Question

The per-phase impedance of a short three-phase transmission line is$\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\angle }\mathbf{53}\mathbf{.}{\mathbf{15}}^{\mathbf{°}}\mathbf{\Omega }$. The three-phase load at the receiving end is1200 kWat0.8p.f. lagging. If the line-to-line sending-end voltage is 3.3kV, determine (a) the receiving-end line-to-line voltage inand (b) the line current. Draw the phasor diagram with the line current l, as reference.

Short answer

(a) The receiving end voltage is .

(b) Theload current is .

Phasor diagram is,

Step-by-step solution

Given data.

The three-phase load absorbs${\mathrm{P}}_{3-\varnothing }=1200\mathrm{kW}$.

The sending end voltage is${\mathrm{V}}_{\mathrm{S}}=3.3\mathrm{KV}$.

The per-phase impedance is$\mathrm{z}=0.5\angle 53.{15}^{°}\mathrm{\Omega }$.

The power factor isp.f = 0.8.

The power angle$\varnothing =53.13$.

Determine the formulas of load current and sending end voltage.

Write the formula of line current.

${\mathbf{l}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{P}}{\left(V_R\right)\left(p.f\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, P is active power,$V_R$is receiving end voltage and p.f is power factor.

Write the formula of sending end voltage.

${\mathbf{V}}_{\mathbf{S}}^{\mathbf{2}}\mathbf{=}{\left(V_R\cos {\varnothing }_R+l_{R}R\right)}^{\mathbf{2}}\mathbf{+}{\left(V_R\sin {\varnothing }_R+l_{R}X\right)}^{\mathbf{2}}$ ……. (2)

Here, $l_R$is line current, ${\varnothing }_{\mathrm{R}}$ is load angle, Ris resistance and X is reactance of load.

Determine the receiving end voltage.

(a)

The per-phase power is,

$$\begin{gathered} P=\frac{P_{3-\varnothing }}{3} \\ =\frac{1200\mathrm{kW}}{3} \\ =400\mathrm{kW} \end{gathered}$$

Determine the load current.

Substitute 400 kW for P and 0.8for p.fin equation (1).

localid="1655288022479" $$\begin{gathered} l_R=\frac{400kW}{V_R\times 0.8} \\ =\frac{500\times {10}^3}{V_R} \end{gathered}$$ ……. (3)

Determine the resistance and reactance.

localid="1655288026825" $$\begin{gathered} z=0.5\angle 53.{15}^{°}\Omega \\ =(0.5\cos 53.{15}^{°}+j0.5\sin 53.{15}^{°})\Omega \\ =(0.3+j0.4)\Omega \\ \mathrm{Hence},\mathrm{R}=0.3\mathrm{\Omega }\mathrm{and}\mathrm{X}=0.4\mathrm{\Omega } \\ \mathrm{Determine}\mathrm{the}\mathrm{load}\mathrm{angle}. \\ {\varnothing }_{\mathrm{R}}={90}^{°}-53.{15}^{°} \\ =36.{85}^{°} \\ \mathrm{Determine}\mathrm{the}\mathrm{sending}\mathrm{end}\mathrm{current}. \end{gathered}$$

Therefore, the line voltage is ,

$$\begin{gathered} V_R=1769\times \sqrt{3} \\ =3063.99V \end{gathered}$$

Therefore, the receiving end voltage is 3.063kV.

Determine the load current and the phasor diagram

(b)

Substitute 3063.99Vfor $V_R$in equation (3).

$$\begin{gathered} l_R=\frac{500\times {10}^3}{3063.99} \\ =163.18A \end{gathered}$$

Therefore, theload current is 163.18A .

Draw the phasor diagram below.