Problem 97-1
Source: Adapted from example 2-55 - page 189 -
SAHDEV, S. K. - Book: Electrical Machines - 1st edition - Cambridge University Press - 2018.
A 500 A load with PF = 0.8 lagging and a terminal voltage of 400 V is supplied by two
transformers connected in parallel. The equivalent impedances of the two transformers,
referenced to the secondary side are 2 + j3 and 2.5 + j5, respectively.
a) Calculate the current and powers supplied by transformer a.
b) Calculate the current and powers supplied by transformer b.
c) What is the power factor at which each transformer is operating?
Solution of the Problem 97-1
Item a
Initially, we will designate the two transformers as a and b. Thus, the impedances referred to the secondary are:
Z2a = 0.2 + j 0.3 = 0.3606 ∠ 56.31°
Z2b = 0.25 + j 0.5 = 0.559 ∠ 63.44°
To calculate the currents, we will use the current divider. To do this, we must determine the sum of the two impedances,
which we will call Zs.
Zs = Z2a + Z2b = 0.45 + j 0.8 = 0.918 ∠ 60.64°
From the data provided in the problem, we know that for the load we have: IL = 500 A and FP = 0.8 lagging.
Thus, we can express the current in the load as:
IL = 500 ∠ - 36.87°
With these values, we can calculate the value of the current I2a in the secondary of the transformer a, using a current divider.
I2a = ( Z2b / Zs ) IL
Replacing the variables with their respective numerical values, we have:
Knowing the voltage and current values in the load, it is possible to calculate the load impedance module.
|ZL| = VL / IL = 400 / 500 = 0.8 Ω
Since we know the power factor of the load, and its value is FP = 0.8,
the result is φL = cos-1 0.8 = 36.87°. Thus, the impedance represented in polar and Cartesian forms is:
ZL = 0.8 ∠ 36.87° = 0.64 + j 0.48 Ω
Separating the impedance ZL into RL and
XL (load resistance and reactance, respectively), we obtain:
RL = 0.64 Ω and
XL = 0.48 Ω
Using RL and IL, we can calculate the actual power dissipated, or:
PL = RL x IL2
= 160,000 W = 160 kW
This value must be equal to the sum of the real power supplied by transformers "a" and "b", or:
PL = P2a + P2b
= 160,025 W = 160.025 kW
Note an error of 25 W due to rounding of values, which corresponds to an error of less than 0.02%.
Similarly, we can calculate the reactive power, or:
QL = XL x IL2
= 120,000 W = 120 kW
This value must be equal to the sum of the reactive powers supplied by transformers "a" and "b", or:
QL = Q2a + Q2b
= 120,990 W = 120.99 kW
Neste caso, um erro inferior a 1%.
Therefore, the values found are within expectations.
Item c
To find the power factors of the transformers, simply analyze the angle of the secondary current of each one. Thus, we have:
FPa = cos (-34.07°) = 0.828
FPb = cos (-41.2°) = 0.752
Note that in both cases the angles are negative. Thus, the power factors are lagging.