Electric Power Generation [Thermal, Hydro, Nuclear, Gas power plants]

By
D.Rajesh Asst. Prof.
EEE Department

 Unit-i Electric Power Generation& Economic
Considerations
 Unit-ii Power Supply Systems& Distribution
Systems
 Unit-iii Inductance & Capacitance Calculations
 Unit-iv Performance Of Transmission Lines
 Unit-v Overhead Line Insulators
7 September 2020 D.Rajesh Asst.Prof. EEE Department 2

 Convenient form
 Easy control
 Greater flexibility
 Cheapness
 Cleanliness
 High transmission efficiency

Renewable energy sources
 The Sun
 The Wind
 Water
Non - Renewable energy sources
 Fuels : Coal, Gas, Oil
 Nuclear energy

Electricity generation - whether from fossil
fuels, nuclear, renewable fuels, or other
sources is usually
based on the:

In September of 1831, Michael Faraday
made the discovery of Electromagnetic Induction.
Faraday attached two wires to a disc and
rotated the disc between the opposing
poles of a horseshoe magnet creating
an electric current.

An electric current is not generated unless the magnetic field is moving
relative to the copper wire, or the copper wire is moving relative to the
magnetic field.
If you place a magnet and a conductor (copper wire), in a room together
there will be no electric current generated.
This is because motion, from our equation for electricity, is missing!

So simple electric generators found in power plants contain, magnets
and copper wire that when put into motion relative to one another
create the electric current that is sent out to homes.

- AC of 60 Hz produced by generator
- Resistance losses are smallest at high voltages and low currents

At home, electric current
that was generated by
generators in the power
plant is used to power
electric appliances.
The electric current,
running through the
copper wire causes
the armature to spin
which is how most
motors generate
motion.

Thermal
Gas
Nuclear
Hydro

❑ Power plants using conventional
(non-renewable) sources of
energy
❖ Steam power plant
❖ Nuclear(Atomic) power plant
❖ Diesel power plant
❖ Gas power plant
❖ Hydro electric(Hydel) power plant
❑ Power plants using Non-
conventional(renewable) sources
of energy
➢ Solar thermal power plant
➢ Wind powered
generation(aerogeneration)
➢ Wave power plant
➢ Tidal power plant
➢ Geothermal power plant
➢ Bio-mass power plant
➢ Oceanthermal power plant

A generating station which
converts heat energy of coal
combustion into electrical
energy is known
as a steam power station.

- Steam power plant is an energy converter
Chemical energy
Thermal energy
Steam energy
Mechanical energy
Electrical energy
Chemical energy of fuel
Heat Energy
Steam Energy
Mechanical Energy
Electrical Energy
Oxidation
Heat energy is transfered to fluid in
Boiler
Heat energy is converted into
Mechanical energy by Prime
mover
Mechanical energy is converted
into Electrical energy by Generator

AIR, GAS
STEAM
WATER
LEGEND

❑ Fuel : Coal or Oil
❑ Main parts : Boiler, Turbine,
Generator , condenser, pump

1
• Pollution
2
• Low efficiency
3
• Position ( near the river) need cooling
water

1. Coal and ash handling arrangement
2. Steam generating plant
3. Steam turbine
4. Alternator
5. Feed water
6. Cooling arrangement

 Turbines perform the energy conversion in two
steps:
 Step 1: Thermal energy of the steam to kinetic
energy of the steam
 Step 2: Kinetic energy of the steam to mechanical
energy of the rotor

 Supply of fuel.
 Availability of water.
 Transportation facilities.
 Cost and type of land.
 Nearness to load centres.
 Distance from populated area.

 (i) Thermal efficiency.
ηthermal =
Heat equivalent of mech.
energy transmitted to turbine shaft
Heat of coal combustion
The thermal efficiency of a modern steam power station is about 30%.
 (ii) Overall efficiency.
ηoverall =
Heat equivalent of electrical ouput
Heat of combustion of coal
The overall efficiency of a steam power station is about 29%
Overall efficiency = Thermal efficiency * Electrical efficiency

2. Superheater
4. Condenser
6. Air preheater
8. Forced Draft Fan
10. Coal Handling Plant
12. Cooling Tower
14. Pumps
1. Boilers
3. Turbine
5. Economizer
7. Induced Draft Fan
9. Chimney
11. Ash Handling Plant
13. Alternator

 Purpose: To produce steam under pressure
Types:
1. Fire Tube Boiler: Hot gases of combustion are inside the
tubes and the tubes are surrounded by water
2. Water Tube Boiler: Water is inside the tube and hot gases
are outside the tube

• It is like a chamber in which fuel is burnt to produce heat
energy
• Furnace walls are made of refractory materials such as
fire clay, silica etc.
Types:
o Plain refractory Walls: Used where furnace
temperature is not very high
o Hollow Refractory Walls: Through which air is circulated
and used for quite high temperatures
o Partially Water Cooled Walls: Similar to plain but a portion
of surface is covered by water tubes. It is used for high
temperature applications

• Device used to remove the traces of moisture from saturated
steam leaving boiler tubes.
• It also increase the temperature above saturation temperature
Classes:
1. Radiant Superheater: Located in the furnace between the
furnace water walls and absorbs heat from the burning fuel
through radiation
2. Convection Superheater: Located well back in boiler tube bank. It
receives its heat from flue gases through convection.

It absorbs heat from outgoing flue gases and used
for raising the temperature of feed water coming
from condenser
Economizer raises efficiency of boiler by 10 -12 %
thus 5 – 15 % of fuel consumption is saved

Air preheater is used to recover heat from flue
gases since entire heat can’t be extracted by
economizer
Boiler Efficiency is increased by 1% if the avg.
air temp. is increased by 20°C

(a) Recuperative type (b) Regenerative type

 A device which
condenses the steam at
the exhaust of turbine.

There are two types of steam turbines
1. Impulse type: Steam expands completely in
the stationary nozzles, the pressure over
the moving blades remains constant.
2. Reaction Type: Steam is expanded both in
fixed blades (nozzles) and moving blades.

 Alternators
 Transformers.
 Switchgear

 A steam power station has an overall efficiency of
20% and 0·6 kg of coal is burnt per kWh of
electrical energy generated. Calculate the calorific
value of fuel.?

 The amount of heat produced by the complete combustion of a unit
weight of fuel is known as its calorific value.
 (i) Electrical and Mechanical
1 kWh = 1 kW × 1 hr
= 1000 watts × 3600 seconds = 36 × 105
watt-sec. or Joules
∴ 1 kWh = 36 × 105
Joules
 (ii) Heat and Mechanical
1 calorie = 4·18 Joules
 (iii) Electrical and Heat
1 kWh = 1000 watts × 3600 seconds = 36 × 105
Joules
=
36 × 105
4.18
calories = 860 × 103
calories
∴ 1 kWh = 860 × 103
calories or 860 kcal

 Let x kcal/kg be the calorific value of fuel.
 Heat produced by 0·6 kg of coal = 0·6 x kcal
 Heat equivalent of 1 kWh = 860 kcal
 Now, ηoverall =
Electrical output in heat units
Heat of combustion
0·2 =
860
0 ⋅6x
∴ x =
860
0 ⋅6 ∗ 0⋅ 2
= 7166·67 kcal/kg

 A thermal station has the following data :
Max. demand = 20,000 kW ;
Load factor = 40%
Boiler efficiency = 85% ;
Turbine efficiency = 90%
Coal consumption = 0·9 kg/kWh ;
Cost of 1 ton of coal = Rs. 300
Determine
(i) thermal efficiency and (ii) coal bill per annum.

(i) Thermal efficiency = ηboiler * ηturbine = 0·85 *·9 = 0·765 or 76·5 %
(ii) Units generated/annum = Max. demand * L.F. * Hours in a year
= 20,000 *0·4 * 8760 = 7008 * 104 kWh
 Coal consumption/annum =
0.9∗7008 ∗ 104
1000
= 63,072 tons
 ∴ Annual coal bill = Rs 300 * 63072 = Rs 1,89,21,600

 A steam power station spends Rs. 30 lakhs per
annum for coal used in the station. The coal has a
calorific value of 5000 kcal/kg and costs Rs. 300 per
ton. If the station has thermal efficiency of 33% and
electrical efficiency of 90%, find the average load on
the station.

 A generating station which utilises the potential
energy of water at a high level for the generation of
electrical energy is known as a hydro-electric
power station.

1 •No fuel
2 •Neat and Clean
3 •Less maintenance
4 •Longer life
5 •Irrigation and Controlling floods.
6 •Few experienced persons

1
•High capital cost
2
•Dependence on weather conditions
3
•High cost of transmission lines

 (i) Availability of water.
 (ii) Storage of water
 (iii) Cost and type of land.
 (iv) Transportation facilities.
Ideal choice of site for such a plant is near a river in
hilly areas where dam can be conveniently built and
large reservoirs can be obtained.

1. Hydraulic
structures.
Dam Spillways Headworks Surge tank Penstock
2.Water
turbines
Impulse
turbines
Reaction
turbines
Francis
Turbines
Kaplan
Turbine
3.Electrical
equipment

 (i) Impulse turbines

 Reaction turbines Francis turbine

 A hydro-electric generating station is supplied
from a reservoir of capacity 5 × 106 cubic
metres at a head of 200 metres. Find the total
energy available in kWh if the overall efficiency
is 75%.

 Weight of water available is
W = Volume of water × density
= (5 × 106
) × (1000)
(mass of 1 𝑚3
of water is 1000 kg)
= 5 × 109
kg = 5 × 109
× 9·81 N
 Electrical energy available = W × H × ηoverall
= (5 × 109 × 9·81) × (200) × (0·75) watt sec
=2.044 × 106 kWh ∴ 1 kWh = 36 × 105 Joules

 It has been estimated that a minimum run off
of approximately 94 𝑚3/sec will be available at
a hydraulic project with a head of 39 m.
Determine (i) firm capacity (ii) yearly gross
output. Assume the efficiency of the plant to be
80%.

 Weight of water available, W = 94 × 1000 = 94000 kg/sec
 Water head, H = 39 m
 Work done/sec = W × H = 94000 × 9·81 × 39 watts
= 35, 963 ×103 W = 35, 963 kW
This is gross plant capacity.
 (i) Firm capacity = Plant efficiency × Gross plant capacity
= 0·80 × 35,963 = 28,770 kW
 (ii) Yearly gross output = Firm capacity × Hours in a year
= 28,770 × 8760 = 252 × 106
kWh

A generating
station in which
nuclear energy is
converted into
electrical energy is
known as a nuclear
power station.

Working principle :
❑ A nuclear power plant works in a similar way as a thermal power plant. The difference
between the two is in the fuel they use to heat the water in the boiler(steam generator).
❑ Inside a nuclear power station, energy is released by nuclear fission in the core of
the reactor.
❑ 1 kg of Uranium U235 can produce as much energy as the burning of 4500 tonnes of
high-grade coal or 2000 tonnes of oil.

❖ Chain Reaction…
❑ Uranium exists as an isotope in the form of U235 which is
unstable.
❑ When the nucleus of an atom of Uranium is split, the neutrons
released hit other atoms and split them in turn. More energy is
released each time another atom splits. This is called a chain
reaction.

proton
neutron
U-235 nucleus
❖ Neutrons released in fission

Nuclear (Atomic) Power Plant…
✓ Nuclear fission:

❑ Nuclear fission: heavy nuclei split into two
smaller parts in order to become more stable
proton
neutron
Kr-92 nucleus
U-235 nucleus
Ba-141 nucleus
energy

Schematic diagram of a nuclear power plant
with PWR
control rods
fuel
rods
reactor
pressure
vessel
water
(cool)
(high
(low
water
pressure)
coolant out
coolant in
steam condenser
steam (low
pressure)
turbine
electric
power
steam
generator
steam (high pressure)
pump
water
pressure)
primary loop secondary loop
reactor
core
water
(hot)
generator
pump
7 September 2020 D.Rajesh Asst.Prof. EEE 90

 (i) Nuclear reactor
 (ii) Heat exchanger
 (iii) Steam turbine
 (iv) Alternator

Core : Here the nuclear fission
process takes place.
Moderator : This reduces the
speed of fast
moving neutrons.
Most moderators
are graphite, water
or heavy water.

Principal parts of a nuclear reactor…
Control rods :
Coolant : They carry the intense heat generated. Water is used as a
coolant, some reactors use liquid sodium as a coolant.
Fuel : The fuel used for nuclear fission is U235 isotope.
Control rods limit the number of fuel
atoms that can split. They are made of
cadmium which absorbs neutrons

1
•Small fuel - saving in the cost of fuel transportation.
2
•Less space
3
•Economical for producing bulk electric power.
4
•large deposits of nuclear fuels - supply of electrical energy for
thousands of years.
5
•Reliability of operation.

1 •Expensive and is difficult to recover.
2 •capital cost on a nuclear plant is very high
3 •High cost of transmission lines
4 •dangerous amount of radioactive pollution.
5 •Not well suited for varying loads
6 •The disposal of the by-products

 (i) Availability of water.
 (ii) Disposal of waste.
 (iii) Distance from populated areas.
 (iv) Transportation facilities.
Ideal choice for a nuclear power station would be near
sea or river and away from thickly populated areas.

 What is the power output of a U235 reactor if it
takes 30 days to use up 2 kg of fuel? Given that
energy released per fission is 200 MeV and
Avogadro’s number = 6·023 ×1026
per
kilomole.

 Number of atoms in 2 kg fuel =
2
235
* 6·023 ×1026 = 5·12 × 1024
These atoms fission in 30 days. Therefore, the fission rate
(i.e., number of fissions per second)
=
5·12 × 1024
30∗24∗60∗60
=1.975* 1018
 Energy released per fission = 200 MeV
= (200 × 106 ) × 1·6 × 10−19
= 3·2 × 10−11
J
 ∴ Energy released per second i.e., power output P is
P = (3·2 × 10−11
) × (1·975 × 1018
) W
= 63·2 × 106 W = 63·2 MW

A generating station
which employs gas
turbine as the prime
mover for the
generation of
known as a gas
turbine power plant

Pressure
Energy
Heat
Energy
Mechanical
Energy
Electrical
Energy
Air
Compressor
Combustion chamber
Heat
Gas turbine Alternator

 (i) Compressor
 (ii) Regenerator
 (iii) Combustion chamber
 (iv) Gas turbine
 (v) Alternator
 (vi) Starting motor

 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 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.

 The curve showing the variation of load on the
power station with respect to (w.r.t) time is
known as a load curve.
1. Daily load curve
2. Monthly load curve
3. Yearly load curve

 (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) Average load =
Area (in kWh) under daily load curve
24 hours
 (v)Load factor =
Average load
Max. demand
=
Average load× 24
Max. demand× 24
=
Area (in kWh) under daily load curve
Total area of rectangle in which the load curve is contained

 (i) Connected load.
 (ii) Maximum demand.
 (iii) Demand factor.
 (iv) Average load.
 (v) Load factor.
 (vi) Diversity factor.
 (vii) Plant capacity factor.
 (viii) Plant use factor.

 (i) Connected load.
It is the sum of continuous ratings of all the
equipments connected to supply system.
❑ (ii) Maximum demand.
It is the greatest demand of load on the power station
during a given period.
 (iii) Demand factor.
=
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒅𝒆𝒎𝒂𝒏𝒅
𝑪𝒐𝒏𝒏𝒆𝒄𝒕𝒆𝒅 𝒍𝒐𝒂𝒅
The value of demand factor is usually less than 1.

❑ (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.
Daily average load =
No. of units (kWh) generated in a day
24 hours
Monthly average load=
No. of units (kWh) generated in a 𝒎𝒐𝒏𝒕𝒉
Number of hours in a month
Yearly average load =
No. of units (kWh) generated in a year
8760 hours

 (v) Load factor.
=
Average load
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
✓ Load factor is always less than 1
✓ 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.
=
Sum of individual max. demands
Max. demand on power station
➢ Always be greater than 1.
❑ (vii) Plant capacity factor.
=
Actual energy produced
Max. energy that could have been produced
=
Average load×T
Plant capacity ×T
=
Average load
Plant capacity
❑ Reserve capacity = Plant capacity − Max. demand

 (viii) Plant use factor.
=
Station output in kWh
Plant capacity × Hours of use

 Find the kWh generated per annum from maximum
demand and load factor.
Load factor =
Average load
Max. demand
Units generated/annum = Average load (in kW) × Hours in a
year
= Max. demand (in kW) × L.F. × 8760

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.

 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.
Demand factor =
=
20
43
=0.465
Load factor=
Average load
Max. demand
average load =
Hours in a year
=
61·5 × 106
𝟖𝟕𝟔𝟎
=7020KW
Load factor=
7020𝑘
20 𝑀
=0.351 or 35.1%

 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.
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
𝑀𝑎𝑥.𝑑𝑒𝑚𝑎𝑛𝑑 𝑖𝑛 𝑀𝑊 × 𝑊𝑜𝑟𝑘𝑖𝑛𝑔 ℎ𝑜𝑢𝑟𝑠
× 100
=
160,000
100 × 320 × 24
× 100 = 20·8%

 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.
Reserve capacity = Plant capacity − Max. demand
Plant capacity factor=
Average load
Plant capacity
Load factor=
Average load
Max. demand
Average load= 25 × 0·60 = 15 MW
Plant capacity=
15
0.5
=30MW
Reserve capacity=30-25=5MW

 (ii) Daily energy produced = Average load × 24
= 15 × 24 = 360 MWh
 (iii) Maximum energy that could be
produced=
Actual energy produced in a day
Plant use factor
=
360
0 ⋅72
= 500 MWh/day

 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.
Reserve capacity = Plant capacity − Max. demand
Plant capacity factor=
Average load
Plant capacity
average load =
Hours in a year
Plant capacity=
𝑃𝑙𝑎𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟𝑋"𝐻𝑜𝑢𝑟𝑠 𝑖𝑛 𝑎 𝑦𝑒𝑎𝑟 "
No. of units (kWh) generated in a year=Max. demand (in kW) × L.F. × 8760
=(15000) × (0·5) × (8760) kWh=65·7 × 106
kWh
Plant capacity=
65·7 × 106 kWh
0 ⋅ 4 × 8760
= 18,750 kW
Reserve capacity= 18,750 − 15000 = 3750 kW

 A power supply is having the following loads :
If the overall system diversity factor is 1·35, determine (i)
the maximum demand and (ii) connected load of each type.
diversity factor=
Max. demand on supply system = 13,500/1·35 = 10,000 kW
Type of load Max. demand (kW) Diversity of group Demand factor
Domestic 1500 1·2 0·8
Commercial 2000 1·1 0·9
Industrial 10,000 1·25 1

Sum of max. demands of different domestic consumers
= Max. domestic demand × diversity factor
= 1500 × 1·2 = 1800 kW
Demand factor =
Connected domestic load = 1800/0·8 = 2250 kW
Connected commercial load = 2000 × 1·1/0·9 = 2444 kW
Connected industrial load = 10,000 × 1·25/1= 12,500 kW

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) 40 50 60 50 70 40
Draw the load curve and find (i) maximum demand (ii) units generated per
day (iii) average load and (iv) load factor.
Daily curve is drawn by taking the load
along Y -axis and time along X-axis
∴ Maximum demand = 70 MW
units generated per day
= Area (in kWh) under the load curve
=[40 × 6 + 50 × 4 + 60 × 2 + 50 × 4 + 70 × 4 + 40 × 4]MWh
= [240 + 200 + 120 + 200 + 280 + 160] MWh
= 12 × 105kWh

 (iii) Average load =
Units generated / day
24 hours
=
12 × 105kWh
24
= 50,000 kW
 (iv) Load factor=
Average load
Max. demand
=
50,000 kW
70𝑀𝑊
= 0·714 = 71·4%

 The daily demands of three consumers are given below :
Plot the load curve and find (i) maximum demand of individual
consumer (ii) load factor of individual consumer (iii) diversity factor
and (iv) load factor of the station.
 (i) Max. demand of consumer 1 = 800 W
Max. demand of consumer 2 = 1000 W
Max. demand of consumer 3 = 1200 W
Time Consumer1 Consumer 2 Consumer 3
12 midnight to 8 A.M. No load 200 W No load
8 A.M. to 2 P.M. 600 W No load 200 W
2 P.M. to 4 P.M. 200 W 1000 W 1200 W
4 P.M. to 10 P.M. 800 W No load No load
10 P.M. to midnight No load 200 W 200 W

Time Consumer1 Consumer 2 Consumer 3
12 midnight to 8 A.M. No load 200 W No load
8 A.M. to 2 P.M. 600 W No load 200 W
2 P.M. to 4 P.M. 200 W 1000 W 1200 W
4 P.M. to 10 P.M. 800 W No load No load
10 P.M. to midnight No load 200 W 200 W
L.F. of consumer 1 =
Energy consumed / day
Max. demand × Hours in a day
× 100
=
600×6 + 200×2 + 800×6
800×24
× 100 = 45·8%
200×8 + 1000×2 + 200×2
1000×24
× 100
= 16·7%
200×6 + 1200×2 + 200×2
1000×24
× 100
= 13·8%

 Diversity factor
=
=
800+1000+1200
2400
=1.25
Station load factor =
Total energy consumed / day
Simultaneous max.demand × 24
× 100
=
8800+40000+4000
2400×24
× 100 = 29·1%

 A power station has the following daily load cycle :
Time in Hours 6—8 8—12 12—16 16—20 20—24 24—6
Load in MW 20 40 60 20 50 20
Plot the load curve and load duratoin curve. Also calculate the energy
generated per day.
Units generated/day = Area (in kWh)
under daily load curve
= [20 × 8 + 40 × 4 + 60 × 4 + 20 × 4 + 50 × 4]
= 840 MWh
Units generated/day = Area (in kWh)
under daily load duration curve
=[60 × 4 + 50 × 4 + 40 × 4 + 20 × 12]
= 840 MWh

The rate at which
supplied to a consumer
is known as tariff.

 (i) Recovery of cost of producing electrical
energy at the power station.
 (ii) Recovery of cost on the capital investment
in transmission and distribution systems.
 (iii) Recovery of cost of operation and
maintenance of supply of electrical energy e.g.,
metering equipment, billing etc.
 (iv) A suitable profit on the capital investment.

 (i) Proper return : The tariff should be such that it ensures the
proper return from each consumer.
 (ii)Fairness : The tariff must be fair so that different types of
consumers are satisfied with the rate of charge of electrical
energy.
 (iii) Simplicity : The tariff should be simple so that an ordinary
consumer can easily understand it.
 (iv) Reasonable profit : The profit element in the tariff should be
reasonable.
 (v) Attractive : The tariff should be attractive so that a large
number of consumers are encouraged to use electrical energy.

1. Simple tariff.
2. Flat rate tariff.
3. Block rate tariff.
4. Two-part tariff.
5. Maximum demand tariff.
6. Power factor tariff.
7. Three-part tariff.

 When there is a fixed rate per unit of energy
consumed, it is called a simple tariff or uniform
rate tariff.
Ex: Rs 3/- unit 20 units Rs 60/-

Disadvantages
▪ Same charge for small and big consumers
▪ Charge is not depend on the load

 When different types of consumers are charged
at different uniform per unit rates, it is called a
flat rate tariff.
 Rate is depend upon the load i.e light load,
heavy load
 Domestic load 60p/unit
 Heavy load 55p/unit

 Disadvantages
 Separate meters are required for lighting load,
power load

When a given block of energy is charged at a
specified rate and the succeeding blocks of energy are
charged at progressively reduced rates, it is called a
block rate tariff.
40
Unit
1
40
Unit
2
40
Unit
3
40
Unit
4
60
P
55
P
50
P
45
P
BLOCK
RATE

When the rate of electrical energy is charged on the basis of
maximum demand of the consumer and the units consumed, it is
called a two-part tariff.
1. Fixed charge ---- Maximum Demand
2. Running charge ----units consumed
 Total charges = Rs (b × kW + c × kWh)
where, b = charge per kW of maximum demand
c = charge per kWh of energy consumed

 Advantages
(i) It is easily understood by the consumers.
(ii) It recovers the fixed charges which depend upon the
maximum demand of the consumer but are independent
of the units consumed.
 Disadvantages
(i) The consumer has to pay the fixed charges irrespective
of the fact whether he has consumed or not consumed the
electrical energy.

 By installing maximum demand meter.
This type of tariff is mostly applied to big
consumers. However, it is not suitable for a small
consumer (e.g., residential consumer) as a
separate maximum demand meter is required.

 The tariff in which power factor of the consumer’s
load is taken into consideration is known as power
factor tariff.
 (i) k VA maximum demand tariff :
maximum demand in kVA and not in kW.
 (ii) Sliding scale tariff (average power factor tariff.)
Additional
charges
Discount is
allowed
0·8 lagging

 (iii) kW and kVAR tariff : In this type, both
active power (kW) and reactive power (kVAR)
supplied are charged separately.

 When the total charge to be made from the consumer is split
into three parts viz., fixed charge, semi-fixed charge and
running charge, it is known as a three-part tariff.
Total charge = Rs (a + b × kW + c × kWh)
where a = fixed charge made during
each billing period.
b = charge per kW of maximum demand,
c = charge per kWh of energy consumed.

A consumer has a maximum demand of 200 kW at 40% load
factor. If the tariff is Rs. 100 per kW of maximum demand plus
10 paise per kWh, find the overall cost per kWh.
Total charges = Rs (b × kW + c × kWh)
where, b = charge per kW of maximum demand
c = charge per kWh of energy consumed
Units consumed/year = Max. demand × L.F. × Hours in a year
= (200) × (0·4) × 8760 = 7,00,800 kWh
Annual charges = Rs (100 × 200 + 0·1 × 7,00,800)
= Rs 90,080
Overall cost/kWh =
𝑅𝑠 90 080
7,00,800
=Re 0·1285 = 12·85 paise

 An electric supply company having a maximum load of 50 MW generates 18 ×107
units per annum
and the supply consumers have an aggregate demand of 75 MW. The annual expenses including
capital charges are :
For fuel = Rs 90 lakhs
Fixed charges concerning generation = Rs 28 lakhs
Fixed charges concerning transmission and distribution = Rs 32 lakhs
Assuming 90% of the fuel cost is essential to running charges and the loss in transmission and
distribution as 15% of kWh generated, deduce a two part tariff to find the actual cost of supply to the
consumers.
Actual cost=Annual fixed charges+Annual running charges
Annual fixed charges
For generation = Rs 28 × 105
For transmission and distribution = Rs 32 × 105
For fuel (10% only) = Rs 0·1 × 90 × 105
= Rs 9 × 105
Total annual fixed charge = Rs (28 + 32 + 9) × 105
= Rs 69 × 105
∴ Cost per kW of maximum demand = Rs
69 × 10^5
75×103 =Rs. 92

 Annual running charges.
Cost of fuel (90%) = Rs 0·9 × 90 × 105
= Rs 81 × 105
Units delivered to consumers = 85% of units generated
`= 0·85 × 18 × 107 = 15·3 × 107 kWh
This cost is to be spread over the units delivered to the
consumers.
∴ Cost/kWh = Rs
81 × 105
15·3 × 107= Rs 0.0537
=5.3 paise
∴ Tariff is Rs 92 per kW of maximum demand plus 5·3 paise
per kWh.

A generating station has a maximum demand of 75 MW and a
yearly load factor of 40%. Generating costs inclusive of station
capital costs are Rs. 60 per annum per kW demand plus 4 paise
per kWh transmitted. The annual capital charges for
transmission system are Rs 20,00,000 and for distribution system
Rs 15,00,000 ; the respective diversity factors being 1·2 and 1·25.
The efficiency of transmission system is 90% and that of the
distribution system inclusive of substation losses is 85%. Find the
yearly cost per kW demand and cost per kWh supplied : (i) at the
substation (ii) at the consumers premises.
Maximum demand = 75 MW = 75,000 kW
Annual load factor = 40% = 0·4

(i) Cost at substation.
The cost per kW of maximum demand is to be determined from the total
annual fixed charges.
The cost per kWh shall be determined from the running charges.
(a) Annual fixed charges
Generation cost = Rs 60 × 75 × 103 = Rs 4·5 × 106
Transmission cost = Rs 2 × 106
Total annual fixed charges at the substation = Rs (4·5 + 2) × 106 = Rs 6·5 × 106
sum of all maximum demands by the various substations
= Max. demand on generating station × Diversity factor
= (75 × 106
) × 1·2 = 90 × 103
kW
Annual cost per kW of maximum demand= Rs
6·5 × 106
90 × 103 = Rs. 72.22

(b) Running Charges.
It is given that cost of 1 kWh transmitted to
substation is 4 paise. As the transmission efficiency is
90%, therefore, for every kWh transmitted, 0·9 kWh
reaches the substation.
∴ Cost/kWh at substation = 4/0·9 = 4·45 paise
Hence at sub-station, the cost is Rs 72·22 per annum
per kW maximum demand plus 4·45 paise per kWh.

(ii) Cost at consumer’s premises. The total annual fixed charges at consumer’s
premises is the sum of annual fixed charges at substation and annual fixed charge for
distribution
∴ Total annual fixed charges at consumer’s premises
= Rs (6·5 + 1·5) × 106 = Rs 8 × 106
sum of maximum demands of all consumers
= Max. demand on Substation × Diversity factor
= (90 × 103) × 1·25 = 112·5 × 103 kW
∴ Annual cost per kW of maximum demand= Rs
8 × 10^6
112·5 × 10^3
= Rs. 71.11
As the distribution efficiency is 85%, therefore, for each kWh delivered from
substation, only 0·85 kWh reaches the consumer’s premises.
∴ Cost per kWh at consumer’s premises=
Cost per kWh at substation
0.85
=
4.45
0.85
= 5.23 paise
Hence at consumer’s premises, the cost is Rs. 71·11 per annum per kW maximum
demand plus 5·23 paise per kWh.

Electric Power Generation [Thermal, Hydro, Nuclear, Gas power plants]

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