Question

Consider three \(\mathrm{AC}\) power lines with phases differing by \(2 \pi / 3\) in the following way \(V_{0} \cos \omega t, V_{0} \cos (\omega t+2 \pi / 3), V_{0} \cos (\omega t+4 \pi / 3)\). Each of the lines has its return current to ground; the loads are organized such that these currents also are in phase \(I_{0} \cos \omega t, I_{0} \cos (\omega t+2 \pi / 3), I_{0} \cos (\omega t+4 \pi / 3)\). (a) Show that the return currents cancel in the ground. (b) Estimate \(I_{0}\) such that the power delivered is the same as in the DC case. (c) Calculate the power loss in the three lines and compare with the DC case.

Short answer

The return currents cancel, \(I_0 = \frac{I_{dc}}{3}\), and AC power loss is \(\frac{P_{loss\, DC}}{3}\).

Step-by-step solution

Express Currents in Complex Form

Convert the given AC currents into their complex representation. Given the phases, the currents can be expressed in polar form:- For line 1: \(I_1 = I_0 e^{j(0)} = I_0\)- For line 2: \(I_2 = I_0 e^{j(2\pi/3)}\) - For line 3: \(I_3 = I_0 e^{j(4\pi/3)}\)

Verify Current Cancellation

Show the sum of the currents, \(I_1 + I_2 + I_3\), is zero.\[I_1 + I_2 + I_3 = I_0 + I_0 e^{j(2\pi/3)} + I_0 e^{j(4\pi/3)}\]Using the identity for a balanced three-phase system, these add up to zero:\[I_0 + I_0 \left(-\frac{1}{2} + j\frac{\sqrt{3}}{2}\right) + I_0 \left(-\frac{1}{2} - j\frac{\sqrt{3}}{2}\right) = 0\]

Calculate Power for AC Case

The total power for each phase of the AC system is given by \(P = V_0 I_0 \cos \phi\), and since they are in phase, \(\phi=0\), hence:\[P_{total} = 3 \cdot V_0 I_0\]

Calculate I₀ for Equal DC Power

For a DC system, the power is \(P = V_{dc} I_{dc}\). To deliver the same power, equate it to the AC power:\[V_{dc} I_{dc} = 3V_0 I_0\]Assume \(V_{dc} = V_0\), we find:\[I_0 = \frac{I_{dc}}{3}\]

Compute Power Loss in AC Lines

Power loss in a line is \(P_{loss} = I^2 R\). For the AC system:\[P_{loss ext{, AC}} = 3 \cdot I_0^2 R = 3 \times \left(\frac{I_{dc}}{3}\right)^2 R = \frac{I_{dc}^2 R}{3}\]

Compare with DC Power Loss

In a DC system, the power loss is:\[P_{loss ext{, DC}} = I_{dc}^2 R\]Comparing the two: \[P_{loss ext{, AC}} = \frac{P_{loss ext{, DC}}}{3}\]The AC system has a lower power loss.

Key concepts

AC Power Calculations

When dealing with alternating current (AC) power systems, understanding how to calculate power is essential. This exercise involves three-phase AC power systems, which are common in electrical power distribution.
AC power is not as straightforward as direct current (DC) power because it involves alternating voltages and currents. In AC systems, power is calculated using complex numbers to represent the sinusoidal voltages and currents.

For each phase in a three-phase system, the current can be expressed using complex exponentials. The detailed breakdown here uses Euler's formula, which provides a way to express sinusoidal functions as complex exponentials. This helps in analyzing currents and voltages in phases with specific angular differences.

• The formula for power in one phase of an AC system is given by:
  \[P = V_0 I_0 \cos \phi\]
  where:
  • \( P \) is the power for one phase,
  • \( V_0 \) is the peak voltage,
  • \( I_0 \) is the peak current,
  • \( \phi \) is the phase angle (which is 0 if voltage and current are in phase).

Overall, the total power in a balanced three-phase system is three times the power of one phase because each phase carries an equal amount of current and experiences the same voltage.

Three-Phase Current Balancing

A three-phase power system is a setup where three AC voltages are spaced out by one-third of a cycle. This unique configuration helps in achieving current balancing where the sum of the return currents cancels out.
In the exercise given, each phase delivers current back through a neutral or ground path. By balancing the three phases, the currents in the neutral wire theoretically cancel out, reducing the need for a large neutral wire.

• Each of the AC power lines carries a current expressed in polar form:
  \( I_1 = I_0 \), \( I_2 = I_0 e^{j(2\pi/3)} \), \( I_3 = I_0 e^{j(4\pi/3)} \).
• The summation of these currents shows cancellation:
  \[ I_1 + I_2 + I_3 = I_0 + I_0 e^{j(2\pi/3)} + I_0 e^{j(4\pi/3)} = 0 \].
• The balanced vector sum of the three currents is zero, illustrating that no current flows through the neutral line, assuming a perfectly balanced load.

This conservation means that in an ideal system with balanced currents, power delivery is efficient with minimal loss in the neutral line, resulting in cost and material savings.

Power Loss Comparison

Comparing power losses between AC and DC systems provides insights into efficiency. Power loss in electrical systems is mainly due to the resistance in the wires.For AC systems, power loss in each line can be calculated using:
\[P_{loss\,\text{AC}} = 3 \, \times \, I_0^2 \, R \]
where:

• \( I_0 \) represents the current in one phase, given as \( \frac{I_{dc}}{3} \) when equivalent AC power is needed as in DC systems,
• \( R \) is the resistance per line.

The power loss in a three-phase AC system is notably lower compared to a DC system because it disperses the power delivery across three separate lines, reducing the current per phase.

• For a DC system, power loss is calculated as:
  \[P_{loss\,\text{DC}} = I_{dc}^2 \, R\]
• The comparison statement here:\[P_{loss\,\text{AC}} = \frac{P_{loss\,\text{DC}}}{3}\]

Thus, the power loss in an AC three-phase system is a third of that in the DC counterpart, showcasing the efficiency advantages of using AC systems especially in long-distance power transmission.