Question

Consider the single-line diagram of the power system shown inFigure 3.38. Equipment ratings are:

Generator 1:$\mathbf{1000}\mathit{M}\mathit{V}\mathit{A}$,$\mathbf{18}\mathit{k}\mathit{V}$,${\mathit{X}}^{\mathbf{\text{'}\text{'}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathit{p}\mathit{u}$ .

Generator 2: $\mathbf{1000}\mathit{M}\mathit{V}\mathit{A}$,$\mathbf{18}\mathit{k}\mathit{V}$, ${\mathit{X}}^{\mathbf{\text{'}\text{'}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathit{p}\mathit{u}$.

Synchronous motor 3: $\mathbf{1500}\mathit{M}\mathit{V}\mathit{A}$,$\mathbf{20}\mathit{k}\mathit{V}$, ${\mathit{X}}^{\mathbf{\text{'}\text{'}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathit{p}\mathit{u}$.

Three-phase$\mathit{\Delta }$–Y transformers$\mathbf{1000}\mathit{M}\mathit{V}\mathit{A}$,$\mathbf{500}\mathit{k}\mathit{V}\mathit{Y}\mathbf{/}\mathbf{20}\mathit{k}\mathit{V}\mathit{\Delta }$, $\mathit{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathit{p}\mathit{u}$.

$\mathit{T}\mathbf{1}\mathbf{,}$$\mathit{T}\mathbf{2}$,$\mathit{T}\mathbf{3}$, $\mathit{T}\mathbf{4}$, :

Three-phase Y–Y transformer${\mathit{T}}_{\mathbf{5}}$: $\mathbf{1500}\mathit{M}\mathit{V}\mathit{A}$, $\mathbf{500}\mathit{k}\mathit{V}\mathbf{/}\mathbf{20}\mathit{k}\mathit{V}$, $\mathit{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathit{p}\mathit{u}$

Neglecting resistance, transformer phase shift, and magnetizing reactance, draw the equivalent reactance diagram. Use a base of $\mathbf{100}\mathit{M}\mathit{V}\mathit{A}$and $\mathbf{500}\mathit{k}\mathit{V}$ for the$\mathbf{50}\mathit{\Omega }$ line. Determine the per-unit reactances.

Short answer

The single line diagram is:

Step-by-step solution

Given data. 

The base power is$100MVA$and base voltage is $500kV$.

Te reactances of generator 1, generator 2, and synchronous motor is$0.2pu$ .

The transformers’reactance are$0.1pu$ .

Determine the formulas ofnew reactance, base impedance and per-unit impedance.

Write the formula of new reactance.

……. (1)

${\mathit{X}}_{\mathbf{new}}\mathbf{=}{\mathit{X}}_{\mathbf{old}}{\mathbf{\left(}\frac{{\mathbf{V}}_{\mathbf{old}}}{{\mathbf{V}}_{\mathbf{new}}}\mathbf{\right)}}^{\mathbf{2}}\frac{{\mathbf{S}}_{\mathbf{new}}}{{\mathbf{S}}_{\mathbf{old}}}$

Here,$S_{new}$is a new base of power, $S_{old}$ is an old base of power, $V_{new}$ is a new base of voltage, $V_{old}$ is an old base of voltage, and $X_{old}$is an old reactance.

Write the formulas of base impedance.

localid="1655727841878" ${\mathit{Z}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{\left(}{\mathbf{V}}_{\mathbf{base}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{S}}_{\mathbf{base}}}$……. (2)

Here, $S_{base}$ is a base power and $V_{base}$is a base voltage.

Write the formula of per-unit impedance.

localid="1655727853482" ${\mathit{Z}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{actual}}}{{\mathbf{Z}}_{\mathbf{base}}}$……. (3)

Here,$Z_{base}$is a base impedance and $Z_{actual}$is an actual impedance.

Determine the new reactances of generators and motor. 

The base power is$100MVA$and the base voltage for transmission is$500kV$and base voltage for motor and generator zones is$20kV$.

Determine the reactance on generator 1 and generator 2.

Substitute$0.2pu$ for$X_{old}$,$1000MVA$for$S_{old}$,$100MVA$for$S_{new}$,$18kV$for$V_{old}$and$20kV$ for$V_{new}$in equation (1).

$\begin{array}{c}{X}_{G_1\left(new\right)}=0.2pu\times {\left(\frac{18kV}{20kV}\right)}^2\left(\frac{100MVA}{1000MVA}\right)\\ X_{G_2\left(new\right)}=0.0162pu\end{array}$

Determine the reactance on motor.

Substitute$0.2pu$ for$X_{old}$,$1500MVA$for$S_{old}$,$100MVA$for$S_{new}$,$20kV$$V_{old}$for$V_{old}$and $20kV$ for$V_{new}$in equation (1).

Determine the reactance of transformers.

Determine the reactance of transformers $T_1$,$T_2$,$T_3$and $T_4$.

Substitute $0.1pu$ for $X_{old}$, $1000MVA$for $S_{old}$, $100MVA$fo$S_{new}$r , $20kV$for$V_{old}$ and $20kV$ for$V_{new}$ in equation (1).

$\begin{array}{c}{X}_{T\left(new\right)}=0.1pu\times {\left(\frac{20kV}{20kV}\right)}^2\left(\frac{100MVA}{1000MVA}\right)\\ =0.01pu\end{array}$

Determine the reactance of transformers$T_5$.

Substitute$0.1pu$ for$X_{old}$,$1500MVA$for$S_{old}$,$100MVA$for$S_{new}$,$20kV$for$V_{old}$and $20kV$ for$V_{new}$in equation (1).

$\begin{array}{c}{X}_{T_5\left(new\right)}=0.1pu\times {\left(\frac{20kV}{20kV}\right)}^2\left(\frac{100MVA}{1500MVA}\right)\\ =0.0067pu\end{array}$

 Step 5: Determine the impedance on transmission lines.

Determine the base impedance.

Substitute $500kV$for $V_{base}$ and$100MVA$ for $S_{base}$in equation (2).

$\begin{array}{c}{Z}_{base}=\frac{{\left(500kV\right)}^2}{100MVA}\\ =2500\Omega \end{array}$

Determine the per-unit impedance on transmission line 1.

Substitute$2500\Omega$for$Z_{base}$and $50\Omega$for$Z_{actual}$in equation (3).

$\begin{array}{c}{Z}_{pu\left(50\right)}=\frac{50\Omega }{2500\Omega }\\ =0.02pu\end{array}$

Determine the per-unit impedance on transmission line 2.

Substitute$2500\Omega$for$Z_{base}$and $25\Omega$for$Z_{actual}$in equation (3).

$\begin{array}{c}{Z}_{pu\left(25\right)}=\frac{25\Omega }{2500\Omega }\\ =0.01pu\end{array}$

Draw the single line diagram: