CHAPTER THREE
Load Curves
The curve showing the variation of load on the power station (power
plant) with reference to time is known as load curve.
Daily Load Curve
Maximum Load (Peak Load)
It represents the maximum power that consumed by the load during a specific
given time. It is also equal to the maximum actual power generated by the plant
when the transmission losses are neglected.
Average Load
It is the average power that consumed by the load during a certain period of time
and it is equal to the average power that generated by the plant during the same
period of time when neglecting transmission line losses.
AverageLoad =
Areaundertheloadcurve(kWh)
(kW)
no. ofhours(h)
Load Factor
It is the ratio of the average load to the maximum load for a certain period of time.
The load factor is called daily load factor if the period of time is a day, and if the
period of time is a month, the load factor is called monthly load factor, and
similarly for the year load factor.
LoadFactor =
AverageLoad
Max.Load
Installed Capacity (Plant Capacity, nameplate capacity)
It represents the maximum possible power that could be produced (generated) from
the power plant. The value of the installed capacity depends on the plant design.
InstalledCapacity = Nominalpowervalueoftheplant(kWorMW)
Reserve Capacity
ReserveCapacity = InstalledCapacity − Max. Demand(kWorMW)
Plant Capacity Factor
The capacity factor of a power plant is the ratio of its average output power over a
period of time, to its maximum possible power that could be produced. It can be
determined as below:
PlantCapacityFactor =
AverageDemand(kW)
InstalledCapacity(kW)
Utilisation Factor
UtilisationFactor =
Max.Demand(kW)
InstalledCapacity(kW)
Plant Use Factor
PlantUseFactor =
Actualenergyproduced(kWh)
InstalledCapacity(kW) × no. ofoperationhours(h)
Or
PlantUseFactor =
AverageDemand × T
InstalledCapacity × no. ofoperatinghours
Where
T = 24h
if the time is a day
T = 24 × 30 h
if the time is a Month
Diversity Factor
DiversityFactor =
SumofIndividualMax. Demand(kW)
Max. DemandonpowerPlant(kW)
Load Duration Curve
Load (MW)
Load (MW)
In this curve, the load elements of a load curve are arranged in order of descending
magnitudes. So, the area under the load duration curve and the area under the load
curve are equal. The following figure shows an example of the load curve and the
load duration curve.
EX: A generation station of 1MW supplied a region which has the following
demands:
From
midnight
5 am
6 pm
7 pm
9 pm
To
5 am
6 pm
7 pm
9 pm
midnight
Demand (kW)
100
No-load
800
900
400
Neglect transmission line losses and find the following:
1. Plot the daily load curve and the load duration curve.
2. Find the load factor, the reserve capacity, plant capacity factor, plant use
factor, the hours that the plant has been off and utilization factor.
Solution:
When the transmission line losses are neglected,
=
, and the demand = load
Installed capacity = 1 MW = 1000 kW and max. load = max. demand = 900 kW
AverageLoad =
Areaundertheloadcurve(kWh)
no. ofhours(h)
AverageLoad =
(5 × 100) + (13 × 0) + (1 × 800) + (2 × 900) + (3 × 400)
24
AverageLoad =
4300kWh
= 179.16kW
24h
900
800
400
100
0
2
4
6
8
10
12
14
16
18
20
22
24
16
18
20
22
24
Time (h)
Load Curve
900
800
400
100
0
2
4
6
8
10
12
14
Time (h)
Load Duration Curve
LoadFactor =
AverageLoad 179.16kW
=
= 0.199 = 19.19%
Max. Load
900kW
TheReserveCapacity = InstalledCapacity − Max. Demand
= 1000 − 900 = 100kW
PlantCapacityFactor =
PlantUseFactor =
AverageDemand 179.16kW
=
= 0.1791 = 17.91%
InstalledCapacity
1000kW
ActualEnergyProducedin(kWh)
4300kWh
=
PlantCapacity × no. ofhours
1000kW × 11h
PlantUseFactor = 0.3909 = 39.09%
UtilisationFactor =
Max. Demand
900kW
=
= 0.9 = 90%
InstalledCapacity 1000kW
EX: A generation station has a maximum demand of 20 MW, a load factor of
60%, plant capacity factor of 48% and plant use factor of 80%. Find:
1.
2.
3.
4.
The daily energy produced
The reserve capacity
The number of operating hours per daily
The maximum energy that could be produced daily if the generation station
was running all the time.
Solution:
1) Find the daily energy produced
LoadFactor =
0.6 =
AverageLoad
Max.Load
AverageLoad
20MW
AverageLoad =
ℎ
AverageLoad = 0.6 × 20MW = 12MW
Areaundertheloadcurve(kWh)
no. ofhours(h)
The time for daily energy produced is 24 h
Areaundertheloadcurve = energyproduced = 12 × 24 = 288MWh
2) Find the reserve capacity
PlantCapacityFactor =
InstalledCapacity =
AverageDemand
InstalledCapacity
AverageDemand
12
=
= 25MW
PlantCapacityFactor 0.48
ReserveCapacity = InstalledCapacity − Max. Demand
ReserveCapacity = 25 − 20 = 5MW
3) Find the number of operating hours per daily
PlantUseFactor =
0.8 =
AverageDemand × T
InstalledCapacity × no. ofoperatinghours
12 × 24
25 × no. ofoperatinghours
no. ofoperatinghours =
12 × 24
= 14.4h
25 × 0.8
4) Find the maximum energy that could be produced daily if the generation
station was running all the time.
Themaximumenergythatcouldbeproduceddaily = InstalledCapacity × 24
= 25 × 24 = 600MWh
= 25 × 24 = 600000kWh
EX: A generation station of 10MW supplied two regions (A and B) which have
the following demands:
From
midnight
9 am
12 noon
5 pm
6 pm
7 pm
Region A
To
9 am
12 noon
5 pm
6 pm
7 pm
midnight
Demand (kW)
600
2500
800
5000
No-load
4000
From
midnight
8 am
1 pm
2 pm
5 pm
Region B
To
8 am
1 pm
2 pm
5 pm
midnight
Demand (kW)
800
5000
800
5000
800
Find the diversity factor.
Solution:
From
midnight
8 am
9 am
12 noon
1 pm
2 pm
5 pm
6 pm
7 pm
Regions A+B
To
Demand (kW)
8 am
600+800=1400
9 am
600+5000=5600
12 noon
2500+5000=7500
1 pm
800+5000=5800
2 pm
800+800=1600
5 pm
800+5000=5800
6 pm
5000+800=5800
7 pm
0+800=800
midnight
4000+800=4800
Max. DemandonpowerPlant = 7500kW
DiversityFactor =
SumofIndividualMax. Demand(kW)
Max. DemandonpowerPlant(kW)
DiversityFactor =
5000(kW) + 5000(kW)
= 1.333
7500(kW)
HW1: for the previous example, find the following:
1) Draw the load curve and load duration curve for regions A and B and for the
total loads
2) The average load and the load factor of the total system
3) The reserve capacity of the plant
4) Plant use factor
5) Utilization factor