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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.
Average	Load =
Area	under	the	load	curve	(kWh)
no. of	hours	(h)
																(kW)
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.
Load	Factor =
Average	Load
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.
Installed	Capacity = Nominal	power	value	of	the	plant																			(kW	or	MW)
Reserve Capacity
Reserve	Capacity = Installed	Capacity − Max. Demand																			(kW	or	MW)
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:
Plant	Capacity	Factor =
Average	Demand		(kW)
Installed	Capacity	(kW)
Utilisation Factor
Utilisation	Factor =
Max.		Demand		(kW)
Installed	Capacity	(kW)
Plant Use Factor
Plant	Use	Factor =
Actual	energy	produced		(kWh)
Installed	Capacity	(kW) × no.of	operation	hours(h)	
Or
Plant	Use	Factor =
Average	Demand × T
Installed	Capacity × no. of	operating	hours	
Where
T = 24	h			 if the time is a day
T = 24 × 30 h if the time is a Month
Diversity Factor
Diversity	Factor =
Sum	of	Individual	Max. Demand		(kW)
Max. Demand	on	power	Plant		(kW)
Load Duration Curve
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.
Load(MW)Load(MW)
EX: A generation station of 1MW supplied a region which has the following
demands:
From To Demand (kW)
midnight 5 am 100
5 am 6 pm No-load
6 pm 7 pm 800
7 pm 9 pm 900
9 pm midnight 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
Average	Load =
Area	under	the	load	curve	(kWh)
no. of	hours	(h)
Average	Load =
(5 × 100) + (13 × 0) + (1 × 800) + (2 × 900) + (3 × 400)
24
Average	Load =
4300	kWh
24	h
= 179.16			kW
180
Time (h)
2 204 6 168 1410 12 22 24
100
400
800
900
180
Time (h)
2 204 6 168 1410 12 22 24
100
400
800
900
Load Curve
Load Duration Curve
Load	Factor =
Average	Load
Max. Load
=
179.16	kW
900	kW
= 0.199 = 19.19%
The	Reserve	Capacity = Installed	Capacity − Max. Demand
																																											= 1000 − 900 = 100	kW
Plant	Capacity	Factor =
Average	Demand
Installed	Capacity
=
179.16	kW
1000	kW
= 0.1791 = 17.91%
Plant	Use	Factor =
Actual	Energy	Produced	in	(kWh)
Plant	Capacity × no. of	hours
=
4300	kWh
1000	kW × 11	h
Plant	Use	Factor = 0.3909 = 39.09%
Utilisation	Factor =
Max. Demand
Installed	Capacity
=
900	kW
1000	kW
= 0.9 = 90%
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. The daily energy produced
2. The reserve capacity
3. The number of operating hours per daily
4. The maximum energy that could be produced daily if the generation station
was running all the time.
Solution:
1) Find the daily energy produced
Load	Factor =
Average	Load
Max.		Load
0.6 =
Average	Load
20	MW
							 	 ℎ 								Average	Load = 0.6 × 20	MW = 12	MW		
Average	Load =
Area	under	the	load	curve	(kWh)
no. of	hours	(h)
The time for daily energy produced is 24 h
Area	under	the	load	curve = energy	produced = 12 × 24 = 288	MWh
2) Find the reserve capacity
Plant	Capacity	Factor =
Average	Demand		
Installed	Capacity	
Installed	Capacity =
Average	Demand		
	Plant	Capacity	Factor
=
12
0.48
= 25	MW
Reserve	Capacity = Installed	Capacity − Max. Demand
Reserve	Capacity = 25 − 20 = 5	MW
3) Find the number of operating hours per daily
Plant	Use	Factor =
Average	Demand × T
Installed	Capacity × no.of	operating	hours	
0.8 =
12 × 24
25 × no. of	operating	hours	
no. of	operating	hours =
12 × 24
25 × 0.8
= 14.4	h
4) Find the maximum energy that could be produced daily if the generation
station was running all the time.
The	maximum	energy	that	could	be	produced	daily	 = Installed	Capacity × 24
																																																																																																				= 25 × 24 = 600	MWh
																																																																																																			= 25 × 24 = 600	000	kWh
EX: A generation station of 10MW supplied two regions (A and B) which have
the following demands:
Region A
From To Demand (kW)
midnight 9 am 600
9 am 12 noon 2500
12 noon 5 pm 800
5 pm 6 pm 5000
6 pm 7 pm No-load
7 pm midnight 4000
Region B
From To Demand (kW)
midnight 8 am 800
8 am 1 pm 5000
1 pm 2 pm 800
2 pm 5 pm 5000
5 pm midnight 800
Find the diversity factor.
Solution:
Regions A+B
From To Demand (kW)
midnight 8 am 600+800=1400
8 am 9 am 600+5000=5600
9 am 12 noon 2500+5000=7500
12 noon 1 pm 800+5000=5800
1 pm 2 pm 800+800=1600
2 pm 5 pm 800+5000=5800
5 pm 6 pm 5000+800=5800
6 pm 7 pm 0+800=800
7 pm midnight 4000+800=4800
Max. Demand	on	power	Plant	 = 7500	kW
Diversity	Factor =
Sum	of	Individual	Max. Demand		(kW)
Max. Demand	on	power	Plant		(kW)
Diversity	Factor =
5000	(kW) + 5000		(kW)
7500		(kW)
= 1.333
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

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Examples on load curve

  • 1. 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
  • 2. 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. Average Load = Area under the load curve (kWh) no. of hours (h) (kW) 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. Load Factor = Average Load Max. Load
  • 3. 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. Installed Capacity = Nominal power value of the plant (kW or MW) Reserve Capacity Reserve Capacity = Installed Capacity − Max. Demand (kW or MW) 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: Plant Capacity Factor = Average Demand (kW) Installed Capacity (kW) Utilisation Factor Utilisation Factor = Max. Demand (kW) Installed Capacity (kW)
  • 4. Plant Use Factor Plant Use Factor = Actual energy produced (kWh) Installed Capacity (kW) × no.of operation hours(h) Or Plant Use Factor = Average Demand × T Installed Capacity × no. of operating hours Where T = 24 h if the time is a day T = 24 × 30 h if the time is a Month Diversity Factor Diversity Factor = Sum of Individual Max. Demand (kW) Max. Demand on power Plant (kW)
  • 5. Load Duration Curve 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. Load(MW)Load(MW)
  • 6. EX: A generation station of 1MW supplied a region which has the following demands: From To Demand (kW) midnight 5 am 100 5 am 6 pm No-load 6 pm 7 pm 800 7 pm 9 pm 900 9 pm midnight 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 Average Load = Area under the load curve (kWh) no. of hours (h) Average Load = (5 × 100) + (13 × 0) + (1 × 800) + (2 × 900) + (3 × 400) 24 Average Load = 4300 kWh 24 h = 179.16 kW
  • 7. 180 Time (h) 2 204 6 168 1410 12 22 24 100 400 800 900 180 Time (h) 2 204 6 168 1410 12 22 24 100 400 800 900 Load Curve Load Duration Curve
  • 8. Load Factor = Average Load Max. Load = 179.16 kW 900 kW = 0.199 = 19.19% The Reserve Capacity = Installed Capacity − Max. Demand = 1000 − 900 = 100 kW Plant Capacity Factor = Average Demand Installed Capacity = 179.16 kW 1000 kW = 0.1791 = 17.91% Plant Use Factor = Actual Energy Produced in (kWh) Plant Capacity × no. of hours = 4300 kWh 1000 kW × 11 h Plant Use Factor = 0.3909 = 39.09% Utilisation Factor = Max. Demand Installed Capacity = 900 kW 1000 kW = 0.9 = 90%
  • 9. 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. The daily energy produced 2. The reserve capacity 3. The number of operating hours per daily 4. The maximum energy that could be produced daily if the generation station was running all the time. Solution: 1) Find the daily energy produced Load Factor = Average Load Max. Load 0.6 = Average Load 20 MW ℎ Average Load = 0.6 × 20 MW = 12 MW Average Load = Area under the load curve (kWh) no. of hours (h) The time for daily energy produced is 24 h Area under the load curve = energy produced = 12 × 24 = 288 MWh
  • 10. 2) Find the reserve capacity Plant Capacity Factor = Average Demand Installed Capacity Installed Capacity = Average Demand Plant Capacity Factor = 12 0.48 = 25 MW Reserve Capacity = Installed Capacity − Max. Demand Reserve Capacity = 25 − 20 = 5 MW 3) Find the number of operating hours per daily Plant Use Factor = Average Demand × T Installed Capacity × no.of operating hours 0.8 = 12 × 24 25 × no. of operating hours no. of operating hours = 12 × 24 25 × 0.8 = 14.4 h
  • 11. 4) Find the maximum energy that could be produced daily if the generation station was running all the time. The maximum energy that could be produced daily = Installed Capacity × 24 = 25 × 24 = 600 MWh = 25 × 24 = 600 000 kWh EX: A generation station of 10MW supplied two regions (A and B) which have the following demands: Region A From To Demand (kW) midnight 9 am 600 9 am 12 noon 2500 12 noon 5 pm 800 5 pm 6 pm 5000 6 pm 7 pm No-load 7 pm midnight 4000 Region B From To Demand (kW) midnight 8 am 800 8 am 1 pm 5000 1 pm 2 pm 800 2 pm 5 pm 5000 5 pm midnight 800 Find the diversity factor.
  • 12. Solution: Regions A+B From To Demand (kW) midnight 8 am 600+800=1400 8 am 9 am 600+5000=5600 9 am 12 noon 2500+5000=7500 12 noon 1 pm 800+5000=5800 1 pm 2 pm 800+800=1600 2 pm 5 pm 800+5000=5800 5 pm 6 pm 5000+800=5800 6 pm 7 pm 0+800=800 7 pm midnight 4000+800=4800 Max. Demand on power Plant = 7500 kW Diversity Factor = Sum of Individual Max. Demand (kW) Max. Demand on power Plant (kW) Diversity Factor = 5000 (kW) + 5000 (kW) 7500 (kW) = 1.333
  • 13. 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