Problem 52-8    Source:  Problem created by the author of the site.

Be a consumer where the voltage supplied by the energy distribution company is   220 ∠0° V (rms).    The impedance representing the entire consumer circuit is given by the complex impedance Z = 5 + j 9 Ω.   See the Figure 52-08. What is the value of the consumer's power factor and how to fix it to FP = 0.95?

Solution of the Problem 52-8   

Initially, let's transform the impedance given in rectangular form to polar form, that is:

Therefore, the lag angle between voltage and current is given by the angle of impedance, that is, θ = 61°. This indicates that the current is 61° delayed in relation to voltage as it is an inductive circuit.

See in the Figure 52-08.a, the circuit with the source and the impedance.

Knowing the lag angle, we can calculate the power factor.

Note that we have a power factor well below that intended. To better understand the solution to the problem, let's calculate the current flowing through the circuit, as well as the powers involved.

We easily calculate the current flowing through the impedance by applying the Ohm's law.

Knowing the value of the current we can calculate the value of the apparent power.

Now let's calculate the average or real power from the apparent power and the cosine of the angle, that is:

In the same way we can proceed to find the reactive power.

With these values we can write the apparent power in rectangular form.

And if there is any skepticism about these results, we can simply check the validity of it by calculating:

Note that we can calculate these powers in various ways, depending on the data that is at our disposal.

In the Figure 52-08.b we can see the power triangle of our problem and how it is a rectangle triangle that, by itself, explains the above equation we used to calculate apparent power.

To answer the second part of the problem, namely, increasing the power factor for 0.95, we must calculate what is the new angle of lag between the voltage and current in the circuit.

See in the Figure 52-08.c, the design of the new circuit with the inclusion of the capacitor in parallel with the load. As we know in the capacitor, the current is advanced in relation to tension. Thus, adding the current I, delayed by 61°, with the current I_C, advanced by 90°, we should get a current I_N delayed by no more than 18.20° in relation to tension.

In order to clarify this situation, we present the Figure 52-08.d with the current diagram. Notice that when we add the capacitor, the current I_C, which is 90° ahead of the voltage, when added to I (backward of θ = 61°), results in a current I_N delayed only from θ_N = 18.20°. Therefore, by placing a capacitor in parallel with the original circuit, we can considerably improve the consumer's power factor.

We can also see that the modulus (length) of I_N is much smaller than the modulus of I. Now we need to calculate the value of C. For this there are several ways. Let's look at two alternative paths.

Alternative 1 -     Since we know the average or real power and the new lag angle (θ_N) we can calculate the new reactive power (Q_N), or:

In order for us to get this new reactive power, it means that the capacitor must introduce a power of value equal to:

Substituting the numerical values, we find the value of Q_C, in module, or:

Alternative 2 -     Another way is to know the real power and the angles θ and θ_N, that is:

Substituting the numerical values, we find the value of Q_C, in module, or:

Therefore, the two alternative paths led to the same result. Thus, we easily calculate the value of C, remembering that:

As X_C = 1 / (2 π f C), after some algebraic arrangements we arrive at:

Substituting the numerical values, performing the calculation and remembering that for Brazil, f= 60 Hz, we have:

Thus, by placing a capacitor in parallel with the load, we can correct the power factor to 0.95. To get a sense of how important this is, let's calculate the new apparent power and the new current supplied by the power company.

This is the complex conjugate value of the current. How do we want the current value just change the sign of the angle, ie:

Note that we were able to reduce both the apparent power and the current electrical, practically at half of their original values. Thus, the power company electric power gains extra power and can increase the number of consumers without overloading the system. This fact alone justifies the requirement to maintain the power factor as close to unity as possible.