Question

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

The total complex power at the sending end is the sum of load and line powers. Percentage of power delivered to loads is calculated from their real power fractions.

Step-by-step solution

Calculate Load 1 Complex Power

Load 1 is Y-connected with an impedance of \(300 + j100 \ \Omega/\phi\). The line-to-neutral voltage \( V_{LN} = 7.2 \ kV = 7200 \ V \). The phase current for Load 1 is \( I_{L1} = \frac{V_{LN}}{Z_1} = \frac{7200}{300+j100} \). Compute \( I_{L1} \):\[I_{L1} = \frac{7200}{300+j100} = 23.0769 - j7.6923 \ \text{A}\]The complex power for Load 1 is \( S_1 = 3 \cdot V_{LN} \cdot I_{L1}^* \), where \(I_{L1}^*\) is the complex conjugate of \(I_{L1}\):\[S_1 = 3 \cdot 7200 \cdot (23.0769 + j7.6923)\]

Calculate Load 2 Complex Power

Load 2 is \(\Delta\)-connected with an impedance of \(5400-j2700\ \Omega/\phi\). Convert to Y-equivalent:\[Z_{2\ (Y)} = \frac{5400-j2700}{3} = 1800-j900\ \Omega/\phi\]The phase current for Load 2 is:\[I_{L2} = \frac{7200}{1800-j900} = 3.6923 + j1.8462 \ \text{A}\]Calculate \( S_2 = 3 \cdot V_{LN} \cdot I_{L2}^* \):\[S_2 = 3 \cdot 7200 \cdot (3.6923 - j1.8462)\]

Convert Load 3 kVA to Complex Power

Load 3 gives apparent power directly as \(112.32 + j95.04\ KVA\). Convert this to complex power:\[ S_3 = (112.32 + j95.04) \ \mathrm{kVA} \]

Calculate Total Load Complex Power

Add the calculated complex powers of all three loads:\[ S_{total\ loads} = S_1 + S_2 + S_3 \]

Include Line Impedance

The line impedance is \(1 + j10 \ \Omega/\phi\). Total current \(I_{total}\) is the sum of currents through loads:\[ I_{total} = I_{L1} + I_{L2} - j \cdot \frac{S_3}{V_{LN}} \]Calculate the complex power loss on the line \( S_{line} = 3 \cdot I_{total}^2 \cdot (1+j10)\)

Calculate Total Complex Power at Sending End

Sum the power delivered to the loads and the line loss:\[ S_{sending} = S_{total\ loads} + S_{line} \]

Calculate Delivered Power Percentage

Calculate the real power portion of \(S_{sending}\):\[ P_{sending} = \text{Re}(S_{sending}) \]Calculate the total delivered real power \(P_{delivered} = \text{Re}(S_{total\ loads})\).Compute the percentage:\[ \text{Percentage} = \frac{P_{delivered}}{P_{sending}} \times 100\]

Simplify and State Final Results

Using previous calculations, finalize the value for total power at the sending end and the percentage delivered to loads.

Key concepts

Complex Power Calculation

Complex power is a fundamental concept in three-phase power systems, representing the combination of real and reactive power. It's crucial for analyzing the power flow in electrical circuits. Complex power is expressed as \( S = P + jQ \), where \( P \) is the real power (measured in watts), and \( Q \) is the reactive power (measured in vars).
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:

• \( S = 3 \, V_{LN} \, I^* \)

Here, \( V_{LN} \) is the line-to-neutral voltage, and \( I^* \) is the complex conjugate of the current. In this exercise, we calculate the complex power for each of the connected loads separately: the \( Y \)-connected load (Load 1), the \( \Delta \)-connected load (Load 2), and directly provided apparent power (Load 3). Understanding how to solve \( S \) accurately facilitates deeper insights into the total power distribution in a circuit.

Impedance in Y and Delta Connections

The arrangement of components in a circuit can greatly influence its behavior. In three-phase systems, connections can be made in two common configurations: \( Y \) (wye) and \( \Delta \) (delta). Understanding these connections is key to efficiently analyzing circuits.

• **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.

Mastering how to handle these configurations allows for better analysis and troubleshooting of electrical networks.

Power Factor and Efficiency

Power factor and efficiency are both vital when assessing the performance of an electrical system. They help determine how effectively the system is converting electrical energy into useful work without unnecessary waste.

• **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.

Monitoring and improving power factor and efficiency reduces wastage, leading to financial and energy savings.

Three-phase Load Analysis

Analysis of three-phase loads involves examining the power consumed, generated, and lost within a three-phase power system. This is crucial for ensuring system reliability and efficiency.

• **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.

By understanding the interactions between loads and the grid, engineers can design better systems to handle demand without excessive energy losses.