Question

The ratings of a three-phase, three-winding transformer are

Primary: $\mathit{Y}$connected,$\mathbf{66}\mathit{k}\mathit{V}\mathbf{,}\mathbf{15}\mathit{M}\mathit{V}\mathit{A}$

Secondary: $\mathit{Y}$connected,$\mathbf{13}\mathbf{.}\mathbf{2}\mathit{k}\mathit{V}\mathbf{,}\mathbf{10}\mathit{M}\mathit{V}\mathit{A}$

Tertiary: $\mathit{\Delta }$ connected,$\mathbf{2}\mathbf{.}\mathbf{3}\mathit{k}\mathit{V}\mathbf{,}\mathbf{5}\mathit{M}\mathit{V}\mathit{A}$

Neglecting resistances and exciting current, the leakage reactance’s are:

${\mathit{X}}_{\mathbf{P}\mathbf{S}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{09}$per unit on a$\mathbf{15}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}\mathbf{66}\mathit{k}\mathit{V}$base

${\mathit{X}}_{\mathbf{P}\mathbf{T}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{08}$per unit on a$\mathbf{15}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}\mathbf{66}\mathit{k}\mathit{V}$base

${\mathit{X}}_{\mathbf{S}\mathbf{T}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$per unit on a $\mathbf{10}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}\mathbf{13}\mathbf{.}\mathbf{2}\mathit{k}\mathit{V}$base

Determine the reactance’s of the per-phase equivalent circuitusing a base of $\mathbf{15}\mathit{M}\mathit{V}\mathit{A}\mathbf{,}$and$\mathbf{66}\mathit{k}\mathit{V}$ for the primary.

Short answer

Therefore, primary, secondary and tertiary per phase reactance’s are$0.0475\text{\hspace{0.17em}p.u,\hspace{0.17em}0.0425\hspace{0.17em}p.u}$ and $0.0325\text{\hspace{0.17em}p.u}$.

Step-by-step solution

Determine the formulas to calculate the reactance value on the given quantities.

The reactance can be converted from one base to another,

${\mathit{X}}_{\mathbf{S}\mathbf{T}}\mathbf{=}{\mathit{X}}_{\mathbf{S}\mathbf{T}\mathbf{\left(}\mathbf{o}\mathbf{l}\mathbf{d}\mathbf{\right)}}\mathbf{\left(}\frac{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}\mathbf{M}\mathbf{V}\mathbf{A}}{\mathbf{A}\mathbf{c}\mathbf{t}\mathbf{u}\mathbf{a}\mathbf{l}\mathbf{M}\mathbf{V}\mathbf{A}}\mathbf{\right)}$…..(1)

The expression for primary per unit reactance is given by,

${\mathit{X}}_{\mathbf{P}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{X}}_{\mathbf{PS}}\mathbf{+}{\mathbf{X}}_{\mathbf{PT}}\mathbf{-}{\mathbf{X}}_{\mathbf{ST}}\mathbf{)}$…..(2)

The expression for secondary per unit reactance is given by,

${\mathit{X}}_{\mathbf{S}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{X}}_{\mathbf{SP}}\mathbf{+}{\mathbf{X}}_{\mathbf{ST}}\mathbf{-}{\mathbf{X}}_{\mathbf{PT}}\mathbf{)}$…..(3)

The expression for tertiary per unit reactance is given by,

${\mathit{X}}_{\mathbf{T}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{X}}_{\mathbf{TP}}\mathbf{+}{\mathbf{X}}_{\mathbf{TS}}\mathbf{-}{\mathbf{X}}_{\mathbf{PS}}\mathbf{)}$…..(4)

Determine the values of the reactance in per unit.

Since the reactance are $X_{PS}\&X_{PT}$already specified on the base values.

The per unit value of leakage reactance$X_{TS}$can be calculated by using equation (1),

$\begin{array}{l}{X}_{ST}=0.05\left(\frac{15}{10}\right)\\ X_{ST}=0.075\text{\hspace{0.17em}p.u}\end{array}$

The primary per unit per phase reactance can be calculated by using the equation (2),

$\begin{array}{l}{X}_P=\frac{1}{2}\left(0.09+0.08-0.075\right)\\ X_P=\frac{1}{2}\times 0.095\\ X_P=0.0475\text{\hspace{0.17em}p.u}\end{array}$

The secondary per unit per phase reactance can be calculated by using the equation (3),

$\begin{array}{l}{X}_S=\frac{1}{2}\left(0.09+0.075-0.08\right)\\ X_S=\frac{1}{2}\times 0.085\\ X_S=0.0425\text{\hspace{0.17em}p.u}\end{array}$
The secondary per unit per phase reactance can be calculated by using the equation (4),

$\begin{array}{l}{X}_T=\frac{1}{2}\left(0.08+0.075-0.09\right)\\ X_T=\frac{1}{2}\times 0.065\\ X_T=0.0325\text{\hspace{0.17em}p.u}\end{array}$