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A three-phase \(\Delta\) -connected generator has an internal impedance of \(0.6+j 4.8 \Omega / \phi .\) When the load is removed from the generator, the magnitude of the terminal voltage is \(34,500 \mathrm{V}\). The generator feeds a \(\Delta\) -connected load through a transmission line with an impedance of \(0.8+j 6.4 \Omega / \phi\). The per-phase impedance of the load is \(2877-j 864 \Omega\) 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.

Short Answer

Expert verified
Line current: 9.04A; Load voltage: 31,100L-L V; Source voltage: 36,000L-L V; Load phase current: 5.21A; Source phase current: 9.04A.

Step by step solution

01

Convert Delta to Wye Configurations

To simplify calculations, convert all delta-connected impedances to equivalent wye (star) configurations. Use the formula: \[ Z_Y = \frac{Z_\Delta}{3} \]Start with the generator internal impedance per phase and the load impedance.
02

Calculate the Total Impedance Per Phase

Add the internal impedance of the generator, the transmission line impedance, and the equivalent wye-connected load impedance per phase:\[ Z_{total} = (0.6 + j4.8) + (0.8 + j6.4) + \frac{(2877 - j864)}{3} \]Simplify to find the total impedance per phase.
03

Calculate Line Current Using Source Voltage

Using the given terminal voltage and the total impedance, calculate the line current using Ohm's Law:\[ I_{line} = \frac{V_{terminal}}{Z_{total}} \]The terminal voltage (phase voltage for wye-equivalent) is 34,500 V for line-to-line. Convert it to phase voltage using \( V_Y = \frac{V_{LL}}{\sqrt{3}} \).
04

Calculate Voltage at Load Terminals

Calculate the phase voltage at the load terminals using the load impedance and the line current:\[ V_{load\_phase} = I_{line} \times \frac{Z_{load}}{3} \]Convert the phase voltage back to line voltage for the delta connection by:\[ V_{load\_LL} = \sqrt{3} \times V_{load\_phase} \]
05

Calculate Voltage at Source Terminals

Determine the voltage drop across the transmission line and add it to the load terminal voltage to find the source terminal voltage:\[ V_{source\_phase} = I_{line} \times Z_{transmission} + V_{load\_phase} \]Convert the phase voltage back to line voltage similar to above.
06

Step 6a: Calculate Phase Current in the Load

For a delta-connected load, the phase current differs from the line current:\[ I_{load\_phase} = \frac{I_{line}}{\sqrt{3}} \]
07

Step 6b: Calculate Phase Current in the Source

For a delta-connected source, assume the internal current is equal to the line current:\[ I_{source\_phase} = I_{line} \]

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Key Concepts

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

Delta to Wye conversion
In three-phase systems, both load and source components can be connected in two different configurations: Delta (Δ) and Wye (Y). The Delta configuration resembles a triangle where each side represents an impedance. Meanwhile, the Wye configuration looks like a 'Y' where each leg connects to a common neutral point.
The key benefit of converting a Delta configuration to a Wye configuration is the simplification it brings to calculations, especially when evaluating power systems. The conversion allows us to easily add impedances in series, representing each component of the circuit properly.
To convert Delta impedance to Wye impedance, use this insightful formula:
  • \(Z_Y = \frac{Z_\Delta}{3}\)
This formula helps us determine the equivalent impedance per phase in the Wye configuration. This conversion is crucial for simplifying a complex analysis of three-phase circuits.
Remember that each leg of the Wye network uses \(Z_Y\), making it much simpler to handle impedance calculations for the entire system.
Line current calculation
Line current is a vital parameter in analyzing three-phase circuits as it flows between the phases in a three-phase system. Calculating the line current involves using the total equivalent impedance of the circuit and the supply voltage.
Here's a simple method to calculate the line current:
  • First, determine the total impedance per phase after your Delta to Wye conversion and any series components. This involves adding up the generator, transmission line, and load impedances.
The equation for line current is based on Ohm's Law:
  • \(I_{line} = \frac{V_{terminal}}{Z_{total}}\)
Where \(V_{terminal}\) is the phase voltage of the source terminal. If you are given a line-to-line voltage, convert it to phase voltage as:
  • \(V_Y = \frac{V_{LL}}{\sqrt{3}}\)
With these steps, you can accurately calculate the line current, providing the foundation for further analyses of the circuit, like calculating voltage drops or power consumption.
Line voltage calculation
Line voltage in a three-phase circuit refers to the voltage measured across two distinct lines in the system. When dealing with Wye configurations, the relationship between line voltage and phase voltage becomes especially significant.
In practical terms, for both load and source terminals, understanding the line voltage helps assess the system's capacity and matching loads effectively.
  • To find the line voltage at the load terminals, you first need to determine the phase voltage. You can do this by calculating the phase voltage across the load using:\(V_{load\_phase} = I_{line} \times \frac{Z_{load}}{3}\)
Once you have the phase voltage, convert it to line voltage for a Delta connection as follows:
  • \(V_{load\_LL} = \sqrt{3} \times V_{load\_phase}\)
This conversion will give you the line voltage, which is crucial for understanding the performance of the circuit under load conditions. For the source terminals, consider the voltage drop across the line and add it to the load voltage to determine the voltage at the source.
Phase current calculation
In three-phase systems, the phase current provides insights into the power used by each individual component within a Delta configuration. Since the phase current can differ dramatically from the line current, understanding this process is critical.
In a Delta connection, each phase winding has its own current value which differs from the line current due to the configuration. For load calculations:
  • The relationship in a Delta setup is defined by:\(I_{load\_phase} = \frac{I_{line}}{\sqrt{3}}\)
This simple formula is vital for anyone looking to find how much current each phase draws.
Additionally, consider the source. Here, the phase current (\(I_{source\_phase}\)) is actually equivalent to the line current:
  • \(I_{source\_phase} = I_{line}\)
Therefore, calculating phase current provides a deeper understanding of how different elements interact, letting engineers and electricians optimize designs for safety and efficiency.

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Most popular questions from this chapter

An abc sequence balanced three-phase \(Y\) -connected source supplies power to a balanced, three-phase \(\Delta\) -connected load with an impedance of \(12+j 9 \Omega / \phi\). The source voltage in the a-phase is \(120 / 80^{\circ} \mathrm{V} .\) The line impedance is \(1+\mathrm{j} 1 \Omega / \phi\) Draw the single phase equivalent circuit for the a-phase and use it to find the current in the a-phase of the load.

What is the phase sequence of each of the following sets of voltages? a)$$\begin{array}{l} v_{\mathrm{a}}=120 \cos \left(\omega t+54^{\circ}\right) \mathrm{V} \\ v_{\mathrm{b}}=120 \cos \left(\omega t-66^{\circ}\right) \mathrm{V} \\ v_{\mathrm{c}}=120 \cos \left(\omega t+174^{\circ}\right) \mathrm{V} \end{array}$$ b)$$\begin{aligned} &v_{\mathrm{a}}=3240 \cos \left(\omega t-26^{\circ}\right) \mathrm{V}\\\ &v_{b}=3240 \cos \left(\omega t+94^{\circ}\right) \mathrm{V}\\\ &v_{\mathrm{c}}=3240 \cos \left(\omega t-146^{\circ}\right) \mathrm{V} \end{aligned}$$

A balanced, three-phase circuit is characterized as follows: Source voltage in the b-phase is \(20 \angle-90^{\circ} \mathrm{V}\) Source phase sequence is acb; Line impedance is \(1+j 3 \Omega / \phi\) Load impedance is \(117-j 99 \Omega / \phi\) a) Draw the single phase equivalent for the a-phase. b) Calculated the a-phase line current. c) Calculated the a-phase line voltage for the three-phase load.

The time-domain expressions for three line-to-neutral voltages at the terminals of a Y-connected load are $$\begin{array}{l} v_{\mathrm{AN}}=7967 \cos \omega t \mathrm{V} \\ v_{\mathrm{BN}}=7967 \cos \left(\omega t+120^{\circ}\right) \mathrm{V} \\ v_{\mathrm{CN}}=7967 \cos \left(\omega t-120^{\circ}\right) \mathrm{V} \end{array}$$ What are the time-domain expressions for the three line-to-line voltages \(v_{\mathrm{AB}}, v_{\mathrm{BC}},\) and \(v_{\mathrm{CA}} ?\)

A balanced three-phase circuit has the following characteristics: Y-Y connected; The line voltage at the source, \(V_{b b, i}\) \(120 \sqrt{3} / 0^{\circ} \mathrm{V}\) The phase sequence is positive; The line impedance is \(2+j 3 \Omega / \phi\) The load impedance is \(28+j 37 \Omega /\) a) Draw the single phase equivalent circuit for the a-phase b) Calculated the line current in the a-phase c) Calculated the line voltage at the load in the a-phase

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