Question

A factory has an electrical load of \(1800 \mathrm{kW}\) at a lag ging power factor of \(0.6 .\) An additional variable power factor load is to be added to the factory, The new load will add \(600 \mathrm{kW}\) to the real power load of the factory. The power factor of the added load is to be adjusted so that the overall power factor of the factory is 0.96 lagging. a) Specify the reactive power associated with the added load. b) Does the added load absorb or deliver magnet izing vars? c) What is the power factor of the additional load? d) Assume that the voltage at the input to the fac tory is \(4800 \mathrm{V}(\mathrm{rms})\). What is the rms magnitud of the current into the factory before the val able power factor load is added? c) What is the rms magnitude of the current in 100 the factory after the variable power factor has been added?

Short answer

a) -1400 kVAR, b) Delivers vars (capacitive), c) 0.395 (leading), d) 0.625 kA, e) 0.521 kA.

Step-by-step solution

Calculate Initial Reactive Power

First, calculate the initial reactive power (Q) of the factory's original load using the power factor and real power. Use the formula for power factor: \( \text{Power Factor} = \frac{P}{S} \), where \( P \) is real power and \( S \) is apparent power. Since \( S = \frac{P}{\text{Power Factor}} \) and \( Q = \sqrt{S^2 - P^2} \),\[ S_\text{initial} = \frac{1800}{0.6} = 3000 \text{ kVA} \]\[ Q_\text{initial} = \sqrt{3000^2 - 1800^2} = 2400 \text{ kVAR} \].

Calculate Total Real Power After Load Addition

Add the new real power load to the initial load to find the total real power (P_total).\[ P_\text{total} = 1800 + 600 = 2400 \text{ kW} \]

Calculate Total Apparent Power with Desired Power Factor

Calculate the total apparent power (S_total) required for the overall desired power factor.Use the formula: \( \text{Power Factor} = \frac{P}{S} \).\[ S_\text{total} = \frac{2400}{0.96} = 2500 \text{ kVA} \]

Calculate Total Reactive Power for Overall Power Factor

Calculate the total reactive power (Q_total) for the desired overall power factor.\[ Q_\text{total} = \sqrt{S_\text{total}^2 - P_\text{total}^2} = \sqrt{2500^2 - 2400^2} = 1000 \text{ kVAR} \]

Determine Reactive Power for Added Load

Subtract the initial reactive power from the total reactive power to find the reactive power of the added load.\[ Q_\text{added} = Q_\text{total} - Q_\text{initial} = 1000 - 2400 = -1400 \text{ kVAR} \] Since the reactive power is negative, the load is capacitive (it delivers vars).

Calculate Power Factor of the Added Load

Determine the power factor of the added load using its real and reactive power.Using \( \text{Power Factor} = \frac{P_\text{added}}{S_\text{added}} \), and \( S_\text{added}^2 = P_\text{added}^2 + Q_\text{added}^2 \),\[ S_\text{added} = \sqrt{600^2 + (-1400)^2} = 1520 \text{ kVA} \]\[ \text{Power Factor} = \frac{600}{1520} = 0.395 \].Since \( Q_\text{added} \) is negative, the power factor is leading.

Calculate RMS Current Magnitude Before Load Addition

Determine the current before adding the load using \( S_\text{initial} = V \cdot I \).\[ I_\text{before} = \frac{S_\text{initial}}{V} = \frac{3000}{4800} = 0.625 \text{ kA} \].

Calculate RMS Current Magnitude After Load Addition

Determine the current after adding the load.\[ I_\text{after} = \frac{S_\text{total}}{V} = \frac{2500}{4800} = 0.521 \text{ kA} \].

Key concepts

Reactive Power

Reactive power, often represented by the symbol $Q$, is a key concept in AC electrical systems. Unlike real power, which performs actual work, reactive power oscillates between the source and the load. This oscillation occurs because reactive components, such as inductors and capacitors, store energy temporarily before releasing it. Thus, reactive power supports the electromagnetic fields required for the functioning of devices. Units of reactive power are Volt-Amperes Reactive (VAR). It's worth noting that reactive power does not contribute to the net energy required by electrical devices but is crucial for maintaining voltage levels necessary for stable operation.
In our scenario, the calculation of reactive power before and after adding load reveals whether additional components deliver or absorb energy. When the additional load introduced a negative reactive power, it indicated a capacitive nature, meaning it delivered vars and helped correct the power factor.

Real Power

Real power, denoted as $P$, is the component of electrical power that performs actual work, such as lighting up a bulb or driving an electric motor. Measured in kilowatts (kW), it represents the rate at which energy is converted into useful forms. Unlike reactive power, real power doesn't oscillate back and forth; instead, it steadily flows from the source to the load.
In the factory exercise, the initial real power is 1800 kW, with an additional 600 kW added later. It's important to calculate how these changes affect the overall system. Real power helps in understanding the efficiency with which an electrical system delivers energy. The increase in real power is directly associated with the improved work performance of the factory's equipment.

Apparent Power

Apparent power combines both real power and reactive power, symbolized as \(S\). It signifies the total power flowing from the source to the load and is measured in kilovolt-amperes (kVA). Apparent power is calculated using the formula \( S = \sqrt{P^2 + Q^2} \).
This value represents the total demand on an electrical system and is crucial in designing equipment that can handle specific loads effectively. For our exercise, determining the initial and final apparent power (3000 kVA and 2500 kVA, respectively) allowed us to assess how well the system could accommodate changes in load while maintaining stability. Apparent power is key in sizing electrical components such as transformers and generators.

Electrical Loads

Electrical loads are the devices or systems that consume electrical power. Loads can be categorized as resistive, inductive, or capacitive based on their composition. A mix of these loads influences the power factor of the entire system.

• Resistive loads, like heaters, have a power factor close to 1.0 as they use mainly real power.
• Inductive loads, like motors and transformers, need reactive power, thereby lowering the power factor.
• Capacitive loads store and release energy, which can help correct the power factor.

In our problem, the introduction of an additional load required adjusting its power factor to improve overall system efficiency. This act of correcting power factors is essential to reduce losses and avoid potential charges from utility companies for low power factors.

Voltage and Current Calculations

Voltage and current are the fundamental aspects of AC power systems. In our example, we used RMS (Root Mean Square) values, which are typical in AC circuits, to calculate currents before and after load changes. The RMS magnitude provides a measure of the continuous equivalent DC value.
Initially, the current before adding the load was determined as \( 0.625 \text{ kA} \) based on the apparent power and voltage level. After adding the additional load and correcting the power factor, the current decreased to \( 0.521 \text{ kA} \), indicating better efficiency and reduced losses. Thus, maintaining optimal RMS values through adjustments in load and power factor can lead to a more reliable and cost-effective electrical system.