3.3 The Per-Unit (pu) System: Bases, Conversions & System Modeling

3.3 The Per-Unit (pu) System: Bases, Conversions & System Modeling

Executive Overview: Power systems encompass vast networks spanning multiple voltage levels—from generator terminals ($13.8\text{ kV}$) to bulk transmission ($115\text{ kV} - 765\text{ kV}$) down to industrial distribution ($480\text{ V} - 13.8\text{ kV}$). Analyzing such systems in physical ohms requires continually reflecting impedances across transformer turns ratios ($a^2$). The Per-Unit (pu) System normalizes all voltages, currents, powers, and impedances to dimensionless ratios, eliminating ideal transformers and condensing equipment parameters into predictable, standardized ranges. Mastery of per-unit base transformations is among the highest-yield competencies on the NCEES PE Power examination.

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1. Core Principles & Advantages of the Per-Unit System

The per-unit value of any electrical quantity is defined as:

$$\text{Quantity}_{\text{pu}} = \frac{\text{Actual Quantity (in physical engineering units)}}{\text{Base Value of Quantity (in identical physical units)}}$$ $$\text{Quantity}_{\%} = \text{Quantity}_{\text{pu}} \times 100\%$$

+---------------------------------------------------------------------------------------------------+
| KEY ENGINEERING ADVANTAGES OF THE PER-UNIT SYSTEM                                                 |
+---------------------------------------------------------------------------------------------------+
| 1. Eliminates Ideal Transformers: Transformer turns ratios become 1:1 in per-unit, reducing       |
|    multi-voltage networks to simple connected impedance diagrams without turns-ratio scaling.     |
| 2. Standardized Parameter Ranges: Equivalent impedances of similar apparatus fall within narrow   |
|    numerical bands (e.g., power transformers typically have X_leakage = 0.06 - 0.12 pu), making   |
|    erroneous data immediately detectable.                                                         |
| 3. Equal High-Side and Low-Side pu Impedance: A transformer's per-unit impedance is identical     |
|    whether calculated from the primary or secondary winding terminals.                            |
| 4. Intuitive Voltage Profiles: Operating voltages near nominal are close to 1.00 pu (e.g.,         |
|    0.95 pu indicates a 5% undervoltage; 1.05 pu indicates a 5% overvoltage).                      |
+---------------------------------------------------------------------------------------------------+

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2. Selection & Derivation of System Base Quantities

A three-phase electrical network is defined by four fundamental interrelated variables: Apparent Power ($S$), Voltage ($V$), Current ($I$), and Impedance ($Z$). In standard practice, two independent base quantities are arbitrarily selected, which mathematically dictate the remaining two derived bases.

Standard Base Selection Protocol

1. System Three-Phase Apparent Power Base ($S_{\text{base},3\phi}$): Chosen globally for the entire study (standard utility convention is $100\text{ MVA}$ or $10\text{ MVA}$; industrial studies often select $10\text{ MVA}$ or $1\text{ MVA}$).
   • Single-phase power base: $S_{\text{base},1\phi} = \frac{S_{\text{base},3\phi}}{3}$.
2. System Line-to-Line Voltage Base ($V_{\text{base},LL}$): Selected for one reference voltage zone, and then propagated across each transformer according to its rated voltage ratio.
   • Line-to-neutral voltage base: $V_{\text{base},LN} = \frac{V_{\text{base},LL}}{\sqrt{3}}$.

Derivation of Derived Base Current ($I_{\text{base}}$)

$$I_{\text{base}} = \frac{S_{\text{base},3\phi}}{\sqrt{3} V_{\text{base},LL}} = \frac{S_{\text{base},1\phi}}{V_{\text{base},LN}} \quad [\text{Amperes}]$$

Derivation of Derived Base Impedance ($Z_{\text{base}}$)

$$Z_{\text{base}} = \frac{V_{\text{base},LN}}{I_{\text{base}}} = \frac{V_{\text{base},LL} / \sqrt{3}}{S_{\text{base},3\phi} / (\sqrt{3} V_{\text{base},LL})} = \frac{(V_{\text{base},LL})^2}{S_{\text{base},3\phi}} \quad [\Omega]$$

When line-to-line base voltage is expressed in kilovolts ($V_{\text{base},\text{kV}}$) and three-phase base power in megavolt-amperes ($S_{\text{base},\text{MVA}}$):

$$Z_{\text{base}} = \frac{(V_{\text{base},\text{kV}} \times 10^3)^2}{S_{\text{base},\text{MVA}} \times 10^6} = \frac{(V_{\text{base},\text{kV}})^2}{S_{\text{base},\text{MVA}}} \quad [\Omega]$$

Derivation of Derived Base Admittance ($Y_{\text{base}}$)

$$Y_{\text{base}} = \frac{1}{Z_{\text{base}}} = \frac{S_{\text{base},\text{MVA}}}{(V_{\text{base},\text{kV}})^2} \quad [\text{Siemens}]$$

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| PER-UNIT SYSTEM FUNDAMENTAL BASE EQUATIONS (PE FORMULA MATRIX)                                    |
+---------------------------------------------------------------------------------------------------+
| Base Current:    |  I_base = S_base,3φ / (sqrt(3) * V_base,LL)  [A]                              |
| Base Impedance:  |  Z_base = (V_base,LL_kV)^2 / S_base,3φ_MVA  [Ω]                               |
| Physical Ohms:   |  Z_pu = Z_actual_ohms / Z_base  = Z_actual_ohms * (S_base,MVA / V_base,kV^2) |
| Actual Current:  |  I_actual = I_pu * I_base  [A]                                                 |
| Actual Voltage:  |  V_actual = V_pu * V_base  [V]                                                 |
+---------------------------------------------------------------------------------------------------+

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3. Changing Per-Unit Impedance Bases (The Master Scaling Law)

Manufacturer nameplates state equipment impedance in per-unit on the equipment's own nominal ratings ($S_{\text{base},\text{old}}$ and $V_{\text{base},\text{old}}$). Before assembling a system-wide reactance network on a common system base ($S_{\text{base},\text{new}}$ and $V_{\text{base},\text{new}}$), every per-unit impedance must be rescaled using the Master Base Conversion Formula:

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

Analytical Derivation

1. Convert the old per-unit impedance to physical ohms ($Z_\Omega$): $Z_\Omega = Z_{\text{pu},\text{old}} \times Z_{\text{base},\text{old}} = Z_{\text{pu},\text{old}} \times \frac{(V_{\text{base},\text{old}})^2}{S_{\text{base},\text{old}}}$
2. Convert physical ohms to the new per-unit base: $Z_{\text{pu},\text{new}} = \frac{Z_\Omega}{Z_{\text{base},\text{new}}} = \frac{Z_{\text{pu},\text{old}} \times \frac{(V_{\text{base},\text{old}})^2}{S_{\text{base},\text{old}}}}{\frac{(V_{\text{base},\text{new}})^2}{S_{\text{base},\text{new}}}} = Z_{\text{pu},\text{old}} \times \left(\frac{S_{\text{base},\text{new}}}{S_{\text{base},\text{old}}}\right) \times \left(\frac{V_{\text{base},\text{old}}}{V_{\text{base},\text{new}}}\right)^2$

Exam Warning on Voltage Squaring: In the base conversion formula, power scales linearly ($\frac{S_{\text{new}}}{S_{\text{old}}}$), whereas voltage scales with the square of the inverse ratio ($\left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2$). Omitting the square exponent is one of the most common calculation mistakes on the PE Power exam!

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4. Multi-Voltage Zone Power System Partitioning

In a multi-voltage power network, transformers establish distinct voltage zones. Once a base voltage is chosen for any one zone, the base voltages for all other zones are strictly locked by the nominal rated turns ratios of the intervening transformers.

MULTI-ZONE VOLTAGE PARTITIONING:
 [Zone 1: Generation]       [Zone 2: Transmission]       [Zone 3: Distribution]
   V_base1 = 13.8 kV           V_base2 = 138 kV             V_base3 = 4.16 kV
+---------------------+     +---------------------+     +---------------------+
|  (~) Generator      | T1  |   Transmission      | T2  |    Motor Load       |
|  13.8 kV Nominal    |==#==|   Line              |==#==|    4.16 kV Nominal  |
+---------------------+     +---------------------+     +---------------------+
        T1 Ratio:                   T2 Ratio:
     13.8 kV / 138 kV            138 kV / 4.16 kV

Step-by-Step Zone Propagation Rules

1. Set $V_{\text{base}1}$ in Zone 1 (e.g., $V_{\text{base}1} = 13.8\text{ kV}$).
2. Across Transformer $T_1$ (ratio $N_1 : N_2 = 13.8\text{ kV} : 138\text{ kV}$): $V_{\text{base}2} = V_{\text{base}1} \times \left(\frac{V_{T1,\text{sec}}}{V_{T1,\text{pri}}}\right) = 13.8\text{ kV} \times \left(\frac{138\text{ kV}}{13.8\text{ kV}}\right) = 138\text{ kV}$
3. Across Transformer $T_2$ (ratio $N_2 : N_3 = 138\text{ kV} : 4.16\text{ kV}$): $V_{\text{base}3} = V_{\text{base}2} \times \left(\frac{V_{T2,\text{sec}}}{V_{T2,\text{pri}}}\right) = 138\text{ kV} \times \left(\frac{4.16\text{ kV}}{138\text{ kV}}\right) = 4.16\text{ kV}$
4. Off-Nominal Transformer Voltage Ratings: If a motor or generator operates at a nominal nameplate voltage that differs from the zone base voltage (e.g., a $4.0\text{ kV}$ motor installed in a $4.16\text{ kV}$ zone), the voltage correction term $\left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2 = \left(\frac{4.0}{4.16}\right)^2 = 0.9245$ must be included in that equipment's base transformation.

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5. One-Line Diagram to Per-Unit Reactance Diagram Conversion

To perform fault analysis or load flow on a power system, the physical one-line diagram is systematically transformed into a single-phase per-unit reactance network:

PHYSICAL ONE-LINE DIAGRAM:
  [Gen G1]----(Bus 1)====[Trans T1]====(Bus 2)----[Trans Line]----(Bus 3)====[Trans T2]====(Bus 4)----[Motor M1]

PER-UNIT REACTANCE DIAGRAM (ALL VALUES ON COMMON 100 MVA BASE):
    jX_G1          jX_T1           Z_line,pu          jX_T2          jX_M1
  +-[ZZZZ]- Bus 1 -[ZZZZ]- Bus 2 --[ZZZZZZZZ]-- Bus 3 -[ZZZZ]- Bus 4 -[ZZZZ]-+
  |                                                                          |
 (+)[E_G1]                                                                 (+)[E_M1]
  |                                                                          |
 === Reference Bus (Ground Neutral) ==========================================

Systematic 5-Step Network Modeling Algorithm

1. Select Global System Bases: Establish a common three-phase power base $S_{\text{base},3\phi}$ (e.g., $100\text{ MVA}$) and a voltage base for one designated zone.
2. Determine Base Voltages for All Zones: Calculate $V_{\text{base}}$ for each zone using transformer nominal turns ratios.
3. Compute Base Impedances ($Z_{\text{base}}$) and Base Currents ($I_{\text{base}}$): Determine $Z_{\text{base}} = (V_{\text{base},\text{kV}})^2 / S_{\text{base},\text{MVA}}$ for every zone.
4. Convert All Component Impedances to System Base:
   • Generators & Motors: Rescale per-unit subtransient reactances $X_d''$ via the Master Base Conversion Formula.
   • Transformers: Rescale per-unit leakage reactances $X_T$.
   • Transmission Lines & Cables: Convert physical series ohms ($Z_\Omega$) to per-unit: $Z_{\text{pu}} = Z_\Omega / Z_{\text{base},\text{zone}}$.
5. Assemble the Reactance Diagram & Solve: Combine all per-unit impedances into a single interconnected network. Solve for per-unit fault currents or bus voltages, then multiply by zone base quantities to recover actual physical values.

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6. Comprehensive Step-by-Step Worked System Example

Problem Statement

A balanced three-phase power system has the following equipment specifications:

• Generator $G_1$: $50\text{ MVA}$, $13.8\text{ kV}$, subtransient reactance $X_d'' = 0.15\text{ pu}$.
• Step-Up Transformer $T_1$: $60\text{ MVA}$, $13.8\text{ kV (}\Delta\text{)} - 115\text{ kV (Y)}$, leakage reactance $X_{T1} = 0.10\text{ pu}$.
• Transmission Line: $20\text{ miles}$ in length, series impedance $z = 0.15 + j0.80,\Omega/\text{mile}$, operating at $115\text{ kV}$.
• Step-Down Transformer $T_2$: $40\text{ MVA}$, $115\text{ kV (Y)} - 13.8\text{ kV (}\Delta\text{)}$, leakage reactance $X_{T2} = 0.08\text{ pu}$.
• Motor $M_1$: $30\text{ MVA}$, $13.2\text{ kV}$, subtransient reactance $X_m'' = 0.20\text{ pu}$.

System Base Specifications: Select a common system power base of $S_{\text{base},3\phi} = 100\text{ MVA}$ and a generator zone voltage base of $V_{\text{base}1} = 13.8\text{ kV}$.

Calculate:

1. The voltage base ($V_{\text{base}}$) and base impedance ($Z_{\text{base}}$) in all three zones.
2. The per-unit impedance of each system component on the common $100\text{ MVA}$ base.
3. The total series Thevenin impedance between the internal generator EMF and motor EMF.
4. The symmetrical three-phase short-circuit current in Amperes at the motor bus (Bus 4) for a solid three-phase fault, assuming a prefault voltage of $V_f = 1.0\angle 0^\circ\text{ pu}$ and neglecting prefault load current and line resistance.

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CALCULATION WORKFLOW & DETAILED STEP-BY-STEP SOLUTION:
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Step 1: Determine Voltage and Impedance Bases for All Zones
  System Power Base: S_base,3ph = 100 MVA

  Zone 1 (Generator Zone):
    V_base1 = 13.8 kV
    Z_base1 = (V_base1)^2 / S_base = (13.8)^2 / 100 = 190.44 / 100 = 1.9044 ohms
    I_base1 = S_base / (sqrt(3) * V_base1) = 100,000 / (1.73205 * 13.8) = 4,183.7 A

  Zone 2 (Transmission Line Zone):
    V_base2 = V_base1 * (115 / 13.8) = 13.8 * (115 / 13.8) = 115.0 kV
    Z_base2 = (V_base2)^2 / S_base = (115.0)^2 / 100 = 13225.0 / 100 = 132.25 ohms
    I_base2 = S_base / (sqrt(3) * V_base2) = 100,000 / (1.73205 * 115.0) = 502.04 A

  Zone 3 (Motor Zone):
    V_base3 = V_base2 * (13.8 / 115.0) = 115.0 * (13.8 / 115.0) = 13.8 kV
    Z_base3 = (V_base3)^2 / S_base = (13.8)^2 / 100 = 1.9044 ohms
    I_base3 = S_base / (sqrt(3) * V_base3) = 4,183.7 A

Step 2: Convert Component Impedances to Common 100 MVA Base
  1. Generator G1 (50 MVA, 13.8 kV, X" = 0.15 pu):
     X_G1,new = X_G1,old * (S_base,new / S_base,old) * (V_base,old / V_base,new)^2
              = 0.15 * (100 / 50) * (13.8 / 13.8)^2
              = 0.15 * 2.0 * 1.0
              = 0.3000 pu

  2. Step-Up Transformer T1 (60 MVA, 13.8/115 kV, X = 0.10 pu):
     X_T1,new = 0.10 * (100 / 60) * (115 / 115)^2
              = 0.10 * 1.6667 * 1.0
              = 0.1667 pu

  3. Transmission Line (20 miles * (0.15 + j0.80 ohms/mi) = 3.0 + j16.0 ohms):
     Z_line,pu = Z_actual / Z_base2
               = (3.0 + j16.0) / 132.25
               = (3.0 / 132.25) + j(16.0 / 132.25)
               = 0.02268 + j0.12098 pu

  4. Step-Down Transformer T2 (40 MVA, 115/13.8 kV, X = 0.08 pu):
     X_T2,new = 0.08 * (100 / 40) * (115 / 115)^2
              = 0.08 * 2.50 * 1.0
              = 0.2000 pu

  5. Motor M1 (30 MVA, 13.2 kV, X" = 0.20 pu):
     Notice: Motor nominal voltage (13.2 kV) differs from Zone 3 base voltage (13.8 kV)!
     X_M1,new = X_M1,old * (S_base,new / S_base,old) * (V_base,old / V_base,new)^2
              = 0.20 * (100 / 30) * (13.2 / 13.8)^2
              = 0.20 * 3.3333 * (0.95652)^2
              = 0.66667 * 0.91493
              = 0.60995 pu ≈ 0.6100 pu

Step 3: Total Series Impedance Calculation (Generator to Motor)
  Z_series = jX_G1 + jX_T1 + Z_line + jX_T2 + jX_M1
           = j0.3000 + j0.1667 + (0.0227 + j0.1210) + j0.2000 + j0.6100
           = 0.0227 + j(0.3000 + 0.1667 + 0.1210 + 0.2000 + 0.6100)
           = 0.0227 + j1.3977 pu

Step 4: Symmetrical Short-Circuit Fault Current Calculation at Motor Bus (Bus 4)
  For a solid 3-phase fault at Bus 4 (upstream Thevenin reactance excluding motor):
    X_th,upstream = X_G1 + X_T1 + X_line + X_T2
                  = 0.3000 + 0.1667 + 0.1210 + 0.2000
                  = 0.7877 pu

  Per-unit fault current contributed from utility/generator source:
    I_f,pu = V_f / X_th,upstream
           = 1.0 / 0.7877
           = 1.2695 pu

  Convert per-unit fault current to actual Amperes in Zone 3:
    I_f,actual = I_f,pu * I_base3
               = 1.2695 * 4,183.7 A
               = 5,311.2 A = 5.311 kA
=========================================================================================

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7. Common Exam Traps & Tactical Pitfalls

• Omitting the Square on the Voltage Conversion Ratio: Using $\frac{V_{\text{old}}}{V_{\text{new}}}$ instead of $\left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2$. Because impedance is proportional to voltage squared ($Z \propto V^2$), the voltage adjustment ratio must always be squared.
• Inverting the Voltage Ratio: Multiplying by $\left(\frac{V_{\text{new}}}{V_{\text{old}}}\right)^2$ instead of $\left(\frac{V_{\text{old}}}{V_{\text{new}}}\right)^2$. An apparatus with a lower rated voltage operating in a higher base voltage zone will have a smaller per-unit impedance on the new base.
• Single-Phase vs. Three-Phase Base Power Mismatch: Using single-phase MVA in $Z_{\text{base}} = \frac{V_{LL}^2}{S}$ with line-to-line voltage. Remember: $Z_{\text{base}} = \frac{(V_{\text{base},LL})^2}{S_{\text{base},3\phi}} = \frac{(V_{\text{base},LN})^2}{S_{\text{base},1\phi}}$. Mixing $V_{LL}$ with $S_{1\phi}$ causes a factor of 3 error!
• Neglecting Off-Nominal Equipment Nameplates: Assuming equipment rated voltage always matches the zone base voltage. Always check the motor and transformer nameplates against the zone base.