Solutions

ECE 476 – Power System Analysis Fall 2014
Exam #1, Thursday, October 2, 2014. 9:30AM - 10:50AM
Name:
Problem 1 (25 p)
Two balanced 3-phase loads are connected in parallel. One is Y-connected and draws 75 kW (3-phase) at 0.8
power factor lagging. The other is ∆-connected and draws 60 kVA (3-phase) at 0.9 power factor leading. The
supply is 480 V (line-line), 60 Hz. Determine:
(a) The magnitude of the source line current.
Solution:
For Y-connected load,
SY =
75
∠ cos−1 (0.8)kVA.
0.8
For ∆-connected load,
S∆ = 60∠ − cos−1 (0.9)kVA.
Total complex power (3 phase):
ST = SY + S∆ = 129 + j30.1 = 132∠13.33o kVA
The magnitude of the current:
|ST |
= 159A
|I| = √
3|V |
(b) The amount of capacitive VARS (3-phase) needed to make the overall power factor to be 1.0.
Solution:
From Part (a), without compensation:
ST = 129 + j30.1kVA
.
With compensation:
STc = 129 + j(30.1 + Qc )kVA
.
To make the power factor to be 1.0, the total reactive power has to be 0. Therefore,
Qc = −30.1kVar
(c) The source line current when the capacitor have been added.
Solution:
|S c |
129 × 103
|I| = √ T = √
= 155.2A
3|V |
3 · 480
Problem 2 (25 p)
A three-phase transmission line is mounted on the tower as shown below.
A
B
C
The distance between A and B is d meters. The distance between A and C is d meters. The radius of each conductor is r meters. The three line currents are defined as positive into the paper and they sum to zero. Determine:
(a) The distributed flux (per meter) linking conductor A in terms of IA , d, and r.
Solution:
λA =
1
1
1
µ0
[IA ln 0 + IB ln + IC ln ]
2π
r
d
d
Since IA + IB + IC = 0,
µ0
d
IA ln 0
2π
r
(b) The distributed flux (per meter) linking conductor B in terms of IA , IB , d, and r.
Solution:
λA =
1
1
1
µ0
[IA ln + IB ln 0 + IC ln √
2π
d
r
2d
µ0
1
1
1
= [IA ln + IB ln 0 − (IA + IB ) ln √
2π
d
r
2d
√
√
µ0
2d
= [IA ln 2 + IB ln 0 ]
2π
r
λB =
(c) The distributed flux (per meter) linking conductor C in terms of IA , IB , d, and r.
Solution:
1
1
1
µ0
[IA ln + IB ln √ + IC ln 0
2π
d
r
2d
µ0
1
1
1
= [IA ln + IB ln √ − (IA + IB ) ln 0
2π
d
r
2d
µ0
r0
r0
= [IA ln + IB ln √ ]
2π
d
2d
λC =
Problem 3 (20 p)
A 3φ, 300 mile, 345-kV line has series impedance z=j0.48 Ω/mile and shunt admittance y=j6.0×10−6 siemens/mile.
(a) Calculate the line’s characteristic impedance Zc and the propagation constant γ.
Solution:
r
z
Zc =
= 282Ω
y
γ=
√
z · y = j1.7 × 10−3 mile−1
(b) Rated line voltage is applied to the sending end of this line. Calculate the receiving-end voltage when the
receiving end is terminated by one-half of the surge (characteristic) impedance.
Solution:
Vs = cosh(γl)VR + jZc sinh(γl)IR
and
IR =
VR
Zc
2
. Therefore,
VS = cosh(γl)VR + jZc sinh(γl)IR
=[cosh(γl) + j2 sinh(γl)]VR
VR =
Vs
= 263.4∠ − 48.2o kV
cosh(γl) + j2 sinh(γl)
Question 4 (10 p)
√
√
The voltage and current of a single-phase load are v(t) = 2 sin(2πf t) and i(t) = 2 sin(2πf t), where f is the
standard value of frequency in the US. Denote the instantaneous power consumed by the load as p(t). Among
the four different power snapshots below, circle the one that corresponds to p(t). Explain your answer with
a couple of sentences.
(b)
1
1.5
0.5
p(t) [W]
p(t) [W]
(a)
2
1
0.5
−0.5
0.01
0.02
0.03
time [s]
(c)
0.04
−1
0
0.05
2
2
1.5
1.5
p(t) [W]
p(t) [W]
0
0
0
1
0.5
0
0
0.01
0.02
0.03
time [s]
(d)
0.04
0.05
0.01
0.02
0.03
time [s]
0.04
0.05
1
0.5
0.01
Explanation:
0.02
0.03
time [s]
0.04
0.05
0
0
The forth option is correct.
p(t) = v(t)i(t) = 2 sin(2πf t) sin(2πf t) = 1 − cos(4πf t)
Therefore, the frequency of p(t) is two times the standard voltage frequency, i.e., 120 Hz.
Question 5 (10 p)
The terminals of a three-phase balanced delta connected inductive load are labeled a, b, and c. The load is
supplied by a three phase abc positive
√ sequence balanced voltage source. The voltage across terminals a and
b of the this three-phase
load
is
v
=
2V cos(ωt), while the current flowing into terminal a of this three-phase
ab
√
load is ia (t) = 2I cos(ωt − θ). Let V ab and I a denote the phasors associated with vab (t) and ia (t). Let
S = P + jQ be the complex power consumed by the load. Among the options below, circle the correct one.
Explain your choice and what is wrong with the other options.
(1) S =
(2) S =
(3) S =
(4) S =
√
√
√
√
3V I and S =
3V I and P =
3V I and Q =
3V I and P =
√
∗
3V ab I a
√
3V I cos θ
√
√
3V I sin(θ − 30)
3V I cos(θ + 30)
Explanation:
(3) is correct.
I
Iab = √ ∠(−θ + 30o ).
3
√
∗
S = 3V¯ab I¯ab = 3V I∠(θ − 30o ).
√
√
Therefore, Q = 3V I sin(θ − 30) and P = 3V I cos(θ − 30)
Question 6 (10 p)
A three-phase Y-connected inductive load is supplied by a three phase abc positive sequence balanced
voltage source. The load is balanced and the per-phase impedance of this load is Z∠θ. Among the choices
below, pick the one that is correct. Explain your choice.
V ca
V ab
Vc
V bc
Ib
θ
θ
Va
θ
θ
Ia
θ
Ic
Vc
Vb
V bc
(a)
(b)
V ab
Vb
V ca
V ca
V ab
Vc
Ic
Ia
θ
θ
Ib
θ
θ
Va
Ia
Va
θ
θ
Ib
Ic
Vb
Vc
(c)
Va
θ
Ia
Ib
V bc
V ab
Vb
Ic
V ca
(d)
V bc
Explanation:
(a) is correct.
Since the load is inductive, current lags voltage. Therefore, it should be (a) or (b). It is positive sequence.
Therefore, it should be (a) or (d).

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