Problem 106-1    Source:   Example 6-1 - page 283 - SEN, P. C. - Book: Principles of Electric Machines and Power Eletronics - ed. Wilei - 3ª edition - 2013.

In a factory, a three-phase synchronous motor of 400 kVA, 4 kV is installed, together with induction motors whose power is 500 kVA with a PF = 0.8 delayed. Furthermore, a synchronous motor is installed 300 kVA with PF = 1. Find:

a) The total power factor of the system.

b) To improve the power factor, the synchronous motor works overexcited to provide leading current, there is no increase in load. Without overloading the engine, how much could we improve the power factor?

c) The current of the synchronous motor operating in the above condition.

Solution of the Problem 106-1   

Item a

As we know the installed power of induction motors (S_ind) and their power factor, then it is possible to calculate the real or active power (P_ind) and the reactive power (Q_ind) of the set. So:

In the case of a synchronous motor, with its PF = 1, then there is no reactive power and the real or active power is equal to the apparent power. Soon:

With this data, we can calculate the total real power and the total reactive power consumed by the system. Like this:

With this data, we can calculate the apparent power module of the system. Then:

Replacing the variables with their respective numerical values ​​and carrying out the calculation, we have:

Now, applying the definition of power factor, we obtain:

Item b

We saw in the previous item that Q_ind = 300 kVAr. As the 400 kVA synchronous motor can generate, at most 400 kVAr in advance, then the amount of reactive, Q_R, is:

Then, the reactant that will be available at the factory, Q_f, is given by

Now we can calculate the new apparent power, S_f, that the factory consumes, remembering that real or active power does not change. So:

So, the new factory power factor is:

Item c

To find the current flowing through the synchronous motor in an overexcited condition, we have: