Transmission & Distribution
for the Power P.E. Exam

by Justin Kauwale, P.E.

Introduction

The section, Transmission & Distribution, accounts for approximately 10 questions on the Power Engineering, Electrical PE exam. Transmission covers the extra-high (>230 kV), and high (115 kV to 230 kV) voltage lines that transport electricity from the electric power plants (generators) to the substations. These lines can travel miles and miles between substations and are typically under the jurisdiction of the electric utility. These transmission lines are also governed by the NES (under Section 11.0 Codes and Standards). The distribution system consists of the medium (69 kV to 2.4kV) and low voltage (600V and less) lines that transmit electricity between substations and also transmit electricity to consumers. These consumers can be residential, commercial voltages at 480 V and below. But these consumers can also be industrial at medium voltages (69 kV to 2.4kV). Motors and generators are not included under this section, since they are covered under Rotating Machines. Transformers are also included in a separate section, but are technically covered under Transmission and Distribution. Protection devices are also technically part of Transmission and Distribution but are excluded from this section and are included in its own section.



The information shown on this website is a sample of the material provided in the technical study guide and sample exam. See the STORE to purchase these items.

Transmission Analysis

Transmission lines may seem very simple, since transmission lines are just very long conductors. However, the long length and high voltages on transmission lines make the analysis of transmission lines a little more difficult than a simple wire running from your residential panel to lights. On the PE exam, you should understand the following three parameters of transmission lines, (1) resistance, (2) inductance and (3) capacitance.

Transmission lines are typically constructed of either aluminum or copper. Aluminum conductors are lighter and cheaper and copper conductors are more expensive and heavier. However, aluminum is less conductive than copper. This means that larger diameter aluminum conductors are needed to transmit the same amount of current than a smaller diameter copper conductor. Conductors are also provided with insulation and possibly strengthening steel to make the conductor more robust.

The purpose of transmission systems and distribution systems is to maintain power at all times and to ensure stable voltage and frequency.

Equivalent Circuits

An equivalent circuit is typically used to understand the three parameters of transmission lines, however before the equivalent circuit is presented, you should first understand the details of the three parameters (1) Resistance, (2) Inductance and (3) Capacitance. Each different transmission line will have an equivalent circuit that consists of these three parameters, which are uniformly distributed the length of the transmission line.

Resistance

The first voltage drop is the drop due to resistance in the transmission line. The resistance of the transmission line is directly related to the length and resistivity of the line. The voltage drop is inversely related to the cross sectional area of the line. In addition, the transmission line material will also affect the resistance of the line, which will affect the voltage drop.



Make sure you use the correct units for these variables, because the exam may try to confuse you with varying units that require converting. A circular mil is the area of a circle with a diameter of 1 mil, where 1 mil is equal to .010 inches. Sometimes the exam will give you conductor sizes in terms of American Wire Gauge (AWG). AWG is a wire size rating system used in North America and common throughout the NEC. The increasing gauge number corresponds to decreasing conductor diameter. The AWG and corresponding diameters may appear to be random, but the numbers follow the below formula. Each diameter is a function of the AWG number (N) and the reference point (N = 36), which corresponds to a diameter of 0.005 inches.



The CMIL, KCMIL and area terms are found with the following equations.



1 Mil is equal to .001 inches.







The last four AWG values, (1/0, 2/0, 3/0 and 4/0) are to describe gauges as “0”, “00”, “000” and “0000”. Once the conductor diameters are larger than this point, the AWG scale is no longer and used. The conductor diameter scale switches to KCMIL as shown below.



Now that you understand the length and area term, the last remaining term is the resistivity term. The resistivity values for the various types of copper and aluminum should be provided to you in the exam, otherwise you will have to use the values in the NEC. This shall be discussed in the Codes and Standards section. In this section, you will see that resistivity is dependent not only on materials but also on temperature.

On the PE exam, you will either be asked to solve for the resistance value of the transmission line like in this process above or you will be given the resistance value and will asked to find the voltage drop in the transmission line. Another important concept to understand is that power will be lost due to the resistance in the line. The amount of power lost will be determined by the following equation.



The information shown on this website is a sample of the material provided in the technical study guide and sample exam. See the STORE to buy the products for continuation on Transmission Analysis including the following topics:

  • Inductance
  • Capacitance
  • Geometric Mean Distance
  • Impedance
  • Short Transmission Line Equivalent Circuit
  • Medium Transmission Line Equivalent Circuit
  • Long Transmission Line
  • Voltage Drop
  • Voltage Regulation
  • Power Factor Correction
  • Correcting a Lagging Power Factor
  • Correcting a Leading Power Factor
  • Power Factor Correction Tables
  • Power Quality
  • Harmonics
  • Power Flow Between Voltage Sources

Distribution Analysis

Short-circuit faults can occur on transmission lines as well as on distribution lines. However for purposes of this book, fault current analysis is included in the distribution section. Fault current analysis involves finding the short circuit current values for various types of faults. This information is used to determine the ratings of circuit breakers, interrupting capacity.

There are 6 types of faults that you should understand for the PE exam.



The three phase fault is a symmetrical fault that is used for short circuit, over-current protection calculations. Single line-to-ground faults are the most common faults. This fault is an unsymmetrical fault and occurs when a short is made between a single phase and ground. The next fault is the line-to-line fault, this occurs when a short occurs between two phases. A double-line-to-ground fault occurs when two phases are shorted together and connected to ground. Similarly, three-line-to-ground fault occurs when three phases are shorted together and connected to ground. Finally, the line-to-ground fault with resistance is a variation on the line to ground fault. A higher resistance in the line to ground fault limits the fault current.

Symmetrical Faults

A balanced three-phase short circuit fault is a symmetrical fault. A three phase fault occurs when the three phases are shorted together and the three phases are balanced.

Three Phase to Ground Fault

Since the three phases are balanced, then there is no current on the neutral and the sum of the three phases is equal to zero.



Next, you can convert these phase values to symmetrical components and you will find that there is no zero sequence components and no negative sequence components.

Since the sequences are balanced, the sum of the three phases is equal to zero.



Next, you know that the phases are 120 degrees off of the previous phase in the current A-B-C sequence.



Next, substitute into the negative sequence equation and you will get the following.



Finally, you have the positive sequence equation, which results in the following.



This can all be shown in a figure, please see below. This allows you to solve for the short circuit current, which is found by dividing the full load voltage by the positive sequence impedance. This impedance is the same impedance as in the short circuit calculations. This impedance will be given to you in questions and can be disguised as the percent impedance value.



Three phase fault to ground fault shown with positive, negative and zero sequence components.

Sometimes, the circuit that leads to the ground may have a impedance and this impedance is called fault impedance. This value will reduce the short circuit current and should be added to the positive sequence impedance aka short circuit impedance aka percent impedance.



Most often, you will not include a fault impedance unless specifically part of the grounding system. This is because the fault impedance reduces the short circuit current and you want the worst case scenario and don’t want to underestimate the short circuit current.

The information shown on this website is a sample of the material provided in the technical study guide and sample exam. See the STORE to buy the products for continuation on Distribution Analysis including the following topics:

  • Three Phase Faults
  • Line to Ground Fault
  • Line to Line Fault
  • Double Line to Ground Fault

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