Per Unit System MCQ Quiz - Objective Question with Answer for Per Unit System - Download Free PDF

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:

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:

Let's consider

Zpu old = x

MVAold = M

kVb old = V

As base kV and base MVA are both doubled

MVAnew = 2M

kVb new = 3V

⇒ Xpu(new) = x × (2M / M) × (V / 3V)2 = x/9

= \(\frac{2}{9} X\) p.u

Concept:

In symmetrical fault calculations, the correct option is 1) base kVA/rated kVA × % reactance at rated kVA.

• Symmetrical fault calculations involve analyzing faults or short circuits that occur in a power system. When a fault occurs, the fault current flows through the system, and the system impedance affects the magnitude of the fault current.
• In these calculations, the base kVA refers to the base power or apparent power used as a reference for the calculations. The rated kVA represents the rated power or apparent power of the equipment or system under consideration.
• The percentage reactance at rated kVA represents the reactance of the system expressed as a percentage of the rated kVA.

To determine the percentage reactance at the base kVA, you need to scale the reactance based on the ratio of the base kVA to the rated kVA. Therefore, the correct formula is:

Percentage reactance at base kVA = \(\frac{base\; kVA}{ rated\;kVA}\times percentage\; reactance\; at\; rated\; kVA\)

This formula allows you to calculate the reactance at the base kVA level, taking into account the scaling factor based on the ratio of the base kVA and the rated kVA.

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 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

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

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

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

The correct answer is option 4):(2.5 Ω)

Concept:

For an AC network,

VA Rating = \(V^2\over Z\)

Where V is voltage in volt

Z is the impedance (base) in Ω

Calculation:

VA rating = 40 MVA = 40 × 10⁶ VA

V = 10 × 10³ volts

= \(\frac{(10\times 10^3)^2}{40\times 10^6}= \frac{10^2}{40}= 2.5 \ \Omega\)

Concept:

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

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

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

Where,

(Xpu)new = New per unit value of reactance

(Xpu)old = Old per unit value of reactance

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:

Given that,

Xd(old) = 1 pu

MVA(new) = 200

MVA(old) = 100

kV(old) = 20

kV(new) = 20

∴ \({X_{d\left( {new} \right)}} = 1 \times \frac{{200}}{{100}} \times {\left( {\frac{{20}}{{20}}} \right)^2} = 2\;pu\)

Per unit (p.u.) quantity:

The per-unit value of any quantity is defined as the ratio of actual value in any unit to the base or reference value in the same unit.

The per-unit value is dimensionless.

​Per unit impedance (Zpu):

Zpu is defined as the ratio of actual value to base impedance.

It is also defined as The ratio of full-load volt-amperes to short-circuit volt-amperes.

\({Z_{pu}} = \frac{{Z{_a}}}{{Z_B}}\)

\({Z_{pu}} = \frac{{{Z_{a}} \times {{\left( {kVA} \right)}_B}}}{{\left( {kV} \right)_B^2 \times 1000}}\)

Za = Actual impedance

ZB = Base impedance

(kVA)B = Base kVA

(kV)B = Base voltage in kV

Important points:

• Per unit voltage (Vpu) = (kV)actual / (kV)B
• Short circuit current in pu = Actual current / Base current

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

Per Unit System:

The per-unit value of any quantity is defined as the ratio of actual value in any unit to the base or reference value in the same unit.

Any quantity is converted into per unit quantity by dividing the numeral value by the chosen base value of the same dimension. The per-unit value is dimensionless.

PU Value (PU) = \(\frac{A}{B}\)

Here, A is the actual value in any unit
B is the Base or reference value in the same unit

Consider the base value of voltage is (kV_B), the Base value of current is (I_B) and the base value of apparent power is (kVA_B)

And, Base value of current (I_B) = \(\frac{kVA_B}{kV_B}\) .... (1)

Now, the Base value of the impedance (Z_B) will be the ratio of the base value of voltage to the current.

⇒ Z_B = \(\frac{kV_B}{I_B}\)

From equation (1),

Z_B = \(\frac{kV_B}{(kVA_B/kV_B)}=\frac{(kV_B)^2}{kVA_B}\)

Hence, per unit value of impedance can be written as if Z is the actual value of impedance,

Z_{PU (1000)} = \(\frac{Z_A}{Z_B}=Z\times \frac{kVA_B}{(kV_B)^2}\)

Since the above expression is given infraction of 1000 units,

Hence per unit value will be given as,

Z_PU = \(Z\times kVA_B\times \frac{1000}{(kV_B)^2}\)

The percentage reactance of a circuit is defined as the percentage of the total phase voltage dropped in the circuit when full-load current is flowing through it.

% X = \({I X\over V}\times 100\)

where, X = Reactance in ohms per phase

I = Full load current

V = Phase Voltage