Question

The synchronous generator in Figure 11.4 delivers 0.8 per unit real power at 1.05 per unit terminal voltage. Determine: (a) the reactive power output of the generator; (b) the generator internal voltage; and (c) an equation for the electrical power delivered by the generator versus power angle $\mathit{\delta }$.

Short answer

(a)

The value of reactive power output of the generator is 0.306 per unit.

(b)

The value of generator internal voltage is$1.16\angle 21.01°\text{\hspace{0.33em}per\hspace{0.33em}unit}$.

(c)

The value of real power delivered by the generator to the infinite bus is$2.231\sin \delta$.

Step-by-step solution

Write the given data from the question.

As a reference of figure 11.4.

Consider the value of infinite bus receives 0.8 per unit of real power.

Consider the value of infinite bus receives 1.05 per unit terminal voltage.

All the reactances are given in per-unit on a common system base.

Step 2: Determine the formula of ωsyn and ωmsyn, per unit swing equation and load angle.

Write the formula of reactive power output of the generator.

$\mathbf{Q}=\mathbf{Im}\left(S\right)$ …… (1)

Here, I is current flowing in the infinite bus and S is complex power at the terminal voltage.

Write the formula ofgenerator internal voltage.

${\mathbf{E}}^{\mathbf{\text{'}}}\angle \delta ={\mathbf{V}}_{\mathrm{bus}}\mathbf{+}\mathbf{j}\left({X^{\text{'}}}_4\mathbf{+}\mathbf{X}\right)\mathbf{I}$ …… (2)

Here, V_bus is bus voltage, X represent between the terminal voltage and the infinite bus voltage and I is current flowing in the infinite bus.

Write the formula ofreal power delivered by the generator to the infinite bus.

${\mathbf{P}}_e=\frac{{\mathbf{E}}^{\mathbf{\text{'}}}{\mathbf{V}}_{\mathrm{bus}}}{{X^{\text{'}}}_4\mathbf{+}\mathbf{X}}\mathbf{sin}\delta$ …… (3)

Here, E' is generator internal voltage, V_bus is bus voltage, X represent between the terminal voltage and the infinite bus voltage and $\mathit{\delta }$is generator power angle.

(a) Determine the value of reactive power output of the generator.

As illustrated in figure 1, create a single-line schematic of a straight forward generating and distribution unit with a problem.

Draw an impedance diagram that corresponds to the figure 1 diagram.

The equivalent reactance between the terminal voltage and the infinite bus voltage is shown in figure 2.

$X=X_{\text{TR}}+X_{12}\parallel \left(X_{13}+X_{23}\right)$

Here, X_TR represent the transformer reactance; X₁₂ represent the reactance between the bus 1 and bus 2, X₁₃ represent the reactance between the bus 1 and bus 3 and X₂₃ represent the reactance between the bus 2 and bus 3.

Substitute 0.10 for X_TR , 0.20 for X₁₂, 0.10 for X₁₃ and 0.20 for X₂₃ into above equation.

$\begin{array}{c}X=0.10+0.20\parallel \left(0.10+0.20\right)\\ =0.10+0.12\\ =0.22\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Therefore, the equivalent reactance between the terminal voltage and the infinite bus voltage is 0.22 per unit.

Determine the expression for real power P_e flowing by the synchronous generator.

$P_e=\frac{V_T{V}_{bus}}{\text{X}}\sin \delta$

Here, V_T is terminal voltage, V_bus is voltage bus, X represent the reactance between the terminal voltage and the infinite bus voltage and $\delta$is machine power angle.

Substitute 0.8 per unit for P_e , 1.05 per unit for V_T, 1.0 per unit for V_bus and 0.22 per unit for X into above equation.

$\begin{array}{c}0.8=\frac{\left(1.05\right)\left(1.0\right)}{0.22}\sin \delta \\ \sin \delta =0.168\\ \delta ={\sin }^{-1}\left(0.168\right)\\ \delta =9.649°\end{array}$

Therefore, the power angle with respect to the infinite bus $9.649°$.

Write an expression for the current entering the infinite bus.

$I=\frac{V_{\text{T}}-V_{\text{bus}}}{jX}$

Here, V_T is terminal voltage, V_bus is voltage bus, X represent the reactance between the terminal voltage and the infinite bus voltage

Substitute $1.05\angle 9.649°\text{\hspace{0.33em}per\hspace{0.33em}unit}$ for V_T , $1.0\angle 0°\text{\hspace{0.33em}per\hspace{0.33em}unit}$ for V_bus and $0.22\text{\hspace{0.33em}per\hspace{0.33em}unit}$for X into above equation.

role="math" localid="1655970852614" $\begin{array}{c}I=\frac{1.05\angle 9.649°-1.0\angle 0°}{j0.22}\\ =\frac{0.035+j0.176}{j0.22}\\ =0.8-j0.16\\ =0.816\angle -11.29°\text{\hspace{0.33em}per\hspace{0.17em}unit}\end{array}$

Therefore, the current flowing into the infinite bus is $0.816\angle -11.29°\text{\hspace{0.33em}per\hspace{0.33em}unit}$.

Determine the formula of complex power at the terminal voltage.

$S=V_{\text{T}}{I}^{*}$

Here, V_T is terminal voltage and I^* is current entering the infinite bus.

Substitute $1.05\angle 9.649°\text{\hspace{0.33em}per\hspace{0.33em}unit}$for V_T and $0.816\angle 11.29°\text{\hspace{0.33em}per\hspace{0.33em}unit}$for I^* into above equation.

$\begin{array}{c}S=\left(1.05\angle 9.649°\right)\left(0.816\angle 11.29°\right)\\ =0.857\angle 20.94°\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Determine thereactive power output from the complex power.

Substitute $0.857\angle 20.94°\text{\hspace{0.33em}per\hspace{0.33em}unit}$for S into equation (1).

$\begin{array}{c}Q=\left(0.816\angle -11.29°\right)\left(0.857\right)\\ =0.306\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Therefore, the value of reactive power output of the generator is 0.306 per unit.

(b) Determine the value of generator internal voltage.

Determine the expression for generator internal voltage.

Substitute $1.0\angle 0°\text{\hspace{0.33em}per\hspace{0.33em}unit}$ for V_bus, $0.816\angle -11.29°\text{\hspace{0.17em}per\hspace{0.33em}unit}$ for I, $0.22\text{\hspace{0.33em}per\hspace{0.33em}unit}$ for X and $0.30\text{\hspace{0.33em}per\hspace{0.33em}unit}$for X^'_d into equation (2).

$\begin{array}{c}{E}^{\text{'}}\angle \delta =1.0\angle 0°+j\left(0.30+0.22\right)\left(0.816\angle -11.29°\right)\\ =1.0\angle 0°+0.424\angle 78.71°\\ =1.16\angle 21.01°\text{\hspace{0.33em}per\hspace{0.33em}unit}\end{array}$

Therefore, the value of generator internal voltage is $1.16\angle 21.01°\text{\hspace{0.33em}per\hspace{0.33em}unit}$.

(c) Determine the value of value of real power delivered by the generator to the infinite bus.

Determine the expression for real power delivered by generator to the infinite bus.

Substitute $1.16\angle 21.01°\text{\hspace{0.33em}per\hspace{0.33em}unit}$ for E', $1.0\angle 0°\text{\hspace{0.33em}per\hspace{0.33em}unit}$ for V_bus, $0.22\text{\hspace{0.33em}per\hspace{0.33em}unit}$ for X and $\text{0.30\hspace{0.33em}per\hspace{0.33em}unit}$for X^'_d into equation (3).

$\begin{array}{c}{P}_e=\frac{\left(1.160\right)\left(1.0\right)}{0.30+0.22}\sin \delta \\ =\frac{1.160}{0.52}\sin \delta \\ =2.231\sin \delta \end{array}$

Therefore, the value of real power delivered by the generator to the infinite bus is $2.231\sin \delta$.