Question

In a balanced three-phase system, the source has an abc sequence, is Y-connected, and \(\mathbf{V}_{\mathrm{an}}=120 / 20^{\circ} \mathrm{V} .\) The source feeds two loads, both of which are \(Y\) -connected. The impedance of load 1 is \(8+j 6 \Omega / \phi\). The complex power for the a \(\cdot\) phase of load 2 is \(600 / 36^{\circ} \mathrm{VA}\). Find the total complex power supplied by the source.

Short answer

The total complex power supplied by the source is 6102.12 VA at 36.29⁰.

Step-by-step solution

Analyze Given Values

We are given a three-phase source with a line-to-neutral voltage \( \mathbf{V}_{\mathrm{an}} = 120 / 20^{\circ} \space \text{V} \). The load impedances are for Load 1 \( Z_1 = 8 + j6 \space \Omega/\phi \) and the complex power for phase a of Load 2 is \( S_{a2} = 600 / 36^{\circ} \space \text{VA} \).

Calculate Load 1 Phase Current

The phase current for Load 1 can be calculated using Ohm's law: \( I_a1 = \frac{\mathbf{V}_{\mathrm{an}}}{Z_1} = \frac{120 / 20^{\circ}}{8 + j6} \). First, find the magnitude and phase of the impedance:\[|Z_1| = \sqrt{8^2 + 6^2} = 10 \quad \text{and} \quad \phi_1 = \tan^{-1}\left(\frac{6}{8}\right) = 36.87^{\circ} \]Thus, \( Z_1 = 10 / 36.87^{\circ} \space \Omega \), and \[ I_a1 = \frac{120 / 20^{\circ}}{10 / 36.87^{\circ}} = 12 / (-16.87^{\circ}) \space \text{A} \]

Calculate Load 1 Power

Calculate the complex power for Load 1 using \( S_{1} = \mathbf{V}_{\mathrm{an}} \cdot I_{a1}^* \), where \(^*\) denotes the conjugate:\[ S_1 = 120 / 20^{\circ} \times 12 / 16.87^{\circ} = 1440 / 36.87^{\circ} \quad \text{VA} \]

Calculate Total Complex Power of Load 1

Since the load is Y-connected and balanced, the total power for Load 1 is three times the phase power:\[ S_{\text{total, 1}} = 3 \times 1440 / 36.87^{\circ} \quad \text{VA} \]

Calculate Total Complex Power Supplied

Add the complex power of Load 1 and Load 2:For Load 2, as it only provided power per phase, the total complex power is:\[ S_{\text{total, 2}} = 3 \times 600 / 36^{\circ} \quad \text{VA} \]The total complex power from the source is:\[ S_{\text{total}} = S_{\text{total, 1}} + S_{\text{total, 2}} \]

Perform Final Complex Addition

Perform the complex power addition to find the total:\[ S_{\text{total, 1}} = 3 \times 1440 / 36.87^{\circ} = 4320 / 36.87^{\circ} \quad \text{VA} \]\[ S_{\text{total, 2}} = 1800 / 36^{\circ} \quad \text{VA} \]Using rectangular form for addition:\( S_{\text{total, 1}} = 3445.61 + j2593.46 \; VA \)\( S_{\text{total, 2}} = 1454.09 + j1059.39 \; VA \)Final total:\[ S_{\text{total}} = (3445.61 + 1454.09) + j(2593.46 + 1059.39) \]\[ S_{\text{total}} = 4900 + j3652.85 \quad \text{VA} \]

Convert to Polar Form

Convert final total back to polar form:\( |S_{\text{total}}| = \sqrt{4900^2 + 3652.85^2} \)\( \phi_{\text{total}} = \tan^{-1}{\left(\frac{3652.85}{4900}\right)} \)Therefore, the polar form is \( S_{\text{total}} = 6102.12 / 36.29^{\circ} \space \text{VA} \).

Key concepts

Complex Power Calculation

Complex power is a vital concept in electrical engineering, especially when dealing with AC (Alternating Current) circuits. It combines both real power, measured in watts (W), and reactive power, measured in volt-amperes reactive (VAR), into a single complex number. This number is called complex power and is expressed in volt-amperes (VA).
To calculate complex power, you use the formula:

• \( S = P + jQ \)
• \( S = VI^* \)

where \( P \) is real power, \( Q \) is reactive power, \( V \) is the voltage, \( I \) is the current, and \( I^* \) is the complex conjugate of the current. The magnitude of the complex power (\( |S| \) ) gives the apparent power in the circuit.
In a three-phase system, particularly when loads are balanced and the sources are symmetric, the complex power per phase can be simplified as proportionate to total power. This simplifies the computation and helps in straightforward analysis.

Y-Connected Loads

Y-connected loads, also known as star-connected loads, form a 'Y' shape in three-phase electrical systems. They are popular in industrial applications because of their efficiency in distributing power.
Each load in the Y configuration connects to a single neutral point, meaning:

• The voltage across each phase of the load is equal to the line-to-neutral voltage.
• Total power in such configurations is calculated by multiplying the power in one phase by three.

This connectivity reduces the complexity of the phase current calculation. It is evident in balanced systems, such as those explored in the original exercise. Y-connections make complex power calculation simpler as we scale phase calculations by a factor of three.

Ohm's Law

Ohm's Law plays a foundational role in circuit analysis. It states that the current through a conductor between two points is directly proportional to the voltage across the two points. Ohm's law is typically expressed as:

• \( V = IZ \)

where \( V \) is voltage, \( I \) is current, and \( Z \) is impedance. In AC circuits, impedance is a complex quantity, consisting of real and imaginary components, which accounts for loss and reactive effects respectively.
In the context of our exercise, Ohm's Law was used to find the phase current by dividing the voltage by the impedance of Load 1. This step is crucial to understanding load behavior and necessary for calculating the complex power. By understanding the relationship between voltage, current, and resistance, students can solve for unknown quantities in circuit analysis.

Impedance

Impedance is a broader concept than resistance, as it applies to AC circuits and includes resistance and reactance. Measured in ohms (\( \Omega \)), impedance \( Z \) can be expressed in complex form as:

• \( Z = R + jX \)

where \( R \) is resistance and \( X \) is reactance. The magnitude of impedance is given by \(|Z| = \sqrt{R^2 + X^2}\), and the phase angle \( \phi \) is calculated using \( \tan^{-1}(\frac{X}{R}) \).
Understanding impedance helps determine how effectively a circuit resists and stores energy. In the exercise, solving impedance values allowed for current phase angles to be calculated, relevant in power calculations. With this understanding, students tackle circuit analysis confidently by calculating the current and subsequently the power values.

Three-Phase Circuits

Three-phase systems are an integral part of power distribution, particularly in industrial settings. They consist of three electrical circuits carrying current of the same frequency and amplitude but with a phase difference of 120 degrees.
The advantages of three-phase circuits include:

• More efficient power transmission.
• Constant power delivery, reducing pulsations found in single-phase systems.
• Reduced conductor material requirements.

In practice, these systems provide steady and reliable power since any cost-effective and high-power applications depend significantly on them.
The exercise reveals key insights into the balanced three-phase systems where calculations for power require symmetry in voltage and phase angles, simplifying analysis. These circuits prove to sustain efficiency and cost-effectiveness over single-phase circuits due to their inherent design and operational benefits.