Design a transmission line that can transmit 300 MW of electricity across 200 km with only a 29% loss.
Assume the wires are to be made of aluminum with 2.82×10-8 .m of resistivity and the transmission
voltage is 600 kV. (
To design a transmission line that can transmit 300 MW of electricity across 200 km with only a 29% loss, we need to calculate the parameters of the transmission line that will meet these requirements.
1. Determine the power loss:
The power loss is given as 29% of the transmitted power, which is 0.29 * 300 MW = 87 MW.
2. Calculate the resistance of the transmission line:
The resistance of the transmission line can be calculated using the formula:
Resistance (R) = (resistivity * length) / cross-sectional area
To find the cross-sectional area, we need to determine the required current carrying capacity of the transmission line.
Current (I) = Power / Voltage
I = 300 MW / 600 kV = 500 A
To calculate the cross-sectional area (A), we can use the formula:
A = (Current / Current density)
Assuming a current density of 5 A/mm^2, we have:
A = 500 A / (5 A/mm^2) = 100 mm^2
Converting mm^2 to m^2, we get:
A = 0.0001 m^2
Now, we can calculate the resistance of the transmission line:
R = (2.82 x 10^-8 Ω.m) * (200,000 m) / (0.0001 m^2)
R = 5.64 Ω
3. Calculate the total impedance of the transmission line:
The total impedance (Z) is the sum of the resistance and the reactance of the transmission line.
Z = √(R^2 + X^2)
Assuming a negligible reactance (X), we can omit it.
Z = √(5.64 Ω^2 + 0 Ω^2)
Z = 5.64 Ω
4. Calculate the voltage drop:
Voltage drop (Vd) = Power loss / Current
Vd = 87 MW / 500 A
Vd = 174 kV
Now, we have the necessary parameters for the design of the transmission line:
Resistance (R) = 5.64 Ω
Impedance (Z) = 5.64 Ω
Voltage drop (Vd) = 174 kV
By using these parameters, we can design a transmission line that can transmit 300 MW of electricity across 200 km with a 29% loss.