Problem 92-6
Source: Problem developed by the site author.
In relation to the previous problem find:
a) If you change the load to 150 KVA with FP = 0.85, what is the new efficiency?
b) Determine the apparent power of the transformer so that maximum efficiency occurs and calculate the value of this efficiency, considering FP = 1.
Solution of the Problem 92-6
Note that the power is not the rated power, so the joule losses are different from the one given in the problem. Thus, we must first calculate this loss. Let's determine the current in the primary for this power, that is:
With the value of I1 we can calculate the losses, or
Note that since we are not using the rated power of the transformer and the losses depend on the current,
the value found is lower than the losses in the short-circuit test. On the other hand, iron losses do not depend on the
current, so its value remains 2,000 W.
As was done in Problem 92-5 to find the efficiency we will use eq. 92-21. Remembering that
V2 I2 cos φ ≈ V1 I1 cos φ = 13,200 x 11.36 x 0.85 = 127,459.2 W. Thus, substituting these numeric values in the eq. 92-21, we get:
To answer this item we must remember the item 8.5 - Maximizing Efficiency
See here!, where we state that to achieve maximum efficiency in a transformer we must have the joule losses equal to the iron losses. In this way, we must have:
Now we must determine which current in the transformer generates a joule loss equal to 2,000 W. Using the value of R'eq = 13.94 Ω calculated in the previous problem, we get:
Calculating the value of V2 I2 cos φ ≈ V1 I1 cos φ = 13,200 11.98 = 158,136 W, where we consider cos φ = 1 as stated in the problem. So, using eq. 92-21 and substituting for numerical values we find an efficiency of:
