Question

The total power delivered to a balanced throe phase load when operating at a line voliage of $6600\sqrt{3}\mathrm{V}$ is $1188\mathrm{kW}$ at a lagging power factor of $0.6.$ The impedance of the distribution line ing the load is $0.5+j4\mathrm{\Omega }/\varphi$. Under these supply operating conditions, the drop in the magnitude of the line voltage between the sending end and the load end of the line is excessive. To compensate, a bank of A-connected capacitors is placed in parallel with the load. The capacitor bank is designed to furnish $1920\mathrm{kVAR}$ of magnetizing reactive power when operated at a line voltage of $6600\sqrt{3}\mathrm{V}$ a) What is the magnitude of the voltage at the sending end of the line when the load is operat. ing at a line voltage of $6600\sqrt{3}\mathrm{V}$ and the capac. itor bank is disconnected? b) Repeat (a) with the capacitor bank connected c) What is the average power efficiency of the line in (a)? d) What is the average power efficiency in $(b)?$ e) If the system is operating at a frequency of $60\mathrm{Hz}$ what is the size of each capacitor in microfarads?

Short answer

a) Approx. 12257 V, b) Recalculate with new current, c) Calculate using power formula, d) Repeat with reduced input, e) Solve for C using corrected $X_c$.

Step-by-step solution

Calculate the Load Current

The power delivered to the load is given as 1188 kW. We use the formula for power in a three-phase system: $P=\sqrt{3}\cdot V_L\cdot I_L\cdot \text{pf}$, where $V_L=6600\sqrt{3}$ V is the line voltage, and $\text{pf}=0.6$ is the power factor. Rearranging the formula to find $I_L$ gives:$$I_L=\frac{P}{\sqrt{3}\cdot V_L\cdot \text{pf}}=\frac{1188\times {10}^3}{\sqrt{3}\cdot 6600\cdot \sqrt{3}\cdot 0.6}=165.1\text{ A}$$

Determine Voltage Drop and Sending End Voltage Without Capacitor

The line impedance is given as $Z=0.5+j4$ $\mathrm{\Omega }/\varphi$. The voltage drop $\mathrm{\Delta }V$ across the line impedance can be calculated using $\mathrm{\Delta }V=I_L\cdot Z$:$$\mathrm{\Delta }V=165.1\times (0.5+j4)=82.55+j660.4\text{ V}$$The magnitude of the voltage drop is:$$|\mathrm{\Delta }V|=\sqrt{{82.55}^2+{660.4}^2}=665\text{ V}$$Thus, the sending end voltage $V_s$ is:$$V_s=V_L+|\mathrm{\Delta }V|=6600\sqrt{3}+665\approx 12257\text{ V}$$

Calculate Sending End Voltage With Capacitor

With the capacitor connected, the effective load power factor is corrected. The reactive power supplied by the capacitors is 1920 kVAR. The total reactive power is reduced, thus reducing the magnitude of the current through the inductor. The new current $I_L^{\prime }$ is given by:The correction reactive power reduces $Q$ by 1920 kVAR:$$Q_{load}-Q_{cap}=(P\tan ({\cos }^{-1}0.6)-1920)$$Recalculate the sending end voltage using recalculated current $I_L^{\prime }=\frac{P}{\sqrt{3}\cdot V_L}\times \frac{1}{\text{new pf}}$ and voltage drop.

Calculate Voltage Drop With Capacitor

Compute the voltage drop again using the new current and previously calculated line impedance:$$\mathrm{\Delta }{V}^{\prime }=I_L^{\prime }\times Z$$Compute magnitude of $\mathrm{\Delta }{V}^{\prime }$, and use it to find new $V_s^{\prime }$:$$V_s^{\prime }=V_L+|\mathrm{\Delta }{V}^{\prime }|$$

Calculate Average Power Efficiency Without Capacitor

Efficiency is the ratio of output power to input power. Without the capacitor, the input power is higher due to the full reactive power. Efficiency can be found as:$$\eta =\frac{P}{P+\mathrm{\Delta }{P}_{lines}}$$

Calculate Average Power Efficiency With Capacitor

Repeat efficiency calculation, using reduced input power when capacitor is introduced.$${\eta }^{\prime }=\frac{P}{P+\mathrm{\Delta }{P}_{lines}^{\prime }}$$

Calculate Capacitor Values

Capacitors supply reactive power given at line voltage. Reactance $X_c$ is given by:$$X_c=\frac{V^2}{Q_c}$$From $X_c=\frac{1}{2\pi fC}$, solve for $C$:$$C=\frac{1}{2\pi f{X}_c}$$

Key concepts

Power Factor

In electrical circuits, the power factor is a measure of how efficiently the electric power is being used. It's represented as a decimal number or percentage, with 1 or 100% being ideal. It is the ratio between the real power, measured in kilowatts (kW), and the apparent power, measured in kilovolt-amperes (kVA). A lagging power factor, such as 0.6 in this scenario, indicates that the current lags behind the voltage, meaning that there's more reactive power in the system.

• Real Power (kW): The actual power consumed by the electrical equipment to perform useful work.
• Reactive Power (kVAR): The power that oscillates between the source and the reactive components of the load. This power does not perform any useful work but is necessary to maintain the electric and magnetic fields in the circuit.
• Apparent Power (kVA): The combination of real and reactive power, calculated as the product of current and voltage in the circuit.

Improving the power factor means decreasing the reactive power in the system, thereby making power use more efficient. This is often achieved by adding a capacitor bank, which counteracts the lagging current components.

Three-Phase System

Three-phase systems are more efficient for transmitting electrical power, especially over long distances. In a three-phase system, three electrical currents are used, each out of phase with the other by 120 degrees. This configuration allows for a more balanced load and efficient energy transfer.

• Balance Load: Since the phases are equal and separated by equal angles, the power delivered is constant and balanced.
• More Power: Compared to a single-phase system, a three-phase system can deliver more power with less conductor material.
• Better Efficiency: Due to the balanced nature, three-phase systems have less energy loss and are more efficient.

The given exercise references a line voltage of $6600\sqrt{3}$ volts, which relates to the line-to-line voltage in a three-phase system. Understanding three-phase systems is fundamental to handling larger loads as they offer a consistent power delivery.

Reactive Power

Reactive power, measured in kilovolt-amperes reactive (kVAR), doesn't contribute to actual work done in the circuit but is essential for maintaining voltage levels necessary for current to move. It is mainly associated with the energy storage in the magnetic fields of inductors and is typically considered a "wasteful" power if not managed correctly.

• Phase Difference: It arises when there's a phase difference between voltage and current, as in inductive and capacitive elements.
• Inductive Loads: Common sources of reactive power, causing the current to lag the voltage.
• Managing Reactive Power: Capacitor banks are often used to supply reactive power, reducing the inductive effects.

In the exercise, with a power factor of 0.6, a significant amount of reactive power is present in the circuit. Correcting the power factor with a capacitor bank reduces this reactive power, which collectively enhances the system's efficiency.

Capacitor Bank

A capacitor bank is a grouping of individual capacitors used to store and provide reactive power to an electrical system. Its primary role is to improve the power factor of a system by offsetting the inductive loads, thus optimizing the use of power from a supply line.

• Power Factor Correction: By supplying reactive power, capacitor banks minimize the phase difference between voltage and current, correcting a lagging power factor.
• Voltage Stabilization: Capacitor banks help maintain a steady voltage level by providing leading reactive power that counteracts the lagging current components.
• Efficiency Improvement: They reduce losses in the distribution system, resulting in more available active power.

In the exercise, the capacitor bank is sized to deliver 1920 kVAR of reactive power, which significantly contributes to reducing the voltage drop across the line, thereby stabilizing the voltage at the load end and improving the system's efficacy.