Respuesta :
Our values are given as,
[tex]P_{plant} = 1000MW[/tex]
[tex]V = 115kV = 115*10^3V[/tex]
[tex]L = 40Km[/tex]
PART A)
Power is defined as the product between the current and the voltage, therefore the current would be
[tex]I = \frac{P}{V}[/tex]
[tex]I = \frac{1000}{115*10^3}[/tex]
[tex]I = 8704.34A[/tex]
PART B) The resistance unit length is,
[tex]\frac{R}{L} = 0.05\Omega /km[/tex]
[tex]R = 0.05*40[/tex]
[tex]R = 2\Omega[/tex]
Therefore for the calculation of the loss of Power we would have to apply the relation:
[tex]P_{l} = I^2 R[/tex]
[tex]P_l = (8704.34)^2*2[/tex]
[tex]P_l = 151.53MW[/tex]
Fraction loss power transmission is,
[tex]\frac{P_l}{P_{plant}} = \frac{151.53}{1000}[/tex]
[tex]\frac{P_l}{P_{plant}} = 0.151[/tex]
[tex]\frac{P_l}{P_{plant}} = 15.1\%[/tex]
(a) The current in the wire is [tex]8704\;A[/tex]
(b) The fraction of power lost in transmission is 15.1%
Electrical power:
Given that the electrical power produced is [tex]P=1000MW=1000\times10^6W[/tex],
the distance of the city is d = 40km,
resistance of wire per kilometer length is [tex]r=0.05\Omega /km[/tex] ,
and the potential difference [tex]V=115kV=115\times10^3V[/tex]
(a) The current is given by the following relation:
[tex]P=VI\\\\I=\frac{1000\times10^6}{115\times10^3}A=8704\;A[/tex]
(b) Power lost in the transmission:
the total resistance of the wire in 40 km is:
[tex]R=0.05\times40=2\Omega[/tex]
so the power lost in transmission is:
[tex]P_t=I^2R=(8704)^2\times2=151MW[/tex]
So the fraction of power lost is:
[tex]f=\frac{P_t}{P}\times100\%=\frac{151}{1000}\times100\% \\\\f=15.1\%[/tex]
Learn more about electrical power:
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