Energy, economic and environmental issues of power plants

POWER PLANT ENGINEERING
S.BALAMURUGAN - M.E
ASSISTANT PROFESSOR
MECHANICAL ENGINEERING
AAA COLLEGE OF ENGINEERING & TECHNOLOGY
UNIT 5 – ENERGY, ECONOMIC &
ENVIRONMENTAL ISSUES OF POWER PLANTS

SITE SELECTION FOR POWER PLANTS
ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET

TERMS AND DEFINITON OF LOAD
CONNECTED LOAD – It is the sum of rating in kilowatts of each equipment's installed by the
consumer. If consumer connects four tube lights of 40 watts, 2 fan of 60 watts. Then total
connected load = (4 × 40) + (2 × 60) = 280 watts
MAXIMUM LOAD – It is the maximum possible load which a customer uses at any time.
DEMAND FACTOR =
𝑀𝐴𝑋𝐼𝑀𝑈𝑀 𝐿𝑂𝐴𝐷
𝐶𝑂𝑁𝑁𝐸𝐶𝑇𝐸𝐷 𝐿𝑂𝐴𝐷
, less than one, Connected load always high
Example – Maximum demand 85 MW, Connected load 100 MW, Demand factor = 85/100 = 0.85.
It is vital in determining the capacity of the plant equipment.

LOAD FACTOR =
𝐴𝑉𝐸𝑅𝐴𝐺𝐸 𝐿𝑂𝐴𝐷
𝑃𝐸𝐴𝐾 𝐿𝑂𝐴𝐷
=
𝐸𝑁𝐸𝑅𝐺𝑌 𝐶𝑂𝑁𝑆𝑈𝑀𝐸𝐷 𝐼𝑁 𝐴 𝐺𝐼𝑉𝐸𝑁 𝑃𝐸𝑅𝐼𝑂𝐷 𝑂𝐹 𝑇𝐼𝑀𝐸
𝑃𝐸𝐴𝐾 𝐿𝑂𝐴𝐷 × 𝐻𝑂𝑈𝑅𝑆 𝑂𝐹 𝑂𝑃𝐸𝑅𝐴𝑇𝐼𝑂𝑁 𝐼𝑁 𝑇𝐻𝐸 𝐺𝐼𝑉𝐸𝑁 𝑃𝐸𝑅𝐼𝑂𝐷
It is always less than one. It plays key role in determining the per unit generation cost.
Higher load factor of the power station, lesser will be the cost per unit generated.
DIVERSITY FACTOR =
𝑆𝑈𝑀 𝑂𝐹 𝐼𝑁𝐷𝐼𝑉𝐼𝐷𝑈𝐴𝐿 𝑀𝐴𝑋𝐼𝑀𝑈𝑀 𝐷𝐸𝑀𝐴𝑁𝐷𝑆
𝑀𝐴𝑋𝐼𝑀𝑈𝑀 𝐷𝐸𝑀𝐴𝑁𝐷 𝑂𝐹 𝑇𝐻𝐸 𝑊𝐻𝑂𝐿𝐸
Example – Four individual feeder circuits with connected loads 250 kVA, 200 kVA, 150 kVA,
400 kVA 7 its demand factors of 90%, 80%, 75% & 85%. Use diversity factor of 1.5.
Maximum demand of customer 1 = 250 kVA × 90% = 225 kVA
Maximum demand of customer 3 = 150 kVA × 75% = 112.5 kVA
Sum of individual maximum demands = 225 + 160 + 112.5 + 340 = 837.5 kVA
Diversity factor = 1.5, maximum demand on the feeder circuits = 837.5 / 1.5 = 558 kVA
This situation requires 600 kVA Transformer.
If the diversity factor 1 means, it requires 850 kVA Transformer.
Higher the diversity factor of their loads, the smaller will be the capacity of the power
plant required & leads to reduce the fixed charges due to capital investment.

PLANT CAPACITY FACTOR =
𝐸𝑛𝑒𝑟𝑔𝑦 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑝𝑒𝑟𝑖𝑜𝑑
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑡 × 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑕𝑜𝑢𝑟𝑠 𝑖𝑛 𝑔𝑖𝑣𝑒𝑛 𝑝𝑒𝑟𝑖𝑜𝑑
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 greater than the
maximum demand of the plant.
Reserve Capacity = Plant capacity – Maximum Demand
PLANT USE FACTOR =
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑡 ×𝐴𝑐𝑡𝑢𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑕𝑜𝑢𝑟𝑠 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑡 𝑖𝑛 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛
A plant having installed capacity of 20 MW produces annual output of 7.8 × 106 kWh &
remains in operation for 2190 hours in a year.
Then Plant use factor = 7.8 × 106 / (20 × 106 × 2190) = 0.178 = 17.8 %
UTILIZATION FACTOR =
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑚𝑎𝑛𝑑 𝑜𝑛 𝑡𝑕𝑒 𝑝𝑜𝑤𝑒𝑟 𝑝𝑙𝑎𝑛𝑡
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡𝑕𝑒 𝑝𝑙𝑎𝑛𝑡

LOAD CURVE & LOAD DURATION CURVE
Load curve – A graphical record showing the power demands for every instant during a
certain time interval.
Load Duration Curve – It represents re-arrangements of all the load elements of
chronological load curve in order of descending magnitude.
Significance of load curve
• It shows the variations of load on the power station during different hours of the day
• The area under the daily load curve gives the number of units generated in the day
• The highest point on the daily load curve represents the maximum demand on the
station on that day
• It helps in selecting the size & the number of generating units
• It helps in preparing the operation schedule of the station
• The area under the curve divided by the total number of hours gives the average load.
AVERAGE LOAD =
𝑇𝑂𝑇𝐴𝐿 𝐸𝑁𝐸𝑅𝐺𝑌
𝑇𝐼𝑀𝐸 𝑃𝐸𝑅𝐼𝑂𝐷
=
𝐸𝑁𝐸𝑅𝐺𝑌 𝐶𝑂𝑁𝑆𝑈𝑀𝐸𝐷 𝐼𝑁 24 𝐻𝑂𝑈𝑅𝑆
24

POWER FACTOR
The power factor plays a major role in the plant economics. The low power factor increases
the load current which increases the losses in the system. Thus, the regulation becomes
poor. For improving the power factor, the power factor correction equipment is installed at
the generating station. Thus, the cost of the generation increases.

POWER TARIFF TYPES• The actual tariffs that the customer pay depends on the consumption of the electricity.
• The consumer bill varies according to their requirements.
• The industrial consumers pay more tariffs because they use more power for long times
than the domestic consumers.
The electricity tariffs depends on the following factors
• Type of load
• Time at which load is required.
• The power factor of the load.
• The amount of energy used.
The total bill of the consumer has three parts, namely, fixed charge C, semi-fixed charge Ax
and running charge By.
Z = A X + B Y + C
where, Z – total charge for a period
x – maximum demand during the period (kW), A – cost per kW
y – Total energy consumed during the period (kW),
B – cost per kWh of energy consumed, C – fixed charge during each billing period
Power Factor is a measure of how effectively incoming power is used in your electrical
system and is defined as the ratio of Real (working) power to Apparent (total) power

FLAT DEMAND TARIFF
TOTAL AMOUNT Z = A X
• In this type of tariff, the bill of the power
consumption depends only on the maximum
demand of the load.
• The generation of the bill is independent of the
normal energy consumption.
• This type of tariff is used on the street light, sign
lighting, irrigation, etc., where the working hours
of the equipment are known.
• The metering system is not used for calculating
such type of tariffs.
STRAIGHT LINE METER RATE
TOTAL AMOUNT Z = B Y
• The generation of the bills depends on the
energy consumption of the load.
• The customer need not pay anymore even
though the plant incurred some expenditure to
provide readiness to serve the customers
• It doesn’t encourage the customers to use
electricity
POWER TARIFF TYPES

BLOCK METER RATE
TOTAL AMOUNT Z = B1 Y1 + B2 Y2 + B3 Y3
• In this type of tariff, a particular amount per unit is
charged for the particular block of each unit
• The reduced price per unit are charged for the
succeeding block of units.
• This type of tariff reduces the unit energy charges
with increasing consumptions.
• Demerit – Lacks a measure of the customer
demand.
TWO PART TARIFF
TOTAL AMOUNT Z = A X + B Y
• The generation of the bills depends on maximum
demand & the energy consumption of the load.
• It requires two meters to record maximum
demand & energy consumption of the consumer.
• Used for industrial customers.
POWER TARIFF TYPES

THREE PART TARIFF
TOTAL AMOUNT Z = A X + B Y + C
• In this type of tariff, the total charge to be made
from the consumer is split into three parts.
• By adding fixed charges to two part tariff, it
becomes three part tariff.
• This type of tariff is applied to big consumers.
POWER TARIFF TYPES

Time hours
A power station supplies the following loads are necessary to the customers
Find 1) Draw the load curve & Estimate the load factor of the plant
2) What is the load factor of a standby equipment of 30 MW capacity if it takes up all loads
above 70 MW 3) What is its Use factor ?
Solution
Energy Generated = Area under the curve
= 30 × 6 + 70 × 4 + 90 × 2 + 60 × 4 + 100 × 4 + 80 × 2 + 60 × 2
= 1560 MWh
Load factor =
𝑨𝑽𝑬𝑹𝑨𝑮𝑬 𝑳𝑶𝑨𝑫
𝑷𝑬𝑨𝑲 𝑳𝑶𝑨𝑫
, Average Load =
𝟏𝟓𝟔𝟎
𝟐𝟒
= 65 MW
Maximum Demand = 100 MW, Load factor =
𝟔𝟓
𝟏𝟎𝟎
= 0.65
Load Factor of a Standby Equipment
The standby equipment's supplies the power at three situations (more than 70 MW)
90 – 70 = 20 MW for 2 hours, 100 – 70 = 30 MW for 4 hours, 80-70 = 10 MW for 2 hours
Energy Generated = Area under the curve = 20 × 2 + 30 × 4 + 10 × 2 = 180 MWh
Total Operation hours for standby equipment = 2 + 4 + 2 = 8 hours
Average Load =
𝟏𝟖𝟎
𝟖
= 22.5 MW, Load factor =
𝟐𝟐.𝟓
𝟑𝟎
= 0.75
Use Factor =
=
𝟏𝟖𝟎
𝟑𝟎 × 𝟖
= 0.75

The maximum demand of a power station is 96000 kW & daily load curve is described as follows
(i) Determine the load factor of power station
(ii) What is the load factor of standby equipment rated at 30 MW that takes up all load in
excess of 72 MW? Also calculate its Use factor.
Solution
Energy Generated = Area under the curve = 48 × 6 + 60 × 2 + 72 × 4 + 60 × 2 + 84 × 4 + 96 × 4 + 48 × 2
= 1632 MWh
Load factor =
, Average Load =
𝟏𝟔𝟑𝟐
𝟐𝟒
= 68 MW
Maximum Demand = 96000 kW = 96 MW, Load factor =
𝟔𝟖
𝟗𝟔
= 0.71
Load Factor of a Standby Equipment
The standby equipment's supplies the power at two situations (more than 72 MW)
84 – 72 = 12 MW for 4 hours, 96 – 72 = 24 MW for 4 hours
Energy Generated = Area under the curve = 12 × 4 + 24 × 4 = 144 MWh
Total Operation hours for standby equipment = 4 + 4 = 8 hours
Average Load =
𝟏𝟒𝟒
𝟖
= 18 MW, Load factor =
𝟏𝟖
𝟐𝟒
= 0.75
Use Factor = 𝐸𝑛𝑒𝑟𝑔𝑦 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑝𝑒𝑟𝑖𝑜𝑑
=
𝟏𝟒𝟒
𝟑𝟎 × 𝟖
= 0.6
Time hours

A peak load on the thermal power plant is 75MW. The loads having maximum demands of
35MW, 20MW, 15MW & 18MW are connected to the power plant. The capacity of the plant is
90MW and annual load factor is 0.53. Calculate the average load on power plant, energy
supplied per year, demand factor and Diversity factor.
Given
Max. demand or Peak load = 75 MW, Capacity of the plant = 90 MW, Annual Load factor = 0.53
Solution
Load factor =
, 0.53 =
𝟕𝟓
, Average Load = 39.75 MW
Energy Supplied per Year = Average load × Number of hours in one year
= 39.75 × 365 × 24 = 348210 MWh
Demand Factor =
𝑀𝐴𝑋𝐼𝑀𝑈𝑀 𝐿𝑂𝐴𝐷
𝐶𝑂𝑁𝑁𝐸𝐶𝑇𝐸𝐷 𝐿𝑂𝐴𝐷
=
75
35+20+15+18
= 0.85
Diversity Factor =
𝑆𝑈𝑀 𝑂𝐹 𝐼𝑁𝐷𝐼𝑉𝐼𝐷𝑈𝐴𝐿 𝑀𝐴𝑋𝐼𝑀𝑈𝑀 𝐷𝐸𝑀𝐴𝑁𝐷𝑆
𝑀𝐴𝑋𝐼𝑀𝑈𝑀 𝐷𝐸𝑀𝐴𝑁𝐷 𝑂𝐹 𝑇𝐻𝐸 𝑊𝐻𝑂𝐿𝐸
=
35+20+15+18
75
= 1.17

The following data for a 2200 kW diesel power station is given. The peak load on the plant is
1600kW and its load factor is 45%. Capacity cost / kW installed = Rs.15000, Annual costs = 15%
of capital, Annual Maintenance cost = Fixed Rs. 1,00,000 and Variable cost Rs.2,00,000, annual
operating cost = Rs.6,00,000. , Cost of Fuel = Rs.0.8 per kg, Cost of lubricating oil = Rs. 40 per
kg, C.V of Fuel = 40,000 kJ/kg, Consumption of fuel = 0.5 kg/kwh, Consumption of Lubricating
oil = 1/400 kg/kwh. Determine (a) the annual energy produced (b)cost of generation per kwh
(c) Efficiency
Solution
Capital Cost = Rs. 15,000 / kW
Capital cost of 2200 kW plant = 15,000 × 2200 = Rs. 33 × 106
Annual cost = 15 % of Capital =
15
100
× 33 × 106 = Rs. 4.95 × 106
Load Factor =
, 0.45 =
𝟏𝟔𝟎𝟎
, Average Load = 720 kW
Average Load =
𝑇𝑂𝑇𝐴𝐿 𝐸𝑁𝐸𝑅𝐺𝑌
𝑇𝐼𝑀𝐸 𝑃𝐸𝑅𝐼𝑂𝐷
, Total energy produced = 720 × 365 × 24 = 6.31 × 106 kWh
Consumption of fuel = 0.5 kg/kwh
Fuel cost = Rs.0.8 per kg
For 6.31 × 106 kWh,
Mass of Fuel = 0.5 × 6.31 × 106 = 3.154 × 106 kg
Fuel cost = 0.8 × 3.154 × 106 = Rs. 2.523 × 106
Consumption of Lubricating oil = 1/400 kg/kwh
Cost of lubricating oil = Rs. 40 per kg
For 6.31 × 106 kWh,
Mass of Lubricating oil = (1/400) × 6.31 × 106
= 15775 kg
Lubri. oil cost = 0.40 × 15775 = Rs. 0.631 × 106

The following data for a 2200 kW diesel power station is given. The peak load on the plant is
1600kW and its load factor is 45%. Capacity cost / kW installed = Rs.15000, Annual costs = 15%
of capital, Annual Maintenance cost = Fixed Rs. 1,00,000 and Variable cost Rs.2,00,000, annual
operating cost = Rs.6,00,000. , Cost of Fuel = Rs.0.8 per kg, Cost of lubricating oil = Rs. 40 per
kg, C.V of Fuel = 40,000 kJ/kg, Consumption of fuel = 0.5 kg/kwh, Consumption of Lubricating
oil = 1/400 kg/kwh. Determine (a) the annual energy produced (b)cost of generation per kwh
(c) Efficiency.
Solution
Continued from previous slide
Fixed cost = Annual cost + Annual maintenan. fixed cost = 4.95 × 106 + 1,00,000 = Rs. 5.05 × 106
Variable cost = 6,00,000 + 2,00,000 + 2.523 × 106 + 0.631 × 106 = Rs. 3.954 × 106
Total Cost = Fixed cost + Variable cost = 5.05 × 106 + 3.954 × 106 = Rs. 9.004 × 106
Cost of Power =
𝑻𝒐𝒕𝒂𝒍 𝑪𝒐𝒔𝒕
𝑬𝒏𝒆𝒓𝒈𝒚 𝑷𝒓𝒐𝒅𝒖𝒄𝒆𝒅
=
9.004 × 106
6.31 × 106 = Rs. 1.43
Efficiency =
𝑶𝒖𝒕𝒑𝒖𝒕
𝑰𝒏𝒑𝒖𝒕
Output power = E = 6.31 × 106 kWh = 6.31 × 106 × 3600 = 2.2716 × 1010 kW-sec = kJ
Input Power = m’ × CV = 3.154 × 106 × 40,000 = 1.2626 × 1011 kJ
η =
2.2716 × 1010
1.2626 × 1011 = 0.18 = 18%
kW.sec = k(
𝑱
𝑺
).S = kJ

POLLUTION CONTROL TECHNOLOGY –
COAL BASED POWER PLANT
• Environment pollution by thermal power plant is more than other
power plant
• Air pollution is caused by thermal power plant is very high by burning
fuels like coal.
• Combustible elements of fuel are converted into Gaseous products
• Non-Combustible elements to Ash
• Normally the thermal power plant has high sulphur content coal & it
must be removed before burnt.
• Some sulphur oxides & the fly ash will be removed by coal washing.

EMISSION GASEOUS
EMISSION
SULHUR DIOXIDE
OXIDES OF NITROGEN
HYDROGEN SULPHIDE
CARBON MONOXIDE
PARTICULATE
EMISSION
DUST (DIA 1 MICRON)
PARTICLES (SMOKE, FUMES,
FLY-ASH, CINDERS)
(DIA 10 MICRON)
SOLID WASTE
EMISSION
ASH, CALCIUM &
MAGNESIUM SALTS
THERMAL
POLLUTION
DISCHARGE OF THERMAL
ENERGY INTO WATER

REMOVAL OF SULPHUR DIOXIDE (SO2)
WET SCRUBBER
– Polluted gas is sent into the scrubber tangentially.
– Water spray absorbs these gases & particulate matters
– It is collected on the surface of the scrubber are washed down by water
– This water again treated, filtered & reused
– Used to remove the particulate matters.
– Here the gases are cooled, they must be reheated before being sent to the stack
– The used water contain sulphuric & sulphurous acid which corrode the pipeline &
scrubber

POLLUTION CONTROL TECHNOLOGY – COAL BASED POWER PLANT
EMISSION OF NOx
• The combustion of fossil fuels in air is accomplished by the formation of nitric
oxide which is subsequently partly oxidized to Nitrogen Oxide.
• NOx present in stack gases from coal, oil & gas furnaces
METHODS TO REDUCE THE EMISSION OF NOx
• Reduction of residence period in combustion zone
• Reduction of temperature in combustion zone
PARTICULATE EMISSION & ITS CONTROL
• Dust – Dia – 1micron – which do not settle down
• Particles – Dia – 10 microns – which settle down to the ground
EMISSION OF PARTICLES
• Smoke – Stable suspension of particles – Visible – Dia < 10 microns
• Fumes – Very small particles which is obtained from chemical reaction. It is
composed of metals & metallic oxides
• Fly-Ash – Diameter less than or equal to 100 microns
• Cinders – Diameter of particles are more than or equal to 100 microns

CINDER CATCHERS
• The performance parameters for any particulate remover is called
the collector efficiency.
Collector efficiency =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑢𝑠𝑡 𝑟𝑒𝑚𝑜𝑣𝑒𝑑
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑢𝑠𝑡 𝑝𝑟𝑒𝑠𝑒𝑛𝑡
× 100
Collector efficiency varies from
50 – 99 %
Electrostatic precipitator > 90 %
50 – 75 %
Stroker & small cyclone
furnace(Crushed coal)
Centrifugal force
50 – 75 %

BAGHOUSE FILTERS
• Used to remove the
particulate matters where
low sulphur coal is used.
• Filter is cleaned by
applying a forced air in
reversed direction
• Large filter area required –
Cost high
• Life period – 3 years
• Coal firing systems

ELECTRO STATIC PRECIPITATOR
• The medium between the electrodes is air, and due to the high negativity of negative electrodes,
there may be a corona discharge surround the negative electrode rods or wire mesh. The air
molecules in the field between the electrodes become ionized, and hence there will be plenty of
free electrons and ions in the space
• The flue gases enter into the electrostatic precipitator, dust particles in the gases collide with the
free electrons available in the medium between the electrodes and the free electrons will be
attached to the dust particles.

• As a result, the dust particles become negatively charged. Then these negatively
charged particles will be attracted due to electrostatic force of the positive plates.
• Consequently, the charged dust particles move towards the positive plates and
deposited on positive plates. Here, the extra electron from the dust particles will be
removed on positive plates, and the particles then fall due to gravitational force.
• We call the positive plates as collecting plates. The flue gases after travelling through
the electrostatic precipitator become almost free from ash particles and ultimately get
discharged to the atmosphere through the chimney
ELECTRO STATIC PRECIPITATOR

• Large amount of ash is formed by burning the coal
in the furnace
• This ash is removed as bottom ash from furnace
• From discharged solid wastes, calcium &
magnesium can be generated by absorption of SO2
& SO3 by reactant like lime stone.
SOLID WASTE DISPOSAL

• Discharge of thermal energy into water is called thermal pollution
• Water is the main medium to condense the steam. If this heated water is
discharged into lakes or rivers, the water temperature goes up.
• At about 35°C, the dissolved oxygen will be low that the aquatic life will die.
• As per regulation - 1°C more then the atmospheric temperature acceptable
• In very cold regions, discharging the hot water into lakes helps in increasing
fish growth.
THERMAL DISCHARGE INDEX – Estimates the thermal efficiency of plant
TDI =
𝑇𝑕𝑒𝑟𝑚𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑑𝑖𝑠𝑐𝑕𝑎𝑟𝑔𝑒𝑑 𝑡𝑜 𝑒𝑛𝑣𝑖𝑟𝑜𝑛𝑚𝑒𝑛𝑡 𝑖𝑛 𝑀𝑊 𝑡𝑕𝑒𝑟𝑚𝑎𝑙
𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 𝑀𝑊 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙
THERMAL POLLUTION

METHODS TO REDUCE THERMAL POLLUTION
– Construction of a separate lake – once through cooling
the condenser is adopted. If natural cooling of water
from the lake is not sufficient means, floating spray
pumps can be employed. Expensive
– Cooling pond - A cooling pond with continuously
operating fountains can be adopted. Aesthetic look in
the power plant
– Cooling tower – In order to throw heat to atmosphere,
most power stations use cooling tower.
THERMAL POLLUTION

Mining – Depending on the depth and concentration of the uranium source,
and the conditions of the surrounding rock, mining companies will extract
uranium ore in one of three ways: open pit mining, underground mining
or in-situ recovery.
Milling – To extract the uranium, the ore is crushed in a mill and ground to a
fine slurry. The slurry is then leached in sulfuric acid, which produces a
solution of uranium oxide (U3O8). The concentrate of this solution is called
yellowcake.
Refining – A series of chemical processes separate the uranium from
impurities, producing high-purity uranium trioxide (UO3).
Conversion – UO3 is converted to uranium dioxide (UO2) for use in heavy
water reactors, UO3 to uranium hexafluoride (UF6) for enrichment, before it
can be used in light water reactors.
Enrichment (Gas Centrifugation)– Uranium-235 is the uranium isotope
that can be used in fission, but it makes up only 0.7% of naturally occurring
uranium, which is not concentrated enough for light water reactors. So,
enrichment processes increase the concentration of U-235 to about 3% –
5%. After undergoing enrichment, the UF6 is chemically transformed back
into UO2 powder.
NUCLEAR FUEL CYCLE

Fuel manufacturing – Natural or enriched UO2 powder is pressed into
small cylindrical pellets, which are then baked at high temperatures,
and finished to precise dimensions.
Electricity generation – Fuel is loaded into a reactor, and nuclear
fission generates electricity. After fuel is consumed, it is removed from
the reactor and stored onsite for a number of years while its
radioactivity and heat subside.
Optional chemical reprocessing – After a period of storage, residual
uranium or by-product plutonium, both of which are still useful sources
of energy, are recovered from the spent fuel elements and reprocessed.
Disposal – Depending on the design of the disposal facility, the nuclear
fuel may be recovered if needed again, or remain permanently stored.
At some point in the future the spent fuel will be encapsulated in sturdy,
leach-resistant containers and permanently placed deep underground
where it originated.
NUCLEAR FUEL CYCLE

TYPES OF NUCLEAR WASTE
• High Level Waste (3%) – Spent fuel containing 95% of radioactivity
in the nuclear waste
• Intermediate Waste (7%) – Used filters, steel components from
within the reactor & some effluents from reprocessing containing 4%
of radioactivity in the nuclear waste
• Low level waste (90%) – Lightly contaminated items like tools &
clothing containing only 1% of radioactivity in the nuclear waste.
• Both LLW and ILW from a nuclear power station will be isolated from
the environment for typically 300 years so that their radioactivity will
reduce with time to natural levels.
• HLW is highly radioactive. After being extracted from the spent fuel,
it typically needs a period of 20 to 50 years to cool down.
• Technology has been developed to pack the waste in glass in a
process known as vitrification.
• The packaged vitrified waste can then be stored underground in a
stable geological formation to isolate it and prevent its movement for
over thousands of years. Furthermore, the radioactivity will fall over
time and after several thousand years it will fall to natural levelsME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET

• Spent fuel is processed at facilities in Trombay near Mumbai, at
Tarapur on the west coast north of Mumbai, and at Kalpakkam on the
southeast coast of India.
• Plutonium will be used in a fast breeder reactor (under construction) to
produce more fuel, and other waste vitrified at Tarapur and Trombay.
• Interim storage for 30 years is expected, with eventual disposal in a deep
geological repository in crystalline rock near Kalpakkam.
NUCLEAR WASTE PROCESSING IN INDIA

NUCLEAR WASTE - PROCESSING
• Highly reactive materials created during production
of nuclear power.
– Core of the nuclear reactor or nuclear weapon
– Uranium, plutonium, high reactive elements from fission
– Long lives> 100000years
Management of high level waste in India following three stages
Immobilization- Vitrified borosilicate glasses
Engineered interim storage of vitrified waste for passive
cooling & surveillance over a period of time
Ultimate storage / disposal of the vitrified waste a deep
geological depository
deep geological depository requirements
– remoteness from environment
– absence of circulating ground water

Joule Heated Ceramic Melter (JHCM)

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