8.3 Delta-Balanced Load

8.3 Delta-Balanced Load

When we have delta-connected loads (Fig. 8.3), they undergo both of the respective compound voltages. The current flows through every load.

Fig. 8.3 Balanced load in delta. Source own elaboration

V12​, $V_{23}$, $V_{31}$, $I_1$, $I_2$, $I_3$, $I_{12}$, $I_{23}$, $I_{31}$, and their negative counterparts. The vectors are spaced at 120 degrees apart, indicating phase differences. The diagram includes rotational direction marked by $w$.">

Fig. 8.4 Phasor diagram with compound voltages. Source own elaboration

We call $I_{12}$, $I_{23}$, e $I_{31}$, phase currents ($I_F$).

Since this is a balanced system, we can write:

Z₁₂∠φ₁₂ = Z₂₃∠φ₂₃ = Z₃₁∠φ₃₁

Recalling Ohm's law for a balanced system we have:

Ī₁₂ = Ū₁₂ / Z₁₂∠φ₁₂ (8.18)

Ī₂₃ = Ū₂₃ / Z₂₃∠φ₂₃ (8.19)

Ī₃₁ = Ū₃₁ / Z₃₁∠φ₃₁ (8.20)

Keep in mind that because the compound voltages are 120° out of phase with one another, the currents will be at an angle of φ to each of the corresponding compound voltages and will also be 120° out of phase with one another. Since the voltage that we apply and the impedances have the same modular value, the currents will likewise be the same.

I₁₂ = I₂₃ = I₃₁ = I_f (8.21)

Then, three additional currents, I₁, I₂ and I₃ appear in the supply line to the loads. We call these currents as line currents (IL). For every node that appear in the delta load connections, we apply Kirchhoff's first to determine the correlation between these line currents and phase currents.

Node 1: Ī₁ = Ī₁₂ - Ī₃₁ (8.22)

Node 2: Ī₂ = Ī₂₃ - Ī₁₂ (8.23)

Node 3: Ī₃ = Ī₃₁ - Ī₂₃ (8.24)

Now, we draw a vector diagram with the compound voltages set at 120º intervals from each other (Fig. 8.4).

Assuming that the loads are inductive, we draw the phase currents lagging each of their respective compound voltages by an angle φ. If we draw the line currents by performing vector sum operations of their corresponding phase currents (refer to the relationships obtained above) we obtain:

Node 2: Ī₂ = Ī₂₃ - Ī₁₂ (8.25)

Node 3: Ī₃ = Ī₃₁ - Ī₂₃ (8.26)

Then, we have the line current vector diagram (Fig. 8.5). It shows the sum of −Ī₁₂ and −Ī₂₁.

Fig. 8.5 Vector diagram with line currents.

We see that there exists a 30º angle between the phase and line currents.

An isosceles triangle appears where it is satisfied that:

α = β (8.27)

Using simple trigonometric, we have that:

180° = α + β + 120° (8.28)

This implies:

α = 30° (8.29)

Only by applying basic trigonometry we determine the relationship between phase currents and line currents (Fig. 8.5).

cos 30° = I₁ / 2I₁₂ → I₁ = 2·I₁₂·cos 30° (8.30)

I₁ = 2·I₁₂ · √3/2 = I₁₂·√3 (8.31)

Then, we conclude that:

IL = √3·If (8.32)

The line currents have the same modulus value and lag 120° with one another. They are √3 higher than the phase currents.