Copper resistivity (resistivity of copper)

1. Overview

The voltage drop of the copper core wire and its It is related to resistance, and its resistance calculation formula is:

At 20°C: 17.5 cross-sectional area (square millimeter)=resistance value per kilometer ()

At 75°C: 21.7 cross-sectional area (square millimeter) mm) = resistance value per kilometer ()

Voltage drop calculation formula (according to Ohm’s law): V=RA

Line loss is related to the voltage drop and current used.

Line loss calculation formula: P=VA

P-line loss power (watts) V-voltage drop (volts) A-line current (amperes)

2. Calculation method of copper core power line current

1mm copper power line-17a safe carrying capacity.

Safe current carrying capacity of 1.5mm2 copper power cord-21a.

2.5 square millimeter copper power cord-28a safe current carrying capacity.

Safe current carrying capacity of 4mm square copper power cord-35A

Safe current carrying capacity of 6mm copper power cord-48a

10mm2 copper power cord-65A safe carrying capacity.

16mm2 copper power cord-91A safe ampacity

25mm2 copper power cord-120a safe ampacity.

The single-phase load is 4.5A per kilowatt (COS&=1), and then select the conductor after calculating the current.

3. Current comparison method between copper core wire and aluminum core wire

2.5 square millimeters of copper wire is equal to 4 square millimeters of aluminum wire.

A 4 mm square copper wire is equal to a 6 mm square aluminum wire.

A 6 mm square copper wire is equal to a 10 mm square aluminum wire.

That is: 2.5 square millimeter copper core wire = 20 amps = 4400 watts;

4 mm square copper wire = 30a = 6600 W;

6mm copper wire=50 amps=11000 watts

The method is 1 square 1KW copper core wire and 2 square 1KW aluminum core wire. The unit is square millimeters.

is the cross-sectional area (mm2)

The ampacity of the cable varies according to the copper/aluminum core. You can use 2.5 (square millimeter) copper core as standard:

0.75/1.0/1.5/2.5/4/6/10/16/25/35/50/70/95/120/150 /185/240/300/400 …

There are also non-Chinese standards such as: 2.0.

Aluminum core 1 square has a maximum current carrying capacity of 9A, and copper core 1 square has a maximum current carrying capacity of 13.5A

Two point five times nine, one minus one. Let’s go.

Thirty-five multiplied by three point five, minus five points in pairs.

Condition changes and conversions, copper upgrades at high temperatures are 10% off.

The number of pipe roots is two percent, three, four, eight, seven, six and full load flow.

1. “2: 50 multiplied by 9, and then minus a number” said:

2.5 mm’ and the following various cross-sections of aluminum core insulated conductors have an ampacity of approximately 9 times the number. Such as 2.5mm ‘conductor, ampacity is 2.59 = 22.5 (a). The relationship between the ampacity and the multiples of the number of wire sections with a diameter of 4mm’ and above is that the number along the line increases and the multiple decreases L, that is, 48, 67, 106, 165, 254.

2. “Thirty-five times three point five, minus five points in pairs” means:

35mm” wire carrying capacity is 3.5 times the number of sections, That is, 353.5 = 122.5 (a). From 50mm ‘and above, the relationship between the ampacity and the multiple of the number of sections becomes a set of two wire numbers, and the multiples are successively reduced by 0.5. That is, the ampacity of the 50 mm and 70mm ‘conductors is the number of sections 3 times; 95, 120mm”conductor’s ampacity is 2.5 times its cross-sectional area, and so on.

3. “Condition change conversion, high temperature 10% copper upgrade”.

The above formulas are based on aluminum insulated conductors exposed to an ambient temperature of 25°C. If the aluminum core insulated wire is exposed to an area where the ambient temperature is higher than 25°C for a long time, the current carrying capacity of the wire can be calculated according to the above formula, and then a 10% discount will be given. When copper insulated wire is used instead of aluminum wire, its carrying capacity is slightly larger than that of aluminum wire of the same specification, and the carrying capacity of one wire size larger than aluminum wire can be calculated according to the above formula. For example, the carrying capacity of 16mm ‘copper wire can be calculated as 25mm2 aluminum wire.

2.5 square millimeter copper power cord-28a safe current carrying capacity.

4 mm square copper power cord – safe ampacity of 35A.

6mm copper power cord-48a safe current carrying capacity.

10mm2 copper power cord-65a safe current carrying capacity.

16mm2 copper power cord-91A safe current carrying capacity.

25mm2 copper power cord-120a safe current carrying capacity.

If it is an aluminum wire, the wire diameter should be 1.5-2 times that of the copper wire.

If the copper wire current is less than 28A, it is safe to take 10A per square millimeter.

If the copper wire current is greater than 120A, take 5A/square millimeter.

The current that can usually pass through the cross-sectional area of the wire can be selected based on the total current that the wire needs to conduct. Generally, it can be determined according to the following jingle:

15, 100, 20,

35543 worlds,

79.5 times and a half,

p>Copper upgrade calculation.

Explain to you:

It is an aluminum wire under 10 square meters, and the square millimeter is multiplied by5. If it’s copper wire, go up a notch. For example, a copper wire of 2.5 square meters is calculated as 4 square meters. For those above 100, multiply the cross-sectional area by 2, for those below 25 square meters, multiply by 4, for those above 35 square meters, multiply by 3, and for those above 70 square meters and 95 square meters, multiply by 2.5. These formulas should be easy to remember.

15, 100, 225, 35, 43, 70, 952, semi-bare wire plus half, copper wire upgrade counts through the tube, high temperature 20% off, 10% off Description:

1. The cross-section of the “ten down five” wire is less than 10 square millimeters, and the safe current per 1 square millimeter is 5 amperes.

2. “100-up-2” conductors have a cross-section of more than 100 square millimeters, and the safe current per 1 square millimeter is 2 amperes.

3. The cross-sections of the “25, 35, 43 boundary” guide lines are 25 square millimeters and 16 square millimeters, and the safe current per 1 square millimeter is 4 amperes. The cross section of the wire is 35 square millimeters and 50 square millimeters, and the safe current per 1 square millimeter is 3 amperes.

4. The cross-section of the “70, 952 and a half times” guide wire is 70 square millimeters and 95 square millimeters, and the safe current per 1 square millimeter is 2.5 amperes.

“Bare wire plus half, copper wire upgrade calculation” refers to the bare wire of the same cross section, and the safe current can be calculated by multiplying the insulated wire by 1.5 times. Copper conductors of the same cross-section are calculated according to the grade with the larger number of aluminum conductors. Safe use of electricity p=ui42000/220=191A pure resistance element. 150 square meters of cable is recommended.

Safe current carrying value of 1 square plastic insulated conductor:

Open wire-17 amps,

Steel pipe: two-12 A, three-11 A, four- 10 A,

Plastic tube: two 10 A, three 10 A and four 9 A.

Jacketed wire: two cores – 13 amps, three cores and four cores – 9.6 amps.

Rubber insulated wire; open wire-18A, steel pipe: 2-13A, 3-12A, 4-11A.

Note: It can only be used as an estimate, not very accurate.

In addition, if you remember that the indoor wire is less than 6 square millimeters of copper wire, it is safe to have a current of no more than 10A per square meter. From this point of view, you can choose 1.5 square meters of copper wire or 2.5 square meters of aluminum wire.

Within 10 meters, the conductor current density of 6A/ mm2 is more suitable than that of Youyou resource network, 10-50 meters, 3A/ mm2, 50-200 meters, 2A/ mm2, and more than 500 meters, less than 1A / mm2. From this point of view, if it is not very far, you can choose 4 square copper wires or 6 square aluminum wires.

If it is really 150 meters away from the power supply (not to mention tall buildings), you must use 4 square meters of copper wire.

The impedance of a wire is proportional to its length and inversely proportional to its diameter. When using the power supply, pay special attention to the wire and wire gauge of the input and output lines. prevent currentIf it is too large, it will cause the wire to overheat and cause an accident.

The following is a table of copper wire diameter and the higher current it can withstand at different temperatures.

Steel wire diameter (approx.) (mm2)

4. Wire diameter calculation

Steel wire diameter Generally calculated according to the following formula:

Copper wire: s = il/54.4 * u’

Wire: s = il/34 * u’

Type:

I——The greater current through the wire (a).

L——wire length (meter)

U`——allowed power drop (v)

S——wire cross-sectional area (MM2)

Description:

1.u `The voltage drop can be divided into the power supply voltage rating of the system according to the range of equipment (such as detectors) used in the entire system for comprehensive consideration.

2. The calculated cross-sectional area is upward. Estimation of the ampacity of insulated wires.

★Relationship between current carrying capacity and cross-section of aluminum core insulated conductor (25 degrees)

★Copper core insulated conductor ( 25 degrees) relationship between the carrying capacity and the multiple of the cross section

5. Current carrying capacity

Estimation formula:

Two point five times nine, one minus one. Let’s go.

Thirty-five multiplied by three point five, minus five points in pairs.

Condition changes and conversions, copper upgrades at high temperatures are 10% off.

The number of pipe roots is two percent, three, four, eight, seven, six and full load flow.

Description:

The formula in this section does not directly point out the carrying capacity (safety current) of various insulated wires (rubber and plastic insulated wires), but “multiplies the cross section by a certain multiple” , I can figure it out in my head. It can be seen from Table 5-3 that the multiple decreases with the increase of the section.

1. “2: 50 multiplied by 9, and then minus a number” said:

2.5 mm’ and the following various cross-sections of aluminum core insulated conductors have an ampacity of approximately 9 times the number. Such as 2.5mm ‘conductor, ampacity is 2.59 = 22.5 (a). The relationship between the ampacity and the multiples of the number of wire sections with a diameter of 4mm’ and above is that the number along the line increases and the multiple decreases L, that is, 48, 67, 106, 165, 254.

2. “Thirty-five multiplied by 3.5, minus five points to pair” said:

35mm”The carrying capacity of the wire is 3.5 times the number of sections, that is, 353.5 = 122.5 (a). From 50mm’ and above, the relationship between the current-carrying capacity and the multiple of the number of sections becomes a set of two wire numbers, and the multiples are successively reduced by 0.5. That is, the current-carrying capacity of the 50 mm and 70mm’ wires is 3 times the number of sections ; The ampacity of 95, 120mm” conductor is 2.5 times of its cross-sectional area, and so on.

The above formulas are based on aluminum core insulated wires exposed to an ambient temperature of 25°C. If the aluminum core insulated wire is exposed to an area where the ambient temperature is higher than 25°C for a long time, the current carrying capacity of the wire can be calculated according to the above formula, and then a 10% discount will be given. When copper insulated wire is used instead of aluminum wire, its carrying capacity is slightly larger than that of aluminum wire of the same specification, and the carrying capacity of one wire size larger than aluminum wire can be calculated according to the above formula. For example, the ampacity of 16mm ‘copper wire can be calculated according to 25mm2 aluminum wire. The larger the cross-sectional area, the smaller the resistance, and the larger the current passing at the same voltage.

Household wires are generally calculated using the empirical formula in the “Household Circuit System Layout Method”. Taking copper conductors as an example, the empirical formula is: conductor cross section (mm2) ≈ I/4(A).

The rated ampacity of Youyou Resource Network is ≈4A for a copper conductor with a cross-section of 1 square millimeter, and 10A for a copper conductor with a cross-section of 2.5 square millimeters.

In fact, the carrying capacity of the wire mainly depends on the heat dissipation of the wire and the importance of the wiring place.

The heat dissipation conditions of the wires laid on the frame are better, but the heat dissipation conditions of the closed and concealed wires passing through pipes, buried walls, etc. are poor; the heat dissipation conditions of multiple wires are worse than that of a single wire. The more inner wires, the worse the heat dissipation. Wires of the same size can have a larger download flow when the heat dissipation condition is good, and a smaller download flow when the heat dissipation condition is poor.

Wiring is very important, the safety requirements are high, the carrying capacity is small, and the carrying capacity is larger in places that are not very important. The above-mentioned household wires have a very small ampacity.

Under strict requirements, the cross-sectional area of the conductor is multiplied by 4 as the carrying capacity. If the ampacity is small, more wires are needed and the cost is high. In order to reduce costs and ensure safety, many home improvement companies multiply the cross-sectional area of conductors by 5 or 6 times to increase the current-carrying capacity of conductors in non-closed wiring.

This is also the maximum ampacity of home wiring. No matter how big it is, it cannot guarantee the safe and normal use of household appliances. If it is industrial electricity, it can be used by multiplying the cross-sectional area of the conductor by about 9 times the current carrying capacity.

According to the empirical formula of home wiring, the ampacity of 2.5 square copper wires is 2.5*4, which is correct. 2.5*5 is also available, and 2.5*6 is also useful. But in places with good heat dissipation conditions such as factories, 2.5*9 or even 2.5*10 is fine.

In short, the carrying capacity of the same wire is different under different conditions and requirements.

Here is a long-term allowable ammeter for your reference. Its current carrying capacity is conditional and larger.

6. Resistance per kilometer of copper and aluminum wire

1. Nominal cross-sectional area/(copper core wire)

2. Nominal cross-sectional area/(aluminum core wire)

Resistance formula :

Voltage drop = 2X resistance x total equipment current – AC voltage = equipment supply voltage

R=p*l/s

p-resistivity lookup table;

l-resistor length;

s——resistor cross-sectional area perpendicular to the current

The two resistors R connected in series will become larger and become 2R, which means double the length and double the resistance value. Resistance is proportional to length.

Two resistors R are connected in parallel, 1/2R, the cross-sectional area is doubled, and the resistance value becomes 1/2, and the resistance is inversely proportional to the cross-sectional area.

Therefore:

R=pl/S. p=RS/l

The metal wire is the water pipe, and the current is the small fish in the water pipe. The water pipe is long, and the fish find it difficult to resist from one end to the other. Of course, the thinner the pipe (smaller cross-section), the harder it is to move to the other end. Resistivity is the constant resistance of water to fish. No matter how long or thin your pipe is, the resistance of water to fish (electrons) is the same as that of similar water pipes. Copper has a resistivity of 0.0175 and aluminum has a resistivity of 0.026. In terms of ampacity, 4 square meters of copper is equivalent to 4 * 0.026/0.0175 = 4 * 1.485 = 5.94 square meters of aluminum wire.

But because the heat resistance of aluminum is worse than that of copper, the available current of 4 square copper wires can exceed 6 square aluminum wires.

10 square twin-core cables, the allowable carrying capacity at 20°C is 75A, and the voltage drop is 4.67MV/m. At 20°C, the resistance is 0.018 ohms, and the resistance R=P*L/S (P resistivity. Length m.s cross-section square millimeters) is 0.72 ohms.

The line loss at 220V voltage is calculated as follows:

P=I2R(p loss power w, I equipment current, r cable resistance)

The equipment current is At 50A, the power loss is 50*2*0.72=72W.

Resistivity should be

P=0.0175 ω mm 2/m = 1.75 * 10-8 ω m.

L=6000mm=6m,S=6mm^2

R=pL/S=0.0175*6/6=0.0175 Euro

R=PL/ S

r is the resistance (unit), P is the resistivity of the material (the resistivity of copper is 0.0175), L is the length of the conductor in the current direction (the unit is m), and S is the conductor perpendicular to The cross-sectional area in the direction of current flow (in square meters).

M=PV is calculated by weight, M is weight (in kg kg), P is density (copper density = 8.910 kg/m3), and V is volume (in m3).

7. Give an example.

How many square cables are used for 500W power? ? What is this formula?

It depends on whether you use three-phase or single-phase.

According to single-phase P=UI

Three-phase pressure P=1.732*UI

Suppose you are single-phase: P=UI I=P/U=500/220=2.27 A.

Assuming you are three-phase: P = 1.732 ui I = P/1.732 u = 500/1.732 * 380 = 0.75 a.

Voltage drop=current*wire resistance

wire resistance=(wire length*resistivity)/wire cross-sectional area

current=power consumption/voltage.

Resistivity of copper = 1.75*10-8 ohm-meter.

A 1.5 square copper wire has a cross-sectional area of 1.5mm2, so the resistance of a 2000m long copper wire (at room temperature) is:

(2000m*1.75*10-8 power ω m)/1.5 * 10-6 power m2 = 23.33 ohms

So if 1A of current flows through your wire, the voltage drop will be 23.3 volts. The voltage drop per square meter is 0.0024 millivolts

If your electrical appliance has a rated power of 1 kW and a rated operating voltage of 220V, if you use such a wire, the voltage drop will be 71.5V (about one-third One of the electric energy is wasted on the wire). If you connect a 100-watt power supply, the voltage drop is 10.1V (basically negligible and does not affect the normal operation of the electrical appliance).

So it is recommended that the power used should not exceed 200W. If it exceeds 200 W, the actual voltage on your appliance will be less than 200 V.

The calculation is divided into three stages to explain the load situation, and the motor is used as the load for calculation.

1. Wire length: L = 400m,

Motor power: 15kW+12kW+18kW = 45kW,

Current: I≈90A, copper wire Section: S = 25 square meters,

Resistivity of copper wire: = 0.0172

Find the line resistance (single line) at 400 meters:

R = (L/S Youyou Resources *** )= 0.0172 (400/25) ≈ 0.275()

Find the voltage drop at 400 meters (single line): U = RI = 0.27590 ≈ 25 (V)

Segment line voltage is:

80-252 (two-wire)= 330 (V)

2. Wire length: L = 600-400 = 200 m,

Motor power: 12kW+18kW = 30kW, I≈60A

Find the line resistance (single line) at 200 meters:

R=(L /S)=0.0172(200/25)≈0.138()

Find the voltage drop at 200 meters (single line):

U=RI=0.13860≈8(V)

The segment line voltage is:

380-(25+8)2=314(V)

3. Wire length: L = 700-600-400 = 100m,

Motor power: 18KW,

I≈36A

Seeking 100 meters Distance line resistance (single line):

R=(L/S)=0.0172(100/25)≈0.069()

Find the voltage drop at a distance of 100 meters (single line) :

U=RI=0.06936≈2.5(V)

The terminal voltage is:

380-(25+8+2.5)2=309(V )

For example, if a 4 square (mm) copper wire is used, the wire length is 200m (three-phase four-wire). So how much does the 30 day line consume? What is the 6 (mm) square line?

1 square millimeter is equal to 1/100000 square meters.

(1 mm2 = 0.01 cm2 = 0.0001 mm2 = 0.00001 mm2, 1mm = 1/1000m.

1 mm=1/(1000*1000) square meters)

Object resistance formula:

r = liters/second

Type:

r is the resistance of the object (ohms);

is the resistivity of the substance in ohm-meters (.mm/meter).

l is the length, the unit is meter (m)

s is the cross-sectional area, the unit is square meter (mm)

Ohm’s law (I=U /R)

The voltage drop across the wire is equal to the product of the current in the wire and the wire resistance (U=I*R)

The power calculation formula P=U/R (U stands for Voltage, R stands for resistance)

The resistivity of pure silver wire at 0°C is 0.016. mm/m (ohmmeter), the ohm value of pure copper wire at 0°C is 0.0175. mm/m (ohmmeter), the temperature coefficient of resistance is 0.00393/℃. The resistivity of pure aluminum wire at 0°C is 0.0283. mm/m (ohmmeter),

=o(1+a*t)

Medium

——resistivity at t℃.

o——Resistivity at 0°C

Temperature

Resistivity of pure copper wire at 21°C = 0.0175(1+0.00393 * 21) = 0.0189.mm/m) R = 0.0175 x200/4 = 0.875 ohms.

For the power consumption of No. 30, if it is 220V, then the power P=U/R=220/0.875=55.314W is required first.

(In this state, the wires directly form a loop, that is, there are no household appliances connected. If it is so empty, the meeting will trip directly!)

55.314X30X241000=39.826 degrees (1 degree of electricity=1 kWh )

Voltage loss=∑(P*L)/(C*S)

p: total power of the circuit (KW) L: distance between power supply and load (m) C : Material coefficient (at 380V, copper is 77, aluminum is 46.3; at 220V, copper is 12.8, aluminum is 7.75: Conductor cross-sectional area (mm)

8. Resistivity of different metal conductors

Resistivity of several metal conductors at 20°C

(1) Silver 1.65 (Euro) 10-8 (meter)

(2) Copper 1.75 10-8

(3) Aluminum 2.83 10-8

(4) Tungsten 5.48 10-8

(5) Iron 9.78 10-8

(6) Platinum 2.22 10-7

(7) Manganin 4.4 10-7

(8) Mercury 9.6 10-7

(9) Kangtong 5.0 10-7

(10) Nickel-chromium alloy 1.0 10-6

(11) Iron-chromium-aluminum alloy 1.4 10-6