Question

A balanced three-phase distribution line has an impedance of $1+j5\mathrm{\Omega }/\varphi$. This line is used to supply three balanced three-phase loads that are connected in parallel. The three loads are ${\mathrm{L}}_1=75\mathrm{kVA}$ at 0.96 pf $\mathrm{lead},{\mathrm{L}}_2=150\mathrm{kVA}$ at 0.80 pf $\mathrm{lag},$ and ${\mathrm{L}}_3=168\mathrm{kW}$ and $36\mathrm{kVAR}(\mathrm{mag}-$ netizing). The magnitude of the line voltage at the terminals of the loads is $2500\sqrt{3}\mathrm{V}$ a) What is the magnitude of the line voltage at the sending end of the line? b) What is the percent efficiency of the distribution line with respect to average power?

Short answer

Use complex power to find total line losses; add it to load voltages to find sending voltage.

Step-by-step solution

Calculate Load Currents

For each load, we need to find the current drawn from the supply. Convert power factor and apparent power to complex power using the formula: $S=P+jQ$, where $S$ is the complex power, $P$ is real power, and $Q$ is reactive power. Use the equation $S=V_{LL}\times I^{\ast }$ where $I^{\ast }$ is the complex conjugate of the current.1. **Load L1**: $S_1=75\text{kVA}$ at 0.96 pf lead: - $Q_1=S_1\times \sin ({\cos }^{-1}(0.96))=75\times \sin ({\cos }^{-1}(0.96))=75\times 0.28=21\text{kVAR}$ - $P_1=S_1\times \mathrm{pf}=75\times 0.96=72\text{kW}$ - Current: $I_1=\frac{S_1}{\sqrt{3}\times V_{LL}}=\frac{75}{\sqrt{3}\times 2500}=17.3\text{A}\mathrm{\angle }-16.26°$2. **Load L2**: $S_2=150\text{kVA}$ at 0.80 pf lag: - $Q_2=S_2\times \sin ({\cos }^{-1}(0.8))=150\times \sin ({\cos }^{-1}(0.8))=150\times 0.6=90\text{kVAR}$ - $P_2=S_2\times \mathrm{pf}=150\times 0.8=120\text{kW}$ - Current: $I_2=\frac{S_2}{\sqrt{3}\times V_{LL}}=\frac{150}{\sqrt{3}\times 2500}=34.64\text{A}\mathrm{\angle }36.87°$3. **Load L3**: $P_3=168\text{kW},Q_3=36\text{kVAR}$ magnetizing: - $S_3=\sqrt{P_3^2+Q_3^2}=\sqrt{{168}^2+{36}^2}=171.8\text{kVA}$ - Current: $I_3=\frac{S_3}{\sqrt{3}\times V_{LL}}=\frac{171.8}{\sqrt{3}\times 2500}=39.6\text{A}\mathrm{\angle }11.85°$ (lagging due to magnetizing)

Key concepts

Three-Phase Loads

When dealing with electric power systems, three-phase loads are commonly used due to their efficiency and reliability. In a three-phase system, electrical power is delivered through three wires, each carrying alternating current, with a phase difference of 120 degrees between each pair.
This setup offers a more consistent power supply and allows for smaller and more cost-effective conductors compared to single-phase systems.
Three-phase loads often consist of multiple devices connected in parallel, drawing power from a balanced three-phase distribution line. Each load's power demand is represented in either kilovolt-amperes (kVA) or kilowatts (kW), along with their power factor.

• Load L1 draws 75 kVA at a 0.96 leading power factor.
• Load L2 draws 150 kVA at a 0.80 lagging power factor.
• Load L3 draws real power of 168 kW with reactive power of 36 kVAR.

This configuration allows efficient transmission of power to multiple devices or systems running concurrently.

Power Factor

The power factor is a crucial concept in the optimization of electric power systems. It measures how effectively electrical power is converted into useful work output and is typically expressed as a decimal between 0 and 1.
The power factor is calculated by dividing the real power (in kW) by the apparent power (in kVA) used in the circuit. A power factor of 1 (or 100%) indicates that all the energy supplied by the source is being effectively used for productive work.
Loads with a low power factor draw more current than necessary to achieve the same output, leading to energy inefficiencies. Understanding and correcting power factor in a system can help reduce energy costs and increase system capacity.

• Leading power factors (e.g., 0.96 lead) are typically associated with capacitive loads.
• Lagging power factors (e.g., 0.80 lag) are often found in inductive loads.

A high power factor is desired to increase the efficiency of power distribution.

Complex Power Calculation

In electric power systems, complex power is an essential concept that combines real and reactive power into a single phasor quantity. This phasor is represented as $S=P+jQ$, where $P$ represents real power, and $Q$ represents reactive power.
Real power (measured in kilowatts, kW) does the actual work, such as lighting, heating, or turning a motor. In contrast, reactive power (measured in kilovolt-amperes reactive, kVAR) supports the electric field necessary for certain equipment's magnetic functions, like transformers or motors.

• Load L1 has a complex power $S_1$ with 72 kW real power, and 21 kVAR reactive power.
• Load L2 operates with a complex power $S_2$ consisting of 120 kW real power and 90 kVAR reactive power.
• Load L3 uses 168 kW of real power with a notable reactive power component of 36 kVAR.

Calculating complex power for each load helps determine the current drawn from the supply.

Line Voltage

Line voltage is a fundamental parameter in three-phase power systems. It refers to the voltage between any two lines in a three-phase system and is often denoted as $V_{LL}$. For the three-phase loads in the exercise, the line voltage at the terminals is $2500\sqrt{3}\text{ V}$.
The magnitude of line voltage is critical as it influences the current that flows through the system and the overall power distribution efficiency.
High line voltage can reduce current for a fixed power load, which decreases resistive losses and improves efficiency. Understanding line voltage and maintaining it within proper limits ensures the safe and efficient operation of electrical equipment.
The line voltage affects the calculated current for each load, highlighting its role in determining the overall stability of the power delivery system.

Distribution Efficiency

Distribution efficiency in an electric power system refers to the ratio of the average power delivered to the loads to the power supplied by the power station, expressed as a percentage. It measures how effectively the power is transmitted over the distribution line.
Electricity distribution often involves some losses due to the line's impedance and other factors like heat dissipation, which decreases the power system's efficiency.

• The efficiency is calculated by comparing the total real power delivered to the loads to the input power required at the sending end of the distribution line.
• Ensuring a high distribution efficiency minimizes operational costs and enhances energy utilization.

In scenarios where the power factor is high and impedance is low, reduced losses can lead to better efficiency, exemplified by our balanced three-phase loads.