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.
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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: The connected 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%
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
6.
Units generated in year: Units generated in year is given as follows 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
7. Units generated in day: Units generated / Day = Max.demand X Load factor X 24
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
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 106 kWh and remains in operation for 2080 hours inn year then
Plant use factor =
7 X 106
= 0 .168 = 16.8%
20000 X 2080
12. Loss factor: It is defined as average power loss to the peak load power loss during a specified period of time.
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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)
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 20 MW
6849310 20000000
Demand factor =0.5 =50% Load factor= 0.3424 = 34.24 %
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
0.3424
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 103 kW 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 103 kWh
=
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 103 kW Load factor = 40% = 0.4
Energy generated per year
=
Max. Demand x Load factor x 8760
=
(100 x 103) x 0.4 x 8760
=
350400 x 103 kWh
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 = 365 – 60 = 305 days
Annual energy supplied = 600 x 305 = 183000 MWh
Annual Average demand =
= = 20.89 MW
= = 0.2089 = 20.89%
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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:
JUNE 10, 12M
Load factor = 48% = 0.48 Connected load = 15000 + 12000 + 8500 + 6000 + 450 = 41,950 kW Max. Demand = 22,000 kW
22000 41950 0.5244
52.44%
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 105 kWh Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 570.78 MW
= 0.57078 = 57.078% Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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.
JAN 09, 05M
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 Load factor = 60% = 0.6 Max. demand = ? Annual energy supplied = ?
Sum of individual max. demands = 10000 + 5000 + 8000 + 7000 = 30,000 kW
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
Maximum demand on power station
= 20,000 KW
Annual energy supplied = max. Demand x LF x 8760 = 20000 x 0.6 x 8760 = 105120000 kWh = 105.12 x106 kWh
Maximum demand on station = 20000 kW Annual energy supplied
= 105.12 x106 kWh
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 106 kWhr
ii)
Annual base load station output
= 101.0 x 106 kWhr
iii)
Peak load on the stand by station = 120 MW
iv)
Hours of use by stand by station/ year = 3000 hrs JUNE 11, 10 M
Solution: Stand by station:
Given data: Annual stand by station output = 87.35 x 106 kWhr 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%
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
87.35 x 106 50000 x 8760 0.1994 = 19.94% Base load station: Given data: Annual base load station output = 101.0 x 106 kWhr 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.e 8760 hours
101.0 x 106 8760 11529.68 kW
= 0.0288 = 2.88%
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
101.0 x 106 400000 x 8760 0.0288 = 2.88%
Note : Use this formula to find load factor directly
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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
Load factor
= 60% = 0.6
Capacity factor
= 45% = 0.45
Units generated per year = Max.demand x LF x 8760 = 400 x 0.6 x 8760 = 2.10 x 106 MWh
2.10 x 106 MWh 0.45 X 8760 533.33 MW
Reserve capacity
Plant Capacity – Max.Demand 533.33 – 400 133.33 MW
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
Alternate method: Max. Demand
= 400 MW Load factor
= 60% = 0.6
Capacity factor = 45% = 0.45
Average demand = max. Demand X Load factor = 400 X 0.6 = 240 MW
Reserve capacity
Plant Capacity – Max.Demand 533.33 – 400 133.33 MW
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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:
Given data:.
Max. Demand
= 400 MW
Load factor
= 60% = 0.6
Capacity factor
= 45% = 0.45
Units generated per year = max.demand x LF x 8760 = 500 x 0.5 x 8760 = 2.19 x 106 MWh
2.19 x 106 MWh 0.4 X 8760 625 MW
Reserve capacity
Plant Capacity – Max.Demand 625 – 500 125 MW
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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:
DEC 11, 8 M
Given data:
Max. Demand
= 20 MW
Load factor
= 70% = 0.7
Capacity factor
= 50% = 0.5
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 106 kWh
122640 MWh 0.5 X 8760 28 MW Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
Reserve capacity
Plant Capacity – Max.Demand
28 – 20 8 MW
Hours during which the plant is not in service per year = 8760 – 7300 = 1430 hours
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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.
JUNE 08, 12, 10 M
Solution: Given data:.
Max. Demand Load factor
= 20 MW = 60% = 0.6
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 Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
Reserve capacity
Plant Capacity – Max.Demand
25 – 20 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 when running according to operating schedule is given by
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
A power supply is having the following loads Type of load
Maximum demand
Diversity factor of
Demand
group
factor
(kW) 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)
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 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
If diversity factor among the transformer is 1.3 find max demand on the feeder Solution:
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 = 0.5
Transformer 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 feeder DEC 11, 8 M
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 Load factor
Diversity factor of group
(kW) 1) Residential
1000
20
1.2
2) Commercial
2000
25
1.1
10,000
80
1.25
3)Industrial
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:
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
JUNE 09, 12 M
ECONOMIC ASPECTS
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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.
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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.
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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. Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
DEC 11, 7 M
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
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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:
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
DEC 10, 8M
ECONOMIC ASPECTS
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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:
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
JUNE 10, 8M
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%
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ECONOMIC ASPECTS
A power station has the following daily load cycle Time in hours
6–8
8 – 12
12- 16
Load in MW
20
40
60
16- 20 20 -24 20
50
24 -6 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 103 kWh
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 103 kWh
This is same for both the curves
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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
12 – 14
14 - 18
48
60
52
40
84
Load (MW)
18- 22 22 – 24 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
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
JUNE 11, 10 M
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
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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) iv)
Installed capacity Maximum demand
ii) Load factor
iii) plant factor
v) Utilization factor JAN 09, 10 M
Solution:
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
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 factor
iii) capacity factor iv) utilisation factor
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
ii)
load
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
Vinayaka.B.G, Asst.Professor, Dept of E&E, BIET, Davanagere
DEC 08, 8 M