Home » Power in Balanced 3 Phase Loads
Electrical Theory

# Power in Balanced 3 Phase Loads

Balanced loads, in a 3φ system, have identical impedance in each secondary winding (Figure 12). The impedance of each winding in a delta load is shown as Z (Figure 12a), and the impedence in a wye load is shown as Zy (Figure 12b). For either the delta or wye connection, the lines A, B, and C supply a 3φ system of voltages. Figure 12 : 3φ Balanced Loads

In a balanced delta load, the line voltage (VL) is equal to the phase voltage (Vø), and the line current (IL) is equal to the square root of three times the phase current (√3Iø).

The below Equation is a mathematical representation of VL in a balanced delta load.

V= Vø

The below Equation is a mathematical representation of IL in a balanced delta load.

IL =√3Iø

In a balanced wye load, the line voltage (VL) is equal to the square root of three times phase voltage (√3Vø), and line current (IL) is equal to the phase current (Iø).

The below Equation is a mathematical representation of VL in a balanced wye load.

V= √3Vø

The below Equation is a mathematical representation of IL in a balanced wye load.

IL = Iø

Because the impedance of each phase of a balanced delta or wye load has equal current, phase power is one third of the total power.

The below Equation is the mathematical representation for phase power (Pø) in a balanced delta or wye load.

Pø = Vø Iø cosθ

Total power (PT) is equal to three times the single-phase power.

The below Equation is the mathematical representation for total power in a balanced delta or wye load.

P= 3 Vø Iø cosθ  As you can see, the total power formulas for delta- and wye-connected loads are identical.

Total apparent power (ST) in volt-amperes and total reactive power (QT) in volt-amperes-reactive are related to total real power (PT) in watts (Figure 13). Figure 13 : 3φ Power Triangle

A balanced three-phase load has the real, apparent, and reactive powers given by: Example 1:

Each phase of a delta- connected 3φ AC generator supplies a full load current of 200 A at 440 volts with a 0.6 lagging power factor, as shown in Figure 14. Figure 14 : Three-Phase Delta Generator

Find:

1. VL
2. IL
3. PT
4. QT
5. ST

Solution :

1. Calculate VL

V= Vø

V= 440V

2. Calculate IL

IL =√3Iø

IL  = 1.73 x 200

IL  = 346 amps

3. Calculate PT

P= 3 VL IL cosθ

P= 1.73 x 440 x 346 x 0.6

P= 158.2 kW

4. Calculate QT

Q= 3 VL IL sinθ

Q= 1.73 x 440 x 346 x 0.8

Q= 210.7 KVR

5. Calculate ST

S= 3 VL IL

S= 263.4 KVA

Example 2:

Each phase of a wye- connected 3φ AC generator supplies a 100 A current at a phase voltage of 240V and a power factor of 0.9 lagging, as shown in Figure 15. Figure 15 : Three-Phase Wye Generator

Find:

1. VL
2. PT
3. QT
4. ST

Solution :

1. Calculate VL

V= √3Vø

V= 1.73 x 240

V= 415.2 volts

2. Calculate PT

P= 3 VL IL cosθ

P= 1.73 x 415.2 x 100 x 0.9

P= 64.6 kW

3. Calculate QT

Q= 3 VL IL sinθ

Q= 1.73 x 415.2 x 100 x 0.436

Q= 31.3 KVAR

4. Calculate ST

S= 3 VL IL

S= 1.73 x 415.2 x 100

S= 71.8 KVA

#### DC Motor Theory

This website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Accept Read More

Send this to a friend