# Per Unit System in PE Power Exam

In power engineering, understanding the behavior and characteristics of electrical systems is essential for efficient operation, fault analysis, and protection coordination. One powerful tool that simplifies complex calculations and harmonizes various system components is the Per Unit System.

Preparing for the per unit systems in the PE Power exam allows you to discover a standardized and simplified approach to represent electrical quantities, such as voltage, current, impedance, and power, in a normalized form.

By embracing per-unit systems, power engineers gain a comprehensive and standardized framework for analyzing power system components, assessing fault conditions, optimizing equipment ratings, and achieving efficient protection coordination.

In this detailed study guide for per-unit systems in PE Power exam, we delve into the power domain with different per-unit PE Exam practice problems, unraveling the technical complexities of per-unit examples and use cases.

Let’s explore mathematical foundations, conversion techniques, and the benefits of per-unit systems for various use cases in power engineering.

## Basics of the Per-Unit System

A** per-unit system** is a mathematical approach used in power engineering to normalize and simplify electrical calculations by expressing quantities relative to a chosen base value. It is particularly useful for analyzing and comparing electrical systems, equipment, and components.

**Normalization **is expressing quantities relative to a chosen base value. *

**Base values **are predetermined reference values that represent the normal operating conditions of a power system. By using base values, engineers can analyze and compare plans more effectively.*

Check out the following table for the representation or notation of **Normalized **and **base values** that we will use in our guide for the Per-Unit System in the PE Power Exam.

Quantity | Actual Value | Base Value | Normalized Value |

Current (I) | I (in amperes) | Ibase (in A) | Ipu |

Voltage (V) | V (in volts) | Vbase (in V) | Vpu |

Impedance (Z) | Z (in ohms) | Zbase (in Ω) | Zpu |

Power (P) | P (in watts) | Pbase (in W) | Ppu |

Reactance (X) | X (in ohms) | Xbase (in Ω) | Xpu |

By normalizing the values using base values, engineers can perform calculations and comparisons more efficiently and accurately, regardless of the actual magnitudes and units of the quantities involved.

**It is crucial to select appropriate base values that represent the typical operating conditions of the system to ensure meaningful analysis and comparison.*

### Per Unit System Values

In the per-unit system, all quantities, such as voltage, current, power, and impedance, are expressed as ratios or percentages concerning their respective base values.

These base values are typically chosen to represent the typical or rated operating conditions of the system, making it easier to analyze the behavior of different components under varying conditions.

#### Per Unit Current (Ipu)

The **per-unit current **represents the electrical current relative to a chosen base value. It is usually expressed as a decimal or percentage. The per-unit current (Ipu) can be calculated using the formula:

**Ipu = I / Ibase**

Where:

- Ipu = Per unit current
- I = Actual current in amperes
- Ibase = Base current in amperes

Per-unit current is used to compare the current magnitude in different power systems parts, such as transformers, generators, and transmission lines.

#### Per Unit Impedance (Zpu)

Per unit impedance represents impedance (both resistance and reactance) relative to a chosen base value. It is commonly used in power system analysis. The per-unit impedance (Zpu) can be calculated using the formula:

**Zpu = Z / Zbase**

Where

- Zpu = Per unit impedance
- Z = Actual impedance in ohms
- Zbase = Base impedance in ohms

The per-unit impedance simplifies the analysis of electrical circuits by removing the need to consider the actual magnitude of the impedance. It allows for a direct comparison of the impedance of different components.

#### Per Unit Voltage (Vpu)

Per unit voltage represents voltage relative to a chosen base value. It is expressed as a decimal or percentage. The per-unit voltage (Vpu) can be calculated using the formula:

**Vpu = V / Vbase**

Where

- Vpu = Per unit voltage
- V = Actual voltage in volts
- Vbase = Base voltage in volts

The per-unit voltage is often used in power system analysis to determine voltage drop and regulation across different system parts.

### Advantages of Per Unit Systems

The per-unit systems have vast implications in power engineering. The following key points reflect why preparing for the Per Unit System in the PE Power Exam is necessary.

**Simplifies Comparisons**– Per-unit systems allow for easy and direct comparisons of quantities such as current, voltage, and impedance across different components or systems. It simplifies the analysis and evaluation of electrical systems.**Scalability & Flexibility**– Per-unit systems enable scalability, as the same base values can be used to analyze power systems of different scales. This makes applying the same calculations to different scenarios convenient without extensive modifications.**Independence from Unit Conversions**– Using relative values, per-unit systems eliminate the need for unit conversions. This is particularly useful when working with techniques that have different voltage levels or use other units of measurement.**Clear Representation of Magnitudes**– Expressing quantities per unit system values provides a more explicit representation of magnitudes. It allows engineers to focus on the relative importance and behavior of different components without being influenced by absolute values.**Flexibility in Base Value Selection**– Per-unit systems offer flexibility in choosing base values. Engineers can select base values that best represent the normal operating conditions of the system, making the analysis more relevant and accurate.

## Per-Unit System in Power System Analysis

As discussed, the per-unit system is used in power system analysis to simplify calculations and comparisons by expressing quantities relative to chosen base values. It allows engineers to analyze and design electrical systems more efficiently.

Let’s explore the per-unit system’s mathematical techniques with different examples to showcase the fundamental mathematical calculations.

### Conversion to Per-Unit System Values

We must determine the appropriate base values for each quantity to convert actual values to per-unit system values. Once the base values are established, the conversion can be done using the following formulas:

Parameter | Per Unit Formula |

Current | Ipu = I / Ibase |

Voltage | Vpu = V / Vbase |

Impedance | Zpu = Z / Zbase |

Power | Ppu = P / Pbase |

Reactance | Xpu = X / Xbase |

Here, I, V, Z, P, and X represent the actual values of current, voltage, impedance, power, and reactance, respectively. Ibase, Vbase, Zbase, Pbase, and Xbase are the chosen base values for each quantity.

Have a look at the following values:

Parameter | Actual Value | Base Value | Per Unit Value (pu) |

Current | 100 A | 50 A | 2 pu |

Voltage | 240 V | 120 V | 2 pu |

Impedance | 10 Ω | 5 Ω | 2 pu |

Power | 500 W | 250 W | 2 pu |

Reactance | 8 Ω | 4 Ω | 2 pu |

The per-unit system values in the last column are calculated using the formulas mentioned above.

- Ipu = 100 A / 50 A = 2 pu
- Vpu = 240 V / 120 V = 2 pu
- Zpu = 10 Ω / 5 Ω = 2 pu
- Ppu = 500 W / 250 W = 2 pu
- Xpu = 8 Ω / 4 Ω = 2 pu

In power system analysis, components such as transformers, generators, and transmission lines are often represented in per-unit system values. This simplifies calculations and allows for better understanding and comparison of these components.

For example, a transformer with a per unit impedance of 0.05 pu indicates that its impedance is **5%** of the chosen base impedance. Similarly, a generator with a per unit voltage of 1.2 pu implies that its voltage is **20%** higher than the chosen base voltage.

### Per Unit PE Exam Practice Problem

Consider a power system with the following data:

- Generator voltage (Vg) = 13.8 kV
- Generator power (Pg) = 50 MW
- Transmission line impedance (Z) = 0.05 pu
- Load impedance (Zload) = 0.06 pu
- Line length (l) = 100 km

**Step 1 **– Determine the base values.

Consider the following base values for a given power system:

- Base voltage (Vbase) = 10 kV
- Base power (Pbase) = 100 MVA

Base impedance (Zbase) = Vbase^{^}2 / Pbase = 10 kV^{^}2 / 100 MVA = 1 Ω *[From **Watt’s Law**, P = V ^{^}2 / R]*

**Step 2** – Convert the given values to per-unit system values.

Vgpu = Vg / Vbase = 13.8 kV / 10 kV = 1.38 pu (Generator voltage)

Pgpu = Pg / Pbase = 50 MW / 100 MVA = 0.5 pu (Generator power)

Zpu = Z / Zbase = 0.05 pu (Transmission line impedance)

Zloadpu = Zload / Zbase = 0.06 pu (Load impedance)

**Step 3 **– Calculate the apparent power of the generator.

Sg = Pgpu + jQg (complex power)

|Sg| = |Vgpu * Ig| (apparent power)

|Ig| = |Sg| / |Vgpu| (generator current)

|Ig| = |Vgpu| * √(1 – (Pgpu / |Vgpu|)^{^}2) (magnitude of current)

Substituting the given values:

|Ig| = 1.38 * √(1 – (0.5 / 1.38)^{^}2)

|Ig| ≈ 1.38 * √(1 – 0.1825^{^}2)

|Ig| ≈ 1.38 * √(1 – 0.0333)

|Ig| ≈ 1.38 * √(0.9667)

|Ig| ≈ 1.38 * 0.9832

|Ig| ≈ 1.355 pu (generator current)

**Step 4 **– Calculate the transmission line current.

|Il| = |Ig| (since there are no losses in the line)

|Il| = 1.355 pu (transmission line current)

**Step 5** – Calculate the voltage drop across the transmission line.

|Vl| = |Il| * |Zpu| (magnitude of voltage drop)

|Vl| = 1.355 * 0.05

|Vl| = 0.06775 pu (magnitude of voltage drop)

Vl = |Vgpu| – |Vl| (final voltage at the load)

Vl = 1.38 – 0.06775

Vl ≈ 1.31225 pu (final voltage at the load)

**Step 6** – Convert the per unit values back to actual values.

Vl_actual = Vl * Vbase = 1.31225 * 10 kV

Vl_actual ≈ 13.1225 kV (voltage at the load in volts)

The analysis shows the power flow and voltage drop in the power system. Here’s what the analysis demonstrates:

**Generator Current**– The analysis determines the magnitude of the current flowing from the generator, which is approximately 1.355 per unit (pu). This information is crucial in understanding the current levels within the system and ensuring that the generator can supply the required power.**Voltage Drop**– The analysis calculates the magnitude of the voltage drop across the transmission line, which is approximately 0.06775 pu. This voltage drop represents the energy losses and impedance effects experienced during power transmission. It is essential to consider the voltage drop to ensure the load receives the desired voltage level.**Load Voltage**– The analysis determines the final voltage at the load, which is approximately 1.31225 pu. This value represents the voltage level at the load terminals after considering the generator voltage and the voltage drop across the transmission line. It helps assess the voltage regulation and the suitability of the power supply for the connected load.

The analysis provides insights into the power system’s behavior, including the current flow, voltage drop, and load voltage. This information aids in assessing the system’s performance, identifying potential issues such as excessive voltage drop or inadequate voltage regulation, and making informed decisions to optimize the power system’s operation.

## Per Unit System – Equipment Ratings

Let’s consider another Per Unit PE Exam Practice Problem where you are designing a power distribution system with a 480V source. The system consists of a transformer, a motor, and a feeder cable. The equipment ratings are as follows:

**Transformer**

- Primary Voltage: 480V
- Secondary Voltage: 240V
- Apparent Power: 100 kVA

**Motor**

- Voltage: 240V
- Power Factor: 0.8
- Power: 50 hp

**Feeder Cable**

- Voltage Rating: 600V
- Current Rating: 200A

**Step 1** – Calculate the per-unit ratings for the transformer.

**a) Transformer Primary Voltage (Vpu)**

**Vpu = Primary Voltage / Base Voltage**

Vpu = 480V / 480V = 1 pu

**b) Transformer Secondary Voltage (Vpu)**

**Vpu = Secondary Voltage / Base Voltage**

Vpu = 240V / 480V = 0.5 pu

**c) Transformer Apparent Power (Spu)**

**Spu = Apparent Power / Base Apparent Power**

*The Base Apparent Power is typically chosen based on system requirements. Let’s assume it to be @ 100 MVA.*

Spu = 100 kVA / 100 MVA = 0.001 pu

**Step 2** – Calculate the per-unit ratings for the motor.

**a) Motor Voltage (Vpu)**

**Vpu = Voltage / Base Voltage**

Vpu = 240V / 480V = 0.5 pu

**b) Motor Power (Ppu)**

**Ppu = (Power / Base Power) * (Power Factor)**

*The Base Power is typically chosen based on system requirements. Let’s assume it to be rated @ 100 MVA.*

Ppu = (50 hp / 100 MVA) * 0.8 = 0.0004 pu

**Step 3 **– Calculate the per-unit ratings for the feeder cable.

**a) Cable Voltage Rating (Vpu)**

**Vpu = Voltage Rating / Base Voltage**

Vpu = 600V / 480V = 1.25 pu

**b) Cable Current Rating (Ipu)**

**Ipu = Current Rating / Base Current Rating**

*The Base Current Rating is typically chosen based on system requirements. Let’s assume it to be rated @ 100 A.*

Ipu = 200A / 100A = 2 pu

Per-unit Representation of Equipment Parameters for the given system are:

Equipment | Voltage (Vpu) | Power (Ppu) | Current (Ipu) | Apparent Power (Spu) |

Transformer | 1.0 pu | 0.001 pu | – | 0.001 pu |

Motor | 0.5 pu | 0.0004 pu | – | – |

Feeder Cable | 1.25 pu | – | 2 pu | – |

### Analysis of the Above Outcomes

#### Transformer

The per-unit ratings for the transformer show that the primary voltage (Vpu) is 1 pu, indicating that the primary voltage is at the base voltage level. The secondary voltage (Vpu) is 0.5 pu at half the base voltage level.

The apparent power (Spu) is 0.001 pu, indicating that the transformer’s rated apparent power is relatively small compared to the base apparent power.

#### Implications

- The transformer can handle the given voltage levels without requiring any voltage transformation since the primary voltage is at the base voltage level.
- The transformer is designed for a relatively small apparent power compared to the base apparent power, suggesting it is suitable for low-power applications or specific loads.

#### Motor

The per-unit ratings for the motor reveal that the motor voltage (Vpu) is 0.5 pu, indicating that it operates at half of the base voltage level. The power (Ppu) is 0.0004 pu, considering the power factor of 0.8.

#### Implications

- The motor operates at a voltage level lower than the base voltage level, suggesting it is designed for low-voltage applications.
- The motor has a relatively low power rating compared to the base power rating, indicating it is suitable for low-power demand applications.

#### Feeder Cable

The per-unit ratings for the feeder cable show that the cable voltage rating (Vpu) is 1.25 pu, indicating it can handle higher voltage levels than the base voltage. The cable current rating (Ipu) is 2 pu, suggesting it can carry twice the base current rating.

#### Implications

- The feeder cable is rated for higher voltage levels, providing flexibility for connecting it to different voltage sources within the system.
- The cable has a higher current rating compared to the base current rating, indicating it can handle higher current loads without exceeding its capacity.

These per-unit ratings provide valuable information for equipment selection and coordination in a power distribution system. By considering the per-unit values, engineers can ensure that the selected equipment is compatible with the system requirements and adequately coordinated with other components.

Some key considerations include:

**Voltage Levels**– The per-unit system ratings help select equipment that matches the desired voltage levels within the system. It ensures proper voltage transformation and compatibility between different components.**Power Ratings**– The per-unit power rating assists in equipment selection based on the power demand of specific loads. It helps avoid under-sizing or over-sizing equipment, ensuring efficient and reliable operation.**Current Handling**– The per-unit current rating aid in selecting cables and conductors that can handle the expected current loads without exceeding their capacity. It ensures safe and efficient power transmission.**Coordination**– The per-unit system values assist in coordinating the ratings of various equipment within the system to ensure compatibility and efficient operation. It helps avoid mismatched ratings leading to operational issues or equipment damage.

Learn our detailed guide on Demand Calculations in the PE Power exam and discover how to get valuable insights into load analysis, capacity planning, and system optimization.

## Per Unit System in Fault Analysis

Applying per-unit systems in fault analysis brings simplicity and standardization to analyzing fault currents and voltages. It enables efficient fault calculations and helps in power system protection coordination & understanding of the system’s behavior during faults. Here’s how:

**Simplified Fault Analysis**– Using per-unit system values simplifies fault analysis calculations. By expressing system parameters and fault currents in per-unit system values, it becomes easier to compare and analyze the impact of faults on different system components.**Protection Coordination**– Per-unit systems facilitate effective coordination of protection devices. Protection devices such as circuit breakers and relays are often set based on per-unit system values, allowing for consistent and coordinated operation during fault conditions. The per-unit representation enables easy comparison and coordination between protection devices at different voltage levels.**Voltage-Current Relationship**– Per unit values provide a standardized representation of fault currents and voltages. By expressing fault currents and voltages per unit, it becomes easier to understand the relationship between them and analyze the system’s behavior during faults.

Before moving further, read our detailed guide on symmetrical components in the PE Power exam.

Let’s consider another Per Unit PE Exam Practice Problem that involves calculating fault analysis using the per-unit system. For instance, a power system has the following data:

- Base Voltage (Vbase): 11 kV
- Generator Rating (Sbase): 100 MVA
- Transformer Rating (Sbase): 50 MVA
- Fault Impedance (Zf): 0.1 pu

**Step 1** – Fault Calculation

Consider a three-phase symmetrical fault at the terminals of the transformer. The fault impedance is given as 0.1 pu.

**a) Calculate the fault current using the ****symmetrical component method****.**

**Ia = Vbase / (Zf + Zt)**

Ia = 11 kV / (0.1 pu + j0.1 pu)

Ia = 110 kV / (0.1414 pu ∠ 45°) ≈ 779.6 A ∠ -45°

**b) Calculate the fault current (If) per unit (pu) using the base values.**

**If = Ia / (Sbase / Vbase)**

If = 779.6 A ∠ -45° / (50 MVA / 11 kV) ≈ 0.31 pu ∠ -45°

**Step 2** – Per Unit Representation of Fault Current and Voltage

The fault current and voltage are represented relative to their respective base values in the per-unit system. In this case, the fault current If is 0.31 pu ∠ -45°.

Read our detailed guide on Reliability in the PE Power exam and discover how to safeguard against unexpected failures, optimizes efficiency, and ensure an uninterrupted electric supply.

### Conclusion

The uses and benefits of per-unit systems in PE Power exam extend beyond the boundaries of academic studies and serve as an invaluable asset in your power engineering career. Power engineers can gain a competitive edge in their field by mastering the principles and techniques of per-unit representation discussed above.

The ability to simplify complex calculations, analyze system behavior during faults, and optimize equipment ratings is a skill that sets power engineers apart.

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