Classification of Buses MCQ Quiz in मराठी - Objective Question with Answer for Classification of Buses - मोफत PDF डाउनलोड करा

Classification of Buses:

The power system is nothing but the interconnection of the various bus.

Each of these buses is associated with four electrical parameters namely voltage with magnitude and phase angle, active power, and reactive power.

Load Bus:

• For load bus, real power P and reactive power Q are known but the magnitude and phase angle of bus voltage is unknown.
• It is desired to find the bus voltage using load flow analysis. At load bus, voltage is allowed to vary within some specified limit

Generator Bus:

• Generator Bus is also called a voltage-controlled bus.
• The generator is connected to this bus.
• Therefore the bus voltage corresponding to generation voltage and active power generation corresponding to generator rating is specified for this bus.

Slack Bus:

• Slack Bus is also known as Swing or Reference Bus.
• The slack bus does not exist in real rather it is assumed for the consideration of losses occurring during power transmission.
• Actually, there exist only two buses in the power system, Load Bus and Generator Bus for which active power is specified.
• Since active power delivered by Generator Bus and consumed by Load Bus differ
• This means that a power loss equal to the difference between Generator Bus P and Load Bus P is occurring.
• This loss can only be calculated after the solution of Load Flow.
• Therefore to supply power loss, an extra generator bus is considered for which bus magnitude and voltage are specified, and active power and reactive power are to be calculated.
• This active power of slack bus is the equivalent power loss occurring in different systems.
• Generally, the phase angle of the slack bus is taken for reference for the entire load flow solution. Therefore this bus is also called Reference Bus.

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If the reference bus is changed in two load flow runs with the same system data and power obtained for reference bus taken as specified P and Q.

So, the reference bus is taken as a load bus.

Hence, the loses will be same but the complex bus voltage will change.

Calculation:

P = Line impedance angle, δ = 30°, α = 0°, A = D = 1.0, B = 2∠90o 

\(\begin{array}{l} {P_{sending}} = \left| {\frac{D}{B}} \right|{\left| {{N_S}} \right|^2}\cos \left( {β - \alpha } \right) - \frac{{\left| {{V_S}} \right|\left| {{V_r}} \right|}}{{\left| B \right|}}\cos \left( {β + \delta } \right)\\ {Q_{sending}} = \left| {\frac{D}{B}} \right|{\left| {{V_S}} \right|^2}\sin \left( {β - \alpha } \right) - \frac{{\left| {{V_S}} \right|\left| {{V_r}} \right|}}{{\left| B \right|}}\cos \left( {β + \delta } \right)\\ \therefore {P_{sending}} = \left| {\frac{{1.0}}{2}} \right|{50^2}\cos \left( {90 - 0} \right) - \frac{{50 \times 50}}{{\left| 2 \right|}}\cos \left( {90 + 30} \right) \end{array}\)

= 625 W

\({Q_{sending}} = \left| {\frac{{1.0}}{2}} \right|{50^2}\sin \left( {90 - 0} \right) - \frac{{50 \times 50}}{{\left| 2 \right|}}\sin \left( {90 + 30} \right)\)

= 167.5 VAR

Explanation:

To determine the size of the Jacobian matrix in a power system network, we need to understand the components involved. In this network, we have:

• Total buses (n) = 40
• Voltage-controlled buses (PV buses) = 12
• Generator buses (also PV buses) = 5

In power system analysis, the Jacobian matrix is used to solve the power flow equations using the Newton-Raphson method. The size of the Jacobian matrix is determined based on the number of variables in the system. These variables are primarily the voltage magnitudes and angles at different buses, excluding the reference bus (slack bus).

Steps to determine the size of the Jacobian matrix:

• First, identify the total number of buses (n).
• Determine the number of PV buses (voltage-controlled buses).
• Subtract the reference bus (slack bus) from the total number of buses to find the number of unknown variables.

Calculation:

The total number of buses (n) = 40.

Out of these, 12 are PV buses. Since the generator buses are a subset of the PV buses, we do not double-count them. Thus, we have:

• PV buses = 12 (including generator buses)
• PQ buses (load buses) = Total buses - PV buses - 1 (slack bus)
• PQ buses = 40 - 12 - 1 = 27

In power flow analysis, the Jacobian matrix is split into four submatrices:

• J1: Partial derivatives of active power with respect to voltage angles (θ).
• J2: Partial derivatives of active power with respect to voltage magnitudes (V).
• J3: Partial derivatives of reactive power with respect to voltage angles (θ).
• J4: Partial derivatives of reactive power with respect to voltage magnitudes (V).

The size of the Jacobian matrix is determined by the number of equations and the number of variables. For a system with (n-1) buses, we have:

• n-1 voltage angle equations (excluding the reference bus).
• n-m-1 voltage magnitude equations, where m is the number of PV buses.

Thus, the total number of variables (unknowns) is the sum of the voltage angles and magnitudes:

• Number of voltage angle variables = n - 1 = 40 - 1 = 39
• Number of voltage magnitude variables = n - m - 1 = 40 - 12 - 1 = 27

The size of the Jacobian matrix is the sum of these variables:

• Total number of variables = 39 + 27 = 66

Therefore, the size of the Jacobian matrix is 66 * 66.

Correct Option Analysis:

The correct option is:

Option 3: 66 * 66

This option correctly represents the size of the Jacobian matrix for the given power system network. The matrix accounts for all the voltage angles and magnitudes that need to be solved in the power flow analysis.

Additional Information

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

Option 1: 46 * 46

This option is incorrect because it underestimates the total number of variables. The size of the Jacobian matrix should account for both the voltage angles and magnitudes, which result in a larger matrix.

Option 2: 23 * 23

This option is incorrect as it grossly underestimates the number of variables in the system. It does not take into account the total number of buses and the necessary voltage magnitudes and angles.

Option 4: 21 * 21

This option is incorrect as it also underestimates the number of variables. The calculation of the Jacobian matrix size should consider all unknown voltage angles and magnitudes, leading to a larger matrix size.

Conclusion:

Understanding the structure and calculation of the Jacobian matrix is essential for solving power flow equations in electrical networks. The correct size of the Jacobian matrix for a network with 40 buses, including 12 PV buses, is 66 * 66, as it accounts for all the necessary voltage angles and magnitudes. This ensures accurate and efficient analysis of the power system.

Bus-bar Ratings:

A bus-bar is an electrical conductor used to collect and distribute electrical power. It plays a crucial role in electrical power systems, ensuring efficient distribution of electricity. The correct rating of a bus-bar is essential for the safety and reliability of the electrical system.

Factors Determining Bus-bar Ratings:

• Current: The bus-bar must be rated for the maximum current it is expected to carry. This ensures that the bus-bar can handle the load without overheating or causing damage to the system.
• Voltage: The bus-bar must be rated for the maximum voltage of the system. Proper insulation and clearance distances must be maintained to prevent electrical arcing and ensure safe operation.
• Frequency: Although not as critical as current and voltage, frequency can affect the performance of the bus-bar, especially in systems with high harmonic content. The bus-bar must be designed to handle the operating frequency of the system.
• Short Circuit Current: The bus-bar must be capable of withstanding the mechanical and thermal stresses caused by short circuit conditions. This includes the peak short circuit current and the duration for which the bus-bar can sustain this current without damage.

Conclusion:

The correct rating of a bus-bar encompasses current, voltage, frequency, and short circuit current. Ensuring that the bus-bar is appropriately rated for these parameters is critical for the safe and reliable operation of the electrical system. Therefore, the correct option is option 4.

\(I = \frac{{{V_1}L\delta - {V_1}\angle 0}}{{jx}} = \frac{{1\angle 20^\circ - 1\angle 0^\circ }}{{3\angle 90^\circ }} = 0.116\angle 10^\circ \)

S = V₁I* = 1∠20° × .116∠-10 = 0.116∠10

S₁ = SD₁ + s

= 20 + 20j + .114 + 0.020j

S₁ = 20.114 + 20.02 j

Total power = Active power + Reactive power

∴ Reactive power = 20.02 PU