Question

(a) What current (in kA) is needed to transmit 180 MW of power at a voltage of 29.0 kV? kA(b) Find the power loss (in MW) in a 1.50 Ω transmission line. MW(c) What percent loss does this represent? %

Answer 1

Given:

The power is

$\begin{gathered} P=180\text{ MW} \\ =180\times10^6\text{ W} \end{gathered}$

The voltage is

$\begin{gathered} V=29.0\text{ kV} \\ =29.0\times10^3\text{ V} \end{gathered}$

The resistance is

$R=1.50\text{ }\Omega$

To find:

a) the current

b) the power loss

c) the percent loss

Explanation:

a) The current is,

$\begin{gathered} i=\frac{P}{V} \\ =\frac{180\times10^6}{29.0\times10^3} \\ =6.21\times10^3\text{ A} \\ =6.21\text{ kA} \end{gathered}$

Hence, the current is 6.21 kA.

b)

The power loss is,

$\begin{gathered} P_{loss}=i^2R \\ =(6.21\times10^3)^2\times1.50 \\ =57.8\times10^6\text{ W} \\ =57.8\text{ MW} \end{gathered}$

Hence, the power loss is 57.8 MW.

c)

The percentage loss is

$\begin{gathered} \frac{P_{loss}}{P}\times100\text{ \%} \\ =\frac{57.8\times10^6}{180\times10^6}\times100\text{ \%} \\ =32.1\text{ \%} \end{gathered}$

Hence, the power loss percentage is 32.1 %.