Question

Consider a single-phase electric system shown in Figure 3.33. Transformers are rated as follows:

X–Y$\mathbf{15}\mathbf{MVA}$, $\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{/}\mathbf{138}\mathbf{kV}$, leakage reactance $\mathbf{10}\mathbf{\%}$

Y–Z$\mathbf{15}\mathbf{MVA}$,$\mathbf{138}\mathbf{/}\mathbf{69}\mathbf{kV}$, leakage reactance$\mathbf{8}\mathbf{\%}$

With the base in circuit Y chosen as$\mathbf{15}\mathbf{MVA}$,$\mathbf{138}\mathbf{kV}$, determine the per unit impedance of the$\mathbf{500}\mathit{\Omega }$resistive load in circuit Z, referred to circuits

Z, Y, and X. Neglecting magnetizing currents, transformer resistances, and line impedances, draw the impedance diagram in per unit.

Short answer

The per unit load resistance reffered to Z, Y and X is$1.575\text{\hspace{0.17em}pu}$ , $6.3\text{\hspace{0.17em}pu}$ and $0.063\text{\hspace{0.17em}pu}$.

The impedance diagram in per-unit is drawn below:

Step-by-step solution

Determine the formulas of base resistance, per-unit resistance and resistance reffered to secondary 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

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

Write the formula of per-unit load resistance reffered to primary side.

$R^{\text{'}}=R_{pu}{\left(\frac{N_1}{N_2}\right)}^2$ …… (3)

Determine the resistance referred to Z, Y and X.

The base voltage on side Z is ${\mathrm{V}}_{\mathrm{base}}=69\text{\hspace{0.17em}kV}$.

The base power is$S_{base}=15\text{\hspace{0.17em}MVA}$.

Determine the base load resistance.

Substitute $69\text{\hspace{0.17em}kV}$ for $V_{base}$and $15\text{\hspace{0.17em}MVA}$for $S_{base}$ in equation (1).

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

$\begin{array}{c}{R}_{pu}=\frac{500\Omega }{317.4\Omega }\\ =1.575\text{\hspace{0.17em}pu}\end{array}$

Determine the per-unit load-resistance on side Y.

Substitute$1.575\text{\hspace{0.17em}pu}$for $R_{pu}$, $138\text{\hspace{0.17em}kV}$for $N_1$and $69\text{\hspace{0.17em}kV}$for $N_2$in equation (3).

$\begin{array}{c}{R^{\text{'}}}_Y=1.575\text{\hspace{0.17em}pu}\times {\left(\frac{138}{69}\right)}^2\\ =6.3\text{\hspace{0.17em}pu}\end{array}$

Determine the per-unit load-resistance on side X.

Substitute $6.3\text{\hspace{0.17em}pu}$for $R_{pu}$, $13.8\text{\hspace{0.17em}kV}$ for $N_1$ and $138\text{\hspace{0.17em}kV}$ for $N_2$ in equation (3).

$\begin{array}{c}{R^{\text{'}}}_X=6.3\text{\hspace{0.17em}pu}\times {\left(\frac{13.8}{138}\right)}^2\\ =0.063\text{\hspace{0.17em}pu}\end{array}$

Draw the single line diagram.

The leakage reactance in X-Y transformer is$10\text{\hspace{0.17em}\%}$, that is$j0.1\text{\hspace{0.17em}pu}$

The leakage reactance in Y-Z transformer is$8$%, that is$j0.08\text{\hspace{0.17em}pu}$

The single line diagram is drawn below:

The per unit load resistance reffered to Z, Y and X is$1.575\text{\hspace{0.17em}pu}$, $6.3\text{\hspace{0.17em}pu}$ and $0.063\text{\hspace{0.17em}pu}$.