Answer:
The diameter of the wire is 1.45cm
Explanation:
We can obtain the new voltage value using the transformer relationship:
[tex]V=V_o*n_t\\V=30kV*20\\V=600kV[/tex]
Having the new voltage we can obtain the current value going through the wires:
[tex]I=\frac{P}{V}\\I=\frac{330*10^3kW}{600kV}\\I=550A[/tex]
We know that 2.5% of power is loss because the wire, so the power dissipated by the wires is:
[tex]P_w=330MW*(0.025)\\P_w=8.25MW[/tex]
The resistance value of the wire is given by:
[tex]P_w=i^2*R\\R=\frac{P_w}{I^2}\\\\R=\frac{8.25MW}{(550A)^2}\\\\R=27.3ohm[/tex]
We also know that:
[tex]R=\rho*\frac{L}{A}\\A=1.72*10^{-8}*\frac{(260*10^3)}{27.3}\\A=164\µm\\\\A=\pi*(\frac{d}{2})^2\\d=1.45cm[/tex]
Based on the data provided, the diameter of the wire is 1.44 cm
Electrical power is the rate at which electrical energy is expended.
where;
Using the transformer formula to obtain the new voltage V:
V = V₀ * N₁/N₂
V = 30 kV * 20/1
V = 600 kV
Current passing through the wires is calculated thus:
I = 330 * 1000 kW/600
I = 550 A
2.5% of power is dissipated in the wires
Power dissipated, Pw = 0.025 * 330 MW
Pw = 8.25 MW
The resistance in the wires is determined using the formula;
R = Pw/I²
R = 8.25 * 1000 / (550)²
R = 27.3 ohms
The diameter of the wire is determined using the formula;
R = ρL/A
A = ρL/R
where
A = area of wire
ρ = resistivity of wire
L = length of wire
A = 1.72 × 10⁻⁸ Ω.m * 260 * 1000 m / 27.3 Ω
A = 1.64 * 10⁻⁴ m²
Also,
A = πd²/4
d = √4*A/π
d = √(4 * 1.64 * 10⁻⁴ m² * 7/22)
d = 0.0144 m
d = 1.44 cm
Therefore, the diameter of the wire is 1.44 cm
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