Question

Rework Problems 12.15 through 12.16 when the load in area 2 suddenly decreases by $\mathbf{300}\mathbf{MW}$. The load in area 1 does not change.

Short answer

The steady state frequency error is$0.3\text{\hspace{0.17em}Hz}$ . The steady state tie line error for area 1 and 2 are$-180\text{\hspace{0.17em}MW}$ and $180\text{\hspace{0.17em}MW}$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,${\beta }_1=400\text{\hspace{0.17em}MW}/\text{Hz}$

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, $\Delta p_{m1}+\Delta p_{m2}=-300\text{\hspace{0.17em}MW}$

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

The equation to calculate the mechanical power in area 1 is given as follows.

${\mathbf{\Delta p}}_{\mathbf{m}\mathbf{1}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{ref}\mathbf{1}}\mathbf{-}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{\Delta f}$ …… (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 Δptieof each area.  

Calculate the mechanical power in area 1.

Substitute$0$for$\Delta p_{ref1}$into equation (1).

role="math" localid="1656089558163" $\begin{array}{l}\Delta p_{m1}=0-{\beta }_1\Delta f\\ \Delta p_{m1}=-{\beta }_1\Delta f\end{array}$ …… (5)

Calculate the mechanical power in area 2.

Substitute$0$for$AC{E}_2$, $\Delta p_{m2}$and$\Delta p_{tie2}$for into equation (2).

…… (6)

Add equation (5) and (6).

Substitute $-300\text{\hspace{0.17em}MW}$for $\Delta p_{m1}+\Delta p_{m2}$, $400\text{\hspace{0.17em}MW}/\text{Hz}$for ${\beta }_1$,and $600\text{\hspace{0.17em}MW}/\text{Hz}$for ${\beta }_2$into above equation.

$\begin{array}{l}\Delta f=-\frac{-300}{400+600}\\ \Delta f=\frac{300}{1000}\\ \Delta f=0.3\text{\hspace{0.17em}Hz}\end{array}$

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

Substitute$600\text{\hspace{0.17em}MW}/\text{Hz}$ for ${\beta }_{f2}$and$0.3\text{\hspace{0.17em}Hz}$ for$\Delta f$into equation (3).

$\begin{array}{l}\Delta p_{tie2}=0-600\times (-0.3)\\ \Delta p_{tie2}=-180\text{\hspace{0.17em}MW}\end{array}$

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

Substitute$-180\text{\hspace{0.17em}MW}$ for$\Delta p_{tie2}$ into equation (4).

$\begin{array}{l}\Delta p_{tie1}=-(-180)\\ \Delta p_{tie1}=180\text{\hspace{0.17em}MW}\end{array}$

Hence the steady state frequency error is $0.3\text{\hspace{0.17em}Hz}$. The steady state tie line error for area 1 and 2 are$-180\text{\hspace{0.17em}MW}$ and$180\text{\hspace{0.17em}MW}$ respectively.