Question

A $\mathbf{60}\mathbf{Hz}$ power system consists of two interconnected areas. Area 1 has$\mathbf{1200}\mathbf{MW}$ of generation and an area frequency response characteristic${\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}\mathbf{400}\mathbf{MW}\mathbf{/}\mathbf{Hz}$ . Area 2 has $\mathbf{1800}\mathbf{MW}$of generation and ${\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{600}\mathbf{MW}\mathbf{/}\mathbf{Hz}$. Each area is initially operating at one-half its total generation, at ${\mathbf{\Delta p}}_{\mathbf{tie}\mathbf{1}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{tie}\mathbf{2}}\mathbf{=}\mathbf{0}$and at $\mathbf{60}\mathbf{Hz}$, when the load in area 1 suddenly increases by$\mathbf{400}\mathbf{MW}$ . Determine the steady-state frequency error and the steady-state tie-line error${\mathbf{\Delta p}}_{\mathbf{tie}}$ of each area. Assume that the reference power settings of all turbine-governors are fixed. That is, LFC is not employed in any area. Neglect losses and the dependence of load on frequency.

Short answer

The steady state frequency error is$-0.4\text{\hspace{0.17em}Hz}$ . The steady state tie line error for area 1 and 2 are $240\text{\hspace{0.17em}MW}$ and $-240\text{\hspace{0.17em}MW}$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,${\beta }_2=600\text{\hspace{0.17em}MW}/\text{Hz}$

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

The reference power is fixed, therefore change in reference power setting,

$\begin{array}{l}\Delta p_{ref1}=\Delta p_{ref2}\\ \Delta p_{ref1}=0\end{array}$

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 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 mechanical power in area 2 is given as follows.

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

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

$\begin{array}{l}{\mathbf{\Delta p}}_{\mathbf{tie}\mathbf{2}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{m}\mathbf{2}}\\ {\mathbf{\Delta p}}_{\mathbf{tie}\mathbf{2}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{ref}}\mathbf{-}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{\Delta f}\end{array}$ …… (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 steady-state frequency error

Add the equation (1) and (2).

${\mathrm{\Delta p}}_{\mathrm{m}1}+{\mathrm{\Delta P}}_{\mathrm{m}2}={\mathrm{\Delta p}}_{\mathrm{ref}1}+{\mathrm{\Delta p}}_{\mathrm{ref}2}-({\mathrm{\beta }}_1\mathrm{\Delta f}+{\mathrm{\beta }}_2\mathrm{\Delta f})$

Substitute$0$for${\mathrm{\Delta p}}_{\mathrm{ref}1},$and${\mathrm{\Delta p}}_{\mathrm{ref}2}$into above equation.

Substitute $400\text{\hspace{0.17em}MW}$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.

$\begin{array}{l}\Delta f=-\frac{400}{400+600}\\ \Delta f=-\frac{400}{1000}\\ \Delta f=-0.4\text{\hspace{0.17em}Hz}\end{array}$

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

Substitute $0$for $\Delta p_{ref2}$, $600\text{\hspace{0.17em}MW}/\text{Hz}$for ${\beta }_2$and $-0.4\text{\hspace{0.17em}Hz}$for $∆f$ into equation (3).

$\begin{array}{l}\Delta p_{tie2}=0-600\times (-0.4)\\ \Delta p_{tie2}=240\text{\hspace{0.17em}MW}\end{array}$

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

Substitute$240\text{\hspace{0.17em}MW}$ for$\Delta p_{tie2}$ into equation (4).

$\Delta p_{tie1}=-240\text{\hspace{0.17em}MW}$

Hence the steady state frequency error is $-0.4\text{\hspace{0.17em}Hz}$. The steady state tie line error for area 1 and 2 are $240\text{\hspace{0.17em}MW}$ and$-240\text{\hspace{0.17em}MW}$ respectively.