Question

For the power system in Problem 3.41, the synchronous motor absorbs $1500MW$at 0.8 power factor leading with the bus 3 voltage at$18kV$.Determine the bus 1 and bus 2 voltages in$kV$. Assume that generators 1and 2 deliver equal real powers and equal reactive powers. Also assume abalanced three-phase system with positive-sequence sources.

Short answer

The voltages at bus 1 and 2 is .$\text{14.91\hspace{0.33em}V}$

Step-by-step solution

Determine the formulas ofper-unit voltage, base current, actual current, and per-unit current

Write the formula for per-unit voltage:

$V_{3\left(pu\right)}=\frac{V_{base3}}{V_{actual3}}$ …… (1)

Write the formula for the base current in bus 3:

$I_{base3}=\frac{S_{base}}{\sqrt{3}{V}_{base3}}$…… (2)

Write the formula for the actual current in bus 3:

$I_{actual3}=\frac{P}{\sqrt{3}{V}_{actual3}cos\theta }$ …… (3)

Write the formula of the per-unit current in bus 3:

$I_{3\left(pu\right)}=\frac{I_{base3}}{I_{actual3}}$ …… (4)

Determine the per-unit voltage and the current at bus 3.

As it is knows that the base voltage at bus 3 is $V_{base3}=20\text{\hspace{0.33em}kV}$.

The actual voltage is $V_{actual3}=18\text{\hspace{0.33em}kV.}$

Determine the per-unit voltage at bus 3 using Equation (1):

$\begin{array}{c}{V}_{3\left(pu\right)}=\frac{18\text{\hspace{0.33em}kV}}{20\text{\hspace{0.33em}kV}}\\ \text{=0.9\hspace{0.33em}pu}\end{array}$

Determine the base current in bus 3.

Substituting$20\text{\hspace{0.33em}kV}$for $V_{base3}$ and $1000\text{\hspace{0.33em}MVA}$for$S_{base}$ in Equation (2), we have

$\begin{array}{c}{I}_{base3}=\frac{1000\text{\hspace{0.33em}MVA}}{\sqrt{3}\times 20\text{\hspace{0.33em}kV}}\\ \text{=28.87\hspace{0.33em}kA}\end{array}$

Determine the actual current in bus 3.

Substituting$18\text{\hspace{0.33em}kV}$for , $V_{actual3}$ $0.8$for $\cos \theta ,$and $1500\text{\hspace{0.33em}MW}$ for $P$ in Equation (3), we have

$\begin{array}{c}{I}_{actual3}=\frac{1500\text{\hspace{0.33em}MW}}{\sqrt{3}\times 18\text{\hspace{0.33em}kV}\times \text{0.8}}\\ \text{=60.14}\angle \text{36.86}°\text{\hspace{0.33em}kA}\end{array}$

Determine the per-unit current at bus 3 using Equation (4):

$\begin{array}{c}{I}_{3\left(pu\right)}=\frac{\text{60.14}\angle \text{36.86}°\text{\hspace{0.33em}kA}}{\text{28.87\hspace{0.33em}kA}}\\ \text{=2.083}\angle \text{36.86}°\text{\hspace{0.33em}pu}\end{array}$

Determine the voltages at bus 1 and bus 2.

It is known the values are, $X_{T5}=0.0667\text{\hspace{0.33em}pu}$, $X_{T3}=0.1\text{\hspace{0.33em}pu,}$and $X_{Line25}=0.1\text{\hspace{0.33em}pu}$.

Determine the bus voltage.

$\begin{array}{c}{V}_{1\left(pu\right)}=V_{3\left(pu\right)}+j{I}_{3\left(pu\right)}\left(X_{T5}+\frac{X_{Line25}+X_{T3}}{2}\right)\\ =0.9+j\text{2.083}\angle \text{36.86}°\left(0.0667+\frac{0.1+0.1}{2}\right)\\ =0.7454\angle 21.\text{88}°\text{\hspace{0.33em}pu}\end{array}$

Now, the actual voltage at bus 3 using equation (1) is as follows:

$\begin{array}{l}0.7454\text{\hspace{0.33em}pu}=\frac{V_{actual3}}{20\text{\hspace{0.33em}kV}}\\ V_{actual3}=\text{14.91\hspace{0.33em}V}\end{array}$

Therefore, the voltages of bus 1 and bus 2 are .$\text{14.91\hspace{0.33em}V}$