Electrical Power Generation Notes 4

7/25/2019 Electrical Power Generation Notes 4

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Vinayaka.B.G, Asst.Professor, Dept of E& E, BIET, Davanagere

ECONOMIC ASPECTS

Unit.5

ECONOMICS ASPECTS

Introduction. Terms commonly used in system operation. Diversity factor, load factor, plant

capacity factor, plant use factor, plant utilization factor and loss factor, load duration curve.

All the electrical energy generated in a power station must be consumed

immediately as it cannot be stored. So the electrical energy generated in a power

station must be regulated according to the demand. The demand of electrical

energy or load will also vary with the time and a power station must be capable of

meeting the maximum load at any time.

TERMS COMMONLY USED IN SYSTEM OPERATION

Firm power: Firm power is the power intended always to be available even

under emergency conditions.

Cold reserve: Cold reserve is the reserve generating capacity that is available

for service but not in operation.

Hot reserve: Hot reverse is the reserve generating capacity that is in

operation but not in service.

Spinning reserve: Spinning reserve is the reserve generating capacity that is

connected to the bus and ready to take load.


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ECONOMIC ASPECTS

1.

Connected load: The sum of continues ratings of all the equipments which are

connected to the supply system is called connected load.

It is the sum of the continuous ratings of the load consuming

apparatus connected to the system.

Ex: Theconnected loads in the premises of a consumer are shown in Figure below

The total connected load in the consumer's premises

= 60 +500 + 40 + 60 + 500 +100 + 60 + 60 = 1380 watts.

2. Maximum demand : Maximum demand is the greatest of all demands which

have occurred during a given period of time.

3. Demand factor: Demand factor is the ratio of maximum demand to the

connected load of a consumer.

Ex: The lighting installation has 10 bulbs each of rated 100 w and at no time of the day

more than 7 lights are switched on. Fine demand factor for this lighting installation.

Connected load = 10 x 100 =1000W

Maximum demand = 7 x 100 =700W

Demand factor = 700 /1000 = 0.70 =70%


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ECONOMIC ASPECTS

4.

Average demand or Average load: It is defined as the average of the loads

occurring on the power station in a given period. The period may be day or

month or year.

5.

Load factor:It is defined as the ratio of average load to the maximum demand

during a given period.It is always less than 1 because average demand is less than maximum

demand

If the plant operates for T hours, then

That is


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ECONOMIC ASPECTS

6. Units generated in year:

Units generated in year is given as follows

7. Units generated in day:

8. Diversity factor: The diversity factor is thus defined as the ratio of sum of

individual maximum demands to the maximum demand on power station.

Mathematically it is defined as

The diversity factor is always greater than unity because,

Sum of individual maximum demands > maximum demand on power station

Units generated / annum = Average demand X hours in year

Units generated / annum = Average demand X 8760

Units generated / annum = Max.demand X Load factor X 8760

Units generated / Day = Max.demand X Load factor X 24


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ECONOMIC ASPECTS

9. Plant Capacity factor: This is the ratio of actual energy produced to the

maximum possible energy that could have been produced during a given

period.

If the considered period as year, then


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ECONOMIC ASPECTS

10.

Utilization factor: It is defined as the ratio of maximum demand on power

station to the installed capacity (or plant capacity) of the plant.

For reducing the cost utilization factor must be very close to unity.

11.

Plant use factor: It is the ratio of energy (kWh or units) Produced to the

product of plant capacity the number of hours for which the plant was in

operation.

Example: A plant having installed capacity (or plant rated capacity) of 20MW

produces output of 7 X 106kWh and remains in operation for 2080 hours inn

year then

Plant use factor = = 0 .168 = 16.8%

12.

Loss factor: It is defined as average power loss to the peak load power loss

during a specified period of time.

7 X 106

20000 X 2080


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ECONOMIC ASPECTS

Note:

1.

Where

E ------> Energy produced

C --------> Plant capacity

t ---------> hours in that period that is it should be 24 or (30X24) or (8760)

2.

Where

E ------> Energy produced

C --------> Plant capacity

t ---------> hours of operation here is not 24 or (30X24) or (8760)


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ECONOMIC ASPECTS

A generating station has a connected load of 40MW and a

maximum demand of 20MW; the units generated being 60 x 106

calculate i) demand factor ii) load factor

Solution:

Connected load = 40 MW

Maximum demand = 20 MW

Units generated = 60 x 106

60 x 106

8760

6849.31 kW

849.31kW 6849310

20 MW 20000000

Demand factor =0.5 =50%

Load factor= 0.3424 = 34.24 %

0.3424


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ECONOMIC ASPECTS

The maximum demand on a thermal power station is 480 MW. If

the annual load factor is 40%. Calculate the total energy

generated annually.

Solution:

Given data: Maximum Demand = 480 MW = 480 x 103kW

Load factor = 40% = 0.4

Energy generated per year = Max.Demand x Load factor x 8760

= (480 x 103

) x 0.4 x 8760

= 1681920 x 103kWh

= 1681920 MWh

The maximum demand on a power station is 100 MW. If the

annual load factor is 40%. Calculate the total energy generated

per year.

Solution:

Given data: Maximum Demand = 100 MW = 100 x 103kW


Energy generated per year = Max. Demand x Load factor x 8760

= (100 x 103) x 0.4 x 8760

= 350400 x 103kWh


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ECONOMIC ASPECTS

A 100 MW power station deliver 100 MW for 2 hours, 50 MW for

8 hours and it is shut down for the rest of the day. It is also shut

down for maintenance for 60 days each year. Calculate its

annual load factor

Solution:

Energy supplied for each working day = (100 x 2) + (50 x 8)

= 600 MWh

Max.Demand = 100 MW

Station operation days = 36560 = 305 days

Annual energy supplied = 600 x 305 = 183000 MWh

Annual Average demand =

=

= 20.89 MW

=

= 0.2089

= 20.89%


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ECONOMIC ASPECTS

A generating station supplies the following loads: 15000 kW,

12000 kW, 8500 kW, 6000 kW and 450 kW. The station has a

max demand of 22,000 kW. Calculate:

i)Demand factor

ii)Diversity factor

iii)No of units supplied annually, if the load factor is 48%

Solution:


Connected load = 15000 + 12000 + 8500 + 6000 + 450 = 41,950 kW

Max. Demand = 22,000 kW

22000

41950

15000 + 12000 + 8500 + 6000 + 450

22000

41950

22000

1.906

No of units supplied annually = Max. Demand x load factor x 8760

= 22000 x 0.48 x 8760

= 92505600 kWh = 925.056 x 105kWh

JUNE 10, 12M

0.5244 52.44%


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ECONOMIC ASPECTS

A generating station supplies the following loads to various

consumers :

Industrial consumers = 750 MW

Commercial establishment = 350 MW

Domestic power =10 MW

Domestic light = 50 MW

If the maximum demand on station is 1000 MW and the number

of units generated per year is 50 X 105kWh

Determine i) diversity factor ii) annual load factor

5 X105

8760

570.78MW

= 0.57078 = 57.078%


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ECONOMIC ASPECTS

Power station is to supply four regions of load whose peak loads

are 10MW, 5MW, 8MW and 7MW. The delivery factor of the

load at the station is 1.5 and average annual (yearly) load factor

is 0.6. Calculate maximum demand on station and annual energy

supplied from station.

OR

Power station is to supply four regions of load whose peak loads

are 10000kW, 5000kW, 8000kW and 7000kW. The diversity

factor of the load at the station is 1.5 and average annual load

factor is 60%. Calculate maximum demand on station and

annual energy supplied from station.

Solution:

Given data: Diversity factor = 1.5


Max. demand = ?

Annual energy supplied = ?

Sum of individual max. demands = 10000 + 5000 + 8000 + 7000

= 30,000 kW

JAN 09, 05M


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ECONOMIC ASPECTS

= 20,000 KW

Annual energy supplied = max. Demand x LF x 8760

= 20000 x 0.6 x 8760

= 105120000 kWh

= 105.12 x106kWh

Maximum demand on station = 20000 kW

Annual energy supplied = 105.12 x106kWh

Maximum demand on

power station


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ECONOMIC ASPECTS

A base load station having a capacity of 400 MW and stand by

station having capacity of 50 MW share a common load. Find

the annual load factor and capacity factor of two power stations

from the following details

i) Annual stand by station output = 87.35 x 106kWhr

ii) Annual base load station output = 101.0 x 106kWhr

iii) Peak load on the stand by station = 120 MW

iv) Hours of use by stand by station/ year = 3000 hrs

Solution:

Stand by station:

Given data: Annual stand by station output = 87.35 x 106kWhr

Plant capacity = 50 MW = 50,000 kW

Peak load= max load= 120MW = 120,000 kW

Hours of use = 3000 hrs

87.35 x 106

3000

29116.66 kW

= 0.2426 = 24.26%

JUNE 11, 10 M


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ECONOMIC ASPECTS

0.1994 = 19.94%

Base load station:

Given data: Annual base load station output = 101.0 x 106kWhr

Plant capacity = 400 MW = 400,000 kW

Peak load= max load= 400MW = 400,000 kW

Hours of use = 8760 hrs

Because in base load station maximum demand equal to plant capacity

(400MW) and it operates throughout year i.e8760 hours

101.0 x 106

8760

11529.68 kW

= 0.0288 = 2.88%

87.35 x 106

50000 x 8760


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ECONOMIC ASPECTS

0.0288 = 2.88%

Note :

Use this formula to find load factor directly

101.0 x 106

400000 x 8760


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ECONOMIC ASPECTS

A generating station has a maximum demand of 400 MW. An

annual load factor is 60% and capacity factor is 45%. Find the

reserve capacity of the plant.

Solution:

Given data:. Max. Demand = 400 MW


Capacity factor = 45% = 0.45

Units generated per year = Max.demand x LF x 8760

= 400 x 0.6 x 8760

= 2.10 x 106MWh

2.10 x 106MWh

0.45 X 8760

533.33MW

Reserve capacity Plant Capacity Max.Demand

533.33400

133.33 MW


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ECONOMIC ASPECTS

Alternate method:

Max. Demand = 400 MW



Average demand = max. Demand X Load factor

= 400 X 0.6

= 240 MW


533.33400

133.33 MW


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ECONOMIC ASPECTS

A generating station has a maximum demand of 500 MW. An

annual load factor is 50% and capacity factor is 40%. Find the

reserve capacity of the plant

Solution:




Units generated per year = max.demand x LF x 8760

= 500 x 0.5 x 8760

= 2.19 x 106MWh

2.19 x 106MWh

0.4 X 8760

625 MW


625500

125 MW


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ECONOMIC ASPECTS

A power plant has the following annual factors.

Load factor = 70% Capacity factor = 50%

Use factor = 60% Maximum demand = 20MW

Find

i)Annual energy production.

ii)Reserve capacity over and above peak load.

iii)Hours during which the plant is not in service per year

Solution:

Given data: Max. Demand = 20 MW



Use factor = 60% = 0.6

Units generated per year = Max.demand x LF x 8760= 20 x 0.7 x 8760

= 122640 MWh

= 122.64 X 106kWh

122640MWh

0.5 X 8760

28 MW

DEC 11, 8 M


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ECONOMIC ASPECTS


2820

8 MW

Hours during which the plant is not in service per year = 87607300

= 1430 hours


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ECONOMIC ASPECTS

A generating station has a maximum demand of 20 MW, a load

factor of 60%, a plant capacity factor 48%, and plant use factor

0f 80% find

i)Daily energy produced

ii)Reserve capacity of the plant

iii)The maximum energy that could have been produced daily

if the plant were running all the time

iv)The maximum energy that could have been produced if the

plant (when running according to operating schedule) were

fully loaded.

Solution:



Plant Capacity factor = 48% = 0.48

Plant use factor = 80% = 0.8

Daily energy produced = Max.demand x LF x 24

= 20 x 0.6 x 24

= 288 MWh

Average demand = max. Demand X Load factor

= 20 X 0.6

= 12 MW

JUNE 08, 12, 10 M


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ECONOMIC ASPECTS


2520

5 MW

The maximum energy that could have been produced daily if the plant

were running all the time is given by

= installed capacity X hours in day

= 25 X 24

= 600 MWh

The maximum energy that could have been produced if the plant whenrunning according to operating schedule is given by


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ECONOMIC ASPECTS

A power supply is having the following loads

Type of load Maximum demand

(kW)

Diversity factor of

group

Demand

factor

1) Residential 15,000 1.25 0.7

2) Commercial 25,000 1.2 0.9

3)Industrial 50,000 1.3 0.98

If the overall system diversity factor is 1.5, determine

i)

Maximum demand

ii) Connected load of each type

Solution:

Given data: Diversity factor= 1.5

i)


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ECONOMIC ASPECTS

ii)

Note :

1.To find the connected load on power station if demand factor and diversity

factors both are given then

2.To find the maximum demand(load) on power station if demand factor and

diversity factors both are given then


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ECONOMIC ASPECTS

At the end of the power distribution system a certain feeder

supplies three distribution transformers. Each one supplying a

group of customers whose connected load as shown below

If diversity factor among the transformer is 1.3 find max demand

on the feeder

Solution:

Transformer Load Demand factor Diversity factor

Tfr 1 10 0.65 1.5

Tfr 2 12 0.6 3.5

Tfr 3 15 0.7 1.5


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ECONOMIC ASPECTS

Diversity factor among the three transformer is 1.3

A feeder supplies three distribution transformer which feed the

following connected loads:

Transformer 1:

Motor loads = 300 kW; Demand factor = 0.6

Commercial loads = 10 kW; Demand factor = 0.5Transformer 2:

Residential loads = 50 kW; Demand factor = 0.4

Transformer 3:

Residential loads = 50 kW; Demand factor = 0.5

The diversity factors for the loads on three transformers

may be taken as 1.8, 2.5 and 3. The diversity factors between the

transformers may be taken as 1.1 find:

i)Peak load on each transformer

ii)Peak load on feederDEC 11, 8 M


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ECONOMIC ASPECTS

Solution:

i)

ii)

Given : Diversity factor = 1.1

Sum of individual max demands of transformer = 102.77 + 8+ 8.33

=119.1 kW


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ECONOMIC ASPECTS

Find the i) Maximum demand ii) Daily load consumption

iii) Load factor of power supply system having following loads.

Type of load Maximum demand

(kW)

Load factor Diversity factor of group

1) Residential 1000 20 1.2

2) Commercial 2000 25 1.1

3)Industrial 10,000 80 1.25

What are the connected loads under each category if the

Demand factors for residential, commercial and industrial loads

are 80, 90 and 100% respectively?

Solution:

JUNE 09, 12 M


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ECONOMIC ASPECTS


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ECONOMIC ASPECTS


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ECONOMIC ASPECTS

1.

LOAD CURVE

A load curve (or load graph) is a graphical representation of variation of load

with respect to time in chronological order.

It is a graphical record showing the power demands for every instant during

a certain time interval.

If the load curve plotted for 24 hours in a day it is called daily load curve.

If the load curve plotted for hours in a month it is called monthly load curve

If the load curve plotted for hours in a year (8760 hrs) it is called yearly load

curve.


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ECONOMIC ASPECTS

Importance of load curve:

It shows variation of load on the power station.

Area under the load curve represents the number of units (energy)

generated in the period considered.

The area under the curve divided by the total number of hours gives the

average load on the power station.

The peak (highest point) of the load curve indicates the maximum demand

of the power station.

The ratio of area under load curve to the total area of rectangle in which it is

contained gives the load factor.

Load curves give full information about the incoming and help to decide the

installed capacity of the power station and to decide the economical sizes of

various generating units.

2.

LOAD DURATION CURVE

Load duration curve is a rearrangement of all the load elements of the load

curve in descending order with greatest load on left hand side and lesser load

on right hand side.

Load duration curve is obtained from the same data as load curve but

ordinates are arranged in descending order.


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ECONOMIC ASPECTS

Importance of load curve:

The area under the load duration curve is equal to area under the load curve

which gives the total number of units generated in a given period.

Load duration curve gives a clear analysis of generating power economically.

Proper selection of base load power plants and peak load power plants

becomes easier.

3.

ENERGY CURVE( INTEGRATED LOAD CURVE)

It is a graphical representation between load in kW with respect to energy

kWh (or units) is known as energy load curve.

kW is taken in y- axis and kWh in x-axis.

It gives the total number of units generated for a given max. demand or upto

a given demand.

If the energy (kWh) and demand are plotted as percentage quantity the load

curve is called peak percentage load curve.

Importance of energy load curve:

It helps for estimating base load/peak load on a station

This curve helps for variation between rate of water in flow in hydro power

station & that of electrical load.


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ECONOMIC ASPECTS

A residential consumer has 12 lamps of 60 watts connected at

his premises. His demand as follows

From 12 midnight to 5 A.M - 60 watts

5 A.M to 6 P.M - no load

6 P.M to 7 P.M - 240 watts

7 P.M to 9 P.M - 360 watts

9 A.M to Midnight - 180 watts

Plot the load curve and hence determine the maximum load,

load factor, average load and electrical energy consumption

during the day

Solution:

From the above graph

i) Maximum demand = 360 watts

DEC 11, 7 M


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ECONOMIC ASPECTS

ii) No of units generated = (60 X 5) + (0 X13) + (240 X 1) + ( 360 X2) + (180 X 3)

= 1800 watt-hr

iii) Average demand

iv) Load factor


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ECONOMIC ASPECTS

The daily demands of three consumers are given below:

Time Consumer 1 Consumer 2 Consumer 3

12 mid night to 8 A.M No load 200 W No load

8 A.M to 2P.M 600 W No load 200 W

2 P.M to 4P.M 200 W 1000 W 1200 W

4 P.M to 10P.M 800 W No load No load

10 P.M to Mid night No load 200W 200 W

Plot the load curve and find:

i) Maximum demand of individual consumer

ii)Load factor of individual consumer

iii)Diversity factor

iv)Load factor of the station.

Solution:

DEC 10, 8M


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ECONOMIC ASPECTS


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ECONOMIC ASPECTS

A power station has to meet the following load demand:

LOAD A 50kW Between 10 AM and 6 PM

LOAD B 30kW Between 6 PM and 10 PM

LOAD C 20kW Between 4 PM and 10 AM

Plot the daily load curve and determine:

i) Diversity factor

ii) Units generated per day

iii) Load factor

Solution:

JUNE 10, 8M


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ECONOMIC ASPECTS

Sum of individual max.Demand = 50 + 30 + 20

Maximum load on power station = 70

Units generated per day = (20 x 10) +(50 x 6) +(70 x2) +(50 x 4) +(20 x 2)

= 880 kWh

Diversity factor = 1.43

Units generated per day = 880 kWh

Load factor = 52.38%


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ECONOMIC ASPECTS

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 & load duration curve, also calculate energy

generated per day.

Solution: Load curve

Energy generated per day = (20 X 8) + (40 X 4) + (60 X 4) + (20 X 4) +(50 X 4)

= 840 MWh

= 840 X 103kWh


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ECONOMIC ASPECTS

Load duration curve

Energy generated per day = (60 X 4) + (50 X 4) + (40 X 4) + (20 X 12)

= 840 MWh

= 840 X 103kWh

This is same for both the curves


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ECONOMIC ASPECTS

The maximum demand of a power station is 96000 kW and daily

load curve is described as follows:

Time (hours) 0 - 6 6 - 8 8 - 12 1214 14 - 18 18- 22 2224

Load (MW) 48 60 52 40 84 96 48

i) Draw load curve and load duration curve

ii) Determine the load factor and demand factor, energy

supplied per year

Solution:

Load curve

Load duration curve

JUNE 11, 10 M


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ECONOMIC ASPECTS

Maximum demand= 96000 kW= 96 MW

Connected load = 48+ 60+ 52+ 40+ 84+ 96+ 48 = 428 MW

i)

ii) From load duration curve energy supplied per day=

(96X4) + (84X4) + (60X2) + (52X4) + (48X8)+(40X2)= 1512 MWh

iii) Units generated per year= (units generated per day)X 365

= 1512 X 365

=551880 MWh


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ECONOMIC ASPECTS

The yearly load duration curve can be considered as a straight

line from 300MW to 80 MW for a certain power plant. Power is

supplied with one generating unit of 200 MW capacity and two

units of 100 MW capacity each. Determine:

iv) Installed capacity ii) Load factor iii) plant factor

iv) Maximum demand v) Utilization factor

Solution:

JAN 09, 10 M


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ECONOMIC ASPECTS

A 200 MW thermal power plant is to supply power to a system

having maximum and minimum demand of 140 MW and 40

MW respectively during the year, assuming load duration curve

to be straight line, determine i) installed capacity ii) load

factor iii) capacity factor iv) utilisation factor


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ECONOMIC ASPECTS

The load on the power plant on a typical day as under

Time 12 - 5 am 5- 9 am 9 6 pm 6pm- 10pm 10pm 12 pm

Load(MW) 20 40 80 100 20

Draw energy load curve.

Solution:

First draw Load duration curve

To draw energy load curve

DEC 08, 8 M

Electrical Power Generation Notes 4

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