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In three-phase power systems, the measure of effective power that combines both real and reactive power is referred to as apparent power. It's important to grasp this concept because it expresses how much power is "apparently" being used in a circuit, whether or not it does useful work. Apparent power is represented as "S" and measured in volt-amperes (VA).

• For our problem, the total apparent power provided by the source is specified as 90 kVA.
• To find the per-phase apparent power in a three-phase system, we divide this total by three, giving us 30 kVA per phase.

Apparent power is calculated using the formula: \[ S = V imes I^* \] where:

• "S" is the apparent power
• "V" is the voltage
• "I" is the current (conjugated for AC circuits)

Apparent power is not always indicative of power that does useful work due to the presence of reactive power.

Real power, also known as active power, is the portion of power that performs actual work in an electrical circuit. It's the power consumed by resistive elements of the circuit and is measured in watts (W). This is the power that you can directly observe by things like light and heat.

• The real power for the total system, based on a power factor of 0.8, is calculated to be 24 kW per phase.
• In our exercise, Load 1, which is purely resistive, absorbs 60 kW total, or 20 kW per phase.
• The remaining real power for Load 2 is calculated as 4 kW per phase, using the formula \(P = S \times \text{power factor}\).

It's critical to understand that real power considers only those current components that are in phase with the voltage.

Reactive power is the power that does not perform useful work but is important for the functioning of AC systems. It is the power associated with the inductive and capacitive load components in the system. Reactive power is denoted as "Q" and measured in reactive volt-amperes (VAR).

• In the given problem, the reactive power per phase for the total system is calculated as 18 kVAR per phase.
• Since Load 1 is purely resistive, it does not consume any reactive power, all of the system's reactive power is attributed to Load 2.

The relationship between apparent power, real power, and reactive power can be visualized as a right triangle where the hypotenuse represents apparent power \(S\), the base represents real power \(P\), and the height represents reactive power \(Q\). This comes from the Pythagorean theorem:
\[ S = \sqrt{P^2 + Q^2} \]
Understanding reactive power is essential because it affects the amount of current in the system and therefore the sizing of equipment and efficiency.

Power factor is a crucial concept in power systems as it represents the efficiency with which the apparent power is converted into real power. It is defined as the cosine of the phase angle \(\theta\) between the current and voltage. Power factor is a dimensionless number between -1 and 1.

• In our problem, the power factor is given as 0.8 lagging, indicating the presence of inductive loads.
• A lagging power factor means that the current lags behind the voltage, typical in systems with inductive elements.

Using the power factor, we can determine the real power as a proportion of the apparent power: \[ \text{Power Factor} = \frac{P}{S} \] For a power factor of 0.8, 80% of the apparent power is being converted to real power, while the remaining 20% is reactive. Improving the power factor reduces the phase difference, which improves the system's efficiency, minimizing losses.