Question

The following data are obtained when open-circuit and short-circuit tests are performed on a single-phase , $\mathbf{50}\mathbf{}\mathbf{kVA}$, $\mathbf{2400}\mathbf{/}\mathbf{240}\mathbf{}\mathbf{V}$, $\mathbf{60}\mathbf{}\mathbf{MHz}$distribution transformer.

	Voltage (volts)	Current (amperes)	Power (watts)
Measurements on low-voltage side with high-voltage winding open	240	4.85	173
Measurements on high-voltage side with low-voltage winding shorted	52	20.8	650

(a) Neglecting the series impedance, determine the exciting admittance referred to the high-voltage side. (b) Neglecting the exciting admittance, determine the equivalent series impedance referred to the high-voltage side. (c) Assuming equal series impedances for the primary and referred secondary, obtain an equivalent T-circuit referred to the high-voltage side.

Short answer

1. The admittance is $\left(2.02\times {10}^{-4}+\mathrm{j}3.003\times {10}^{-5}\right)\mathrm{S}$.
2. The series impedance is $\left(1.502+\mathrm{j}1.998\right)\mathrm{\Omega }$.
3. The T-circuit is,

Step-by-step solution

Determine the formulas of turns ratio, shunt admittance, conductance, susceptance, leakage reactance, equivalent resistance, series impedance,

Write the formula of turns ratio.

$\mathrm{a}=\frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}$ ……. (1)

Write the formula of admittance refered to primary

$\mathrm{Y}=\frac{{\mathrm{I}}_2}{{\mathrm{V}}_2}\frac{1}{{\mathrm{a}}^2}$ ……. (2)

Write the formula of series impedance.

${\mathrm{G}}_{\mathrm{c}}=\frac{{\mathrm{P}}_{\mathrm{c}}}{{\left({\mathrm{V}}_2\mathrm{a}\right)}^2}$ ……. (3)

Write the formula of susceptance.

${\mathrm{B}}_{\mathrm{m}}=\sqrt{{\mathrm{Y}}^2-{{\mathrm{G}}_{\mathrm{c}}}^2}$ ……. (4)

Write the formula of equivalent impedance

${\mathrm{Z}}_{\mathrm{eq}}=\frac{{\mathrm{V}}_1}{{\mathrm{I}}_1}$ ……. (5)

Write the formula of equivalent resistance

${\mathrm{R}}_{\mathrm{eq}}=\frac{{\mathrm{P}}_{\mathrm{cu}}}{{{\mathrm{I}}_1}^2}$ ……. (6)

Write the formula of leakage reactance.

${\mathrm{X}}_{\mathrm{eq}}=\sqrt{{{\mathrm{Z}}_{\mathrm{eq}}}^2-{{\mathrm{R}}_{\mathrm{eq}}}^2}$ ……. (7)

Determine the conductance, admittance and susceptance of middle branch in a transformer.

(a)

Determine turns ratio.

Substitute $2400$for ${\mathrm{N}}_1$and $240$for ${\mathrm{N}}_2$in equation (1).

$\begin{array}{rcl}\mathrm{a}& =& \frac{2400}{240}\\ & =& 10\end{array}$

Determine the admittance reffered to primary winding.

Substitute $4.85\mathrm{A}$for ${\mathrm{I}}_2$, $240\mathrm{V}$ for ${\mathrm{V}}_2$and $10$for $\mathrm{a}$in equation (2)

$$\begin{gathered} \mathrm{Y}=\frac{4.85}{240}\times \frac{1}{{10}^2} \\ =2.02\times {10}^{-4}\mathrm{S} \end{gathered}$$

Determinethe conductance.

Substitute $173\mathrm{W}$for ${\mathrm{P}}_{\mathrm{c}}$, $240\mathrm{V}$for ${\mathrm{V}}_2$and $10$for $\mathrm{a}$in equation (3)

$$\begin{gathered} {\mathrm{G}}_{\mathrm{c}}=\frac{173}{{\left(240\right)}^2}\times \frac{1}{{10}^2} \\ =3.003\times {10}^{-5}\mathrm{S} \end{gathered}$$

Determinethe susceptance.

Substitute $3.003\times {10}^{-5}\mathrm{S}$for ${\mathrm{G}}_{\mathrm{c}}$and role="math" localid="1655209613515">2.02×10-4Sfor $\mathrm{Y}$in equation (4).

$\begin{array}{rcl}{\mathrm{B}}_{\mathrm{m}}& =& \sqrt{{\left(2.02\times {10}^{-4}\mathrm{S}\right)}^2-{\left(3.003\times {10}^{-5}\mathrm{S}\right)}^2}\\ & =& 1.998\times {10}^{-4}\mathrm{S}\end{array}$

Determine the series impedance, leakage reactance and winding resistance.

(b)

Determinethe series impedance.

Substitute $20.8\mathrm{A}$for ${\mathrm{I}}_1$and $52\mathrm{V}$ for ${\mathrm{V}}_1$ in equation (5)

$\begin{array}{rcl}{Z}_{eq}& =& \frac{52}{20.8}\\ & =& 2.5\Omega \end{array}$

Determine the winding resistance.

Substitute $650\mathrm{W}$for ${\mathrm{P}}_{\mathrm{cu}}$ and $20.8\mathrm{A}$for ${\mathrm{I}}_1$in equation (6)

$\begin{array}{rcl}{\mathrm{R}}_{\mathrm{eq}}& =& \frac{650}{{\left(20.8\right)}^2}\\ & =& 1.502\mathrm{\Omega }\end{array}$

Determinethe leakage reactance.

Substitute $1.502\mathrm{\Omega }$for ${\mathrm{R}}_{\mathrm{eq}}$and $2.5\mathrm{\Omega }$for ${\mathrm{Z}}_{\mathrm{eq}}$in equation (7).

$\begin{array}{rcl}\mathrm{Y}& =& \sqrt{{\left(2.5\right)}^2-{\left(1.502\right)}^2}\\ & =& 1.998\mathrm{\Omega }\end{array}$

Determine an equivalent T-circuit.

(c)

From the given data, the primary and secondary side, winding resistance and leakage resistance is same.

Therefore, the primary and secondary winding resistance is,

$\begin{array}{rcl}{\mathrm{R}}_{\mathrm{eq}1}& =& {\mathrm{R}}_{\mathrm{eq}2}\\ {\mathrm{R}}_{\mathrm{eq}1}& =& \frac{{\mathrm{R}}_{\mathrm{eq}}}{2}\\ & =& \frac{1.502}{2}\\ & =& 0.751\mathrm{\Omega }\end{array}$

The primary and secondary leakage reactance is,

$\begin{array}{rcl}{\mathrm{X}}_{\mathrm{eq}1}& =& {\mathrm{X}}_{\mathrm{eq}2}\\ {\mathrm{X}}_{\mathrm{eq}1}& =& \frac{{\mathrm{X}}_{\mathrm{eq}}}{2}\\ & =& \frac{1.998}{2}\\ & =& 0.999\mathrm{\Omega }\end{array}$

The equivalent T-circuit is drawn below: