Question

Figure 3.32 shows the one line diagram of a three-phase power system. By selecting a common base of$\mathbf{100}\mathbf{MVA}$and$\mathbf{22}\mathbf{kV}$on the generator side, draw an impedance diagram showing all impedances including the load impedance in per-unit. The data are given as follows:

G: $\mathbf{90}\mathbf{MVA}$ $\mathbf{22}\mathbf{kV}$ $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{18}\mathbf{pu}$

T₁:$\mathbf{50}\mathbf{MVA}$ $\mathbf{22}\mathbf{kV}\mathbf{/}\mathbf{220}\mathbf{kV}$ $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{pu}$

T₂:$\mathbf{40}\mathbf{MVA}$ $\mathbf{220}\mathbf{kV}\mathbf{/}\mathbf{11}\mathbf{kV}$ $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{06}\mathbf{pu}$

T₃:$\mathbf{40}\mathbf{MVA}$ localid="1655975246589" $\mathbf{22}\mathbf{kV}\mathbf{/}\mathbf{110}\mathbf{kV}$ $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{064}\mathbf{pu}$

T₄:$\mathbf{40}\mathbf{MVA}$ $\mathbf{\text{110\hspace{0.17em}}}\mathbf{kV}\mathbf{/}\mathbf{11}\mathbf{kV}$ $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{08}\mathbf{pu}$

M: $\mathbf{\text{66.5\hspace{0.17em}}}\mathbf{MVA}$ $\mathbf{10}\mathbf{.}\mathbf{45}\mathbf{}\mathbf{kV}$ $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{185}\mathbf{pu}$

Lines 1 and 2 have series reactane of$\mathbf{48}\mathbf{.}\mathbf{4}\mathbf{\Omega }$and$\mathbf{65}\mathbf{.}\mathbf{43}\mathbf{\Omega }$,respectively. At bus 4, the three-phase load absorbs$\mathbf{57}\mathbf{MVA}$at$\mathbf{10}\mathbf{.}\mathbf{45}\mathbf{kV}$and$\mathbf{0}\mathbf{.}\mathbf{6}$power factor lagging.

Short answer

The single-line diagram is drawn below.

Step-by-step solution

Determine the formulas of the new per-unit value of reactance.

Write the formula for the new per-unit reactance:

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

Write the formula for the base impedance on line:

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

Write the formula for the per-unit impedance on line:

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

Determine the new reactance of the generator.

The base voltage on the generator side is $\text{22kV}$.

The base power is${\mathrm{S}}_{\mathrm{base}}=100\text{\hspace{0.17em}MVA}$.

Determine the new per-unit reactance of the generator.

Substituting$22\text{\hspace{0.17em}kV}$for ${\mathrm{V}}_{\mathrm{base}\left(\mathrm{new}\right)}$ and ${\mathrm{V}}_{\mathrm{base}\left(\mathrm{old}\right)}$, $100\text{\hspace{0.17em}MVA}$ for ${\mathrm{S}}_{\mathrm{base}\left(\mathrm{old}\right)}$, $90\text{\hspace{0.17em}MVA}$ for ${\mathrm{S}}_{\mathrm{base}\left(\mathrm{new}\right)},$ and $0.18\text{\hspace{0.17em}pu}$ for ${\mathrm{X}}_{\mathrm{old}}$ in Equation (1), we have

$\begin{array}{c}{X}_{G\left(new\right)}=0.18\text{\hspace{0.17em}pu}\times {\left(\frac{22\text{\hspace{0.17em}kV}}{22\text{\hspace{0.17em}kV}}\right)}^2\times \frac{100\text{\hspace{0.17em}MVA}}{90\text{\hspace{0.17em}MVA}}\\ =0.2\text{\hspace{0.17em}pu}\end{array}$

Determine the new per-unit reactance of the transformer $T_1$.

Substituting $22\mathrm{kV}$ for ${\mathrm{V}}_{\mathrm{base}\left(\mathrm{new}\right)}$, $22\mathrm{kV}$for ${\mathrm{V}}_{\mathrm{base}\left(\mathrm{old}\right)}$, $100\text{\hspace{0.17em}MVA}$ for ${\mathrm{S}}_{\mathrm{base}\left(\mathrm{old}\right)}$, $50\text{\hspace{0.17em}MVA}$for ${\mathrm{S}}_{\mathrm{base}\left(\mathrm{new}\right)},$ and $0.1\text{\hspace{0.17em}pu}$ for ${\mathrm{X}}_{\mathrm{old}}$ in Equation (1), we have

$\begin{array}{c}{X}_{T_1\left(new\right)}=0.1\text{\hspace{0.17em}pu}\times {\left(\frac{22\text{\hspace{0.17em}kV}}{22\text{\hspace{0.17em}kV}}\right)}^2\times \frac{100\text{\hspace{0.17em}MVA}}{50\text{\hspace{0.17em}MVA}}\\ =0.2\text{\hspace{0.17em}pu}\end{array}$

Determine the new per-unit reactance of the transformer $T_2$.

Substituting $22kV$ for $V_{base\left(new\right)}$ , $22kV$ for $V_{base\left(old\right)}$, $100\text{\hspace{0.17em}MVA}$ for $S_{base\left(old\right)}$, $40\text{\hspace{0.17em}MVA}$ for $S_{base\left(new\right)},$and $0.06\text{\hspace{0.17em}pu}$ for $X_{old}$ in Equation (1), we have

$\begin{array}{c}{X}_{T_2\left(new\right)}=0.06\text{\hspace{0.17em}pu}\times {\left(\frac{22\text{\hspace{0.17em}kV}}{22\text{\hspace{0.17em}kV}}\right)}^2\times \frac{100\text{\hspace{0.17em}MVA}}{40\text{\hspace{0.17em}MVA}}\\ =0.15\text{\hspace{0.17em}pu}\end{array}$

Determine the new per-unit reactance of the transformer $T_3$.

Substituting $22kV$ for $V_{base\left(new\right)}$ , $22kV$for$V_{base\left(old\right)}$, $100\text{\hspace{0.17em}MVA}$ for $S_{base\left(old\right)}$, $40\text{\hspace{0.17em}MVA}$for $S_{base\left(new\right)},$ and $0.064\text{\hspace{0.17em}pu}$for$X_{old}$ in Equation (1), we have

$\begin{array}{c}{X}_{T_3\left(new\right)}=0.064\text{\hspace{0.17em}pu}\times {\left(\frac{22\text{\hspace{0.17em}kV}}{22\text{\hspace{0.17em}kV}}\right)}^2\times \frac{100\text{\hspace{0.17em}MVA}}{40\text{\hspace{0.17em}MVA}}\\ =0.16\text{\hspace{0.17em}pu}\end{array}$

Determine the new per-unit reactance of the transformer $T_4$.

Substituting $22kV$ for $V_{base\left(new\right)}$,$22kV$for $V_{base\left(old\right)}$, $100MVA$ for$S_{base\left(old\right)}$, $40\text{\hspace{0.17em}MVA}$ for${\mathrm{S}}_{\mathrm{base}\left(\mathrm{new}\right)},$ and $0.08\text{\hspace{0.17em}pu}$ for $X_{old}$ in Equation (1), we have

$\begin{array}{c}{X}_{T_4\left(new\right)}=0.08\text{\hspace{0.17em}pu}\times {\left(\frac{22\text{\hspace{0.17em}kV}}{22\text{\hspace{0.17em}kV}}\right)}^2\times \frac{100\text{\hspace{0.17em}MVA}}{40\text{\hspace{0.17em}MVA}}\\ =0.2\text{\hspace{0.17em}pu}\end{array}$

Determine the new per-unit reactance of motor $M$.

Substituting$10.45\text{\hspace{0.17em}kV}$for $V_{base\left(new\right)}$,$11\text{\hspace{0.17em}kV}$ for $V_{base\left(old\right)}$, $100\text{\hspace{0.17em}MVA}$ for $S_{base\left(old\right)}$, $66.5\text{\hspace{0.17em}MVA}$ for $S_{base\left(new\right)},$ and $0.185\text{\hspace{0.17em}pu}$ for $X_{old}$ in Equation (1), we have

$\begin{array}{c}{X}_{M\left(new\right)}=0.185\text{\hspace{0.17em}pu}\times {\left(\frac{10.45\text{\hspace{0.17em}kV}}{11\text{\hspace{0.17em}kV}}\right)}^2\times \frac{100\text{\hspace{0.17em}MVA}}{66.5\text{\hspace{0.17em}MVA}}\\ =0.25\text{\hspace{0.17em}pu}\end{array}$

Draw the single-line diagram.

Determine the base impedance on line 1.

Substituting $220\text{\hspace{0.17em}kV}$ for $V_{base}$ and$100\text{\hspace{0.17em}MVA}$ for $S_{base}$ in Equation (2), we have

localid="1655975256635"

Determine the base impedance on line 2.

Substituting $110\text{\hspace{0.17em}kV}$ for $V_{base}$and $100\text{\hspace{0.17em}MVA}$ for $S_{base}$ in Equation (2), we have

$\begin{array}{c}{Z}_{base2}=\frac{{(110\times {10}^3)}^2}{100\times {10}^6}\\ =121\Omega \end{array}$

Determine the per-unit impedance on line 1.

Substitutinglocalid="1655975067707" $484\Omega$for $Z_{base1}$and $48.4\Omega$ for $Z_1$ in Equation (3), we have

$\begin{array}{c}{Z}_{pu1}=\frac{48.4\Omega }{484\Omega }\\ =0.1\text{\hspace{0.17em}pu}\end{array}$

Determine the per-unit impedance on line 2.

Substitutinglocalid="1655975261438" $121\Omega$for $Z_{base2}$ and $65.43\Omega$ for $Z_2$ in Equation (3), we have

$\begin{array}{c}{Z}_{pu2}=\frac{65.43\Omega }{121\Omega }\\ =0.54\text{\hspace{0.17em}pu}\end{array}$

Determine the base impedance of load.

Substituting $11kV$ for$V_{base}$ and $100MVA$ for $S_{base}$ in Equation (2), we have:

The complex power at the load for the 0.6 power factor is

$S_L=57\angle 53.13°\text{\hspace{0.17em}MVA.}$

Determine the load impedance:

$\begin{array}{c}{Z}_L=\frac{{V_L}^2}{{S_L}^{*}}\\ =\frac{{(10.45\times {10}^3)}^2}{57\angle -53.13°\times {10}^6}\\ =1.1495+j1.532\Omega \end{array}$

Determine the per-unit load impedance:

$\begin{array}{c}{Z}_{L\left(pu\right)}=\frac{1.1495+j1.532\Omega }{1.21\Omega }\\ =0.95+j1.2667\text{\hspace{0.17em}pu}\end{array}$

The single-line diagram is as follows.