S
syn
99.162kVA
=
S
syn
P
in
F
psyn
=
F
psyn
0.8
=
Since the new motors operate at 0.8 power factor, the apparent power will be
P
in
79.33kW
=
P
in
P
out
η
=
P
out
100 hp
⋅
=
For the new motors, the power out is 100 hp or 74.6 kW (2 times 50 hp), so the power in is
the power out divided by efficiency
S
old
166.5
151.337i
+
kVA
=
S
old
S
load
e
j
θ
old
⋅
⋅
=
θ
old
acos F
pold
(
29
=
b.
We can find the total load with the new motors by adding the two loads together as complex
numbers
I
L
270.633A
=
I
L
S
load
3
V
L
⋅
=
a.
The line current can be found from the three-phase power formula
Solution
η
94%
=
V
L
480 V
⋅
=
F
pold
0.74
=
S
load
225 kVA
⋅
=
Given:
kVAR
kW
≡
kVA
kW
≡
Define:
P2-25.
A machine shop is fed by a 300 KVA, 480 Volt, three-phase transformer bank that
has a load consisting of many small motors.
The total shop load is 225 KVA and the power
factor is 0.74 lagging.
(a)
What is the line current in the feeder from the transformer to the facility?

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