01․ Fault calculations using computer are usually done by
Y bus matrix is a sparse matrix, containing more number of zero elements. So that faster calculation is possible. The Y bus matrix is used for the load flow studies. Z bus algorithm or matrix is used for the fault analysis. Inverse of Y bus matrix gives the Z bus matrix, But Z matrix is a full matrix even though Y bus is a sparse matrix. More time is required for inverse of Z bus matrix if the size of the matrix is more than three.
02․ The main advantage of the Decoupled Load Flow (DLF) as compared to the NR method is
The main advantage of the Decoupled Load Flow (DLF) as compared to the NR method is its reduced memory requirements in storing the Jacobian elements. Storing of the Jacobian and matrix triangularisation is saved by a factor 4, that is an overall saving of 30 - 40 % on the formal Newton load flow. Computation time per iteration is less than the Newton method. However, the DLF takes more number of iterations to converge because of approximation made.
03․ The draw back of NR method as compare to Gauss-Seidel method is/are?
Advantages of N-R method: 1. Number of iterations are less, so that it has fast convergence. 2. Convergence is not effected by the choice of slack bus. 3. No need of accelerating factor. 4. Applicable for large power system network. Disadvantages of N-R method: 1. Time taken for each iteration is larger if size of the Jacobian matrix is larger. 2. Computer memory required is larger. 3. Computer programming is difficult.
04․ Which of the following method is/are more reliable?
In FDLF method usually two to five iterations are required for practical accuracies. The method is more reliable than the formal NR method. The speed for iterations is about five times that of the formal NR method or about two thirds of the GS method. Storage requirements are about 60 percent of the formal NR method, but slightly more than the DLF method.
05․ Per unit of any quantity is defined as
Per unit of any quantity is defined as the ratio of an actual value of quantity to the base value of the quantity in same units. Per unit of any quantity = Actual value of quantity /Base value of quantity in same units.
06․ Advantage/s of per unit system as compare to absolute system?
Advantages of per unit system: 1. Only one equation is required 2. Calculation can be done simultaneously 3. Calculation time is less 4. No units for any quantity 5. No need of transformation ratio 6. Memory required is less
07․ A transmission line has (2 + j4)Ω impedance, 100 MVA and base voltage is 10 KV. Find the per unit impedance value of transmission line?
Zpu = ZΩ* (MVA)b/(KV)²b Where, Zpu = Per unit impedance ZΩ = Impedance in Ω (MVA)b = Base MVA (KV)b = Base voltage Zpu = (2 + j4) * 100/10² = (2 + j4) pu
08․ A generator has a rating of 10 MVA, 5 kV has a reactance of 0.02 pu. Find the reactance at a new base values of 50 MVA, 10 kV?
Zpu = ZΩ* (MVA)b/(KV)²b Where, Zpu = Per unit impedance ZΩ = Impedance in Ω (MVA)b = Base MVA (KV)b = Base voltage Zpun = Zpuo * (MVA)bn/(MVA)bo * (KV)²bo/(KV)²bn Here, "n" represents the new values "o" represents the old values Zpun = 0.02 * 50/10 * 5²/10² = 0.025
09․ A 10 MVA generator has reactance of 0.2 pu. Find the new reactance value for 50 MVA base?
Zpun = Zpuo * (MVA)bn/(MVA)bo Where, Zpu = Per unit impedance ZΩ = Impedance in Ω (MVA)b = Base MVA (KV)b = Base voltage Here, "n" represents the new values "o" represents the old values Zpun = 0.2 * 50/10 = 1 pu
10․ A 20 MVA generator has inertia constant of 0.5 pu. Find new value for 50 MVA base?
Inertia constant (H) in pu = H/Sb Sb = Base MVA Hpun = Hpuo* (MVA)bo/(MVA)bn Here, "n" represents the new values "o" represents the old values Hpun = 0.5*20/50 = 0.2 pu