PU System and Symmetrical Components MCQ Quiz - Objective Question with Answer for PU System and Symmetrical Components - Download Free PDF

Explanation:

Per Unit Impedance Calculation

Definition: Per unit impedance is a normalized value used in power system analysis, which simplifies calculations and comparisons. It is expressed as a fraction of the base impedance of the system. When the voltage bases remain the same, the new per unit impedance can be calculated using a specific formula that adjusts the value based on the change in system base powers.

Correct Formula:

The correct formula for calculating the new per unit impedance when the voltage bases are the same is:

Option 1: \(\rm Z_{p u}^{\text {new }}=\frac{Z_{p u}^{\text {old }} S_{B}^{\text {new }}}{S_{B}^{\text {old }}}\)

This formula is derived based on the relationship between the old and new base quantities in the power system. It ensures that the impedance values are properly scaled with respect to the change in the base power values.

Explanation:

The per unit system is a convenient way of representing electrical quantities such as impedance, voltage, and current relative to a set of base values. In the case where the voltage bases remain constant, only the base power values change. The per unit impedance is scaled proportionally to the ratio of the new base power (\(S_{B}^{\text{new}}\)) to the old base power (\(S_{B}^{\text{old}}\)). This adjustment ensures consistency in calculations and comparisons across different systems.

Let’s break down the formula:

• \(Z_{p u}^{\text{old }}\): The per unit impedance value based on the old base quantities.
• \(S_{B}^{\text{new }}\): The new base power value for the system.
• \(S_{B}^{\text{old }}\): The old base power value for the system.
• \(Z_{p u}^{\text{new }}\): The per unit impedance value based on the new base quantities.

The formula effectively scales the old per unit impedance by the ratio of the new base power to the old base power. This is essential for maintaining consistency in system analysis when the base power values are updated.

Advantages of Using Per Unit System:

• Simplifies power system calculations by normalizing values.
• Facilitates comparison between different system components.
• Eliminates the need for unit conversions during calculations.
• Provides a clear representation of electrical quantities relative to the system base values.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 2: \(\rm Z_{p u}^{\text {old }}=\frac{Z_{p u}^{\text {new }} S_{B}^{\text {new }}}{S_{B}^{\text {old }}}\)

This formula is incorrect because it reverses the relationship between the old and new per unit impedance values. Instead of scaling the old impedance to obtain the new impedance, it attempts to derive the old impedance from the new impedance. While mathematically possible, this is not the standard approach used in power system analysis.

Option 3: \(\rm Z_{p u}^{\text {new }}=\frac{Z_{p u}^{\text {old }} S_{B}^{\text {old }}}{S_{B}^{\text {new }}}\)

This formula is incorrect as it inversely scales the old per unit impedance with respect to the ratio of the base power values. It would result in an incorrect representation of the new per unit impedance, leading to errors in system analysis.

Option 4: \(\rm Z_{p u}^{\text {new }}=\frac{Z_{p u}^{\text {old }} S_{B}^{\text {new }} V_{B}^{\text {old }}}{S_{B}^{\text {old }} V_{B}^{\text {new }}}\)

This formula introduces voltage base values (\(V_{B}^{\text{old}}\) and \(V_{B}^{\text{new}}\)) into the calculation, which is unnecessary when the voltage bases are the same. When the voltage bases do not change, these terms cancel out, and the formula simplifies to Option 1. Including voltage base values in this case complicates the calculation unnecessarily.

Conclusion:

The correct formula for calculating the new per unit impedance when the voltage bases remain constant is Option 1. This formula ensures proper scaling of the old per unit impedance value based on the ratio of the new base power to the old base power. It is essential for consistent and accurate power system analysis, especially when transitioning between different base quantities.

Explanation:

Per Unit Impedance in Power Systems

Definition: The per unit (p.u.) system is a method used in power systems to normalize system quantities (impedance, voltage, current, power) to a common base. It simplifies calculations and comparisons by expressing system values as fractions or multiples of base values. The per unit impedance is given by the formula:

Z_pu = Z_actual × (Base MVA / Base Voltage²)

Where:

• Z_actual = Actual impedance in ohms.
• Base MVA = Power base in MVA.
• Base Voltage = Voltage base in kV.

Given Data:

• Initial per unit impedance (Z_{pu, initial}): 0.30
• Base kV and Base MVA are halved.

We need to calculate the new per unit impedance (Z_{pu, new}) after the base values are halved.

Step-by-Step Solution:

Step 1: Relationship between per unit impedance and base values

The per unit impedance is inversely proportional to the square of the base voltage and directly proportional to the base MVA. Mathematically:

Z_pu ∝ Base MVA / (Base Voltage)²

Let the initial base values be:

• Base MVA = S_{base, initial}
• Base Voltage = V_{base, initial}

And the new base values after halving are:

• Base MVA = S_{base, new} = S_{base, initial} / 2
• Base Voltage = V_{base, new} = V_{base, initial} / 2

Step 2: Express the new per unit impedance in terms of the initial impedance

The ratio of the new per unit impedance to the initial per unit impedance is given by:

Z_{pu, new} / Z_{pu, initial} = (S_{base, new} / S_{base, initial}) × (V_{base, initial}² / V_{base, new}²)

Substitute the new base values:

• S_{base, new} = S_{base, initial} / 2
• V_{base, new} = V_{base, initial} / 2

Z_{pu, new} / Z_{pu, initial} = [(S_{base, initial} / 2) / S_{base, initial}] × [V_{base, initial}² / (V_{base, initial} / 2)²]

Simplify the terms:

Z_{pu, new} / Z_{pu, initial} = (1 / 2) × (V_{base, initial}² / [V_{base, initial}² / 4])

Z_{pu, new} / Z_{pu, initial} = (1 / 2) × 4

Z_{pu, new} / Z_{pu, initial} = 2

Therefore:

Z_{pu, new} = 2 × Z_{pu, initial}

Step 3: Calculate the new per unit impedance

Substitute the given initial per unit impedance (Z_{pu, initial} = 0.30):

Z_{pu, new} = 2 × 0.30 = 0.60

Thus, the new per unit impedance is 0.60.

Correct Option: Option 2 (0.6)

Additional Information

To further understand the analysis, let’s evaluate the incorrect options:

Option 1: 0.3

This option assumes that the per unit impedance remains unchanged even after the base values are altered. However, per unit impedance is dependent on the base values, as shown in the formula. When the base kV and base MVA are halved, the per unit impedance changes accordingly. Therefore, this option is incorrect.

Option 3: 0.003

This option represents a drastic reduction in the per unit impedance, which is not possible given the relationship between base values and per unit impedance. Halving the base MVA and base kV results in a doubling of the per unit impedance, not a reduction to such a small value. Hence, this option is incorrect.

Option 4: 0.006

Similar to option 3, this value is unrealistically small and does not align with the mathematical relationship between per unit impedance and the base values. The correct calculation shows that the per unit impedance doubles, not decreases. Thus, this option is also incorrect.

Option 5: (No option provided)

Since no value is specified for this option, it cannot be considered as a valid choice.

Conclusion:

The correct answer is Option 2 (0.6). When the base kV and base MVA are halved, the per unit impedance doubles, as derived using the formula for per unit impedance and its dependence on base values. Understanding the relationship between system quantities and base values in the per unit system is crucial for analyzing and simplifying power system calculations.

The per unit value of any quantity is defined as the ratio of the actual value of that quantity to its base value, both measured in the same unit. Mathematically, it can be expressed as: \({per\ unit\ quantity}=\frac{\text{actual value in any unit}}{\text{base value in the same unit}}\)T

his definition allows for a standardized comparison of quantities by normalizing them to a common reference, which is particularly useful in fields such as electrical engineering and economics.

Concept:

Per unit quantity:

Per unit quantity = Actual quantity in the units / Base (or) reference quantity in the same units

⇒ Per unit impedance Zpu = Zactual / Zbase

⇒ Zpu = ZΩ × MVAb / (kVb)2

Conversion of one per unit impedance into another per unit impedance is given by

\({{\bf{Z}}_{{\bf{pu}}}}\left( {{\bf{new}}} \right) = {{\bf{Z}}_{{\bf{pu}}}}\left( {{\bf{old}}} \right)\left( {\frac{{{\bf{MV}}{{\bf{A}}_{{\bf{new}}}}}}{{{\bf{MV}}{{\bf{A}}_{{\bf{old}}}}}}} \right){\left( {\frac{{{\bf{k}}{{\bf{V}}_{\bf{b}}}_{{\bf{old}}}}}{{{\bf{k}}{{\bf{V}}_{\bf{b}}}_{{\bf{new}}}}}} \right)^2}\)

Calculation:

Given that

Zpu old = 2 pu

MVAold = 1000 MVA

kVb old = 100 kV

MVAnew = 4000 MVA

kVb new = 400 kV

\({{\bf{Z}}_{{\bf{pu}}}}\left( {{\bf{new}}} \right) = {{\bf{Z}}_{{\bf{pu}}}}\left( {{\bf{old}}} \right)\left( {\frac{{{\bf{MV}}{{\bf{A}}_{{\bf{new}}}}}}{{{\bf{MV}}{{\bf{A}}_{{\bf{old}}}}}}} \right){\left( {\frac{{{\bf{k}}{{\bf{V}}_{\bf{b}}}_{{\bf{old}}}}}{{{\bf{k}}{{\bf{V}}_{\bf{b}}}_{{\bf{new}}}}}} \right)^2}\)

\(Z_{pu}(new)=2*\frac{{4000}}{{1000}}*{\left( {\frac{{100}}{{400}}} \right)^2}\)

\({{\bf{Z}}_{{\bf{pu}}}}\left( {{\bf{new}}} \right) = 0.5 \) pu

Important PointsNOTE: The official answer key was given as option 3 with 100 MVA as old MVA. But it gives 5pu as answer which is not possible as pu value is always less than 1. So we have modified the old MVA value as 1000 MVA in  the Question, then we could successfully get the desired & conceptually correct answer.

Concept:

The sequence network of LLG fault is:

The fault current in LLG fault is given by:

I_f = 3 I_ao

where, \(I_{ao} = -I_{a1}({Z_2 \over Z_2+Z_o+3Z_f})\)

\(I_{a1} = {E_{a1} \over Z_1+(Z_2||(Z_o+3Z_f))}\)

Z1 = Positive sequence impedance

Z2 = Negative sequence impedance

Zo = Zero sequence impedance

Calculation:

Given, Z₁ = 0.08 pu

Z₂ = 0.07 pu

Z_o = 0.05 pu

Take Z_f = 0 pu as it is not given.

\(I_{a1} = {1\over 0.08+(0.07||0.05)}\)

I_a1 = 9.1 pu

\(I_{ao} = -9.1({0.07 \over 0.07+0.05})\)

|I_ao| = 5.9 pu

If = 3 × 5.3

If = 15.9 = 16 pu

Purpose of Transposition:

• Transmission lines are transposed to prevent interference with neighbouring telephone lines.
• The transposition arrangement of high voltage lines helps to reduce the system power loss.
• We have developed transposition system for Single circuit tower using same tension tower with reduced deviation angle.
• Transposition arrangement of power line helps to reduce the effect of inductive coupling.
• It is proved more economical Solution, in comparison of the conventional transposition system.

Important:

Transposition arrangement

The transposition arrangement of the conductor can simply show in the following the figure. The conductor in Position 1, Position 2 and Position 3 changes in a specific arrangement to reduce the effect of capacitance and the electrostatic unbalanced voltages.

Z1 = Z2 = ZS - Zm

Zo = ZS + 2Zm

Where,

Z1 = positive sequence impedance

Z2 = positive sequence impedance

Zm= mutual impedance

ZS= self impedance

Positive and negative sequence impedances are equal.

Concept:

The relation between new per-unit value & old per unit value impedance

\({({Z_{pu}})_{new}}\; = {({Z_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)

Also \({Z_{pu}} = \frac{{{Z_{Actual}}}}{{{Z_{base}}}}\)

 \({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

Where,

Z_base = Base value of impedance

Z_Actual = Actual value of impedance 

(Zpu)new = New per unit value of impedance

(Zpu)old = Old per unit value of impedance

kVbase = Old base value of voltage

kVnew = New base value of voltage

MVAnew = New base value of power

MVAold = Old base value of power

Additional Information

Turns ratio for the transformer is given by

\(\frac{{{V_p}}}{{{V_s}}} = \frac{{{I_s}}}{{{I_p}}} = \frac{{{N_p}}}{{{N_s}}}\)

Where,

Vp is the voltage on the primary side

Vs is the voltage on the secondary side

Ip is current on the primary side

Is is current on the secondary side

Np is the number of turns on the primary side

Ns is the number of turns on the secondary side

In the transformer, if \(R_1'\) is the reactance referred to the primary side.

Then the value of the equivalent resistance referred to the secondary side is \(R_2'\)

∴  \(\frac{{R_1'}}{{{R_2'}}} = {\left( {\frac{{{N_p}}}{{{N_s}}}} \right)^2} = {\left( {\frac{{{V_p}}}{{{V_s}}}} \right)^2} = {\left( {\frac{{{I_s}}}{{{I_p}}}} \right)^2}\)

Similarly, if \(X_1'\) is the resistance referred to the primary side.

Then the value of the equivalent reactance referred to the secondary side is \(X_2'\)

∴  \(\frac{{X_1'}}{{{X_2'}}} = {\left( {\frac{{{N_p}}}{{{N_s}}}} \right)^2} = {\left( {\frac{{{V_p}}}{{{V_s}}}} \right)^2} = {\left( {\frac{{{I_s}}}{{{I_p}}}} \right)^2}\)

Calculation:

Base impedance \({Z_B} = \frac{{K{V^2}}}{{MVA}} = \frac{{{{0.4}^2}}}{{0.004}} = 40\;{\rm{\Omega }}\)

Resistance in ohm to HV side = \({r_{pu}} \times {Z_B} = 0.02 \times 40 = 0.8\;{\rm{\Omega }}\)

Reactance in ohm to HV side = \({x_{pu}} \times {Z_B} = 0.06 \times 40 = 2.4\;{\rm{\Omega }}\)

Per unit system:

It is usual to express voltage, current, voltamperes and impedance of an electrical circuit in per unit (or percentage) of base or reference values of these quantities.

The Per Unit value of any quantity is defined as

PU value = actual value/base value

Per unit system in transformers:

The per unit impedance of a transformer is the same whether computed from primary or secondary side so long as the voltage bases on the two sides are in the ratio of transformation (equivalent per phase ratio of a three-phase transformer which is the same as the ratio of line-to-line voltage rating).

Transformation ratio of transformer is given by K = V2/V1 = E2/E1 = N2/N1.

Where N1 is the number of primary turns

V1 is the primary voltage

N2 is the number of secondary turns

V2 is the secondary voltage

Concept:

The sequence network of LLG fault is:

The fault current in LLG fault is given by:

I_f = 3 I_ao

where, \(I_{ao} = -I_{a1}({Z_2 \over Z_2+Z_o+3Z_f})\)

\(I_{a1} = {E_{a1} \over Z_1+(Z_2||(Z_o+3Z_f))}\)

Z1 = Positive sequence impedance

Z2 = Negative sequence impedance

Zo = Zero sequence impedance

Calculation:

Given, Z₁ = 0.08 pu

Z₂ = 0.07 pu

Z_o = 0.05 pu

Take Z_f = 0 pu as it is not given.

\(I_{a1} = {1\over 0.08+(0.07||0.05)}\)

I_a1 = 9.1 pu

\(I_{ao} = -9.1({0.07 \over 0.07+0.05})\)

|I_ao| = 5.9 pu

If = 3 × 5.3

If = 15.9 = 16 pu

Concept:

Zero sequence networks for different 3-phase loads are shown below.

The zero sequence equivalent circuits of 3-phase transformers can be drawn by using the following general circuit.

Z is the zero sequence impedance of the windings of the transformer. These are two series and two shunt switches. One series and one shunt switches are for both sides separately.

The series switch of a particular side is closed if it is star grounded and the shunt switch is closed if that side is delta connected, otherwise they are left open.

Application:

In the given circuit diagram, both the shunt switches are closed. Therefore, both the primary and secondary are delta-connected.

Concept:

The relation between new per-unit value & old per unit value impedance

\({({Z_{pu}})_{new}}\; = {({Z_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)

Also \({Z_{pu}} = \frac{{{Z_{Actual}}}}{{{Z_{base}}}}\)

\( {Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

Where,

(Zpu)new = New per unit value of impedance

(Zpu)old = Old per unit value of impedance

kVbase = Old base value of voltage

kVnew = New base value of voltage

MVAnew = New base value of power

MVAold = Old base value of power

Calculation:

When the voltage bases is increased to 1.1 times  (i.e. kVnew = 1.1kVbase), the new per unit impedance is

\((Z_{pu})_{new}=0.242\times \frac{1}{1.1^2}=0.2\ \ pu\)

The correct answer is option 2):(one third)

Concept:

• In a 3-phase 4 wire unbalanced system, the magnitude of the zero-sequence component is one-third current in the neutral wire.
• Negative sequence currents are involved when an unbalanced fault occurs in the system.
• Zero sequence currents are involved when an unbalanced ground fault occurs in the system.
• For an unbalanced fault with paths for zero sequence currents, at the point of fault both negative and zero sequence voltages are maximum.

Concept:

The relation between the line currents in terms of the symmetrical components of currents is given below.

\(\left[ {\begin{array}{*{20}{c}} {{I_a}}\\ {{I_b}}\\ {{I_c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&{{a^2}}&a\\ 1&a&{{a^2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_{a0}}}\\ {{I_{a1}}}\\ {{I_{a2}}} \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} {{I_{a0}}}\\ {{I_{a1}}}\\ {{I_{a2}}} \end{array}} \right] = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&a&{{a^2}}\\ 1&{{a^2}}&a \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_a}}\\ {{I_b}}\\ {{I_c}} \end{array}} \right]\)

I_a0 = Zero Sequence Component of Current

I_a1 = Positive Sequence Component of Current

I_a2 = Negative Sequence Component of Current

a = 1∠120°, which represents the rotation of 120° in clockwise direction.

a² = 1∠-120° or 1∠240° in anticlockwise direction or in clockwise direction, respectively.

Calculation:

I_a = 4∠60°, I_b­ = 0, I_c = 4∠-120°

Negative sequence component of current,

\({I_{a2}} = \frac{1}{3}\left( {{I_a} + {a^2}{I_b} + a{I_c}} \right) = \frac{1}{3}\left( {4\angle 60^\circ + \left( {1\angle - 120^\circ } \right)\left( 0 \right) + 1\angle 120^\circ \left( {4\angle - 120^\circ } \right)} \right)\)

\({I_{a2}} = \frac{1}{3}\left( {4\angle 60^\circ + 1\angle 120^\circ \left( {4\angle - 120^\circ } \right)} \right)\)

Concept:

In a line to ground fault, the current through the neutral during fault is given by

If = 3Ia0

Where Ia0 is the zero sequence current of generator

Calculation:

Given that, the zero sequence current of generator for line to ground fault = j2.4 pu

Concept:

Per unit quantity:

Per unit quantity = Actual quantity in the units / Base (or) reference quantity in the same units

⇒ Per unit impedance Zpu = Zactual / Zbase

⇒ Zpu = ZΩ × MVAb / (kVb)2

Conversion of one per unit impedance into another per unit impedance is given by

\({{\bf{Z}}_{{\bf{pu}}}}\left( {{\bf{new}}} \right) = {{\bf{Z}}_{{\bf{pu}}}}\left( {{\bf{old}}} \right)\left( {\frac{{{\bf{MV}}{{\bf{A}}_{{\bf{new}}}}}}{{{\bf{MV}}{{\bf{A}}_{{\bf{old}}}}}}} \right){\left( {\frac{{{\bf{k}}{{\bf{V}}_{\bf{b}}}_{{\bf{old}}}}}{{{\bf{k}}{{\bf{V}}_{\bf{b}}}_{{\bf{new}}}}}} \right)^2}\)