S
in
56.569kVA
=
F
p
P
in
S
in
=
F
p
0.85
=
b.
At half load the output is 30 hp
P
out2
30 hp
⋅
=
P
out2
22.371kW
=
P
in2
P
out2
η
=
P
in2
24.055kW
=
We can find the reactive power in part a since we know S and P
Q
mot
S
in
2
P
in
2

=
Q
mot
29.757kVAR
=
with this reactive power and the real power in at halfload, we can calculate the apparent
power:
S
in2
P
in2
2
Q
mot
2
+
=
S
in2
38.264kVA
=
and
F
p2
P
in2
S
in2
=
F
p2
0.629
=
rpm
1
min
≡
kVA
kW
≡
Problem 710:
Solution
kVAR
kW
≡
HP
60 hp
⋅
=
V
L
460 V
⋅
=
I
L
71 A
⋅
=
η
93 %
⋅
=
RPM
1180 rpm
⋅
=
f
60 Hz
⋅
=
a. To find the power factor we need the input power and apparent power.
The input power can
be found from the output power and the efficiency
P
out
HP
=
P
out
44.742kW
=
P
in
P
out
η
=
P
in
48.11kW
=
the apparent power can be found from the input voltage and current.
Since it is a
threephase motor:
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