The voltage drop across the copper wire is related to its resistance.
Its resistance calculation formula:
20 ° C: 17.5 ÷ cross-sectional area (square mm) = resistance per kilometer (Ω)
At 75 ° C: 21.7 ÷ cross-sectional area (square mm) = resistance per kilometer (Ω)
Its pressure drop calculation formula (according to Ohm's law): V = R × A
Line loss is related to the voltage drop and current used.
Its line loss calculation formula: P = V × A
P-line loss power (watts) V-pressure drop value (volts) A-line current (amperes)
Second, copper core line power line current calculation methodThe safe current carrying capacity of the 1 square millimeter copper power cord - -17A.
The safe current carrying capacity of the 1.5 square millimeter copper power cord - 21A.
The safe current carrying capacity of the 2.5 square millimeter copper power cord --28A.
The safe current carrying capacity of the 4 square millimeter copper power cord - 35A.
The safe current carrying capacity of the 6 square millimeter copper power cord - 48A.
The safe current carrying capacity of the 10 square millimeter copper power cord - 65A.
The safe current carrying capacity of the 16 square millimeter copper power cord - 91A.
The safe current carrying capacity of the 25 square millimeter copper power cord - 120A.
The single-phase load is 4.5A per kilowatt (COS & = 1), and the current is calculated and the conductor is selected.
Third, copper core wire and aluminum core wire current comparison method2.5 square millimeter copper core wire equals 4 square millimeter aluminum core wire
4 square millimeter copper core wire equals 6 square millimeter aluminum core wire
6 square millimeter copper core wire equals 10 square millimeter aluminum core wire
<10 square millimeters or less multiplied by five>
Namely: 2.5 square millimeter copper core wire = <4 square millimeter aluminum core wire × 5 > 20 amperes = 4400 kilowatts;
4 mm 2 copper core wire = <6 mm 2 aluminum core wire × 5 > 30 amps = 6600 kW;
6 square millimeter copper core wire = <10 square millimeter aluminum core wire × 5 > 50 amps = 11,000 kW
The soil method is a copper core wire 1 square 1KW, aluminum core 2 square 1KW. The unit is square millimeter
Is the cross-sectional area (square millimeters)
The cable current capacity is different according to the copper core/aluminum core. The copper core can be used with 2.5 (square millimeters).
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 maximum current carrying capacity 9A, copper core 1 square maximum current carrying capacity 13.5A
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.
The number of tubes to be worn is two to three, and eighty-seven percent is full of current.
"Twenty-five times down to nine, go up one minus one" is said to be 2.5mm' and below various cross-section aluminum core insulated wires, the current carrying capacity is about 9 times the number of sections. For example, a 2.5 mm' wire has a current carrying capacity of 2.5 x 9 = 22.5 (A). The multiple of the current carrying capacity and the number of sections from the wire of 4 mm' and above is the upper row along the line number, and the multiple is successively decreased by 1, that is, 4 × 8, 6 × 7, 10 × 6, 16 × 5, 25 × 4.
"Thirty-five times by three-five, two pairs of points minus five", said that the 35mm" wire current carrying capacity is 3.5 times the number of sections, that is, 35 × 3.5 = 122.5 (A). For a wire of 50 mm' and above, the multiple relationship between the current carrying capacity and the number of sections becomes a set of two two wire numbers, and the multiple is sequentially decreased by 0.5. That is, the current carrying capacity of the 50, 70 mm' wire is 3 of the number of sections. Double; 95, 120mm" wire current carrying capacity is 2.5 times the number of its cross-sectional area, and so on.
"The conditions have been changed and converted, and the high temperature ninefold copper upgrade." The above-mentioned port is determined by the aluminum core insulated wire and the bright coating at an ambient temperature of 25 ° C. If the aluminum core insulated wire is applied in 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-mentioned calculation method of the mouth and then 10% off; when the copper wire is insulated, the copper core is used. Its current carrying capacity is slightly larger than that of the same specification aluminum wire. The above-mentioned port method can be used to calculate the current carrying capacity of one wire number than the aluminum wire. For example, the current carrying capacity of a 16mm' copper wire can be calculated as a 25mm2 aluminum wire.
The safe current carrying capacity of the 2.5 square millimeter copper power cord --28A.
The safe current carrying capacity of the 4 square millimeter copper power cord - 35A.
The safe current carrying capacity of the 6 square millimeter copper power cord - 48A.
The safe current carrying capacity of the 10 square millimeter copper power cord - 65A.
The safe current carrying capacity of the 16 square millimeter copper power cord - 91A.
The safe current carrying capacity of the 25 square millimeter copper power cord - 120A.
If it is an aluminum wire, the wire diameter should be 1.5-2 times that of the copper wire.
If the copper current is less than 28A, it is safe to take 10A per square millimeter.
If the copper current is greater than 120A, take it at 5A per square millimeter.
The current through which the cross-sectional area of ​​the wire can normally pass can be selected according to the total number of currents that need to be turned on, and can generally be determined by following the jingle:
Ten to five, one hundred and two,
Two five three five four three worlds,
Pick up two and a half times,
Copper wire upgrade count.
To explain to you, is the aluminum wire below 10 square meters, the square millimeter number is multiplied by 5, if the copper wire, it will rise a file, such as 2.5 square copper wire, it will be calculated according to 4 square. More than one hundred It is the cross-sectional area multiplied by 2, the twenty-five squares or less multiplied by 4, the thirty-five squares or more multiplied by 3, the pick and 95 squares are multiplied by 2.5, so a few words should be well remembered,
Tenth Five-Year, One Hundred and Two Five, Three, Five, Four, Three, Seventy, Nine, Five, and a half, and half of the bare wire plus half, copper wire upgrade counts through the pipe, high temperature eight, 10% off
Note: The “10th Five-Year†guide line has a section of 10 square millimeters or less, and the safety current per square millimeter is 5 amps.
The "100+2" guide line has a cross section of more than 100 square millimeters and a safety current of 2 amps per square millimeter.
The "25, 3, 4, and 3" boundary lines are 25 square millimeters and 16 square millimeters, and the safety current per 1 square millimeter is 4 amps. The conductor cross section is 35 square millimeters and 50 square millimeters, and the safety current per square millimeter is 3 amps.
The "seventy-nine, nine-and-five-half-half-half" guideline sections are 70 square millimeters and 95 square millimeters, with a safe current of 2.5 amps per square millimeter.
“Bare wire plus half, copper wire upgrade calculation†refers to the bare wire of the same section, which can be calculated by multiplying the insulated wire by 1.5 times. The copper wire of the same section is calculated according to the grade of the wire of the aluminum wire. The safety electric p=ui42000/220=191A purely resistive component. A 150 square cable is recommended.
1 square plastic insulated wire safe current carrying value:
Bright line - 17 amps,
Wear steel pipes: two roots - 12 amps, three ties - 11 amps, four ties - 10 amps,
Wear plastic tubes: two - 10 amps, three - 10 amps, four - -9 amps.
Sheathed wire: two cores - 13 amps, three cores and four cores - 9.6 amps.
Rubber insulated wire; bright line - 18 amps, steel pipe: two - 13 amps, three - 12 amps, four - -11 amps
Description: Can only be used as an estimate, not very accurate.
In addition, if you remember the copper wire below 6 square millimeters in the room, it is safe to not exceed 10A per square current. From this point of view, you can choose 1.5 square copper wire or 2.5 square aluminum wire.
Within 10 meters, the conductor current density of 6 A/mm 2 is suitable, 10-50 m, 3 A/mm 2 , 50-200 m, 2 A/mm 2 , and less than 1 A/mm 2 over 500 m. From this perspective, if not very far, you can choose 4 square copper wire or 6 square aluminum wire.
If it is really a distance of 150 meters (not to mention high-rise), it must use 4 square copper wire.
The impedance of a wire is proportional to its length and inversely proportional to its wire diameter. Please pay special attention to the wire and wire diameter of the input and output wires when using the power supply. In order to prevent the current from being too large, the wire is overheated and an accident is caused.
Below is a table of wire diameters and maximum currents that can be withstood at different temperatures.
Wire diameter (approximate value) (mm2)
| Copper wire temperature °C | 60 | 75 | 85 | 90 |
| Current (A) 2.5 level | 20 | 20 | 25 | 25 |
| Current (A) 4.0 level | 25 | 25 | 30 | 30 |
| Current (A) 6.0 level | 30 | 35 | 40 | 40 |
| Current (A) 8.0 level | 40 | 50 | 55 | 55 |
| Current (A) 14.0 level | 55 | 65 | 70 | 75 |
| Current (A) 22 level | 70 | 85 | 95 | 95 |
| Current (A) 30 level | 85 | 100 | 100 | 110 |
| Current (A) 30 level | 95 | 115 | 125 | 130 |
| Current (A) 50 level | 110 | 130 | 145 | 150 |
| Current (A) 60 flat | 125 | 150 | 165 | 170 |
| Current (A) 70 level | 145 | 175 | 190 | 195 |
| Current (A) 80 level | 165 | 200 | 215 | 225 |
| Current (A) 100 level | 195 | 230 | 250 | 260 |
The wire diameter is generally calculated as follows:
Copper wire: S= IL / 54.4*U`
Aluminum wire: S= IL / 34*U`
In the formula:
I——Maximum current passing through the wire (A)
L——the length of the wire (M)
U`——Full power supply drop (V)
S——cross-sectional area of ​​the wire (MM2)
Description:
1. The U` voltage drop can be considered by considering the range of the power supply voltage used for the system power supply in the range of equipment (such as detectors) used in the entire system.
2. The calculated cross-sectional area is up. The current carrying capacity of the insulated wire is estimated.
★ Aluminum core insulated wire current carrying capacity and multiple of the cross section (25 degrees)
| Wire cross-sectional area | Current carrying is the cross-section multiple | Ampacity (A) |
| 1.5mm2 | 9 | 9 |
| 2.5mm2 | 9 | 14 |
| 4mm2 | 8 | twenty three |
| 6mm2 | 7 | 32 |
| 10mm2 | 6 | 48 |
| 16mm2 | 5 | 60 |
| 25mm2 | 4 | 90 |
| 35mm2 | 3.5 | 100 |
| 50mm2 | 3 | 123 |
| 70mm2 | 2.5 | 150 |
| 95mm2 | 210 | |
| 120mm2 | 238~300 |
★The relationship between the current carrying capacity of the copper core insulated wire and the multiple of the section (25 degrees)
| Wire cross-sectional area | Current carrying is the cross-section multiple | Ampacity (A) |
| 1.5mm2 | 9 | 12 |
| 2.5mm2 | 9 | twenty three |
| 4mm2 | 8 | 32 |
| 6mm2 | 7 | 48 |
| 10mm2 | 6 | 60 |
| 16mm2 | 5 | 90 |
| 25mm2 | 4 | 100 |
| 35mm2 | 3.5 | 123 |
| 50mm2 | 3 | 150 |
| 70mm2 | 2.5 | 210 |
| 95mm2 | 250 | |
| 120mm2 | 330 |
Estimated vocabulary:
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. The number of tubes to be worn is two to three, and eighty-seven percent is full of current.
Note: (1) The current carrying capacity (safety current) of various insulated wires (rubber and plastic insulated wire) is not directly indicated, but is expressed by “multiplied by a certain multiple†and calculated by mental arithmetic. It can be seen from Table 5 3 that the multiple decreases as the cross section increases. "Twenty-five times down to nine, go up one minus one" is said to be 2.5mm' and below various cross-section aluminum core insulated wires, the current carrying capacity is about 9 times the number of sections. For example, a 2.5 mm' wire has a current carrying capacity of 2.5 x 9 = 22.5 (A). The multiple of the current carrying capacity and the number of sections from the wire of 4 mm' and above is the upper row along the line number, and the multiple is successively decreased by 1, that is, 4 × 8, 6 × 7, 10 × 6, 16 × 5, 25 × 4. "Thirty-five times by three-five, two pairs of points minus five", said that the 35mm" wire current carrying capacity is 3.5 times the number of sections, that is, 35 × 3.5 = 122.5 (A). For a wire of 50 mm' and above, the multiple relationship between the current carrying capacity and the number of sections becomes a set of two two wire numbers, and the multiple is sequentially decreased by 0.5. That is, the current carrying capacity of the 50, 70 mm' wire is 3 of the number of sections. Double; 95, 120mm" wire current carrying capacity is 2.5 times the number of its cross-sectional area, and so on. "The conditions have been changed and converted, and the high temperature ninefold copper upgrade."
The above-mentioned port is determined by the aluminum core insulated wire and the bright coating at an ambient temperature of 25 ° C. If the aluminum core insulated wire is applied in 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-mentioned calculation method of the mouth and then 10% off; when the copper wire is insulated, the copper core is used. Its current carrying capacity is slightly larger than that of the same specification aluminum wire. The above-mentioned port method can be used to calculate the current carrying capacity of one wire number than the aluminum wire. For example, the current carrying capacity of the 16mm' copper wire can be calculated according to the 25mm2 aluminum wire. The larger the cross-sectional area, the smaller the resistance, and the larger the current passing through the same voltage.
Household wires are generally calculated using empirical formulas in the home circuit system layout method. The choice of wire is based on copper core wire. The empirical formula is:
Conductor cross section (in square millimeters) ≈I/4(A).
A copper conductor with a cross section of about 1 square millimeter has a rated current carrying capacity of A4A, and a copper conductor with a cross section of 2.5 square millimeters has a rated current carrying capacity of about 10 amps.
In fact, the current carrying capacity of the wire depends mainly on the heat dissipation conditions of the wire and the importance of the wiring place. The heat dissipation conditions of the overhead laying conductors are better. The heat dissipation conditions of the closed dark conductors such as the tube and the buried wall are worse. The heat dissipation conditions of the multiple conductors are worse than those of the single conductors. The more the conductors in one tube, the worse the heat dissipation. The same size of the wire can download more traffic when the heat dissipation condition is good, and the download traffic is smaller when the heat dissipation condition is poor.
The wiring place is very important, the safety requirements are high, and the current carrying capacity is small. It is not very important that the current carrying capacity is larger. The above-mentioned home wiring wire carrying capacity is small. For strict requirements, multiply the cross-sectional area of ​​the conductor by 4 as the current carrying capacity. The current carrying capacity is small, the wire is much more, and the cost is high. In order to reduce the cost and ensure the safety, many home appliances use the wire cross-sectional area multiplied by 5 times or 6 times to increase the current carrying capacity of the wire when it is not closed wiring. This is also the maximum current carrying capacity of the home wiring. If it is too large, it will not guarantee the safe and normal use of the home appliances. In the case of industrial electricity, it is possible to use a cross-sectional area of ​​the conductor multiplied by a current carrying capacity of about 9 times. Using the home wiring experience formula to see 2.5 square copper core wire current carrying capacity is 2.5 * 4 is correct, 2.5 * 5 can also be used 2.5 * 6. However, factories with good heat dissipation conditions can also use 2.5*9 or even 2.5*10.
In short, the same wire has different current carrying capacity under different conditions and different requirements.
Let's give you a long-term allowable current-carrying meter for your reference. Its current-carrying capacity is also conditional and the largest.
1. Nominal cross-sectional area / (copper core wire)
| Copper cross-sectional area | Corresponding resistance (Î/Km) |
| 1.0mm2 | 17.5 |
| 1.5mm2 | 11.7 |
| 2.5mm2 | 7.00 |
| 4mm2 | 4.38 |
| 6mm2 | 2.92 |
| 10mm2 | 1.75 |
| 16mm2 | 1.10 |
| 25mm2 | 0.7 |
| 35mm2 | 0.5 |
| 50mm2 | 0.35 |
| 70mm2 | 0.25 |
| 95mm2 | 0.194 |
| 120mm2 | 0.146 |
| 150mm2 | 0.117 |
2. Nominal cross-sectional area / (aluminum core wire)
| Aluminum wire cross-sectional area | Corresponding resistance (Î/Km) |
| 1.0mm2 | 28.3 |
| 1.5mm2 | 18.9 |
| 2.5mm2 | 11.32 |
| 4mm2 | 7.1 |
| 6mm2 | 4.72 |
| 10mm2 | 2.83 |
| 16mm2 | 1.77 |
| 25mm2 | 1.13 |
| 35mm2 | 0.81 |
| 50mm2 | 0.57 |
| 70mm2 | 0.41 |
| 95mm2 | 0.314 |
| 120mm2 | 0.24 |
| 150mm2 | 0.19 |
Resistance formula:
Voltage drop = 2X resistance X device total current - AC voltage = equipment supply voltage
R=p*l/s
P-resistivity check table;
L—resistance length;
S—resistance cross-sectional area perpendicular to the current
The two resistors R become large in series and become 2R, which means that the length is doubled, the resistance value is doubled, and the resistance is proportional to the length.
The two resistors R are connected in parallel, 1/2R, the cross-sectional area is increased by 1 time, 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 wire is a water pipe. The current is the small fish in the water pipe. When the water pipe is long, the fish feels that the resistance from this end to the end is very difficult. Of course, the thinner the pipe (small cross section) feels more moving to the other end. difficult. Resistivity is the constant of the nature of water resistance to fish. No matter how long or thin your pipe is, the resistance of water to fish (electronic) is the same as that of water. The resistivity of copper is 0.0175, and the resistivity of aluminum is 0.026. In terms of current carrying capacity, 4 square copper is equivalent to aluminum wire 4*0.026/0.0175=4*1.485=5.94 square.
However, since the heat resistance of aluminum is worse than that of copper, the current available for 4 square copper wire can exceed 6 square aluminum wire.
10 square two-core cable, the allowable current capacity is 75A at a temperature of 20, the voltage drop is 4.67MV / m. Under the voltage of 220V, the electrical cross-section of 16.5KW can be driven. The length of the copper wire of 1 square millimeter and the length of 1 meter is 20 degrees Celsius. 0.018 ohm, R = P * L / S (P resistivity. L length meter. S section square mm) resistance is 0.72 ohms.
The calculation of the line loss at 220V is:
P=I2R (p loss power W, i device current, r cable resistance)
When the device current is 50A, the power loss is 50*2*0.72=72W.
The resistivity should be
p=0.0175 ohm.mm^2/m =1.75*10^-8 ohm.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 is Ω), P is the resistivity of the material (copper resistivity is 0.0175), L is the length of the conductor in the current direction (in meters m), and S is the conductor perpendicular to the current direction Cross-sectional area (in square meters).
The weight is calculated by M = PV, M is the weight (in kilograms) P is the density (density of copper = 8.9 x 103 kg / cubic meter) V is the volume (unit is cubic meters).
Seven, an exampleHow many square cables do you use for 500W? ? What is this calculation formula?
It depends on whether you use three-phase or single-phase.
Single phase press P=UI
Three-phase press P=1.732*UI
Suppose you are single-phase: P=UI I=P/U=500/220=2.27 安
Suppose you are three-phase: P=1.732UI I=P/1.732U=500/1.732*380=0.75A
Voltage drop = current * wire resistance
Wire resistance = (wire length * resistivity) / wire cross-sectional area
Current = power consumption / power voltage
The resistivity of copper = 1.75 * 10-8 times the euro.
The 1.5 square copper wire is a copper wire with a cross section of 1.5 mm2, so that the 2000 m long copper wire resistance (normal temperature) is:
(2000m*1.75*10-8 power ohms.m)/1.5*10-6 power m2 = 23.33 ohms
So if you pass a current of 1A on the wire, the voltage drop is 23.3V.
That is, the voltage drop per square meter is 0.0024mv.
If you use electric power with a rated power of 1 kW and a working voltage of 220V, then with such a wire, the voltage drop is: 71.5V (about one-third of the power is wasted on the wire). If you pick up A power of 100 watts, the voltage drop is: 10.1V (basically negligible, does not affect the normal operation of the electrical appliances)
Therefore, it is recommended that the power should not exceed 200W. If it exceeds 200W, the actual voltage on your electrical appliance will be less than 200V.
Calculated in 3 steps, indicating the load situation, taking the motor as the load calculation.
ⶠWire length: L=400 meters,
Motor power: 15kw + 12kw + 18kw = 45KW,
Current: I≈90A, copper wire cross section: S=25 square,
Copper wire resistivity: Ï=0.0172
Find a line resistance of 400 meters (single line):
R=Ï×(L/S)=0.0172×(400/25)≈0.275(Ω)
Find the voltage drop of 400 meters (single line): U=RI=0.275×90≈25(V)
The line voltage is:
380-25×2 (double line)=330(V)
â· Wire length: L=600-400=200 meters,
Motor power: 12kw+18kw=30KW, I≈60A
Find a line resistance of 200 meters (single line):
R=Ï×(L/S)=0.0172×(200/25)≈0.138(Ω)
Find the 200m distance voltage drop (single line):
U=RI=0.138×60≈8(V)
The line voltage is:
380-(25+8)×2=314(V)
â¸Wire length: L=700-600-400=100 meters,
Motor power: 18KW,
I≈36A
Find a line resistance of 100 meters (single line):
R=Ï×(L/S)=0.0172×(100/25)≈0.069(Ω)
Find the distance drop of 100 meters (single line):
U=RI=0.069×36≈2.5(V)
The terminal voltage is:
380-(25+8+2.5)×2=309(V)
If 4 square (mm) copper wire is used, the wire length is 200m. (three-phase four-wire). So what is the power consumption of the 30-day line? What is the 6-square (mm) line?
1 square millimeter is equal to 1/1000000 square meters
(1 square millimeter = 0.01 square centimeter = 0.0001 square centimeters = 0.0000 01 square meters, 1 mm = 1 / 1000 meters
1 square mm = 1 / (1000 * 1000) square meters)
Object resistance formula:
R=ÏL/S
In the formula:
R is the resistance of the object (ohms);
Ï is the resistivity of the substance in ohm meters (Ω. mm2/m).
L is the length in meters (m)
S is the cross-sectional area in square meters (mm2)
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 resistance of the wire (U=I*R)
Power calculation formula P=U2/R (U stands for voltage, R stands for resistance)
The resistivity of the pure silver wire at 0 ° C is 0.016 Ω.mm 2 / m (ohm meters), the resistivity of the pure copper wire at 0 ° C is 0.0175 Ω.mm2 / m (ohm meters), the temperature coefficient of resistance is 0.00393 / ° C . The resistivity of pure aluminum wire at 0 ° C is 0.0283 Ω.mm2 / m (ohm meters),
Ï=Ïo(1+a*t)
In the middle
Ρ——resistivity at t degrees Celsius
Ρo - resistivity at 0 ° C
T——temperature
Resistivity of pure copper wire at 21 °C = 0.0175 (1 + 0.00393 * 21) = 0.019 Ω. mm2 / m) R = 0.0175X200 / 4 = 0.875 ohms
The power consumption on the 30th, if it is 220V, then the power is required first P=U2/R=2202/0.875=55.314W
(This state is that the wires directly form the loop, that is to say, no household appliances are connected. If it is so short, the meeting will trip directly!)
55.314X30X24÷1000=39.826 degrees (1 degree electricity = 1 kWh)
Voltage loss = ∑ (P * L) / (C * S)
P: Total power of the circuit (KW) L: Distance of the power supply from the load (m) C: Material coefficient (at 380V, copper is taken as 77, aluminum is 46.3; when 220V, copper is taken as 12.8, aluminum is taken as 7.75) S: Wire cross-sectional area ( Mm)
Eight, different temperature resistivity
Resistivity of several metal conductors at 20 ° C
(1) Silver 1.65 (Europe) × 10-8 (m)
(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) Manganese copper 4.4 × 10-7
(8) Mercury 9.6 × 10-7
(9) Concord 5.0 × 10-7
(10) Nickel-chromium alloy 1.0 × 10-6
(11) Iron chromium aluminum alloy 1.4 × 10-6
(12) AlNi-Fe alloy 1.6 × 10-6
(13) Graphite (8 ~ 13) × 10-6
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