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# Mastering Per Unit Systems: A Comprehensive Guide with 10 Solved Problems
Electrical power systems are intricate networks of generators, transformers, transmission lines, and loads, operating at vastly different voltage and power levels. Analyzing these systems, especially during fault conditions or for stability studies, can become incredibly complex if actual values are used. This is where the Per Unit (PU) system comes in – a powerful normalization technique that simplifies calculations, standardizes equipment representation, and provides an intuitive understanding of system behavior.
In this comprehensive guide, you'll embark on a journey to demystify the Per Unit system. We'll delve into its foundational concepts, explore its historical significance, and equip you with practical tips. Most importantly, we'll solidify your understanding through **ten distinct, solved problems with detailed, step-by-step explanations**, ensuring you gain the confidence to apply this vital tool in your electrical engineering endeavors.
Understanding the Per Unit System: The Foundation
At its core, the Per Unit system is a method of expressing system quantities (voltage, current, power, impedance) as a fraction or multiple of a chosen base quantity of the same dimension.
What is the Per Unit System?
The per unit value of any quantity is defined as:
$$ \text{Per Unit Value} = \frac{\text{Actual Value}}{\text{Base Value}} $$
For example, if the actual voltage is 110 kV and the chosen base voltage is 100 kV, the per unit voltage is 1.1 PU. This dimensionless representation simplifies complex multi-voltage level systems dramatically.
Why Use Per Unit Systems? Historical Context and Benefits
The Per Unit system isn't a modern invention; its roots trace back to the early 20th century as power systems grew in complexity. Engineers needed a more efficient way to analyze interconnected grids. Before its widespread adoption, calculations involved laborious conversions between different voltage levels using transformer turns ratios, prone to errors and time-consuming.
The key benefits that propelled the PU system into standard practice include:
- **Simplification of Calculations:** By normalizing all quantities, calculations become simpler, especially across transformers. The ideal transformer's per unit impedance remains the same regardless of which side it's viewed from.
- **Elimination of Transformer Ratios:** Transformers, which change voltage levels, are automatically accounted for. Per unit impedances "pass through" transformers without needing explicit turns ratio conversions.
- **Standardization:** Manufacturers often specify equipment impedances (e.g., transformers, generators) directly in per unit values based on their own ratings. This allows for easy comparison and integration into system studies.
- **Intuitive Understanding:** A per unit voltage close to 1.0 indicates normal operation, while values significantly different (e.g., 0.8 PU or 1.2 PU) immediately signal under- or over-voltage conditions.
- **Reduced Chance of Error:** Working with dimensionless numbers reduces the likelihood of errors associated with large magnitudes and unit conversions.
Choosing Base Values
The selection of base values is crucial and forms the bedrock of any PU analysis. Typically, two base quantities are chosen, from which others are derived:
1. **Base Power (S_base):** Usually chosen as a single, system-wide base MVA (or kVA) for three-phase systems. Common choices are 100 MVA or 1000 MVA.
2. **Base Voltage (V_base):** Chosen for one specific voltage level (e.g., the generator terminal voltage or a transmission line voltage). Base voltages for other voltage levels are then determined by the transformer turns ratios.
From these two, other base quantities can be derived:
- **Base Current (I_base):**
- For single-phase: $I_{\text{base}} = S_{\text{base}} / V_{\text{base}}$
- For three-phase: $I_{\text{base}} = S_{\text{base}} / (\sqrt{3} \times V_{\text{base}})$ (where $S_{\text{base}}$ is in VA, $V_{\text{base}}$ in Volts)
- **Base Impedance (Z_base):**
- For single-phase: $Z_{\text{base}} = V_{\text{base}} / I_{\text{base}} = (V_{\text{base}})^2 / S_{\text{base}}$
- For three-phase: $Z_{\text{base}} = (V_{\text{base, LL}})^2 / S_{\text{base, 3\phi}}$ (where $V_{\text{base, LL}}$ is line-to-line base voltage in kV, $S_{\text{base, 3\phi}}$ in MVA, resulting $Z_{\text{base}}$ in Ohms)
Practical Tips for Per Unit Calculations
- **Define Bases Clearly:** Always state your chosen system base MVA and the base kV for each voltage level upfront.
- **Convert Everything to PU:** Before performing any calculations, convert all actual values (voltages, currents, impedances, powers) to their per unit equivalents based on your chosen system bases.
- **Master the Base Impedance Formula:** The formula $Z_{\text{base}} = (kV_{\text{base}})^2 / MVA_{\text{base}}$ is your best friend for converting base values.
- **Draw Single-Line Diagrams:** For complex systems, a clear single-line diagram with voltage zones and components labeled in PU is invaluable.
- **Unit Consistency:** Be meticulous with units. Ensure MVA and kV are used consistently for base impedance calculations.
Common Mistakes to Avoid
- **Inconsistent Base Values:** Using different system base MVAs for different parts of the system will lead to incorrect results.
- **Forgetting Base Changes:** When equipment impedance is given on its own rating (MVA, kV), and you need to use it in a system with different bases, remember to convert its PU value to the system base.
- **Incorrect Transformer Ratios:** While PU simplifies transformers, ensure your base voltage selection propagates correctly through transformer turns ratios.
- **Confusing Single-Phase and Three-Phase:** Apply the correct formulas for base current and power depending on whether the system is single-phase or three-phase.
- **Not Converting Back:** If the final answer needs to be an actual value, remember to convert the per unit result back using the appropriate base value.
10 Solved Problems: Step-by-Step Mastery
Here, we'll tackle ten distinct problems, progressively building your understanding of per unit system applications.
**Problem 1: Basic Per Unit Conversion (Voltage & Current)**- **Problem:** A 480V generator supplies a load with 100A. If the base voltage is 500V and base current is 120A, find the per unit voltage and current.
- **Solution:**
- $V_{PU} = 480V / 500V = 0.96 \text{ PU}$
- $I_{PU} = 100A / 120A = 0.833 \text{ PU}$
- **Explanation:** Direct application of the PU definition.
- **Problem:** A 10 Ohm resistor is part of a system with base MVA = 10 MVA and base kV = 10 kV. Calculate the per unit resistance.
- **Solution:**
- First, calculate base impedance: $Z_{base} = (10 \text{ kV})^2 / 10 \text{ MVA} = 100 \text{ kVA}^2 / 10 \text{ MVA} = 100 \text{ Ohms}$.
- $R_{PU} = 10 \text{ Ohms} / 100 \text{ Ohms} = 0.1 \text{ PU}$
- **Explanation:** Crucial to derive the base impedance from the chosen system base MVA and kV.
- **Problem:** A transformer has an impedance of 0.1 PU on its own rating of 20 MVA, 13.8 kV. What is its impedance on a new system base of 50 MVA, 13.8 kV?
- **Solution:**
- $Z_{PU, new} = Z_{PU, old} \times (\text{Old MVA}_{\text{base}} / \text{New MVA}_{\text{base}}) \times (\text{New kV}_{\text{base}} / \text{Old kV}_{\text{base}})^2$
- $Z_{PU, new} = 0.1 \times (20 \text{ MVA} / 50 \text{ MVA}) \times (13.8 \text{ kV} / 13.8 \text{ kV})^2$
- $Z_{PU, new} = 0.1 \times (0.4) \times 1^2 = 0.04 \text{ PU}$
- **Explanation:** This formula is essential for integrating equipment with different nameplate ratings into a unified system base. Notice that if kV base remains the same, the kV ratio cancels out.
- **Problem:** A 3-phase, 100 MVA, 230/34.5 kV transformer has a leakage reactance of 10%. Calculate its per unit reactance on a system base of 100 MVA, 230 kV (HV side).
- **Solution:**
- The transformer's reactance is given as 10% on its own rating, which means $X_{PU, old} = 0.1 \text{ PU}$.
- Since the system base MVA (100 MVA) and HV side base kV (230 kV) match the transformer's own rating, the per unit reactance remains the same.
- $X_{PU, new} = 0.1 \text{ PU}$
- **Explanation:** When the system base matches the equipment's nameplate rating, no conversion is needed. This highlights the standardization benefit.
- **Problem:** A 20 MVA, 13.8 kV synchronous generator has a subtransient reactance of 25%. Express this in per unit on a 100 MVA, 13.8 kV system base.
- **Solution:**
- $X_{PU, new} = X_{PU, old} \times (\text{Old MVA}_{\text{base}} / \text{New MVA}_{\text{base}})$ (kV bases are the same)
- $X_{PU, new} = 0.25 \times (20 \text{ MVA} / 100 \text{ MVA}) = 0.25 \times 0.2 = 0.05 \text{ PU}$
- **Explanation:** Similar to Problem 3, but specifically for a generator.
- **Problem:** A transmission line has an impedance of $Z = (5 + j20)$ Ohms. Calculate its per unit impedance on a 100 MVA, 115 kV system base.
- **Solution:**
- $Z_{base} = (115 \text{ kV})^2 / 100 \text{ MVA} = 13225 / 100 = 132.25 \text{ Ohms}$.
- $Z_{PU} = Z_{actual} / Z_{base} = (5 + j20) / 132.25$
- $Z_{PU} = (5/132.25) + j(20/132.25) = 0.0378 + j0.1512 \text{ PU}$
- **Explanation:** Applies the base impedance calculation to a complex impedance.
- **Problem:** A generator (0.1 PU reactance on 100 MVA, 20 kV) is connected to a transmission line (0.2 PU reactance on 100 MVA, 20 kV) and feeds a load. Find the total per unit impedance from the generator to the load.
- **Solution:**
- Since all components are already on the same 100 MVA, 20 kV base, simply add the per unit reactances.
- $X_{total, PU} = X_{gen, PU} + X_{line, PU} = 0.1 + 0.2 = 0.3 \text{ PU}$
- **Explanation:** One of the greatest benefits of PU: series impedances add directly.
- **Problem:** For the system in Problem 7, if the generator voltage is 1.0 PU and a bolted short circuit occurs at the load, calculate the per unit fault current.
- **Solution:**
- $I_{fault, PU} = V_{gen, PU} / X_{total, PU} = 1.0 \text{ PU} / 0.3 \text{ PU} = 3.333 \text{ PU}$
- **Explanation:** Ohm's Law applies directly in PU. This result immediately tells engineers the severity of the fault without needing to convert to actual amps.
- **Problem:** Using the fault current from Problem 8, and a system base of 100 MVA, calculate the short circuit MVA at the fault point.
- **Solution:**
- Short Circuit MVA = $I_{fault, PU} \times S_{base}$
- Short Circuit MVA = $3.333 \text{ PU} \times 100 \text{ MVA} = 333.3 \text{ MVA}$
- **Explanation:** A direct way to express the fault level, useful for circuit breaker selection.
- **Problem:** A generator (20 MVA, 13.8 kV, $X = 0.2 \text{ PU}$) connects to a step-up transformer ($T_1$: 20 MVA, 13.8/115 kV, $X = 0.1 \text{ PU}$). This connects to a transmission line ($Z = (10 + j50)$ Ohms). Calculate the generator's and line's impedance in PU on a system base of 100 MVA, with 115 kV chosen as the base for the transmission line.
- **Solution:**
- **System Bases:** $S_{base} = 100 \text{ MVA}$.
- Line voltage base: $V_{base, line} = 115 \text{ kV}$.
- Generator voltage base (derived from $T_1$): $V_{base, gen} = 115 \text{ kV} \times (13.8 \text{ kV} / 115 \text{ kV}) = 13.8 \text{ kV}$.
- **Generator Reactance:**
- $X_{gen, PU, new} = X_{gen, PU, old} \times (S_{old} / S_{new}) \times (V_{new} / V_{old})^2$
- $X_{gen, PU, new} = 0.2 \times (20 \text{ MVA} / 100 \text{ MVA}) \times (13.8 \text{ kV} / 13.8 \text{ kV})^2 = 0.2 \times 0.2 \times 1 = 0.04 \text{ PU}$.
- **Transformer Reactance:**
- $T_1$ is 0.1 PU on 20 MVA, 13.8/115 kV. New bases are 100 MVA, 13.8/115 kV.
- $X_{T1, PU, new} = 0.1 \times (20 \text{ MVA} / 100 \text{ MVA}) = 0.02 \text{ PU}$.
- **Transmission Line Impedance:**
- $Z_{base, line} = (115 \text{ kV})^2 / 100 \text{ MVA} = 132.25 \text{ Ohms}$.
- $Z_{line, PU} = (10 + j50) / 132.25 = 0.0756 + j0.378 \text{ PU}$.
- **Explanation:** This problem ties together base selection across transformers and changing bases for different components, showcasing the power of the PU system in multi-voltage systems.
Conclusion
The Per Unit system is an indispensable tool in electrical engineering, particularly in power system analysis. From its historical roots in simplifying manual calculations to its continued relevance in modern software, understanding PU concepts is fundamental. By normalizing complex quantities and standardizing equipment representation, it transforms daunting multi-voltage calculations into manageable, intuitive problems.
Through these ten solved problems, you've gained practical experience in converting values, changing bases, and applying the system to various components like generators, transformers, and transmission lines. Remember to always define your bases clearly, be meticulous with calculations, and appreciate the simplification that the Per Unit system brings to the world of power engineering. Continue practicing, and you'll soon master this essential technique.
