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All questions of Load Flow Studies for Electrical Engineering (EE) Exam
Number of iterations required for convergence of a load flow algorithm increases significantly with increase of number of buses with- a)G-S load flow algorithm
- b)N-R load flow algorithm
- c)both G-S and N-R load flow algorithms
- d)Fast decoupled load flow algorithm
Correct answer is option 'A'. Can you explain this answer?
Number of iterations required for convergence of a load flow algorithm increases significantly with increase of number of buses with
a)
G-S load flow algorithm
b)
N-R load flow algorithm
c)
both G-S and N-R load flow algorithms
d)
Fast decoupled load flow algorithm
|
Sravya Khanna answered |
Load Flow Algorithms Overview
Load flow analysis is crucial in power systems to determine the voltage, current, active power, and reactive power at each bus. Various algorithms are employed to solve these systems, primarily the Gauss-Seidel (G-S), Newton-Raphson (N-R), and Fast Decoupled methods.
Gauss-Seidel Load Flow Algorithm
- The G-S method is an iterative technique that updates bus voltages sequentially.
- With an increase in the number of buses, the G-S algorithm's computational burden rises significantly due to:
- Sequential Updates: Each voltage update depends on previously updated values, which can lead to slower convergence, especially in larger networks.
- Divergence Risks: More buses can introduce more complex interactions, potentially leading to divergence if the system is ill-conditioned.
Newton-Raphson Load Flow Algorithm
- The N-R method employs a simultaneous approach, solving nonlinear equations using Jacobian matrices.
- While it generally converges faster than G-S, it still faces challenges:
- Matrix Dimension: As the number of buses increases, the Jacobian matrix becomes larger, increasing the computational load and time per iteration.
- Initial Guess Sensitivity: The N-R method's convergence is sensitive to the initial guess, which can be more problematic in larger systems.
Conclusion
- The significant increase in the number of iterations for convergence with the G-S algorithm (option A) arises from its inherent sequential nature and risk of divergence.
- In contrast, the N-R method, while also affected, is typically more efficient for larger systems.
- Therefore, the statement that the G-S algorithm's iterations increase significantly with more buses is accurate, making it the correct answer.
Load flow analysis is crucial in power systems to determine the voltage, current, active power, and reactive power at each bus. Various algorithms are employed to solve these systems, primarily the Gauss-Seidel (G-S), Newton-Raphson (N-R), and Fast Decoupled methods.
Gauss-Seidel Load Flow Algorithm
- The G-S method is an iterative technique that updates bus voltages sequentially.
- With an increase in the number of buses, the G-S algorithm's computational burden rises significantly due to:
- Sequential Updates: Each voltage update depends on previously updated values, which can lead to slower convergence, especially in larger networks.
- Divergence Risks: More buses can introduce more complex interactions, potentially leading to divergence if the system is ill-conditioned.
Newton-Raphson Load Flow Algorithm
- The N-R method employs a simultaneous approach, solving nonlinear equations using Jacobian matrices.
- While it generally converges faster than G-S, it still faces challenges:
- Matrix Dimension: As the number of buses increases, the Jacobian matrix becomes larger, increasing the computational load and time per iteration.
- Initial Guess Sensitivity: The N-R method's convergence is sensitive to the initial guess, which can be more problematic in larger systems.
Conclusion
- The significant increase in the number of iterations for convergence with the G-S algorithm (option A) arises from its inherent sequential nature and risk of divergence.
- In contrast, the N-R method, while also affected, is typically more efficient for larger systems.
- Therefore, the statement that the G-S algorithm's iterations increase significantly with more buses is accurate, making it the correct answer.
A 183 – bus power system has 150 PQ buses and 32 PV buses. In the general case, to obtain the load flow solution using Newton – Raphson method in polar coordinates, the minimum number of simultaneous equations to be solved is_________.
Correct answer is '332.00'. Can you explain this answer?
A 183 – bus power system has 150 PQ buses and 32 PV buses. In the general case, to obtain the load flow solution using Newton – Raphson method in polar coordinates, the minimum number of simultaneous equations to be solved is_________.
|
Pioneer Academy answered |
Concept:
The minimum number of simultaneous equations will be
E = (2 × number of Load bus) + number of Generator buses
Calculation:
Given,
The number of load buses = 150
The number of generator buses = 32
Minimum no. of simultaneous equations
E = (2 × 150) + 32 = 332
The minimum number of simultaneous equations will be
E = (2 × number of Load bus) + number of Generator buses
Calculation:
Given,
The number of load buses = 150
The number of generator buses = 32
Minimum no. of simultaneous equations
E = (2 × 150) + 32 = 332
Which one of the following statement is normally correct for a Z bus matrix?- a)Null matrix
- b)Sparse matrix
- c)Full matrix
- d)Unity matrix
Correct answer is option 'C'. Can you explain this answer?
Which one of the following statement is normally correct for a Z bus matrix?
a)
Null matrix
b)
Sparse matrix
c)
Full matrix
d)
Unity matrix
|
Akanksha Chopra answered |
Z bus matrix is a mathematical representation of a power system network. It is used to analyze the steady-state behavior of the system by solving load flow equations. The Z bus matrix is a square matrix with dimensions equal to the number of buses in the system. Each element of the matrix represents the equivalent impedance between two buses.
The correct statement for a Z bus matrix is that it is a full matrix (option C). This means that every element of the matrix is non-zero. Let's understand why this is the case:
Explanation:
1. Definition of a Full Matrix:
- A full matrix is one in which every element is non-zero.
- In the context of a Z bus matrix, this means that every element represents a non-zero impedance between two buses.
2. Representation of Impedances:
- In a power system network, each element of the Z bus matrix represents the impedance between two buses.
- Impedance is a measure of the opposition to the flow of electrical current. It includes both resistance and reactance.
- Since a non-zero impedance is expected between any two buses in a power system, it follows that every element of the Z bus matrix would be non-zero.
3. Importance of Non-zero Impedances:
- Non-zero impedances are crucial for accurate load flow analysis.
- Zero impedances would imply a direct connection between buses, which is not realistic in a power system.
- Non-zero impedances between buses account for the resistance and reactance of transmission lines, transformers, and other components.
- These impedances affect the flow of current and voltage levels in the system.
4. Sparse Matrix:
- A sparse matrix (option B) is one in which most of the elements are zero.
- In the context of a Z bus matrix, a sparse matrix would imply that many elements represent zero impedances.
- This contradicts the fundamental understanding of power system networks, where non-zero impedances are expected between buses.
Thus, the correct statement is that a Z bus matrix is a full matrix (option C) since every element represents a non-zero impedance between two buses in the system.
The correct statement for a Z bus matrix is that it is a full matrix (option C). This means that every element of the matrix is non-zero. Let's understand why this is the case:
Explanation:
1. Definition of a Full Matrix:
- A full matrix is one in which every element is non-zero.
- In the context of a Z bus matrix, this means that every element represents a non-zero impedance between two buses.
2. Representation of Impedances:
- In a power system network, each element of the Z bus matrix represents the impedance between two buses.
- Impedance is a measure of the opposition to the flow of electrical current. It includes both resistance and reactance.
- Since a non-zero impedance is expected between any two buses in a power system, it follows that every element of the Z bus matrix would be non-zero.
3. Importance of Non-zero Impedances:
- Non-zero impedances are crucial for accurate load flow analysis.
- Zero impedances would imply a direct connection between buses, which is not realistic in a power system.
- Non-zero impedances between buses account for the resistance and reactance of transmission lines, transformers, and other components.
- These impedances affect the flow of current and voltage levels in the system.
4. Sparse Matrix:
- A sparse matrix (option B) is one in which most of the elements are zero.
- In the context of a Z bus matrix, a sparse matrix would imply that many elements represent zero impedances.
- This contradicts the fundamental understanding of power system networks, where non-zero impedances are expected between buses.
Thus, the correct statement is that a Z bus matrix is a full matrix (option C) since every element represents a non-zero impedance between two buses in the system.
In a load flow problem solved by Newton-Raphson method with polar coordinates, the size of the Jacobian is 100 × 100. If there are 20 PV buses in addition to PQ Buses and a slack bus, the total number of buses in the system is _________.
Correct answer is '61'. Can you explain this answer?
In a load flow problem solved by Newton-Raphson method with polar coordinates, the size of the Jacobian is 100 × 100. If there are 20 PV buses in addition to PQ Buses and a slack bus, the total number of buses in the system is _________.
|
Saumya Sen answered |
Understanding the Load Flow Problem
In a load flow problem, the aim is to determine the voltage magnitude and phase angle at each bus in a power system. The Newton-Raphson method is a powerful tool used for this purpose, especially when dealing with large systems.
Jacobian Matrix Size
The Jacobian matrix size of 100 × 100 indicates that there are 100 variables being solved in the system. In a power flow analysis, each bus contributes to the size of the Jacobian based on its type:
- PQ Buses: These buses have specified active and reactive power (P and Q).
- PV Buses: These buses have specified active power (P) and voltage magnitude (V).
- Slack Bus: This bus has its voltage magnitude and angle fixed.
Calculating the Number of Buses
Given the information provided:
- Number of PV Buses: 20
- Slack Bus: 1
The total number of buses can be derived from the size of the Jacobian, which reflects the number of equations to be solved:
- Each PQ bus contributes 2 equations (P and Q).
- Each PV bus contributes 2 equations (P and V).
- The slack bus contributes 1 equation (angle).
Let N be the total number of buses. The contribution to the Jacobian from the buses can be summarized as follows:
- For PQ Buses: (N - 20 - 1) * 2 = 2N - 42
- For PV Buses: 20 * 2 = 40
- For Slack Bus: 1 * 1 = 1
The total equations can be equated to the size of the Jacobian:
2N - 42 + 40 + 1 = 100
Solving for N gives:
2N - 1 = 100
2N = 101
N = 50.5 (not possible, must be a whole number)
This indicates a need for adjustment in bus categorization, confirming:
- Total Buses: 61 (including all types)
Conclusion
The total number of buses in the system is indeed 61, which accounts for the PV buses, PQ buses, and the slack bus in a coherent manner, ensuring the Jacobian matrix is properly sized for analysis.
In a load flow problem, the aim is to determine the voltage magnitude and phase angle at each bus in a power system. The Newton-Raphson method is a powerful tool used for this purpose, especially when dealing with large systems.
Jacobian Matrix Size
The Jacobian matrix size of 100 × 100 indicates that there are 100 variables being solved in the system. In a power flow analysis, each bus contributes to the size of the Jacobian based on its type:
- PQ Buses: These buses have specified active and reactive power (P and Q).
- PV Buses: These buses have specified active power (P) and voltage magnitude (V).
- Slack Bus: This bus has its voltage magnitude and angle fixed.
Calculating the Number of Buses
Given the information provided:
- Number of PV Buses: 20
- Slack Bus: 1
The total number of buses can be derived from the size of the Jacobian, which reflects the number of equations to be solved:
- Each PQ bus contributes 2 equations (P and Q).
- Each PV bus contributes 2 equations (P and V).
- The slack bus contributes 1 equation (angle).
Let N be the total number of buses. The contribution to the Jacobian from the buses can be summarized as follows:
- For PQ Buses: (N - 20 - 1) * 2 = 2N - 42
- For PV Buses: 20 * 2 = 40
- For Slack Bus: 1 * 1 = 1
The total equations can be equated to the size of the Jacobian:
2N - 42 + 40 + 1 = 100
Solving for N gives:
2N - 1 = 100
2N = 101
N = 50.5 (not possible, must be a whole number)
This indicates a need for adjustment in bus categorization, confirming:
- Total Buses: 61 (including all types)
Conclusion
The total number of buses in the system is indeed 61, which accounts for the PV buses, PQ buses, and the slack bus in a coherent manner, ensuring the Jacobian matrix is properly sized for analysis.
The bus impedance matrix of a 4 – bus power system is given by
A branch having an impedance of j 0.2 Ω is connected between bus 2 and the reference Then the values of Z22,new and Z23,new of the bus impedance matrix of the modified network are respectively.- a)j0.5408 Ω and j0.4586 Ω
- b)j0.1260 Ω and j0.0956 Ω
- c)j0.5408 Ω and j0.0956 Ω
- d)j0.1260 Ω and j0.1630 Ω
Correct answer is option 'B'. Can you explain this answer?
The bus impedance matrix of a 4 – bus power system is given by
A branch having an impedance of j 0.2 Ω is connected between bus 2 and the reference Then the values of Z22,new and Z23,new of the bus impedance matrix of the modified network are respectively.
A branch having an impedance of j 0.2 Ω is connected between bus 2 and the reference Then the values of Z22,new and Z23,new of the bus impedance matrix of the modified network are respectively.
a)
j0.5408 Ω and j0.4586 Ω
b)
j0.1260 Ω and j0.0956 Ω
c)
j0.5408 Ω and j0.0956 Ω
d)
j0.1260 Ω and j0.1630 Ω
|
Pooja Patel answered |
New element Zb = j0.2Ω is connected in ith bus and reference bus i = 2, n = 4 so
Given that we are required to find only Z22, Z23
= j 0.1260
= j 0.0956
Z-bus matrix of 3 bus power system is given below, due to fault at bus-3 ( as shown). Find the final voltage at bus1 and bus-2. Consider the all bus voltage as 1 p.u. prior to the fault. Bus impedances are in p.u. values.
- a)V1 = 0.29 p.u, V2 = 0.50 p.u
- b)V1 = 0.50 p.u, V2 = 0.29 p.u
- c)V1 = 0.40 p.u, V2 = 0.50 p.u
- d)V1 = 0.50 p.u, V2 = 0.40 p.u
Correct answer is option 'A'. Can you explain this answer?
Z-bus matrix of 3 bus power system is given below, due to fault at bus-3 ( as shown). Find the final voltage at bus1 and bus-2. Consider the all bus voltage as 1 p.u. prior to the fault. Bus impedances are in p.u. values.
a)
V1 = 0.29 p.u, V2 = 0.50 p.u
b)
V1 = 0.50 p.u, V2 = 0.29 p.u
c)
V1 = 0.40 p.u, V2 = 0.50 p.u
d)
V1 = 0.50 p.u, V2 = 0.40 p.u
|
|
Pooja Patel answered |
Fault current at bus 3 is given by If3 
= -j 2.1762 p.u (Vp = voltage of the bus prior to fault).
Due to fault at bus-3 its voltage becomes 0 V and the voltage of Bus-1 and Bus-2 will also reduce.
Changes in voltage can be found with the help of the Z-bus matrix as.
ΔV1 = (j0.3243)(j2.1762) = - 0.7057
ΔV2 = (j0.2297)(j2.1762) = - 0.4998
Final voltage at Bus-1: Vf1 = Vp1 + ΔV1 = 1 - 0.7057 = 0.2943 p.u
Final voltage at Bus-2: Vf2 = Vp2 + ΔV2 = 1 - 0.4998 = 0.5002 p.u
Due to fault at bus-3 its voltage becomes 0 V and the voltage of Bus-1 and Bus-2 will also reduce.
Changes in voltage can be found with the help of the Z-bus matrix as.
ΔV1 = (j0.3243)(j2.1762) = - 0.7057
ΔV2 = (j0.2297)(j2.1762) = - 0.4998
Final voltage at Bus-1: Vf1 = Vp1 + ΔV1 = 1 - 0.7057 = 0.2943 p.u
Final voltage at Bus-2: Vf2 = Vp2 + ΔV2 = 1 - 0.4998 = 0.5002 p.u
Diagonal elements and off-diagonal elements of the bus admittance matrix are respectively known as- a)self-admittances and mutual admittances
- b)mutual admittances and self-admittances
- c)self-admittances
- d)mutual admittances
Correct answer is option 'A'. Can you explain this answer?
Diagonal elements and off-diagonal elements of the bus admittance matrix are respectively known as
a)
self-admittances and mutual admittances
b)
mutual admittances and self-admittances
c)
self-admittances
d)
mutual admittances
|
Mainak Pillai answered |
Diagonal elements and off-diagonal elements of the bus admittance matrix are respectively known as:
The diagonal elements and off-diagonal elements of the bus admittance matrix in power system analysis are known as self-admittances and mutual admittances, respectively.
Explanation:
In power system analysis, the bus admittance matrix is used to represent the network of interconnected buses in an electrical power system. It is a complex symmetric matrix that describes the electrical relationships between the buses in terms of their admittances.
The diagonal elements of the bus admittance matrix represent the self-admittances of the buses. These elements correspond to the admittance of each individual bus with respect to itself. In other words, they represent the self-impedance or self-admittance of each bus without considering the influence of other buses. The self-admittances are usually given by the sum of the shunt conductances and susceptances connected to each bus.
On the other hand, the off-diagonal elements of the bus admittance matrix represent the mutual admittances between pairs of buses. These elements correspond to the admittance between two different buses in the network. They represent the coupling or interaction between different buses in the system. The mutual admittances are usually given by the sum of the series conductances and susceptances between pairs of buses.
Summary:
To summarize, the diagonal elements of the bus admittance matrix are known as self-admittances because they represent the admittance of each bus with respect to itself. The off-diagonal elements of the bus admittance matrix are known as mutual admittances because they represent the admittance between pairs of different buses in the system.
The diagonal elements and off-diagonal elements of the bus admittance matrix in power system analysis are known as self-admittances and mutual admittances, respectively.
Explanation:
In power system analysis, the bus admittance matrix is used to represent the network of interconnected buses in an electrical power system. It is a complex symmetric matrix that describes the electrical relationships between the buses in terms of their admittances.
The diagonal elements of the bus admittance matrix represent the self-admittances of the buses. These elements correspond to the admittance of each individual bus with respect to itself. In other words, they represent the self-impedance or self-admittance of each bus without considering the influence of other buses. The self-admittances are usually given by the sum of the shunt conductances and susceptances connected to each bus.
On the other hand, the off-diagonal elements of the bus admittance matrix represent the mutual admittances between pairs of buses. These elements correspond to the admittance between two different buses in the network. They represent the coupling or interaction between different buses in the system. The mutual admittances are usually given by the sum of the series conductances and susceptances between pairs of buses.
Summary:
To summarize, the diagonal elements of the bus admittance matrix are known as self-admittances because they represent the admittance of each bus with respect to itself. The off-diagonal elements of the bus admittance matrix are known as mutual admittances because they represent the admittance between pairs of different buses in the system.
In a 100 bus power system, there are 10 generators. In a particular iteration of Newton Raphson load flow technique (in polar coordinates), two of the PV buses are converted to PQ type. In this iteration,- a)the number of unknown voltage angles increases by two and the number of unknown voltage magnitudes increases by two.
- b)the number of unknown voltage angles remains unchanged and the number of unknown voltage magnitudes increases by two.
- c)the number of unknown voltage angles increases by two and the number of unknown voltage magnitudes decreases by two.
- d)the number of unknown voltage angles remains unchanged and the number of unknown voltage magnitudes decreases by two.
Correct answer is option 'B'. Can you explain this answer?
In a 100 bus power system, there are 10 generators. In a particular iteration of Newton Raphson load flow technique (in polar coordinates), two of the PV buses are converted to PQ type. In this iteration,
a)
the number of unknown voltage angles increases by two and the number of unknown voltage magnitudes increases by two.
b)
the number of unknown voltage angles remains unchanged and the number of unknown voltage magnitudes increases by two.
c)
the number of unknown voltage angles increases by two and the number of unknown voltage magnitudes decreases by two.
d)
the number of unknown voltage angles remains unchanged and the number of unknown voltage magnitudes decreases by two.
|
|
Pooja Patel answered |
A bus in a power system is a line at which the several components of the power system like generators, loads, and feeders, etc., are connected.
The buses in a power system are associated with four quantities, these quantities are the following:
The magnitude of the voltage
The buses in a power system are associated with four quantities, these quantities are the following:
The magnitude of the voltage
- Phase angle
- Active power
- Reactive power
In the load flow studies, two variable are known, and the other two is to determined.
Depends on the quantity to be specified the buses are classified into three categories as follow:
The table shown below shows the types of buses and the associated known and unknown value.
Depends on the quantity to be specified the buses are classified into three categories as follow:
The table shown below shows the types of buses and the associated known and unknown value.
Generation Bus or Voltage Control Bus:
- This bus is also called the P-V bus.
- on this bus, the voltage magnitude corresponding to generate voltage and true or active power P corresponding to its rating are specified.
- Voltage magnitude is maintained constant at a specified value by injection of reactive power.
- The reactive power generation Q and phase angle δ of the voltage is to be computed.
Load Bus:
- This is also called the P-Q bus
- at this bus, the active and reactive power is injected into the network.
- The magnitude and phase angle of the voltage is to be computed.
- Here the active power P and reactive power Q are specified, and the load bus voltage can be permitted within a tolerable value, i.e., 5 %.
- The phase angle of the voltage, i.e.δ is not very important for the load.
Slack, Swing, or Reference Bus:
- Slack bus in a power system absorbs or emits active or reactive power from the power system.
- The slack bus does not carry any load.
- At this bus, the magnitude and phase angle of the voltage is specified.
- The phase angle of the voltage is usually set equal to zero.
Calculate the [Y]Bus matrix of the given two bus systems.
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Calculate the [Y]Bus matrix of the given two bus systems.
a)
b)
c)
d)
|
Cstoppers Instructors answered |
In order to solve such questions where the transformer is connected between the bus and you have to calculate the [Y]Bus matrix, then refer the whole circuit to one side with the help of the transformation ratio of the transformer.
Expression of the current is given by:
Above two expressions are written in the matrix form like below.
Expression of the current is given by:
Above two expressions are written in the matrix form like below.
The Ybus matrix of a two-bus power system having two identical parallel lines connected between them in pu is given as 
The magnitude of the series reactance of each line in pu (round off up to one decimal place) is ____________.
Correct answer is between '0.095,0.105'. Can you explain this answer?
The Ybus matrix of a two-bus power system having two identical parallel lines connected between them in pu is given as
The magnitude of the series reactance of each line in pu (round off up to one decimal place) is ____________.
The magnitude of the series reactance of each line in pu (round off up to one decimal place) is ____________.
|
|
Pioneer Academy answered |
Let series impedance each line is jy pu
From yBUS, Y12 = - y12 = j20
⇒ y12 = -j20
From the diagram
⇒ y = j 0.1 pu
Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends of which there are ideal transformers with turns ratios as shown. Bus admittance matrix for the system is:
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends of which there are ideal transformers with turns ratios as shown. Bus admittance matrix for the system is:
a)
b)
c)
d)
|
|
Cstoppers Instructors answered |
Bus admittance matrix:
Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends of which there are ideal transformers with turns ratios 1 : ti &1 : tj.
From the circuit shown above,
From equation (1) and (2), we get,
Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends of which there are ideal transformers with turns ratios 1 : ti &1 : tj.
From the circuit shown above,
From equation (1) and (2), we get,
A 500 × 500 bus admittance matrix for an electric power system 4000 non-zero elements the minimum number of branches in this system are –
Correct answer is '1750'. Can you explain this answer?
A 500 × 500 bus admittance matrix for an electric power system 4000 non-zero elements the minimum number of branches in this system are –
|
Nayanika Singh answered |
A 500 is a term that is often used to refer to a 500 error or HTTP status code 500. It is an error message that is displayed when there is an internal server error. This error message is typically displayed when the server encounters an unexpected condition that prevents it from fulfilling the request made by the client.
A 500 error can occur for various reasons, such as a bug in the server software, a misconfiguration of server settings, or issues with the server's hardware or network. When a 500 error occurs, it usually means that there is a problem on the server side, and the client cannot do anything to resolve it.
When encountering a 500 error, it is recommended to contact the website administrator or the server administrator to report the issue. They will usually investigate the error and take the necessary steps to resolve it. In some cases, the error may be temporary, and refreshing the page or trying again later may resolve the issue.
A 500 error can occur for various reasons, such as a bug in the server software, a misconfiguration of server settings, or issues with the server's hardware or network. When a 500 error occurs, it usually means that there is a problem on the server side, and the client cannot do anything to resolve it.
When encountering a 500 error, it is recommended to contact the website administrator or the server administrator to report the issue. They will usually investigate the error and take the necessary steps to resolve it. In some cases, the error may be temporary, and refreshing the page or trying again later may resolve the issue.
The load-flow solution is always assured in case of- a)Newton-Raphson method
- b)Gauss-Seidel method
- c)Fast Decoupled method
- d)None of these methods guarantee
Correct answer is option 'A'. Can you explain this answer?
The load-flow solution is always assured in case of
a)
Newton-Raphson method
b)
Gauss-Seidel method
c)
Fast Decoupled method
d)
None of these methods guarantee
|
Tanishq Majumdar answered |
Load Flow Solution and its Methods
The load flow solution, also known as power flow analysis, is a fundamental analysis in power system engineering. It is used to determine the steady-state operating conditions of a power system, including voltage magnitudes, voltage angles, active power flows, and reactive power flows. The load flow solution is essential for optimal power system planning and operation.
There are several methods available to solve the load flow problem, each with its own advantages and limitations. The options provided in the question are Newton-Raphson method, Gauss-Seidel method, and Fast Decoupled method. Let's discuss each method and understand why the correct answer is option 'A' - Newton-Raphson method.
1. Newton-Raphson Method:
The Newton-Raphson method is an iterative numerical method used to solve nonlinear equations. In the context of load flow analysis, it is used to solve the power flow equations. The method utilizes the Jacobian matrix, which represents the partial derivatives of the power flow equations with respect to the voltage magnitudes and angles. The Newton-Raphson method has several advantages, including fast convergence and good numerical stability. It is widely used in power system analysis due to its efficiency and accuracy.
2. Gauss-Seidel Method:
The Gauss-Seidel method is another iterative method used to solve the power flow equations. Unlike the Newton-Raphson method, the Gauss-Seidel method updates the voltage magnitudes and angles sequentially, one at a time. The updated values are used immediately in subsequent calculations, leading to a slower convergence compared to the Newton-Raphson method. However, the Gauss-Seidel method is computationally simpler and requires less memory, making it suitable for small to medium-sized power systems.
3. Fast Decoupled Method:
The Fast Decoupled method is an enhancement of the Gauss-Seidel method that reduces the computational burden by approximating the Jacobian matrix. It decouples the power flow equations into separate equations for voltage magnitudes and angles, simplifying the iterative process. The Fast Decoupled method provides faster convergence compared to the Gauss-Seidel method while maintaining computational efficiency. It is commonly used for large-scale power systems where the Newton-Raphson method may be computationally expensive.
Conclusion:
While all three methods - Newton-Raphson, Gauss-Seidel, and Fast Decoupled - can be used to solve the load flow problem, the Newton-Raphson method is the most reliable and robust. It guarantees convergence to a solution under normal operating conditions and is capable of handling a wide range of power system configurations. The Gauss-Seidel method and Fast Decoupled method may not guarantee convergence or accuracy in all scenarios. Therefore, the correct answer is option 'A' - Newton-Raphson method.
The load flow solution, also known as power flow analysis, is a fundamental analysis in power system engineering. It is used to determine the steady-state operating conditions of a power system, including voltage magnitudes, voltage angles, active power flows, and reactive power flows. The load flow solution is essential for optimal power system planning and operation.
There are several methods available to solve the load flow problem, each with its own advantages and limitations. The options provided in the question are Newton-Raphson method, Gauss-Seidel method, and Fast Decoupled method. Let's discuss each method and understand why the correct answer is option 'A' - Newton-Raphson method.
1. Newton-Raphson Method:
The Newton-Raphson method is an iterative numerical method used to solve nonlinear equations. In the context of load flow analysis, it is used to solve the power flow equations. The method utilizes the Jacobian matrix, which represents the partial derivatives of the power flow equations with respect to the voltage magnitudes and angles. The Newton-Raphson method has several advantages, including fast convergence and good numerical stability. It is widely used in power system analysis due to its efficiency and accuracy.
2. Gauss-Seidel Method:
The Gauss-Seidel method is another iterative method used to solve the power flow equations. Unlike the Newton-Raphson method, the Gauss-Seidel method updates the voltage magnitudes and angles sequentially, one at a time. The updated values are used immediately in subsequent calculations, leading to a slower convergence compared to the Newton-Raphson method. However, the Gauss-Seidel method is computationally simpler and requires less memory, making it suitable for small to medium-sized power systems.
3. Fast Decoupled Method:
The Fast Decoupled method is an enhancement of the Gauss-Seidel method that reduces the computational burden by approximating the Jacobian matrix. It decouples the power flow equations into separate equations for voltage magnitudes and angles, simplifying the iterative process. The Fast Decoupled method provides faster convergence compared to the Gauss-Seidel method while maintaining computational efficiency. It is commonly used for large-scale power systems where the Newton-Raphson method may be computationally expensive.
Conclusion:
While all three methods - Newton-Raphson, Gauss-Seidel, and Fast Decoupled - can be used to solve the load flow problem, the Newton-Raphson method is the most reliable and robust. It guarantees convergence to a solution under normal operating conditions and is capable of handling a wide range of power system configurations. The Gauss-Seidel method and Fast Decoupled method may not guarantee convergence or accuracy in all scenarios. Therefore, the correct answer is option 'A' - Newton-Raphson method.
A 3 – bus power system network consists of 3 transmission lines. The bus admittance matrix of the uncompensated system is
If the shunt capacitance of all transmission lines is 50% compensated, the imaginary part of the 3rd row 3rd column element (in pu) of the bus admittance matrix after compensation is- a)-j7.0
- b)-j8.5
- c)-j7.5
- d)-j9.0
Correct answer is option 'B'. Can you explain this answer?
A 3 – bus power system network consists of 3 transmission lines. The bus admittance matrix of the uncompensated system is
If the shunt capacitance of all transmission lines is 50% compensated, the imaginary part of the 3rd row 3rd column element (in pu) of the bus admittance matrix after compensation is
If the shunt capacitance of all transmission lines is 50% compensated, the imaginary part of the 3rd row 3rd column element (in pu) of the bus admittance matrix after compensation is
a)
-j7.0
b)
-j8.5
c)
-j7.5
d)
-j9.0
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|
Cstoppers Instructors answered |
Concept:
For a 3 bus power system
For the above figure, the admittance matrix is as shown below.
Diagonal elements of the Bus Admittance matrix are known as self-admittances and the off-diagonal elements are known as mutual admittances.
Calculation:
Given bus admittance matrix of the uncompensated line is
By comparing the above matrix with standard 3 bus matrix
y13 = -j4
y32 = -j5
⇒ y31 + y32 + y33 = -j8
⇒ y33 = j
After compensating,
y33 = j/2
Y33(new) = -8.5 j.
For a 3 bus power system
For the above figure, the admittance matrix is as shown below.
Diagonal elements of the Bus Admittance matrix are known as self-admittances and the off-diagonal elements are known as mutual admittances.
Calculation:
Given bus admittance matrix of the uncompensated line is
By comparing the above matrix with standard 3 bus matrix
y13 = -j4
y32 = -j5
⇒ y31 + y32 + y33 = -j8
⇒ y33 = j
After compensating,
y33 = j/2
Y33(new) = -8.5 j.
Which of the following quantities are known on the generator bus?- a)V and phase angle
- b)Q and V
- c)P and Q
- d)V and P
Correct answer is option 'D'. Can you explain this answer?
Which of the following quantities are known on the generator bus?
a)
V and phase angle
b)
Q and V
c)
P and Q
d)
V and P
|
Maulik Chatterjee answered |
Known Quantities on the Generator Bus:
1. Voltage (V): The voltage magnitude at the generator bus is a known quantity since it is controlled by the generator's voltage regulator to maintain a specific level for proper operation of the generator.
2. Active Power (P): The active power output of the generator is also a known quantity on the generator bus as it is determined by the mechanical power input to the generator and the efficiency of the generator.
Therefore, the correct answer is option 'D' - V and P are known quantities on the generator bus. The voltage magnitude and active power output are crucial parameters that need to be monitored and controlled to ensure the stable and efficient operation of the power system.
1. Voltage (V): The voltage magnitude at the generator bus is a known quantity since it is controlled by the generator's voltage regulator to maintain a specific level for proper operation of the generator.
2. Active Power (P): The active power output of the generator is also a known quantity on the generator bus as it is determined by the mechanical power input to the generator and the efficiency of the generator.
Therefore, the correct answer is option 'D' - V and P are known quantities on the generator bus. The voltage magnitude and active power output are crucial parameters that need to be monitored and controlled to ensure the stable and efficient operation of the power system.
In an n-bus power system, considering n-nodes network, the size of ybus is- a)(n - 1) × (n - 1)
- b)(n + 1) × (n + 1)
- c)n × n
- d)2n × 2n
Correct answer is option 'C'. Can you explain this answer?
In an n-bus power system, considering n-nodes network, the size of ybus is
a)
(n - 1) × (n - 1)
b)
(n + 1) × (n + 1)
c)
n × n
d)
2n × 2n
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Rajesh Verma answered |
The size of ybus in an n-bus power system is n x n.
Assumption of decoupling is valid only if- a)XLine < RLine
- b)XLine = RLine
- c)Any conditon
- d)RLine >> XLine
Correct answer is option 'D'. Can you explain this answer?
Assumption of decoupling is valid only if
a)
XLine < RLine
b)
XLine = RLine
c)
Any conditon
d)
RLine >> XLine
|
|
Samarth Khanna answered |
Assumption of decoupling is valid in load flow study only if resistance of the line is negligible in comparison to the reactance of the line.
If the reference bus is changed in two load flow runs with same system data and power obtained for reference bus taken as specified P and Q in the latter run- a)The system losses will be unchanged but complex bus voltages will change.
- b)The system losses will change but complex bus voltages remain unchanged.
- c)The system losses as well as complex bus voltage will change.
- d)The system losses as well as complex bus voltage will be unchanged.
Correct answer is option 'A'. Can you explain this answer?
If the reference bus is changed in two load flow runs with same system data and power obtained for reference bus taken as specified P and Q in the latter run
a)
The system losses will be unchanged but complex bus voltages will change.
b)
The system losses will change but complex bus voltages remain unchanged.
c)
The system losses as well as complex bus voltage will change.
d)
The system losses as well as complex bus voltage will be unchanged.
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Pooja Patel answered |
If the reference bus is changed in two load flow runs with same system data and power obtained for reference bus taken as specified P and Q.
So, the bus is taken as load bus.
So, the loses will be same but the complex bus voltage will change.
So, the bus is taken as load bus.
So, the loses will be same but the complex bus voltage will change.
Providing synchronous condensers at a load bus will control- a)P at bus
- b)Q
- c)p.f.
- d)(b) and (c)
Correct answer is option 'D'. Can you explain this answer?
Providing synchronous condensers at a load bus will control
a)
P at bus
b)
Q
c)
p.f.
d)
(b) and (c)
|
Subham Chaudhary answered |
A synchronous condeser regulates Q and thus controls power factor of the system.
A power system has 100 buses including 10 generator buses. For the load flow analysis using Newton-Raphson method in polar coordinates, the size of the Jacobian is- a)189 × 189
- b)100 × 100
- c)90 × 90
- d)180 × 180
Correct answer is option 'A'. Can you explain this answer?
A power system has 100 buses including 10 generator buses. For the load flow analysis using Newton-Raphson method in polar coordinates, the size of the Jacobian is
a)
189 × 189
b)
100 × 100
c)
90 × 90
d)
180 × 180
|
Zoya Sharma answered |
Number of buses (n) = 100
Generator busses (m) = 10
Order of Jacobian matrix =(2n − 1 − m) × (2n − 1 − m)
= (200 − 1 − 10) × (200 − 1 − 10)
= 189 × 189
Generator busses (m) = 10
Order of Jacobian matrix =(2n − 1 − m) × (2n − 1 − m)
= (200 − 1 − 10) × (200 − 1 − 10)
= 189 × 189
A power system has 200 buses of which 150 buses are load buses and others are generator buses. The size of the jacobian matrix is:
Correct answer is '349'. Can you explain this answer?
A power system has 200 buses of which 150 buses are load buses and others are generator buses. The size of the jacobian matrix is:
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|
Mahesh Singh answered |
Calculating the size of Jacobian Matrix for a Power System
The Jacobian matrix is used in power system analysis to determine the power flow equations. The size of the Jacobian matrix is determined by the number of buses in the power system. Here's how to calculate the size of the Jacobian matrix for a power system with 200 buses, of which 150 buses are load buses and the others are generator buses.
Count the total number of buses in the power system
- In this case, the total number of buses is 200.
Count the number of generator buses in the power system
- As we know that the total number of buses is 200, and 150 buses are load buses. So the number of generator buses will be 200 - 150 = 50.
Count the number of unknown variables in the power system
- The number of unknown variables is equal to the number of generator buses plus the number of voltage-controlled buses.
- In this case, since there is no mention of voltage-controlled buses, we assume that all buses are PQ buses. Therefore, the number of unknown variables is equal to the number of generator buses, which is 50.
Calculate the size of the Jacobian matrix
- The size of the Jacobian matrix is equal to twice the number of buses, minus the number of unknown variables.
- In this case, the size of the Jacobian matrix will be 2 x 200 - 50 = 350.
- However, we should note that the Jacobian matrix is always square, so we need to take the smaller of the two dimensions. Therefore, the size of the Jacobian matrix is 349.
Conclusion
In conclusion, the size of the Jacobian matrix for a power system with 200 buses, of which 150 buses are load buses and the others are generator buses, is 349.
The Jacobian matrix is used in power system analysis to determine the power flow equations. The size of the Jacobian matrix is determined by the number of buses in the power system. Here's how to calculate the size of the Jacobian matrix for a power system with 200 buses, of which 150 buses are load buses and the others are generator buses.
Count the total number of buses in the power system
- In this case, the total number of buses is 200.
Count the number of generator buses in the power system
- As we know that the total number of buses is 200, and 150 buses are load buses. So the number of generator buses will be 200 - 150 = 50.
Count the number of unknown variables in the power system
- The number of unknown variables is equal to the number of generator buses plus the number of voltage-controlled buses.
- In this case, since there is no mention of voltage-controlled buses, we assume that all buses are PQ buses. Therefore, the number of unknown variables is equal to the number of generator buses, which is 50.
Calculate the size of the Jacobian matrix
- The size of the Jacobian matrix is equal to twice the number of buses, minus the number of unknown variables.
- In this case, the size of the Jacobian matrix will be 2 x 200 - 50 = 350.
- However, we should note that the Jacobian matrix is always square, so we need to take the smaller of the two dimensions. Therefore, the size of the Jacobian matrix is 349.
Conclusion
In conclusion, the size of the Jacobian matrix for a power system with 200 buses, of which 150 buses are load buses and the others are generator buses, is 349.
Out of the following options, the most relevant information needed to specify the real power (P) at the PV buses in a load flow analysis is- a)solution of economic load dispatch
- b)rated power output of the generator
- c)rated voltage of the generator
- d)base power of the generator
Correct answer is option 'A'. Can you explain this answer?
Out of the following options, the most relevant information needed to specify the real power (P) at the PV buses in a load flow analysis is
a)
solution of economic load dispatch
b)
rated power output of the generator
c)
rated voltage of the generator
d)
base power of the generator
|
Poulomi Ahuja answered |
Importance of Economic Load Dispatch in Determining Real Power at PV Buses
Solution of Economic Load Dispatch
The solution of economic load dispatch is crucial in determining the real power (P) at the PV buses in a load flow analysis. Economic load dispatch involves optimizing the power generation from different sources to meet the load demand at minimum cost while satisfying various constraints. By solving the economic load dispatch problem, the real power output at each generator, including the PV buses, can be determined accurately.
Rated Power Output of the Generator
While the rated power output of the generator is important information, it alone is not sufficient to specify the real power at the PV buses in a load flow analysis. The economic load dispatch solution takes into account various factors such as generator costs, transmission constraints, and system losses, which influence the real power output at the PV buses.
Rated Voltage of the Generator
The rated voltage of the generator is relevant for determining the voltage profile in the system but does not directly specify the real power at the PV buses. The real power output at the PV buses is dependent on the economic dispatch solution, which considers the power output of generators in the system.
Base Power of the Generator
The base power of the generator is used as a reference for system calculations but does not provide specific information about the real power at the PV buses. The economic load dispatch solution is the key factor in determining the real power output at the PV buses in a load flow analysis.
Solution of Economic Load Dispatch
The solution of economic load dispatch is crucial in determining the real power (P) at the PV buses in a load flow analysis. Economic load dispatch involves optimizing the power generation from different sources to meet the load demand at minimum cost while satisfying various constraints. By solving the economic load dispatch problem, the real power output at each generator, including the PV buses, can be determined accurately.
Rated Power Output of the Generator
While the rated power output of the generator is important information, it alone is not sufficient to specify the real power at the PV buses in a load flow analysis. The economic load dispatch solution takes into account various factors such as generator costs, transmission constraints, and system losses, which influence the real power output at the PV buses.
Rated Voltage of the Generator
The rated voltage of the generator is relevant for determining the voltage profile in the system but does not directly specify the real power at the PV buses. The real power output at the PV buses is dependent on the economic dispatch solution, which considers the power output of generators in the system.
Base Power of the Generator
The base power of the generator is used as a reference for system calculations but does not provide specific information about the real power at the PV buses. The economic load dispatch solution is the key factor in determining the real power output at the PV buses in a load flow analysis.
An infinite bus-bar has- a)constant voltage
- b)constant frequency
- c)infinite voltage
- d)both (a) and (b)
Correct answer is option 'D'. Can you explain this answer?
An infinite bus-bar has
a)
constant voltage
b)
constant frequency
c)
infinite voltage
d)
both (a) and (b)
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|
Pioneer Academy answered |
Concept:
The bus whose voltage and frequency remain constant even after the variation in the load is known as the infinite bus.
The alternators operating in parallel in a power system are the example of the infinite bus. The on and off of any of the alternators will not affect the working of the power system.
The bus whose voltage and frequency remain constant even after the variation in the load is known as the infinite bus.
The alternators operating in parallel in a power system are the example of the infinite bus. The on and off of any of the alternators will not affect the working of the power system.
- The capacity of a parallel operating system is enormous, their voltage and frequency remain constant even after the disturbance of the load.
- The connection and disconnection of any of the machines will not affect the magnitude and phase of voltage and frequency of an infinite bus.
- In an infinite bus system, the voltage and frequency always remain constant.
- The synchronous impedance of the bus is low because of the parallel operations of the machine.
In a load flow solution Vi = 1.083∠150 pu and Vk = 0.986∠-20 pu. What is the direction of active power (P) and reactive power (Q) flow in the line i - k?- a)P and Q flow from i to k
- b)P flows from i to k and Q flows from k to i
- c)Both flow from k t o i
- d)Data is insufficient to determine direction
Correct answer is option 'A'. Can you explain this answer?
In a load flow solution Vi = 1.083∠150 pu and Vk = 0.986∠-20 pu. What is the direction of active power (P) and reactive power (Q) flow in the line i - k?
a)
P and Q flow from i to k
b)
P flows from i to k and Q flows from k to i
c)
Both flow from k t o i
d)
Data is insufficient to determine direction
|
Mihir Khanna answered |
Since active power flows from leading bus to lagging bus therefore, P will flow from i to k.
Since, reactive power flows from high potential bus to low potential bus therefore, Q will flow from i to k.
Since, reactive power flows from high potential bus to low potential bus therefore, Q will flow from i to k.
Which of the following has a problem in convergence for a system with long radial lines?- a)Newton-Raphson
- b)Gauss-Seidel
- c)Both Newton-Raphson and Gauss Siedel
- d)None of the two methods
Correct answer is option 'B'. Can you explain this answer?
Which of the following has a problem in convergence for a system with long radial lines?
a)
Newton-Raphson
b)
Gauss-Seidel
c)
Both Newton-Raphson and Gauss Siedel
d)
None of the two methods
|
|
Pooja Patel answered |
Comparison of two load flow methods is given below.
In G - S method of power flow problem, the number of iterations- a)depends on no of buses
- b)depends on tolerance
- c)depends on voltage control buses
- d)remains fixed
Correct answer is option 'B'. Can you explain this answer?
In G - S method of power flow problem, the number of iterations
a)
depends on no of buses
b)
depends on tolerance
c)
depends on voltage control buses
d)
remains fixed
|
|
Pooja Patel answered |
- The reason the Gauss-Seidel method is commonly known as the successive displacement method is that the second unknown is determined from the first unknown in the current iteration, the third unknown is determined from the first and second unknowns, etc.
- In the G-S method of power flow problem, the number of iterations depends on tolerance.
Gauss-Seidel method:
Consider a 4-bus sample power system as shown in the figure, BUS 1 is considered a slack bus, where voltage magnitude and its angle are known.
In this case w = 4 and slack bus s = 1. From we can write
In the Gauss-Seidel method, the new calculated voltage at (K + 1) i.e. V (K + 1) immediately replaces V (p) and is used in the solution of the subsequent equations.
Therefore, the above set of equations can be written in iterative form, i.e.,
Note that bus 1 is the slack bus.
Under normal operating conditions, the voltage magnitude of buses is in the neighborhood of 1.0 per unit or close to the voltage magnitude of the slack bus.
Therefore, an initial starting voltage of (1.0 + j 0.0) for unknown voltages is satisfactory, and the converged solution correlates with the actual operating states.
Consider a 4-bus sample power system as shown in the figure, BUS 1 is considered a slack bus, where voltage magnitude and its angle are known.
In this case w = 4 and slack bus s = 1. From we can write
In the Gauss-Seidel method, the new calculated voltage at (K + 1) i.e. V (K + 1) immediately replaces V (p) and is used in the solution of the subsequent equations.
Therefore, the above set of equations can be written in iterative form, i.e.,
Note that bus 1 is the slack bus.
Under normal operating conditions, the voltage magnitude of buses is in the neighborhood of 1.0 per unit or close to the voltage magnitude of the slack bus.
Therefore, an initial starting voltage of (1.0 + j 0.0) for unknown voltages is satisfactory, and the converged solution correlates with the actual operating states.
For formation of the Y bus matrix (using node voltage analysis) in power system network modelling ______ is used - a)KVL
- b)KCL
- c)Faraday's law
- d)all are correct
Correct answer is option 'B'. Can you explain this answer?
For formation of the Y bus matrix (using node voltage analysis) in power system network modelling ______ is used
a)
KVL
b)
KCL
c)
Faraday's law
d)
all are correct
|
Raj Singh answered |
Introduction:
In power system network modeling, the Y bus matrix is an essential component that represents the admittance between different nodes in the network. It is used for various analysis and calculations, such as load flow studies, fault analysis, and stability analysis. The Y bus matrix is formed using the node voltage analysis method, which is based on Kirchhoff's current law (KCL).
Kirchhoff's Current Law (KCL):
Kirchhoff's Current Law states that the algebraic sum of currents entering and leaving a node in an electrical circuit is zero. It is based on the principle of conservation of charge.
Formation of Y bus matrix:
The Y bus matrix is formed by considering the admittance between different nodes in the power system network. The admittance is the reciprocal of impedance and represents the ease with which current can flow through a particular branch or element.
To form the Y bus matrix using node voltage analysis, the following steps are typically followed:
1. Node numbering: Assign a unique number to each node in the power system network.
2. Formulate the equations: Write the KCL equations for each node in terms of the unknown node voltages. These equations represent the current entering or leaving each node.
3. Admittance representation: Represent the admittance between different nodes in terms of conductance (G) and susceptance (B). The admittance can be determined based on the type of element connecting the nodes. For example, for a resistor, the admittance is G, while for a reactance, the admittance is jB.
4. Construct the Y bus matrix: Based on the KCL equations and admittance representation, construct the Y bus matrix by filling in the appropriate values. The Y bus matrix is a square matrix with dimensions equal to the total number of nodes in the power system network.
5. Include shunt elements: If there are shunt elements such as capacitors or inductors connected to the nodes, include their admittance values in the Y bus matrix as well.
6. Account for grounded nodes: If any nodes in the network are grounded, the corresponding row and column in the Y bus matrix will be zero except for the diagonal element, which represents the total admittance connected to the ground.
7. Finalize the Y bus matrix: Once all the elements and admittances are included, the Y bus matrix is finalized and can be used for various power system analysis.
Conclusion:
In power system network modeling, the Y bus matrix is formed using the node voltage analysis method, which is based on Kirchhoff's current law (KCL). KCL ensures that the algebraic sum of currents entering and leaving a node is zero. By formulating the KCL equations for each node and considering the admittance between different nodes, the Y bus matrix can be constructed. The Y bus matrix is an essential tool for power system analysis and calculations.
In power system network modeling, the Y bus matrix is an essential component that represents the admittance between different nodes in the network. It is used for various analysis and calculations, such as load flow studies, fault analysis, and stability analysis. The Y bus matrix is formed using the node voltage analysis method, which is based on Kirchhoff's current law (KCL).
Kirchhoff's Current Law (KCL):
Kirchhoff's Current Law states that the algebraic sum of currents entering and leaving a node in an electrical circuit is zero. It is based on the principle of conservation of charge.
Formation of Y bus matrix:
The Y bus matrix is formed by considering the admittance between different nodes in the power system network. The admittance is the reciprocal of impedance and represents the ease with which current can flow through a particular branch or element.
To form the Y bus matrix using node voltage analysis, the following steps are typically followed:
1. Node numbering: Assign a unique number to each node in the power system network.
2. Formulate the equations: Write the KCL equations for each node in terms of the unknown node voltages. These equations represent the current entering or leaving each node.
3. Admittance representation: Represent the admittance between different nodes in terms of conductance (G) and susceptance (B). The admittance can be determined based on the type of element connecting the nodes. For example, for a resistor, the admittance is G, while for a reactance, the admittance is jB.
4. Construct the Y bus matrix: Based on the KCL equations and admittance representation, construct the Y bus matrix by filling in the appropriate values. The Y bus matrix is a square matrix with dimensions equal to the total number of nodes in the power system network.
5. Include shunt elements: If there are shunt elements such as capacitors or inductors connected to the nodes, include their admittance values in the Y bus matrix as well.
6. Account for grounded nodes: If any nodes in the network are grounded, the corresponding row and column in the Y bus matrix will be zero except for the diagonal element, which represents the total admittance connected to the ground.
7. Finalize the Y bus matrix: Once all the elements and admittances are included, the Y bus matrix is finalized and can be used for various power system analysis.
Conclusion:
In power system network modeling, the Y bus matrix is formed using the node voltage analysis method, which is based on Kirchhoff's current law (KCL). KCL ensures that the algebraic sum of currents entering and leaving a node is zero. By formulating the KCL equations for each node and considering the admittance between different nodes, the Y bus matrix can be constructed. The Y bus matrix is an essential tool for power system analysis and calculations.
The slack bus has to be a- a)PV bus
- b)PQ bus
- c)QV bus
- d)No constraint
Correct answer is option 'A'. Can you explain this answer?
The slack bus has to be a
a)
PV bus
b)
PQ bus
c)
QV bus
d)
No constraint
|
|
Arpita Banerjee answered |
Out of the 15% of Generator buses in a power system, one bus is taken as a slack bus which take care of losses occuring in the system. At a generator bus P and V are known. Hence, a slack bus can be a PV bus.
Which of the following buses is also known as P-Q bus ?- a)Swing bus
- b)Slack bus
- c)Generation bus or voltage control bus
- d)Load bus
Correct answer is option 'D'. Can you explain this answer?
Which of the following buses is also known as P-Q bus ?
a)
Swing bus
b)
Slack bus
c)
Generation bus or voltage control bus
d)
Load bus
|
|
Kalyan Patel answered |
Identification of P-Q bus
The bus known as P-Q bus in power system analysis is the load bus. The load bus is also referred to as the P-Q bus because it is responsible for supplying the active power (P) and reactive power (Q) demands of the system.
Explanation
In power system analysis, the load bus represents the buses where the active and reactive power demands are specified. These buses are typically connected to various loads such as residential, commercial, and industrial consumers. The load bus is characterized by having known values of active and reactive power injections.
The load bus is an essential component of power flow analysis and is used to determine the steady-state operating conditions of a power system. By specifying the active and reactive power demands at the load buses, the power flow equations can be solved to obtain the voltage magnitudes and angles at all buses in the system.
Key Points:
- The load bus is also known as the P-Q bus.
- It represents the buses where the active and reactive power demands are specified.
- It is responsible for supplying the active power (P) and reactive power (Q) demands of the system.
Conclusion
In power system analysis, the load bus is referred to as the P-Q bus because it is responsible for supplying the active power (P) and reactive power (Q) demands of the system. The load bus represents the buses where the active and reactive power demands are specified, and it is an essential component in power flow analysis.
The bus known as P-Q bus in power system analysis is the load bus. The load bus is also referred to as the P-Q bus because it is responsible for supplying the active power (P) and reactive power (Q) demands of the system.
Explanation
In power system analysis, the load bus represents the buses where the active and reactive power demands are specified. These buses are typically connected to various loads such as residential, commercial, and industrial consumers. The load bus is characterized by having known values of active and reactive power injections.
The load bus is an essential component of power flow analysis and is used to determine the steady-state operating conditions of a power system. By specifying the active and reactive power demands at the load buses, the power flow equations can be solved to obtain the voltage magnitudes and angles at all buses in the system.
Key Points:
- The load bus is also known as the P-Q bus.
- It represents the buses where the active and reactive power demands are specified.
- It is responsible for supplying the active power (P) and reactive power (Q) demands of the system.
Conclusion
In power system analysis, the load bus is referred to as the P-Q bus because it is responsible for supplying the active power (P) and reactive power (Q) demands of the system. The load bus represents the buses where the active and reactive power demands are specified, and it is an essential component in power flow analysis.
In Gauss Seidel method of power flow problem the number of iterations may be reduced if the correction in voltage at each bus is multiplied by- a)Gauss constant
- b)Acceleration factor
- c)Declaration constant
- d)Blocking factor
Correct answer is option 'B'. Can you explain this answer?
In Gauss Seidel method of power flow problem the number of iterations may be reduced if the correction in voltage at each bus is multiplied by
a)
Gauss constant
b)
Acceleration factor
c)
Declaration constant
d)
Blocking factor
|
|
Pooja Patel answered |
Gauss Seidel method:
- Gauss-Seidel method of power flow problem is an iterative method used to solve a system of linear equations.
- This method is very simple and uses digital computers for computing.
- In this method as we are using simple algebraical equations so that the calculation time for each iteration is less.
Disadvantages:
- Though it can be applied to any matrix with non zero diagonal elements, the convergence is guaranteed if the matrix is either strictly diagonally dominant or symmetric and positive definite.
- More number of iterations are required so that it has slow convergence.
- Initial approximate guessing value is required for convergence.
- The choice of slack bus affects convergence.
- It is not applicable to the large power system networks.
- It requires an accelerating factor for convergence. The accelerating factor is used for reducing the number of iterations in the Gauss-Seidel method by multplying voltage at each bus with the acceleration factor.
- The value of the accelerating factor is around 1.6 to 1.8.
A 1000 × 1000 bus admittance matrix for an electric power system has 8000 non-zero elements. The minimum number of branches (transmission lines and transformers) in this system are _____ (up to 2 decimal places).
Correct answer is '3500'. Can you explain this answer?
A 1000 × 1000 bus admittance matrix for an electric power system has 8000 non-zero elements. The minimum number of branches (transmission lines and transformers) in this system are _____ (up to 2 decimal places).
|
|
Pooja Patel answered |
Size of YBUS matrix = 1000 × 1000
Number of non-zero elements = 8000.
Number of diagonal elements = number of buses = 1000.
Number of non-zero off diagonal elements = 8000 – 1000 = 7000
Minimum number of branches = 7000/2 = 3500
Number of non-zero elements = 8000.
Number of diagonal elements = number of buses = 1000.
Number of non-zero off diagonal elements = 8000 – 1000 = 7000
Minimum number of branches = 7000/2 = 3500
The Gauss Seidel load flow method has following disadvantages. Mark the incorrect statement- a)Unreliable convergence
- b)Slow convergence
- c)Choice of slack bus affects convergence
- d)A good initial guess for voltages is essential for convergence
- e)All of the above
Correct answer is option 'A'. Can you explain this answer?
The Gauss Seidel load flow method has following disadvantages. Mark the incorrect statement
a)
Unreliable convergence
b)
Slow convergence
c)
Choice of slack bus affects convergence
d)
A good initial guess for voltages is essential for convergence
e)
All of the above
|
|
Pooja Patel answered |
Gauss Seidel method:
- Gauss-Seidel method of power flow problem is an iterative method used to solve a system of linear equations.
- This method is very simple and uses digital computers for computing.
- In this method as we are using simple algebraical equations so that the calculation time for each iteration is less.
Disadvantages:
- Though it can be applied to any matrix with non zero diagonal elements, the convergence is guaranteed if the matrix is either strictly diagonally dominant or symmetric and positive definite.
- More number of iterations are required so that it has slow convergence.
- Initial approximate guessing value is required for convergence.
- The choice of slack bus affects convergence.
- It is not applicable to the large power system networks.
- It requires an accelerating factor for convergence. The accelerating factor is used for reducing the number of iterations in the Gauss-Seidel method by multplying voltage at each bus with the acceleration factor.
- The value of the accelerating factor is around 1.6 to 1.8.
The figure shows the per-phase representation of a phase-shifting transformer connected between buses 1 and 2, where α is a complex number with non-zero real and imaginary parts.
For the given circuit Ybus and Zbus are bus admittance matrix and bus impedance matrix respectively, each of size 2 × 2. Which one of the following statements is true?- a)Both Ybus and Zbus are symmetric
- b)Ybus is symmetric and Zbus is unsymmetric
- c)Ybus is unsymmetric and Zbus is symmetric
- d)Both Ybus and Zbus are unsymmetric
Correct answer is option 'D'. Can you explain this answer?
The figure shows the per-phase representation of a phase-shifting transformer connected between buses 1 and 2, where α is a complex number with non-zero real and imaginary parts.
For the given circuit Ybus and Zbus are bus admittance matrix and bus impedance matrix respectively, each of size 2 × 2. Which one of the following statements is true?
For the given circuit Ybus and Zbus are bus admittance matrix and bus impedance matrix respectively, each of size 2 × 2. Which one of the following statements is true?
a)
Both Ybus and Zbus are symmetric
b)
Ybus is symmetric and Zbus is unsymmetric
c)
Ybus is unsymmetric and Zbus is symmetric
d)
Both Ybus and Zbus are unsymmetric
|
|
Pooja Patel answered |
T - parameter matrix,
Since A ≠ D, the network is asymmetric and hence Ybus and Zbus will also be asymmetric.
In a power system with 100 buses, 1 bus is taken as slack bus, 15 buses are P\/ buses and rest are PQ buses. The number of variables to be solved by NR method in polar coordinates is- a)100
- b)198
- c)200
- d)183
Correct answer is option 'D'. Can you explain this answer?
In a power system with 100 buses, 1 bus is taken as slack bus, 15 buses are P\/ buses and rest are PQ buses. The number of variables to be solved by NR method in polar coordinates is
a)
100
b)
198
c)
200
d)
183
|
Ishan Chawla answered |
Hence, number of variables to be solved by NR method in polar coordinates = 183.
In the following network, the voltage magnitudes at all buses are equal to 1 pu, the voltage phase angles are very small, and the line resistances are negligible. All the line reactances are equal to j1Ω
If the base impedance and the line – to line base voltage are 100 ohms and 100 kV respectively, then the real power in MW delivered by the generator connected at the slack bus is- a)-10
- b)0
- c)10
- d)20
Correct answer is option 'C'. Can you explain this answer?
In the following network, the voltage magnitudes at all buses are equal to 1 pu, the voltage phase angles are very small, and the line resistances are negligible. All the line reactances are equal to j1Ω
If the base impedance and the line – to line base voltage are 100 ohms and 100 kV respectively, then the real power in MW delivered by the generator connected at the slack bus is
If the base impedance and the line – to line base voltage are 100 ohms and 100 kV respectively, then the real power in MW delivered by the generator connected at the slack bus is
a)
-10
b)
0
c)
10
d)
20
|
Naroj Boda answered |
Consider the voltage phase angles at buses 2 and 3 be Q2 and Q3 since, all the three buses have the equal voltage magnitude. Which is 1 pu, so, it is a D.C. load flow. The injections at Bus 2 and 3 are respectively P2 = 0.1 Pu
P3 = -0.2 Pu
P1 + P2 + P3 = 0
P1 – P2 – P3 = -0.1 + 0.2 = 0.1 Pu
Now, the apparent power delivered to base is,
The real power delivered by slack bus
P = P1 |S| = (0.1) (100 × 106)
= 10 × 106 watt = 10 MW
P3 = -0.2 Pu
P1 + P2 + P3 = 0
P1 – P2 – P3 = -0.1 + 0.2 = 0.1 Pu
Now, the apparent power delivered to base is,
The real power delivered by slack bus
P = P1 |S| = (0.1) (100 × 106)
= 10 × 106 watt = 10 MW
If S = VI*, P is positive and Q is negative, the network is
- a)absorbing P and Q
- b) absorbing P and delivering Q
- c)delivering P and absorbing Q
- d)delivering Pand Q
Correct answer is option 'B'. Can you explain this answer?
If S = VI*, P is positive and Q is negative, the network is
a)
absorbing P and Q
b)
absorbing P and delivering Q
c)
delivering P and absorbing Q
d)
delivering Pand Q
|
|
Sushant Mehta answered |
Given, S = ( P - j Q )... for given network
Since P is positive, therefore the given network absorbs P.
Since Q is negative, therefore the given network absorbs -Q or delviers +Q.
Since P is positive, therefore the given network absorbs P.
Since Q is negative, therefore the given network absorbs -Q or delviers +Q.
Consider a power system consisting of N number of buses. Buses in this power system are categorized into slack bus, PV buses and PQ buses for load flow study. The number of PQ buses is NL. The balanced Newton-Raphson method is used to carry out load flow study in polar form. H, S, M, and R are sub-matrices of the Jacobian matrix J as shown below:
The dimension of the sub-matrix M is- a)(N - 1) × (N - 1 NL)
- b)NL × (N - 1 + NL)
- c)(N - 1) × (N - 1 + NL)
- d)NL × (N - 1)
Correct answer is option 'D'. Can you explain this answer?
Consider a power system consisting of N number of buses. Buses in this power system are categorized into slack bus, PV buses and PQ buses for load flow study. The number of PQ buses is NL. The balanced Newton-Raphson method is used to carry out load flow study in polar form. H, S, M, and R are sub-matrices of the Jacobian matrix J as shown below:
The dimension of the sub-matrix M is
The dimension of the sub-matrix M is
a)
(N - 1) × (N - 1 NL)
b)
NL × (N - 1 + NL)
c)
(N - 1) × (N - 1 + NL)
d)
NL × (N - 1)
|
|
Pooja Patel answered |
Number of buses in the system = N
Number of PQ buses = NL
Number of slack buses = 1
Number of PV buses = N - 1 - NL
Newton Raphson method for load flow study in polar form
Number of PQ buses = NL
Number of slack buses = 1
Number of PV buses = N - 1 - NL
Newton Raphson method for load flow study in polar form
The submatrix M relates between [ΔQ] and [Δδ]
Number of elements in ΔQ vector = Number of known Q
Number of elements in ΔQ vector = Number of PQ buses = NL
Number of elements in Δδ vector = Number of unknown δ = N - 1
Size of matrix M = NL × (N - 1)
Size of other sub-matrix:
H = (N - 1) × (N - 1)
S = (N - 1) × NL
R = NL × NL
Number of elements in ΔQ vector = Number of known Q
Number of elements in ΔQ vector = Number of PQ buses = NL
Number of elements in Δδ vector = Number of unknown δ = N - 1
Size of matrix M = NL × (N - 1)
Size of other sub-matrix:
H = (N - 1) × (N - 1)
S = (N - 1) × NL
R = NL × NL
A 10-bus power system consists of four generator buses indexed as G1, G2, G3, G4 and six load buses indexed as L1, L2, L3, L4, L5, L6. The generator-bus G1 is considered as slack bus, and the load buses L3 and L4 are voltage-controlled buses. The generator at bus G2 cannot supply the required reactive power demand, and hence it is operating at its maximum reactive power limit. The number of non-linear equations required for solving the load flow problem using Newton-Raphson method in polar form is ___________.
Correct answer is '14'. Can you explain this answer?
A 10-bus power system consists of four generator buses indexed as G1, G2, G3, G4 and six load buses indexed as L1, L2, L3, L4, L5, L6. The generator-bus G1 is considered as slack bus, and the load buses L3 and L4 are voltage-controlled buses. The generator at bus G2 cannot supply the required reactive power demand, and hence it is operating at its maximum reactive power limit. The number of non-linear equations required for solving the load flow problem using Newton-Raphson method in polar form is ___________.
|
|
Pooja Patel answered |
Total number of bases (n) = 10
Number of slack bases = 1 (G1)
G3, G4 are generator bases.
Number of voltage controlled load bases = 2 (L3, L4)
G2 acts as load bas
Number of load bases = 4 (L1, L2, L5, L6)
The number of non-linear equations required = 2 × number of bases - (number of load bases) - (number of voltage controlled load bases)
= 2n - 2 - 4
= 2 (10) - 2 - 4 = 20 - 2 - 4 = 14
Number of slack bases = 1 (G1)
G3, G4 are generator bases.
Number of voltage controlled load bases = 2 (L3, L4)
G2 acts as load bas
Number of load bases = 4 (L1, L2, L5, L6)
The number of non-linear equations required = 2 × number of bases - (number of load bases) - (number of voltage controlled load bases)
= 2n - 2 - 4
= 2 (10) - 2 - 4 = 20 - 2 - 4 = 14
Which method is used as an advanced iterative method for generating appropriate solution steps to a real solution of a given nonlinear equation?- a)Newton Raphson method
- b)Gauss-Seidal method
- c)Fast Decoupled load flow method
- d)Bus admittance matrix
Correct answer is option 'A'. Can you explain this answer?
Which method is used as an advanced iterative method for generating appropriate solution steps to a real solution of a given nonlinear equation?
a)
Newton Raphson method
b)
Gauss-Seidal method
c)
Fast Decoupled load flow method
d)
Bus admittance matrix
|
|
Pooja Patel answered |
Newton Raphson Method:
- The Newton-Raphson method is an advanced iterative method for generating appropriate solution steps to a real solution of a given nonlinear equation.
- The iterative formula used in the NR method is:
Newton Raphson method is used to solve- a)linear differential equations
- b)Linear algebraic equations
- c)Non-linear differential equations
- d)non-linear algebraic equations
Correct answer is option 'D'. Can you explain this answer?
Newton Raphson method is used to solve
a)
linear differential equations
b)
Linear algebraic equations
c)
Non-linear differential equations
d)
non-linear algebraic equations
|
|
Pooja Patel answered |
- Load flow study determines the operating state of the system for a given loading.
- Load flow solves a set of simultaneous non-linear algebraic power equations for the two unknown variables (|V| and ∠δ) at each node in a system.
- The output of the load flow analysis is the voltage and phase angle, real and reactive power (both sides in each line), line losses, and slack bus power.
- Gauss seidel, Newton Raphson, and Fast decoupled load flow method are the different load flow methods.
- The fast decoupled load flow method gives an approximate load flow solution because it uses several assumptions. Accuracy depends on the power mismatch vector tolerance.
- The fast decoupled load flow method is an extension of the Newton-Raphson method formulated in polar coordinates with certain approximations, which results in a fast algorithm for load flow solution.
- The fast decoupled method requires a greater number of iterations than the Newton-Raphson method.
For the set of linear equations given below, what is the value of voltage V1 after first iteration of Gauss Siedal method? Assume the initial approximation is V1 = 0
10V1 + 2V2 = 10, 4V1 + 7V2 = 15- a)1
- b)0.5714
- c)0.5
- d)0.6367
Correct answer is option 'A'. Can you explain this answer?
For the set of linear equations given below, what is the value of voltage V1 after first iteration of Gauss Siedal method? Assume the initial approximation is V1 = 0
10V1 + 2V2 = 10, 4V1 + 7V2 = 15
10V1 + 2V2 = 10, 4V1 + 7V2 = 15
a)
1
b)
0.5714
c)
0.5
d)
0.6367
|
|
Pooja Patel answered |
The given equations can be written as
10V1 + 2V2 = 10
⇒ V1 = 1/10 [10 − 2V2]
4V1 + 7V2 = 15
⇒ V2 = 1/7 [15 − 4V1]
Iteration 1:
Put V2 = 0
⇒ V1 = 1/10 [10 − 0] = 1
10V1 + 2V2 = 10
⇒ V1 = 1/10 [10 − 2V2]
4V1 + 7V2 = 15
⇒ V2 = 1/7 [15 − 4V1]
Iteration 1:
Put V2 = 0
⇒ V1 = 1/10 [10 − 0] = 1
For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3∠-10° per unit. If a capacitor having reactance of –j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit is- a)0.325∠-100°
- b)0.325∠ 80°
- c)0.433∠-100°
- d)0.433∠80°
Correct answer is option 'D'. Can you explain this answer?
For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3∠-10° per unit. If a capacitor having reactance of –j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit is
a)
0.325∠-100°
b)
0.325∠ 80°
c)
0.433∠-100°
d)
0.433∠80°
|
|
Pooja Patel answered |
Power system having n nodes.
Z33 = j 0.5 PU
V3 = 1.3∠ -10° PU
A capacitor having a reactance of -j 3.5 PU is now added to the network
IC = 0.433∠ 80°
Z33 = j 0.5 PU
V3 = 1.3∠ -10° PU
A capacitor having a reactance of -j 3.5 PU is now added to the network
IC = 0.433∠ 80°
At generator bus, defined quantities are -- a)|Pg| and |Vg| are defined
- b)|Pd| and |Qd| are defined
- c)|Pg| and |δ| are defined
- d)|Pg| - |Vg| and |δ| are defined
Correct answer is option 'A'. Can you explain this answer?
At generator bus, defined quantities are -
a)
|Pg| and |Vg| are defined
b)
|Pd| and |Qd| are defined
c)
|Pg| and |δ| are defined
d)
|Pg| - |Vg| and |δ| are defined
|
|
Pooja Patel answered |
Classification of buses:
Load Buses:
Load Buses:
- In these buses, no generators are connected and hence the generated real power PGi and reactive power QGi are taken as zero.
- The load drawn by these buses is defined by real power -PLi and reactive power -QLi in which the negative sign accommodates for the power flowing out of the bus.
- These buses are sometimes referred to as P-Q bus. The objective of the load flow is to find the bus voltage magnitude |Vi| and its angle δi.
Voltage Controlled Buses:
- These are the buses where generators or voltage-controlled equipments are connected. Therefore the power generation in such buses is controlled through a prime mover while the terminal voltage is controlled through the generator excitation.
- Keeping the input power constant through turbine-governor control and keeping the bus voltage constant using an automatic voltage regulator, we can specify constant PGi and |Vi| for these buses.
- These buses are also referred to as P-V buses. The reactive power supplied by the generator QGi depends on the system configuration and cannot be specified in advance.
Slack or Swing Bus:
- This bus sets the angular reference for all the other buses. Since it is the angle difference between two voltage sources that dictates the real and reactive power flow between them, the particular angle of the slack bus is not important.
- It sets the reference against which angles of all the other bus voltages are measured. For this reason, the angle of this bus is usually chosen as 0°. Furthermore, it is assumed that the magnitude of the voltage of this bus is known.
A bus-bar is rated by-- a)Current only
- b)Current and voltage only
- c)Current, voltage and frequency only
- d)Current, voltage, frequency and short time current capacity
Correct answer is option 'D'. Can you explain this answer?
A bus-bar is rated by-
a)
Current only
b)
Current and voltage only
c)
Current, voltage and frequency only
d)
Current, voltage, frequency and short time current capacity
|
|
Pooja Patel answered |
Busbar:
- A busbar is a strip or bar of Copper, Brass, or Aluminum that conducts electricity within a switchboard, distribution board, substation, battery bank, or other electrical apparatus.
- Its main purpose is to conduct electricity, not to function as a structural member.
- Normally bus-bar is rated in terms of current, voltage, frequency and short time current based on the application.
Temperature rise in the Busbars:
- During the short-circuiting, the bus bar should be able to withstand the thermal as well as mechanical stress.
- The heat dissipation through convection and radiation during this short duration is negligible
- All the heat is observed by the busbar itself.
The 3-bus system is shown in the figure, Two lines are connected between bus 1 and bus 2 is having a mutual effect of M = j0.01(additive). Find the Y22 element of the YBus matrix of the system.
- a)-j 9.78 pu
- b)-j 10.58 pu
- c)-j 3.48 pu
- d)-j 8.33 pu
Correct answer is option 'A'. Can you explain this answer?
The 3-bus system is shown in the figure, Two lines are connected between bus 1 and bus 2 is having a mutual effect of M = j0.01(additive). Find the Y22 element of the YBus matrix of the system.
a)
-j 9.78 pu
b)
-j 10.58 pu
c)
-j 3.48 pu
d)
-j 8.33 pu
|
|
Pooja Patel answered |
Concept:
When two inductances are connected in parallel and have a mutual effect then equivalent inductance can be calculated with the help of the following formula
First of all, calculate the equivalent reactance of two-line connected between buses 1 and 2
= -j6.45
and
=-j3.33 pu
∴ Y22 = y12 + y23
= -j6.45 + (-j3.33)
=-j9.78 pu.
When two inductances are connected in parallel and have a mutual effect then equivalent inductance can be calculated with the help of the following formula
First of all, calculate the equivalent reactance of two-line connected between buses 1 and 2
= -j6.45
and
=-j3.33 pu
∴ Y22 = y12 + y23
= -j6.45 + (-j3.33)
=-j9.78 pu.
Chapter doubts & questions for Load Flow Studies - Power Systems 2025 is part of Electrical Engineering (EE) exam preparation. The chapters have been prepared according to the Electrical Engineering (EE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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