Question

Given the open-delta VT connection shown in Figure 10.38, both VTs having a voltage rating of 240 kV : 120 V, the voltages are specified as ${\mathit{V}}_{\mathbf{A}\mathbf{B}}\mathbf{=}\mathbf{230}\mathbf{\angle }{\mathbf{0}}^{\mathbf{\circ }}\mathbf{,}\mathbf{}{\mathit{V}}_{\mathbf{B}\mathbf{C}}\mathbf{=}\mathbf{230}\mathbf{\angle }\mathbf{-}{\mathbf{120}}^{\mathbf{\circ }}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{V}}_{\mathbf{B}\mathbf{C}}\mathbf{=}\mathbf{230}\mathbf{\angle }{\mathbf{120}}^{\mathbf{\circ }}$. Determine V_ab, V_bc and V_ca for the following cases: (a) The dots are shown in Figure 10.38. (b) The dot near c is moved to b in Figure 10.38.

Short answer

(a) The voltages V_ab, V_bc and V_ca are $115\angle 0^{\circ }V,115\angle -{120}^{\circ }V$and $230\angle {120}^{\circ }V$.

(b) The voltages V_ab, V_bc and V_ca are $115\angle 0^{\circ }V,115\angle {60}^{\circ }V$and $199.18\angle {150}^{\circ }V$.

Step-by-step solution

Write the given data from the question:

The voltage rating of the open delta VTs is 240 kV : 120 V.

Write the line voltages,

$$\begin{gathered} V_{AB}=230\angle 0^{\circ }kV \\ {V}_{BC=}230\angle -{120}^{\circ }kV \\ {V}_{CA}=230\angle {120}^{\circ }kV \end{gathered}$$

Determine the formula to calculate the Vab, Vbc and Vca.

The equation to calculate secondary line voltage V_ab of PT is given as follows.

${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{=}\left(\frac{1}{n}\right){\mathit{V}}_{\mathbf{A}\mathbf{B}}$ …… (1)

The equation to calculate secondary line voltage V_bc of PT is given as follows.

${\mathit{V}}_{\mathbf{b}\mathbf{c}}\mathbf{=}\left(\frac{1}{n}\right){\mathit{V}}_{\mathbf{B}\mathbf{C}}$ …… (2)

The equation to calculate secondary line voltage V_ca of PT is given as follows.

${\mathit{V}}_{\mathbf{c}\mathbf{a}}\mathbf{=}\left(\frac{1}{n}\right){\mathit{V}}_{\mathbf{C}\mathbf{A}}$ …… (3)

When dot c moved to b, then secondary voltage V_ab remains the same but V_bc gets reversed.

The equation to calculate the secondary voltage V_ca is given as follows.

V_CA = -(V_ab + V_bc) …… (4)

Calculate the voltages Vab, Vbc and Vca .

(a)

Calculate the potential transformation ratio.

$$\begin{gathered} n=\frac{240\times {10}^3}{120} \\ n=2000 \end{gathered}$$

Calculate secondary line voltage V_ab of PT.

Substitute $230\angle 0^{\circ }kV$ for V_AB and 2000 for into equation (1).

$$\begin{gathered} V_{ab}=\frac{1}{2000}\times 230\times {10}^3\angle 0^{\circ } \\ {V}_{ab}=115\angle 0^{\circ }V \end{gathered}$$

Calculate secondary line voltage V_bc of PT.

Substitute $230\angle -{120}^{\circ }kV$ for V_BC and 2000 for n into equation (2).

$$\begin{gathered} V_{bc}=\frac{1}{2000}\times 230\times {10}^3\angle -{120}^{\circ } \\ {V}_{bc}=115\angle -{120}^{\circ }V \end{gathered}$$

Calculate secondary line V_bc voltage of PT.

Substitute $230\angle {120}^{\circ }kV$ for V_CA and 2000 for n into equation (3).

$$\begin{gathered} V_{ca}=\frac{1}{2000}\times 230\times {10}^3\angle {120}^{\circ } \\ {V}_{ca}=115\angle {120}^{\circ }V \end{gathered}$$

Hence the voltages V_ab, V_bc and V_ca are$115\angle 0^{\circ }V,115\angle -{120}^{\circ }V$ and $230\angle {120}^{\circ }V$.

Calculate the voltages Vab, Vbc and Vca when dot c moved to b.

(b)

For dot c moved to b, then secondary voltage V_ab remains the same but V_bc gets reversed.

$$\begin{gathered} V_{ab}=115\angle 0^{\circ }V \\ {V}_{bc}=115\angle -{120}^{\circ }V \\ {V}_{bc}=115\angle {60}^{\circ }V \end{gathered}$$

Calculate secondary line voltage V_ca.

Substitute $115\angle 0^{\circ }V$ for V_ab and $115\angle {60}^{\circ }$ for V_bc into equation (4).

$$\begin{gathered} V_{ca}=-(115\angle 0^{\circ }+115\angle {60}^{\circ })\le \\ {V}_{ca}=-199.18\angle {30}^{\circ }V \\ {V}_{ca}=199.18\angle {150}^{\circ }V \end{gathered}$$

Hence the voltages V_ab, V_bc and V_ca are$115\angle 0^{\circ }V,115\angle {60}^{\circ }V$ and $199.18\angle {150}^{\circ }V$.