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A 3.0 horsepower induction motor is connected to 240 VAC rms, 60 Hz. The efficiency of this motor is equal to 70% and the motor has a FP-0.6 lagging when it is delivering rated power of 3.0 hp. a. Construct the power triangle for this motor at 3.0 hp ( 1.0 hp = 746 W), NOTE: This 3.0 hp is the mechanical output power, not the electrical input power to the motor. b. Calculate the amount of capacitance connected in parallel with this motor to obtain unity Fp c. Determine the AWG wire size to run this motor before and after Fp correction.

User MacakM
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a. To construct the power triangle, we first need to calculate the real power (P) and the reactive power (Q) of the motor.

Real power:

P = 3.0 hp x 746 W/hp = 2238 W

Reactive power:

Q = P x tan(cos⁻¹(FP)) = 2238 x tan(cos⁻¹(0.6)) = 1481 VAR

The apparent power (S) can be calculated using the power triangle equation:

S = P / power factor = P / cos(cos⁻¹(FP)) = P / cos(arccos(0.6)) = 3726 VA

Now we can draw the power triangle using the values we have calculated:

/|

S / | Q

/ / |

/ /θ |

/___/_____|

P

where θ is the angle between S and P.

b. To correct the power factor to unity, we need to add a capacitor in parallel with the motor. The capacitive reactive power (Qc) needed to cancel out the inductive reactive power of the motor is given by:

Qc = Q = 1481 VAR

The capacitance (C) needed can be calculated using the formula:

Qc = 1 / (2πfC)

where f is the frequency (60 Hz) and C is the capacitance in farads. Solving for C:

C = 1 / (2πfQc) = 1 / (2π x 60 x 1481) = 0.0000045 F or 4.5 µF

c. To determine the wire size needed, we first need to calculate the full-load current (FLC) of the motor:

FLC = P / (efficiency x line-to-line voltage) = 2238 / (0.7 x 240) = 13.1 A

Before power factor correction, the wire size needed can be determined using the NEC ampacity tables based on the FLC and the installation conditions (e.g. ambient temperature, number of conductors, etc.).

After power factor correction, the motor will draw less current because the reactive power drawn by the motor has been reduced by the capacitor. The new current can be calculated using the following formula:

I = P / (efficiency x power factor x line-to-line voltage) = 2238 / (0.7 x 1.0 x 240) = 13.4 A

The new wire size needed can be determined using the same process as before, using the new current value.

User Ville M
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