Question

A three-phase transformer is rated $\mathbf{1000}\mathbf{MVA}$,$\mathbf{220}\mathbf{Y}\mathbf{/}\mathbf{22}\mathbf{\Delta }\mathbf{kV}$. The Y-equivalent short-circuit impedance, considered equal to the leakage reactance, measured on the low-voltage side is$\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{\Omega }$.Compute the per-unit reactance of the transformer. In a system in which the base on the high-voltage side of the transformer is$\mathbf{100}\mathbf{MVA}$,$\mathbf{230}\mathbf{kV}$, what value of the per-unit reactance should be used to represent this transformer?

Short answer

The old per-unit reactance is $0.206\text{\hspace{0.17em}pu}$ and for new bases, the per-unit reactance is $0.0188\text{\hspace{0.17em}pu.}$

Step-by-step solution

Determine the formulas for base impedance, per-unit impedance, and new value of per-unit impedance.

Write the formula for base impedance:

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

Write the formula for per-unit impedance:

${\mathrm{Z}}_{\mathrm{pu}}=\frac{{\mathrm{Z}}_{\mathrm{original}}}{{\mathrm{Z}}_{\mathrm{base}}}$ ……. (2)

Write the formula for the new per-unit impedance:

${\mathrm{Z}}_{\mathrm{pe}\left(\mathrm{new}\right)}={\mathrm{Z}}_{\mathrm{pe}\left(\mathrm{old}\right)}{\left(\frac{{\mathrm{V}}_{\mathrm{base}\left(\mathrm{old}\right)}}{{\mathrm{V}}_{\mathrm{base}\left(\mathrm{new}\right)}}\right)}^2\frac{{\mathrm{S}}_{\mathrm{base}\left(\mathrm{new}\right)}}{{\mathrm{S}}_{\mathrm{base}\left(\mathrm{old}\right)}}$ …… (3)

Determine the per-unit reactance of the transformer.

The base voltage on the low side is${\mathrm{V}}_{\mathrm{base}}=22\text{\hspace{0.17em}kV}$.

The base power is${\mathrm{S}}_{\mathrm{base}}=1000\text{\hspace{0.17em}MVA}$.

Determine the base reactance.

Substituting $22\text{\hspace{0.17em}kV}$ for ${\mathrm{V}}_{\mathrm{base}}$and $1000\text{\hspace{0.17em}MVA}$ for $S_{base}$ in Equation (1), we have

Determine the per-unit impedance using Equation (2):

$\begin{array}{c}{Z}_{pu}=\frac{0.1\Omega }{0.484\Omega }\\ =0.206\text{\hspace{0.17em}pu}\end{array}$

Draw the per-unit impedance for the new base voltage and the base power.

The new base voltage on the high side is${\mathrm{V}}_{\mathrm{base}\left(\mathrm{new}\right)}=230\text{\hspace{0.17em}kV}$ and the old base voltage is ${\mathrm{V}}_{\mathrm{base}\left(\mathrm{new}\right)}=220\text{\hspace{0.17em}kV.}$

The old base power is ${\mathrm{S}}_{\mathrm{base}\left(\mathrm{old}\right)}=1000\text{\hspace{0.17em}MVA}$ and the new base power is ${\mathrm{S}}_{\mathrm{base}\left(\mathrm{new}\right)}=100\text{\hspace{0.17em}MVA.}$

Determine the new impedance.

Substituting $0.206\text{\hspace{0.17em}pu}$ for $Z_{pu\left(old\right)}$,$230\text{\hspace{0.17em}kV}$for ${\mathrm{V}}_{\mathrm{base}\left(\mathrm{new}\right)}$, $220\text{\hspace{0.17em}kV}$ for ${\mathrm{V}}_{\mathrm{base}\left(\mathrm{old}\right)}$, $100\text{\hspace{0.17em}MVA}$ for ${\mathrm{S}}_{\mathrm{base}\left(\mathrm{new}\right)},$ and $1000\text{\hspace{0.17em}MVA}$ for $S_{base\left(old\right)}$ in Equation (3), we have

Therefore, the old per-unit reactance is $0.206\text{\hspace{0.17em}pu}$ and the per-unit reactance for new bases is $0.0188\text{\hspace{0.17em}pu}$.