Power system analysis, per unit system help

In summary, the conversation discusses a homework problem involving drawing a reactance diagram and calculating per unit impedances for a system with specified voltage and power values. The equations used include P = VI, s = p + jq, and apparent power = (reactive power^2 + real power^2)^(1/2). It is suggested to work with actual quantities rather than the per unit system for better accuracy.
  • #1
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Homework Statement


https://dl.dropboxusercontent.com/u/88040310/powerhw3.png
1. Draw the reactance diagram.
2. Calculate the per unit impedances
3. If bms #1 is operating at 30 kV, find the current
Assume base of 100 MVA, 300 kV on the generator side.

Homework Equations


P = VI
s = p +jq
Apparent power = (reactive power^2 + real power^2)^(1/2)

per phase equations, tap transformer equations

The Attempt at a Solution


Starting it slowly, is this how you draw the reactance diagram?
https://dl.dropboxusercontent.com/u/88040310/hw3xdiagram1.png
 
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  • #2
At the risk of being called a heretic, I suggest that the best thing you can do is abandon the "per unit" system. Work with actual quantities, so that your equations make sense and are dimensionally correct. Per unit is useful only in that for certain system classes, it enables estimates of parameters before the actual values are know. Thereafter, it is simply a confusion factor (and a large one at that!).
 

1. What is power system analysis?

Power system analysis is the study of electrical power systems, including their components, operations, and behavior. It involves analyzing the performance and characteristics of power systems to ensure efficient and reliable delivery of electricity.

2. What is the per unit system in power system analysis?

The per unit system is a method used in power system analysis to simplify and standardize calculations. It involves representing all quantities, such as voltage, current, and power, in a system in terms of their percentage of a base value. This allows for easier comparison and analysis of different systems.

3. Why is the per unit system used in power system analysis?

The per unit system is used in power system analysis because it allows for easy comparison of different systems with varying voltage and power levels. It also simplifies calculations and makes them less prone to errors, as all quantities are represented in a standardized unit.

4. How is the per unit system calculated in power system analysis?

The per unit system is calculated by dividing the actual value of a quantity by its base value. For example, if the base voltage is 100 kV and the actual voltage is 200 kV, the per unit voltage would be 2 pu (200 kV / 100 kV = 2 pu).

5. What are the benefits of using the per unit system in power system analysis?

The benefits of using the per unit system in power system analysis include simplifying and standardizing calculations, allowing for easy comparison of different systems, and making the analysis less prone to errors. It also helps in identifying the relative importance of different components in a system and their impact on its overall performance.

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