Question

The three pieces of computer equipment described below are installed as part of a computation center. Each piece of equipment is a balanced three-phase load rated at \(208 \mathrm{V}\). Calculate (a) the magnitude of the line current supplying these three devices and (b) the power factor of the combined load. Hard Drive: \(5.742 \mathrm{kW}\) at 0.82 pf lag CD/DVD drive: \(18.566 \mathrm{kVA}\) at 0.93 pf lag CPU: line current \(81.6 \mathrm{A}, 11.623 \mathrm{kVAR}\)

Short answer

Line current is approximately 118.35 A with a power factor of 0.79 lagging.

Step-by-step solution

Convert the Hard Drive Load to Complex Power

The formula to convert real power (P) and power factor (\text{pf}) to complex power (S) is \[ S = \frac{P}{\text{pf}} \text{ (kVA)} \]. For the hard drive: \[ S_{\text{hard drive}} = \frac{5.742}{0.82} = 7.0049 \text{ kVA} \]. Real power (P) = 5.742 kW, reactive power (Q) = \( Q_{\text{hard drive}} = \sqrt{S^2 - P^2} = \sqrt{7.0049^2 - 5.742^2} \approx 3.92 \\text{ kVAR}\).

Calculate Total Load

Calculate the total complex power by adding the complex power of all devices. Total complex power (S_t) is computed as: \[ S_{\text{Total}} = S_{\text{hard drive}} + S_{\text{CD/DVD}} + S_{\text{CPU}} \].Where:\( S_{\text{CD/DVD}} = 18.566\text{ kVA}\),\( S_{\text{CPU}} = \sqrt{(208 * 81.6)^2 - (11.623)^2} + j11.623 \approx 16.9744 \text{ kVA}\).Hence:\[ S_{\text{Total}} = \sqrt{(5.742 + 16.9744)^2 + (3.92 + 11.623)^2} \approx 42.5453 \text{ kVA}\].

Calculate Line Current

Use the formula \( I = \frac{S_t}{\sqrt{3} \times V} \) to find the line current.\[ I = \frac{42.5453 \times 10^3}{\sqrt{3} \times 208} \approx 118.35 \text{ A} \].

Determine Power Factor

Calculate total real power (P_t) and reactive power (Q_t) of the system:\[ P_t = 5.742 + (18.566 \times 0.93) + \text{(CPU Real Power)} \approx 33.68 \text{ kW} \]. \[ Q_t = 3.92 + 18.566 - (18.566 \times 0.93) + 11.623 \approx 19.49 \text{ kVAR} \].The power factor is calculated as:\[ \text{Power Factor} = \frac{P_t}{S_t} = \frac{33.68}{42.5453} \approx 0.79 \text{ lagging} \].

Key concepts

Complex Power

When dealing with three-phase loads, it is important to understand the concept of complex power, denoted often as \( S \). Complex power is a combination of real power (P), which is the actual power consumed by the device, and reactive power (Q), which is the power stored and released by the device's reactive components, such as inductors and capacitors. These two components combine to form the complex power using the equation \[ S = P + jQ \]. Here \( j \) denotes the imaginary unit. This relationship shows how both real and reactive power affect the power consumption of an electrical device.

In practice, complex power is represented in kilo-volt-amperes (kVA). To convert from real power (measured in kilowatts, kW) and power factor to complex power, we use the equation \[ S = \frac{P}{\text{pf}} \], where \( \text{pf} \) is the power factor. An example calculation for a hard drive shows this application; if the real power is 5.742 kW at a power factor of 0.82, the complex power is calculated as \( 7.0049 \text{ kVA} \). This step is crucial for understanding the demand each piece of equipment places on the system.

Power Factor

Power factor is an essential concept when examining electrical loads. It is a measure of how effectively electrical power is being converted into useful work output. In more technical terms, it is the ratio of real power (\( P \)) to apparent power (\( S \)), expressed as: \[ \text{pf} = \frac{P}{S} \]. The power factor value ranges between 0 and 1, where a value closer to 1 indicates more efficient usage of electrical power.

For example, in a system with total real power \( \approx 33.68 \text{ kW} \) and total apparent power \( \approx 42.5453 \text{ kVA} \), the power factor can be calculated as \( \frac{33.68}{42.5453} \approx 0.79 \). This indicates a lagging power factor due to the presence of inductive loads, commonly found in motors and other similar equipment.

Correcting a low power factor can be beneficial as it reduces the load on the power system, leading to economic savings and more efficient operation. However, understanding the existing power factor is the first necessary step before considering correction strategies.

Line Current

Understanding line current is crucial when designing or analyzing electrical systems as it impacts everything from the sizing of wires to safety protocols. Line current is the current flowing through one conductor of a three-phase power system and is typically measured in amperes (A).

The formula to determine the line current in a three-phase system is given by \[ I = \frac{S_t}{\sqrt{3} \times V} \], where \( S_t \) is the total complex power of the system, and \( V \) is the line voltage. For a system with a total complex power of 42.5453 kVA at a voltage of 208 V, the line current can be calculated as \( \approx 118.35 \text{ A} \).

This calculated current ensures that the power delivery to the loads is adequate without causing overload conditions. Proper current levels also maintain the efficiency and reliability of the electrical power distribution system.

Reactive Power

Reactive power is a key concept in the world of AC electrical circuits, especially in systems involving inductors and capacitors. It is measured in kilo-volt-amperes reactive (kVAR) and represents the power that does not perform any work but is necessary for maintaining the magnetic fields in inductive loads.

The formula to calculate reactive power from complex power involves using the expression \( Q = \sqrt{S^2 - P^2} \). For instance, the hard drive discussed earlier has a reactive power of approximately 3.92 kVAR, calculated from its complex power of 7.0049 kVA and real power of 5.742 kW.

Reactive power leads to a phase difference between voltage and current, hence affecting the power factor. While it does not contribute to actual work output, managing reactive power is crucial for ensuring efficient power system operation, minimizing losses, and reducing the burden on transmission and distribution systems.