Question

Repeat Problem 12.14 if LFC is employed in both areas. The frequency bias coefficients are ${\mathbf{\beta }}_{\mathbf{f}\mathbf{1}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}\mathbf{400}\mathbf{MW}\mathbf{/}\mathbf{Hz}$ and${\mathbf{\beta }}_{\mathbf{f}\mathbf{2}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{600}\mathbf{MW}\mathbf{/}\mathbf{Hz}$ .

Short answer

The steady state frequency error is $0\text{\hspace{0.17em}Hz}$. The steady state tie line error for area 1 and 2 are$0\text{\hspace{0.17em}MW}$ and$0\text{\hspace{0.17em}MW}$ respectively.

Step-by-step solution

Write the given data from the question.

The generation of the area 1 is$1200\text{\hspace{0.17em}MW}$ .

The area frequency response characteristic of area 1,

$\begin{array}{c}{\beta }_{f1}={\beta }_1\\ {\beta }_1=400\text{\hspace{0.17em}MW}/\text{Hz}\end{array}$

The generation of the area 2 is $1800\text{\hspace{0.17em}MW}$.

The area frequency response characteristic of area 2,

$\begin{array}{l}{\beta }_{f2}={\beta }_2\\ {\beta }_{f2}=600\text{\hspace{0.17em}MW}/\text{Hz}\end{array}$

The frequency of the system,$f=60\text{\hspace{0.17em}Hz}$

The load increase in area 1, $\Delta p_{m1}+\Delta p_{m2}=400\text{\hspace{0.17em}MW}$

Determine the equation to calculate the steady-state frequency error and the steady-state tie-line error Δptieof each area.  

The equation to calculate the area control error in area 1 is given as follows.

${\mathbf{ACE}}_{\mathbf{1}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{tie}\mathbf{1}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{f}\mathbf{1}}\mathbf{\Delta f}$ …… (1)

Here $\mathbf{\Delta f}$is the steady state frequency error.

The equation to calculate the area control error in area 2 is given as follows.

${\mathbf{ACE}}_{\mathbf{2}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{tie}\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{f}\mathbf{2}}\mathbf{\Delta f}$ …… (2)

The equation to calculate the steady state tie-line error of area 2 is given as follows.

${\mathbf{\Delta p}}_{\mathbf{tie}\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{\beta }}_{\mathbf{f}\mathbf{2}}\mathbf{\Delta f}$ …… (3)

The equation to calculate the steady state tie-line error of area 1 is given as follows.

${\mathbf{\Delta p}}_{\mathbf{tie}\mathbf{1}}\mathbf{=}\mathbf{-}{\mathbf{\Delta p}}_{\mathbf{tie}\mathbf{2}}$ …… (4)

Calculate the steady-state frequency error and the steady-state tie-line error Δptie of each area.  

Calculate the mechanical power in area 1.

Substitute$0$for${\text{ACE}}_{\text{1}}$into equation (1).

$\begin{array}{c}0=\Delta p_{tie1}+{\beta }_{f1}\Delta f\\ \Delta p_{tie1}=-{\beta }_{f1}\Delta f\end{array}$ …… (5)

Calculate the mechanical power in area 1.

Substitute$0$for${\text{ACE}}_{\text{2}}$into equation (2).

$\begin{array}{c}0=\Delta p_{tie2}+{\beta }_{f2}\Delta f\\ \Delta p_{tie2}=-{\beta }_{f2}\Delta f\end{array}$ …… (6)

Add equation (5) and (6).

The sum of the net tie line power is equal to zero.

Substitute$0$for $\Delta p_{tie1}+\Delta p_{tie2}$, $400\text{\hspace{0.17em}MW}/\text{Hz}$for${\beta }_1$,and$600\text{\hspace{0.17em}MW}/\text{Hz}$ for${\beta }_2$into above equation.

Calculate the steady state tie-line error of area 2.

Substitute $600\text{\hspace{0.17em}MW}/\text{Hz}$for ${\beta }_{f2}$and $\text{0\hspace{0.17em}Hz}$for $\Delta f$into equation (3).

$\begin{array}{l}\Delta p_{tie2}=0-600\times \left(0\right)\\ \Delta p_{tie2}=0\text{\hspace{0.17em}MW}\end{array}$

Calculate the steady state tie-line error of area 1.

Substitute$0\text{\hspace{0.17em}MW}$ for$\Delta p_{tie2}$ into equation (4).

$\Delta p_{tie1}=0\text{\hspace{0.17em}MW}$

Hence the steady state frequency error is $0\text{\hspace{0.17em}Hz}$. The steady state tie line error for area 1 and 2 are $0\text{\hspace{0.17em}MW}$and$0\text{\hspace{0.17em}MW}$ respectively.