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Simple calculation of total cable diameter and cable flow current è¯€çª and resistivity analysis
I. Longitudinal
The voltage drop across a copper wire is directly related to its resistance. The resistance of the wire can be calculated using the following formulas:
At 20°C: Resistance per kilometer (Ω) = 17.5 ÷ Cross-sectional area (mm²)
At 75°C: Resistance per kilometer (Ω) = 21.7 ÷ Cross-sectional area (mm²)
According to Ohm's Law, the voltage drop (V) can be calculated as V = R × I, where R is the resistance and I is the current.
Line loss is closely related to both the voltage drop and the current flowing through the wire. The power loss (P) in the line can be calculated using the formula P = V × I, where:
- P is the power loss in watts
- V is the voltage drop in volts
- I is the current in amperes
II. Copper Core Wire Current Calculation Method
The safe current-carrying capacity for different cross-sectional areas of copper power cables is as follows:
- 1 mm²: 17 A
- 1.5 mm²: 21 A
- 2.5 mm²: 28 A
- 4 mm²: 35 A
- 6 mm²: 48 A
- 10 mm²: 65 A
- 16 mm²: 91 A
- 25 mm²: 120 A
For single-phase loads, each kilowatt corresponds to approximately 4.5 A when the power factor (cosφ) is 1. This allows for the calculation of current and selection of the appropriate conductor size.
III. Comparison of Current-Carrying Capacities Between Copper and Aluminum Wires
- 2.5 mm² copper core wire is equivalent to 4 mm² aluminum core wire.
- 4 mm² copper core wire is equivalent to 6 mm² aluminum core wire.
- 6 mm² copper core wire is equivalent to 10 mm² aluminum core wire.
There are also some general rules for estimating current-carrying capacities based on wire size. For example:
- 2.5 mm² copper wire = 4 mm² aluminum wire × 5 = 20 A = 4.4 kW
- 4 mm² copper wire = 6 mm² aluminum wire × 5 = 30 A = 6.6 kW
- 6 mm² copper wire = 10 mm² aluminum wire × 5 = 50 A = 11 kW
In terms of current carrying capacity, a 1 mm² copper wire can carry about 1 kW, while a 2 mm² aluminum wire can carry 1 kW. The current-carrying capacities of copper and aluminum wires vary depending on their cross-sectional areas.
IV. Wire Diameter Calculation
The wire diameter can be calculated using the following formulas:
- For copper wire: S = IL / (54.4 × U`)
- For aluminum wire: S = IL / (34 × U`)
Where:
- I = Maximum current passing through the wire (A)
- L = Length of the wire (m)
- U` = Voltage drop (V)
- S = Cross-sectional area of the wire (mm²)
When calculating the cross-sectional area, it's important to consider the voltage drop range based on the system's power supply requirements. The estimated current-carrying capacity should also be taken into account.
V. Current Carrying Capacity and Wire Size
There are several rules of thumb for estimating the current-carrying capacity of insulated wires. For instance:
- "Multiply by 9.5 and multiply by nine."
- "Thirty-five by three-five, and the two groups are reduced by five."
- "The conditions have been changed and converted, and the high temperature ninefold copper upgrade."
These rules help estimate the current-carrying capacity of wires based on their cross-sectional area and ambient temperature. For example, a 2.5 mm² copper wire has a current-carrying capacity of 22.5 A, and a 16 mm² copper wire is calculated as if it were a 25 mm² aluminum wire.
VI. Copper and Aluminum Wire Resistance Per Kilometer
Below are the resistances per kilometer for copper and aluminum wires at different cross-sectional areas:
**Copper Wire Resistance (Ω/km):**
- 1.0 mm²: 17.5
- 1.5 mm²: 11.7
- 2.5 mm²: 7.00
- 4.0 mm²: 4.38
- 6.0 mm²: 2.92
- 10.0 mm²: 1.75
- 16.0 mm²: 1.10
- 25.0 mm²: 0.7
- 35.0 mm²: 0.5
- 50.0 mm²: 0.35
- 70.0 mm²: 0.25
- 95.0 mm²: 0.194
- 120.0 mm²: 0.146
- 150.0 mm²: 0.117
**Aluminum Wire Resistance (Ω/km):**
- 1.0 mm²: 28.3
- 1.5 mm²: 18.9
- 2.5 mm²: 11.32
- 4.0 mm²: 7.1
- 6.0 mm²: 4.72
- 10.0 mm²: 2.83
- 16.0 mm²: 1.77
- 25.0 mm²: 1.13
- 35.0 mm²: 0.81
- 50.0 mm²: 0.57
- 70.0 mm²: 0.41
- 95.0 mm²: 0.314
- 120.0 mm²: 0.24
- 150.0 mm²: 0.19
The resistance of a wire is proportional to its length and inversely proportional to its cross-sectional area. Therefore, longer wires or thinner wires will have higher resistance, which can lead to greater voltage drops and energy losses.
VII. Example
To determine the appropriate cable size for a 500W load, you need to consider whether it is a single-phase or three-phase system:
- Single-phase: I = P / U = 500 / 220 ≈ 2.27 A
- Three-phase: I = P / (1.732 × U) = 500 / (1.732 × 380) ≈ 0.75 A
Voltage drop can be calculated as V = I × R, where R is the resistance of the wire. The resistance of a 1.5 mm² copper wire over 2000 meters would be approximately 23.33 Ω. If a current of 1A flows through this wire, the voltage drop would be 23.3V, which could significantly affect the performance of the connected devices.
VIII. Different Temperature Resistivity
The resistivity of various metal conductors at 20°C is as follows:
- Silver: 1.65 × 10â»â¸ Ω·m
- Copper: 1.75 × 10â»â¸ Ω·m
- Aluminum: 2.83 × 10â»â¸ Ω·m
- Tungsten: 5.48 × 10â»â¸ Ω·m
- Iron: 9.78 × 10â»â¸ Ω·m
- Platinum: 2.22 × 10â»â· Ω·m
- Manganese copper: 4.4 × 10â»â· Ω·m
- Mercury: 9.6 × 10â»â· Ω·m
- Nichrome: 5.0 × 10â»â· Ω·m
- Nickel-chromium alloy: 1.0 × 10â»â¶ Ω·m
- Iron-chromium-aluminum alloy: 1.4 × 10â»â¶ Ω·m
- AlNi-Fe alloy: 1.6 × 10â»â¶ Ω·m
- Graphite: (8–13) × 10â»â¶ Ω·m
Understanding the resistivity of different materials is essential for selecting the right wire for a specific application. Higher resistivity means more energy is lost as heat, which can affect the efficiency and safety of the electrical system.