Question

An area of an interconnected $\mathbf{60}\mathbf{Hz}$ power system has three turbine generator units rated$\mathbf{200}\mathbf{,}\mathbf{300}\mathbf{,}$ and $\mathbf{500}\mathbf{MVA}$. The regulation constants of the units are$\mathbf{0}\mathbf{.}\mathbf{03}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{,}$ and$\mathbf{0}\mathbf{.}\mathbf{05}$ per unit, respectively, based on their ratings. Each unit is initially operating at one-half its own rating when the load suddenly decreases by$\mathbf{150}\mathbf{MW}$ . Determine (a) the unit area frequency response characteristic$\mathbf{\beta }$ on a$\mathbf{100}\mathbf{MVA}$ base, (b) the steady-state increase in area frequency, and (c) the MW decrease in mechanical power output of each turbine. Assume that the reference power setting of each turbine-governor remains constant. Neglect losses and the dependence of load on frequency.

Short answer

(a) The unit area frequency response characteristics $\beta$is $243.58\text{\hspace{0.17em}pu}$.

(b) The steady state increase in the area frequency is $0.372\text{\hspace{0.17em}Hz}$.

(c) The value of the mechanical power output for turbine 1, 2 and 3 are$-41.33\text{\hspace{0.17em}MVA,}$$-\text{47.69\hspace{0.17em}MVA}$ and$-62\text{\hspace{0.17em}MVA}$ respectively.

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.

$\begin{array}{l}{S}_{1,old}=200\text{\hspace{0.17em}MVA}\\ {\text{S}}_{2,old}=300\text{\hspace{0.17em}MVA}\\ S_{3,old}=500\text{\hspace{0.17em}MVA}\end{array}$

Write the regulation constant of three generator units.

$\begin{array}{l}{R}_{1,old}=0.03\text{\hspace{0.17em}pu\hspace{0.17em}}\\ R_{2,old}=\text{0.04\hspace{0.17em}pu}\\ R_{3,old}=0.05\text{\hspace{0.17em}pu}\end{array}$

The base rating, $S_{new}=100\text{\hspace{0.17em}MVA}$

The decrease value of load, ${\mathrm{P}}_{\mathrm{L}}=150\text{\hspace{0.17em}MW}$

Change in the reference setting power, ${\mathrm{\Delta p}}_{\mathrm{ref}}=0$.

Determine the equation to calculate the unit area frequency response characteristic, steady state decrease in the area frequency and MW decrease in the mechanical power output.

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 new value of the regulation constant for 3^rd unit is given as follows.

${\mathbf{R}}_{\mathbf{3}\mathbf{new}}\mathbf{=}{\mathbf{R}}_{\mathbf{3}\mathbf{old}}\frac{{\mathbf{S}}_{\mathbf{new}}}{{\mathbf{S}}_{\mathbf{3}\mathbf{old}}}$ …… (3)

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

$\mathbf{\beta }\mathbf{=}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}\mathbf{new}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}\mathbf{new}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{3}\mathbf{new}}}$ …… (4)

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

${\mathbf{\Delta p}}_{\mathbf{m}}\mathbf{=}\mathbf{-}\frac{{\mathbf{P}}_{\mathbf{L}}}{{\mathbf{S}}_{\mathbf{new}}}$ …… (5)

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

${\mathbf{\Delta p}}_{\mathbf{m}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{ref}}\mathbf{-}\mathbf{\beta \Delta f}$ …… (6)

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

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

${\mathbf{\Delta f}}_{\mathbf{hertz}}\mathbf{=}\mathbf{\Delta f}\mathbf{\times }\mathbf{f}$ …… (7)

The equation to calculate the per unit value of mechanical power is given as follows.

${\mathbf{\Delta p}}_{\mathbf{m}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{ref}}\mathbf{-}\frac{\mathbf{\Delta f}}{{\mathbf{R}}_{\mathbf{new}}}$ …… (8)

The equation to calculate the per unit value of mechanical power in MW is given as follows.

${\mathbf{\Delta p}}_{\mathbf{m}\mathbf{,}\mathbf{MW}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{m}}{\mathbf{S}}_{\mathbf{new}}$ …… (9)

Calculate the value of unit area frequency response characteristics.

(a)

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

Substitute$0.03\text{\hspace{0.17em}pu}$for ${\mathrm{R}}_{1\mathrm{old}}$,$200\text{\hspace{0.17em}MVA}$for$S_{1old}$and$100\text{\hspace{0.17em}MVA}$for $S_{new}$into equation (1).

$\begin{array}{l}{R}_{1new}=0.03\times \frac{100}{200}\\ R_{1new}=0.015\text{\hspace{0.17em}pu}\end{array}$

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

Substitute$0.04\text{\hspace{0.17em}pu}$for$R_{2old}$,$300\text{\hspace{0.17em}MVA}$for$S_{2old}$and $100\text{\hspace{0.17em}MVA}$for$S_{new}$into equation (2).

$\begin{array}{l}{R}_{2new}=0.04\times \frac{100}{300}\\ R_{2new}=0.013\text{\hspace{0.17em}pu}\end{array}$

Calculate the new value of the regulation constant for 3^rd unit.

Substitute$0.05\text{\hspace{0.17em}pu}$for $R_{3old}$,$500\text{\hspace{0.17em}MVA}$for$S_{3old}$and$100\text{\hspace{0.17em}MVA}$ for$S_{new}$into equation (3).

$\begin{array}{l}{R}_{3new}=0.05\times \frac{100}{500}\\ R_{3new}=0.01\text{\hspace{0.17em}pu}\end{array}$

Calculate the unit area frequency response characteristics.

Substitute$0.015\text{\hspace{0.17em}pu}$ for $R_{1new}$,$0.013\text{\hspace{0.17em}pu}$ for$R_{2new}$ and$0.01\text{\hspace{0.17em}pu}$ for$R_{3new}$ into equation (4).

$\begin{array}{l}\beta =\frac{1}{0.015}+\frac{1}{0.013}+\frac{1}{0.01}\\ \beta =66.66+75.92+100\\ \beta =243.58\text{\hspace{0.17em}pu}\end{array}$

Hence, the unit area frequency response characteristics $\beta$ is $243.58\text{\hspace{0.17em}pu}$.

Calculate the steady state decrease in the area frequency.

(b)

Calculate the total change in turbine mechanical power in per unit

Substitute $150\text{\hspace{0.17em}MW}$for$P_L$and$100\text{\hspace{0.17em}MVA}$for$S_{new}$into equation (5).

$\begin{array}{l}\Delta p_m=-\frac{150}{100}\\ \Delta p_m=-1.5\text{\hspace{0.17em}pu}\end{array}$

The negative sign represents the decrease in the power.

Calculate the steady state increase in the area frequency.

Substitute $0$for role="math" localid="1656073922186" ${\mathrm{\Delta p}}_{\mathrm{ref}}$, $-1.5\text{\hspace{0.17em}pu}$for $\Delta p_m$, and$243.58\text{\hspace{0.17em}pu}$ for$\beta$into equation (6).

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

Substitute$0.0062\text{\hspace{0.17em}pu}$ for$\Delta f$ and $60\text{\hspace{0.17em}Hz}$for$f$ into equation (7).

$\begin{array}{l}\Delta f_{hertz}=0.0062\times 60\\ \Delta f_{hertz}=0.372\text{\hspace{0.17em}Hz}\end{array}$

Hence the steady state increase in the area frequency is$0.372\text{\hspace{0.17em}Hz}$ .

Calculate the MW decrease in the mechanical power output of each turbine.

(c)

Calculate the per unit value of mechanical power output for 1^st unit.

Substitute $0$for $\Delta p_{ref}$, $0.0062\text{\hspace{0.17em}pu}$for $\Delta f$ and $0.015\text{\hspace{0.17em}pu}$for $R_{1new}$into equation (8).

$\begin{array}{l}\Delta p_{m1}=0-\frac{0.0062}{0.015}\\ \Delta p_{m1}=-0.4133\text{\hspace{0.17em}pu}\end{array}$

Calculate the per unit value of mechanical power output for 1^st unit in MW.

Substitute $-0.4133\text{\hspace{0.17em}pu}$for $\Delta p_{m1}$and $100\text{\hspace{0.17em}MVA}$for $S_{new}$into equation (9).

$\begin{array}{l}\Delta p_{m1,MW}=(-0.4133)\left(100\right)\\ \Delta p_{m1,MW}=-41.33\text{\hspace{0.17em}MVA}\end{array}$

Calculate the per unit value of mechanical power output for 2^nd unit.

Substitute $0$for role="math" localid="1656074447895" $\Delta p_{ref}$, $0.0062\text{\hspace{0.17em}pu}$for $\Delta f$and $0.0133\text{\hspace{0.17em}pu}$ for $R_{2new}$into equation (8).

$\begin{array}{l}\Delta p_{m2}=0-\frac{0.0062}{0.013}\\ \Delta p_{m2}=-0.4769\text{\hspace{0.17em}pu}\end{array}$

Calculate the per unit value of mechanical power output for 2^nd unit in MW.

Substitute$-0.4769\text{\hspace{0.17em}pu}$for$\Delta p_{m2}$and $100\text{\hspace{0.17em}MVA}$for $S_{new}$into equation (9).

$\begin{array}{l}\Delta p_{m2,MW}=(-0.4769)\left(100\right)\\ \Delta p_{m2,MW}=-47.69\text{\hspace{0.17em}MVA}\end{array}$

Calculate the per unit value of mechanical power output for 3^rd unit.

Substitute$0$for $\Delta p_{ref}$, $0.0062\text{\hspace{0.17em}pu}$ for $\Delta f$ and$0.01\text{\hspace{0.17em}pu}$ for$R_{3new}$into equation (8).

$\begin{array}{l}\Delta p_{m2}=0-\frac{0.0062}{0.01}\\ \Delta p_{m2}=-0.62\text{\hspace{0.17em}pu}\end{array}$

Calculate the per unit value of mechanical power output for 3^rd unit in MW.

Substitute$-0.62\text{\hspace{0.17em}pu}$ for$\Delta p_{m3}$ and$100\text{\hspace{0.17em}MVA}$ for $S_{new}$into equation (9).

$\begin{array}{l}\Delta p_{m3,MW}=(-0.62)\left(100\right)\\ \Delta p_{m3,MW}=-62\text{\hspace{0.17em}MVA}\end{array}$

Hence, the value of the mechanical power output for turbine 1, 2 and 3 are$-41.33\text{\hspace{0.17em}MVA,}$$-\text{47.69\hspace{0.17em}MVA}$ and$-62\text{\hspace{0.17em}MVA}$ respectively.