Question

Three single-phase two-winding transformers, each rated $25MVA$,$\frac{54.2}{5.42kV}$ , are connected to form a three-phase$Y-\Delta$ bank with a balanced Y-connected resistive load of $0.6\Omega$per phase on the low-voltage side. By choosing a base of $75MVA$ (three phase) and $94kV$ (line-to-line) for the high-voltage side of the transformer bank, specify the base quantities for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then determine the load resistance in ohms referred to the high-voltage side and the per-unit value of this load resistance on the chosen base

Short answer

The base quantities at the low-voltage side are $75\text{\hspace{0.33em}MVA}$ and $\text{5.41\hspace{0.33em}kV}$.

The per-unit resistance on the base for the low-voltage side is $1.538\text{\hspace{0.33em}pu}$.

The load resistance in ohms and per-unit reistance reffered to the high-voltage side is

$181.13\Omega$and .$1.537\text{\hspace{0.33em}pu}$

Step-by-step solution

Determine the formulasfor line voltage,base resistance, per-unit resistance, and resistance reffered to primary side.

Write the formula for line voltage.

$V_{line}=\sqrt{3}{V}_{phase}$ …… (1)

Write the formula for base resistance.

$R_{base}=\frac{V_{base}^2}{S_{base}}$ …… (2)

Write the formula for per-unit resistance.

$R_{pu}=\frac{Z_{actual}}{Z_{base}}$ …… (3)

Write the formula for load resistance reffered to the high-voltage side:

${R_L}^{\text{'}}=R_L{\left(\frac{V_1}{V_2}\right)}^2$ …… (4)

Determine the base quantities for the low-voltage side.

The phase voltage at the Y-connection of the transformer is .$V_{phase}=54.2\text{\hspace{0.33em}kV}$

Determine the line voltage using Equation (1)

$\begin{array}{c}{V}_{line}=\sqrt{3}\times 54.2\text{\hspace{0.33em}kV}\\ \text{=93.87\hspace{0.33em}kV}\\ \approx \text{94\hspace{0.33em}kV}\end{array}$

Therefore, the rating of the three-phase transformer is .$\text{94\hspace{0.33em}kV/5.41\hspace{0.33em}kV}$

So, on the low-voltage side, the base power is $75\text{\hspace{0.33em}MVA}$and the base voltage at the low-voltage side is $\text{5.41\hspace{0.33em}kV}$.

Determine the per-unit load resistance at the base of the low-voltage side.

The new base voltage is $V_{base}=5.41\text{\hspace{0.33em}kV.}$

The base power is.$S_{base}=75\text{\hspace{0.33em}MVA}$

Determine the base resistance.

Substituting$5.41\text{\hspace{0.33em}kV}$for $V_{base}$ and $75\text{\hspace{0.33em}MVA}$ for $S_{base}$ in Equation (2), we have

$\begin{array}{c}{R}_{base}=\frac{{\left(5.41\times {10}^3\right)}^2}{75\times {10}^6}\\ =0.39\Omega \end{array}$

Determine the per-unit load resistance.

Substituting$0.39\Omega$for $Z_{base}$ and $0.6\Omega$ for $Z_{actual}$ in Equation (3), we have

$\begin{array}{c}{R}_{\left(pu\right)}=\frac{0.6\Omega }{0.39\Omega }\\ =1.538\text{\hspace{0.33em}pu}\end{array}$

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

Determine the load resistance in ohms reffered to high-voltage side.

Substituting$5.41\text{\hspace{0.33em}kV}$for$V_2$, $94\text{\hspace{0.33em}kV}$for $V_1,$and $0.6\Omega$ for$R_L$ in Equation (4), we have

$\begin{array}{c}{R_L}^{\text{'}}=0.6{\left(\frac{94}{5.41}\right)}^2\\ =181.13\Omega \end{array}$

Determine the load resistancein per-unit reffered to high-voltage side.

Substituting$94\text{\hspace{0.33em}kV}$for $V_{base}$and $75\text{\hspace{0.33em}MVA}$for $S_{base}$in Equation (2), we have

$\begin{array}{c}{R}_{base}=\frac{{\left(94\times {10}^3\right)}^2}{75\times {10}^6}\\ =117.81\Omega \end{array}$

Determine the per-unit load resistance.

Substituting$117.81\Omega$for $Z_{base}$and $181.13\Omega$ for $Z_{actual}$in Equation (3), we have

$\begin{array}{c}{R}_{\left(pu\right)}=\frac{181.13\Omega }{117.81\Omega }\\ =1.537\text{\hspace{0.33em}pu}\end{array}$

Therefore, the base quantities at the low-voltage side are$75\text{\hspace{0.33em}MVA}$and $\text{5.41\hspace{0.33em}kV.}$

The per-unit resistance on the base for the low-voltage side is .$1.538\text{\hspace{0.33em}pu}$

The load resistance in ohms and the per-unit resistance reffered to high-voltage side are $181.13\Omega$ and $1.537\text{\hspace{0.33em}pu}$.