Chapter 21, Problem 33P

To determine

(a)The emf at the generates end of the lines

(b)Fraction power loss in the lines.

Answer to Problem 33P

Solution:

(a)Emf = 49 Kv

(b)Fractional power loss = 0.36

Explanation of Solution

Given info:

Power = 35MW = ${\text{35×10}}^{\text{6}}\text{ W}$ (arrives at the town)

Voltage (rms)= ${\text{45×10}}^3\text{ v}$

$\text{= 45 Kv}$[Since rms = root square for defining current of AC wave]

Resistance(ohm)= 4.6Ώ.

(a) Formula to be used:

Using the formula of power,

P = Workdone by unit time

P = VI

Calculation:

Applying values in power formula,as we need current first.

$\begin{array}{l}\text{I=}\frac{\text{P}}{\text{V}}\\ =\frac{{\text{35×10}}^6}{{\text{45×10}}^3}\\ =0.77\times {10}^3\\ \text{=777 A}\end{array}$

The current in wire, (ohm’s law)

E = IR + v [E =Emf ]

$\begin{array}{l}\text{= 777×4}{\text{.6+45×10}}^{\text{3}}\\ \text{= 48574}\text{.2 v}\\ \text{= 49 Kv }\left(\text{Approximately}\right)\end{array}$

(b) Formula to be used:

The power loss in the lines $P_L$,

$\begin{array}{l}{\text{P}}_{\text{L}}{\text{= I}}^{\text{2}}\text{R}\\ \end{array}$

Calculation:

$\begin{array}{l}{\text{P}}_{\text{L}}{\text{= I}}^{\text{2}}\text{R}\\ {\text{ =(777)}}^{\text{2}}\text{×(4}\text{.6)}\\ \text{ = 12}\text{.8 Mw }\left(\text{Approximately}\right)\end{array}$

Fractional power loss $\text{=}\frac{{\text{P}}_{\text{L}}}{{\text{P}}_{\text{T}}}$

$\begin{array}{l}\text{=}\frac{\text{12}\text{.8}}{\text{35}}\\ \text{=0}\text{.36}\end{array}$

% fractional power loss=$36\%$