Problem 52-11
Source: Problem prepared by the website author.
A 200 V network is supplying 300 A to a given load. Connecting a capacitor bank
15 kVAR in parallel with the load the line current reduces to 250 A. Determine:
a) the active or real power of the load;
b) the PF of the load.
Solution of the Problem 52-11
Item a
Initially, we will calculate the apparent power before the capacitor bank is connected. So:
Sold = V . I = 200 x 300 = 60,000 VA = 60 kVA
Now let's calculate the new apparent power after turning on the capacitor bank.
Snew = V . I = 200 x 250 = 50,000 VA = 50 kVA
When inserting the capacitor bank, the current decreases. This means that the payload has an inductive character. In this way, we can draw a diagram with the indication of the potencies involved in the solution of the problem, as can be seen
in Figure 52-11.1.
Figure 52-11.1
Now our goal is to calculate the φ and θ angles. To do this, we will use what we learned in the
Chapter 51 in item 3.2 (the reader can review Here!). See that it is possible to calculate the angle φ, since we know the three sides of the triangle. Let's use the eq. 51-03 repeated below for clarity.
eq. 51-03
In this case, the variable x will be represented by the leg opposite the angle φ, that is, 15 kVAR. So, replacing the variables by their respective values, we have:
cos φ = 602 + 502 - 152 / 2 x 60 x 50 = 0.97912
Applying the inverse function of cos, that is, cos-1, we will find the value of the angle φ.
φ = cos-1 0.97912 = 11.716°
At this point, we must calculate the angle θ. Note that from Figure 52-11.1 we can write the following
relationship, based on the fact that P = S cos α.
P = Snew cos θ = Sold cos (θ + φ) ❶
To solve the second member of the equation, let's recall item 2.4 of Chapter 51 where we studied
Important Trigonometric Relationships (the reader can review Here!).
From this item, we extract the following relation:
cos (θ + φ) = cos φ cos θ - sen θ sen φ ❷
So, replacing ❷ in ❶ and performing a little algebraic work,
we get:
Snew / Sold = cos φ - sen φ tg θ
In the above equation, replacing the variables by their respective values, we have:
50 / 60 = 0.97912 - 0.2031 tg θ
Carrying out the calculation, we get:
tg θ = 0.7178
So, we easily get the value of θ by applying the inverse function of the tangent , that is,
tg-1. Soon:
θ = 35.67°
Therefore, the active or real power of the load is given by:
