ECE421/521 POWER SYSTEM ANALYSIS
Homework #1
Score Allocation
1.1
ECE421 20’
ECE521 10’
1.2
1.3a
Optional 20’
4’
10’
10’
1.3b
20’
10’
1.4a
Optional
4’
10’
1.4b
1.5
Optional 20’
4’
10’
10’
1.6
20’
10’
1.7
Optional
4’
10’
1.7pic
total
Optional 100’+20’
4’
10’
100’
*Score allocation by steps is shown in brackets, for example, (10’/20’) means this step is allocated 10’ for
students of ECE521 and 20’ for students of ECE421.
1.1 Given,
h=85ft=85*0.3048m=25.9m
q=3000ft^3/s=3000*0.3048^3m^3/s=85.0m^3/s
η=69.5%
g=9.81m/s^2
ρ=1000kg/m^3
we have,
(10’/20’)
1.2 Given,
Overall P=64*8+42*2=596MW
Overall h=20*8+15*2=190m
Q=400m^3/s
We have,
(10’/4’)
1.3 (a)Given,
D=64m
v=7.5m/s
ρ=1.18kg/m^3
η =42.8%
We have,
power output:
(
( )
)
annual energy produced:
(b)Given,
total installed cost=capital cost=1.2*10^6
n=20years
i=0.05
(5’/10’)
(5’/10’)
We have,
(2’/4’)
(2’/4’)
(2’/4’)
(2’/4’)
(2’/4’)
1.4(a)Given,
h=5m
S=21.5km^2=2.15*10^7m^2
g=9.8m/s^2
ρ=1025kg/m^3
period=12.5h=45000s
We have,
(5’/2’)
(5’/2’)
(b)Given
η=30%
t=365*24h
We have,
(10’/4’)
1.5 The following commands
a = 0.034;
P0 = 480;
t0 = 1984;
t = [1984:1999];
P=P0*exp(a*(t-t0));
plot(t,P,'*');
xlabel('Time, year'),ylabel('P, GW');
The plot of predicated peak demand in GW from 1984 to 1999 is shown as
The variable P shows that the peak power demand for the year 1999 is 799GW.
1.6 When t=0,
When t=10,
Let
Then,
(10’/20’)
1.7 The following commands
data = [ 0 1 8
1 2 6
2 3 4
3 4 2
4 5 6
5 6 12
6 7 16
7 8 14
8 9 10
9 10 4
10 11 6
11 12 8];
P = data(:,3);
Dt = data(:,2) - data(:,1);
W = P'*Dt;
Pavg = W/sum(Dt);
Peak = max(P);
AnnualLF = W/(Peak*12)*100;
barcycle(data);
xlabel('Time, month'),ylabel('P,MW');
The variable output is shown in workspace
(10’/20’)
The average load is Pavg = 8MW
(5’/2’)
The annual load factor is AnnualLF = 50%
(5’/2’)
The annual load curve is shown as:
(10’/4’)