DATA:
V ≔ 240 V
P ≔ 1.6 kW
f ≔ 50 Hz
PF ≔ 0.76
lorentz lov
a. Find the current I for the motor
P
S ≔ ――
= ⎛⎝2.105 ⋅ 10 3 ⎞⎠ W
PF
S
I ≔ ―= 8.772 A
V
(apparent power)
(indput)
b. Find the phase angle between current and voltage and draw it in a phasor diagram
solve , ϕ
ϕ ≔ PF ＝ cos ((ϕ)) ―――
→ 0.70748321177934297162 = 40.536 deg
c. Find apparent power S and reactive power Q for the motor
P
S ≔ ――
= ⎛⎝2.105 ⋅ 10 3 ⎞⎠ W
PF
(apparent power)
1
⎡
―― ⎤
2.0
⎢
⎥
solve , Q ⎢ ⎛⎝4432132.963988932964 ⋅ W 2 - 2.56 ⋅ kW 2 ⎞⎠
2
2
2
⎥ = ⎡ 1.368 ⎤ kW
Q ≔ Q + P ＝ S ―――
→
1
⎢
⎥ ⎢⎣ -1.368 ⎥⎦
――
2.0
⎢
⎥
2
2
⎣ -1.0 ⋅ ⎛⎝4432132.963988932964 ⋅ W - 2.56 ⋅ kW ⎞⎠ ⎦
Q ≔ Q = 1.368 kW
0
(reactive power)
d. Due to utility demands, the Power Factor need to be improved to a value of PF =
0,91 Find the capacitance of the capacitor required to reduce the reactive power Q
PFim ≔ 0.91
P
Sim ≔ ――= 1.758 kW
PFim
Qim ≔ Qim 2 + P 2 ＝ Sim 2
⎡
solve , Qim ⎢ ⎛
3.0914140804250692892 ⋅ kW 2 - 2.56 ⋅ kW 2
―――→ ⎢ ⎝
...
⎢
⎢
2
⎣ -1.0 ⋅ ⎛⎝3.0914140804250692892 ⋅ kW - 2.56 ⋅ kW
Qim ≔ Qim = 0.729 kW
0
Q - Qim
C ≔ ――――
= 35.328 μF
2 ⋅ π ⋅ f ⋅ V2
e. Draw the power triangle before and after the power factor correction