Let us now consider the power in a balanced three-phase system. We begin by examining the instantaneous power absorbed by the load.

where the factor √2 is necessary because V

where I

Applying the trigonometric identity

gives

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 P

and the reactive power per phase is

The apparent power per phase is

The complex power per phase is

where

For a Y-connected load, I

and the total complex power is

where

Remember that V

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 V

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

For the three-wire three-phase system in Figure.(1b), I'

The power loss in the three wires is

Equations.(14) and (15) show that for the same total power delivered P

R = ρ

If the same power loss is tolerated in both systems, then r

since r

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.

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.

It is sufficient to consider one phase, as the system is balanced. For phase

Thus, at the source, the complex power supplied is

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

where

Hence

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

which is the difference between

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.

The apparent power is

Since the real power is

the power factor is

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

(a) For load 1, given that P

Hence,

and Q

For load 2, if Q

and P

From Equations.(3.1) and (3.2), the total complex power absorbed by the load is

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 = √3V

We apply this to each load, keeping in mind that for both loads, V

Since the power factor is lagging, the line current lags the line voltage by θ

For load 2,

and the line current lags the line voltage by θ

Hence,

The total line current is

Alternatively, we could obtain the current from the total complex power using Equation.(3.4) as

and

which is the same as before. The other line currents,

(c) We can find the reactive power needed to bring the power factor to 0.9 lagging,

where P = 90 kW, θ

Hence,

This reactive power is for the three capacitors. For each capacitor, the rating Q'

Since the capacitors are ∆-connected as shown in Figure.(3b), V

Thus,

*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(1) |

_{p}has been defined as the rms value of the phase voltage. If**Z**_{Y}= Z∠θ, the phase currents lag behind their corresponding phase voltages by θ. Thus,(2) |

_{p}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,(3) |

(4) |

(5) |

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 P

_{p}for either the ∆-connected load or the Y-connected load is p/3, or(6) |

(7) |

(8) |

(9) |

**V**_{p}and**I**_{p}are the phase voltage and phase current with magnitudes V_{p}and I_{p}, respectively. The total average power is the sum of the average powers in the phases:(10) |

_{L}= I_{p}but V_{L}= √3V_{p}, whereas for a ∆-connected load, I_{L}= √3I_{p}but V_{L}= V_{p}. Thus, Equation.(10) applies for both Y-connected and ∆-connected loads. Similarly, the total reactive power is(11) |

(12) |

**Z**_{p}= Z_{p}∠θ is the load impedance per phase. (**Z**_{p}could be**Z**_{Y}or**Z**∆) Alternatively, we may write Equation.(12) as(13) |

_{p}, I_{p}, V_{L}, and I_{L}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 V

_{L}and the same absorbed power P_{L}.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), I_{L}= P_{L}/V_{L}, so the power loss in the two wires is(14) |

_{L}= |**I**_{a}| = |**I**_{b}| = |**I**_{c}| = P_{L}/√3V_{L}from Equation.(10)Figure 1. Comparing the power loss in (a) a single-phase system, and (b) a three-phase system. |

(15) |

_{L}and same line voltage V_{L},(16) |

*l*/πr^{2}and R' = ρ*l*/πr^{2}, where r and r' are the radii of the wires. Thus,(17) |

^{2}= 2r'^{2}. The ratio of material required is determined by the number of wires and their volumes, so(18) |

^{2}= 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.

Figure 2 |

*Solution:*It is sufficient to consider one phase, as the system is balanced. For phase

*a*,

Thus, at the source, the complex power supplied is

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

where

**Z**

_{p}= 10 + j8 = 12.81∠38.66◦ and

**I**

_{p}=

**I**

_{a}= 6.81∠−21.8◦.

Hence

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

which is the difference between

**S**

_{s}and

**S**

_{L}, that is,

**S**

_{s}+

**S**

_{l}+

**S**

_{L}= 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

Since the real power is

the power factor is

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.

Figure 3 |

*Solution:*(a) For load 1, given that P

_{1}= 30 kW and cos θ

_{1}= 0.6, then sin θ

_{1}= 0.8.

Hence,

and Q

_{1}= S

_{1}sin θ

_{1}= 50(0.8) = 40 kVAR. Thus, the complex power due to load 1 is

(3.1) |

_{2}= 45 kVAR and cos θ

_{2}= 0.8, then sin θ

_{2}= 0.6. We find

and P

_{2}= S

_{2}cos θ

_{2}= 75(0.8) = 60 kW. Therefore the complex power due to load 2 is

(3.2) |

(3.3) |

(b) Since S = √3V

_{L}I

_{L}, the line current is

(3.4) |

_{L}= 240 kV. For load 1,

Since the power factor is lagging, the line current lags the line voltage by θ

_{1}= cos

^{−1}0.6 = 53.13 ◦. Thus,

For load 2,

and the line current lags the line voltage by θ

_{2}= cos

^{−1}0.8 = 36.87 ◦.

Hence,

The total line current is

Alternatively, we could obtain the current from the total complex power using Equation.(3.4) as

and

which is the same as before. The other line currents,

**I**

_{b2}and

**I**

_{ca}, can be obtained according to the

*abc*sequence (i.e.,

**I**

_{b}= 297.82∠−163.36◦ mA and

**I**

_{c}= 297.82∠76.64◦ mA).

(c) We can find the reactive power needed to bring the power factor to 0.9 lagging,

where P = 90 kW, θ

_{old}= 43.36◦, and θ

_{new}= cos

^{−1}0.9 = 25.84◦.

Hence,

This reactive power is for the three capacitors. For each capacitor, the rating Q'

_{C}= 13.8 kVAR. The required capacitance is

Since the capacitors are ∆-connected as shown in Figure.(3b), V

_{rms}in the above formula is the line-to-line or line voltage, which is 240 kV.

Thus,

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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