Problem 82-2
Fonte:
Problem elaborated by the author of the site.
In the circuit of the Figure 82-02.1, assuming a direct sequence, we have:
VCA = 220 ∠30° , and Z = 10∠45°. Calculate:
a) The line and phase voltages.
b) The phase and line currents.
c) The total apparent power in the load.
d) The total actual power at the load.
e) The total reactive power in the load.
Solution of the Problem 82-2
This problem is a repeat of the previous one, just changing the star load setting to delta.
Item a
Since the data is equal to that of problem 82-1, the graph shown there is also valid here. Notice that the information given is VCA = 220 ∠30°. As the problem informs that the sequence is direct, then:
VAB = 220 ∠ - 90°
VBC = 220 ∠ 150°
Item b
Since the load is in a delta configuration and the line voltages are known, then to find the phase currents simply divide the voltage by the impedance. Soon:
IFCA = VCA / Z = 220∠30° /10∠45° = 22∠-15° A
IFAB = VAB / Z = 220∠-90° /10∠45° = 22∠-135° A
IFBC = VBC / Z = 220∠150° /10∠45° = 22∠105° A
These are the phase currents of the circuit. Note that there is a 120° lag between them. In order to calculate the line currents, we must use the transforming equation, or:
Therefore, the line currents are:
IA = √3 IFAB ∠-135° - 30° = 38.11∠-165° A
IB = √3 IFBC ∠+105° - 30° = 38.11∠+75° A
IC = √3 IFCA ∠-15° - 30° = 38.11∠-45° A
Another way to solve this is to applyKirchhoff to currents to node A and
we find the same result. Repeat the method for the node B and node C.
IA = IFAB - IFCA
Item c
To calculate the total apparent power in the load, we use the eq. 82-03, already studied, or:
In this equation the effective values are of the line voltage and current. E φ is the angle of the impedance, whose value is φ = 45°. Thus:
S = √3 VAB IA∠45° = √3 x 220 x 38.11∠+45°
Performing the calculation:
S = 14,521.86 ∠+45° VA
It should be noted that the 45° angle can be used to calculate the power factor of the circuit, or:
PF = cos 45° = 0.707
Item d
To find the real power P, just use the eq. 82-04, or:
P = 14,521.86 cos 45° = 10,268.51 W
Item e
To find the reactive power Q, we use the eq. 82-05, or:
Q = 14,521.86 sen 45° = 10,268.51 VAr
Addendum
Having the value of P and Q, we can write the total power total in its Cartesian form.
S = 10,268.51 + j 10,268.51 VA
Note that by modifying only the load configuration, there was a substantial increase in power.