Power PE Technical Study Guide Errata webpage and email ...Power PE Technical Study Guide Errata -1...

Power PE Technical Study Guide Errata -1 www.engproguides.com

Power PE Technical Study Guide Errata

This product has been updated to incorporate all changes shown in the comments on the webpage and email comments as of January 1, 2020. If you have purchased this product prior to this date and wish for the latest version then please email Justin Kauwale at [email protected].

Circuits Analysis -15 www.engproguides.com 9 out of 80 problems

that voltage drop. After the charging time, then the voltage drop across the inductor will be zero. This corresponds to a “short”. In the discharge section, the rate of change in current will be negative but will be large. This corresponds to a large negative voltage across the inductor.

2.5.2 Inductor Characteristics Each inductor is rated based on its ability to store energy by creating a magnetic field. The unit of inductance is henries. Often times, the units will be represented in a sub unit of henries, as shown below. Just remember that when completing problems, henries, volts and currents must be used and not any of the subunits.

𝑚𝑖𝑙𝑖 ℎ𝑒𝑛𝑟𝑖𝑒𝑠 → 𝑚𝐻 10

𝑚𝑖𝑐𝑟𝑜 ℎ𝑒𝑛𝑟𝑖𝑒𝑠 → 𝜇𝐻 10

𝑛𝑎𝑛𝑜 ℎ𝑒𝑛𝑟𝑖𝑒𝑠 → 𝑛𝐻 10

In order to answer any conceptual type questions on capacitors, you should know how the inductance of an inductor can change.

First, the magnetic field will create a back emf (voltage). This voltage will be a function of the inductance of the inductor and the change in current.

𝑉 𝐿 ∗𝑑𝐼𝑑𝑇

;

𝐿 ℎ𝑒𝑛𝑟𝑖𝑒𝑠; 𝑉 𝑣𝑜𝑙𝑡𝑎𝑔𝑒; 𝑑𝐼𝑑𝑇

𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡;

The inductance will vary based on the construction of the inductor, which as a reminder consists of wire wound around a core.

Characteristics Inductance (C)

Number of turns An increase in turns will increase the inductance by square Permeability An increase in permeability of core will increase the inductance Area An increase in area of core will increase the inductance Length An increase in the length of the coil will decrease the inductance

The equation below summarizes most of the characteristics. Permeability describes how well a material can produce a magnetic field.

𝐿𝑁 ∗ 𝜇 ∗ 𝐴

𝑙;

𝐿 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝐻 ; 𝜇 𝑝𝑒𝑟𝑚𝑒𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑟𝑒 𝐻𝑚

;

𝑁 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑢𝑟𝑛𝑠; 𝐴 𝑐𝑜𝑟𝑒 𝑎𝑟𝑒𝑎 𝑚 ; 𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑟𝑒 𝑚


In the previous figure, the phasor diagram shows inductance as 𝑋 ∠90°, which is characterized as a positive vector on the imaginary axis.

3.4.3 Capacitance or Capacitive Reactance (Capacitors) A capacitor is made up of two conductors separated by a dielectric. As current is supplied to the capacitor on the positive side, charge is built up, which restricts current. Then the current alternates and charge is built up on the negative side, which also restricts current. Capacitors are used in the next section, 3.0 Devices, to smooth waveforms because of its timing component. It takes time to fill the capacitor with charge, which delays any sudden inrush of voltage.

𝑋 = − 1𝑤𝐶 = − 12𝜋𝑓𝐶

𝑍 = −𝑗𝑋

In the previous figure, the phasor diagram shows inductance as 𝑋 ∠ − 90°, which is characterized as a negative vector on the imaginary axis.

3.4.4 Impedance The term impedance combines inductive and resistive loads into a single term. Impedance is defined as the complex resistance or the resistance with both a magnitude and angle term in the polar format. In the equation below, impedance, “Z” is given as a function of the resistance, “R” and the reactance, “X”. 𝐶𝑜𝑚𝑝𝑙𝑒𝑥 𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 → 𝑍 = 𝑅 + 𝑗𝑋

Impedance can also be shown in its phasor form. 𝐶𝑜𝑚𝑝𝑙𝑒𝑥 𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 → 𝑍∠𝜃° Now you will also have new equations for power and for Ohm’s law, since resistance is no longer, “R”, it is now replaced with impedance. When you have complex loads, remember to use these equations. 𝑂ℎ𝑚 𝑠 𝐿𝑎𝑤 → 𝑉 = 𝐼𝑍

𝑃𝑜𝑤𝑒𝑟 → 𝑃 = 𝐼 𝑅 ; 𝑃 = 𝑉𝑅

3.5 SINGLE-PHASE VS. THREE-PHASE In power engineering, you will encounter both single-phase and three-phase circuits and you should be familiar with the differences in both single-phase and three-phase for the exam. In


Figure 56: A split phase system consists of two phases, A & B. Each phase is 180 degrees offset from one another. Small loads can be fed by a circuit consisting of a single phase (A or

B) and the common neutral line. Large loads can be fed by a circuit consisting of two phases (A & B). There is no neutral for these loads, because there is no return current. Similar to how three phase circuits, “cancel out” when added together, phase A & B current will “cancel out”

when added together.

𝐼 10∠0°; 𝐼 10∠180°

𝐼 𝐼 10∠0° 10∠180° 0 𝐴

Figure 57: This figure shows graphically the alternating currents for phases A & B of a split phase arrangement. This shows that the phases are 180 degrees offset from one another. It

also shows how the voltage between A & B is twice the voltage of a single phase.


𝑤ℎ𝑒𝑟𝑒 𝑍 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑖𝑛 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑣𝑎𝑙𝑢𝑒

5.2 APPLICATION OF PER-UNIT

The application of per-unit analysis can be most readily seen when you have transformers and generators in a one-line diagram. Generators and motors are given in terms of transient reactance. These values correspond to a per-unit impedance of the fractional value.

Figure 73: Per unit example

𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟: 𝑍 0.12 𝐵𝑎𝑠𝑒 → 𝑆 1,500 𝐾𝑉𝐴; 𝑉 13.8 𝐾𝑉

𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝐿𝑖𝑛𝑒: 𝑍 0.15 𝐵𝑎𝑠𝑒 → 𝑆 1,500 𝐾𝑉𝐴; 𝑉 13.8 𝐾𝑉

𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐻𝑖𝑔ℎ 𝑆𝑖𝑑𝑒 : 𝑍 0.07 𝐵𝑎𝑠𝑒 → 𝑆 1,000 𝐾𝑉𝐴; 𝑉 13.8 𝐾𝑉

𝑇𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝐿𝑜𝑤 𝑆𝑖𝑑𝑒 : 𝑍 0.07 𝐵𝑎𝑠𝑒 → 𝑆 1,000 𝐾𝑉𝐴; 𝑉 480 𝑉

𝑀𝑜𝑡𝑜𝑟 1: 𝑍 0.07 𝐵𝑎𝑠𝑒 → 𝑆 150 𝐾𝑉𝐴; 𝑉 480 𝑉

𝑀𝑜𝑡𝑜𝑟 2: 𝑍 0.05 𝐵𝑎𝑠𝑒 → 𝑆 100 𝐾𝑉𝐴; 𝑉 480 𝑉

Next, you need to choose a base. Just to make things a little difficult, let us choose the base shown below. Typically, you will choose a base at the analysis point.

𝑆𝑦𝑠𝑡𝑒𝑚 𝐵𝑎𝑠𝑒: 1,000 𝐾𝑉𝐴 & 10,000 𝑉

Now, you have to establish voltage bases, because you have a transformer. If you didn’t have a transformer, then the system base will apply for the entire circuit. The high side of the transformer will have a voltage base of 10,000 V and the low side of the transformer will have the following voltage base.

Rotating Machines -24 www.engproguides.com 8 out of 80 problems

3.3 EQUIVALENT CIRCUITS

Equivalent circuits help you to better understand the flow of electricity in an induction motor. You should understand the equivalent circuits for all equipment (transformers, motors, generators, transmission lines) and how current flows through the circuit during various loading conditions.

The equivalent circuit below shows the losses due to the stator winding resistances and the rotor winding resistances as “Rstator and Rrotor”. Winding resistance describes the losses due to resistance of the flow of current through the windings. The losses due to stator and rotor leakage reactances are shown as “jXstator and jXrotor”. Leakage reactance describes the flux that escapes the insulation of the motor at the air gap. There is also a branch parallel to the stator that consists of the currents, “I1 and I2”. The sum of these currents is the exciting current and this branch is called the excitation branch. The excitation branch describes the process of generating a magnetic field through an electric current. When power enters the induction motor, an equal amount of voltage is applied to the excitation branch and the motor branch. Some of the current enters the excitation branch (I1) to generate the magnetic field and the remaining current (I0) powers the motor.

Figure 21: Equivalent circuit of an induction motor

In the equivalent circuit above, the voltage drop across b1 to b2 is equal to the voltage drop from b1 to c2, which is also equal to the voltage drop across a1 to a2.

𝑉 𝑉 𝑉

𝑉 𝑖 ∗ 𝑅 𝑜𝑟 𝑖 ∗ 𝑗𝑋 𝑖 ∗ 𝑅 𝑖 ∗ 𝑗𝑋

In an induction motor, the speed of the motor slips and is slightly less than the synchronous speed. This is shown as the speed difference between the stator and rotor.

𝑓 𝑓

This same slip factor will also affect the induced voltage in the rotor and the speed of the motor.

𝐸 ∗ 𝑠𝑙𝑖𝑝% 𝐸 ; 𝑛 ∗ 1 𝑠𝑙𝑖𝑝% 𝑛

Electric Power Devices -9 www.engproguides.com 8 out of 80 problems

First, transform the secondary impedances to the primary side.

𝑉

𝑍

𝑉

𝑍

480𝑍

10,000500 Ω

→ 𝑍 1.152 Ω∠90°

480𝑍

10,0001,000 Ω

→ 𝑍 2.304 Ω∠0°

Next, re-draw the circuit but since you only have primary impedances, you only need the primary side.

Now, add up all the impedances, but make sure you use the angles for the capacitor and inductor.

𝑍 3∠ 90° 1.152 Ω∠90° 2.304 Ω∠0° 2.95∠ 38°

Next, solve for the current. You should take the voltage as the zero point.

𝑉 480 𝑉∠0° 𝐼 2.95∠ 38°

𝐼 162.5∠38.7°

The current is lagging the voltage, because the load is primarily inductive.

Next, multiply the current by the impedance to find the primary voltage drop for each component.


𝑉 , 162.5∠38.7° ∗ 3∠ 90° 487.5∠ 51.3°

𝑉 , 162.5∠38.7° ∗ 1.152∠90° 187.2∠128.7°

𝑉 , 162.5∠38.7° ∗ 2.304∠0° 374.4∠38.7°

You can double check your work by adding all of the voltage drops. The sum should be equal to 480 V angle 0 degrees.

Finally, you can transform these voltage drops with the voltage ratio to the secondary side. Only magnitudes will be used for simplicity.

𝑉 , 487.5 ∗10,000

48010,156 𝑉

𝑉 , 187.2 ∗10,000

4803,900 𝑉

𝑉 , 374.4 ∗10,000

4807,800 𝑉

2.4 REAL TRANSFORMERS

The real transformer incorporates the efficiency losses within the equipment. There are four major losses: Coil, Leakage Flux, Eddy Current, and Hysteresis. These losses can be grouped as losses through the core or through the coil of the transformer. You will commonly hear the terms “Coil (Copper) Losses” and “Core Losses” when analyzing a real transformer. The four losses are summarized in the losses section below for your reference. However, these terms are not as important as being able to analyze core losses versus coil losses for the test.

2.4.1 Equivalent Circuit The equivalent circuit showing the losses through the transformer is illustrated below. The resistors and inductors represent the losses within the transformer. Current enters through the power source at a, is transferred between the windings from E1 to E2, and finally leaves to the distribution system at terminal “e”.

Figure 3: Equivalent Circuit of a Transformer


The various construction types are broken up into low voltage, medium voltage and high voltage. The transformers can either be liquid filled or dry type. Each transformer will have inefficiencies and these inefficiencies will cause heat within the equipment. The dry/liquid types describe the method in which this heat is dissipated. A liquid type uses a liquid to cool the equipment, while a dry type uses air. The dry type will be much larger since it takes a higher volume of air to provide the same amount of cooling as a liquid.

2.1.2 Tap Setting Tap settings are used to slightly adjust the voltage ratings on transformers. In real world conditions, voltage drops will occur in transmission lines or the distribution will provide higher voltages than required, causing the voltage at the transformer to be different than it is rated for. A tap will compensate for this variance and allow the secondary side to deliver the appropriate voltage. For example, a 480V/120V transformer may actually be receiving 450V, causing the secondary voltage to be 112V. Using a -6.25% tap at the primary side will change the 480/120 V ratio to 450/120V, effectively raising the secondary voltage to 120V, closer to its rated 120V. You could also raise the secondary voltage by +6.25% to change the ratio to 480V/127.5V. This would result in the transformation of the 450 V to 120 V.

Taps are terminals that are connected to the windings of the transformer. It can either be located on the primary or secondary winding, but is usually located on the high voltage side, where the current is lower. They are installed such that a whole number turn is either added or subtracted from the windings to adjust the turns ratio, and therefore the secondary voltage of the transformer. Depending on the size of a transformer, there will be a certain amount of windings per voltage, which will determine the % of tap settings available. Taps will typically not change the power rating of the transformer, but will vary the current. This is known as a full capacity tap.

2.3 IDEAL TRANSFORMERS

In an ideal transformer, it is assumed there are no losses: the apparent power that enters the primary side of the transformer is equal to the apparent power that leaves the secondary side of the transformer. This conservation of energy is the same for the real power and reactive power.

𝑆 𝑆 ; 𝑃 𝑃 ; 𝑄 𝑄 𝑖𝑑𝑒𝑎𝑙 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟

The primary purpose of the transformer is to change the voltage. The change in voltage is related to how many times the wires wrap around the coil, as shown in the equation below, where N is equal to the number of windings (also known as the turns) around the coil.

𝑉𝑉

𝑁𝑁

, where N turns

The figure below is a typical schematic of a transformer used in circuit analysis.

Transmission & Distribution -8 www.engproguides.com 11 out of 80 problems

2.1.2 Inductance

The second term is the inductance and this is the transmission systems impedance due to inductive reactance. As alternating current flows through the transmission lines, the change in current will cause an electromagnetic field to be generated around each transmission line. The force created by this field in one line will impede the flow of current in another line. Thus, the impedance value due to inductance will be dependent on the distance between the lines and the radius of the conductors. Since there are a number of arrangements for the conductors, it would be difficult to test and in practice these calculations are completed with a computer program for more exact results. However, you should be familiar with the general equation for calculating impedance of a transmission line.

𝐿𝜇𝜋

∗ ln𝐷𝐷

; 𝐿𝜇2𝜋

∗ ln𝐷𝐷

; 𝜇 𝑝𝑒𝑟𝑚𝑒𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎𝑖𝑟 1.256 𝑥 10 6𝐻𝑚

𝐿 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑠𝑖𝑛𝑔𝑙𝑒 𝑝ℎ𝑎𝑠𝑒 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑙𝑖𝑛𝑒 𝑖𝑛𝐻𝑚

;

𝐿 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 𝑝ℎ𝑎𝑠𝑒 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑙𝑖𝑛𝑒 𝑖𝑛𝐻𝑚

;

𝐷 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑒𝑎𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠;

𝐷 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑒𝑎𝑛 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠

2.1.3 Capacitance

When multiple transmission lines run next to each other, there will be a potential difference between the multiple conductors. This is because the conductors are operated at different phases from each other and thus when one has a high positive voltage another may have a low negative voltage. This separation and large voltage difference causes the conductors to develop a charge between each other, similar to a capacitor.

Figure 1: The difference in voltage between the phases causes a potential that acts like a capacitor at very large voltages. At certain points the potential difference is 0 (Phase A = Phase C) and at certain points there is a large difference. This change in potential causes an increase

and decrease in charge.


Figure 26: The negative and zero sequence components are equal to zero and only the positive sequences are connected to the voltage source.

First, solve for the sequence currents. Remember to take the magnitude of Z.

𝑉 𝐼 ∗ |𝑍|

1.05 𝑝𝑢 𝐼 ∗ |0.12𝑗 0.06𝑗 𝑝𝑢| 𝐼 ∗ 0.18

𝐼 5.833 𝑝𝑢

𝐼 𝐼 0 𝑝𝑢

Now, solve for the phase currents.

𝐼 𝐼 𝐼 𝐼 5.833 𝑝𝑢 0 0 5.833 𝑝𝑢

𝐼 𝑎 𝐼 𝑎𝐼 𝐼 5.833∠ 120° 𝑝𝑢 0 0 5.833∠ 120° 𝑝𝑢

𝐼 𝑎𝐼 𝑎 𝐼 𝐼 5.833∠ 120° 𝑝𝑢 0 0 5.833∠ 120° 𝑝𝑢

Next, solve for the sequence voltages. The fault voltage is equal to 1.05 pu. Since there is no other impedances on the circuit, then the voltage drop through the positive sequence component will equal the fault voltage.

𝑉 𝑉 𝐼 ∗ |𝑍 |

𝑉 1.05 𝑝𝑢 1.05 𝑝𝑢 0


The negative and zero sequence voltages are zero as well.

𝑉 𝑉 0

For completeness, here are the phase voltage calculations too. But it should be obvious that the phase voltages at the fault will be equal to zero.

𝑉 𝑉 𝑉 𝑉 0 𝑝𝑢 0 0 0 𝑝𝑢

𝑉 𝑎 𝑉 𝑎𝑉 𝑉 0 𝑝𝑢 0 0 0 𝑝𝑢

𝑉 𝑎𝑉 𝑎 𝑉 𝑉 0 𝑝𝑢 0 0 0 𝑝𝑢

Phase A to Ground Fault:

In a phase to ground fault, all sequence impedances are included in the calculation. Also the currents in the ungrounded phases will be equal to zero

𝐼 𝐼 0

At the location of the fault, the grounded phase voltage will be equal to zero. This is based on the assumption that is no impedance between the phase and the ground.

𝑉 0

The phase sequence diagram will look like the one shown below.


Figure 27: The positive, negative and zero sequence currents are equal to each other and all components are arranged in series.

First, solve for the sequence currents. Remember to take the magnitude of the impedance.

𝑉 𝐼 ∗ | 𝑍 𝑍 𝑍 |

1.05 𝑝𝑢 𝐼 ∗ |0.12𝑗 0.06𝑗 𝑝𝑢 0.08𝑗 0.05𝑗 𝑝𝑢 0.08𝑗 0.04𝑗 𝑝𝑢| 𝐼 ∗ 0.43

𝐼 2.44 𝑝𝑢

𝐼 𝐼 𝐼 2.44 𝑝𝑢


𝐼 𝐼 𝐼 𝐼 2.44 𝑝𝑢 2.44 2.44 7.32 𝑝𝑢

𝐼 𝑎 𝐼 𝑎𝐼 𝐼 2.44∠ 120° 𝑝𝑢 2.44∠120° 𝑝𝑢 2.44∠0° 0 𝑝𝑢

𝐼 𝑎𝐼 𝑎 𝐼 𝐼 2.44∠120° 𝑝𝑢 2.44∠ 120° 𝑝𝑢 2.44∠0° 0 𝑝𝑢

Next, solve for the sequence voltages. The fault voltage is equal to 1.05 pu.

𝑉 𝑉 𝐼 ∗ |𝑍 |

𝑉 1.05 𝑝𝑢 2.44 𝑝𝑢 ∗ | 0.12𝑗 0.06𝑗 𝑝𝑢 |

𝑉 1.05 𝑝𝑢 2.44 𝑝𝑢 ∗ 0.18 𝑝𝑢 0.611𝑝𝑢


𝑉 2.44 𝑝𝑢 ∗ | 0.08𝑗 0.05𝑗 𝑝𝑢 | 2.44 ∗ .13 0.318 𝑝𝑢

𝑉 2.44 𝑝𝑢 ∗ | 0.08𝑗 0.04𝑗 𝑝𝑢 | 2.44 ∗ .12 0.293 𝑝𝑢

Finally, solve for the phase voltages

𝑉 𝑉 𝑉 𝑉 0.611 𝑝𝑢 0.318 𝑝𝑢 0.293 𝑝𝑢 0 𝑝𝑢

𝑉 𝑎 𝑉 𝑎𝑉 𝑉 0.611∠ 120° 0.318∠ 120° 0.293 0.917∠ 118.7° 𝑝𝑢

𝑉 𝑎𝑉 𝑎 𝑉 𝑉 0.611∠120° 0.318∠ 120° 0.293 0.917∠118.7° 𝑝𝑢

Phase B to Phase C Fault:

In a phase to phase fault, only the positive and negative sequences are included in the calculation. Also the currents in the connected phases will be equal and opposite to one another. The current in the other phase will be equal to zero.

𝐼 𝐼 ; 𝐼 0

At the location of the fault, the voltage of the two connected phases will be equal to one another.

𝑉 𝑉



Figure 28: The positive and negative sequence currents are opposite, but equal to one another in magnitude. The zero sequence is not connected.

First, solve for the sequence currents. Remember to take the magnitude of the impedance.

𝑉 𝐼 ∗ | 𝑍 𝑍 |

1.05 𝑝𝑢 𝐼 ∗ | 0.12𝑗 0.06𝑗 𝑝𝑢 0.08𝑗 0.05𝑗 𝑝𝑢 |

𝐼 3.387 𝑝𝑢

𝐼 𝐼 3.387 𝑝𝑢; 𝐼 0


𝐼 𝐼 𝐼 𝐼 3.387 𝑝𝑢 3.387 0 0 𝑝𝑢

𝐼 𝑎 𝐼 𝑎𝐼 𝐼 3.387∠ 120° 𝑝𝑢 3.387∠120° 𝑝𝑢 0∠0° 5.867∠ 90° 𝑝𝑢

𝐼 𝑎𝐼 𝑎 𝐼 𝐼 3.387∠120° 𝑝𝑢 3.387∠ 120° 𝑝𝑢 0∠0° 5.867∠90° 𝑝𝑢

Next, solve for the sequence voltages. The fault voltage is equal to 1.05 pu.

𝑉 𝑉 𝐼 ∗ |𝑍 |


𝑉 1.05 𝑝𝑢 3.387 𝑝𝑢 ∗ | 0.12𝑗 0.06𝑗 𝑝𝑢 | 0.440 𝑝𝑢

𝑉 3.387 𝑝𝑢 ∗ | 0.08𝑗 0.05𝑗 𝑝𝑢 | 0.440 𝑝𝑢

𝑉 0 𝑝𝑢

Finally, solve for the phase voltages

𝑉 𝑉 𝑉 𝑉 0.440 𝑝𝑢 0.440 𝑝𝑢 0 𝑝𝑢 0.880 𝑝𝑢

𝑉 𝑎 𝑉 𝑎𝑉 𝑉 0.440∠ 120° 0.440∠ 120° 0 0.44∠180° 𝑝𝑢

𝑉 𝑎𝑉 𝑎 𝑉 𝑉 0.440∠120° 0.440∠ 120° 0 0.44∠180° 𝑝𝑢

Phase B to Phase C to Ground Fault:

In a double phase to ground fault, all the sequences are included in the calculation. Also the voltages on the connected phases at the location of the fault will be equal to zero.

𝑉 𝑉 0; 𝐼 0


Figure 29: The negative and zero sequence currents are arranged in parallel.


First, solve for the positive sequence current.

𝑉 𝐼 ∗ | 𝑍 𝑍 ||𝑍 |

1.05 𝑝𝑢 𝐼 ∗ | 0.12𝑗 0.06𝑗 𝑝𝑢1

0.08𝑗 0.05𝑗 𝑝𝑢1

. 08𝑗 0.04𝑗 𝑝𝑢|

1.05 𝑝𝑢 𝐼 ∗ | 0.18𝑗 𝑝𝑢 7.692𝑗 𝑝𝑢 8.33𝑗 𝑝𝑢 |

1.05 𝑝𝑢 𝐼 ∗ | 0.18𝑗 𝑝𝑢 0.0624𝑗 𝑝𝑢 |

𝐼 4.332 𝑝𝑢

Now, solve for the positive sequence voltage.

𝑉 𝑉 𝐼 ∗ |𝑍 |

𝑉 1.05 𝑝𝑢 4.332 𝑝𝑢 ∗ | 0.12𝑗 0.06𝑗 𝑝𝑢 | 0.270 𝑝𝑢

𝑉 0.270 𝑝𝑢

𝑉 0.270 𝑝𝑢

Next, solve for the phase voltages

𝑉 𝑉 𝑉 𝑉 0.270 𝑝𝑢 0.270 𝑝𝑢 0.270 𝑝𝑢 0.810 𝑝𝑢

𝑉 𝑎 𝑉 𝑎𝑉 𝑉 0.270∠ 120° 0.270∠ 120° 0.270 0 𝑝𝑢

𝑉 𝑎𝑉 𝑎 𝑉 𝑉 0.270∠120° 0.270∠ 120° 0.210 0 𝑝𝑢

Next, find the negative and zero sequence currents.

𝑉 0.270 𝑝𝑢 𝐼 ∗ | 0.08𝑗 0.05𝑗 𝑝𝑢 |

𝐼 2.077 𝑝𝑢

𝑉 0.270 𝑝𝑢 𝐼 ∗ | 0.08𝑗 0.04𝑗 𝑝𝑢 |

𝐼 2.255 𝑝𝑢

Finally, solve for the phase currents.

𝐼 𝐼 𝐼 𝐼 4.332 𝑝𝑢 2.077 2.255 0 𝑝𝑢

𝐼 𝑎 𝐼 𝑎𝐼 𝐼 4.332∠ 120° 𝑝𝑢 2.077∠120° 𝑝𝑢 2.255∠0° 6.5∠ 121° 𝑝𝑢

𝐼 𝑎𝐼 𝑎 𝐼 𝐼 4.332∠120° 𝑝𝑢 2.077∠ 120° 𝑝𝑢 2.255∠0° 6.5∠7121° 𝑝𝑢


Figure 42: The generator voltage and the infinite bus voltage are in phase and of equal magnitude in the original condition. This results in no current transfer and no power flow

between the generator and infinite bus. In the right condition, the phase angle of the generator voltage is increased. This creates positive power flow, meaning power flows from the infinite

bus to the generator. Reactive Power: Reactive power is controlled by the excitation current, also known as the field current. By increasing the field current, the armature current increases. The armature current flows through reactive impedance, XA (or ZL), which causes the current to lag the internal voltage by 90 degrees, where it falls on the reactive axis.

Reactive Power Supplied to Grid: When the internal voltage (real component) of the generator, Vsend, is larger than the terminal voltage (real component), VReceive, the voltage potential difference, Ed=IAZL, across the reactor will be positive, causing reactive power to be supplied to the grid.

Figure 43: The generator (sending) magnitude is increased to be larger than the terminal voltage (receiving). This creates a positive reactive power, meaning reactive power flows from

the generator to the infinite bus.


Reactive Power Absorbed from the Grid: When the internal voltage (real component) of the generator, Vsend, is less than the terminal voltage (VReceive), the voltage potential difference, Ed=IAXL, across the reactor will be negative, causing reactive power to be absorbed from the grid.

Figure 44: The generator voltage magnitude is decreased less than the voltage terminal. This creates a negative reactive power, which means reactive power flows to the generator from the

infinite bus.

5.2 PARALLEL TRANSFORMERS


𝑆 𝐼∗𝑥 𝑉𝑥 √3

The difficulty is finding the angle of the current. Luckily, you know from this section that the angle should be -90 degrees.

𝑆 200𝐴∠90° ∗ 13.4𝑘𝑉∠0° ∗ √3

𝑆 4,642∠90°𝐾𝑉𝐴

The correct answer is most nearly, (c) 4,642∠90°𝐾𝑉𝐴

8.7 SOLUTION 7 – PARALLEL GENERATORS

A 3-phase generator is connected to a very large power distribution system. It is assumed the generator is connected to an infinite bus. The infinite bus is operating at a voltage of 13.2 KV∠0°. The generator is operating at a voltage of 13.0 KV∠0° and 200 A. The internal reactance is Xa and the internal resistance of the generator is negligible. What is the power consumed by the generator?

𝑆 𝐼∗𝑥 𝑉𝑥 √3

The difficulty is finding the angle of the current. Luckily, you know from this section that the angle should be +90 degrees.


𝑆 200𝐴∠ 90° ∗ 13.0𝑘𝑉∠0° ∗ √3

𝑆 4,503∠ 90°𝐾𝑉𝐴

The correct answer is most nearly, (b) 4,503∠ 90°𝐾𝑉𝐴

8.8 SOLUTION 8 – FAULT CURRENT ANALYSIS

A three phase bolted fault occurs at the location in the diagram below. What is the short circuit current?

First, you need to convert all the percent reactance values to the same base in per unit. In this solution, the generator is selected as the base.

𝑆 75 𝐾𝑉𝐴; 𝑉 480 𝑉;

480/3000V

Protection - 45 www.engproguides.com 13 out of 80 problems

6.16 PROBLEM 16 – TCC

A 3-ph, wye-delta, 2,400 V/480 V, 1,000 KVA, transformer with 5% impedance has a thermal damage table as shown below.

Current (p.u.) Time (sec) 2 1,800 3 300 5 60

10 30 20 10

What is most nearly the maximum current in the primary side of the transformer that can be accepted for a duration of 10 seconds before thermal damage occurs to the transformer?

(A) 300 A

(B) 1,230 A

(C) 2,780 A

(D) 4,811 A

Protection - 60 www.engproguides.com 13 out of 80 problems

Figure 29: This shows how a right angle triangle is formed and how the sides of the triangle are found.

The load impedance can reach 5.5 before the distance relay is tripped.

The correct answer is most nearly, (B) 5.5∠15° Ω

7.16 SOLUTION 16 – TCC

First, you need to calculate the current flowing in the phase of the transformer on the primary side. This can be done through the use of the three phase apparent power equation.

𝑆 1,000 𝐾𝑉𝐴 𝐼 ∗ 𝑉 ∗ √3

1,000 𝐾𝑉𝐴 ∗ 1,000 𝐼 ∗ 2,400 𝑉 ∗ √3

𝐼 241 𝐴

Now you can convert this line current to phase current.

𝐼 𝐼 240.56 𝐴

The maximum current is 20x the base value of 241 A.

𝐼 20 ∗ 241 𝐴 4,811.25 𝐴

The correct answer is most nearly, (D) 4,811 A.

7.17 SOLUTION 17 – TCC

Codes & Standards - 22 www.engproguides.com 12 out of 80 problems

Temperature of 40°C (104°F)* Table 310.15(B)(17) covers Type THHW, THW, THWN, UF, RHW, etc. conductors.

These are normal rated temperature conductors from 140 °F to 194 °F.

Table 310.15(B)(19) covers Type Z, FEP, FEPB, PFA, SA, PFAH, TFE & Z conductors.

These are higher rated temperature conductors from 302 °F to 482 °F.

As an example, assume you have a six conductors carrying 210 A each. The conductors and all connected components are rated at 75 C. The conductors are located in the same raceway on a roof with a design ambient temperature of 100 F. What size conductors should be used? Assume 480 V system, copper and THHW.

If there was no adjustment factors then the size would be based on Table 310.15 (B) (16). However, you have a temperature adjustment factor and multiple conductor adjustment factor that will change Table 310.15(B) (16). The new table will have all of the values multiplied by the following adjustment factors.

𝐴𝑑𝑗𝑢𝑠𝑡𝑚𝑒𝑛𝑡 𝐹𝑎𝑐𝑡𝑜𝑟 0.88 0.8 0.704

A snippet of the table will look like the below table and the correct answer would be 350 KCMIL.

Size Copper (60 C) Copper (75 C) Copper (90 C) 4/0 195 x (0.704) = 137 A 230 x (0.704) = 161 A 260 x (0.704) = 183 A 250 215 x (0.704) = 151 A 255 x (0.704) = 179 A 290 x (0.704) = 204 A 300 240 x (0.704) = 168 A 285 x (0.704) = 200 A 320 x (0.704) = 225 A 350 260 x (0.704) = 183 A 310 x (0.704) = 218 A 350 x (0.704) = 246 A 400 280 x (0.704) = 197 A 335 x (0.704) = 235 A 380 x (0.704) = 267 A 500 320 x (0.704) = 225 A 380 x (0.704) = 267 A 430 x (0.704) = 302 A

2.11.5 Grounding Conductor or Bonding Jumper Sizing – Article 250 A grounding conductor, serves as the return path for any line to ground faults. This conductor carries unbalanced current back to the overcurrent protection device, where the overcurrent protection device can sense the fault and open the circuit. A bonding jumper is the connection point between metal parts. The bonding jumper ensures that metal parts cannot be at varying voltages, which will create an unsafe condition. NEC requires that metal parts that normally do not carry current, need to be connected electrically in order to ensure that any current during a fault condition can be transferred from these metal parts back to the grounding conductor and to the overcurrent protection device.

A grounding conductor is sized based on Table 250.102 (C) (1). In this table you use the largest ungrounded conductor or equivalent area for parallel conductors and find the corresponding grounding conductor size.

Table 250.102(C)(1) Grounded Conductor, Main Bonding Jumper, System Bonding Jumper, and Supply-Side Bonding Jumper for Alternating-Current Systems

Size of Largest Ungrounded Conductor or Equivalent Area for Parallel Conductors

Size of Grounded Conductor or Bonding Jumper

Codes & Standards - 24 www.engproguides.com 12 out of 80 problems

2.11.7 Conduit Sizing – Article 314 Conduits provide a path for conduits and a way to protect conductors from damage. Conduits are sized in accordance with Table 1 in Chapter 9 Tables. This table indicates that conduits with 1 conductor must occupy less than 53% of the cross sectional area of the conduit. Conduits with 2 conductors must occupy less than 31% of the cross sectional area and conduits with more than 2 conductors must occupy less than 40% of the cross sectional area of the conduit.

Table 4 has the areas of various conductors at 40%, 60%, 53%, 31% and 100%. The 60% value is used to ensure that conductors with nipples fit within the 60% area requirement.

Table 5 contains the dimensions of insulated conductors and table 5A contains compact versions of some conductors. The compact versions have smaller areas but are typically more expensive.

Typically, you have your conductors sized and then you determine what conduit size you need to house those conductors. For example, you have (3) phases and (1) ground conductor in a conduit. Each phase consists of (3) THHN #1 conductors. The ground conductor is a 1/0 THHN conductor. What size EMT conduit should be used?

First, calculate the total area of the conductors.

𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝐴𝑟𝑒𝑎 → 3 ∗ 3 ∗ 0.1562 𝑖𝑛 1 ∗ 0.1855 𝑖𝑛 1.5913 𝑖𝑛

Since, you have more than 2 conductors, you must find an EMT conduit that has a 40% area that is greater than 1.5913 in2.

𝑇𝑎𝑏𝑙𝑒 4 → 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑀𝑒𝑡𝑎𝑙 𝑇𝑢𝑏𝑖𝑛𝑔 𝐸𝑀𝑇 →

2" ℎ𝑎𝑠 𝑎 40% 𝑎𝑟𝑒𝑎 𝑜𝑓 1.342 𝑖𝑛

2 1/2" ℎ𝑎𝑠 𝑎 40% 𝑎𝑟𝑒𝑎 𝑜𝑓 2.343 𝑖𝑛

So you must choose a 2-1/2” EMT conduit.

2.11.8 Junction Box Sizing – Article 314 Junction boxes are used to house the connection between two or more wired. Junction boxes are also required at certain intervals. Junction boxes are sized based on the number of conductors that are passing through, terminated and spliced in the box. Junction boxes are also sized to accommodate equipment & devices like cable clamps, receptacles, switches, timers, etc.

Each conductor that passes through or terminated is counted once. Each conductor that is spliced is also counted once. For example, if a #12 and #10 wire are spliced together in the junction box, then that would count as two conductors, since the #12 and #10 is being spliced.

Equipment is sized based on the number of each equipment, based on the largest wire in the box. Only devices are counted twice for each yoke or mounting strap. Clamps and support fittings are counted once.

AC

+

-

SOURCE

DC LOAD

ACSOURCE

DC LOAD

SOURCEDC LOAD

V

max, line

V

max, line

Single Phase - Full Wave Rectifier (No capacitor)

Section 2 Devices & Power Electronic Circuits: 8 out of 80 Problems

Subsection(s): Power supplies and converters

V

max, phase

V

DC load

Three Phase - Full Wave Rectifier (No capacitor)

V

max, phase

= √2 x V

RMS, phase

V

max, line

= √2 x V

RMS, line

V

line

= √3 x V

phase

r = √(V

rms

/V

avg

)

2

- 1

V

rms, phase

V

max, phase

V

max, phase

= √2 x V

RMS, phase

V

RMS, phase

V

DC, max

r = √(V

rms

/V

avg

)

2

- 1

Ripple Factor

V

avg, phase

V

avg, phase

Ripple Factor

V

rms, phase

V

DC, load

= 1.35 x V

RMS, line

V

DC, max

= √2 x V

RMS, line

V

DC load

= 6 x x (0.5+ cos(δ°))V

AC, MAX

Delayed firing angle ->

V

DC, load

= 0.9 x V

RMS, phase

V

DC, max

= √2 x V

RMS, phase

V

DC, load

V

RMS, line

V

DC, load

= 2.33 x V

RMS, phase

V

DC, load

= 0.95 x V

max, line

V

DC, load

= 1.65 x V

max, phase

3-Phase Full Wave Rectifier

Cheat Sheets - 45

V

DC, load

= 1.35 x V

RMS, line

1

2pi

V

DC, load

= x (1+cos(δ°)) x V

AC, Max

pi

1

Delayed firing angle ->

Date post:	25-Mar-2020
Category:	Documents
Upload:	others
View:	3 times
Download:	0 times

Power PE Technical Study Guide Errata webpage and email ...Power PE Technical Study Guide Errata -1...

Documents