Problem 91-5
Source:
Adapted from example 2-1 - page 73 -
CHAPMAN, Stephen J. - Book: Fundamentals of Electric Machines - 5th Edition - Ed. McGrawHill - 2013.
A single-phase power system consists of a generator with voltage VG = 480∠0° V that supplies an ideal transformer whose transformation ratio is 1:10. This transformer is connected to another transformer
ideal with a transform ratio of 10:1 across a transmission line that has an impedance
Zline = 0.18 + j0.24 Ω. The secondary of the second transformer is connected to a load of
value ZL = 4 + j3 Ω, as shown in Figure 91-05.1.
It is asked:
a) Calculate the currents in the circuit as shown in Figure 91-05.1.
b) Calculate the real and reactive power that the load consumes.
c) Calculate the losses in the transmission line.
Solution of the Problem 91-5
Note that this problem is similar to the previous one, however, two transformers were added, one as an elevator
voltage and another as voltage step-down. Let's analyze this circuit and see the practical results of this configuration.
Item a
The best way to solve this circuit is to reflect the impedance of the load, which is connected to the
secondary from the second transformer to its primary. And then use the same principle for the first
transformer. Let's start by doing the reflection of the load. We will call this impedance Z2P, that is,
the impedance of the second transformer referred to the primary.
Z2P = a2 ZL = 102 x (4 + j3) = 400 + j300 Ω
Now we must add this value to that of the Zline, finding the impedance Z1S, that is,
the impedance of the secondary of the first transformer.
Z1S = Z2P + Zlinha = 400.18 + j 300.24 Ω
And finally, reflecting Z1S to the primary and finding Z1P, we get:
We are now able to calculate the value of IG, obtaining:
IG = VG / Z1P = 95.94 ∠-36.88° A
As the two transformers have the same transformation ratio, making the due calculation, via reflection of the currents,
let's conclude that:
IL = IG = 95.94 ∠-36.88° A
And the current IZ can be calculated as:
IZ = IG / a = 9.594 ∠-36.88° A
Item b
Before calculating the powers, we must calculate the voltage across the load. So:
VL = IL ZL = 95.94∠-36.88° x 5∠36.86°
Carrying out the calculation, we find:
VL = IL ZL = 479.7∠-0.02° V
To calculate the requested powers in the problem, the equations below are used:
PL = VL IL cos φ = 479.7∠-0.02° x 95.94∠-36.88° x cos 36.86°
Carrying out the calculation, we find:
PL = 36,755 W
And the reactive power is:
QL = VL IL sen φ = 479.7∠-0,02° x 95.94∠-36.88° x sen 36.86°
Carrying out the calculation, we find:
QL = 27,607 VAr
Item c
The line losses are due to the electrical resistance of the transmission line and its value is
Rline = 0.18
Ω
. Therefore:
Lossline = Rlinha IZ2 = 0.18 x 9.5942
Carrying out the calculation, we find:
Lossline = 16.57 W
Historical Addendum
Note that the losses in the line of the previous problem, when no transformers were used, was 1,484 W.
With the use of a voltage step-up transformer next to the generator and another voltage step-down transformer next to the load,
the losses in the line has dropped to an insignificant value of 16.57 W. This was the reason why, at the
end of the 19th century,
Nikolas Tesla, proponent of alternating current theory, won a resounding victory over direct current proponent,
Thomas Edison. While a direct current power source needed to be located no further than
800 meters from the consumer, alternating current using voltage step-up and step-down transformers could be
located thousands of kilometers away without major losses. This technique is still used today.