How to calculate the length of nichrome wire. Electric heating elements, heating elements, types, designs, connection and testing Open nichrome spiral
Calculation of a nichrome spiral. Ready for you to make a nichrome spiral. Nichrome length at 220 volts
Calculation of a nichrome spiral. Ready for you to make a nichrome spiral
When winding a nichrome spiral for heating elements, the operation is often performed by trial and error, and then voltage is applied to the spiral and by heating the nichrome wire, the threads select the required number of turns.
Usually, such a procedure takes a long time, and nichrome loses its characteristics with multiple kinks, which leads to rapid burnout in places of deformation. In the worst case, nichrome scrap is obtained from business nichrome.
With its help, you can accurately determine the length of the winding turn to turn. Depending on the Ø of the nichrome wire and the Ø of the rod on which the nichrome spiral is wound. It is not difficult to recalculate the length of a nichrome spiral to a different voltage using a simple mathematical proportion.
|
Ø nichrome 0.2 mm |
Ø nichrome 0.3 mm | nichrome 0.4 mm | Ø nichrome 0.5 mm | Ø nichrome 0.6 mm | Ø nichrome 0.7 mm | ||||||
| rod Ø, mm | spiral length, cm |
rod, mm |
spiral length, cm |
rod, mm |
spiral length, cm |
rod, mm |
spiral length, cm |
rod, mm |
spiral length, cm |
rod, mm |
spiral length, cm |
| 1,5 | 49 | 1,5 | 59 | 1,5 | 77 | 2 | 64 | 2 | 76 | 2 | 84 |
| 2 | 30 | 2 | 43 | 2 | 68 | 3 | 46 | 3 | 53 | 3 | 64 |
| 3 | 21 | 3 | 30 | 3 | 40 | 4 | 36 | 4 | 40 | 4 | 49 |
| 4 | 16 | 4 | 22 | 4 | 28 | 5 | 30 | 5 | 33 | 5 | 40 |
| 5 | 13 | 5 | 18 | 5 | 24 | 6 | 26 | 6 | 30 | 6 | 34 |
| 6 | 20 | 8 | 22 | 8 | 26 | ||||||
For example, it is required to determine the length of a nichrome spiral for a voltage of 380 V from a wire Ø 0.3 mm, a winding rod Ø 4 mm. The table shows that the length of such a spiral for a voltage of 220 V will be 22 cm. Let's make a simple ratio:
220 V - 22 cm
380 V - X cm
X = 380 22 / 220 = 38 cm
Calculation of electric heating elements from nichrome wire
The length of the nichrome wire for the manufacture of the spiral is determined based on the required power.
Example: Determine the length of a nichrome wire for a tile heating element with a power of P = 600 W at Umains = 220 V.
1) I = P/U = 600/220 = 2.72 A
2) R \u003d U / I \u003d 220 / 2.72 \u003d 81 ohms
3) According to these data (see table 1), we choose d=0.45; S=0.159
then the length of nichrome
l \u003d SR / ρ \u003d 0.159 81 / 1.1 \u003d 11.6 m
where l - wire length (m)
S - wire section (mm2)
R - wire resistance (Ohm)
ρ - resistivity (for nichrome ρ=1.0÷1.2 Ohm mm2/m)
Our Company PARTAL is ready to produce a nichrome spiral according to the specifications and sketches of the customer
It is profitable to buy a nichrome spiral in the PARTAL company
Nichrome for spiral High Quality only Russian production. Strict compliance with quality and brand
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Calculation of nichrome spiral | Useful
The calculation of a nichrome spiral is, in fact, a very important process. Very often, in factories, industries, factories, this is neglected and the calculation is made “by eye”, after which the spiral is connected to the network, and then the required number of turns is selected depending on the heating of the nichrome wire. Perhaps this procedure is very simple, but it takes a long time and part of the nichrome is simply wasted.
However, this procedure can be performed much more accurately, easier and faster. In order to rationalize your work, to calculate a nichrome spiral for a voltage of 220 volts, you can use the table below. Based on the fact that the specific resistance of nichrome is (Ohm mm2 / m) C, you can quickly calculate the winding length turn to turn depending on the diameter of the rod on which the nichrome thread is wound, and directly on the very thickness of the nichrome wire. And using a simple mathematical proportion, you can easily calculate the length of the spiral for a different voltage.
For example, you need to determine the length of a nichrome spiral for a voltage of 127 volts from a wire whose thickness is 0.3 mm, and a winding rod 4 mm in diameter. Looking at the table, it can be seen that the length of this spiral for a voltage of 220 volts will be 22 cm. We make a simple ratio: 220 V - 22 cm 127 V - X cm then: X \u003d 127 22 / 220 \u003d 12.7 cm
Having wound a nichrome spiral, carefully connect it, without cutting it, to a voltage source and make sure in your calculations, or rather in the calculations of the correct winding. And it is worth remembering that for closed spirals, the winding length is increased by a third of the value given in this table.
Nichrome wire, nichrome weight calculation, nichrome application
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We manufacture electric spirals from NICHROMA according to the specifications and sketches of the customer
Nichrome spiral
Everyone knows what a nichrome spiral is. This is a heating element in the form of a wire coiled with a screw for compact placement.
This wire is made from nichrome, a precision alloy whose main components are nickel and chromium.
The "classic" composition of this alloy is 80% nickel, 20% chromium.
The composition of the names of these metals formed the name that denotes the group of chromium-nickel alloys - "nichrome".
The most famous brands of nichrome are X20H80 and X15H60. The first of them is close to the "classics". It contains 72-73% nickel and 20-23% chromium.
The second is designed to reduce the cost and improve the machinability of the wire.
On the basis of these alloys, their modifications with higher survivability and resistance to oxidation at high temperatures were obtained.
These are the Kh20N80-N (-N-VI) and Kh15N60 (-N-VI) brands. They are used for heating elements in contact with air. The recommended maximum operating temperature is from 1100 to 1220 °C
The use of nichrome wire
The main quality of nichrome is high resistance electric current. It defines the scope of the alloy.
The nichrome spiral is used in two qualities - as a heating element or as a material for the electrical resistance of electrical circuits.
For heaters, an electric spiral made of Kh20N80-N and Kh15N60-N alloys is used.
Application examples:
- household thermal reflectors and fan heaters;
- Heating elements for household heating appliances and electric heating;
- heaters for industrial furnaces and thermal equipment.
Alloys Kh15N60-N-VI and Kh20N80-N-VI obtained in vacuum induction furnaces are used in industrial equipment increased reliability.
Spiral made of nichrome grades X15N60, X20N80, X20N80-VI, N80HYUD-VI differs in that its electrical resistance changes little with temperature.
Resistors, connectors of electronic circuits, critical parts of vacuum devices are made from it.
How to wind a spiral from nichrome
A resistive or heating coil can be made at home. To do this, you need a nichrome wire of a suitable brand and the correct calculation of the required length.
The calculation of a nichrome spiral is based on the resistivity of the wire and the required power or resistance, depending on the purpose of the spiral. When calculating the power, it is necessary to take into account the maximum allowable current at which the coil heats up to a certain temperature.
Temperature accounting
For example, a wire with a diameter of 0.3 mm at a current of 2.7 A will heat up to 700 ° C, and a current of 3.4 A will heat it up to 900 ° C.
To calculate the temperature and current, there are reference tables. But you still need to consider the operating conditions of the heater.
When immersed in water, heat transfer increases, then the maximum current can be increased by up to 50% of the calculated one.
A closed tubular heater, on the contrary, impairs heat dissipation. In this case, the permissible current must also be reduced by 10-50%.
The intensity of heat removal, and hence the temperature of the heater, is affected by the winding pitch of the spiral.
Tightly spaced coils provide more heat, larger pitch enhances cooling.
It should be noted that all tabular calculations are given for a heater located horizontally. When the angle to the horizon changes, the conditions for heat removal worsen.
Calculation of the resistance of a nichrome spiral and its length
Having decided on the power, we proceed to the calculation of the required resistance.
If the determining parameter is power, then first we find the required current according to the formula I \u003d P / U.
Having the strength of the current, we determine the required resistance. To do this, we use Ohm's law: R=U/I.
The designations here are generally accepted:
- P is the released power;
- U is the voltage at the ends of the spiral;
- R is the resistance of the coil;
- I - current strength.
The calculation of the resistance of nichrome wire is ready.
Now let's determine the length we need. It depends on the resistivity and wire diameter.
You can make a calculation based on the resistivity of nichrome: L=(Rπd2)/4ρ.
- L is the desired length;
- R is the resistance of the wire;
- d is the wire diameter;
- ρ is the resistivity of nichrome;
- π is the constant 3.14.
But it is easier to take a ready-made linear resistance from the tables of GOST 12766.1-90. You can also take temperature corrections there, if you need to take into account the change in resistance during heating.
In this case, the calculation will look like this: L=R/ρld, where ρld is the resistance of one meter of wire with a diameter of d.
Spiral winding
Now let's make a geometric calculation of the nichrome spiral. We have chosen the wire diameter d, determined the required length L and have a rod with a diameter D for winding. How many turns do you need to make? The length of one turn is: π(D+d/2). The number of turns is N=L/(π(D+d/2)). Calculation completed.
practical solution
In practice, rarely anyone is engaged in independent winding of wire for a resistor or heater.
It is easier to buy a nichrome spiral with the required parameters and, if necessary, separate the required number of turns from it.
To do this, you should contact the PARTAL company, which since 1995 has been a major supplier of precision alloys, including nichrome wire, tape and coils for heaters.
Our company is able to completely remove the question of where to buy a nichrome spiral, since we are ready to make it to order according to the sketches and specifications of the customer.
partalstalina.ru
Calculation and repair of the heating winding of the soldering iron
When repairing or when making an electric soldering iron or any other heating device on your own, you have to wind the heating winding from nichrome wire. The initial data for the calculation and selection of wire is the resistance of the winding of the soldering iron or heater, which is determined based on its power and supply voltage. You can calculate what the resistance of the winding of a soldering iron or heater should be using the table.
Knowing the supply voltage and measuring the resistance of any heating appliance, such as a soldering iron, electric kettle, electric heater or electric iron, you can find out the power consumed by this household appliance. For example, the resistance of a 1.5 kW electric kettle will be 32.2 ohms.
| 12 | 48,0 | 108 | 1344 | 4033 |
| 6,0 | 24,0 | 54 | 672 | 2016 |
| 4,0 | 16,0 | 36 | 448 | 1344 |
| 3,4 | 13,7 | 31 | 384 | 1152 |
| 2,4 | 9,6 | 22 | 269 | 806 |
| 1.9 | 7.7 | 17 | 215 | 645 |
| 1,4 | 5,7 | 13 | 161 | 484 |
| 0,96 | 3,84 | 8,6 | 107 | 332 |
| 0,72 | 2,88 | 6,5 | 80,6 | 242 |
| 0,48 | 1,92 | 4,3 | 53,8 | 161 |
| 0,36 | 1,44 | 3,2 | 40,3 | 121 |
| 0,29 | 1,15 | 2,6 | 32,3 | 96,8 |
| 0,21 | 0,83 | 1,85 | 23,0 | 69,1 |
| 0,16 | 0,64 | 1,44 | 17,9 | 53,8 |
| 0,14 | 0,57 | 1,30 | 16,1 | 48,4 |
| 0,10 | 0,38 | 0,86 | 10,8 | 32,3 |
| 0,07 | 0,29 | 0,65 | 8,06 | 24,2 |
| 0,06 | 0,23 | 0,52 | 6,45 | 19,4 |
| 0,05 | 0,19 | 0,43 | 5,38 | 16,1 |
Let's look at an example of how to use the table. Let's say you need to rewind a 60 W soldering iron designed for a supply voltage of 220 V. Select 60 W from the leftmost column of the table. On the upper horizontal line, select 220 V. As a result of the calculation, it turns out that the resistance of the soldering iron winding, regardless of the material of the winding, should be equal to 806 ohms.
If you needed to make a soldering iron with a power of 60 W, designed for a voltage of 220 V, a soldering iron for power from a 36 V network, then the resistance of the new winding should already be 22 ohms. You can independently calculate the winding resistance of any electric heater using an online calculator.
After determining the required resistance value of the soldering iron winding, from the table below, the appropriate diameter of the nichrome wire is selected based on the geometric dimensions of the winding. Nichrome wire is a chromium-nickel alloy that can withstand heating temperatures up to 1000˚C and is marked X20H80. This means that the alloy contains 20% chromium and 80% nickel.
To wind a soldering iron spiral with a resistance of 806 ohms from the example above, you will need 5.75 meters of nichrome wire with a diameter of 0.1 mm (you need to divide 806 by 140), or 25.4 m of wire with a diameter of 0.2 mm, and so on.
When winding the soldering iron spiral, the turns are stacked close to each other. When heated, the red-hot surface of the nichrome wire oxidizes and forms an insulating surface. If the entire length of the wire does not fit on the sleeve in one layer, then the wound layer is covered with mica and the second one is wound.
For electrical and thermal insulation of the heating element winding the best materials is mica, fiberglass cloth and asbestos. Asbestos has an interesting property, it can be soaked with water and it becomes soft, allows you to give it any shape, and after drying it has sufficient mechanical strength. When insulating the winding of the soldering iron with wet asbestos, it should be taken into account that wet asbestos conducts eclectic current well and it will be possible to turn on the soldering iron in the mains only after the asbestos has completely dried.
felstar.mypage.ru
HOW TO CALCULATE A SPIRAL FROM NICHROME?
Post written by admin at 18.01.2015 23:23Categories: 3. Home electrical, Electrical workshop
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The winding of a nichrome spiral for heating devices is often performed “by eye”, and then, including the spiral in the network, the required number of turns is selected by heating the nichrome wire. Usually such a procedure takes a lot of time, and nichrome is wasted.
When using a spiral for a voltage of 220 V, you can use the data given in the table, on the basis that the resistivity of nichrome ρ = (Ohm mm2 / m). Using this formula, you can quickly determine the length of the winding turn to turn, depending on the thickness of the nichrome wire and the diameter of the rod on which the spiral is wound.
For example, if it is required to determine the length of a spiral for a voltage of 127 V from a nichrome wire 0.3 mm thick, a winding rod with a diameter of. 4 mm. The table shows that the length of such a spiral for a voltage of 220 V will be 22 cm.
Let's make a simple ratio:
220 V - 22 cm
X \u003d 127 * 22 / 220 \u003d 12.7 cm.
After winding the spiral, connect it without cutting it to a voltage source and make sure that the winding is correct. For closed spirals, the winding length is increased by 1/3 of the value given in the table.
Conventions in the table: D - rod diameter, mm; L is the length of the spiral, cm.
| diam. nichrome 0.2 mm | diam. nichrome 0.3 mm | diam. nichrome 0.4 mm | diam. nichrome 0.5 mm | diam. nichrome 0.6 mm | diam. nichrome 0.7 mm | diam. nichrome 0.8 mm | diam. nichrome 0.9 mm | diam. nichrome 1.0 mm | |||||||||
| D | L | D | L | D | L | D | L | D | L | D | L | D | L | D | L | D | L |
| 1,5 | 49 | 1,5 | 59 | 1,5 | 77 | 2 | 64 | 2 | 76 | 2 | 84 | 3 | 68 | 3 | 78 | 3 | 75 |
| 2 | 30 | 2 | 43 | 2 | 68 | 3 | 46 | 3 | 53 | 3 | 62 | 4 | 54 | 4 | 72 | 4 | 63 |
| 3 | 21 | 3 | 30 | 3 | 40 | 4 | 36 | 4 | 40 | 4 | 49 | 5 | 46 | 6 | 68 | 5 | 54 |
| 4 | 16 | 4 | 22 | 4 | 28 | 5 | 30 | 5 | 33 | 5 | 40 | 6 | 40 | 8 | 52 | 6 | 48 |
| 5 | 13 | 5 | 18 | 5 | 24 | 6 | 26 | 6 | 30 | 6 | 34 | 8 | 31 | 8 | 33 | ||
| 6 | 20 | 8 | 22 | 8 | 26 | 10 | 24 | 10 | 30 | ||||||||
| 10 | 22 | ||||||||||||||||
elctricvs.ru
Electrical resistance is one of the most important characteristics of nichrome. It is determined by many factors, in particular, the electrical resistance of nichrome depends on the size of the wire or tape, the grade of the alloy. The general formula for active resistance is: R = ρ l / S R - active electrical resistance (Ohm), ρ - electrical resistivity (Ohm mm), l - conductor length (m), S - cross-sectional area (mm2)
To rationalize this work when using a nichrome spiral for a voltage of 220 V, I propose to use the data given in the table, on the basis that the resistivity of nichrome = (Ohm mm2 / m) C. With its help, you can quickly determine the length of the winding turn to turn, depending on the thickness of the nichrome wire and the diameter of the rod on which the nichrome spiral is wound. It is not difficult to recalculate the length of a nichrome spiral to a different voltage using a simple mathematical proportion.
For example, it is required to determine the length of a nichrome spiral for a voltage of 380 V from a wire 0.3 mm thick, a winding rod Ø 4 mm. The table shows that the length of such a spiral for a voltage of 220 V will be 22 cm. Let's make a simple ratio: 220 V - 22 cm 380 V - X cm then: X = 380 22 / 220 = 38 cm Having wound a nichrome spiral, connect it without cutting it to a voltage source and make sure that the winding is correct. For closed spirals, the winding length is increased by 1/3 of the value given in the table. This table shows the theoretical weight of 1 meter of nichrome wire and tape. It varies depending on the size of the product.
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www.metotech.ru
| Heating element calculation Calculation example. Given: U=220V, t=700°C, type Х20Н80, d=0.5mm-----------L,P-? corresponds to S = 0.196 mm², and the current at 700 ° C I = 5.2 A. The type of alloy X20H80 is nichrome, the specific resistance of which is ρ = 1.11 μOhm m. We determine the resistance R = U / I = 220 / 5.2 = 42.3 Ohm. From here we calculate the wire length: L = RS / ρ = 42.3 0.196 / 1.11 = 7.47 m. Determine the power of the heating element: P = U I = 220 5.2 = 1.15 kW .When winding the spiral, the following relationship is observed: D=(7÷10)d, where D is the diameter of the spiral, mm, d is the diameter of the wire, mm. Note: - if the heaters are inside the heated liquid, then the load (current) can be increased by 1 ,1-1.5 times; - in the closed version of the heater, the current should be reduced by 1.2-1.5 times. A smaller coefficient is taken for a thicker wire, a larger one for a thin one. For the first case, the coefficient is chosen exactly the opposite. I will make a reservation: we are talking about a simplified calculation of the heating element. Perhaps someone will need a table of electrical resistance values \u200b\u200bfor 1 m of nichrome wire, as well as its weight Table 1. Permissible current strength of nichrome wire at normal temperature
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www.elektrikii.ru
When winding a nichrome spiral for heating elements, the operation is often performed by trial and error, and then voltage is applied to the spiral and by heating the nichrome wire, the threads select the required number of turns.
Usually, such a procedure takes a long time, and nichrome loses its characteristics with multiple kinks, which leads to rapid burnout in places of deformation. In the worst case, nichrome scrap is obtained from business nichrome.
With its help, you can accurately determine the length of the winding turn to turn. Depending on the Ø of the nichrome wire and the Ø of the rod on which the nichrome spiral is wound. It is not difficult to recalculate the length of a nichrome spiral to a different voltage using a simple mathematical proportion.
The length of the nichrome spiral depending on the diameter of the nichrome and the diameter of the rod
|
Ø nichrome 0.2 mm |
Ø nichrome 0.3 mm | nichrome 0.4 mm | Ø nichrome 0.5 mm | Ø nichrome 0.6 mm | Ø nichrome 0.7 mm | ||||||
| rod Ø, mm | spiral length, cm |
Ø rod, mm |
spiral length, cm |
Ø rod, mm |
spiral length, cm |
Ø rod, mm |
spiral length, cm |
Ø rod, mm |
spiral length, cm |
Ø rod, mm |
spiral length, cm |
| 1,5 | 49 | 1,5 | 59 | 1,5 | 77 | 2 | 64 | 2 | 76 | 2 | 84 |
| 2 | 30 | 2 | 43 | 2 | 68 | 3 | 46 | 3 | 53 | 3 | 64 |
| 3 | 21 | 3 | 30 | 3 | 40 | 4 | 36 | 4 | 40 | 4 | 49 |
| 4 | 16 | 4 | 22 | 4 | 28 | 5 | 30 | 5 | 33 | 5 | 40 |
| 5 | 13 | 5 | 18 | 5 | 24 | 6 | 26 | 6 | 30 | 6 | 34 |
| 6 | 20 | 8 | 22 | 8 | 26 | ||||||
For example, it is required to determine the length of a nichrome spiral for a voltage of 380 V from a wire Ø 0.3 mm, a winding rod Ø 4 mm. The table shows that the length of such a spiral for a voltage of 220 V will be 22 cm. Let's make a simple ratio:
220 V - 22 cm
380 V - X cm
Then:
X = 380 22 / 220 = 38 cm
Having wound a nichrome spiral, connect it without cutting it to a voltage source and make sure that the winding is correct. For closed spirals, the winding length is increased by 1/3 of the value given in the table.
Calculation of electric heating elements from nichrome wire
The length of the nichrome wire for the manufacture of the spiral is determined based on the required power.
Example: Determine the length of nichrome wire for a tile heating element with a power P= 600 W at U networks = 220 V.
Solution:
1) I=P/U= 600/220 = 2.72 A
2) R = U/I= 220 / 2.72 = 81 ohms
3) Based on these data (see Table 1), we select d=0,45; S=0,159
then the length of nichrome
l = SR / ρ\u003d 0.159 81 / 1.1 \u003d 11.6 m
Where l- wire length (m)
S- wire section (mm 2)
R- wire resistance (Ohm)
ρ - resistivity (for nichrome ρ=1.0÷1.2 Ohm mm 2 /m)
| Permissible current (l), A | |||||||
| Ø nichrome at 700 °C , mm |
0,17 |
0,45 |
0,55 |
0.65 It is convenient and profitable to buy a nichrome spiral in the PARTAL company - online order Delivery of orders in Russia, Kazakhstan and Belarus |
In the practice of a home master, one has to repair or design heating devices. These can be various furnaces, heaters, soldering irons and cutters. Most often, spirals or nichrome wire are used for this. The main task in this case is to determine the length and cross section of the material. In this article, we will talk about how to calculate the length of a nichrome wire or spiral from power, resistance and temperature.
Basic information and brands of nichrome
Nichrome is an alloy of nickel and chromium with the addition of manganese, silicon, iron, aluminum. For this material, the parameters depend on the specific ratio of substances in the alloy, but on average they lie within:
- specific electrical resistance - 1.05-1.4 Ohm * mm 2 / m (depending on the brand of alloy);
- temperature coefficient of resistance - (0.1-0.25) 10 −3 K −1;
- working temperature - 1100 °C;
- melting point - 1400°C;
In tables, resistivity is often given in μOhm * m (or 10 -6 Ohm * m) - the numerical values are the same, the difference is in dimension.
Currently, there are two most common brands of nichrome wire:
- X20H80. It consists of 74% nickel and 23% chromium, as well as 1% iron, silicon and manganese. Conductors of this brand can be used at temperatures up to 1250 ᵒ C, melting point - 1400 ᵒ C. It also has an increased electrical resistance. The alloy is used for the manufacture of elements of heating devices. Resistivity - 1.03-1.18 μOhm m;
- X15H60. Composition: 60% nickel, 25% iron, 15% chromium. Operating temperature is not more than 1150 ᵒ С. Melting temperature is 1390 ᵒ С. Contains more iron, which increases the magnetic properties of the alloy and increases its anti-corrosion resistance.
You will learn more about the grades and properties of these alloys from GOST 10994-74, GOST 8803-89, GOST 12766.1-90 and others.
As already mentioned, nichrome wire is used everywhere where needed. heating elements. High resistivity and melting point make it possible to use nichrome as a basis for various heating elements, from a kettle or a hair dryer to a muffle furnace.
Calculation methods
By resistance
Let's figure out how to calculate the length of a nichrome wire in terms of power and resistance. The calculation begins with the determination of the required power. Let's imagine that we need a nichrome filament for a small-sized soldering iron with a power of 10 watts, which will work from a 12V power supply. For this we have a wire with a diameter of 0.12 mm.
The simplest calculation of the length of nichrome in terms of power without taking into account heating is performed as follows:
Let's determine the current strength:
I=P/U=10/12=0.83A
The calculation of the resistance of nichrome wire is carried out according to:
R=U/I=12/0.83=14.5 Ohm
The length of the wire is:
l=SR/ ρ ,
where S is the area cross section, ρ – resistivity.
Or with this formula:
l= (Rπd2)/4 ρ
L=(14.5*3.14*0.12^2)/4*1.1=0.149m=14.9cm
The same can be taken from GOST 12766.1-90 tab. 8, where the value of 95.6 Ohm / m is indicated, if you recalculate it, you get almost the same thing:
L=R required / R table = 14.4 / 95.6 = 0.151m = 15.1cm
For a 10 watt heater powered by 12V, you need 15.1cm.
If you need to calculate the number of turns of a spiral in order to twist it from nichrome wire of this length, then use the following formulas:
Length of one turn:
Number of turns:
N=L/(π(D+d/2)),
where L and d are the length and diameter of the wire, D is the diameter of the rod on which the spiral will be wound.
Let's say we wind a nichrome wire on a rod with a diameter of 3 mm, then we carry out the calculations in millimeters:
N=151/(3.14(3+0.12/2))=15.71 turns
But at the same time, it must be taken into account whether nichrome of such a cross section is capable of withstanding this current at all. Detailed tables for determining the maximum allowable current at a certain temperature for specific sections are given below. In simple words- you determine how many degrees the wire should be heated and choose its cross section for the rated current.
Also note that if the heater is inside the liquid, then the current can be increased by 1.2-1.5 times, and if in a confined space, then vice versa - reduced.
By temperature
The problem with the above calculation is that we calculate the resistance of the cold coil by the diameter of the nichrome filament and its length. But it depends on the temperature, and at the same time it is necessary to take into account under what conditions it will be possible to achieve it. If for cutting foam or for a heater such a calculation is still applicable, then for a muffle furnace it will be too rough.
Let us give an example of nichrome calculations for a furnace.
First, its volume is determined, let's say 50 liters, then the power is determined, for this there is a rule of thumb:
- up to 50 liters - 100W / l;
- 100-500 liters - 50-70 W / l.
Then in our case:
I=5000/220=22.7 Amps
R=220/22.7=9.7 ohm
For 380V when connecting the spirals with a star, the calculation will be as follows.
We divide the power into 3 phases:
Pf=5/3=1.66 kW per phase
When connected in a star, 220V is applied to each branch (phase voltage, may differ depending on your electrical installation), then the current is:
I=1660/220=7.54 A
Resistance:
R=220/7.54=29.1 ohm
For a triangle connection, we calculate from a linear voltage of 380V:
I=1660/380=4.36 A
R=380/4.36=87.1 ohm
To determine the diameter, the specific surface power of the heater is taken into account. We calculate the length, take the resistivity from the table. 8. GOST 12766.1-90, but first let's determine the diameter.
To calculate the specific surface power of the furnace, use the formula.
Bef (depends on the heat-receiving surface) and a (radiation efficiency coefficient) are selected according to the following tables.
So, to heat the furnace to 1000 degrees, let's take the temperature of the spiral at 1100 degrees, then according to the selection table V eff we select a value of 4.3 W / cm 2, and according to the selection table of the coefficient a - 0.2.
V add \u003d V ef * a \u003d 4.3 * 0.2 \u003d 0.86 W / cm 2 \u003d 0.86 * 10 ^ 4 W / m 2
The diameter is determined by the formula:
p t - specific resistance of the heater material at a given t, determined according to GOST 12766.1, table 9 (given below).
For nichrome Х80Н20 - 1.025
p t \u003d p 20 * p 1000 \u003d 1.13 * 10 ^ 6 * 1.025 \u003d 1.15 * 10 ^ 6 Ohm / mm
Then, to connect to a three-phase network according to the "Star" scheme:
The length is calculated by the formula:
Let's check the values:
L=R/(p*k)=29.1/(0.82*1.033)=34m
The values differ due to the high temperature of the coil, the check does not take into account a number of factors. Therefore, we will take for the length of 1 spiral - 42m, then for three spirals you need 126 meters of nichrome 1.3 mm.
Conclusion
- environmental conditions;
- location of heating elements;
- spiral temperature;
- the temperature to which the surface must be heated and other factors.
Even the above calculation, despite its complexity, cannot be called sufficiently accurate. Because the calculation of heating elements is a continuous thermodynamics and a number of other factors can be cited that affect its results, for example, the thermal insulation of the furnace and so on.
In practice, after the estimated calculations, the spirals are added or removed depending on the result obtained, or temperature sensors and devices are used to adjust it.
materials
If home master Because of the nature of the work he performs, a muffle furnace is necessary, he, of course, can purchase a finished device in a store or through advertisements. However, such factory-made equipment is very expensive. Therefore, many craftsmen take up the manufacture of such furnaces on their own.
The main "working unit" of an electric muffle furnace is a heater, which in handicraft production is usually made in the form of a spiral of special wire with high resistance and thermal efficiency. Its characteristics must strictly correspond to the power of the equipment being created, the expected temperature conditions of operation, and also meet some other requirements. If planned independent production device, we recommend using the algorithm proposed below and convenient calculators for calculating the muffle furnace heater.
The calculation requires certain explanations, which we will try to state as clearly as possible.
Algorithm and calculators for calculating the heater of a muffle furnace
What are heating coils made of?
To begin with, just a few words about the wire that is used for winding heating coils. Usually, nichrome or fechral is used for such purposes.
- Nichrome(from abbreviations nickel + chromium) is most often represented by alloys Kh20N80-N, Kh15N60 or Kh15N60-N.
muffle furnace prices
muffle furnace
Her dignity :
- high margin of safety at any heating temperature;
- plastic, easy to process, amenable to welding;
- durability, resistance to corrosion, lack of magnetic qualities.
Flaws :
— high price;
- lower heating rates and thermal stability compared to Fechraleva.
- Fekhraleva(from abbreviations ferrum, chromium, aluminum) - in our time, material from the Kh23Yu 5T alloy is more often used.
Advantages fehral:
- much cheaper than nichrome, due to which the material is mainly popular;
- has more significant indicators of resistance and resistive heating;
- high heat resistance.
Flaws :
- low strength, and after even a single heating over 1000 degrees - pronounced fragility of the spiral;
- outstanding durability;
- the presence of magnetic qualities, susceptibility to corrosion due to the presence of iron in the composition;
- unnecessary chemical activity - it is able to react with the material of the fireclay lining of the furnace;
- excessively large thermal linear expansion.
Each of the masters is free to choose any of the listed materials, having analyzed their pros and cons. The calculation algorithm takes into account the features of such a choice.
Step 1 - determining the power of the furnace and the strength of the current passing through the heater.
In order not to go into unnecessary given case of details, we will immediately say that there are empirical compliance standardsvolumemuffle furnace working chamber and her power. They are shown in the table below:
If there are design sketches of the future device, then the volume of the muffle chamber is easy to determine - the product of height, width and depth. Then the volume is converted to liters and multiplied by the recommended power rates indicated in the table. So we get the power of the furnace in watts.
Table values are in some ranges, so either use interpolation or take an approximate average value.
The found power, with a known mains voltage (220 volts), allows you to immediately determine the strength of the current that will pass through the heating element.
I=P/U.
I- current strength.
R– the power of the muffle furnace determined above;
U- supply voltage.
This entire first step of the calculation can be done very easily and quickly with the help of a calculator: all tabular values \u200b\u200bare already entered into the calculation program.
Calculator of muffle furnace power and current through the heater
Very often, if you want to make or repair heater do-it-yourself electric furnaces, a person has many questions. For example, what diameter to take the wire, what should be its length, or what power can be obtained using a wire or tape with given parameters, etc. With the right approach to solving this issue, it is necessary to take into account quite a lot of parameters, for example, the strength of the current passing through heater, operating temperature, type of electrical network and others.This article provides reference data on the materials most common in the manufacture of heaters. electric ovens, as well as the methodology and examples of their calculation (calculation of heaters for electric furnaces).
Heaters. Materials for the manufacture of heaters
Directly heater- one of the most important elements furnace, it is he who performs heating, has the highest temperature and determines the performance of the heating installation as a whole. Therefore, heaters must meet a number of requirements, which are listed below.Requirements for heaters
Basic requirements for heaters (heater materials):- Heaters must have sufficient heat resistance (scaling resistance) and heat resistance. Heat resistance - mechanical strength at high temperatures. Heat resistance - resistance of metals and alloys to gas corrosion at high temperatures (the properties of heat resistance and heat resistance are described in more detail on the page).
- Heater in an electric furnace must be made of a material with high electrical resistivity. talking plain language The higher the electrical resistance of the material, the more it heats up. Therefore, if you take a material with less resistance, then you need a heater of greater length and with a smaller cross-sectional area. Not always a sufficiently long heater can be placed in the furnace. It should also be taken into account that the larger the diameter of the wire from which the heater is made, the longer its service life . Examples of materials with high electrical resistance are chromium-nickel alloy, iron-chromium-aluminum alloy, which are precision alloys with high electrical resistance.
- A low temperature coefficient of resistance is an essential factor when choosing a material for a heater. This means that when the temperature changes, the electrical resistance of the material heater doesn't change much. If the temperature coefficient of electrical resistance is large, to turn on the furnace in a cold state, it is necessary to use transformers that initially give a reduced voltage.
- The physical properties of the heater materials must be constant. Some materials, such as carborundum, which is a non-metallic heater, can change their properties over time. physical properties, in particular electrical resistance, which complicates the conditions of their operation. To stabilize the electrical resistance, transformers with a large number of steps and a voltage range are used.
- metal materials must have good technological properties, namely: ductility and weldability, so that they can be used to make wire, tape, and from the tape - heating elements of complex configuration. Also heaters can be made from non-metals. Non-metallic heaters are pressed or molded into a finished product.
Materials for the manufacture of heaters
The most suitable and most used in the production of heaters for electric furnaces are precision alloys with high electrical resistance. These include alloys based on chromium and nickel ( chromium-nickel), iron, chromium and aluminum ( iron-chromium-aluminum). Grades and properties of these alloys are discussed in “Precision alloys. Marks». Representatives of chromium-nickel alloys are grades Kh20N80, Kh20N80-N (950-1200 °C), Kh15N60, Kh15N60-N (900-1125 °С), iron-chromoaluminum - grades Kh23Yu5T (950-1400 °С), Kh27Yu5T (950-1350 °С ), X23Yu5 (950-1200 °C), X15Yu5 (750-1000 °C). There are also iron-chromium-nickel alloys - Kh15N60Yu3, Kh27N70YuZ.The alloys listed above have good heat resistance and heat resistance properties, so they can work at high temperatures. good heat resistance provides a protective film of chromium oxide, which forms on the surface of the material. The melting temperature of the film is higher than the melting temperature of the alloy itself; it does not crack when heated and cooled.
Let's bring comparative characteristic nichrome and fechral.
Advantages of nichrome:
- good mechanical properties at both low and high temperatures;
- the alloy is creep-resistant;
- has good technological properties - ductility and weldability;
- well processed;
- does not age, non-magnetic.
- high cost of nickel - one of the main components of the alloy;
- lower operating temperatures compared to Fechral.
- cheaper alloy compared to nichrome, tk. does not contain ;
- has better heat resistance than nichrome, for example, Fechral X23Yu5T can operate at temperatures up to 1400 ° C (1400 ° C is the maximum operating temperature for a heater made of wire Ø 6.0 mm or more; Ø 3.0 - 1350 ° C; Ø 1.0 - 1225 °С; Ø 0.2 - 950 °С).
- brittle and fragile alloy, these negative properties are especially pronounced after the alloy has been at a temperature of more than 1000 ° C;
- because fechral has iron in its composition, then this alloy is magnetic and can rust in a humid atmosphere at normal temperatures;
- has low creep resistance;
- interacts with fireclay lining and iron oxides;
- Fechral heaters elongate significantly during operation.
Recently, alloys of the Kh15N60Yu3 and Kh27N70YuZ types have been developed; with the addition of 3% aluminum, which significantly improved the heat resistance of alloys, and the presence of nickel virtually eliminated the disadvantages of iron-chromium-aluminum alloys. Alloys Kh15N60YuZ, Kh27N60YUZ do not interact with chamotte and iron oxides, they are quite well processed, mechanically strong, not brittle. The maximum operating temperature of the X15N60YUZ alloy is 1200 °C.
In addition to the alloys listed above based on nickel, chromium, iron, aluminum, other materials are also used for the manufacture of heaters: refractory metals, as well as non-metals.
Among non-metals for the manufacture of heaters, carborundum, molybdenum disilicide, coal, and graphite are used. Carborundum and molybdenum disilicide heaters are used in high-temperature furnaces. In furnaces with a protective atmosphere, carbon and graphite heaters are used.
Among refractory materials, tantalum and niobium can be used as heaters. In high temperature vacuum and protective atmosphere furnaces, molybdenum heaters And tungsten. Molybdenum heaters can operate up to a temperature of 1700 °C in vacuum and up to 2200 °C in a protective atmosphere. This temperature difference is due to the evaporation of molybdenum at temperatures above 1700 °C in vacuum. Tungsten heaters can operate up to 3000 °C. In special cases, tantalum and niobium heaters are used.
Calculation of heaters of electric furnaces
Usually, the input data for are the power that the heaters must provide, the maximum temperature that is required to implement the corresponding technological process(tempering, hardening, sintering, etc.) and dimensions of the working space of the electric furnace. If the furnace power is not set, then it can be determined by the rule of thumb. During the calculation of heaters, it is required to obtain the diameter and length (for wire) or cross-sectional area and length (for tape), which are necessary for manufacture of heaters.It is also necessary to determine the material from which to make heaters(this item is not considered in the article). In this article, as a material for heaters, a high electrical resistance nickel-chromium precision alloy is considered, which is one of the most popular in the manufacture of heating elements.
Determining the diameter and length of the heater (nichrome wire) for a given furnace power (simple calculation)
Perhaps the most simple option heater calculation of nichrome is the choice of diameter and length at a given power of the heater, the supply voltage of the network, as well as the temperature that the heater will have. Despite the simplicity of the calculation, it has one feature, which we will pay attention to below.An example of calculating the diameter and length of the heating element
Initial data:
Device power P
= 800 W; mains voltage U
= 220 V; heater temperature 800 °C. Nichrome wire X20H80 is used as a heating element.
1. First you need to determine the current strength that will pass through the heating element:
I=P/U
\u003d 800 / 220 \u003d 3.63 A.
2. Now you need to find the resistance of the heater:
R=U/I
= 220 / 3.63 = 61 ohms;
3. Based on the value obtained in paragraph 1 of the current passing through nichrome heater, you need to select the diameter of the wire. And this moment is important. If, for example, at a current strength of 6 A, a nichrome wire with a diameter of 0.4 mm is used, then it will burn out. Therefore, having calculated the current strength, it is necessary to select the appropriate value of the wire diameter from the table. In our case, for a current strength of 3.63 A and a heater temperature of 800 ° C, we select a nichrome wire with a diameter d = 0.35 mm and cross-sectional area S \u003d 0.096 mm 2.
General rule wire diameter selection can be formulated as follows: it is necessary to choose a wire whose permissible current strength is not less than the calculated current strength passing through the heater. In order to save the material of the heater, you should choose a wire with the nearest higher (than the calculated) allowable current.
Table 1
| Diameter, mm | Cross-sectional area of nichrome wire, mm 2 | Heating temperature of nichrome wire, °C | ||||||
|---|---|---|---|---|---|---|---|---|
| 200 | 400 | 600 | 700 | 800 | 900 | 1000 | ||
| Maximum allowable current, A | ||||||||
| 5 | 19,6 | 52 | 83 | 105 | 124 | 146 | 173 | 206 |
| 4 | 12,6 | 37,0 | 60,0 | 80,0 | 93,0 | 110,0 | 129,0 | 151,0 |
| 3 | 7,07 | 22,3 | 37,5 | 54,5 | 64,0 | 77,0 | 88,0 | 102,0 |
| 2,5 | 4,91 | 16,6 | 27,5 | 40,0 | 46,6 | 57,5 | 66,5 | 73,0 |
| 2 | 3,14 | 11,7 | 19,6 | 28,7 | 33,8 | 39,5 | 47,0 | 51,0 |
| 1,8 | 2,54 | 10,0 | 16,9 | 24,9 | 29,0 | 33,1 | 39,0 | 43,2 |
| 1,6 | 2,01 | 8,6 | 14,4 | 21,0 | 24,5 | 28,0 | 32,9 | 36,0 |
| 1,5 | 1,77 | 7,9 | 13,2 | 19,2 | 22,4 | 25,7 | 30,0 | 33,0 |
| 1,4 | 1,54 | 7,25 | 12,0 | 17,4 | 20,0 | 23,3 | 27,0 | 30,0 |
| 1,3 | 1,33 | 6,6 | 10,9 | 15,6 | 17,8 | 21,0 | 24,4 | 27,0 |
| 1,2 | 1,13 | 6,0 | 9,8 | 14,0 | 15,8 | 18,7 | 21,6 | 24,3 |
| 1,1 | 0,95 | 5,4 | 8,7 | 12,4 | 13,9 | 16,5 | 19,1 | 21,5 |
| 1,0 | 0,785 | 4,85 | 7,7 | 10,8 | 12,1 | 14,3 | 16,8 | 19,2 |
| 0,9 | 0,636 | 4,25 | 6,7 | 9,35 | 10,45 | 12,3 | 14,5 | 16,5 |
| 0,8 | 0,503 | 3,7 | 5,7 | 8,15 | 9,15 | 10,8 | 12,3 | 14,0 |
| 0,75 | 0,442 | 3,4 | 5,3 | 7,55 | 8,4 | 9,95 | 11,25 | 12,85 |
| 0,7 | 0,385 | 3,1 | 4,8 | 6,95 | 7,8 | 9,1 | 10,3 | 11,8 |
| 0,65 | 0,342 | 2,82 | 4,4 | 6,3 | 7,15 | 8,25 | 9,3 | 10,75 |
| 0,6 | 0,283 | 2,52 | 4 | 5,7 | 6,5 | 7,5 | 8,5 | 9,7 |
| 0,55 | 0,238 | 2,25 | 3,55 | 5,1 | 5,8 | 6,75 | 7,6 | 8,7 |
| 0,5 | 0,196 | 2 | 3,15 | 4,5 | 5,2 | 5,9 | 6,75 | 7,7 |
| 0,45 | 0,159 | 1,74 | 2,75 | 3,9 | 4,45 | 5,2 | 5,85 | 6,75 |
| 0,4 | 0,126 | 1,5 | 2,34 | 3,3 | 3,85 | 4,4 | 5,0 | 5,7 |
| 0,35 | 0,096 | 1,27 | 1,95 | 2,76 | 3,3 | 3,75 | 4,15 | 4,75 |
| 0,3 | 0,085 | 1,05 | 1,63 | 2,27 | 2,7 | 3,05 | 3,4 | 3,85 |
| 0,25 | 0,049 | 0,84 | 1,33 | 1,83 | 2,15 | 2,4 | 2,7 | 3,1 |
| 0,2 | 0,0314 | 0,65 | 1,03 | 1,4 | 1,65 | 1,82 | 2,0 | 2,3 |
| 0,15 | 0,0177 | 0,46 | 0,74 | 0,99 | 1,15 | 1,28 | 1,4 | 1,62 |
| 0,1 | 0,00785 | 0,1 | 0,47 | 0,63 | 0,72 | 0,8 | 0,9 | 1,0 |
Note :
- if the heaters are inside the heated liquid, then the load (permissible current) can be increased by 1.1 - 1.5 times;
- when the heaters are closed (for example, in chamber electric furnaces), it is necessary to reduce the load by 1.2 - 1.5 times (a smaller coefficient is taken for a thicker wire, a larger one for a thin one).
R = ρ l/S ,
Where R - electrical resistance of the conductor (heater) [Ohm], ρ - electrical resistivity of the heater material [Ohm mm 2 / m], l - conductor (heater) length [mm], S - cross-sectional area of the conductor (heater) [mm 2 ].
Thus, we get the length of the heater:
l = R S / ρ
\u003d 61 0.096 / 1.11 \u003d 5.3 m.
In this example, nichrome wire Ø 0.35 mm is used as a heater. In accordance with "Wire made of precision alloys with high electrical resistance. Specifications" the nominal value of the electrical resistivity of nichrome wire brand Kh20N80 is 1.1 Ohm mm 2 / m ( ρ \u003d 1.1 Ohm mm 2 / m), see table. 2.
The result of the calculations is the required length of the nichrome wire, which is 5.3 m, diameter - 0.35 mm.
table 2
Determining the diameter and length of the heater (nichrome wire) for a given furnace (detailed calculation)
The calculation presented in this paragraph is more complex than the one above. Here we will take into account the additional parameters of the heaters, we will try to figure out the options for connecting heaters to a three-phase current network. Calculation of the heater will be carried out on the example of an electric furnace. Let the initial data be the internal dimensions of the furnace.1. The first thing to do is to calculate the volume of the chamber inside the furnace. In this case, let's take h = 490 mm, d = 350 mm and l = 350 mm (height, width and depth, respectively). Thus, we get the volume V = h d l \u003d 490 350 350 \u003d 60 10 6 mm 3 \u003d 60 l (a measure of volume).
2. Next, you need to determine the power that the furnace should give out. Power is measured in Watts (W) and is determined by rule of thumb: for an electric oven with a volume of 10 - 50 liters, the specific power is 100 W / l (Watts per liter of volume), with a volume of 100 - 500 liters - 50 - 70 W / l. Let us take the specific power of 100 W/l for the furnace under consideration. Thus, the power of the electric furnace heater should be P \u003d 100 60 \u003d 6000 W \u003d 6 kW.
It should be noted that with a power of 5-10 kW heaters are usually made in single phase. At high powers, for uniform loading of the network, the heaters are made three-phase.
3. Then you need to find the strength of the current passing through the heater I=P/U , Where P - heater power, U - the voltage on the heater (between its ends), and the resistance of the heater R=U/I .
There may be two options for connecting to the electrical network:
- to a household single-phase current network - then U = 220 V;
- to the industrial network of three-phase current - U = 220 V (between neutral wire and phase) or U = 380 V (between any two phases).
I=P/U
\u003d 6000 / 220 \u003d 27.3 A - the current passing through the heater.
Then it is necessary to determine the resistance of the furnace heater.
R=U/I
\u003d 220 / 27.3 \u003d 8.06 ohms.
Figure 1 Wire heater in a single-phase current network
The desired values of the wire diameter and its length will be determined in paragraph 5 of this paragraph.
With this type of connection, the load is distributed evenly over three phases, i.e. 6 / 3 = 2 kW per phase. So we need 3 heaters. Next, you need to choose the method of connecting the heaters (load) directly. There can be 2 ways: “STAR” or “TRIANGLE”.
It is worth noting that in this article the formulas for calculating the current strength ( I ) and resistance ( R ) for a three-phase network are not written in the classical form. This is done in order not to complicate the presentation of the material on the calculation of heaters with electrical terms and definitions (for example, phase and linear voltages and currents and the relationship between them are not mentioned). The classical approach and formulas for calculating three-phase circuits can be found in the specialized literature. In this article, some mathematical transformations carried out on classical formulas are hidden from the reader, and this does not have any effect on the final result.
When connecting type “STAR” the heater is connected between phase and zero (see Fig. 2). Accordingly, the voltage at the ends of the heater will be U
= 220 V.
I=P/U
\u003d 2000 / 220 \u003d 9.10 A.
R=U/I
= 220 / 9.10 = 24.2 ohms.
Figure 2 Wire heater in a three-phase current network. Connection according to the "STAR" scheme
When connecting type “TRIANGLE” the heater is connected between two phases (see fig. 3). Accordingly, the voltage at the ends of the heater will be U
= 380 V.
The current passing through the heater is
I=P/U
\u003d 2000 / 380 \u003d 5.26 A.
Resistance of one heater -
R=U/I
\u003d 380 / 5.26 \u003d 72.2 ohms.
Figure 3 Wire heater in a three-phase current network. Connection according to the scheme "TRIANGLE"
4. After determining the resistance of the heater with an appropriate connection to the electrical network choose the diameter and length of the wire.
When determining the above parameters, it is necessary to analyze specific surface power of the heater, i.e. power dissipated per unit area. The surface power of the heater depends on the temperature of the heated material and on the design of the heaters.
Example
From the previous calculation points (see paragraph 3 of this paragraph), we know the resistance of the heater. For a 60 liter oven with a single-phase connection, it is R
= 8.06 ohms. As an example, take a diameter of 1 mm. Then, in order to obtain the required resistance, it is necessary l = R / p
\u003d 8.06 / 1.4 \u003d 5.7 m of nichrome wire, where ρ
- the nominal value of the electrical resistance of 1 m of the wire in [Ohm / m]. The mass of this piece of nichrome wire will be m = l μ
\u003d 5.7 0.007 \u003d 0.0399 kg \u003d 40 g, where μ
- weight of 1 m of wire. Now it is necessary to determine the surface area of a piece of wire 5.7 m long. S = l π d
\u003d 570 3.14 0.1 \u003d 179 cm 2, where l
– wire length [cm], d
– wire diameter [cm]. Thus, 6 kW should be allocated from an area of 179 cm 2. Solving a simple proportion, we get that power is released from 1 cm 2 β=P/S
\u003d 6000 / 179 \u003d 33.5 W, where β
- surface power of the heater.
The resulting surface power is too high. Heater will melt if it is heated to a temperature that would provide the obtained value of surface power. This temperature will be higher than the melting point of the heater material.
The example given is a demonstration of the incorrect choice of the wire diameter that will be used to make the heater. In paragraph 5 of this paragraph, an example will be given with the correct selection of the diameter.
For each material, depending on the required heating temperature, the permissible value of the surface power is determined. It can be determined using special tables or graphs. Tables are used in these calculations.
For high temperature furnaces(at a temperature of more than 700 - 800 ° C) the allowable surface power, W / m 2, is equal to β add \u003d β eff α , Where β eff - surface power of heaters depending on the temperature of the heat-receiving medium [W / m 2 ], α is the radiation efficiency factor. β eff is selected according to table 3, α - according to table 4.
If low temperature oven(temperature less than 200 - 300 ° C), then the allowable surface power can be considered equal to (4 - 6) · 10 4 W / m 2.
Table 3
| Heat-receiving surface temperature, °C | β eff, W/cm 2 at heater temperature, °C | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 | 1300 | 1350 | ||
| 100 | 6,1 | 7,3 | 8,7 | 10,3 | 12,5 | 14,15 | 16,4 | 19,0 | 21,8 | 24,9 | 28,4 | 36,3 | |
| 200 | 5,9 | 7,15 | 8,55 | 10,15 | 12,0 | 14,0 | 16,25 | 18,85 | 21,65 | 24,75 | 28,2 | 36,1 | |
| 300 | 5,65 | 6,85 | 8,3 | 9,9 | 11,7 | 13,75 | 16,0 | 18,6 | 21,35 | 24,5 | 27,9 | 35,8 | |
| 400 | 5,2 | 6,45 | 7,85 | 9,45 | 11,25 | 13,3 | 15,55 | 18,1 | 20,9 | 24,0 | 27,45 | 35,4 | |
| 500 | 4,5 | 5,7 | 7,15 | 8,8 | 10,55 | 12,6 | 14,85 | 17,4 | 20,2 | 23,3 | 26,8 | 34,6 | |
| 600 | 3,5 | 4,7 | 6,1 | 7,7 | 9,5 | 11,5 | 13,8 | 16,4 | 19,3 | 22,3 | 25,7 | 33,7 | |
| 700 | 2 | 3,2 | 4,6 | 6,25 | 8,05 | 10,0 | 12,4 | 14,9 | 17,7 | 20,8 | 24,3 | 32,2 | |
| 800 | - | 1,25 | 2,65 | 4,2 | 6,05 | 8,1 | 10,4 | 12,9 | 15,7 | 18,8 | 22,3 | 30,2 | |
| 850 | - | - | 1,4 | 3,0 | 4,8 | 6,85 | 9,1 | 11,7 | 14,5 | 17,6 | 21,0 | 29,0 | |
| 900 | - | - | - | 1,55 | 3,4 | 5,45 | 7,75 | 10,3 | 13 | 16,2 | 19,6 | 27,6 | |
| 950 | - | - | - | - | 1,8 | 3,85 | 6,15 | 8,65 | 11,5 | 14,5 | 18,1 | 26,0 | |
| 1000 | - | - | - | - | - | 2,05 | 4,3 | 6,85 | 9,7 | 12,75 | 16,25 | 24,2 | |
| 1050 | - | - | - | - | - | - | 2,3 | 4,8 | 7,65 | 10,75 | 14,25 | 22,2 | |
| 1100 | - | - | - | - | - | - | - | 2,55 | 5,35 | 8,5 | 12,0 | 19,8 | |
| 1150 | - | - | - | - | - | - | - | - | 2,85 | 5,95 | 9,4 | 17,55 | |
| 1200 | - | - | - | - | - | - | - | - | - | 3,15 | 6,55 | 14,55 | |
| 1300 | - | - | - | - | - | - | - | - | - | - | - | 7,95 | |
Table 4
Wire spirals, half-closed in the grooves of the lining
Wire spirals on shelves in tubes
Wire zigzag (rod) heaters
Let us assume that the temperature of the heater is 1000 °C, and we want to heat the workpiece to a temperature of 700 °C. Then, according to table 3, we select β eff \u003d 8.05 W / cm 2, α = 0,2, β add \u003d β eff α \u003d 8.05 0.2 \u003d 1.61 W / cm 2 \u003d 1.61 10 4 W / m 2.
5. After determining the permissible surface power of the heater, it is necessary find its diameter(for wire heaters) or width and thickness(for tape heaters), as well as length.
The wire diameter can be determined using the following formula: d
- wire diameter, [m]; P
- heater power, [W]; U
- voltage at the ends of the heater, [V]; β add
- allowable surface power of the heater, [W/m 2 ]; ρt
- resistivity of the heater material at a given temperature, [Ohm m].
ρ t = ρ 20 k
, Where ρ 20
- electrical resistivity of the heater material at 20 °C, [Ohm m] k
- correction factor for calculating the change in electrical resistance depending on temperature (by ).
The length of the wire can be determined by the following formula:
l
- wire length, [m].
We select the diameter and length of the wire from nichrome Х20Н80. The specific electrical resistance of the heater material is
ρ t = ρ 20 k
\u003d 1.13 10 -6 1.025 \u003d 1.15 10 -6 Ohm m.
Household single-phase current network
For a 60 liter stove connected to a household single-phase current network, it is known from the previous calculation steps that the power of the stove is P
\u003d 6000 W, voltage at the ends of the heater - U
= 220 V, permissible surface heater power β add
\u003d 1.6 10 4 W / m 2. Then we get
The resulting size must be rounded up to the nearest larger standard. Standard sizes for nichrome and fechral wire can be found in. Appendix 2, Table 8. In this case, the nearest large standard size is Ø 2.8 mm. Heater diameter d = 2.8 mm.
Heater length l = 43 m.
It is also sometimes required to determine the mass of the required amount of wire.
m = l μ
, Where m
- mass of a piece of wire, [kg]; l
- wire length, [m]; μ
- specific gravity (mass of 1 meter of wire), [kg/m].
In our case, the mass of the heater m = l μ \u003d 43 0.052 \u003d 2.3 kg.
This calculation gives the minimum wire diameter at which it can be used as a heater under given conditions.. From the point of view of material savings, such a calculation is optimal. In this case, wire of a larger diameter can also be used, but then its quantity will increase.
Examination
Calculation results can be checked in the following way. A wire diameter of 2.8 mm was obtained. Then the length we need is
l = R / (ρ k)
\u003d 8.06 / (0.179 1.025) \u003d 43 m, where l
- wire length, [m]; R
- heater resistance, [Ohm]; ρ
- nominal value of electrical resistance of 1 m of wire, [Ohm/m]; k
- correction factor for calculating the change in electrical resistance depending on temperature.
This value is the same as the value obtained from another calculation.
Now it is necessary to check whether the surface power of the heater we have chosen will not exceed the allowable surface power, which was found in step 4. β=P/S \u003d 6000 / (3.14 4300 0.28) \u003d 1.59 W / cm 2. Received value β \u003d 1.59 W / cm 2 does not exceed β add \u003d 1.6 W / cm 2.
Results
Thus, the heater will require 43 meters of X20H80 nichrome wire with a diameter of 2.8 mm, which is 2.3 kg.
Industrial three-phase current network
You can also find the diameter and length of the wire required for the manufacture of furnace heaters connected to a three-phase current network.
As described in point 3, each of the three heaters has 2 kW of power. Find the diameter, length and mass of one heater.
STAR connection(see fig. 2)
In this case, the nearest larger standard size is Ø 1.4 mm. Heater diameter d = 1.4 mm.
Length of one heater l
= 30 m.
Weight of one heater m = l μ
\u003d 30 0.013 \u003d 0.39 kg.
Examination
A wire diameter of 1.4 mm was obtained. Then the length we need is
l = R / (ρ k)
\u003d 24.2 / (0.714 1.025) \u003d 33 m.
β=P/S \u003d 2000 / (3.14 3000 0.14) \u003d 1.52 W / cm 2, it does not exceed the permissible.
Results
For three heaters connected according to the “STAR” scheme, you will need
l
\u003d 3 30 \u003d 90 m of wire, which is
m
\u003d 3 0.39 \u003d 1.2 kg.
Connection type “TRIANGLE”(see fig. 3)
In this case, the nearest larger standard size is Ø 0.95 mm. Heater diameter d = 0.95 mm.
Length of one heater l
= 43 m.
Weight of one heater m = l μ
\u003d 43 0.006 \u003d 0.258 kg.
Examination
A wire diameter of 0.95 mm was obtained. Then the length we need is
l = R / (ρ k)
\u003d 72.2 / (1.55 1.025) \u003d 45 m.
This value almost coincides with the value obtained as a result of another calculation.
Surface power will be β=P/S \u003d 2000 / (3.14 4300 0.095) \u003d 1.56 W / cm 2, it does not exceed the permissible.
Results
For three heaters connected according to the “TRIANGLE” scheme, you will need
l
\u003d 3 43 \u003d 129 m of wire, which is
m
\u003d 3 0.258 \u003d 0.8 kg.
If we compare the 2 options discussed above for connecting heaters to a three-phase current network, we can see that “STAR” requires a larger diameter wire than “TRIANGLE” (1.4 mm vs. 0.95 mm) in order to achieve a given furnace power of 6 kW. Wherein the required length of the nichrome wire when connected according to the “STAR” scheme is less than the length of the wire when connecting the “TRIANGLE” type(90 m vs. 129 m), and the required mass, on the contrary, is more (1.2 kg vs. 0.8 kg).
Spiral calculation
During operation, the main task is to place the heater of the estimated length in the limited space of the furnace. Nichrome and fechral wire are wound in the form of spirals or bent in the form of zigzags, the tape is bent in the form of zigzags, which makes it possible to accommodate large quantity material (lengthwise) into the working chamber. The most common option is the spiral.The ratios between the pitch of the spiral and its diameter and the diameter of the wire are chosen in such a way as to facilitate the placement of heaters in the furnace, ensure their sufficient rigidity, to the maximum extent possible exclude local overheating of the turns of the spiral itself and at the same time not hinder the heat transfer from them to the products.
The larger the diameter of the spiral and the smaller its pitch, the easier it is to place heaters in the furnace, but with an increase in diameter, the strength of the spiral decreases, and the tendency of its turns to lie on top of each other increases. On the other hand, with an increase in the frequency of winding, the shielding effect of the part of its turns facing the products on the rest increases and, consequently, the use of its surface deteriorates, and local overheating may also occur.
Practice has established well-defined, recommended ratios between the wire diameter ( d ), step ( t ) and the diameter of the spiral ( D ) for wire Ø 3 to 7 mm. These ratios are as follows: t ≥ 2d And D = (7÷10) d for nichrome and D = (4÷6) d - for less durable iron-chromium-aluminum alloys, such as fechral, etc. For thinner wires, the ratio D And d , and t usually take more.
Conclusion
The article discussed various aspects related to calculation of electric furnace heaters- materials, calculation examples with the necessary reference data, references to standards, illustrations.In the examples, methods for calculating only wire heaters. In addition to wire from precision alloys, tape can also be used for the manufacture of heaters.
The calculation of heaters is not limited to the choice of their sizes. Also it is necessary to determine the material from which the heater should be made, the type of heater (wire or tape), the type of location of the heaters and other features. If the heater is made in the form of a spiral, then it is necessary to determine the number of turns and the pitch between them.
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Bibliography
- Dyakov V.I. "Typical calculations for electrical equipment".
- Zhukov L.L., Plemyannikova I.M., Mironova M.N., Barkaya D.S., Shumkov Yu.V. "Alloys for heaters".
- Sokunov B.A., Grobova L.S. "Electrothermal installations (electric resistance furnaces)".
- Feldman I.A., Gutman M.B., Rubin G.K., Shadrich N.I. "Calculation and design of heaters for electric resistance furnaces".
- http://www.horss.ru/h6.php?p=45
- http://www.electromonter.info/advice/nichrom.html
