ENGR_400_Module_1_Assignment
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Electrical Engineering
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Jan 9, 2024
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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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