engineering

1. POWER PLANT ECONOMICS
Contents:
• Type of power plant.
• Requirement of plant Design.
• Terms and Definition.
• Tariffs.
• Requirement of a Tariffs.
• Economics of power Generation.
• Power plant site Selection.
• Problem-3 nos.

2. POWER PLANT ECONOMICS
Power Plant:
Base upon the various factors the power plant are classified as
follows-
1. On the Basis of fuel used:
(a) Condensing power plant.
(b) Non –condensing power plant.
i) Diesel power plant.
ii) Nuclear power plant.
iii) Hydro-electric power plant.
iv) Gas-turbine power plant.
2. On the basis of Nature of load:
(a) Base load power plant.
(b)Peak load power plant.
.

3. POWER PLANT ECONOMICS
3. On the Basis of Location:
(a) Central power station
(b)Isolated power station
4. On the Basis of service Rendered-
(a) Stationary power plant
(b) Locomotive power plant
Requirement of plant Design:
The factor to be kept in view while designing a power station are as
fallows-
1. Economy of expenditure, I.e., minimum-
(i) Capital cost
(ii) Operating& maintenance cost
2. Safety of plant personnel
3. Reliability.
4. Efficiency
5. Ease of maintenance

4. POWER PLANT ECONOMICS
6. Good working condition
7. Minimum transmission lose.
Terms & Definition:
1. Connecting Load:
It is the sum of ratings in kilowatts(kw) of equipment installed in
the consumers premises.
For example – if a consumer has connection for 4 lamps of 60w
each and point of 500w and a radio consuming 60w, then the
total connected load of the consumer =4 X 60 + 500 +60 = 800 w
4. Demand factor: it is defined as the ratio of maximum demand to
connecting load.
Demand factor = Maximum demand / Connecting Load.
5. Load Factor : It is defined as the ratio of average load to the
maximum demand load factor & demand factor always less than
unity.

5. POWER PLANT ECONOMICS
Load factor = Average load / Maximum demand
Fig. Load curve
6. Plant capacity factor: It is defined as the ratio of actual energy
that could have been produced during the same period.

6. POWER PLANT ECONOMICS
Plant capacity factor = E / (c x t)
Where,
E = Energy produced (kwh) in a given period.
C = Capacity of the plant in kw.
t = total number of hours in the given period.
7. Plant use Factor: It is defined as the ratio of actual energy
product to the maximum possible energy that could have
been produced during the actual number of hours the
plant was in operation.
Plant use factor = E / (c x t1)

7. POWER PLANT ECONOMICS
Where,
E = Energy produced (kwh) in a given period
C = Capacity of the plant in kw.
t1 = Actual number of hours the plant has been operation.
Maximum Demand:
It is the maximum load which a consumer uses at any time. It can be
less than or equal to connected load. If all the devices fitted in
consumer’s house run to their fullest extent simultaneously then the
maximum demand will be equal to connected load.
Load curve:
It is a representation between load in kilowatt and time in hours. It
shows variation of load on the power station.

8. POWER PLANT ECONOMICS
Tariffs:
Energy rates are the different methods of charging the consumers for the
consumption of electricity. It is desirable to charge the consumer according to
his maximum demand (kw) and energy consumed (kwh). The tariff chosen
should recover the fixed cost, operating cost and profit etc.
Requirement of a tariff:
The tariff should satisfy the following requirements :-
1. It should be easier to understand.
2. It should provide law rates for high consumption.
3. It should encourage the consumers having high load factors.
4. It should take into account maximum demand charges and energy charges.
5. It should provide less charges for power connection than for lighting.
6. It should avoid the complication of separate wiring and metering
connections.

9. POWER PLANT ECONOMICS
Economic of power Generation:
Economy is the main principle design of power plant. Power
plant economics is important in controlling the total power
costs to the consumer. The cost of power generation can
be reduced by:
1. Selecting equipment of longer life and proper capacities.
2. Running the power station at high load factor.
3. Increasing the efficiency of the power plant
4. Carrying out proper maintenance of plant- equipment to avoid
plant break downs.
5. Keeping proper supervision, because a good supervision is
reflected in lesser break downs and extended plant life.
6. Using a plant of simple design that does not need highly
skilled personnel.

10. POWER PLANT ECONOMICS
Power plant site selection:
The power plant site should satisfy the following requirements:-
1. Cost of land as well as taxes on land should be law.
2. It should be nearer to load centre, so that cost of
transmitting energy is low.
3. It should be accessible by road rail or sea. So that
transportation of fuel etc is easier.
4. Sufficient quantity of cooling water should be available near
the site. 5.the site selected should be away from thickly
populated areas in order to avoid atmospheric pollution and
to reduce noise.
5. Sufficient space for future expansion of power station should
be available near the site.
6. Site sub-soil conditions should be such that foundation can
be made at reasonable depth

11. POWER PLANT ECONOMICS
Problem: 1
A power station has a Maximum demand of 80×108 kw and daily
load curve is defined as follows:-
Time
(Hours)
0 - 6 6 - 8 8 - 12 12 - 14 14 -18 18 - 22 22- 24
Load
(MW)
40 50 60 50 70 80 40
Determine e the load factor of power station.

12. POWER PLANT ECONOMICS
Solution:
Given,
Maximum demand = 80 × 103 kw
To find, Load factor = ?
Here,
Energy generated
= (40×6) + (50×2) + (60×40) + (50×2) + (70×4) + (80×4) + (40×2)
= 1360 Mw-h
= 1360 X 103 kw-h
Average load = Energy generated /24
= (1360 X 103) / 24
= 56,666 kw
Load factor = Average load / Maximum demand
= 56,666 / 80,000
= 0.71

13. POWER PLANT ECONOMICS
Problem: 2
A central power station has annual factor as follows-
Load factor = 60%
Capacity factor = 40%
Use factor = 45%
Maximum demand = 15,000 kw
Determine-
(a) annual energy production.
(b) reserve capacity over and above peak load.
(c) Hours per year not in service.
Solution:
Given,
load factor = 60% = 0.6
Capacity factor = 40% = 0.4
Use factor = 45% = 0.45
Maximum demand = 15,000 kw.

14. POWER PLANT ECONOMICS
To find,
(a) Annual energy production = ?
(b) Reserve capacity over and above peak load = ?
(c) Hours per year not in service = ?
Load factor = Average load / Maximum demand
=> Average load = load factor × Maximum demand
= 0.6 × 15,000
= 9000 kw
Annual energy produced, E = Average load × Total hours in one year (t)
= 900 × 8760
= 78.84 × 106 kwh
I.e. Annual energy production is 78.84×106 kw-h [Answer] .

15. POWER PLANT ECONOMICS
Capacity factor = E / (C x t)
=> C = E / (Capacity factor x t)
= 78.84 x 106 / (0.4 x 8760)
= 22500 kw
Reserve capacity = capacity of the plant(c) – Maximum demand
= 22,500 - 15000
= 7500kw
i.e. Reserve capacity over and above peak load is 7500 kw
(answer)

16. POWER PLANT ECONOMICS
Use factor = E / (C X t1)
=> T1 = E / (Use factor x C)
= 78.84 x 106 / (0.45 x 22,500)
= 7786 hrs
Where,
t1= Actual number of hours of the year for
which the plant remains in operation.
Hours per year not in service = t - t1
= 8760 -7786
= 974 hours
I.e. 974 hours per year not in service [answer].

17. POWER PLANT ECONOMICS
Problem: 3
For a power station the yearly load duration curve is a straight line
from 30,000 to 4000 kw. To meet the load three turbo-generators
are installed. The capacity of two generators is 15,000 kw each and
the 3rd is rated at 5,000 kw. Determine the following :-
a) Load factor.
b) Capacity factor.
c) Maximum demand.

18. POWER PLANT ECONOMICS
Here,
Energy generated per year = Area under the curve
= (4000 × 8760) + {(1/2) × 8760 × 2600 }
= 14.89 × 107 kw-h
Average load = Energy generated year/t
= 14.89 x 107/ 8760
= 17,000 kw
Maximum demand = 30,000kw
Load factor = Average load / Maximum demand
= 17,000 / 3,000
= 0.57

19. POWER PLANT ECONOMICS
Capacity of the plant, C = (15,000 × 2) + 5000
= 35,000 kw
Capacity factor = E / (c X t)
= 14.89 x 107/ 35,000 x 8760
= 0.48
i.e. the capacity factor is 0.48 or 48%(answer)
(C) maximum demand = 30,000 kw (From curve)
i.e. the maximum demand is 30,000 kw (answer)