Question

A single-phase $\mathbf{100}\mathbf{}\mathit{k}\mathit{V}\mathit{A}$, $\mathbf{2400}\mathbf{/}\mathbf{240}\mathbf{}\mathit{v}\mathit{o}\mathit{l}\mathit{t}\mathit{s}$, $\mathbf{60}\mathbf{}\mathit{H}\mathit{z}$distribution transformer is used as a step-down transformer. The load, which is connected to the $\mathbf{240}\mathbf{}\mathit{v}\mathit{o}\mathit{l}\mathit{t}\mathit{s}$secondary winding, absorbs role="math" localid="1655831901768" $\mathbf{60}\mathbf{}\mathit{k}\mathit{V}\mathit{A}$at 0.8 power factor lagging and is at $\mathbf{230}\mathbf{}\mathit{v}\mathit{o}\mathit{l}\mathit{t}\mathit{s}$. Assuming an ideal transformer, calculate the following (a) Primary voltage, (b) load impedance, (c) load impedance referred to the primary, and (d) the real and reactive power supplied to the primary winding.

Short answer

Answer

(a)The primary voltage is $2300V$.

(b)The load impedance is $0.881\Omega$

(c)The load impedance referred to the primary is $88.1\Omega$

(d)The real and reactive power is $48kW$and $36kVAR$.

Step-by-step solution

Determine the formulas of transformer ratio, load impedance, active power, reactive power and apparent power.

Write the formula of transformer ratio.

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

Write the formula of apparent power

$\mathit{S}\mathbf{=}{\mathit{V}}_{\mathbf{2}}\left(\frac{V_2}{Z_2}\right)$ ……. (2)

Write the formula of load impedance referred to the primary

……. (3)

Write the formula of active power

$\mathit{P}\mathbf{=}\mathit{S}\mathbf{·}\mathit{p}\mathit{f}$ ……. (4)

Write the formula of active power

$\mathit{Q}\mathbf{=}\mathit{S}\mathbf{·}\mathit{s}\mathit{i}\mathit{n}\left(co{s}^{-1}pf\right)$ ……. (5)

Determine the primary voltage.

(a)

Determine the primary voltage.

Substitute 230 V for $V_2$, 2400 for $N_1$and 240 for $N_2$ in equation (1).

$$\begin{gathered} \frac{230}{V_1}=\frac{240}{24000} \\ {V}_1=2300V \end{gathered}$$

Determine the load impedance.

(b)

Determine the load impedance

Substitute 60 kVA for S and 230 V for $V_2$in equation (2).

$$\begin{gathered} 60\times {10}^3=230\times \left(\frac{230}{Z_2}\right) \\ {Z}_2=\frac{230\times 230}{60\times {10}^3} \\ {Z}_2=0.881\Omega \end{gathered}$$

Determine the secondary load impedance referred to the primary.

(c)

Determine the load impedance referred to primary.

Substitute $0.881\Omega$for $Z_2$, 2400 for $N_1$and 240 for $N_2$ in equation (3).

localid="1655832830214" $$\begin{gathered} {Z_2}^{\text{'}}=0.881\Omega \times {\left(\frac{2400}{240}\right)}^2 \\ {Z_2}^{\text{'}}=88.1\Omega \end{gathered}$$

Determine the real and reactive power.

(d)

As the ideal transformer is considered, the real and reactive powers of primary and secondary windings will be same.

Determine the active power at primary side.

Substitute 60 kVA for S and 0.8 for $p·f$in equation (4).

$$\begin{gathered} P=60\times 0.8 \\ =48kW \end{gathered}$$

Determine the reactive power at primary side.

Substitute 60 kVA for S and 0.8 for $p·f$in equation (5).

$$\begin{gathered} Q=60\times \sin \left({\cos }^{-1}0.8\right) \\ =60\times 0.6 \\ =36kVAR \end{gathered}$$