Chapter 11: Problem 27
Three balanced three-phase loads are connected in parallel. Load 1 is \(Y\) -connected with an impedance of \(300+j 100 \Omega / \phi\); load 2 is \(\Delta\) -connected with an impedance of \(5400-j 2700 \Omega / \phi\); and load 3 is \(112.32+j 95.04 \mathrm{kVA} .\) The loads are fed from a distribution line with an impedance of \(1+j 10 \Omega / \phi\) The magnitude of the line-to- neutral voltage at the load end of the line is \(7.2 \mathrm{kV}\) a) Calculate the total complex power at the sending end of the line. b) What percentage of the average power at the sending end of the line is delivered to the loads?
Short Answer
Step by step solution
Calculate Load 1 Complex Power
Calculate Load 2 Complex Power
Convert Load 3 kVA to Complex Power
Calculate Total Load Complex Power
Include Line Impedance
Calculate Total Complex Power at Sending End
Calculate Delivered Power Percentage
Simplify and State Final Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Power Calculation
To find the complex power in a circuit, you need to know the voltage across the load and the current flowing through it. For a three-phase system, complex power \( S \) is often calculated using the formula:
- \( S = 3 \, V_{LN} \, I^* \)
Impedance in Y and Delta Connections
- **Y-connection**: The components are connected at a central point. This configuration allows the line voltage to be higher than the phase voltage, making it effective for long-distance power transmission. Impedance in this configuration is simpler to analyze as seen with Load 1 in our example.
- **Delta-connection**: Each component is connected in a loop resembling a triangle. Here, the line voltage equals the phase voltage, which is useful in different, usually shorter, transmission scenarios. To simplify calculations, \( \Delta \)-connected impedances can be converted to \( Y \) equivalents using the formula \( Z_{(Y)} = \frac{Z_{(\Delta)}}{3} \). This conversion simplifies the process of computing currents and power, as seen in Load 2 of the exercise.
Power Factor and Efficiency
- **Power Factor**: It is the ratio of real power \( P \) to apparent power \( S \). It's expressed as \( \cos(\theta) \), where \( \theta \) is the phase angle between voltage and current. A higher power factor indicates a more efficient system because more of the electricity is doing useful work rather than circulating reactively. In calculations, pay attention to the real component of complex power \( S \) when finding power factor.
- **Efficiency**: This is commonly analyzed by comparing the output power to the input power. In the context of the exercise, you might seek to determine what percentage of generated power gets delivered to and used by loads as opposed to being lost in transmission as heat in the lines. Calculate efficiency by subtracting line losses from the total power and dividing by total power to give clarity on system performance.
Three-phase Load Analysis
- **Balancing Loads**: It is essential to analyze each phase separately but also ensure overall balance across the system to prevent issues such as neutral conductor overheating.
- **Apparent Power**: For Load 3, apparent power is given directly as \( 112.32 + j95.04 \) kVA, making it a bit simpler when calculating the load's influence on the system's power draw.
- **Total Load Complex Power**: This involves summing individual powers from each load. Each load's contribution often starts with calculating the individual line currents and using them to find their complex power.
- **Line Losses**: Beyond the loads, it's important to account for power lost due to the line impedance. Analysing the entire system includes combining line and load powers to understand the sending-end requirements accurately.