Consider the protection system shown in the figure below. The circuit breakers numbered from $$1$$ to $$7$$ are of identical type. A single line to ground fault with zero fault impedance occurs at the midpoint of the line (at point F), but circuit breaker $$4$$ fails to operate (''Stuck breaker''). If the relays are coordinated correctly, a valid sequence of circuit breaker operation is b

Suppose we define a sequence transformation between ''a-b-c'' and ''p-n-0''' variables as follows:
$$\left[ {\matrix{ {{f_a}} \cr {{f_b}} \cr {{f_c}} \cr } } \right] = k\left[ {\matrix{ 1 & 1 & 1 \cr {{\alpha ^2}} & \alpha & 1 \cr \alpha & {{\alpha ^2}} & 1 \cr } } \right]\left[ {\matrix{ {{f_p}} \cr {{f_n}} \cr {{f_o}} \cr } } \right]$$ where $$\,\alpha = {e^{j{{2\pi } \over 3}}}\,\,$$ and $$k$$ is a constant
Now, if it is given that:
$$\left[ {\matrix{ {{V_p}} \cr {{V_n}} \cr {{V_o}} \cr } } \right] = k\left[ {\matrix{ {0.5} & 0 & 0 \cr 0 & {0.5} & 0 \cr 0 & 0 & {2.0} \cr } } \right]\left[ {\matrix{ {{i_p}} \cr {{I_n}} \cr {{i_o}} \cr } } \right]\,\,$$ and $$\left[ {\matrix{ {{V_a}} \cr {{V_b}} \cr {{V_c}} \cr } } \right] = z\left[ {\matrix{ {{i_a}} \cr {{I_b}} \cr {{i_c}} \cr } } \right]\,\,$$ then,

Consider a synchronous generator connected to an infinite bus by two identical parallel transmission lines. The transient reactance $$x'$$ of the generator $$0.1$$ pu. Due to some previous disturbance, the rotor angle $$(d)$$ is undergoing an undamped oscillation, with the maximum value of $$\delta \left( t \right)$$ equal to $$\,{130^ \circ }\,.$$ One of the parallel lines trip due to relay mal-operation at an instant $$\,\,\,\,\,$$ when $$\,\delta \left( t \right)\,\, = {130^ \circ }\,\,$$ as shown in the figure. The maximum value of the per unit line reactance $$x,$$ such that the system does not lose synchronism subsequent to this tripping is

The incremental cost curves in Rs/MWhr for two generators supplying a common load of $$700$$ MW are shown in the figures. The maximum and minimum generation limits are also indicated. The optimum generation schedule is: