MCQs on Power Systems

01․ Incremental cost of two generators I_c1 = 0.2 P₁+60, I_c2 = 0.3 P₂+40 and the ratings of the generator are 150 and 250 MW. Find the load sharing of each generator for a load of 200 MW?

Given that P₁ + P₂ = 200 MW Optimum condition of economic load dispatch is I_c1 = I_c2 0.2 P₁ + 60 = 0.3 P₂ + 40 0.2 P₁ - 0.3 P₂ = -20 By solving of above equations P₁ = 80 MW, P₂ = 120 MW

02․ The Y bus matrix of 100 bus interconnected system is 90% sparse. Hence the number of transmission lines in the system must be

Total number of transmission lines Where, n = number of buses x = sparsity

03․ A power system consists of 300 buses out of which 20 buses are generator buses, 25 buses are the ones with reactive power support buses and 15 buses are ones with fixed shunt capacitors. All the other buses are load buses. It is proposed to perform a load flow analysis for the system using Newton-Raphson Jacobian matrix is

Size of the Jacobian matrix = (2n-m-2)×(2n-m-2) Where, n = Total number of buses = 300 m = Total number of PV buses m = Voltage control buses + Reactive power support buses + generator buses except slack bus = 44 Fixed shunt capacitors are supplying constant amount of reactive power, so that fixed shunt capacitors are considered as load buses or PQ buses. Size of the Jacobian matrix = (2×300-44-2)×(2×300-44-2) = 554×554

04․ The Gauss Seidel load flow method has following disadvantages. Which of the following statement is wrong?

Advantages of gauss siedel method: 1. It is a simple algebraical equation, so that calculation time for each iteration is less. 2. More suitable for small size networks. Disadvantages of gauss siedel method: 1. More number of iterations are required, so that it has slow convergence. 2. Initial approximate guessing value is required for convergence. 3. It required accelerating factor for convergence. 4. The choice of slack bus affects the convergence. 5. It is not applicable for the large power system networks.

05․ The incremental fuel costs for two generating units G₁ and G₂ are given by I_c1 = 25 + 0.2 P_G1 and I_c2 = 32 + 0.2 P_G2, where P_G1 and P_G2 are real powers generated by the units. The economic allocation for a total load of 250 MW, neglecting transmission loss, is given by

Given that P₁ + P₂ = 250 MW Optimum condition of economic load dispatch is I_c1 = I_c2 25 + 0.2 P_G1 = 32 + 0.2 P_G2 0.2 P_G1 - 0.2 P_G2 = 7 By solving of above equations P_G1 = 142.5 MW and P_G2 = 107.5 MW

06․ A lossless power system has to serve a load of 250 MW. There are two generators(G₁ and G₂) in the system with cost curves C₁ and C₂ respectively defined as follows: C₁ = P_G1+0.055P²_G1 C₂ = 3 P_G2 + 0.03 P²_G2 Where P_G1 and P_G2 are the MW injections from generators G₁ and G₂ respectively. Thus, the minimum cost dispatch will be

Given that P₁ + P₂ = 250 MW........(1) Optimum condition of economic load dispatch is After solving of above equations P_G1 = 100 MW and P_G2 = 150 MW

07․ A load centre is at an equidistant from the two thermal generating stations G1 and G2. The fuel cost characteristics of the generating stations are given by F₁ = a + bP₁ + cP²₁ F₂ = a + b P₂ + 2cP²₂ Where P₁ and P₂ are the generation in MW of G₁ and G₂, respectively. For most economic generation to meet 300 MW of load, P₁ and P₂ respectively, are

Given that P₁ + P₂ = 200 MW Optimum condition of economic load dispatch is I_c1 = I_c2 b + 2cP₁ = b + 4cP₂ 2cP₁ = 4cP₂ P₁ = 2P₂ Therefore, P₁ = 200 MW and P₂ = 100 MW

08․ The power generated by two plants are; P₁ = 50 MW, P₂ = 40 MW. If the loss coefficients are B₁₁ = 0.001, B₂₂ = 0.0025 and B₁₂ = -0.0005, then power loss will be

Power loss of a transmission line P_L = B₁₁×P²₁ + B₂₂×P²₂ + 2×B₁₂×P₁×P₂ Where, B₁₁,B₂₂,B₁₂ and B₂₁ are B coefficients P1 and P2 are power delivered by generating stations. Given that, P₁ = 50 MW, P₂ = 40 MW B₁₁ = 0.001, B₂₂ = 0.0025 and B₁₂ = -0.0005 By substituting the above values, Power loss = 4.5 MW

09․ If, for a given alternator in economic operation mode, the incremental cost is given by (0.012 P + 8) Rs/MWh, dP_L/dP = 0.2 and plant λ = 25, then power generation is

Given that,

10․ The economics of power plant is greatly influenced by

Plant load factor is defined as the ratio of average power generation to the maximum power generation. Plant load factor = Average power/ Maximum power Therefore, ideal plant load factor is one and practical load factor is less than 1. Diversity factor is defined as the ratio of sum of individual maximum demand to the station maximum demand. Diversity factor = Sum of individual maximum demand/ Station maximum demand Both load factor and diversity factor depends on maximum demand. Therefore, economics of power plant is greatly influenced by both load factor and diversity factor.