ECE 320
Law #1, 2, & 3
Energy Systems I
HW #1
Session 03
Spring 2005
j :=
Law Problem 1
VA := W
Var := W
I. Define the Situation
kVA := kW
The voltage across and the current into the positive terminal
of a single port network are given below.


v( t) = 2 ⋅ 120 ⋅ cos ω⋅ t +


i( t) = 2⋅ 10⋅ cos ω⋅ t −
π
⋅V
6
π
⋅A
6
II. Goals
a) Detemine p(t), S, P, and Q into the network
b) Find a simple (two element) series circuit consistent with the prescribed
terminal behavior as described above.
III. Generate Ideas
a)

S = V⋅ I
p ( t) = v( t) ⋅ i( t)
b)
Z=
Vp
Ip
S = P + j⋅ Q
Z = R + j⋅ X
IV. Make a Plan
a)
1) Write the expression for p(t) directly from p(t) = v(t) i(t) and simplify
2) Determine phasor representation of the voltage and current.
3) Compute the complex power.
4) Set P = to the real part of S.
5) Set Q = to the imaginary part of S
b)
1) Determine Z.
2) R equal to the real part of Z
3) X equal to the imaginary part of Z
4) calculate L or C
C:\JLAW\CLASSES\S05320\HW
\HW01.mcd
−1
Page 1 of 6
ECE 320
Law #1, 2, & 3
Energy Systems I
HW #1
Session 03
Spring 2005
V. Take Action
a)
1) Write the expression for p(t) directly from p(t) = v(t) i(t) and simplify


p ( t) = 2 ⋅ 120 ⋅ cos ω⋅ t +
π

 ⋅ 2 ⋅ 10⋅ cos ω⋅ t −
6



p ( t) = 2 ⋅ 120 ⋅ 10⋅ cos ω⋅ t +


p ( t) = 2400⋅ cos ω⋅ t +
π

 ⋅ cos ω⋅ t −
6

π

 ⋅ cos ω⋅ t −
6



π
⋅W
6
π
⋅W
6
π
⋅W
Acceptable Answer
6
 π ⋅W

 3 
p ( t) =  1200⋅ cos( 2 ⋅ ω⋅ t) + 1200⋅ cos
p ( t) = ( 1200 cos( 2 ⋅ ω⋅ t) + 600 ) ⋅ W
Better Answer
2) Determine phasor representation of the voltage and current.
j⋅
Vp := 120 ⋅ e
π
6
− j⋅
⋅V
Ip := 10⋅ e
π
6
⋅A
3) Compute the complex power.

S := Vp ⋅ Ip
S = 1.2 kW
arg( S) = 60 deg
P := Re( S)
P = 600 W
4) Set P = to the real part of S.
3
S = 600 + 1.039j × 10 VA
5) Set Q = to the imaginary part of S
3
S = 600 + 1.039j × 10 VA
b)
Q := Im( S)
3
Q = 1.039 × 10 Var
1) Determine Z.
Z :=
Vp
Ip
Z = 6 + 10.392j Ω
2) R equal to the real part of Z
R := Re( Z)
C:\JLAW\CLASSES\S05320\HW
\HW01.mcd
R = 6Ω
Page 2 of 6
ECE 320
Law #1, 2, & 3
Energy Systems I
HW #1
Session 03
Spring 2005
3) X equal to the imaginary part of Z
X := Im( Z)
X = 10.392 Ω
4) calculate L or C
X is positive is it is inductive
L=
or assume:
L :=
ω := 377 ⋅
X
ω
rad
sec
X
L = 0.0276 H
ω
VI. Review, Reflect, and Refine
Alternate Method
2
P=
( Ip ) ⋅ R
Q=
( Ip )
2
R :=
⋅X
X :=
P
( Ip )
2
Q
(
Ip
)
2
R = 6Ω
X = 10.392 Ω
The current lags the voltage, so Q is positive and the reactance is inductive.
Note that:
p ( t) = ( 1200 cos( 2 ⋅ ω⋅ t) + 600 ) ⋅ W
and
S = 1200 W
That is:
(
)
p ( t) = S ⋅ cos 2 ⋅ ω⋅ t + θ v + θ i + P
θ v :=
π
6
θ i := −
C:\JLAW\CLASSES\S05320\HW
\HW01.mcd
P = 600 W
Page 3 of 6
π
6
ECE 320
Law #1, 2, & 3
Energy Systems I
HW #1
Session 03
Spring 2005
Law Problem 2
I. Define the Situation
The rms magnitude of the voltage across a single port network is 100V.
The instantaneous power into the network has a maximum value of 1707 W and
a minimum of -293 W.
II. Goals
a) Find a series RL circuit equivalent to the network.
b) Determine S = P + jQ into the network.
c) Determine the maximum instantaneous power into L and compare with Q.
III. Generate Ideas
Pmax = P + S
P=
( Ip )2⋅ R
Q=
( Ip )2⋅ X
Pmin = P − S
(
S
)
2
2
2
=P + Q
IV. Make a Plan
Pmax + Pmin = 2 ⋅ P
P=
(Pmax + Pmin)
S = Pmax − P
2
V. Take Action
Pmax := 1707⋅ W
P :=
Pmin := −293 ⋅ W
(Pmax + Pmin)
2
Smag := Pmax − P
Q :=
2
2
Smag − P
C:\JLAW\CLASSES\S05320\HW
\HW01.mcd
P = 707 W
3
Smag = 1 × 10 VA
Q = 707.214 Var
Page 4 of 6
Q=
(
S
) 2 − P2
ECE 320
Law #1, 2, & 3
Energy Systems I
HW #1
S := P + j⋅ Q
Vp := 100 ⋅ e
S = 707 + 707.214j VA
j⋅ 0
⋅V

S
Ip :=
Vp
R :=
X :=
L :=
)
2
( Ip )
2
(
( )
Ip = 10 A
P
Ip
Session 03
Spring 2005
arg Ip = −45.009 deg
( )
θ vL := arg Ip
R = 7.07 Ω
Q
X = 7.072 Ω
X
L = 0.0188 H
ω
a)
R = 7.07 Ω
b)
S = 707 + 707.214j VA
L = 0.0188 H
or
3
S = 1 × 10 VA
c)
arg( S) = 45.009 deg
VpL := j⋅ X⋅ Ip
PL := 0
VpL = 70.721 V
(
P = 707 W
(
)
arg VpL = 44.991 deg
PL := VpL ⋅ Ip ⋅ cos θ vL − θ i
)
Q = 707.214 Var
(
PL = 0 W
θ vL − θ i = −90 deg
SmagL := VpL ⋅ Ip
(
SmagL = 707.214 W
)
p L( t) := SmagL⋅ cos 2 ⋅ ω⋅ t + θ vL + θ i + PL
p maxL := SmagL
p maxL = 707.214 W
Maximum instantaneous power into L is equal to Q.
C:\JLAW\CLASSES\S05320\HW
\HW01.mcd
Page 5 of 6
)
θ i := arg VpL
ECE 320
Law #1, 2, & 3
Energy Systems I
HW #1
Session 03
Spring 2005
Law Problem 3
I. Define the Situation
A certain single phase load draws 5 MW at 0.7 power factor lagging.
II. Goals
a) Determine the reactive power required from a parallel capacitor to bring the power factor
of the parallel combination up to 0.9 lagging.
III. Generate Ideas
S1 = P1 + j⋅ Q1
S2 = P2 + j⋅ Q2
P2 = P1
(
P = S ⋅ pf
S
) 2 = P2 + Q2
6
IV. Take Action
MW := 10 ⋅ W
MVA := MW
P1 := 5 ⋅ MW
pf1 := 0.7
P1
Smag1 :=
pf1
Smag1 = 7.143 MVA
Q1 :=
2
Smag1 − P1
2
Q1 = 5.101 MVar
P2 := P1
P2 = 5 MW
P2
Smag2 :=
pf2
Smag2 = 5.556 MVA
Q2 :=
2
Smag2 − P2
2
QC := Q2 − Q1
C:\JLAW\CLASSES\S05320\HW
\HW01.mcd
pf2 := 0.9
Q2 = 2.422 MVar
QC = −2.679 MVar
Page 6 of 6
MVar := MW