Simple calculation of total cable diameter and cable flow current 诀窍 and resistivity analysis

I. Longitudinal Voltage Drop and Resistance The voltage drop across a copper wire is directly related to its resistance. The resistance of a copper wire can be calculated using the following formulas: At 20°C: 17.5 ÷ cross-sectional area (in square mm) = resistance per kilometer (Ω) At 75°C: 21.7 ÷ cross-sectional area (in square mm) = resistance per kilometer (Ω) According to Ohm's Law, the pressure drop calculation formula is V = R × I, where V represents the voltage drop, R is the resistance, and I is the current. Line loss is associated with both the voltage drop and the current used. The line loss power can be calculated as P = V × I, where P is the power loss in watts, V is the voltage drop in volts, and I is the current in amperes. II. Copper Core Wire Current Carrying Capacity Calculation Method The safe current carrying capacity for copper wires varies depending on their cross-sectional area: - 1 mm²: 17A - 1.5 mm²: 21A - 2.5 mm²: 28A - 4 mm²: 35A - 6 mm²: 48A - 10 mm²: 65A - 16 mm²: 91A - 25 mm²: 120A For single-phase loads, each kilowatt corresponds to approximately 4.5A when the power factor (cosφ) is 1. This allows for the calculation of current and the selection of appropriate conductor sizes. III. Comparison of Current Carrying Capacities Between Copper and Aluminum Wires There are standard equivalencies between copper and aluminum wires based on their cross-sectional areas: - 2.5 mm² copper ≈ 4 mm² aluminum - 4 mm² copper ≈ 6 mm² aluminum - 6 mm² copper ≈ 10 mm² aluminum These comparisons help in selecting the right size of wire for specific applications. For instance, 2.5 mm² copper wire is equivalent to 4 mm² aluminum wire in terms of current carrying capacity. Similarly, 4 mm² copper is equivalent to 6 mm² aluminum, and so on. IV. Wire Diameter Calculation Wire diameter can be calculated using the following formulas: For copper wires: S = IL / (54.4 × U') For aluminum wires: S = IL / (34 × U') Where: - I: Maximum current through the wire (in A) - L: Length of the wire (in meters) - U': Voltage drop (in volts) - S: Cross-sectional area of the wire (in mm²) V. Current Carrying Capacity and Cross-Sectional Area Relationship The current carrying capacity of insulated wires depends on their cross-sectional area and environmental conditions. The general rule is that for wires below 10 mm², the current carrying capacity is about five times the cross-sectional area. For wires above 10 mm², the current carrying capacity decreases gradually. For example, a 2.5 mm² copper wire has a current carrying capacity of around 22.5A, while a 4 mm² copper wire has a current carrying capacity of about 32A. VI. Resistance Per Kilometer for Copper and Aluminum Wires Below are the resistances per kilometer for copper and aluminum wires at different cross-sectional areas: Copper Wires: - 1.0 mm²: 17.5 Ω/km - 1.5 mm²: 11.7 Ω/km - 2.5 mm²: 7.00 Ω/km - 4 mm²: 4.38 Ω/km - 6 mm²: 2.92 Ω/km - 10 mm²: 1.75 Ω/km - 16 mm²: 1.10 Ω/km - 25 mm²: 0.7 Ω/km - 35 mm²: 0.5 Ω/km - 50 mm²: 0.35 Ω/km - 70 mm²: 0.25 Ω/km - 95 mm²: 0.194 Ω/km - 120 mm²: 0.146 Ω/km - 150 mm²: 0.117 Ω/km Aluminum Wires: - 1.0 mm²: 28.3 Ω/km - 1.5 mm²: 18.9 Ω/km - 2.5 mm²: 11.32 Ω/km - 4 mm²: 7.1 Ω/km - 6 mm²: 4.72 Ω/km - 10 mm²: 2.83 Ω/km - 16 mm²: 1.77 Ω/km - 25 mm²: 1.13 Ω/km - 35 mm²: 0.81 Ω/km - 50 mm²: 0.57 Ω/km - 70 mm²: 0.41 Ω/km - 95 mm²: 0.314 Ω/km - 120 mm²: 0.24 Ω/km - 150 mm²: 0.19 Ω/km VII. Example Calculation To determine the required cable size for a 500W load, consider whether it is single-phase or three-phase: Single-phase: I = P / U = 500 / 220 ≈ 2.27A Three-phase: I = P / (1.732 × U) = 500 / (1.732 × 380) ≈ 0.75A Voltage drop is calculated as V = I × R, where R is the resistance of the wire. The resistance of a copper wire can be determined using the formula R = ρ × L / S, where ρ is the resistivity, L is the length, and S is the cross-sectional area. VIII. Resistivity of Different Materials at 20°C Resistivity values for various materials at 20°C are as follows: - Silver: 1.65 × 10⁻⁸ Ω·m - Copper: 1.75 × 10⁻⁸ Ω·m - Aluminum: 2.83 × 10⁻⁸ Ω·m - Tungsten: 5.48 × 10⁻⁸ Ω·m - Iron: 9.78 × 10⁻⁸ Ω·m - Platinum: 2.22 × 10⁻⁷ Ω·m - Manganese Copper: 4.4 × 10⁻⁷ Ω·m - Mercury: 9.6 × 10⁻⁷ Ω·m - Constantan: 5.0 × 10⁻⁷ Ω·m - Nichrome: 1.0 × 10⁻⁶ Ω·m - Iron-Chromium-Aluminum Alloy: 1.4 × 10⁻⁶ Ω·m - AlNi-Fe Alloy: 1.6 × 10⁻⁶ Ω·m - Graphite: (8–13) × 10⁻⁶ Ω·m Understanding these resistivity values helps in selecting the appropriate material for specific electrical applications, ensuring optimal performance and safety.

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