Long distance electrical transmission efficiency

In summary, a coil with a 1/2 inch amorphous magnetic core and a resistance of 0.1 ohms per kilometer can generate a current of 2000 A and lose 1.2 * 10^7 W of power. The wire would reach a temperature of 3.17 degrees Celsius.
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MrSargeant
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TL;DR Summary: Just need help with verification of data submitte

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Efficiency distance over 3000 km +, plus Temperature over that distance​


  • 1-inch amorphous magnetic core
  • 0.05 microns graphene
  • 0.10 microns metallic PZT
  • Helical copper wire with 15 mm diameter spaced 0.01 mm apart
  • Pulsed current of 5000 A
B = μ₀ * μ_r * N * I / L

  • B is the magnetic field strength (in Tesla)
  • μ₀ is the vacuum permeability constant (4π x 10^-7 T·m/A)
  • μ_r is the relative permeability of the core material
  • N is the number of turns in the coil
  • I is the current passing through the coil (in amperes)
  • L is the length of the coil (in meters)
Assuming the amorphous magnetic core has a relative permeability (μ_r) of 100,000, a current (I) of 5000 A (pulsed), and approximately 225 turns (N), we need to determine the length (L) of the coil.
If we assume that the coil is wrapped tightly around the 1inch (0.0508 m) core and the wire thickness is 15 mm (0.015 m) with a spacing of 0.01 mm (0.00001 m) between the turns, the length of the coil will be approximately:

L = (0.015 + 0.00001) * 225 = 3.3765 m
Now, we can calculate the magnetic field strength:
B = (4π x 10^-7 T·m/A) * 100,000 * 225 * 5000 A / 3.3765 m ≈ 13,187 T

Assuming a 1/2 inch core and a wire with a resistance of 0.1 ohms per kilometer, the total resistance of the wire over a distance of 3000 km would be:

R_total = R * L = 0.1 ohms/km * 3000 km = 300 ohms
Using the power (P) and voltage (V) levels given, we can calculate the current (I) flowing through the transmission line using Ohm's law:
I = P / V = 1,000,000,000 W / 500,000 V = 2000 A
The power lost due to the resistance of the transmission line can be calculated using Joule's law:
P_loss = I^2 * R_total = (2000 A)^2 * 300 ohms = 1.2 * 10^7 W
To calculate the efficiency of the transmission, we need to subtract the power lost from the total power transmitted:
P_transmitted = P - P_loss = 1,000,000,000 W - 1.2 * 10^7 W = 9.88 * 10^8 W
The efficiency of the transmission is then:
efficiency = P_transmitted / P = 9.88 * 10^8 W / 1,000,000,000 W = 0.988 = 98.8
Below is a hypothetical calculation
  1. Resistance per unit length of the wire (R): 0.1 ohms per kilometer
  2. Power (P): 1 GW (1,000,000,000 W)
  3. Voltage level (V): 500 kV (500,000 V)
  4. Length of the transmission line (L): 3000 km
Now we can calculate the energy efficiency using the following steps:
  1. Calculate the total resistance of the wire (R_total): R_total = R * L = 0.1 ohms/km * 3000 km = 300 ohms
  2. Calculate the current (I) flowing through the wire: I = P / V = 1,000,000,000 W / 500,000 V = 2000 A
  3. Calculate the total power loss in the transmission line due to resistance (P_loss): P_loss = I^2 * R_total = (2000 A)^2 * 300 ohms = 1,200,000,000 W
  4. Calculate the energy efficiency (η) of the transmission line: η = (P - P_loss) / P = (1,000,000,000 W - 1,200,000,000 W) / 1,000,000,000 W = -0.2
Temperature of transmission wire..

Assuming a 1/2 inch core and a wire with a resistance of 0.1 ohms per kilometer, the total resistance of the wire over a distance of 3000 km would be:
R_total = R * L = 0.1 ohms/km * 3000 km = 300 ohms
Using the power (P) and voltage (V) levels given, we can calculate the current (I) flowing through the transmission line using Ohm's law:
I = P / V = 1,000,000,000 W / 500,000 V = 2000 A
The power lost due to the resistance of the transmission line can be calculated using Joule's law:
P_loss = I^2 * R_total = (2000 A)^2 * 300 ohms = 1.2 * 10^7 W
Assuming that the heat generated from the power loss is dissipated into the surrounding environment, the temperature rise of the wire can be estimated using the equation:
ΔT = P_loss / (m * c)
where ΔT is the temperature rise, P_loss is the power loss, m is the mass of the wire, and c is the specific heat capacity of the wire material.
Assuming a copper wire with a diameter of 15 mm and a length of 3000 km, the mass of the wire can be estimated using the density of copper (8.96 g/cm^3):
m = π * (d/2)^2 * L * ρ = π * (1.5 cm / 2)^2 * 3000 km * 8.96 g/cm^3 ≈ 8.88 * 10^9 g
Assuming a specific heat capacity of copper of 0.385 J/g·K, the temperature rise of the wire can be estimated as:
ΔT = P_loss / (m * c) = 1.2 * 10^7 W / (8.88 * 10^9 g * 0.385 J/g·K) ≈ 3.17 K
Therefore, assuming that the heat generated from the power loss is dissipated into the surrounding environment, the temperature rise of the wire over a distance of 3000 km would be approximately 3.17 degrees Celsius.[Moderator's note: moved from a technical forum.]
 
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What's the initial calculation of some coil doing? How is that connected to the rest of the thread?
Why do you calculate some things two or three times?
These things make the post needlessly harder to follow.

Transmission over 3000 km needs 6000 km of wire so your losses will be doubled.
ΔT = P_loss / (m * c) = 1.2 * 10^7 W / (8.88 * 10^9 g * 0.385 J/g·K) ≈ 3.17 K
Check your units, they don't match.

If you don't consider how the wire is cooled, how could you possibly get a temperature difference instead of a rate of temperature increase?
 

What is long distance electrical transmission efficiency?

Long distance electrical transmission efficiency refers to the amount of electricity that is lost during the process of transmitting it over long distances. This can include losses due to resistance in the transmission lines, as well as losses from other factors such as weather and environmental conditions.

Why is long distance electrical transmission efficiency important?

Efficient transmission of electricity over long distances is important because it allows for the delivery of electricity from power plants to homes and businesses. This helps to ensure a reliable and consistent supply of electricity, which is essential for modern society to function.

What factors affect long distance electrical transmission efficiency?

Several factors can affect long distance electrical transmission efficiency, including the type and condition of the transmission lines, the voltage of the electricity being transmitted, and the distance the electricity needs to travel. Weather conditions, such as high temperatures or strong winds, can also impact efficiency.

How is long distance electrical transmission efficiency measured?

Long distance electrical transmission efficiency is typically measured using a metric called the transmission and distribution loss factor. This factor is calculated by dividing the total amount of electricity generated by the amount of electricity delivered to end users.

What are some ways to improve long distance electrical transmission efficiency?

There are several ways to improve long distance electrical transmission efficiency, including using higher voltage transmission lines, implementing better insulation and weatherproofing techniques, and regularly maintaining and upgrading transmission infrastructure. Additionally, using renewable energy sources closer to where the electricity is needed can reduce the need for long distance transmission and improve overall efficiency.

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