Question

A bank of three single-phase transformers, each rated$\mathbf{30}\mathbf{MVA}$,$\mathbf{38}\mathbf{.}\mathbf{1}\mathbf{/}\mathbf{3}\mathbf{.}\mathbf{81}\mathbf{kV}$, are connected in Y–$\mathbf{∆}$with a balanced load of three$\mathbf{1}\mathbf{\Omega }$, Y-connected resistors. Choosing a base of$\mathbf{90}\mathbf{MVA}$,$\mathbf{66}\mathbf{kV}$for the high-voltage side of the three-phase transformer, specify the base for the low-voltage side. Compute the per-unit resistance of the load on the base for the low-voltage side. Also, determine the load resistance in ohms referred to the high-voltage side and the per-unit value on the chosen base.

Short answer

Therefore, the base voltage at low-voltage side is $3.81\mathrm{kV}$. The per-unit load resistance at low voltage side is $6.2\mathrm{pu}$. The load resistance in ohms reffered to high voltage side is $300\Omega$. The per unit value of resistance reffered to H.V side is $6.2\text{\hspace{0.17em}pu}$.

Step-by-step solution

Determine the formulas of base resistance, pe-unit resistance and resistance reffered to high voltage side.

Write the formula of base resistance

${\mathrm{R}}_{\mathrm{base}}=\frac{{{\mathrm{V}}_{2\mathrm{base}}}^2}{{\mathrm{S}}_{\mathrm{base}}}$ ……. (1)

Write the formula of per-unit resistance

${\mathrm{R}}_{\mathrm{per}-\mathrm{unit}}=\frac{{\mathrm{Z}}_{\mathrm{original}}}{{\mathrm{Z}}_{\mathrm{base}}}$ ……. (2)

Write the formula of load resistance reffered to H.V.

${\mathrm{R}}^{\text{'}}=\mathrm{R}{\left(\frac{{\mathrm{V}}_{1\mathrm{base}}}{{\mathrm{V}}_{2\mathrm{base}}}\right)}^2$ …… (3)

Determine the base for low voltage side and per-unit resistance.

The voltage rating of the transformer is$38.1/3.81\text{\hspace{0.17em}kV}$.

Therefore the base on low voltage side is $3.81\text{\hspace{0.17em}kV}$.

The base kVA power is$S_{base}=90\text{\hspace{0.17em}MVA}$.

Determine the base load resistance.

Substitute$3.81\text{\hspace{0.17em}kV}$for $V_{2base}$ and $90\text{\hspace{0.17em}MVA}$ for $S_{base}$ in equation (1).

Determine the per-unit load-resistance on low-voltage side using equation (2)

$\begin{array}{c}{R}_{pu}=\frac{1\Omega }{0.1613\Omega }\\ =6.2\text{\hspace{0.17em}pu}\end{array}$

Determine the load resistance in ohms reffered to high voltage side and also in per-unit.

Determine the load resistance in ohms referred to high voltage side.

Substitute$3.81\text{\hspace{0.17em}kV}$for $V_{2base}$, $66\text{\hspace{0.17em}kV}$for$V_{1base}$ and $1\Omega$ for $R$ in equation (3).

$\begin{array}{c}{R}^{\text{'}}=\left(1\right){\left(\frac{66\times {10}^3}{3.81\times {10}^3}\right)}^2\\ =300\Omega \end{array}$

Determine the base load resistance reffered to H.V side.

Substitute$66\text{\hspace{0.17em}kV}$for $V_{1base}$and $90\text{\hspace{0.17em}MVA}$ for $S_{base}$ in equation (1).

Determine the per-unit load-resistance on high-voltage side using equation (2)

$\begin{array}{c}{R}_{pu}=\frac{300\Omega }{48.4\Omega }\\ =6.2\text{\hspace{0.17em}pu}\end{array}$

Therefore, the base voltage at low-voltage side is $3.81\text{\hspace{0.17em}kV}$. The per-unit load resistance at low voltage side is $6.2\text{\hspace{0.17em}pu}$. The load resistance in ohms reffered to high voltage side is $300\Omega$. The per unit value of resistance reffered to H.V side is $6.2\text{\hspace{0.17em}pu}$.