Question

Each unit in Problem 12.5 is initially operating at one-half its own rating when the frequency increases by $\mathbf{0}\mathbf{.}\mathbf{005}$ per unit. Determine the MW decrease of each unit. The reference power setting of each turbine governor is fixed. Neglect losses and the dependence of load on frequency.

Short answer

The decrease 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,$\Delta p_{ref}=0$

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

Determine the equation to calculate the MW decrease 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.

${\mathrm{R}}_{1\mathrm{new}}={\mathrm{R}}_{1\mathrm{old}}\frac{{\mathrm{S}}_{\mathrm{new}}}{{\mathrm{S}}_{1\mathrm{old}}}$ …… (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 decrease 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 $R_{1old}$, $200\text{\hspace{0.17em}MVA}$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 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).

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

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).

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

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).

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

Hence, the decrease 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.