Chapter 3: Q64P (page 158)

The equivalent circuit of two transformers $T_{a}$ and $T_{b}$ connected in parallel, With the same nominal voltage ratio and the same reactance of $0.1$ per unit on the same base, is shown in Figure 3.43. Transformer $T_{b}$ has a voltage magnitude Step-up toward the load of $1.05$ times that of $T_{a}$ (that is, the tap on the secondary winding ofis set to $1.05$ ). The load is represented by $0.8 + j 0.6$ unit at a voltage $V_{2} = 1.0 ∠ 0 °$ per unit. Determine the complex power in per unit transmitted to the load througheach transformer. Comment on how the transformers share the real and reactive powers.

Short Answer

Expert verified

The power transferred by the transformer $T_{a}$ and $T_{b}$ are $0.4 + j 0.05 pu$ and $0.4 + j 0.55 pu$ respectively.

The active power transferred by both the transformer is the same but the reactive power transferred by the transformer is $T_{b}$ more than $T_{a}$

Step by step solution

Write the given data by the question:

The nominal ratio of the transformer $T_{a}$ and $T_{b}$ is the same.

The reactance of the transformer $T_{a} & T_{b}$ , $X = j 0.1 pu$ .

The load impedance $Z_{L} = 0.8 + j 0.6 pu$ .

The voltage is $Z_{L} = 0.8 + j 0.6 pu$ .

Transformer $T_{b}$ has a voltage magnitude Step-up toward the load of $1.05$ times that of $T_{a}$ .

Determine the formulas to calculate the complex power delivered to the load by each transformer.

The expression to calculate the load current when the switch is closed is given by,

$I_{L} = \frac{V_{2}}{Z_{L}}$ $I_{L} = \frac{V_{2}}{Z_{L}}$ …… (1)

The expression to calculate the current is $I_{a}$ given by,

$I_{a} = \frac{V_{1} - V_{2}}{X}$ …… (2)

The expression to calculate the current is $I_{b}$ given by,

$I_{b} = \frac{V_{1} + Δ V - V_{2}}{X}$ …… (3)

The expression for the load current is given by,

$I_{L} = I_{a} + I_{b}$ …… (4)

The expression to calculate the complex power transferred by the transformer $T_{a}$ is given by,

$S_{a} = V_{2} I_{a}^{*}$ …… (5)

The expression to calculate the complex power transferred by the transformer $T_{b}$ is given by,

$S_{b} = V_{2} I_{b}^{*}$ …… (6)

Calculate the complex power transferred by the transformer to load.

Consider the circuit when the switch is closed.

Calculate the load current $I_{L}$ .

Substitute $1 ∠ 0 ° pu$ for $V_{2}$ and $0.8 + j 0.6 pu$ for $Z_{L}$ into equation (1),

$\begin{array}{l} I_{L} = \frac{1 ∠ 0 °}{0.8 + j 0.6} \\ I_{L} = 0.8 - j 0.6 pu \end{array}$

Calculate the current $I_{a}$ .

Substitute $1 ∠ 0 ° pu$ for $V_{2}$ and $j 0.1 pu$ for $X$ into equation (2).

$I_{a} = \frac{V_{1} - 1 ∠ 0 °}{j 0.1}$

Calculate the current $I_{b}$ .

Substitute $1 ∠ 0 ° pu$ for $V_{2}$ and $0.05 ∠ 0 ° pu$ for $Δ V$ into equation (3).

$\begin{array}{l} I_{b} = \frac{V_{1} + 0.05 ∠ 0 ° - 1 ∠ 0 °}{j 0.1} \\ I_{b} = \frac{V_{1} - 0.95 ∠ 0 °}{j 0.1} \end{array}$

Calculate voltage $V_{1}$ .

Substitute $0.8 - j 0.6 pu$ for $I_{L}$ , $I_{a} = \frac{V_{1} - 1 ∠ 0 °}{j 0.1}$ for $I_{a} = \frac{V_{1} - 1 ∠ 0 °}{j 0.1}$ and $I_{b} = \frac{V_{1} - 0.95 ∠ 0 °}{j 0.1}$ for into equation (4).

$\begin{matrix} 0.8 - j 0.6 = \frac{V_{1} - 1 ∠ 0 °}{j 0.1} + \frac{V_{1} - 0.95 ∠ 0 °}{j 0.1} \\ 0.06 + j 0.08 = 2 V_{1} - 1.95 \\ 2 V_{1} = 0.06 + j 0.08 + 1.95 \\ 2 V_{1} = 2.01 + j 0.08 \end{matrix}$

Solve further as,

$\begin{array}{l} V_{1} = \frac{2.01 + j 0.08}{2} \\ V_{1} = 1.005 + j 0.04 pu \end{array}$

Calculate the current $I_{a}$

Substitute $1.005 + j 0.04 pu$ for $V_{1}$ , $1 ∠ 0 ° V$ for $V_{2}$ and $j 0.1 pu$ for $X$ into equation (2).

$\begin{array}{l} I_{a} = \frac{1.005 + j 0.04 - 1}{j 0.1} \\ I_{a} = 0.4 - j 0.05 pu \end{array}$

Calculate the current $I_{b}$ .

Substitute $1.005 + j 0.04 pu$ for $V_{1}$ , $1 ∠ 0 ° pu$ for $V_{2}$ and $0.05 ∠ 0 ° pu$ for $Δ V$ into equation (3).

$\begin{array}{l} I_{b} = \frac{1.005 + j 0.04 + 0.05 ∠ 0 ° - 1 ∠ 0}{j 0.1} \\ I_{b} = \frac{1.005 + j 0.04 + 0.95}{j 0.1} \\ I_{b} = \frac{0.055 + j 0.04}{j 0.1} \\ I_{b} = 0.4 - j 0.55 pu \end{array}$

Calculate the complex power transferred by the transformer $T_{a}$ .

Substitute $1 ∠ 0 ° V$ for $V_{2}$ and $0.4 - j 0.05 pu$ for $I_{a}$ into equation (5).

$\begin{array}{l} S_{2} = 1 ∠ 0 ° {(0.4 - j 0.05)}^{*} \\ S_{2} = 1 (0.4 + j 0.05) \\ S_{2} = 0.4 + j 0.05 pu \end{array}$

Calculate the complex power transferred by the transformer $T_{b}$ .

Substitute $1 ∠ 0 ° V$ for $V_{2}$ and $0.4 - j 0.55 pu$ for $I_{b}$ into equation (6).

$\begin{array}{l} S_{2} = 1 ∠ 0 ° {(0.4 - j 0.55)}^{*} \\ S_{2} = 1 (0.4 + j 0.55) \\ S_{2} = 0.4 + j 0.55 pu \end{array}$

Hence the power transferred by the transformer $T_{a}$ and $T_{b}$ are $0.4 + j 0.05 pu$ and $0.4 + j 0.55 pu$ respectively.

From the above, the active power transferred by both the transformer is the same but the reactive power transferred by the transformer $0.4 + j 0.55 pu$ is more than $T_{a}$ .