generators.
 Input – output curves of unit X and Y

This input-output carries for two generators within a power plant which are operating in parallel
and supply a common load is shown in fig. From the fig, it is evident that the generator X is
more efficient than Y throughout its load range as the output of X is more than output of Y for
the same input. Therefore, the engineers may think that they can load generator X first to its
full capacity and then for generator Y for the remaining load. But it is not proper distribution
as the overall efficiency of the system would not be highest with the distribution of loads as
mentioned above. Therefore, it is essential to find out load distribution between generators to
give the highest efficiency for the system.

Variation of combined input with varying load sharing between X and Y

This problem can be resolved by plotting the sum of the inputs of X and Y against the
load X for a given constant load on the two limits. It is shown in fig.
Say total load is 4MW. If the load on X = 0 and on Y = 4MW, then the load on X+Y =
4MW. The required input for X and Y can be calculated from fig. if the load on X = 1 and on
Y 3MW then again the combined load supplied by X+Y = 4MW and the corresponding required
input for this load division can also be calculated from fig. different outputs of X as shown in
fig. Such different curves for different constant value of (Ox + Oy) can also be drawn using the
same procedure as described above. In the curves drawn, it is clear that one point on the curve
the combined input is minimum for a given total load. Corresponding to this point (A) of
minimum input to the system for the required output, we can find out the load on generator X.
the load on generator Y is the difference of load sharing, highest efficiency can be achieved.

This method is useful only for the system having two generators. If the number of
generators supplying the load is more than two, this method becomes cumbersome process.
The condition of minimum input for any combined constant output is

d (I x + I y )
=0
dOx
dI dI y
 x + =0
dOx dC x
dI x dI y
=− ------------ (1)
dOx dOx

dI y dI y dOy
But = . ------------ (2)
dOx dOy dOx
But O y = Oc − O x
Where, Oc = O x + O y Combined output of X and Y
dOy dOc dOx
= .
dOx dOx dOx
Where Oc= constant
dOy
 = −1
dOx
Substituting this value in equation (2)
dI y dI y
=− -------------- (3)
dOx dOy
This is the condition for the maximum input for the combined constant output. If there
are n units, supplying a constant load, then the required condition for the minimum input or
maximum system efficiency is

dI1 dI dI dI
= 2 = 3 = .......... .. = n
dO1 dO2 dO3 dOn