ENGR_400_Module_1_Assignment
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Jason L. Wilson
10/21/2023
ENGR 400 - Module 1
Chapter 8 - Three-Phase Systems
8.1
A Y-connected balanced three-phase source feeds a balanced three-phase load. The
voltage and current of the source coil are:
v(t) = 340sin(377t+0.5236) V
i(t) = 100sin(377t+0.87266) A
Calculate the following:
a.
The rms phase voltage:
V
rms
(
phase
)
=
340
√
2
V
rms
(
phase
)
=
240.42
V
b.
The rms line-to-line voltage:
V
rms
(
line
⟶
line
)
=
240.42
∗
√
3
V
rms
(
phase
)
=
416.42
V
c.
The rms current in the source:
I
rms
(
source
)
=
I
rms
(
phase
)
=
100
√
2
I
rms
(
source
)
=
70.71
A
d.
The rms current in the transmission line:
I
rms
(
transmission
)
=
70.71
A
e.
The frequency of the supply:
F
=
w
2
π
F
=
377
2
π
Jason L. Wilson
10/21/2023
F
=
60
Hz
f.
The power factor at the source side, state leading or lagging:
θ
=
0.87266
−
0.5236
θ
=
0.34906
∗
(
180
π
)
θ
=
20
°
λ
=
cos
(
20
°
)
λ
=
0.9397
leading
g.
The three-phase real power delivered to the load:
P
=
√
3
∗
V
rms
(
phase
)
∗
I
rms
(
phase
)
∗
cos
(
θ
)
P
=
√
3
∗
416.42
∗
70.71
∗
0.9397
P
=
47.925
kW
h.
The three-phase reactive power delivered to the load:
Q
=
¿
√
3
∗
V
rms
(
phase
)
∗
I
rms
(
phase
)
∗
sin
(
θ
)
Q
=
√
3
∗
416.42
∗
70.71
∗
sin
(
20
)
Q
=−
17.443
kVAR
i.
If the load is connected in delta configuration, calculate the load impedance:
V
rms
(
line
)
=
√
3
∗
V
rms
(
phase
)
∠
30
°
¿
(
√
3
∠
30
°
)(
340
∠
30
°
)
V
rms
(
line
)
=
588.89
∠
60
° V
I
phase
Δ
=
I
rms
(
phase
)
√
3
∠
−
30
°
¿
100
∠
50
°
√
3
∠
−
30
°
Jason L. Wilson
10/21/2023
I
phase
Δ
=
57.74
∠
80
° A
Z
Δ
(
phase
)
=
V
rms
(
phase
)
Δ
I
rms
(
phase
)
Δ
¿
588.89
∠
60
°
57.74
∠
80
° A
Z
Δ
(
phase
)
=
(
9.58
−
j
3.49
)
Ω
8.3
A three-phase, 480 V system is connected to a balanced three phase load. The transmission line current
I
a
is 10 A and is in phase with the line-to-line voltage
V
bc
.
Calculate the impedance of the load
for the following cases:
a.
If the load is wye-connected:
V
ab
=
480
∠
0
°
i
a
=
10
∠
−
120
°
V
ab
=
√
3
V
an
V
an
=
V
ab
√
3
V
an
=
480
∠
0
°
√
3
∠
−
30
°
V
an
=
277.13
∠
−
30
°
Z
a
=
277.13
∠
−
30
°
10
∠
−
120
°
Z
a
=
27.71
∠
90
° Ω
b.
If the load is delta-connected:
i
a
=
√
3
i
pha
i
pha
=
i
a
√
3
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Jason L. Wilson
10/21/2023
i
pha
=
10
∠
−
120
°
+
30
°
√
3
i
pha
=
5.77
∠
−
90
°
Z
a
=
480
∠
0
°
0
°
5.77
∠
−
90
°
Z
a
=
83.15
∠
90
° Ω
8.5
A balanced wye-connected load of (4 + j3) Ω is connected across a three-phase source of 173 V(line-to-
line
a.
Find the transmission line current:
V
ph
=
173
√
3
V
ph
=
99.88
V
I
L
=
99.88
4
+
j
3
I
l
=
19.75
∠
−
36.87
° A
b.
Find the power factor:
cos
(
θ
)
=
4
√
4
2
+
3
2
λ
=
0.8
c.
The complex power:
¿
√
3
∗
171
∗
19.75
∠
−
36.86
°
S
=
(
4678.56
+
j
3508.92
)
VA
Jason L. Wilson
10/21/2023
|
s
|
=
5948.2
VA
=
5.848
kVVA
d.
The real power:
P
=
4.68
kW
e.
The reactive power of the load:
Q
=
3.51
kVAR
f.
With
V
ab
as the reference, sketch the phasor diagram that shows all voltages and currents:
Jason L. Wilson
10/21/2023
8.7
The line-to-line voltage of a three-phase transmission line is
V
bc
=
340
∠
20
°V
.
V
bn
=
340
√
3
V
bn
=
196.30
∠
−
10
°
a.
Calculate the phase voltage
V
an
:
V
an
=−
10
°
+
120
°
V
an
=
196.30
∠
110
°
b.
If a wye load impedance of
Z
=
10
∠
60
° Ω
is connected to the transmission line, calculate
the transmission line current
I
b
.
I
b
=
196.3
∠
−
10
°
10
∠
60
°
I
b
=
19.63
∠
−
70
°
c.
Calculate the current in the neutral line
I
n
.
Since the loadisbalanced ,
I
n
=
0
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10/21/2023
8.11
A three- phase Y-connected source is energizing a delta-d connected load. The phase volage of the
transmission line is
V
bn
=
120
∠
0
° V ,
and the load impedance is
Z
=
9
∠
30
°
. Compute
I
a
and the power consumed by the delta load.
Z
Y
=
9
∠
30
3
Z
Y
=
3
∠
30
Z
Y
=
2.598
+
j
1.5
Ω
I
a
=
120
∠
120
°
3
∠
30
°
I
a
=
40
∠
90
° A
P
=
3
∗
40
2
∗
2.598
P
=
12.47
kW
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