Question

Reconsider Problem 3.64 with the change that now $T_b$includes both a transformer of the same turns ratio as $T_a$and a regulating transformer with a$4°$ phase shift. On the base of$T_a$ the impedance of the two components of $T_b$per unit. Determine the complex power in per unit transmitted to the load through each transformer. Comment on how the transformers share the real and reactive powers.

Short answer

The power transferred by the transformer $T_a$and $T_b$are $0.05+j0.3\text{\hspace{0.17em}pu}$and $0.12+j0.27\text{\hspace{0.17em}pu}$respectively.Comparing the complex power of both the transformers, the transformer $T_b$transferred more real and reactive power compare to transformer $T_a$.

Step-by-step solution

Write the given data by the question:

The nominal ratio of the transformer $T_a\&T_b$is the same.

The reactance of the transformer$T_a$ and $T_b$, $X=j0.1\text{\hspace{0.17em}pu}$.

The load impedance $Z_L=0.8+j0.6\text{\hspace{0.17em}pu}$.

The voltage, $V_2=1\angle 0°\text{\hspace{0.17em}pu}$.

The tap setting of the transformer $C=1\angle 4°$.

Transformer $T_b$has a voltage magnitude Step-up toward the load of times$1.05$ that of $T_a$.

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

The expression to calculate the value of is given by,

$\Delta V=C-1$ …… (1)

Here, C represent the tap setting of the transformer.

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

$I_L=\frac{V_2}{Z_L}$ …… (2)

The expression to calculate the current $I_a$ is given by,

$I_a=\frac{V_1-V_2}{X}$ …… (3)

The expression to calculate the current $I_b$is given by,

…… (4)

$I_b=\frac{V_1+\Delta V-V_2}{X}$

The expression for the load current is given by,

$I_L=I_a+I_b$ …… (5)

The expression to calculate the complex power transferred by the transformer $T_a$ is given by,

$S_a=V_2{I}_a^{*}$ …… (6)

The expression to calculate the complex power transferred by the transformer is given by,

$S_b=V_2{I}_b^{*}$…… (7)

Calculate the complex power transferred by the transformer to load.

Consider the circuit when the switch is closed.

Calculate the value of the voltage $\Delta V$.

Substitute $1\angle 4°$for into equation (1).

$\begin{array}{l}\Delta V=1\angle 4°-1\\ \Delta V=0.07\angle 92°\text{\hspace{0.17em}V}\end{array}$

Calculate the load current $I_L$.

Substitute$1\angle 0°\text{\hspace{0.17em}pu}$for$V_2$and $0.8+j0.6\text{\hspace{0.17em}pu}$for$Z_L$into equation (2).

$\begin{array}{l}{I}_L=\frac{1\angle 0°}{0.8+j0.6}\\ I_L=0.8-j0.6\text{\hspace{0.17em}pu}\end{array}$

Calculate the current $I_a$.

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$for $V_2$and $V_2$for $X$into equation (3).

$I_a=\frac{V_1-1\angle 0°}{j0.1}$

Calculate the current$I_b$

Substitute$1\angle 0°\text{\hspace{0.17em}pu}$for $1\angle 0°\text{\hspace{0.17em}pu}$and $0.07\angle 92°\text{\hspace{0.17em}pu}$for $\Delta V$into equation (4).

$\begin{array}{l}{I}_b=\frac{V_1+0.07\angle 92°-1\angle 0°}{j0.1}\\ I_b=\frac{V_1-1.005\angle 176°}{j0.1}\end{array}$

Calculate voltage $V_1$.

Substitute$0.8-j0.6\text{\hspace{0.17em}pu}$for $I_L$,$\frac{V_1-1\angle 0°}{j0.1}$for $I_a$and$\frac{V_1-1.005\angle 176°}{j0.1}$for$I_b$into equation (5).

$\begin{array}{l}0.8-j0.6=\frac{V_1-1\angle 0°}{j0.1}+\frac{V_1-1.005\angle 176°}{j0.1}\\ 0.06+j0.08=2V_1-2\angle 178°\\ 2V_1=0.1\angle 53.13°-2\angle 178°\\ 2V_1=2.059\angle 0.284°\end{array}$

Solve further as,

$\begin{array}{l}{V}_1=1.029\angle 0.0284°\\ V_1=1.03+j0.005\text{\hspace{0.17em}pu}\end{array}$

Calculate the current$I_a$

$1.03+j0.005\text{\hspace{0.17em}pu}$for$V_1$,$1\angle 0°\text{V}$for $V_2$and$j0.1\text{\hspace{0.17em}pu}$for $X$into equation (3),

$\begin{array}{l}{I}_a=\frac{1.03+j0.005-1}{j0.1}\\ I_a=\frac{0.03+j0.005}{j0.1}\\ I_a=0.05-j0.3\text{\hspace{0.17em}pu}\end{array}$

Calculate the current $I_b$.

Substitute $1.03+j0.005\text{\hspace{0.17em}pu}$for$V_1$,$1\angle 0°\text{\hspace{0.17em}pu}$for $V_2$and $0.07\angle 92°\text{\hspace{0.17em}pu}$for$\Delta V$into equation (4).

$\begin{array}{l}{I}_b=\frac{1.03+j0.005+0.07\angle 92°-1\angle 0}{j0.1}\\ I_b=\frac{1.005+j0.04+1.005\angle 176°}{j0.1}\\ I_b=\frac{0.027+j0.012}{j0.1}\\ I_b=0.12-j0.27\text{\hspace{0.17em}pu}\end{array}$

Calculate the complex power transferred by the transformer$T_a$

Substitute$1\angle 0°\text{V}$for $V_2$and$0.05-j0.3\text{\hspace{0.17em}pu}$for $I_a$into equation (6),

$\begin{array}{l}{S}_2=1\angle 0°{\left(0.05-j0.3\text{\hspace{0.17em}pu}\right)}^{*}\\ S_2=1\left(0.05+j0.3\text{\hspace{0.17em}pu}\right)\\ S_2=0.05+j0.3\text{\hspace{0.17em}pu}\end{array}$

Calculate the complex power transferred by the transformer$T_b$

Substitute$1\angle 0°\text{V}$for $V_2$and$0.12-j0.27\text{\hspace{0.17em}pu}$for $I_b$into equation (7),

$\begin{array}{l}{S}_2=1\angle 0°{\left(0.12-j0.27\text{\hspace{0.17em}pu}\right)}^{*}\\ S_2=1\left(0.12+j0.27\text{\hspace{0.17em}pu}\right)\\ S_2=0.12+j0.27\text{\hspace{0.17em}pu}\end{array}$

Hence the power transferred by the transformer $T_a$and $T_b$are $0.05+j0.3\text{\hspace{0.17em}pu}$and $0.12+j0.27\text{\hspace{0.17em}pu}$respectively.

Comparing the complex power of both the transformers, the transformer $T_b$transferred more real and reactive power compare to transformer $T_a$.