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The total power delivered to a balanced throe phase load when operating at a line voliage of 66003V is 1188kW at a lagging power factor of 0.6. The impedance of the distribution line ing the load is 0.5+j4Ω/ϕ. 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 1920kVAR of magnetizing reactive power when operated at a line voltage of 66003V 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 66003V 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 60Hz what is the size of each capacitor in microfarads?

Short Answer

Expert verified
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 Xc.

Step by step solution

01

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=3VLILpf, where VL=66003 V is the line voltage, and pf=0.6 is the power factor. Rearranging the formula to find IL gives:IL=P3VLpf=1188×1033660030.6=165.1 A
02

Determine Voltage Drop and Sending End Voltage Without Capacitor

The line impedance is given as Z=0.5+j4 Ω/ϕ. The voltage drop ΔV across the line impedance can be calculated using ΔV=ILZ:ΔV=165.1×(0.5+j4)=82.55+j660.4 VThe magnitude of the voltage drop is:|ΔV|=82.552+660.42=665 VThus, the sending end voltage Vs is:Vs=VL+|ΔV|=66003+66512257 V
03

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 IL is given by:The correction reactive power reduces Q by 1920 kVAR:QloadQcap=(Ptan(cos10.6)1920)Recalculate the sending end voltage using recalculated current IL=P3VL×1new pf and voltage drop.
04

Calculate Voltage Drop With Capacitor

Compute the voltage drop again using the new current and previously calculated line impedance:ΔV=IL×ZCompute magnitude of ΔV, and use it to find new Vs:Vs=VL+|ΔV|
05

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:η=PP+ΔPlines
06

Calculate Average Power Efficiency With Capacitor

Repeat efficiency calculation, using reduced input power when capacitor is introduced.η=PP+ΔPlines
07

Calculate Capacitor Values

Capacitors supply reactive power given at line voltage. Reactance Xc is given by:Xc=V2QcFrom Xc=12πfC, solve for C:C=12πfXc

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 66003 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.

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Most popular questions from this chapter

A three-phase Δ -connected generator has an internal impedance of 0.6+j4.8Ω/ϕ. When the load is removed from the generator, the magnitude of the terminal voltage is 34,500V. The generator feeds a Δ -connected load through a transmission line with an impedance of 0.8+j6.4Ω/ϕ. The per-phase impedance of the load is 2877j864Ω a) Construct a single-phase equivalent circuit. b) Calculate the magnitude of the line current. c) Calculate the magnitude of the line voltage at the terminals of the load. d) Calculate the magnitude of the line voltage at the terminals of the source. e) Calculate the magnitude of the phase current in the load. f ) Calculate the magnitude of the phase current in the source.

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At full load, a commercially available 200 hp, threephase induction motor operates at an efficiency of 96% and a power factor of 0.92 lag. The motor is supplied from a three-phase outlet with a line-voltage rating of 208V a) What is the magnitude of the line current drawn from the 208V outlet? (1hp=746W.) b) Calculate the reactive power supplied to the motor.

A balanced three-phase source is supplying 90kVA at 0.8 lagging to two balanced Y-connected parallel loads. The distribution line connecting the source to the load has negligible impedance. Load 1 is purely resistive and absorbs 60kW. Find the per-phase impedance of Load 2 if the line voltage is 415.69V and the impedance components are in series.

A balanced three-phase source is supplying 1800kVA at 0.96 pf lead to two balanced Y-connected parallel loads. The distribution line connecting the source to the load has negligible impedance. The power associated with load 1 is 192+j1464kVA. a) Determine the impedance per phase of load 2 if the line voltage is 64003V and the impedance components are in series. b) Repeat (a) with the impedance components in parallel.

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