Respuesta :
Answer:
(a) 30.174 kW
(b) [tex]6.035\times 10^{- 3}\ N[/tex]
(c) I = 1.112 kA
Solution:
As per the question:
Resistance per unit length, [tex]\sigma = 2.40\times 10^{- 4}\Omega/ m[/tex]
Power transmitted, P = 5.00 MW = [tex]5\times 10^{6}\ W[/tex]
Length, L = [tex]6.44\times 10^5\ m[/tex]
Output voltage, V = 4.50 kV
Stepped-up Voltage, V' = 506 kV
Now,
(a) The line loss can be calculated as:
Current, I = [tex]\frac{P}{V'} = \frac{5\times 10^{6}}{506\times 10^{3}} = 9.88\ A[/tex]
To calculate the power loss:
[tex]P_{L} = I^{2}R[/tex]
where
R = Resistance in the transmission line
R = [tex]2\sigma L = 2\times 2.40\times 10^{- 4}\times 6.44\times 10^5 = 309.12\Omega [/tex]
Thus
[tex]P_{L} = 9.88^{2}\times 309.12 = 30.174\ kW[/tex]
(b) Fraction of the input power lost is given by:
[tex]f = \frac{P_{L}}{P} = \frac{30.174\times 10^{3}}{5\times 10^{6}} = 6.035\times 10^{- 3}\ W[/tex]
(c) To calculate the current at the sending end at ther source voltage of 4.50 kV:
[tex]I = \frac{P}{V} = \frac{5\times 10^{6}}{4.50\times 10^{3}} = 1.112\ kA[/tex]
This value of current is very high and sending such a high watt power is impossible, even if the circuit at the other end is open then also this current can be destructive.
