Question

Two \(660 \mathrm{V}\) (rms) loads are connected in parallel. The two loads draw a total average power of \(52,800 \mathrm{W}\) a power factor of 0.8 leading. One of the loads draws \(40 \mathrm{kVA}\) at a power factor of 0.96 lagging. What is the power factor of the other load?

Short answer

The power factor of the other load is 0.5538 leading.

Step-by-step solution

Calculate Total Apparent Power

First, we calculate the total apparent power drawn by the loads. Given the total average power is \(52,800 \mathrm{W}\) and the power factor is 0.8 (leading), we can use the formula: \[ S_{\text{total}} = \frac{P_{\text{total}}}{\text{Power Factor}} = \frac{52,800}{0.8} = 66,000 \mathrm{VA}. \]

Identify Apparent Power for Load 1

One of the loads consumes \(40 \mathrm{kVA}\) at a power factor of 0.96 lagging. This means \(S_1 = 40,000 \mathrm{VA}\), and the active power (\(P_1\)) can be calculated using: \[ P_1 = S_1 \times \text{Power Factor} = 40,000 \times 0.96 = 38,400 \mathrm{W}. \]

Calculate Apparent and Active Power for Load 2

The apparent power of Load 2 \(S_2\) can be found by subtracting \(S_1\) from \(S_{\text{total}}\): \[ S_2 = S_{\text{total}} - S_1 = 66,000 - 40,000 = 26,000 \mathrm{VA}. \] The active power for Load 2 \(P_2\) is \(P_2 = P_{\text{total}} - P_1 = 52,800 - 38,400 = 14,400 \mathrm{W}.\)

Determine Power Factor of Load 2

The power factor of Load 2 (\( \text{Power Factor}_2 \)) can be determined by taking the ratio of the active power to the apparent power: \[ \text{Power Factor}_2 = \frac{P_2}{S_2} = \frac{14,400}{26,000} = 0.5538. \] Since the total power factor is leading while Load 1 has a lagging power factor, Load 2 must counteract with a leading power factor of approximately 0.5538.

Key concepts

Apparent Power

Apparent power is an important concept in electrical engineering, especially when dealing with AC circuits. It combines both active and reactive power, representing the total power flow in a circuit. Apparent power is measured in Volt-Amperes (\mathrm{VA}). In simple terms, it helps identify the capacity an electrical system can handle without considering the phase difference between voltage and current.

• Apparent power (S) is calculated using the voltage (V) and the current (I): \( S = V imes I \) .
• For systems with a given power factor, apparent power can also be determined using the formula \( S = \frac{P}{\text{Power Factor}} \), where P is the active power.

Understanding apparent power is crucial as it determines how much load a system can support. In the given exercise, the total apparent power was calculated to ensure the loads are within the system's capabilities.

Power Factor

Power factor is a measure of how effectively electrical power is being used. It's a ratio between active power, which does useful work, and apparent power. The power factor value ranges between -1 and 1 and it indicates the phase difference between voltage and current.

• A power factor of 1 (or 100%) means all supplied energy is used effectively for work.
• A low power factor suggests that more apparent power is required to perform the same amount of work.
• If the current lags behind the voltage, the power factor is said to be lagging. If the current leads the voltage, the power factor is leading.

For the exercise, the power factor played a critical role in determining the unknown values. Knowing the power factor of one load and the total helped calculate the unknown power factor of the other load.

Active Power

Active power is the real power consumed in an electrical circuit. It's the portion of power that performs actual, useful work and is measured in Watts (\mathrm{W}). Active power does not get wasted on components like inductors and capacitors, which would otherwise draw power but not use it effectively for work.

• The active power is directly used to perform the intended task in the circuit, like lighting or heating.
• It can be calculated using the apparent power and the power factor as \( P = S \times \text{Power Factor} \).

Understanding active power assists in allocating adequate power for useful applications without surpassing designed limits. In the given scenario, calculating the active power for the second load, after knowing the first, played a vital role in determining the power factor for efficiency assessment.