Question

Repeat Problem 12.7 if the frequency decreases by $\mathbf{0}\mathbf{.}\mathbf{005}$ per unit. Determine the MW increase of each unit.

Short answer

The increase in the mechanical power output for turbine 1, 2 and 3 are$33.33\text{\hspace{0.17em}MVA}$ , $38.46\text{\hspace{0.17em}MVA}$and$50\text{\hspace{0.17em}MVA}$ 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, $S_{new}=100\text{\hspace{0.17em}MVA}$

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

decrease in the frequency,$\Delta f=-0.005\text{\hspace{0.17em}pu}$

Determine the equation to calculate the MW increase in the mechanical power output of each unit.

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 per unit value of mechanical power is given as follows.

${\mathbf{\Delta p}}_{\mathbf{m}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{ref}}\mathbf{-}\frac{\mathbf{\Delta f}}{{\mathbf{R}}_{\mathbf{new}}}$ …… (4)

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

${\mathbf{\Delta p}}_{\mathbf{m}\mathbf{,}\mathbf{MW}}\mathbf{=}{\mathbf{\Delta p}}_{\mathbf{m}}{\mathbf{S}}_{\mathbf{new}}$ …… (5)

Calculate the MW increase in the mechanical power output of each unit.

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${\mathrm{S}}_{1\mathrm{old}}$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 per unit value of mechanical power output for 1^st unit.

Substitute $0$for $\Delta p_{ref}$, $-0.005\text{\hspace{0.17em}pu}$ for $\Delta f$and $0.015\text{\hspace{0.17em}pu}$for $R_{1new}$into equation (4).

$\begin{array}{l}\Delta p_{m1}=0-\frac{(-0.005)}{0.015}\\ \Delta p_{m1}=0.3333\text{\hspace{0.17em}pu}\end{array}$

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

Substitute$0.3333\text{\hspace{0.17em}pu}$ for$\Delta p_{m1}$and$100\text{\hspace{0.17em}MVA}$for$S_{new}$into equation (5).

$\begin{array}{l}\Delta p_{m1,MW}=\left(0.3333\right)\left(100\right)\\ \Delta p_{m1,MW}=33.33\text{\hspace{0.17em}MVA}\end{array}$

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

Substitute$0$for$\Delta p_{ref}$,$-0.005\text{\hspace{0.17em}pu}$for$\Delta f$and$0.013\text{\hspace{0.17em}pu}$for$R_{2new}$into equation (4).

$\begin{array}{l}\Delta p_{m2}=0-\frac{(-0.005)}{0.013}\\ \Delta p_{m2}=0.3846\text{\hspace{0.17em}pu}\end{array}$

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

Substitute $0.3846\text{\hspace{0.17em}pu}$for$\Delta p_{m2}$and$100\text{\hspace{0.17em}MVA}$for$S_{new}$into equation (5).

$\begin{array}{l}\Delta p_{m2,MW}=\left(0.3846\right)\left(100\right)\\ \Delta p_{m2,MW}=38.46\text{\hspace{0.17em}MVA}\end{array}$

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

Substitute $0$for$\Delta p_{ref}$,$-0.005\text{\hspace{0.17em}pu}$for$\Delta f$and$0.01\text{\hspace{0.17em}pu}$for$R_{3new}$into equation (4).

$\begin{array}{l}\Delta p_{m2}=0-\frac{(-0.005)}{0.01}\\ \Delta p_{m2}=0.5\text{\hspace{0.17em}pu}\end{array}$

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

Substitute$0.5\text{\hspace{0.17em}pu}$ for$\Delta p_{m3}$ and$100\text{\hspace{0.17em}MVA}$ for$S_{new}$ into equation (5).

$\begin{array}{l}\Delta p_{m3,MW}=\left(0.5\right)\left(100\right)\\ \Delta p_{m3,MW}=50\text{\hspace{0.17em}MVA}\end{array}$

Hence the increase in the mechanical power output for turbine 1, 2 and 3 are $33.33\text{\hspace{0.17em}MVA}$, $38.46\text{\hspace{0.17em}MVA}$and $50\text{\hspace{0.17em}MVA}$respectively.