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INSE 6640 – Smart Grids and Control System Security (Fall 2022)
Abhishek Satasiya | Student ID – 40197772
Assignment #1
(
100 points
): Due 11:59 pm, Friday, October 21, 2022
1.
(
10 points
) The peak load
of a power grid refers to the highest load demand seen during a certain period.
The table below shows the weekly peak load in a regional power grid, in terms of the percentage with
respect to the peak load of the entire year: for example, the peak load observed in Week 1 is 86.2% of the
peak load observed over the entire year. Weeks 1-8 and 44-52 are in the winter, weeks 9-17 are in the spring,
weeks 18-30 are in the summer, and weeks 31-43 are in the fall. Answer the following questions. Week
Peak load (%)
Week
Peak load (%)
Week
Peak load (%)
Week
Peak load (%)
1
86.2
14
75.0
27
75.5
40
72.4
2
90.0
15
72.1
28
81.6
41
74.3
3
87.8
16
80.0
29
80.1
42
74.4
4
83.4
17
75.4
30
88.0
43
80.0
5
88.0
18
83.7
31
72.2
44
88.1
6
84.1
19
87.0
32
77.6
45
88.5
7
83.2
20
88.0
33
80.0
46
90.9
8
80.6
21
85.6
34
72.9
47
94.0
9
74.0
22
81.1
35
72.6
48
89.0
10
73.7
23
90.0
36
70.5
49
94.2
11
71.5
24
88.7
37
78.0
50
97.0
12
72.7
25
89.6
38
69.5
51
100.0
13
70.4
26
86.1
39
72.4
52
95.2
(Reference: IEEE Reliability Test System (RTS) - 1996)
a.
Which weeks have the highest and lowest peak load of the year? If the annual peak load is 2,850 MW,
what are the average peak loads (in MW) of the winter weeks
and the summer weeks
, respectively?
b.
If the annual peak load is 2,850 MW and we have two power plants whose capacities (maximal
generation power) are 1,500 MW and 1,000 MW, respectively. During the week with the highest peak
load, how much extra power generation capacity (in MW) do we need to meet the peak load of the
year? During the week with the lowest peak load, at least how much power generation capacity (in MW)
would be in idle? Answer
:
a)
Week 51 has the highest (100%) peak load of the year. Week 38 has the lowest (69.5%) peak load of the year.
The annual peak load is 2,850 MW.
Average peak load for winter weeks, = Weeks
1
¿
∑ peak load
(
8
+
Weeks
44
¿
52
)
∑
(
N umberof weeks
)
= 1520.2
17
= 89.42% of 2850
INSE 6640 – Smart Grids and Control System Security (Fall 2022)
Abhishek Satasiya | Student ID – 40197772
= 2548.47 MW is the average peak load for winter weeks. Average peak load for summer weeks,
= Weeks
18
∑ peak load
(
¿
30
)
∑
(
N umberof weeks
)
= 1105
13
= 85% of 2850 = 2422.5 MV is the average peak load of summer weeks b)
Total power plant generation capacity = 1500 MW + 1000 MW= 2500 MW
a.
The highest peak load is in week 51 which is 100%.
Therefore, the highest peak load is 100% * 2850 = 2850 MW, which requires 2850 – 2500 = 350 MW extra power capacity.
b.
c.
The lowest peak load is in week 38 which is 69.5%.
Therefore, the lowest peak load is 69.5% * 2850 = 1980.75 MW, which leaves 2500 – 1980.75 =
519.25 MW generation capacity in idle.
INSE 6640 – Smart Grids and Control System Security (Fall 2022)
Abhishek Satasiya | Student ID – 40197772
2.
(
15 points
) The figure on the right shows an abstract model of power
transmission grids in Quebec, where the ultra-high voltage transmission
line has a rated voltage of 735 kilovolts (kV). The Ohm’s law states that
the active power consumption at the end of the line is P
=
UI
and
the active power loss along the line P
loss
=
¿
, where U
,
I
, and
R
are the rated voltage, line current, and line resistance,
respectively. Assume that the active power consumption at substation
#2 is P
=
1,000
MW
. a.
If the 735-kV line is 100 kilometers long with a resistance of 0.5Ω per
kilometer, what is the active power loss along the line according to
the Ohm’s law? Show your result in megawatts (MW). If the rated
voltage of this line is reduced from 735 kV to 315 kV (the length and
resistance remain the same) what is the active power loss along the
new 315-kV line?
b.
According to recent data, the average power consumption of a
household in Quebec is 1.9 kilowatts (kW); the average charging
demand of an all-electric vehicle over a common 120 V outlet is 1.4
kW. When increasing the rated voltage from 315 kV to 735 kV, the
saved active power loss can power up how many households in Quebec? How many all-electric vehicles?
Answer
: a)
Total Resistance R = 0.5Ω × 100 = 50 Ω
For 735-kV Line the Line Current,
I = P
U
= 1000
×
10
6
735
×
10
3
= 1.36 ×
10
3
A
Active power loss = I
2
R
= (
1.36
×
10
3
)
2
×
50
= 92.48 ×
10
6
= 92.48 MW
The rated voltage of this line is reduced from 735 kV to 315 kV.
For 315-kV Line the Line Current,
INSE 6640 – Smart Grids and Control System Security (Fall 2022)
Abhishek Satasiya | Student ID – 40197772
I = P
U
= 1000
×
10
6
315
×
10
3
= 3.17 ×
10
3
A
Active power loss = I
2
R
= (
3.17
×
10
3
)
2
×
50
= 502.4 ×
10
6
= 502.4 MW
b)
Saved active power loss = (502.4 – 92.48) MW
= 409.9 MW
For households in Quebec, = 409.9
×
10
6
1.9
×
10
3
= 215.7 ×
10
3
= 215.7 thousand homes
So, it can power up 215.7 thousand households in Quebec. For Electric vehicles, = 409.9
×
10
6
1.4
×
10
3
= 292.8 ×
10
3
= 292.8 thousand Electric vehicles
So, it can power up 292.8 thousand Electric vehicles.
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