Question

Reconsider Problem 5.7 and find the following: (a) sending-end power factor,(b) sending-end three-phase power, and (c) the three-phase line loss.

Short answer

(a) Thesource power factor is0.768.

(b) The three phase power is 746kW .

(c) The power loss is 23.96kW.

Step-by-step solution

Write the given data.

The source voltage is $V_S=\frac{3.3\times {10}^3}{\sqrt{3}}V$.

The line to line receiving end voltage is 3063.99V and phase voltage is 1769V.

The resistance and reactance is $\mathrm{R}=0.3\mathrm{\Omega }$and $\mathrm{X}=0.4\mathrm{\Omega }$.

Determine the load angle is${\varnothing }_R=36.{85}^{°}$.

The load current is $l_R=163.18A$.

Determine the formulas of source power factor, sending end three-phase power and line power loss.

Write the formula of source power factor.

$\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{R}}\mathbf{cos}{\mathbf{\varnothing }}_{\mathbf{R}}\mathbf{+}{\mathbf{l}}_{\mathbf{R}}\mathbf{R}}{{\mathbf{V}}_{\mathbf{S}}}\mathbf{}...........\left(1\right)$

Here, $l_R$is line current, ${\varnothing }_R$is load angle, $V_R$is receiving end voltage,$V_S$is sending end voltage and R is resistance.

Write the formula of sending end power.

${\mathbf{P}}_{\mathbf{S}}\mathbf{=}\mathbf{3}{\mathbf{V}}_{\mathbf{S}}{\mathbf{l}}_{\mathbf{R}}\mathbf{cos}{\mathbf{\varnothing }}_{\mathbf{R}}\mathbf{}............\left(2\right)$

Here, $l_R$is line current, ${\varnothing }_R$is load angle, and $V_S$is sending end voltage.

Write the formula of power loss in line.

${\mathbf{P}}_{\mathbf{L}}\mathbf{=}\mathbf{3}{\mathbf{l}}_{\mathbf{R}}^{\mathbf{2}}\mathbf{R}\mathbf{}................\left(3\right)$

Determine the source power factor.

(a)

$$\begin{gathered} \mathrm{Substitute}1769\mathrm{V}\mathrm{for}{\mathrm{V}}_{\mathrm{R}},36.{85}^{°}\mathrm{for}{\varnothing }_{\mathrm{R}},163.18\mathrm{Afor}{\mathrm{l}}_{\mathrm{R}},\frac{3.3\times {10}^3}{\sqrt{3}}\mathrm{for}{\mathrm{V}}_{\mathrm{S}}\mathrm{and} \\ 0.3\mathrm{\Omega }\mathrm{for}\mathrm{R}\mathrm{in}\mathrm{equation}\left(1\right). \\ \mathrm{p}.\mathrm{f}=\frac{\left(1769\mathrm{V}\right)\cos 36.{85}^{°}+(163.18\mathrm{A})(0.3\mathrm{\Omega })}{{\displaystyle \frac{3.3\times {10}^3}{\sqrt{3}}}} \\ =0.768 \end{gathered}$$

Therefore, the source power factor is0.768.

Determine the three phase power.

(b)

$$\begin{gathered} \mathrm{Substitute}36.{85}^{°}\mathrm{for}{\varnothing }_{\mathrm{R}},163.18\mathrm{A}\mathrm{for}{\mathrm{l}}_{\mathrm{R}},\frac{3.3\times {10}^3}{\sqrt{3}}\mathrm{for}{\mathrm{V}}_{\mathrm{S}}\mathrm{in}\mathrm{equation}\left(2\right). \\ {\mathrm{P}}_{\mathrm{S}}=3\left[\frac{3.3\times {10}^3}{\sqrt{3}}\right]\left(163.18\mathrm{A}\right)\cos 36.{85}^{°} \\ =746\mathrm{kW} \end{gathered}$$

Therefore, the three phase power is 746 kW.

Determine the three phase power.

$$\begin{gathered} \left(\mathrm{c}\right) \\ \mathrm{Substitute}163.18\mathrm{A}\mathrm{for}{\mathrm{l}}_{\mathrm{R}},\mathrm{and}0.3\mathrm{\Omega }\mathrm{for}\mathrm{R}\mathrm{in}\mathrm{equation}\left(3\right). \\ {\mathrm{P}}_{\mathrm{L}}=3{(163.18\mathrm{A})}^2(0.3\mathrm{\Omega }) \\ =23.96\mathrm{kW} \\ \mathrm{Therefore},\mathrm{the}\mathrm{power}\mathrm{loss}\mathrm{is}23.96\mathrm{kW}. \end{gathered}$$