Per Unit System in Power Systems: Complete Calculation Guide with Examples

The per unit system is one of the most powerful calculation methods in power system analysis, transforming complex multi-voltage networks into simplified normalized calculations. Whether you're analyzing fault currents, calculating transformer impedances, or designing three-phase power systems, mastering the per unit system is essential for accurate and efficient power system engineering.

This normalization technique eliminates the confusion of working with multiple voltage levels simultaneously and makes equipment comparisons straightforward. Instead of juggling 480V, 13.8kV, and 138kV in the same calculation, everything becomes dimensionless quantities near 1.0, dramatically simplifying analysis while maintaining accuracy.

Understanding the Per Unit System Fundamentals

What is the Per Unit System?

The per unit (pu) system expresses power system quantities as decimal fractions of defined base values rather than in their actual units. Any electrical quantity can be expressed in per unit by dividing its actual value by an appropriate base value:

$$\text{Per Unit Value} = \frac{\text{Actual Value}}{\text{Base Value}}$$

Key Concept: All per unit values are dimensionless (no units), making calculations cleaner and errors easier to spot.

Why Power Engineers Use Per Unit System

Advantages of Per Unit Analysis:

Benefit	Practical Impact
Simplifies multi-voltage calculations	Analyze entire power grid as single system
Eliminates unit conversions	No confusion between kV and V, MW and kW
Equipment comparison	Compare generators and transformers directly
Error detection	Values far from 1.0 indicate problems
Computer simulation	Reduces numerical errors in iterative calculations
Standard representation	Universal language for power system engineers

When to Use Per Unit System

Ideal Applications:

• Fault current calculations
• Power flow analysis
• Transformer impedance studies
• Generator and motor modeling
• Protection coordination
• System stability analysis

Less Useful For:

• Simple single-voltage circuits
• Wire sizing calculations
• Basic load calculations
• Consumer electrical installations

Base Value Selection and Calculation

The Four Base Quantities

In power systems, we work with four fundamental quantities. Choose any two as base values, and the other two are derived:

$$\begin{align} S_{\text{base}} &= \text{Base Power (VA, kVA, or MVA)} \\ V_{\text{base}} &= \text{Base Voltage (V, kV)} \\ I_{\text{base}} &= \text{Base Current (A)} \\ Z_{\text{base}} &= \text{Base Impedance (Î)} \end{align}$$

Base Value Relationships

Once you select $S_{base}$ and $V_{base}$, calculate the others:

Single-Phase Systems:

$$I_{\text{base}} = \frac{S_{\text{base}}}{V_{\text{base}}}$$ $$Z_{\text{base}} = \frac{V_{\text{base}}^2}{S_{\text{base}}} = \frac{V_{\text{base}}}{I_{\text{base}}}$$

Three-Phase Systems:

$$I_{\text{base}} = \frac{S_{\text{base}}}{\sqrt{3} \times V_{\text{base(L-L)}}}$$ $$Z_{\text{base}} = \frac{V_{\text{base(L-L)}}^2}{S_{\text{base}}}$$

Selecting Appropriate Base Values

Standard Practice:

• Base Power ($S_{base}$): Choose a round number like 10 MVA, 100 MVA, or equipment rating
• Base Voltage ($V_{base}$): Use nominal system voltage at each level (13.8 kV, 138 kV, etc.)

Example Selection: For a 100 MVA, 13.8 kV system:

• $S_{base}$ = 100 MVA
• $V_{base}$ = 13.8 kV

Step-by-Step Per Unit Calculations

Example 1: Three-Phase Generator Analysis

Given System:

• Generator: 50 MVA, 13.8 kV, Xd" = 15%
• Operating at: 52 MVA, 14.2 kV
• Find: Per unit power and voltage on generator's base

Step 1: Identify Equipment Base Values

Generator nameplate ratings become its base values:

• $S_{base}$(gen) = 50 MVA
• $V_{base}$(gen) = 13.8 kV

Step 2: Calculate Per Unit Values on Generator Base

$$S_{\text{pu}} = \frac{52 \text{ MVA}}{50 \text{ MVA}} = 1.04 \text{ pu}$$ $$V_{\text{pu}} = \frac{14.2 \text{ kV}}{13.8 \text{ kV}} = 1.029 \text{ pu}$$

Interpretation: Generator operates at 104% of rated power and 102.9% of rated voltage - both within acceptable limits (typically ±10%).

Example 2: Transformer Impedance Conversion

Problem: A transformer has 8% impedance on its own 25 MVA base. Convert to system base of 100 MVA.

Given:

• $Z_{transformer}$ = 0.08 pu on 25 MVA base
• $V$ same on both bases (voltage doesn't change across transformer)
• Find: Impedance on 100 MVA system base

Base Change Formula:

$$Z_{\text{pu,new}} = Z_{\text{pu,old}} \times \frac{S_{\text{base,new}}}{S_{\text{base,old}}} \times \left(\frac{V_{\text{base,old}}}{V_{\text{base,new}}}\right)^2$$

Since voltage bases are the same:

$$Z_{\text{pu,new}} = 0.08 \times \frac{100 \text{ MVA}}{25 \text{ MVA}} \times (1)^2 = 0.32 \text{ pu}$$

Result: On the 100 MVA system base, transformer impedance is 0.32 pu (32%).

Example 3: Complete System Analysis

System Description:

• Generator: 100 MVA, 13.8 kV, X = 0.15 pu
• Step-up Transformer: 100 MVA, 13.8/138 kV, X = 0.10 pu
• Transmission Line: X = 50 Ω at 138 kV
• System Base: 100 MVA, 13.8 kV at generator side

Step 1: Convert Line Impedance to Per Unit

First, calculate base impedance at 138 kV level:

$$Z_{\text{base}} = \frac{V_{\text{base}}^2}{S_{\text{base}}} = \frac{(138 \text{ kV})^2}{100 \text{ MVA}} = \frac{19044}{100} = 190.44 \text{ Î}$$

Then convert actual impedance:

$$X_{\text{line,pu}} = \frac{50 \text{ Î}}{190.44 \text{ Î}} = 0.2625 \text{ pu}$$

Step 2: Calculate Total System Impedance

All components now on same base (100 MVA):

$$X_{\text{total}} = X_{\text{gen}} + X_{\text{trans}} + X_{\text{line}} = 0.15 + 0.10 + 0.2625 = 0.5125 \text{ pu}$$

Step 3: Calculate Short Circuit Current

If fault occurs at remote end (assuming 1.0 pu voltage source):

$$I_{\text{fault,pu}} = \frac{V_{\text{pu}}}{X_{\text{total}}} = \frac{1.0}{0.5125} = 1.951 \text{ pu}$$

Convert to actual current at fault location (138 kV):

$$I_{\text{base}} = \frac{100 \text{ MVA}}{\sqrt{3} \times 138 \text{ kV}} = 418.4 \text{ A}$$ $$I_{\text{fault}} = 1.951 \times 418.4 = 816.3 \text{ A}$$

Advanced Per Unit Concepts

Changing Between Voltage Levels

When analyzing systems with transformers, base voltage changes proportionally with transformer turns ratio:

$$\frac{V_{\text{base,secondary}}}{V_{\text{base,primary}}} = \frac{N_{\text{secondary}}}{N_{\text{primary}}}$$

Example:

• Primary base: 13.8 kV
• Transformer ratio: 13.8/138 kV (1:10)
• Secondary base: 138 kV

Critical Rule: Base power ($S_{base}$) remains constant throughout the system, only base voltage changes at transformers.

Per Unit Impedance Diagrams

Creating per unit impedance diagrams simplifies complex networks:

Step-by-Step Process:

1. Select system-wide base values ($S_{base}, V_{base}$ at one point)
2. Calculate base voltages at all voltage levels using transformer ratios
3. Convert all impedances to per unit on system base
4. Draw single-line diagram with per unit impedances
5. Perform analysis using simple series/parallel combinations

Power Flow in Per Unit

Active and reactive power in per unit:

$$P_{\text{pu}} = \frac{P_{\text{actual}}}{S_{\text{base}}}$$ $$Q_{\text{pu}} = \frac{Q_{\text{actual}}}{S_{\text{base}}}$$

Power Balance Check: In a lossless system:

$$\sum P_{\text{generated,pu}} = \sum P_{\text{load,pu}}$$

Practical Applications in Power Systems

Fault Current Analysis

Per unit system excels in short circuit calculations:

Three-Phase Fault Current:

$$I_{\text{fault,pu}} = \frac{V_{\text{prefault,pu}}}{Z_{\text{total,pu}}}$$

Typical Impedance Values:

• Generators: 0.10 - 0.30 pu (subtransient)
• Transformers: 0.05 - 0.15 pu
• Transmission lines: 0.01 - 0.50 pu (length dependent)
• Motors: 0.15 - 0.25 pu

Generator Synchronization

When connecting generators in parallel, per unit system helps verify:

• Voltage magnitude matching (both near 1.0 pu)
• Frequency matching
• Phase angle alignment
• Impedance compatibility

Protection Relay Settings

Protective relays often use per unit settings:

• Overcurrent relays: Trip at 1.2-1.5 pu continuous
• Overvoltage relays: Trip at 1.10-1.20 pu
• Undervoltage relays: Trip at 0.80-0.90 pu
• Frequency relays: Trip at 0.98-1.02 pu

Common Mistakes and How to Avoid Them

Mistake 1: Mixing Base Values

Problem: Using different base values for connected equipment Solution: Convert everything to common system base before calculations

Mistake 2: Incorrect Voltage Base

Problem: Using phase voltage instead of line-to-line voltage for three-phase Solution: Always use $V_{L-L}$ for three-phase base voltage

Mistake 3: Forgetting Base Changes at Transformers

Problem: Using same base voltage on both sides of transformer Solution: Apply transformer turns ratio to change base voltage

Mistake 4: Wrong Transformer Impedance Base

Problem: Using transformer impedance without converting to system base Solution: Always convert using base change formula

Mistake 5: Mixing Per Unit and Actual Values

Problem: Adding per unit impedance to ohmic impedance Solution: Convert all to same system (either all pu or all actual)

Per Unit System Quick Reference

Essential Formulas

Base Values (Three-Phase):

$$\begin{align} Z_{\text{base}} &= \frac{V_{\text{base}}^2}{S_{\text{base}}} \\ I_{\text{base}} &= \frac{S_{\text{base}}}{\sqrt{3} \times V_{\text{base}}} \end{align}$$

Per Unit Conversion:

$$\text{Per Unit} = \frac{\text{Actual Value}}{\text{Base Value}}$$

Base Change:

$$Z_{\text{pu,new}} = Z_{\text{pu,old}} \times \frac{S_{\text{base,new}}}{S_{\text{base,old}}} \times \left(\frac{V_{\text{base,old}}}{V_{\text{base,new}}}\right)^2$$

Typical Per Unit Ranges

Quantity	Normal Range	Action Required
Voltage	0.95 - 1.05 pu	Outside range: voltage regulation needed
Current	0.5 - 1.0 pu	Above 1.0: overload condition
Generator Impedance	0.1 - 0.3 pu	-
Transformer Impedance	0.05 - 0.15 pu	-
Fault Current	3 - 20 pu	Determines breaker ratings

Conclusion: Mastering Per Unit Analysis

The per unit system transforms complex power system calculations into manageable analysis by normalizing all quantities to dimensionless values near 1.0. Whether you're calculating fault currents, analyzing voltage drops, or designing protection schemes, this method provides clarity and accuracy across multi-voltage networks.

Key Takeaways:

• Choose consistent base values (typically $S_{base}$ and $V_{base}$)
• Convert all equipment impedances to common system base
• Base voltage changes at transformers, base power stays constant
• All calculations in per unit, convert back to actual values for final results
• Per unit values near 1.0 indicate normal operation

Mastering per unit calculations is essential for power system engineers working with transformers, generators, and transmission networks. Practice with progressively complex systems to build confidence and proficiency.

Ready to apply per unit analysis to real systems? Start with single-generator configurations, then progress to multi-transformer networks and finally complete power systems with multiple voltage levels.

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🔗 Related Posts

• Transformer Sizing: Complete Guide with Calculations
• Three Phase Generators: How They Work and Applications
• Complete Guide to Voltage Drop Calculations
• Circuit Breaker Sizing: Complete Guide
• Understanding Electric Current

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• Voltage Drop Calculator
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Credits

• Photo by David Vives on Unsplash

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