Question

A manufacturing plant has a 25 KVA single phase motor with a lagging power factor of 0.85 and this motor gets its power from a nearby a.c. voltage supply. A power factor correction capacitor of 12 kVar is also connected parallel to the motor.

(a) Calculate the real power (kW) consumed by the motor (3)
(b) Calculate the input apparent power (S) taken from the supply (14)
(c) The power factor is to be corrected or improved from 0.85 to 0.99 lagging. Calculate the rating (in Vars) of the capacitor required for this improvement. (8)

Answer 1

Sure, I can help you with those calculations.

(a) To calculate the real power consumed by the motor, we can use the formula:

Real Power (kW) = Apparent Power (kVA) * Power Factor

Given that the motor has a power factor of 0.85 and an input power of 25 KVA, we can find the real power consumption as follows:

Real Power (kW) = 25 * 0.85 = 21.25 kW


(b) To calculate the input apparent power taken from the supply, we can use the formula:

Apparent Power (kVA) = Voltage (V) * Current (A) / 1000

However, since we are not given the current drawn by the motor, we cannot calculate the apparent power directly. If we are given the voltage though, we can use Ohm's law to find the current and then use the above formula. Let me know if you have that information.

(c) To improve the power factor from 0.85 to 0.99 lagging, we need to add a capacitor that will generate reactive power that cancels out the reactive power from the motor. The formula for the reactive power generated by a capacitor is:

Reactive Power (vars) = Capacitance (F) * Voltage (V)^2 * 2 * pi * Frequency (Hz)

Since we are not given the frequency, we cannot directly solve for the capacitance. However, we can use the following formula to relate the new power factor to the old power factor, with the help of the required reactive power (in vars) to improve the power factor:

cos(phi1) = cos(phi2) - Qp / S

where cos(phi1) = initial power factor (0.85), cos(phi2) = final power factor (0.99), Qp = required reactive power, and S = input apparent power.

We were not given S, but let's say that it is equal to the real power consumption of the motor (21.25 kW), since we know that the power factor is lagging. Then we can solve for Qp as follows:

Qp = (cos(phi2)-cos(phi1)) * S = (0.99 - 0.85) * 21.25 = 2.98 kvar

Now that we have the required reactive power, we can use the formula for the reactive power generated by a capacitor to solve for the required capacitance:

Capacitance (F) = Qp / (Voltage (V)^2 * 2 * pi * Frequency (Hz))

Again, we don't have the frequency, but if we assume a typical value of 50 Hz for the power supply frequency, we can solve for the capacitance:

Capacitance (F) = 2.98 / (Voltage (V)^2 * 2 * pi * 50) = 1.9E-5 Vars / V^2

So the required capacitance is 1.9E-5 farads, or approximately 19 microfarads.

Answer 2

(a) The real power (kW) consumed by the motor can be calculated using the formula:

P = S x pf

where P is the real power in kilowatts (kW), S is the apparent power in kilovolt-amperes (kVA), and pf is the power factor.

Given that the motor has a rating of 25 kVA and a power factor of 0.85 lagging, we have:

P = 25 kVA x 0.85 = 21.25 kW

Therefore, the real power consumed by the motor is 21.25 kW.

(b) The input apparent power (S) taken from the supply can be calculated using the formula:

S = P / pf

where P is the real power in kilowatts (kW), and pf is the power factor.

Given that the motor has a rating of 25 kVA and a power factor of 0.85 lagging, we have:

S = 21.25 kW / 0.85 = 25 kVA

Therefore, the input apparent power taken from the supply is 25 kVA.

(c) The rating (in Vars) of the capacitor required to improve the power factor from 0.85 to 0.99 lagging can be calculated using the formula:

C = (P x (tan θ1 - tan θ2)) / V^2

where C is the capacitance in farads (F), P is the real power in watts (W), θ1 is the original power factor angle (cos^-1(pf1)), θ2 is the final power factor angle (cos^-1(pf2)), and V is the voltage in volts (V).

Given that the motor has a rating of 25 kVA, a power factor of 0.85 lagging, and the desired power factor is 0.99 lagging, we have:

P = 21.25 kW = 21,250 W
θ1 = cos^-1(0.85) = 31.78°
θ2 = cos^-1(0.99) = 8.11°
V = unknown

To find the voltage, we need to use the apparent power formula:

S = V x I

where S is the apparent power in volt-amperes (VA), V is the voltage in volts (V), and I is the current in amperes (A).

Given that the input apparent power is 25