# Long distance electrical transmission efficiency

• MrSargeant
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.
MrSargeant
TL;DR Summary: Just need help with verification of data submitte

1 of 94

## 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.]

Last edited by a moderator:
DeBangis21
1 of 94?

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 factors affect the efficiency of long-distance electrical transmission?

The efficiency of long-distance electrical transmission is influenced by several factors, including the resistance of the transmission lines, the voltage level, the quality of the conductors, the distance the electricity must travel, and the presence of any transformers or substations along the route. Higher voltage levels typically reduce losses, while higher resistance and longer distances increase them.

## How do high-voltage transmission lines reduce energy loss?

High-voltage transmission lines reduce energy loss by lowering the current for a given amount of power. Since power loss due to resistance in the wires (I²R losses) is proportional to the square of the current, reducing the current by increasing the voltage significantly decreases these losses. This is why electricity is transmitted at high voltages over long distances and then stepped down to lower voltages for local distribution.

## What role do transformers play in long-distance electrical transmission?

Transformers are crucial in long-distance electrical transmission as they enable the efficient transfer of electricity at different voltage levels. Step-up transformers increase the voltage for transmission over long distances, which reduces current and minimizes losses. Step-down transformers then decrease the voltage to safer levels for distribution and use in homes and businesses.

## Why is alternating current (AC) typically used for long-distance transmission instead of direct current (DC)?

Alternating current (AC) is typically used for long-distance transmission because it is easier to transform between different voltage levels using transformers. This capability allows for efficient high-voltage transmission and subsequent reduction to lower voltages for end use. However, high-voltage direct current (HVDC) is also used in specific scenarios where it can be more efficient over very long distances or underwater cables due to lower losses and the absence of reactive power issues.

## What are some modern technologies improving the efficiency of long-distance electrical transmission?

Several modern technologies are improving the efficiency of long-distance electrical transmission, including high-voltage direct current (HVDC) systems, which offer lower losses over very long distances; advanced conductor materials like high-temperature superconductors, which have near-zero resistance; and smart grid technologies that optimize the flow of electricity and reduce losses. Additionally, innovations in power electronics and grid management systems are contributing to more efficient and reliable transmission networks.

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