Question

The following parameters are based on a preliminary line design: $V_s=1$per unit, $V_R=0.9$per unit, $\lambda =5000km$,$Z_c=320\Omega ,\delta =36.8°$ . A three-phase power of$700MW$ is to be transmitted to a substation located$315km$ from the source of power. (a) Determine a nominal voltage level for the three-phase transmission line, based on the practical line load ability equation. (b) For the voltage level obtained in part (a), determine the theoretical maximum power that can be transferred by the line.

Short answer

(a) The nominal voltage level for three phase transmission line is $400\text{\hspace{0.33em}kV}$.

(b) The maximum power that can be transmitted by the line$1167.9\text{\hspace{0.33em}MW}$ is .

Step-by-step solution

 Write the given data from the question.

The sending end voltage $V_S=1\text{\hspace{0.33em}pu}$.

The receiving end voltage.$V_R=0.9\text{\hspace{0.33em}pu}$

Wavelength of the line $\lambda =5000\text{\hspace{0.33em}km}$.

Characteristics impedance $Z_c=320\Omega$.

Three phase power $P=700\text{\hspace{0.33em}MW}$.

The length of the line $315\text{\hspace{0.33em}km}$.

The phase angle $\delta =36.8°$.

Determine the equation to calculate the nominal voltage level and maximum power that can be transmit by the line.

The equation to calculate the line phase constant is given as follows.

$\beta l=\frac{2\pi }{\lambda }l$ …… (1)

The equation to calculate the power transmit by the line is given as follows.

$P_{3f}=\frac{V_s{V}_R\left(SIL\right)}{sin\left(\beta l\right)}sin\delta$ …… (2)

Derive the expression for surge impedance loading as,

$SIL=\frac{P_{3f}sin\left(\beta l\right)}{V_S{V}_{R}sin\delta }$ …… (3)

The equation to calculate the rated voltage is given as follows.

$V_{rated}=\sqrt{Z_c\left(SIL\right)}$ …… (4)

The equation to calculate the equivalent reactance for the losses line is given as follows.

$X\text{'}=Z_{c}sin\left(\beta l\right)$ …… (5)

The equation to calculate the maximum power that can be transmitted by the line is given as follows.

$P_R=\frac{V_R{V}_S}{X\text{'}}sin\delta$ …… (6)

Calculate the nominal voltage level for three phase transmission line.

(a)

Calculate the line phase constant.

Substitute $5000\text{\hspace{0.33em}km}$for $\lambda$and$315\text{\hspace{0.33em}km}$for $l$into equation (1).

$\begin{array}{l}\beta l=\frac{2\pi }{5000\times {10}^3}\times 315\times {10}^3\\ \beta l=\frac{630\pi }{5000}\\ \beta l=0.3958\text{\hspace{0.33em}rad}\\ \beta l=22.68°\end{array}$

Calculate the surge impedance loading.

Substitute $1\text{\hspace{0.33em}pu}$for $V_S$, data-custom-editor="chemistry" $0.9\text{\hspace{0.33em}pu}$for $V_R$, $36.87°$for $\delta$,$700\text{\hspace{0.33em}MW}$for $P_{3\varphi }$ and $22.68°$for $\beta l$into equation (3).

$\begin{array}{c}SIL=\frac{700\times {10}^6\times \sin \left(22.68\right)}{1\times 0.9\times \sin \left(36.87\right)}\\ =\frac{700\times {10}^6\times 0.385}{0.9\times 0.6}\\ =\frac{269.5\times {10}^6}{0.54}\\ =499\text{\hspace{0.33em}MW}\end{array}$

Calculate the rated voltage.

Substitute $499\text{\hspace{0.33em}MW}$for $\text{SIL}$and$320\Omega$into equation (4).

$\begin{array}{l}{V}_{rated}=\sqrt{320\times 499\times {10}^6}\\ V_{rated}=\sqrt{159.68\times {10}^9}\\ V_{rated}=400\text{\hspace{0.33em}kV}\end{array}$

Hence, the nominal voltage level for three phase transmission line is $400\text{\hspace{0.33em}kV}$.

Calculate the maximum power that can be transmit by the line.

Calculate the equivalent reactance.

Substitute $320\Omega$for $Z_c$and $22.68°$ for $\beta l$ into equation (5).

$\begin{array}{l}X\text{'}=320\times \sin \left(22.68\right)\\ X\text{'}=320\times 0.3855\\ X\text{'}=123.39\Omega \end{array}$

The maximum power is transmitted under steady state condition which occurs at $90°$load angle.

Calculate the maximum power transmit by the line.

Substitute$400\text{\hspace{0.33em}kV}$ for $V_S$and$360\text{\hspace{0.33em}kV}$ for $V_R$ and $123.39\Omega$ for $X\text{'}$into equation (6).

$\begin{array}{l}{P}_{R,\max }=\frac{360\times {10}^3\times 400\times {10}^3}{123.29}\\ P_{R,\max }=\frac{144\times {10}^9}{123.29}\\ P_{R,\max }=1.1679\times {10}^6\\ P_{R,\max }=1167.9\text{\hspace{0.33em}MW}\end{array}$

Therefore, the maximum power that can be transmitted by the line is $1167.9\text{\hspace{0.33em}MW}$.