Question

The following data are obtained when open-circuit and short-circuit tests are performed on a single-phase $\mathbf{50}\mathbf{\text{kVA}}$,$\mathbf{2400}\mathbf{/}\mathbf{240}\mathbf{\text{V}}$,$\mathbf{60}\mathbf{\text{Hz}}$, 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 $(2.02\times {10}^{-4}+j3.003\times {10}^{-5})S$.
2. The series impedance is $(1.502+j1.998)\Omega$.
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.

$\mathbf{a}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{1}}}{{\mathbf{N}}_{\mathbf{2}}}$ ……. (1)

Write the formula of admittance reffered to primary

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

Write the formula of series impedance.

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

Write the formula of susceptance.

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

Write the formula of equivalent impedance

${\mathbf{Z}}_{\mathbf{eq}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{I}}_{\mathbf{1}}}$ ……. (5)

Write the formula of equivalent resistance

${\mathbf{R}}_{\mathbf{eq}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{cu}}}{{{\mathbf{I}}_{\mathbf{1}}}^{\mathbf{2}}}$ ……. (6)

Write the formula of leakage reactance.

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

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

(a)

Determine turns ratio.

Substitute $2400$ for $N_1$ and $240$ for $N_2$ in equation (1).

role="math" localid="1655985316115" $\begin{array}{c}a=\frac{2400}{240}\\ =10\end{array}$

Determinethe admittance reffered to primary winding.

Substitute $4.85\text{\hspace{0.17em}A}$ for $I_2$, $240V$ for $V_2$ and $10$for $a$ in equation (2)

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

Determinethe conductance.

Substitute$173\text{\hspace{0.17em}W}$for $P_C$, $240\text{\hspace{0.17em}V}$ for $v_2$ and $10$ for $a$ in equation (3)

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

Determinethe susceptance.

Substitute$3.003\times {10}^{-5}S$for $G_c$and $2.02\times {10}^{-4}S$ for $Y$ in equation (4).

$\begin{array}{c}{B}_m=\sqrt{{(2.02\times {10}^{-4}S)}^2-{(3.003\times {10}^{-5}S)}^2}\\ =1.998\times {10}^{-4}S\end{array}$

Determine the series impedance, leakage reactance and winding resistance. 

(b)

Determinethe series impedance.

Substitute$20.8\text{\hspace{0.17em}A}$for $I_1$ and $52\text{\hspace{0.17em}V}$ for $V_1$in equation (5)

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

Determine the winding resistance.

Substitute$650\text{\hspace{0.17em}W}$for $P_{cu}$ and $20.8\text{\hspace{0.17em}A}$for $I_1$ in equation (6)

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

Determinethe leakage reactance.

Substitute$1.502\Omega$for $R_{eq}$and $2.5\Omega$ for $Z_{eq}$ in equation (7).

$\begin{array}{c}{X}_{eq}=\sqrt{{\left(2.5\right)}^2-{\left(1.502\right)}^2}\\ =1.998\Omega \end{array}$

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,

The primary and secondary leakage reactance is,

$\begin{array}{c}{X}_{eq1}=X_{eq2}\\ =\frac{X_{eq}}{2}\\ =\frac{1.998}{2}\\ =0.999\Omega \end{array}$

The equivalent T-circuit is drawn below: