Problem 97-2
Source: Adapted from example 2-55 - page 189 -
SAHDEV, S. K. - Book: Electrical Machines - 1st edition - Cambridge University Press - 2018.
Taking into account the results of the previous problem, find:
a) The load impedance.
b) The value of the voltages E1 and E2, that is, the voltages of the transformer secondaries.
Solution of the Problem 97-2
Initially, we will transcribe the necessary data from the previous problem.
VL = 400 ∠0° V and IL = 500 A
Z2a = 0.2 + j 0.3 Ω and Z2b = 0.25 + j 0.5 Ω
I2a = 304.5 ∠ - 34.07° A and I2b = 196.45 ∠ - 41.2° A
Item a
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
Item b
With the value of I2a* = 304.5 ∠ + 34.07°,
we can calculate the powers involved in the transformer "a", or:
S2a = VL I2a*
Replacing the variables with their respective numerical values and performing the calculation, we obtain:
S2a = 400 x 304.5 = 121,800 VA = 121.8 kVA
Real and reactive power are:
P2a = 400 x 304.5 x cos 34.07° = 100,900 W = 100.9 kW
Q2a = 400 x 304.5 x sen 34.07° = 69,230 VA = 69.23 kVA
Item c
With the value of I2b* = 196.45 ∠ + 41.2°,
we can calculate the powers involved in the transformer "b", or:
S2b = VL I2b*
Replacing the variables with their respective numerical values and performing the calculation, we obtain:
S2b = 400 x 196.45 = 78,580 VA = 78.58 kVA
Real and reactive power are:
P2b = 400 x 196.45 x cos 41.2° = 59,125 W = 59.125 kW
Q2b = 400 x 196.45 x sen 41.2° = 51,760 VA = 51.76 kVA
Item d
Ea is the EMF induced in the secondary of transformer "a" and Eb
is the EMF induced in the secondary of transformer "b". Therefore, the EMF
will be the terminal voltage at the load plus the voltage drop across the internal impedance of each transformer. Therefore, we can write:
Ea = VL + I2a x Z2a = 400 + 304.5 ∠ - 34.07° x 0.3606 ∠ 56.31°
Performing the calculation, we find:
Ea = 501.63 ∠ 4.74°
And for the calculation of Eb, we have:
Eb = VL + I2b x Z2b = 400 + 196.45 ∠ - 41.2° x 0.559 ∠ 63.44°
