Question

An interconnected $\mathbf{60}\mathbf{Hz}$ power system consisting of one area has two turbine-generator units, rated$\mathbf{500}$ and $\mathbf{750}\mathbf{MVA}$, with regulation constants of $\mathbf{0}\mathbf{.}\mathbf{04}$and$\mathbf{0}\mathbf{.}\mathbf{06}$ per unit, respectively, based on their respective ratings. When each unit carries a$\mathbf{300}\mathbf{MVA}$ steady-state load, let the area load suddenly increase by$\mathbf{250}\mathbf{MVA}$ . (a) Compute the area frequency response characteristic$\mathbf{\beta }$ on a$\mathbf{1000}\mathbf{MVA}$ base. (b) Calculate$\mathbf{\Delta f}$ in per unit on a$\mathbf{60}\mathbf{Hz}$ base and in Hz.

Short answer

(a) The area frequency response characteristic$\mathrm{\beta }$is$25\text{\hspace{0.17em}pu}$ .

(b) The steady state value of the area frequency in per unit is$0.1\text{\hspace{0.17em}pu}$ and in Hz is $-6\text{\hspace{0.17em}Hz}$.

Step-by-step solution

Write the given data from the question.

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

Write the rating of three turbine generator

$\begin{array}{l}{S}_{1,old}=500\text{\hspace{0.17em}MVA}\\ {\text{S}}_{2,old}=750\text{\hspace{0.17em}MVA}\end{array}$

Write the regulation constant of three generator units

$\begin{array}{l}{R}_{1,old}=0.04\text{\hspace{0.17em}pu\hspace{0.17em}}\\ R_{2,old}=\text{0.06\hspace{0.17em}pu}\end{array}$

The base rating, $S_{new}=1000\text{\hspace{0.17em}MVA}$

The increase value of load, $P_L=250\text{\hspace{0.17em}MW}$

Change in the reference setting power,$\Delta p_{ref}=0$

Determine the equation to calculate the area frequency response characteristic β on a 1000 MVAbase and Δf in per unit on a 60 Hz base and in Hz.

The equation to calculate the new value of the regulation constant for 1^st unit is given as follows.

${\mathbf{R}}_{\mathbf{1}\mathbf{new}}\mathbf{=}{\mathbf{R}}_{\mathbf{1}\mathbf{old}}\frac{{\mathbf{S}}_{\mathbf{new}}}{{\mathbf{S}}_{\mathbf{1}\mathbf{old}}}$ …… (1)

The equation to calculate the new value of the regulation constant for 2^nd unit is given as follows.

${\mathbf{R}}_{\mathbf{2}\mathbf{new}}\mathbf{=}{\mathbf{R}}_{\mathbf{2}\mathbf{old}}\frac{{\mathbf{S}}_{\mathbf{new}}}{{\mathbf{S}}_{\mathbf{2}\mathbf{old}}}$ …… (2)

The equation to calculate the unit area frequency response characteristics is given as follows.

$\mathbf{\beta }\mathbf{=}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}\mathbf{new}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}\mathbf{new}}}$ …… (3)

The equation to calculate the total change in turbine mechanical power in per unit is given as,

${\mathbf{\Delta p}}_{\mathbf{m}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{L}}}{{\mathbf{S}}_{\mathbf{new}}}$ …… (4)

The equation for the steady state response power relation is given as follows.

${\mathbf{\Delta p}}_{\mathbf{m}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{ref}}\mathbf{-}\mathbf{\beta \Delta f}$ …… (5)

Here $\mathbf{\Delta f}$ is the steady state increase in the area frequency.

The equation to calculate the steady state decrease in the value of frequency in hertz is given as follows.

${\mathbf{\Delta f}}_{\mathbf{hertz}}\mathbf{=}\mathbf{\Delta f}\mathbf{\times }\mathbf{f}$ …… (6)

calculate the area frequency response characteristic β on a 1000 MVA base.

Calculate the new value of the regulation constant for 1^stunit.

Substitute $0.04\text{\hspace{0.17em}pu}$for $R_{1old}$, $500\text{\hspace{0.17em}MVA}$for $S_{1old}$and $1000\text{\hspace{0.17em}MVA}$ for $S_{new}$into equation (1).

$\begin{array}{l}{R}_{1new}=0.04\times \frac{1000}{500}\\ R_{1new}=0.08\text{\hspace{0.17em}pu}\end{array}$

Calculate the new value of the regulation constant for 2^nd unit.

Substitute $0.06\text{\hspace{0.17em}pu}$for$R_{2old}$,$750\text{\hspace{0.17em}MVA}$ for$S_{2old}$and$1000\text{\hspace{0.17em}MVA}$ for$S_{new}$into equation (2).

$\begin{array}{l}{R}_{2new}=0.06\times \frac{1000}{750}\\ R_{2new}=0.08\text{\hspace{0.17em}pu}\end{array}$

Calculate the unit area frequency response characteristics.

Substitute$0.08\text{\hspace{0.17em}pu}$ for$R_{1new}$ , and $0.08\text{\hspace{0.17em}pu}$for $R_{2new}$into equation (3).

$\begin{array}{l}\mathrm{\beta }=\frac{1}{0.08}+\frac{1}{0.08}\\ \mathrm{\beta }=12.5+12.5\\ \mathrm{\beta }=25\text{\hspace{0.17em}pu}\end{array}$

Hence the area frequency response characteristic$\beta$is$25\text{\hspace{0.17em}pu}$ .

Calculate Δf in per unit on a 60 Hz base and in Hz.

Calculate the total change in turbine mechanical power in per unit

Substitute$250\text{\hspace{0.17em}MW}$for$P_L$and$1000\text{\hspace{0.17em}MVA}$for $S_{new}$into equation (4).

$\begin{array}{l}\Delta p_m=\frac{250}{1000}\\ \Delta p_m=0.25\text{\hspace{0.17em}pu}\end{array}$

Calculate the steady state decrease in the area frequency

Substitute 0 for $\Delta p_{ref}$, $0.25\text{\hspace{0.17em}pu}$for $\Delta p_m$and $25\text{\hspace{0.17em}pu}$ for $\beta$into equation (5).

$\begin{array}{c}0.25=0-25\times \Delta f\\ \Delta f=-\frac{0.25}{25\text{\hspace{0.17em}pu}}\\ \Delta f=-0.1\text{\hspace{0.17em}pu}\end{array}$

Calculate the steady state decrease in the area frequency in hertz.

Substitute$-0.1\text{\hspace{0.17em}pu}$for$∆f$ and$60\text{\hspace{0.17em}Hz}$ for$f$ into equation (6).

$\begin{array}{l}\Delta f_{hertz}=-0.1\times 60\\ \Delta f_{hertz}=-6\text{\hspace{0.17em}Hz}\end{array}$

The negative sign represents the decrease in the value of the frequency.

Hence, the steady state value of the area frequency in per unit is$0.01\text{\hspace{0.17em}pu}$ and in Hz is $-6\text{\hspace{0.17em}Hz}$.