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Balanced Three-Phase System Power Formula Explained

Let us now consider the power in a balanced three-phase system. We begin by examining the instantaneous power absorbed by the load.
Power Formula for Balanced System
It is better to read about what is three phase circuit first.

Power Formula for Balanced System

This requires that the analysis be done in the time domain. For a Y-connected load, the phase voltages are
Power Formula for Balanced System
(1)
where the factor √2 is necessary because Vp has been defined as the rms value of the phase voltage. If ZY = Z∠θ, the phase currents lag behind their corresponding phase voltages by θ. Thus,
Power Formula for Balanced System
(2)
where Ip is the rms value of the phase current. The total instantaneous power in the load is the sum of the instantaneous powers in the three phases; that is,
Power Formula for Balanced System
(3)
Applying the trigonometric identity
Power Formula for Balanced System
(4)
gives
Power Formula for Balanced System
(5)
Thus the total instantaneous power in a balanced three-phase system is constant—it does not change with time as the instantaneous power of each phase does. This result is true whether the load is Y- or ∆-connected.

This is one important reason for using a three-phase system to generate and distribute power. We will look into another reason a little later.

Since the total instantaneous power is independent of time, the average power per phase Pp for either the ∆-connected load or the Y-connected load is p/3, or
Power Formula for Balanced System
(6)
and the reactive power per phase is
Power Formula for Balanced System
(7)
The apparent power per phase is
Power Formula for Balanced System
(8)
The complex power per phase is
Power Formula for Balanced System
(9)
where Vp and Ip are the phase voltage and phase current with magnitudes Vp and Ip, respectively. The total average power is the sum of the average powers in the phases:
Power Formula for Balanced System
(10)
For a Y-connected load, IL = Ip but VL = √3Vp, whereas for a ∆-connected load, IL = √3Ip but VL = Vp. Thus, Equation.(10) applies for both Y-connected and ∆-connected loads. Similarly, the total reactive power is
Power Formula for Balanced System
(11)
and the total complex power is
Power Formula for Balanced System
(12)
where Zp = Zp∠θ is the load impedance per phase. (Zp could be ZY or Z∆) Alternatively, we may write Equation.(12) as
Power Formula for Balanced System
(13)
Remember that Vp, Ip, VL, and IL are all rms values and that θ is the angle of the load impedance or the angle between the phase voltage and the phase current.

A second major advantage of three-phase systems for power distribution is that the three-phase system uses a lesser amount of wire than the single-phase system for the same line voltage VL and the same absorbed power PL.

We will compare these cases and assume in both that the wires are of the same material (e.g., copper with resistivity ρ), of the same length l, and that the loads are resistive (i.e., unity power factor). For the two-wire single-phase system in Figure.(1a), IL = PL/VL, so the power loss in the two wires is
Power Formula for Balanced System
(14)
For the three-wire three-phase system in Figure.(1b), I'L = |Ia| = |Ib| = |Ic| = PL/√3VL from Equation.(10)
Power Formula for Balanced System
Figure 1. Comparing the power loss in (a) a single-phase system, and (b) a three-phase system.
The power loss in the three wires is
Power Formula for Balanced System
(15)
Equations.(14) and (15) show that for the same total power delivered PL and same line voltage VL,
Power Formula for Balanced System
(16)
R = ρl/πr2 and R' = ρl/πr2, where r and r' are the radii of the wires. Thus,
Power Formula for Balanced System
(17)
If the same power loss is tolerated in both systems, then r2 = 2r'2. The ratio of material required is determined by the number of wires and their volumes, so
Power Formula for Balanced System
(18)
since r2 = 2r'2. Equation.(18) shows that the single-phase system uses 33 percent more material than the three-phase system or that the threephase system uses only 75 percent of the material used in the equivalent single-phase system.

In other words, considerably less material is needed to deliver the same power with a three-phase system than is required for a single-phase system.

Power Formula for Balanced System Examples

For better understanding let us review the examples below:
1. Refer to the circuit in Figure.(2). Determine the total average power, reactive power, and complex power at the source and at the load.
Power Formula for Balanced System
Figure 2
Solution:
It is sufficient to consider one phase, as the system is balanced. For phase a,
Power Formula for Balanced System
Thus, at the source, the complex power supplied is
Power Formula for Balanced System
The real or average power supplied is −2087 W and the reactive power is −834.6 VAR.
At the load, the complex power absorbed is
Power Formula for Balanced System
where Zp = 10 + j8 = 12.81∠38.66◦ and Ip = Ia = 6.81∠−21.8◦.
Hence
Power Formula for Balanced System
The real power absorbed is 1391.7 W and the reactive power absorbed is 1113.3 VAR.

The difference between the two complex powers is absorbed by the line impedance (5 − j2) Ω. To show that this is the case, we find the complex power absorbed by the line as
Power Formula for Balanced System
which is the difference between Ss and SL, that is, Ss + Sl  + SL = 0, as expected.

2. A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor.
Solution:
The apparent power is
Power Formula for Balanced System
Since the real power is
Power Formula for Balanced System
the power factor is
Power Formula for Balanced System

3. Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown in Figure.(3a). Load 1 draws 30 kW at a power factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging.

Assuming the abc sequence, determine: (a) the complex, real, and reactive powers absorbed by the combined load, (b) the line currents, and (c) the kVAR rating of the three capacitors ∆-connected in parallel with the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor.
Power Formula for Balanced System
Figure 3
Solution:
(a) For load 1, given that P1 = 30 kW and cos θ1 = 0.6, then sin θ1 = 0.8.
Hence,
Power Formula for Balanced System
and Q1 = S1 sin θ1 = 50(0.8) = 40 kVAR. Thus, the complex power due to load 1 is
Power Formula for Balanced System
(3.1)
For load 2, if Q2 = 45 kVAR and cos θ2 = 0.8, then sin θ2 = 0.6. We find
Power Formula for Balanced System
and P2 = S2 cos θ2 = 75(0.8) = 60 kW. Therefore the complex power due to load 2 is
Power Formula for Balanced System
(3.2)
From Equations.(3.1) and (3.2), the total complex power absorbed by the load is
Power Formula for Balanced System
(3.3)
which has a power factor of cos 43.36◦ = 0.727 lagging. The real power is then 90 kW, while the reactive power is 85 kVAR.

(b) Since S = √3VLIL, the line current is
Power Formula for Balanced System
(3.4)
We apply this to each load, keeping in mind that for both loads, VL = 240 kV. For load 1,
Power Formula for Balanced System
Since the power factor is lagging, the line current lags the line voltage by θ1 = cos−1 0.6 = 53.13 ◦. Thus,
Power Formula for Balanced System
For load 2,
Power Formula for Balanced System
and the line current lags the line voltage by θ2 = cos−1 0.8 = 36.87 ◦.
Hence,
Power Formula for Balanced System
The total line current is
Power Formula for Balanced System
Alternatively, we could obtain the current from the total complex power using Equation.(3.4) as
Power Formula for Balanced System
and
Power Formula for Balanced System
which is the same as before. The other line currents, Ib2 and Ica, can be obtained according to the abc sequence (i.e., Ib = 297.82∠−163.36◦ mA and Ic = 297.82∠76.64◦ mA).

(c) We can find the reactive power needed to bring the power factor to 0.9 lagging,
Power Formula for Balanced System
where P = 90 kW, θold = 43.36◦, and θnew = cos−1 0.9 = 25.84◦.
Hence,
Power Formula for Balanced System
This reactive power is for the three capacitors. For each capacitor, the rating Q'C = 13.8 kVAR. The required capacitance is
Power Formula for Balanced System
Since the capacitors are ∆-connected as shown in Figure.(3b), Vrms in the above formula is the line-to-line or line voltage, which is 240 kV.
Thus,
Power Formula for Balanced System

Have you understood what is power formula for balanced system? Don't forget to share and subscribe! Happy learning!
Reference:  Fundamentals of electric circuits by Charles K. Alexander and Matthew N. O. Sadiku
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