Question

Using the transformer ratings as base quantities, work Problem 3.14 in per-unit.

Short answer

1. The voltage at primary terminal of transformer is$2448\angle 0.{683}^0\mathrm{V}$.
2. The source voltage is $2490\angle 1.{1505}^0\mathrm{V}$.
3. The real and reactive power supplied by source is $40.87\mathrm{kW}$ and role="math" localid="1655294870070" $31.95\mathrm{kVAR}$.

Step-by-step solution

Determine the formulas of base impedance, per-unit impedance, per-unit voltage, original voltage, apparent power in per-unit and original source power.

Write the formula of base impedance.

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

Write the formula of per-unit impedance.

${\text{Z}}_{\mathrm{per}-\mathrm{unit}}=\frac{{\mathrm{Z}}_{\mathrm{original}}}{{\mathrm{Z}}_{\mathrm{base}1}}$ ……. (2)

Write the formula of per-unit voltage.

${\mathrm{V}}_{\mathrm{per}-\mathrm{unit}}=\frac{{\mathrm{V}}_{\mathrm{original}}}{{\mathrm{V}}_{\mathrm{base}}}$ ……. (3)

Write the formula of voltage at primary terminal of transformer.

${\mathrm{V}}_{1\mathrm{per}-\mathrm{unit}}={\mathrm{V}}_{2\mathrm{per}-\mathrm{unit}}+{\mathrm{I}}_{2\mathrm{per}-\mathrm{unit}}{\mathrm{Z}}_{\mathrm{eq}\left(\mathrm{per}-\mathrm{unit}\right)}$…… (4)

Write the formula of original terminal voltage.

${\mathrm{V}}_1={\mathrm{V}}_{1\mathrm{per}-\mathrm{unit}}\times {\mathrm{V}}_{\mathrm{base}1}$ …… (5)

Write the formula of apparent power in per-unit.

${\mathrm{S}}_{\mathrm{pu}}={\mathrm{V}}_{\mathrm{sper}-\mathrm{unit}}\times {\mathrm{I}}_{2\mathrm{per}-\mathrm{unit}}$ …… (6)

Write the formula of source apparent power.

$\mathrm{S}={\mathrm{S}}_{\mathrm{pu}}\times {\mathrm{S}}_{\mathrm{base}}$ …… (7)

Draw the per-unit single line diagram.

The base kVA power is ${\mathrm{S}}_{\mathrm{base}}=50\mathrm{kVA}$.

The base voltage for primary side is ${\mathrm{V}}_{\mathrm{base}1}=2400\mathrm{V}$and the base voltage at secondary side is ${\mathrm{V}}_{\mathrm{base}2}=240\mathrm{V}$.

Determine the base impedance.

Substitute $2400\mathrm{V}$for localid="1655279179016" ${\mathrm{V}}_{\mathrm{base}1}$and $50\mathrm{kVA}$ for ${\mathrm{S}}_{\mathrm{base}}$ in equation (1).

$\begin{array}{rcl}{\mathrm{Z}}_{\mathrm{base}1}& =& \frac{{2400}^2}{50\times {10}^3}\\ & =& 115.2\mathrm{\Omega }\end{array}$

The original diagram is drawn below:

The feeder impedance in per-unit is calculated using equation (2)

$\begin{array}{rcl}{\mathrm{Z}}_{\mathrm{feed}\left(\mathrm{per}-\mathrm{unit}\right)}& =& \frac{\left(1+\mathrm{j}2\right)\mathrm{\Omega }}{115.2\mathrm{\Omega }}\\ & =& 8.6806\times {10}^{-3}+\mathrm{j}1.7361\times {10}^{-2}\end{array}$

The transformer impedance in per-unit is calculated using equation (2)

$\begin{array}{rcl}{\mathrm{Z}}_{\mathrm{eq}\left(\mathrm{per}-\mathrm{unit}\right)}& =& \frac{\left(1+\mathrm{j}2.5\right)\mathrm{\Omega }}{115.2\mathrm{\Omega }}\\ & =& 8.6806\times {10}^{-3}+\mathrm{j}2.1701\times {10}^{-2}\end{array}$

The load voltage in per-unit is calculated using equation (3)

$\begin{array}{rcl}{\mathrm{V}}_{2\mathrm{per}-\mathrm{unit}}& =& \frac{240}{240}\\ & =& 1\angle 0^0\end{array}$

Therefore, the per-unit secondary current is ${\mathrm{I}}_{2\mathrm{per}-\mathrm{unit}}=1\angle -36.{86}^0$.

The single-line diagram is drawn below:

Detremine the voltage at transformer primary terminal.

(a)

Determine the voltage at transformer primary terminal.

Substitute $1\angle 0^0$ for $V_{2per-unit}$, $8.6806\times {10}^{-2}+j2.1701\times {10}^{-2}$for ${\mathrm{Z}}_{\mathrm{eq}\left(\mathrm{per}-\mathrm{unit}\right)}$and $1\angle -36.{86}^0$ for ${\mathrm{I}}_{2\mathrm{per}-\mathrm{unit}}$ in equation (4).

$\begin{array}{rcl}{\mathrm{V}}_{1\mathrm{per}-\mathrm{unit}}& =& 1\angle 0^0+\left(8.6806\times {10}^{-3}+\mathrm{j}2.1701\times {10}^{-2}\right)\left(1\angle -36.{86}^0\right)\\ & =& 1.02\angle 0.{683}^0\mathrm{pu}\end{array}$

Determine the original terminal voltage.

Substitute $1.02\angle 0.{683}^0$ for ${\mathrm{V}}_{1\mathrm{per}-\mathrm{unit}}$ and $2400\mathrm{V}$for ${\mathrm{V}}_{\mathrm{base}1}$ in equation (5).

$\begin{array}{rcl}{\mathrm{V}}_1& =& \left(1.02\angle 0.{683}^0\right)\left(2400\mathrm{V}\right)\\ & =& 2448\angle 0.{683}^0\end{array}$

Therefore, the voltage at the primary terminal is $2448\angle 0.{683}^0$.

Detremine the voltage at transformer primary terminal.

(b)

Determine the source voltage.

Substitute $1.02\angle 0.{683}^0\mathrm{pu}$for ${\mathrm{V}}_{1\mathrm{per}-\mathrm{unit}}$,$8.6806\times {10}^{-3}+\mathrm{j}1.7361\times {10}^{-2}$for ${\mathrm{Z}}_{\mathrm{feed}\left(\mathrm{per}-\mathrm{unit}\right)}$and $1\angle -36.{86}^0$ for ${\mathrm{I}}_{2\mathrm{per}-\mathrm{unit}}$in equation (4).

$\begin{array}{rcl}{\mathrm{V}}_{\mathrm{s}\left(\mathrm{per}-\mathrm{unit}\right)}& =& {\mathrm{V}}_{1\mathrm{per}-\mathrm{unit}}+{\mathrm{I}}_{2\mathrm{per}-\mathrm{unit}}{\mathrm{Z}}_{\mathrm{feed}\left(\mathrm{per}-\mathrm{unit}\right)}\\ & =& 1.02\angle 0.{683}^0+\left(8.6806\times {10}^{-3}+1.7361\mathrm{j}\times {10}^{-2}\right)\left(1\angle -36.{86}^0\right)\\ & =& 1.037\angle 1.{1505}^0\mathrm{pu}\end{array}$

Determine the original source voltage.

Substitute$1.037\angle 1.{1505}^0\mathrm{pu}$for ${\mathrm{V}}_{\mathrm{sper}-\mathrm{unit}}$ and $2400\mathrm{V}$ for ${\mathrm{V}}_{\mathrm{base}1}$ in equation (5).

$\begin{array}{rcl}{\mathrm{V}}_{\mathrm{s}}& =& \left({\mathrm{V}}_{\mathrm{sper}-\mathrm{unit}}\right)\left({\mathrm{V}}_{\mathrm{base}1}\right)\\ & =& \left(1.037\angle 1.{1505}^0\mathrm{pu}\right)\left(2400\mathrm{V}\right)\\ & =& 2490\angle 1.{1505}^0\mathrm{V}\end{array}$

Therefore, the voltage at the sending end is $2490\angle 1.{1505}^0\mathrm{V}$.

Detremine the real and reactive power supplied by source.

(c)

Determine the apparent power

Substitute $1.037\angle 1.{1505}^0$for ${\mathrm{V}}_{\mathrm{sper}-\mathrm{unit}}$ and $1\angle -36.{86}^0$for ${\mathrm{I}}_{2\mathrm{per}-\mathrm{unit}}$in equation (6).

$\begin{array}{rcl}{\mathrm{S}}_{\mathrm{pu}}& =& \left({\mathrm{V}}_{\mathrm{sper}-\mathrm{unit}}\right)\left({\mathrm{I}}_{2\mathrm{per}-\mathrm{unit}}\right)\\ & =& \left(1.037\angle 1.{1505}^0\right)\left(1\angle 36.{86}^0\right)\\ & =& 1.037\angle 38.{02}^0\mathrm{pu}\end{array}$

Determine the original source power.

Substitute role="math" localid="1655293594611" $1.037\angle 38.{02}^0\mathrm{pu}$for ${\mathrm{S}}_{\mathrm{pu}}$ and $50\mathrm{kVA}$ for ${\mathrm{S}}_{\mathrm{base}}$ in equation (7).

$\begin{array}{rcl}\mathrm{S}& =& \left(50\mathrm{kVA}\right)\left(1.037\angle 38.{02}^0\mathrm{pu}\right)\\ & =& 50\angle 38.{02}^0\mathrm{kVA}\\ & =& 40.87\mathrm{kW}+\mathrm{j}31.95\mathrm{kVAR}\end{array}$