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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 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)
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 - Resistance per unit length of the wire (R): 0.1 ohms per kilometer
- Power (P): 1 GW (1,000,000,000 W)
- Voltage level (V): 500 kV (500,000 V)
- Length of the transmission line (L): 3000 km
- Calculate the total resistance of the wire (R_total): R_total = R * L = 0.1 ohms/km * 3000 km = 300 ohms
- Calculate the current (I) flowing through the wire: I = P / V = 1,000,000,000 W / 500,000 V = 2000 A
- 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
- 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
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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