Question

For the system shown in Figure 3.34, draw an impedance diagram in per unit by choosing$\mathbf{100}\mathbf{kVA}$to be the base$\mathbf{kVA}$and$\mathbf{2400}\mathbf{V}$as the base voltage for the generators.

Short answer

The per-unit single-line diagram is as follows.

Step-by-step solution

Determine the formulas for base impedance, per-unit impedance, and new value of per-unit impedance.

Write the formula for new per-unit impedance:

${\mathrm{Z}}_{\mathrm{pe}\left(\mathrm{new}\right)}={\mathrm{Z}}_{\mathrm{pe}\left(\mathrm{old}\right)}{\left(\frac{{\mathrm{V}}_{\mathrm{base}\left(\mathrm{old}\right)}}{{\mathrm{V}}_{\mathrm{base}\left(\mathrm{new}\right)}}\right)}^2\frac{{\mathrm{S}}_{\mathrm{base}\left(\mathrm{new}\right)}}{{\mathrm{S}}_{\mathrm{base}\left(\mathrm{old}\right)}}$ …… (1)

Write the formula for base impedance:

${\mathrm{Z}}_{\mathrm{base}}=\frac{{\mathrm{V}}_{\mathrm{base}}^2}{{\mathrm{S}}_{\mathrm{base}}}$ …… (2)

Write the formula for per-unit impedance”

${\mathrm{Z}}_{\mathrm{pu}}=\frac{{\mathrm{Z}}_{\mathrm{actual}}}{{\mathrm{Z}}_{\mathrm{base}}}$ …… (3)

Determine the per-unit impedance of generator 1, generator 2, transformer 1, and transformer 2.

The base voltage on generator 1 is ${\mathrm{V}}_{\mathrm{base}\left(\mathrm{old}\right)}=2400\text{\hspace{0.17em}V}$.

The new base voltage is ${\mathrm{V}}_{\mathrm{base}\left(\mathrm{new}\right)}=2400\text{\hspace{0.17em}V.}$

The old base power on generator 1 is ${\mathrm{S}}_{\mathrm{base}\left(\mathrm{old}\right)}=10\text{\hspace{0.17em}kVA}$ and the new base power is role="math" localid="1655964231545" $S_{base\left(new\right)}=100\text{\hspace{0.17em}kVA.}$

Determine the new impedance on generator 1.

Substituting $j0.2\text{\hspace{0.17em}pu}$ for $Z_{G1\left(old\right)}$, $2400\text{\hspace{0.17em}V}$ for role="math" localid="1655964239799" $V_{base\left(new\right)}$, $2400\text{\hspace{0.17em}V}$ for role="math" localid="1655964245880" $V_{base\left(old\right)}$, $100\text{\hspace{0.17em}kVA}$ forrole="math" localid="1655964213660" $S_{base\left(new\right)},$ and $10\text{\hspace{0.17em}kVA}$ for role="math" localid="1655964219753" $S_{base\left(old\right)}$ in Equation (1), we have

Determine the new impedance on generator 2.

Substituting $j0.2\text{\hspace{0.17em}pu}$ for $Z_{G2\left(old\right)}$, $2400\text{\hspace{0.17em}V}$ for $V_{base\left(new\right)}$, $2400\text{\hspace{0.17em}V}$ for $V_{base\left(old\right)}$, $100\text{\hspace{0.17em}kVA}$for $S_{base\left(new\right)},$ and $20\text{\hspace{0.17em}kVA}$for ${\mathrm{S}}_{\mathrm{base}\left(\mathrm{old}\right)}$ in Equation (1), we have

Determine the new impedance on generator 2.

Substituting $j0.2\text{\hspace{0.17em}pu}$for ${\mathrm{Z}}_{\mathrm{G}2\left(\mathrm{old}\right)}$, $2400\text{\hspace{0.17em}V}$ for $V_{base\left(new\right)}$, $2400\text{\hspace{0.17em}V}$ for $V_{base\left(old\right)}$ , $100\text{\hspace{0.17em}kVA}$ for $S_{base\left(new\right)},$ and $20\text{\hspace{0.17em}kVA}$ for $S_{base\left(old\right)}$ in Equation (1), we have

$\begin{array}{c}{Z}_{G2\left(new\right)}=j0.2\text{\hspace{0.17em}pu}{\left(\frac{2400\text{\hspace{0.17em}V}}{2400\text{\hspace{0.17em}V}}\right)}^2\left(\frac{100\text{\hspace{0.17em}kVA}}{20k\text{VA}}\right)\\ =j1\text{\hspace{0.17em}pu}\end{array}$

Determine the new impedance on transformer 1.

Substituting $j0.1\text{\hspace{0.17em}pu}$ for $Z_{T1\left(old\right)}$, $2400\text{\hspace{0.17em}V}$ for $V_{base\left(new\right)}$, $2400\text{\hspace{0.17em}V}$ for $V_{base\left(old\right)}$, $100\text{\hspace{0.17em}kVA}$ for $S_{base\left(new\right)},$ and $40\text{\hspace{0.17em}kVA}$for $S_{base\left(old\right)}$in Equation (1), we have

Determine the new impedance on transformer 2.

Substituting $j0.1\text{\hspace{0.17em}pu}$ for $Z_{T2\left(old\right)}$,$9.6\text{\hspace{0.17em}kV}$for $V_{base\left(new\right)}$, $10\text{\hspace{0.17em}kV}$ for $V_{base\left(old\right)}$, $100\text{\hspace{0.17em}kVA}$ for $S_{base\left(new\right)},$ and $80\text{\hspace{0.17em}kVA}$ for $S_{base\left(old\right)}$ in Equation (1), we have

$\begin{array}{c}{Z}_{T2\left(new\right)}=j0.1\text{\hspace{0.17em}pu}{\left(\frac{10\text{\hspace{0.17em}kV}}{9.6\text{\hspace{0.17em}kV}}\right)}^2\left(\frac{100\text{\hspace{0.17em}kVA}}{80k\text{VA}}\right)\\ =j0.136\text{\hspace{0.17em}pu}\end{array}$

Draw the per-unit single-line diagram.

Determine the base impedance on line.

Substituting$9.6\text{\hspace{0.17em}kV}$for $V_{base}$ and for in Equation (2), we have

Determine the per-unit impedance on line.

Substituting $921.6\Omega$ for $Z_{base}$ and$(50+j200)\Omega$ for $Z_{actual}$ in Equation (3), we have

$\begin{array}{c}{\mathrm{Z}}_{\mathrm{line}\left(\mathrm{pu}\right)}=\frac{(50+\mathrm{j}200)\mathrm{\Omega }}{921.6\mathrm{\Omega }}\\ =0.054+\mathrm{j}0.217\text{\hspace{0.17em}\hspace{0.17em}pu}\end{array}$

Determine the per-unit power and the per-unit voltage on the motor side.

The per-unit voltage is as follows:

$\begin{array}{c}{V}_{M\left(pu\right)}=\frac{4\text{\hspace{0.17em}kV}}{5\text{\hspace{0.17em}kV}}\\ =0.8\text{\hspace{0.17em}pu}\end{array}$

The per-unit power is as follows:

$\begin{array}{c}{S}_{M\left(pu\right)}=\frac{25\text{\hspace{0.17em}kVA}}{100k\text{VA}}\\ =0.25\text{\hspace{0.17em}pu}\end{array}$

Therefore, the single-line diagram is as follows: