Question

Consider a bank of three single-phase two-winding transformers whose high-voltage terminals are connected to a three-phase,$\mathbf{13}\mathbf{.}\mathbf{8}\mathit{k}\mathit{V}$feeder. The low-voltage terminals are connected to a three-phase substation load rated$\mathbf{2}\mathbf{.}\mathbf{0}\mathit{M}\mathit{V}\mathit{A}$and$\mathbf{2}\mathbf{.}\mathbf{5}\mathit{k}\mathit{V}$. Determine the required voltage, current, and$\mathit{M}\mathit{V}\mathit{A}$ratings of both windings of each transformer, when the high-voltage/low-voltage windings are connected (a)$\mathit{Y}\mathbf{-}\mathit{\Delta }$, (b)$\mathit{Y}\mathbf{-}\mathit{\Delta }$, (c)$\mathit{Y}\mathbf{-}\mathit{Y}$, and (d)$\mathit{\Delta }\mathbf{-}\mathit{\Delta }$.

Short answer

(a)Therating of the$Y-\Delta$transformer is $7.97kV/2.5kV$ and $0.667MVA$.

(b)Therating of the$\Delta -Y$transformer is $13.8kV/1.443kV$ and $0.667MVA$.

(c)Therating of the$Y-Y$transformer is and $0.667MVA$.

(d) Therating of the$\Delta -\Delta$transformer is and $0.667MVA$.

Step-by-step solution

Determine the formulas of phase voltage, per-phase power and line current.

Write the formula of phase voltage.

${\mathit{V}}_{\mathbf{ph}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{L}}}{\sqrt{\mathbf{3}}}$…… (1)

Write the formula of per-phase apparent power.

${\mathit{S}}_{\mathbf{1}\mathbf{-}\mathbf{\varphi }}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{3}\mathbf{-}\mathbf{\varphi }}}{\mathbf{3}}$…… (2)

Determine the   phase voltage and single phase apparent power inY-Δ  transformer. 

(a)

Determine the phase voltage at primary side.

Substitute $13.8kV$for$V_{ph}$in equation (1).

$\begin{array}{c}{V}_{ph}=\frac{13.8kV}{\sqrt{3}}\\ =7.97kV\end{array}$

Determine the single phase power.

Substitute$2MVA$ for$S_{3-\varphi }$ in equation (2).

$\begin{array}{c}{S}_{1-\varphi }=\frac{2MVA}{3}\\ =0.667MVA\end{array}$

Therefore, the rating of the $Y-\Delta$transformer is $7.97kV/2.5kV$ and $0.667MVA$.

Determine the   phase voltage and single phase apparent power in Δ-Y transformer.

(b)

Determine the phase voltage at secondary side.

Substitute $2.5kV$for$V_{ph}$ in equation (1).

$\begin{array}{c}{V}_{ph}=\frac{2.5kV}{\sqrt{3}}\\ =1.443kV\end{array}$

Therefore, the rating of the$\Delta -Y$ transformer is$13.8kV/1.443kV$ and$0.667MVA$ .

Determine the   phase voltage and single phase apparent power in Y-Y transformer. 

(c)

Determine the phase voltage at secondary side.

Substitute $2.5kV$for$V_{ph}$ in equation (1).

$\begin{array}{c}{V}_{ph}=\frac{2.5kV}{\sqrt{3}}\\ =1.443kV\end{array}$

Determine the phase voltage at primary side.

Substitute $13.8kV$for$V_{ph}$ in equation (1).

$\begin{array}{c}{V}_{ph}=\frac{13.8kV}{\sqrt{3}}\\ =7.97kV\end{array}$

Therefore, the rating of th$Y-Y$etransformer is $7.97kV/1.443kV$ and $0.667MVA$.

Determine the   phase voltage and single phase apparent power in Δ-Δ transformer.  

(d)

The voltage at high-voltage side is$13.8kV$ and the voltage at low-voltage side is $2.5kV$.

Therefore, the voltage rating of transformer is $13.8kV/2.5kV$.

The single-phase power load is $0.667MVA$.

Therefore, the rating of the$\Delta -\Delta$transformer is $13.8kV/2.5kV$ and $0.667MVA$.