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Lecture 11
Utility Load Analysis2010
ME 430
Thermal Systems Design
Utility Loads and Demand
• utility loading varies with time of day, week, and
season
• the capacity of a plant depends upon maximum power
demand by energy consuming devices on the system,
• loads include motors, lights, heating etc.,
“ the success of the central-station system of energy supply stems
from the diversity of the demand of the various components of the
total connected load”
2
Prime Movers
Distribution Bus
Feeders
Station Bus
Generators
Elementary Power System
Principal elements of a power system (from Skrotzki & Vopat, Power Station Engineering and Economy)
Loads
Examples of utility loads
that vary over time
3
Residential electricity load distribution over year (this is a house
with natural gas heating, augmented by a little electric heat in the
loft) and summer air conditioning.
Summer Peaking Utility
Jan Mar Jun Sept Dec
Toronto, Electricity Demand, 2001
0
2500
5000
7500
10000
Ele
ctr
icit
y (
MW
)
4
• the utility must be able to meet this load and account for:
– plant shut-downs for repairs or maintenance
– unscheduled plant shut-downs due to equipment failure
• the utility must operate to ensure that they produce the least cost electricity and that they get the best return on the investment in their generation assets (power plants, hydro electric, etc)
• the cost of electricity generated depends on the type of power plant, its fuel, its operational life span, and the maintenance costs associated with the plant
• electricity must be transmitted from the power plant to the load and this results in transmission power losses and additional costs
• in a free market system, the cost of electricity is determined by supply and demand
• to meet Ontario’s electricity demands, a non-profit organization buys and sells electricity capacity (power & energy) and sets the price,
• see www.ieso.ca.
Utility Loads
5
Part of the supply may be made up by Ontario Power Generation or by
independent electricity producers within the province – some electricity
will be bought or sold across the provincial borders and to/from the
USA.
6
Load Duration Curves
Load Duration Curves define the energy/power requirement of a load in terms of
i) maximum demand
ii) total energy requirement
iii) distribution of energy demand
- there are also a number of other indices used to describe the characteristics of a utility load.
Load Duration CurveChronological Load Curve
Load Indices
7
Load Duration Curve
Quebec Load Data (Winter Peak)
0
5000
10000
15000
20000
25000
30000
1 1001 2001 3001 4001 5001 6001 7001 8001
Hour of year
Lo
ad
, M
W
8
Load Factor
Lavg E/h
Load Factor = -------- = --------
Lmax Lmax
where:Lavg = average load for period
Lmax= peak load for period, for period ~ or <1 hour
E = total energy under the load cure, e.g., the integral
h = total number of hours in period
Load Indices
Load Factor: L/D Curves with same Max demand
and Load Factors
kW
hours
kW
hours
kW
hours
kW
hours
9
Capacity Factor
Lavg
Capacity Factor = --------
Cap
where: Cap = rated capacity of plant
Utilization Factor
Lmax
Utilization Factor = --------
Cap
Load Indices
• the devices making up the load each have a maximum capacity for
absorbing (consuming) power
• if all devices ran at full load then the “maximum demand” would
equal the “connected load”.
• however, experience shows that the maximum demand of a
consumer is less that their connected load at the same time
• maximum demand is related to connected load by the “demand
factor”
Maximum demand
Demand factor = ------------------------------
Connected load
The Demand Factor depends on the type of the load
e.g., hotels 25%, refrigeration plants 90%
Load Indices
10
• although each device may reach its own maximum demand at sometime,
the “demand factor” measures the extent it contributes toward the maximum
demand of the “group” of devices that it belong to.
• the maximum demand of individual consumers does not occur
simultaneously but is spread out over a period of time.
• this holds even for consumers whose activities and energy requirements
are very similar
• the time distribution of maximum demands for similar types of consumers is
measured by the “diversity factor”
Sum of individual maximum demands
Group Diversity Factor = ---------------------------------------------------
Actual maximum demand of group
• This is always greater than 1
– For residential customers it can be 5
– For industrial consumers it can be as low as 1.3
Load Indices
• the ”PEAK DEMAND” of a system is the sum of the loads of individual devices that are functioning at that time.
• at the time of the system peak demand, the demand of a particular group of similar consumers is seldom at the maximum value that it can reach at other times of the year.
• this effect is measured by the “Diversity Factor”
maximum demand of a consumer group
Diversity factor = -------------------------------------------------------
demand of the consumer group at the timeof the system peak demand
• the utility must ensure that it has sufficient power generation capacity for planned and unplanned Outages
• this may be in two forms – Reserve capacity (extra capacity available on the grid)
– Spinning reserve (idling capacity ready to operate on the grid)
• if necessary the utility may decide to purchase energy on the market to meet short term demands.
Load Indices
11
Pricing and Economics
• electrical energy is generally referred to as:– Base Load
– Intermediate Load
– Peak Demand
• base load is usually the least expensive to produce, and
peak load the most expensive.
• generation (electrical energy) cost is a function of:
– Plant Cost (Construction, financing, insurance, administration)
– Fuel Cost
– Operation and Maintenance Cost
– Distribution Costs
Simple Example of a
Load/Duration Curve
Example based on average hourly utility load values for a 24 hour day. Finer resolution data is typically
used (e.g., 10 minute averages, etc.)
Utilities often complete load /duration curves on an annual or seasonal basis to establish generation
capacity, dispatching and costing
12
Example Load Duration Curve
Hour of Day Load (GW)
1 3
2 2
3 4
4 7
5 9
6 10
7 11
8 14
9 13
10 12
11 12
12 13
13 12
14 11
15 13
16 17
17 20
18 23
19 19
20 14
21 12
22 11
23 9
24 6
Daily Utility Load
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Hour of Day
To
tal L
oa
d (
GW
)
Energy = 277 GWh
Example Load Duration Curve
Hour of Day Load (GW)
1 3
2 2
3 4
4 7
5 9
6 10
7 11
8 14
9 13
10 12
11 12
12 13
13 12
14 11
15 13
16 17
17 20
18 23
19 19
20 14
21 12
22 11
23 9
24 6
Daily Utility Load
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Hour of Day
To
tal L
oa
d (
GW
)
Energy = 277 GWh
Average Load
Peak Load
Base Load
13
Example Load Duration Curve
Daily Utility Load
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Hour of Day
To
tal L
oa
d (
GW
)Number of Hours Load (GW) Energy (GWh)
1 23 23
1 20 20
1 19 19
1 17 17
2 14 28
3 13 39
4 12 48
3 11 33
1 10 10
2 9 18
1 7 7
1 6 6
1 4 4
1 3 3
1 2 2
24 277
Energy = 277 GWh
A major concern of a utility is whether
they have enough capacity to meet
their loads at different times of the
day and the cost of generating that
electricity. The cost charged to a
consumer may reflect the average
cost or “time-of-use” cost .
Example Load Duration Curve
Daily Utility Load
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Hour of Day
To
tal L
oa
d (
GW
)
Number of Hours Load (GW) Energy (GWh)
1 23 23
1 20 20
1 19 19
1 17 17
2 14 28
3 13 39
4 12 48
3 11 33
1 10 10
2 9 18
1 7 7
1 6 6
1 4 4
1 3 3
1 2 2
24 277
Energy = 277 GWh
19 hours above 7 GW
4 hours
above 14
GWh
6 hours
When dispatching certain generation
assets, a utility must consider the cost
energy produced from the device and
its duration of operation. Some plants
need long lead times to start and stop.
Others have high capital costs or
operation and fuel costs. A typical
large thermal plant may take days to
start and stop and 7- 10 years to
construct (longer to get approvals).
14
Example Load Duration Curve
Daily Utility Load
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Hour of Day
To
tal L
oa
d (
GW
)
Daily Utility Load Duration Curve
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Number of Hours at Load
To
tal L
oa
d (
GW
)
(Duration)
Example Load Duration Curve
Daily Utility Load
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Hour of Day
To
tal L
oa
d (
GW
)
Daily Utility Load Duration Curve
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Number of Hours at Load
To
tal L
oa
d (
GW
)
Energy = 277 GWhEnergy = 277 GWh
Base Load
Peak Load
Average Load
4 hours above 14 GWh
(Duration)
4 hours
above 14
GWh
19 hours above 7 GW 19 hours above 7 GW
6 hours above 13 GWh6 hours
15
Example Load Duration Curve
Daily Utility Load
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Hour of Day
To
tal L
oa
d (
GW
)
Daily Utility Load Duration Curve
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Number of Hours at Load
To
tal L
oa
d (
GW
)
Energy = 277 GWh
Base Load Region (Lowest cost electricity)
Peak Load Region(defined by Utility,
e.g., Load > 14 GW)
Intermediate Load Region
(Duration)
The duration of a load level determines the type of generation asset used to meet that current load
(Most expensive electricity)
Daily Utility Load
0
5
10
15
20
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Hour of Day
To
tal
Lo
ad
(G
W)
Example Question: a power generation utility with a capacity of 25 GW
has the following daily demand schedule:
Required:A) Plot the Load/Duration Curve for the 24 period period
B) Indicate the average load, “peak load” and “base load“ on the graph
C) Determine the annual “Capacity Factor”
D) Determine the annual “Load Factor”
E) Determine the annual “Utilization Factor”:
F) If the utility supplies: the base load with existing hydro-electric capacity at a cost of
$0.03/kWh; intermediate load with medium capacity thermal plants at a cost of
$0.06/kWh; and, peak demands with combustion turbines at a cost of $0.50/kWh, what
is the average annual cost of electricity produced by the utility in $/kWh? NOTE:
Note: For this question, loads >14 GW are considered Peak Loads.
16
Solution:
A) Plot the Load/Duration Curve for the 24 period period
B) Indicate the average load, “peak load” and “base load“ on the graph
Average daily Load = 11.54 GW
Hour of Day Load (GW)
1 3
2 2
3 4
4 7
5 9
6 10
7 11
8 14
9 13
10 12
11 12
12 13
13 12
14 11
15 13
16 17
17 20
18 23
19 19
20 14
21 12
22 11
23 9
24 6
Daily Utility Load Duration Curve
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Number of Hours at Load
To
tal L
oa
d (
GW
)
Peak Load
Base Load
Average Load = 11.54 GW
Solution Cont’d:
From:
• Utility Capacity = 25 GW
• Peak Load = 23 GW
• Average Daily Load = 11.54 GW
______________________________________________________________
C) Determine the annual “Capacity Factor”: = 11.54 / 25 = 0.46
D) Determine the annual “Load Factor”: = 11.54 / 23 = 0.50
E) Determine the annual “Utilization Factor”: = 23 / 25 = 0.92
Daily Utility Load Duration Curve
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Number of Hours at Load
To
tal L
oa
d (
GW
)
Peak Load
Base Load
Average Load = 11.54 GW
17
Solution Cont’d:
F) If the utility supplies:• the base load with existing hydro-electric capacity at a cost of $0.03/kWh;
• intermediate load with medium capacity thermal plants at a cost of $0.06/kWh; and,
• peak demands with combustion turbines at a cost of $0.50/kWh,
what is the average daily cost of electricity produced by the utility in $/kWh?
Note: for this calculation, assume that the utility considers loads over 14 GW to be peak demand
______________________________________________________________________________________________
AVERAGE COST OF ELECTRICAL GENERATION
= Total Cost of Energy Generation for day/number kW hours delivered
= (EnergyPL X CostPL + EnergyIL X CostIL + EnergyBL X costBL)/(total GWh)
= 1x106 kW/GWx[(23 GWh) x $0.50/kWh) + (206 GWh x $0.06/kWh)
+ (48 GWh x $0.03/kWh] / [(23+206+48 GWh) x 1x106 kW/GW]
= $ 0.091 / KWh
Or simply (by area weighting)
= [23 x 0.5 + 206 x 0.06 + 48 x 0.03]/277 = 25.3/277 = $0.091/kWh
ANSWER to Part F: Average annual cost of electricity generation is $0.091 $/kWh
Bonus question 1: If the utility was able to shift the peak loads to hours
with lower loads such that for the same total energy demand the hourly
load never exceeded the intermediate load limit, the average cost of
generation would be:
AVERAGE COST WITH NO PEAK
= [(206+23) x 0.06 + (48 x 0.03)]/277 = 15.18/277 = $0.055/kWh
(a 40% reduction in average daily generation cost)
Bonus question 2:
If all the demand was provided by Base Load Generation:
AVERAGE COST FOR ALL BASE LOAD GENERATION
= [277 x 0.03]/277 = $0.03/kWh
(a 67% reduction in average daily generation cost)
