Question

Given the open-delta VT connection shown in Figure 10.38, both VTs having a voltage rating of $\mathbf{240}\mathbf{}\mathit{k}\mathit{V}\mathbf{:}\mathbf{120}\mathbf{}\mathit{V}$, the voltages are specified as ${\mathit{V}}_{\mathbf{A}\mathbf{B}}\mathbf{=}\mathbf{230}\mathbf{\angle }\mathbf{0}\mathbf{°}\mathbf{,}\mathbf{}{\mathit{V}}_{\mathbf{B}\mathbf{C}}\mathbf{=}\mathbf{230}\mathbf{°}\mathbf{\angle }\mathbf{120}\mathbf{°}\mathbf{,}$role="math" localid="1655396371549" ${\mathit{V}}_{\mathbf{B}\mathbf{C}}\mathbf{=}\mathbf{230}\mathbf{\angle }\mathbf{-}\mathbf{120}\mathbf{°}$and ${\mathit{V}}_{\mathbf{B}\mathbf{C}}\mathbf{=}\mathbf{230}\mathbf{\angle }\mathbf{120}\mathbf{°}$. Determine ${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{,}{\mathit{V}}_{\mathbf{b}\mathbf{c}}$and ${\mathit{V}}_{\mathbf{c}\mathbf{a}}$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 $115\angle 0°V,115\angle -120°V$and $230\angle 120°V$.

(b) The voltages$V_{ab},V_{bc}$and $V_{ca}$are $115\angle 0°V,115\angle 60°V$and $199.18\angle 150°V$.

Step-by-step solution

Write the given data from the question:

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

Write the line voltages,

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

Determine the formula to calculate the Vab, Vbcand Vca

The equation to calculate secondary line voltage $V_{ab}$of PT is given as follows.

role="math" localid="1655400402811" ${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{=}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{\right)}{\mathit{V}}_{\mathit{A}\mathit{B}}$ ........................(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.

${\mathit{V}}_{\mathbf{c}\mathbf{a}}\mathbf{=}\mathbf{-}\left(V_{ab}+V_{bc}\right)$ …… (4)

Calculate the voltages Vab, Vbcand 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 $230°\angle 0°kV$for $V_{AB}$and $2000$for $n$into equation (1).

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

Calculate secondary line voltage $V_{bc}$of PT.

Substitute $230\angle -120°kV$for $V_{BC}$and $2000$for$n$into equation (2).

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

Calculate secondary line voltage $V_{bc}$of PT.

Substitute $230\angle 120°kV$for $V_{CA}$and $2000$for $n$into equation (3).

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

Hence the voltages $V_{ab,}{V}_{bc}$and $V_{ca}$are $115\angle 0°V,115\angle -120°V$and $230\angle 120°V$.

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.

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

Calculate secondary line voltage$V_{ca}$

Substitute$115\angle 0°V$ for $V_{ab}$and $115\angle 60°V$for $V_{bc}$into equation (4).

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

Hence the voltages $V_{ab},V_{bc}$and $V_{ca}$are $115\angle 0°V,115\angle 60°V$and $199.18\angle 150°V$.

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