Y-Δ transform

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For application in statistical mechanics see Yang-Baxter equation.

For the device which transforms three-phase electric power without a neutral wire into three-phase power with a neutral wire, see delta-wye transformer.

The Y-Δ transform, also written Y-delta, Wye-delta, Kennelly’s delta-star transformation, star-mesh transformation, T-Π or T-pi transform, is a mathematical technique to simplify the analysis of an electrical network. The name derives from the shapes of the circuit diagrams, which look respectively like the letter Y and the Greek capital letter Δ. In the United Kingdom, the wye diagram is sometimes known as a star. This circuit transformation theory was published by Arthur Edwin Kennelly in 1899. ^[1]

1 Basic Y-Δ transformation
- 1.1 Equations for the transformation from Δ-load to Y-load 3-phase circuit
- 1.2 Equations for the transformation from Y-load to Δ-load 3-phase circuit
2 Graph theory
3 Demonstration
- 3.1 Δ-load to Y-load transformation equations
- 3.2 Y-load to Δ-load transformation equations
4 See also
5 Notes
6 References
7 External links

Basic Y-Δ transformation

Δ and Y circuits with the labels which are used in this article.

The transformation is used to establish equivalence for networks with 3 terminals. Where three elements terminate at a common node and none are sources, the node is eliminated by transforming the impedances. For equivalence, the impedance between any pair of terminals must be the same for both networks. The equations given here are valid for real as well as complex impedances.

Equations for the transformation from Δ-load to Y-load 3-phase circuit

The general idea is to compute the impedance $R y$ at a terminal node of the Y circuit with impedances $R'$ , $R''$ to adjacent nodes in the Δ circuit by

$R_y = \frac{R'R''}{\sum R_\Delta}$

where $R Δ$ are all impedances in the Δ circuit. This yields the specific formulae

$R_1 = \frac{R_aR_b}{R_a + R_b + R_c},$

$R_2 = \frac{R_bR_c}{R_a + R_b + R_c},$

$R_3 = \frac{R_aR_c}{R_a + R_b + R_c}.$

Equations for the transformation from Y-load to Δ-load 3-phase circuit

The general idea is to compute an impedance $R Δ$ in the Δ circuit by

$R_\Delta = \frac{R_P}{R_\mathrm{opposite}}$

where $R P = R 1 R 2 + R 2 R 3 + R 3 R 1$ is the sum of the products of all pairs of impedances in the Y circuit and $R opposite$ is the impedance of the node in the Y circuit which is opposite the edge with $R Δ$ . The formula for the individual edges are thus

$R_a = \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_2},$

$R_b = \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_3},$

$R_c = \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_1}.$

Graph theory

In graph theory, the Y-Δ transform means replacing a Y subgraph of a graph with the equivalent Δ subgraph. The transform preserves the number of edges in a graph, but not the number of vertices or the number of cycles. Two graphs are said to be Y-Δ equivalent if one can be obtained from the other by a series of Y-Δ transforms in either direction. For example, the Petersen graphs are a Y-Δ equivalence class.

Demonstration

Δ-load to Y-load transformation equations

Δ and Y circuits with the labels which are used in this article.

Given the values of $R b$ , $R c$ and $R a$ from the Δ configuration, we want to obtain the values of $R 1$ , $R 2$ and $R 3$ in the equivalent Y configuration. In order to do that, we will calculate the equivalent impedances of both configurations in N₁N₂, N₁N₃ and N₂N₃, supposing in each case that the omitted node is unconnected, and we will equal both expressions, since the resistance must be the same.

The resistance between N₁ and N₂ when N₃ is not connected in the Δ configuration is

$R(N_1, N_2) = R_b \parallel (R_a+R_c) = \frac{R_b(R_a+R_c)}{R_b+R_c+R_a} = \frac{R_bR_a+R_bR_c}{R_b+R_c+R_a}.$

In the Y configuration, we have

R (N 1, N 2) = R 1 + R 2;

hence we have

$R_1+R_2 = \frac{R_bR_a+R_bR_c}{R_b+R_c+R_a}$ (1)

By similar calculations we obtain

$R_2+R_3 = \frac{R_cR_a+R_cR_b}{R_b+R_c+R_a}$ (2)

and

$R_1+R_3 = \frac{R_aR_b+R_aR_c}{R_b+R_c+R_a}.$ (3)

The impedances for the Y configuration can be derived from these equations by adding two equations and subtracting the third. For example, adding (1) and (3), then subtracting (2) yields

$R_1+R_2+R_1+R_3-R_2-R_3 = \frac{R_bR_a+R_bR_c}{R_b+R_c+R_a} + \frac{R_aR_b+R_aR_c}{R_b+R_c+R_a} - \frac{R_cR_a+R_cR_b}{R_b+R_c+R_a}$