Question

For Problem 3.18, the motor operates at full load, at 0.8 power factor leading, and at a terminal voltage of $\mathbf{10}\mathbf{.}\mathbf{45}\mathbf{kV}$. Determine (a) the voltage at bus 1, which is the generator bus, and (b) the generator and motor internal EMFs.

Short answer

(a) The per-unit voltage at bus 1 is $22\angle 15.91°\text{\hspace{0.17em}kV}$.

(b) The generator and motor internal EMFs are $1.0826\angle 25.14°\text{\hspace{0.17em}pu}$ and $1.064\angle -7.54°\text{\hspace{0.17em}pu.}$

Step-by-step solution

Determine the formulas of the per-unit voltage, impedance, current, voltage at bus 1, and internal EMFs of the generator and the motor.

Write the formula for per-unit voltage:

${\mathrm{V}}_{4\left(\mathrm{pu}\right)}=\frac{{\mathrm{V}}_{4\left(\mathrm{actual}\right)}}{{\mathrm{V}}_{4\left(\mathrm{base}\right)}}$ ……. (1)

Write the formula for the per-unit power at the motor:

${\mathrm{S}}_{4\left(\mathrm{pu}\right)}=\frac{{\mathrm{S}}_{4\left(\mathrm{actual}\right)}}{{\mathrm{S}}_{4\left(\mathrm{base}\right)}}$ ……. (2)

Write the formula for motor current:

${\mathrm{I}}_{\mathrm{m}\left(\mathrm{pu}\right)}={\left(\frac{{\mathrm{S}}_{4\left(\mathrm{pu}\right)}}{{\mathrm{V}}_{4\left(\mathrm{pu}\right)}}\right)}^{*}$ ……. (3)

Write the formula for load current:

role="math" localid="1655968315976" ${\mathrm{I}}_{\mathrm{L}\left(pu\right)}=\frac{{\mathrm{V}}_{4\left(pu\right)}}{{\mathrm{Z}}_{\mathrm{L}\left(pu\right)}}$ ……. (4)

Write the formula for generator terminal voltage:

${\mathrm{V}}_{1\left(\mathrm{pu}\right)}={\mathrm{V}}_{4\left(\mathrm{pu}\right)}+{\mathrm{I}}_{\left(\mathrm{pu}\right)}{\mathrm{X}}_{\mathrm{eq}\left(\mathrm{pu}\right)}$ ……. (5)

Write the formula for generator internal EMF:

${\mathrm{E}}_{\mathrm{g}\left(\mathrm{pu}\right)}={\mathrm{V}}_{1\left(\mathrm{pu}\right)}+{\mathrm{Z}}_{\mathrm{g}\left(\mathrm{pu}\right)}{\mathrm{I}}_{\left(\mathrm{pu}\right)}$ ……. (6)

Write the formula for motor internal EMF:

${\mathrm{E}}_{\mathrm{m}\left(\mathrm{pu}\right)}={\mathrm{V}}_{4\left(\mathrm{pu}\right)}-{\mathrm{Z}}_{\mathrm{m}\left(\mathrm{pu}\right)}{\mathrm{I}}_{\left(\mathrm{pu}\right)}$ ……. (7)

Determine the voltage at bus 1.

(a)

The single line diagram in Problem 3.23 is as follows.

Determine the per-unit voltage at bus 4.

Substituting$11\mathrm{kV}$for ${\mathrm{V}}_{4\left(\mathrm{base}\right)}$and $10.45\text{\hspace{0.17em}kV}$ for ${\mathrm{V}}_{4\left(\mathrm{actual}\right)}$ in Equation (1), we have

$\begin{array}{c}{\mathrm{V}}_{4\left(\mathrm{pu}\right)}=\frac{10.45\text{\hspace{0.17em}kV}}{11\text{\hspace{0.17em}kV}}\\ =0.95\text{\hspace{0.17em}pu}\end{array}$

Determine the per-unit power at the motor load:

Substituting $66.5\angle -36.86°\text{\hspace{0.17em}MVA}$for ${\mathrm{S}}_{\mathrm{m}\left(\mathrm{actual}\right)}$ and $100\text{\hspace{0.17em}MVA}$ for ${\mathrm{S}}_{\mathrm{m}\left(\mathrm{base}\right)}$ in Equation (2), we have

Determine the per-unit current at the motor.

Substituting$0.665\angle -36.86°\text{\hspace{0.17em}pu}$for ${\mathrm{S}}_{\mathrm{m}\left(\mathrm{pu}\right)}$and $0.95\text{\hspace{0.17em}pu}$for ${\mathrm{V}}_{4\left(\mathrm{pu}\right)}$in Equation (3), we have

$\begin{array}{c}{I}_{m\left(pu\right)}={\left(\frac{0.665\angle -36.86°\text{\hspace{0.17em}pu}}{0.95\text{\hspace{0.17em}pu}}\right)}^{*}\\ =0.56+j0.42\text{\hspace{0.17em}pu}\end{array}$

Determine the per-unit current at the load.

Substituting$0.95+j1.2667\text{\hspace{0.17em}pu}$for $Z_{L\left(pu\right)}$ and $0.95\text{\hspace{0.17em}pu}$ for $V_{4\left(pu\right)}$ in Equation (4), we have

$\begin{array}{c}{I}_{L\left(pu\right)}=\frac{0.95\text{\hspace{0.17em}pu}}{0.95+j1.2667\text{\hspace{0.17em}pu}}\\ =0.36-j0.48\text{\hspace{0.17em}pu}\end{array}$

So, the total current at bus 4 is as follows:

$\begin{array}{c}{I}_{pu}=I_{m\left(pu\right)}+I_{L\left(pu\right)}\\ =0.56+j0.42\text{\hspace{0.17em}pu+}0.36-j0.48\text{\hspace{0.17em}pu}\\ \text{=}0.92-j0.06\text{\hspace{0.17em}pu}\end{array}$

Determine the equivalent reactance:

$\begin{array}{c}{X}_{eq\left(pu\right)}=j(0.2+0.1+0.15)\left|\right|j(0.16+0.54+0.2)\\ =j0.3\text{\hspace{0.17em}pu}\end{array}$

Determine the generator voltage.

Substituting$0.92-j0.06\text{\hspace{0.17em}pu}$for ${\mathrm{I}}_{\left(\mathrm{pu}\right)}$,$j0.3\text{\hspace{0.17em}pu}$ for $X_{eq\left(pu\right)},$ and $0.95\text{\hspace{0.17em}pu}$ for $V_{4\left(pu\right)}$ in Equation (5), we have

$\begin{array}{c}{V}_{1\left(pu\right)}=0.95\text{\hspace{0.17em}pu}+(0.92-j0.06\text{\hspace{0.17em}pu})\left(j0.3\text{\hspace{0.17em}pu}\right)\\ =1\angle 15.91°\text{\hspace{0.17em}pu}\end{array}$

Determine the generator and motor internal EMFs.

(b)

Determine the generator’s internal voltage.

Substituting $0.92-j0.06\text{\hspace{0.17em}pu}$ for $I_{\left(pu\right)}$, $j0.2\text{\hspace{0.17em}pu}$ for $Z_{g\left(pu\right)},$ and $1\angle 15.91°\text{\hspace{0.17em}pu}$ for $V_{1\left(pu\right)}$ in Equation (6), we have

$\begin{array}{c}{E}_{g\left(pu\right)}=1\angle 15.91°+(0.92-j0.06\text{\hspace{0.17em}pu})\left(j0.2\text{\hspace{0.17em}pu}\right)\\ =1.0826\angle 25.14°\text{\hspace{0.17em}pu}\end{array}$

Determine the motor’s internal voltage.

Substituting$0.56+j0.42\text{\hspace{0.17em}pu}$for $I_{m\left(pu\right)}$, $j0.25\text{\hspace{0.17em}pu}$ for $Z_{m\left(pu\right)},$ and $0.95\text{\hspace{0.17em}pu}$ for $V_{4\left(pu\right)}$ in Equation (7), we have

$\begin{array}{c}{E}_{m\left(pu\right)}=0.95\angle 0°-(0.56+j0.42\text{\hspace{0.17em}pu})\left(j0.25\text{\hspace{0.17em}pu}\right)\\ =1.064\angle -7.54°\text{\hspace{0.17em}pu}\end{array}$