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*10^6\text{ W} \end{gathered}`

The voltage is

`\begin{gathered} V=29.0\text{ kV} \\ =29.0*10^3\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

Step-by-step explanation:

a) The current is,

`\begin{gathered} i=(P)/(V) \\ =(180*10^6)/(29.0*10^3) \\ =6.21*10^3\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*10^3)^2*1.50 \\ =57.8*10^6\text{ W} \\ =57.8\text{ MW} \end{gathered}`

Hence, the power loss is 57.8 MW.

c)

The percentage loss is

`\begin{gathered} (P_(loss))/(P)*100\text{ \%} \\ =(57.8*10^6)/(180*10^6)*100\text{ \%} \\ =32.1\text{ \%} \end{gathered}`

Hence, the power loss percentage is 32.1 %.