Dear all,
After some quick calculations one can see that it is possible to replace a constant power load with a constant impedance load of exactly equal power consumption. Here is an example, say bus 1 absorbs S1 MVAs under V1 volts. Bus 1 is constant power. This would be equal to a constant impedance load, with impedance conjugate(Z1) = (Vn/V1)^2 * S1, where Vn is the nominal voltage. Thus, we can set the PD and QD entries of the case to zero, and replace them with the right values for GS and BS. Specifically, after solving a case with all constant power loads we will obtain the Vm vector (voltage amplitudes) and the Pd and Qd vectors. Then, we can set all PD and QD values to zero and place (1/Vm).^2 * Pd and (1/Vm).^2 * Qd in the GS and BS columns respectively. >From my understanding the output should be exactly the same. The solved case >should remain a perfectly good solution to the PF equations. Yet, when I do >so, the solver iterates. Can someone shed light on this? I apologise if I am asking the question in the wrong place. Best regards, Dionysios Georgiadis
