Calculation of a nichrome spiral. Ready for you to make a nichrome spiral. Method for calculating a nichrome wire heater How to calculate a nichrome spiral
A nichrome coil is a heating element in the form of a wire coiled with a screw for compact placement. The wire is made from nichrome- precision alloy, the main components of which 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 nichrome brands - Х20Н80 and Х15Н60. 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. The content of nickel and chromium in it is reduced - up to 61% and up to 18%, respectively. But the amount of iron is increased - 17-29% versus 1.5 in X20H80.
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 its high resistance to electric current. It defines the scope of the alloy. Nichrome spiral It is used in two qualities - as a heating element or as a material for the electrical resistance of electrical circuits.
Used for heaters electric spiral from Kh20N80-N and Kh15N60-N alloys. 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, used in industrial equipment increased reliability.
Nichrome spiral grades Х15Н60, Х20Н80, Kh20N80-VI differs in that its electrical resistance changes little with temperature. Resistors, connectors are made from it electronic circuits, critical parts of vacuum devices.
How to wind a spiral from nichrome
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.
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 nichrome wire for heating element tiles with power P = 600 W at Unetwork = 220 V.
1) I = P/U = 600/220 = 2.72 A
2) R \u003d U / I \u003d 220 / 2.72 \u003d 81 ohms
3) Based on 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 high quality coil 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 resistivity 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 produce 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 its high resistance to 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 of 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.
Closely spaced coils produce more heat, bigger step 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.
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 sketches and specifications customer.
partalstalina.ru
Calculation and repair of the heating winding of the soldering iron
When repairing or self-manufacturing an electric soldering iron or any other heating device, you have to wind a heating winding made of 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 supply 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 Kh20N80. 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.
Symbols 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, alloy grade. 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 specific resistance 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 length of the wire: L = RS / ρ = 42.3 0.196 / 1.11 = 7.47 m. We determine the power of the heating element: P = U I = 220 5.2 = 1.15 kW .When winding the spiral, the following ratio 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 closed version heater 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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Calculation of the wire heater of an electric furnace.
This article reveals the biggest secrets of the design of electric furnaces - the secrets of heater calculations.
How are the volume, power and rate of heating of the furnace related.
As mentioned elsewhere, there are no conventional ovens. In the same way, there are no ovens for firing faience or toys, red clay or beads. There is just an oven (and here we are talking exclusively about electric ovens) with a certain volume usable space made of some refractories. In this kiln, you can put one large or small vase for firing, or you can put a whole stack of plates on which thick fireclay tiles will lie. You need to fire a vase or tiles, maybe at 1000 o C, or maybe at 1300 o C. For many industrial or household reasons, firing should take 5-6 hours or 10-12.
No one knows what you need from an oven better than yourself. Therefore, before proceeding with the calculation, you need to clarify all these questions for yourself. If there is already a furnace, but it is necessary to install heaters in it or change old ones for new ones, there is no need for designing. If the oven is being built from scratch, you need to start by finding out the dimensions of the chamber, that is, from the length, depth, width.
Let's assume you already know these values. Suppose you need a chamber with a height of 490 mm, width and depth of 350 mm. Further in the text, we will call a furnace with such a chamber a 60-liter one. At the same time, we will design a second furnace, larger, with a height H=800 mm, a width D=500 mm and a depth L=500 mm. We will call this oven a 200-liter oven.
Furnace volume in liters = H x D x L,
where H, D, L are expressed in decimeters.
If you correctly converted millimeters to decimeters, the volume of the first furnace should be 60 liters, the volume of the second - really 200! Do not think that the author is being sarcastic: the most common errors in calculations are errors in dimensions!
We proceed to the next question - what are the walls of the furnace made of. Almost all modern furnaces are made of light refractories with low thermal conductivity and low heat capacity. Very old stoves are made from heavy chamotte. Such furnaces are easily recognizable by the massive lining, the thickness of which is almost equal to the width of the chamber. If you have this case, you are out of luck: during firing, 99% of the energy will be spent on heating the walls, not the products. We assume that the walls are made of modern materials (MKRL-08, ShVP-350). Then only 50-80% of the energy will be spent on heating the walls.
Loading mass remains very uncertain. Although it is generally less than the mass of refractory walls (plus bottom and roof) of the furnace, this mass will certainly contribute to the rate of heating.
Now about power. Power is how much heat the heater releases in 1 second. The unit of power is watts (abbreviated W). A bright incandescent bulb is 100 W, an electric kettle is 1000 W, or 1 kilowatt (abbreviated 1 kW). If you turn on a heater with a power of 1 kW, it will release heat every second, which, according to the law of conservation of energy, will go to heat the walls, products, fly away with air through the cracks. Theoretically, if there are no losses through the slots and walls, 1 kW is able to heat anything up to an infinite temperature in an infinite time. In practice, real (approximate average) heat losses are known for furnaces, so there is the following rule-recommendation:
For a normal oven heating rate of 10-50 liters, power is needed
100 watts per liter of volume.
For a normal heating rate of the furnace of 100-500 liters, power is needed
50-70 W for each liter of volume.
The value of specific power must be determined not only taking into account the volume of the furnace, but also taking into account the massiveness of the lining and load. The larger the load mass, the larger the value to be chosen. Otherwise, the oven will heat up, but for a longer time. Let's choose for our 60 liter specific power 100 W/l, and for the 200 liter - 60 W/l. Accordingly, we obtain that the power of the heaters of a 60-liter bottle should be 60 x 100 = 6000 W = 6 kW, and a 200-liter bottle should be 200 x 60 = 12000 W = 12 kW. See how interesting it is: the volume has increased by more than 3 times, and the power - only by 2. Why? (Question for independent work).
It happens that there is no 6 kW outlet in the apartment, but there is only 4. But you need exactly a 60-liter! Well, you can calculate the heater at 4 kilowatts, but come to terms with the fact that the heating stage during firing will last 10-12 hours. It happens that, on the contrary, heating is necessary for 5-6 hours of a very massive load. Then you will have to invest 8 kW in a 60-liter furnace and not pay attention to red-hot wiring ... For further reasoning, we will restrict ourselves to classical powers - 6 and 12 kW, respectively.
Power, amps, volts, phases.
Knowing the power, we know the need for heat for heating. According to the inexorable law of conservation of energy, we must take the same power from the electrical network. Recall the formula:
Heater power (W) = Heater voltage (V) x Current (A)
or P = U x I
There are two catches in this formula. First: the voltage must be taken at the ends of the heater, and not at all in the outlet. Voltage is measured in volts (abbreviated V). Second: we mean the current that flows through this heater, and not through the machine at all. Current is measured in amperes (abbreviated as A).
We are always given the voltage in the network. If the substation is operating normally and it is not rush hour, the voltage in a regular household outlet will be 220 V. The voltage in an industrial three-phase network between any phase and neutral wire is also equal to 220V, and the voltage between any two phases- 380 V. Thus, in the case of a household, single-phase network, we have no choice in voltage - only 220 V. In the case of a three-phase network, there is a choice, but a small one - either 220 or 380 V. But what about amperes? They will be obtained automatically from the voltage and resistance of the heater according to the great Ohm's law:
Ohm's law for a section of an electrical circuit:
Current (A) \u003d Voltage in the section (V) / Resistance of the section (Ohm)
or I=U/R
In order to get 6 kW from a single-phase network, you need a current I=P/U= 6000/220 = 27.3 amps. This is a large, but the real current of a good household network. For example, such a current flows in an electric stove, in which all burners are turned on at full power and the oven too. To get 12 kW in a single-phase network for a 200-liter, you need twice as much current - 12000/220 = 54.5 amperes! This is unacceptable for any home network. It is better to use three phases, i.e. distribute power to three lines. 12000/3/220 = 18.2 amps will flow in each phase.
Let's take a look at the last calculation. At the moment we DO NOT KNOW what heaters will be in the furnace, we DO NOT KNOW what voltage (220 or 380 V) will be applied to the heaters. But we KNOW for sure that 12 kW must be taken from the three-phase network, the load should be distributed evenly, i.e. 4 kW in each phase of our network, i.e. for each phase wire the input (general) automaton of the furnace will flow 18.2A, and it is not at all necessary that such a current will flow through the heater. By the way, 18.2 A will also pass through the electricity meter. (And by the way: there will be no current through the neutral wire due to the features of a three-phase power supply. These features are ignored here, since we are only interested in the thermal work of the current). If you have any questions at this point in the presentation, read it again. And think: if 12 kilowatts are released in the volume of the furnace, then according to the law of conservation of energy, the same 12 kilowatts pass through three phases, each - 4 kW ...
Let's return to the single-phase 60-liter stove. It is easy to find that the resistance of the furnace heater should be R=U/I\u003d 220 V / 27.3 A \u003d 8.06 Ohm. Therefore, in the very general view the wiring diagram of the furnace will look like this:
A heater with a resistance of 8.06 ohms should carry a current of 27.3 A
A three-phase oven will require three identical heating circuits: in the figure - the most common electrical circuit of a 200-liter.
The power of a 200-liter oven must be evenly distributed over 3 circuits - A, B and C.
But each heater can be turned on either between phase and zero, or between two phases. In the first case, there will be 220 volts at the ends of each heating circuit, and its resistance will be R=U/I\u003d 220 V / 18.2 A \u003d 12.08 Ohm. In the second case, there will be 380 volts at the ends of each heating circuit. To obtain a power of 4 kW, it is necessary that the current be I=P/U= 4000/380 = 10.5 amps, i.e. resistance should be R=U/I\u003d 380 V / 10.5 A \u003d 36.19 Ohm. These connection options are called "star" and "delta". As can be seen from the values of the required resistance, it will not work to simply change the power circuit from a star (heaters of 12.08 Ohm) to a triangle (heaters of 36.19 Ohm) - in each case, you need your own heaters.
In the "star" scheme, each heating circuit
switched between phase and zero for a voltage of 220 volts. A current of 18.2 A flows through each heater with a resistance of 12.08 Ohm. No current flows through the N wire.
In the "delta" scheme, each heating circuit
connected between two phases for a voltage of 380 volts. For each heater with a resistance of 36.19 Ohm, a current of 10.5 A flows. A current of 18.2 A flows through the wire connecting point A1 to the automatic power supply (point A), so that 380 x 10.5 \u003d 220 x 18.2 \u003d 4 kilowatt! Similarly with lines B1 - B and C1 - C.
Homework. There was a star in the 200-liter bottle. The resistance of each circuit is 12.08 ohms. What will be the power of the furnace if these heaters are connected to a triangle?
Limit loads of wire heaters (Kh23Yu5T).
Complete victory! We know the resistance of the heater! It remains just to unwind a piece of wire of the desired length. Let's not get tired of calculations with resistivity - everything has long been calculated with sufficient accuracy for practical needs.
| Diameter, mm | Meters to 1 kg | Resistance 1 meter, Ohm |
| 1,5 | 72 | 0.815 |
| 2,0 | 40 | 0.459 |
| 2,5 | 25 | 0.294 |
| 3,0 | 18 | 0.204 |
| 3,5 | 13 | 0.150 |
| 4,0 | 10 | 0.115 |
For a 60-liter oven, you need 8.06 ohms, we choose a 1.5 and we get that the required resistance will give only 10 meters of wire, which will weigh only 140 grams! Amazing result! Let's check again: 10 meters of wire with a diameter of 1.5 mm have a resistance of 10 x 0.815 = 8.15 ohms. The current at 220 volts will be 220/8.15 = 27 amps. The power will be 220 x 27 = 5940 watts = 5.9 kW. We wanted 6 kW. They didn’t make a mistake anywhere, the only thing that is alarming is that there are no such furnaces ...
A lone red-hot heater in a 60-liter oven.
The heater is very small. Such a feeling is created when considering the above picture. But we are engaged in calculations, not philosophy, so we will move on from sensations to numbers. The numbers say the following: 10 linear meters of wire with a diameter of 1.5 mm have an area S = L x d x pi = 1000 x 0.15 x 3.14 = 471 sq. cm. From this area (and where else?) 5.9 kW is radiated into the volume of the furnace, i.e. per 1 sq. cm area accounts for the radiated power of 12.5 watts. Omitting details, we will point out that the heater must be heated to an enormous temperature before the temperature in the furnace rises significantly.
The overheating of the heater is determined by the value of the so-called surface load p, which we calculated above. In practice, there are limit values for each type of heater p depending on the heater material, diameter and temperature. With a good approximation for wire from the domestic alloy X23Yu5T of any diameter (1.5-4 mm), you can use the value of 1.4-1.6 W / cm 2 for a temperature of 1200-1250 o C.
Physically, the overheating can be associated with the temperature difference on the surface of the wire and inside it. Heat is released throughout the volume, so the higher the surface load, the more these temperatures will differ. When the surface temperature is close to the limiting operating temperature, the temperature in the core of the wire may approach the melting point.
If the furnace is designed for low temperatures, a larger surface load can be chosen, for example, 2 - 2.5 W / cm 2 for 1000 o C. Here you can make a sad remark: real kanthal (this is an original alloy, the analogue of which is the Russian fechral X23Yu5T) allows p up to 2.5 at 1250 o C. This kanthal is made by the Swedish company Kanthal.
Let's go back to our 60-liter tank and choose a thicker wire from the table - a deuce. It is clear that the deuces will have to take 8.06 Ohm / 0.459 Ohm / m = 17.6 meters, and they will already weigh 440 grams. We consider the surface load: p\u003d 6000 W / (1760 x 0.2 x 3.14) cm 2 \u003d 5.43 W / cm 2. A lot of. For a wire with a diameter of 2.5 mm, you get 27.5 meters and p= 2.78. For the troika - 39 meters, 2.2 kilograms and p= 1.66. Finally.
Now we will have to wind 39 meters of the troika (if it bursts, start winding again). But you can use TWO heaters connected in parallel. Naturally, the resistance of each should no longer be 8.06 Ohm, but twice as much. Therefore, for a deuce, you get two heaters of 17.6 x 2 = 35.2 m, each will have 3 kW of power, and the surface load will be 3000 W / (3520 x 0.2 x 3.14) cm 2 = 1, 36 W/cm2. And the weight is 1.7 kg. Saved half a kilo. We got a lot of turns in total, which can be evenly distributed over all the walls of the furnace.
Well distributed heaters in a 60 liter oven.
| Diameter, mm | Current limit for p\u003d 2 W / cm 2 at 1000 o C | Current limit for p\u003d 1.6 W / cm 2 at 1200 o C |
| 1,5 | 10,8 | 9,6 |
| 2,0 | 16,5 | 14,8 |
| 2,5 | 23,4 | 20,7 |
| 3,0 | 30,8 | 27,3 |
| 3,5 | 38,5 | 34,3 |
| 4,0 | 46,8 | 41,9 |
Calculation example for a 200 liter oven.
Now that the basic principles are known, we will show how they are used in the calculation of a real 200 liter oven. All stages of the calculation, of course, can be formalized and written into a simple program that will do almost everything itself.
Let's draw our furnace "in sweep". We seem to be looking at it from above, in the center - under, on the sides of the wall. We calculate the areas of all the walls, so that later, in proportion to the area, organize the heat supply.
"Scan" 200-liter oven.
We already know that when connected in a star, a current of 18.2A must flow in each phase. From the above table on current limits, it follows that for a wire with a diameter of 2.5 mm, you can use one heating element (limit current 20.7A), and for a wire of 2.0 mm, you need to use two elements connected in parallel (because the limit current is only 14.8A), in total there will be 3 x 2 = 6 in the furnace.
According to Ohm's law, we calculate the required resistance of the heaters. For wire diameter 2.5 mm R\u003d 220 / 18.2 \u003d 12.09 ohms, or 12.09 / 0.294 \u003d 41.1 meters. It will take 3 such heaters, approximately 480 turns each, if wound on a 25 mm mandrel. The total weight of the wire will be (41.1 x 3) / 25 = 4.9 kg.
For a 2.0 mm wire, there are two parallel elements in each phase, so the resistance of each should be twice as much - 24.18 Ohm. The length of each will be 24.18 / 0.459 = 52.7 meters. Each element will have 610 turns with the same winding. The total weight of all 6 heating elements (52.7 x 6) / 40 = 7.9 kg.
Nothing prevents us from dividing any spiral into several pieces, which are then connected in series. What for? First, for ease of installation. Secondly, if a quarter of the heater fails, only that quarter will need to be replaced. In the same way, no one bothers to put a whole spiral into the oven. Then the door will require a separate spiral, and we, in the case of a diameter of 2.5 mm, have only three of them ...
We put one phase of 2.5 mm wire. The heater was divided into 8 independent short coils, all connected in series.
When we put all three phases in the same way (see the figure below), the following becomes clear. We forgot about the pod! And it occupies 13.5% of the area. In addition, the spirals are in dangerous electrical proximity to each other. Especially dangerous is the proximity of the spirals on the left wall, where there is a voltage of 220 volts between them (phase - zero - phase - zero ...). If, due to something, the neighboring spirals of the left wall touch each other, a large short circuit cannot be avoided. We offer to independently optimize the location and connection of the spirals.
All phases are set.
In case we decide to use a deuce, the diagram is shown below. Each element 52.7 meters long is divided into 4 consecutive spirals of 610 / 4 = 152 turns (winding on a 25 mm mandrel).
Option for the location of the heaters in the case of wire 2.0 mm.
Features of winding, installation, operation.
The wire is convenient in that it can be wound into a spiral, and then the spiral can be stretched as it is convenient. It is believed that the winding diameter should be more than 6-8 wire diameters. Optimal step between turns is 2-2.5 wire diameters. But it is necessary to wind the coil to coil: stretching the spiral is very easy, compressing it is much more difficult.
Thick wire may break during winding. It is especially disappointing if 5 out of 200 turns are left to be wound. It is ideal to wind on a lathe at a very slow speed of rotation of the mandrel. The Kh23Yu5T alloy is produced tempered and untempered. The latter bursts especially often, so if you have a choice, be sure to purchase wire released for winding.
How many turns are needed? Despite the simplicity of the question, the answer is not obvious. First, the diameter of the mandrel and, consequently, the diameter of one turn is not exactly known. Secondly, it is known for sure that the diameter of the wire varies slightly along the length, so the resistance of the spiral will also vary. Thirdly, the resistivity of an alloy of a particular melting may differ from the reference one. In practice, a spiral is wound 5-10 turns more than calculated, then its resistance is measured - with a VERY ACCURATE device that can be trusted, and not with a soap dish. In particular, you need to make sure that with short-circuited probes, the device shows zero, or a number of the order of 0.02 Ohm, which will need to be subtracted from the measured value. When measuring resistance, the spiral is slightly stretched to eliminate the influence of interturn short circuits. Extra coils bite off.
It is best to place the spiral in the furnace on a mullite-silica tube (MKR). For a winding diameter of 25 mm, a tube with an outer diameter of 20 mm is suitable, for a winding diameter of 35 mm - 30 - 32 mm.
It is good if the oven is heated evenly from five sides (four walls + under). Significant power must be concentrated on the hearth, for example, 20 -25% of the total calculated power of the furnace. This compensates for the intake of cold air from outside.
Unfortunately, absolute uniformity of heating is still impossible to achieve. You can approach it using ventilation systems with a LOWER air extraction from the furnace.
During the first heating, or even the first two or three heatings, scale forms on the surface of the wire. We must not forget to remove it both from the heaters (with a brush), and from the surface of slabs, bricks, etc. Scale is especially dangerous if the spiral simply lies on the bricks: iron oxides with aluminosilicates at high temperatures (the heater is one millimeter!) Form fusible compounds, due to which the heater can burn out.
You will need
- Spiral, caliper, ruler. It is necessary to know the material of the spiral, the values of current I and voltage U at which the spiral will work, and what material it is made of.
Instruction
Find out what resistance R your coil should have. To do this, use Ohm's law and substitute the value of the current I in the circuit and the voltage U at the ends of the spiral into the formula R = U / I.
Using the reference book, determine the electrical resistivity of the material ρ from which the spiral will be made. ρ must be expressed in Ohm m. If the value of ρ in the reference book is given in Ohm mm² / m, then multiply it by 0.000001. For example: copper resistivity ρ = 0.0175 Ohm mm² / m, when converted to SI we have ρ = 0 .0175 0.000001=0.0000000175 Ohm m.
Find the length of the wire using the formula: Lₒ=R S/ρ.
Measure an arbitrary length l on the spiral with a ruler (for example: l \u003d 10cm \u003d 0.1m). Count the number of turns n coming to this length. Determine the helix pitch H=l/n or measure it with a caliper.
Find how many turns N can be made from a wire of length Lₒ: N= Lₒ/(πD+H).
Find the length of the spiral itself using the formula: L \u003d Lₒ / N.
A spiral scarf is also called a boa scarf, a wave scarf. The main thing here is not at all the type of yarn, not the pattern of knitting and not the colors of the finished product, but the technique of execution and the originality of the model. Spiral scarf embodies festivity, splendor, solemnity. It looks like an elegant lace jabot, an exotic boa, and an ordinary, but very original scarf.
How to knit a spiral scarf with knitting needles
To knit a spiral scarf, dial 24 loops on the knitting needles and knit the 1st row:
- 1 edge loop;
- 11 facial;
- 12 purl loops.
The yarn quality and color for this spiral scarf pattern is up to you.
1st row: first 1 edge loop, then 1 yarn over, then 1 front loop, then 1 yarn over and 8 front loops. Remove one on the right knitting needle as a purl, pull the thread between the knitting needles forward. Return the removed loop to the left knitting needle, pull the thread between the knitting needles back (in this case, the loop will turn out to be a wrapped thread). Turn work and knit 12 purl stitches.
2nd row: First knit 1 edge stitch, then yarn over 1, then knit 3 stitches, knit 1 yarn over and knit 6 stitches. Remove one on the right knitting needle as a purl, pull the thread between the knitting needles forward. Next, return the loop to the left knitting needle, pull the thread between the knitting needles back, then turn the work and knit 12 purl loops.
3rd row: knit 1 edge loop, then knit 2 together, then knit 1, then knit 2 together and knit 4. Slip one on the right needle as a purl, pull the thread between the needles forward, return the loop to the left needle, then pull the thread between the needles back. After that, turn the work and knit 8 purl loops.
Row 4: Knit 1 hem, then knit 3 together, then knit 4, *remove the wrapped st from below and knit together with the next knit, knit 1* (repeat from * to * 3 times). Without turning the work, tie the wrong loops.
In this way, knit the spiral scarf to the desired length in blocks of these 4 rows.
Almost all women face the issue of contraception. One of the reliable and proven methods is the intrauterine device, which is still in demand today.
Types of spirals
Intrauterine devices are made of plastic and come in two varieties: copper-containing (silver)-containing devices and hormone-containing devices. Their size is 3X4 cm. The choice of the method of contraception and the spiral itself takes place at the appointment with the gynecologist. You should not do this on your own. The intrauterine device is installed by a gynecologist during menstruation. It is small in size and resembles the shape of the letter T.
Copper spiral is made of copper wire. Its feature is the ability to act on the uterus in such a way that the egg cannot attach to it. This is facilitated by two copper antennae.
The hormone coil has a container that contains progestin. This hormone prevents the onset of ovulation. In the case of using a hormonal intrauterine device, sperm cannot fertilize the egg. As women note, when using such a spiral, menstruation becomes more scarce and less painful. However, this does not bring harm, because it is associated with the action of hormones that are inside the spiral. Gynecologists recommend that women suffering from painful periods install a hormonal spiral.
Spiral selection
Gynecological intrauterine devices are different brands both domestic and foreign production. In addition, their cost can vary from 250 rubles to several thousand. Many factors influence this.
The Juno Bio spiral is quite popular among Russian women. It attracts, first of all, low cost. However, the low efficiency of this spiral entails a high risk of pregnancy.
The Mirena intrauterine device has proven itself well, but it is one of the most expensive in its series. At the same time, the use of an intrauterine device is considered the cheapest and most affordable type of contraception.
This is a hormonal spiral. Its manufacturers promise that the Mirena spiral is less likely to shift in the uterus or fall out. Namely, this leads to the onset of pregnancy, therefore, patients are advised to regularly check the presence of an intrauterine contraceptive in the right place.
Standard voltage in the household power supply U=220V. The current strength is limited by fuses in the electrical panel and is usually equal to I \u003d 16A.
Sources:
- tables physical quantities, I.K. Kikoin, 1976
- spiral length formula
An electric soldering iron is a hand tool designed to fasten parts together with soft solders by heating the solder to liquid state and filling the gap between the parts to be soldered.
Electric soldering irons are available for 12, 24, 36, 42 and 220 V mains voltage, and there are reasons for this. The main thing is human safety, the second is the mains voltage in place soldering work is done. In production, where all equipment is grounded and there is high humidity, it is allowed to use soldering irons with a voltage of not more than 36 V, while the body of the soldering iron must be grounded. The on-board network of a motorcycle has a DC voltage of 6 V, a car - 12 V, a truck - 24 V. In aviation, a network with a frequency of 400 Hz and a voltage of 27 V is used. There are also design limitations, for example, it is difficult to make a 12 W soldering iron on the supply voltage 220 V, since the spiral will need to be wound from a very thin wire and therefore many layers will be wound, the soldering iron will turn out to be large, not convenient for small work. Since the winding of the soldering iron is wound from nichrome wire, it can be powered with both alternating and constant voltage. The main thing is that the supply voltage matches the voltage for which the soldering iron is designed.
Power electric soldering irons are 12, 20, 40, 60, 100 W and more. And this is not accidental either. In order for the solder to spread well over the surfaces of the soldered parts during soldering, they need to be heated to a temperature slightly higher than the melting point of the solder. Upon contact with the part, heat is transferred from the tip to the part and the temperature of the tip drops. If the diameter of the soldering iron tip is not sufficient or the power of the heating element is low, then having given off heat, the tip will not be able to heat up to the set temperature, and it will be impossible to solder. At best, you get a loose and not strong solder. A more powerful soldering iron can solder small parts, but there is a problem of inaccessibility to the soldering point. How, for example, to solder in printed circuit board a microcircuit with a leg pitch of 1.25 mm with a soldering iron tip of 5 mm in size? True, there is a way out, several turns of copper wire with a diameter of 1 mm are wound onto such a sting and soldered with the end of this wire. But the bulkiness of the soldering iron makes the job almost impossible. There is one more limitation. With high power, the soldering iron will quickly warm up the element, and many radio components do not allow heating above 70 ° C, and therefore, the allowable time for their soldering is no more than 3 seconds. These are diodes, transistors, microcircuits.
Soldering iron device
The soldering iron is a red copper rod that is heated by a nichrome spiral to the melting temperature of the solder. The soldering iron rod is made of copper due to its high thermal conductivity. After all, when soldering, you need to quickly transfer heat to the soldering iron tip from the heating element. The end of the rod has a wedge shape, is the working part of the soldering iron and is called a sting. The rod is inserted into a steel tube wrapped in mica or fiberglass. Mica is wound with nichrome wire, which serves as a heating element.
A layer of mica or asbestos is wound over the nichrome, which serves to reduce heat loss and electrical insulation of the nichrome spiral from the metal body of the soldering iron.
The ends of the nichrome spiral are connected to the copper conductors of an electric cord with a plug at the end. To ensure the reliability of this connection, the ends of the nichrome spiral are bent and folded in half, which reduces heating at the junction with copper wire. In addition, the connection is crimped metal plate, it is best to make a crimp from an aluminum plate, which has a high thermal conductivity and will more effectively remove heat from the junction. For electrical insulation, tubes made of heat-resistant insulating material, fiberglass or mica are put on the junction.
The copper rod and the nichrome spiral are closed by a metal case consisting of two halves or a solid tube, as in the photo. The body of the soldering iron on the tube is fixed with cap rings. To protect a person's hand from burns, a handle made of a material that does not see heat well, wood or heat-resistant plastic is mounted on the tube.
When the soldering iron plug is inserted into the socket, electric current flows to the nichrome heating element, which heats up and transfers heat to the copper rod. The soldering iron is ready for soldering.
Low-power transistors, diodes, resistors, capacitors, microcircuits and thin wires are soldered with a 12 W soldering iron. Soldering irons 40 and 60 W are used for soldering powerful and large radio components, thick wires and small parts. For soldering large parts, for example, gas column heat exchangers, you will need a soldering iron with a power of one hundred or more watts.
As you can see in the drawing, the electrical circuit of the soldering iron is very simple, and consists of only three elements: a plug, a flexible electrical wire and a nichrome spiral.
As can be seen from the diagram, the soldering iron does not have the ability to adjust the tip heating temperature. And even if the power of the soldering iron is chosen correctly, it is still not a fact that the temperature of the tip will be required for soldering, since the length of the tip decreases over time due to its constant refilling, solders also have different melting temperatures. Therefore, to maintain the optimum temperature of the soldering tip, it is necessary to connect it through thyristor power controllers with manual adjustment and automatic maintenance of the set temperature of the soldering tip.
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.
The most significant part of the electrothermal installation is the heating element. The main component of indirect heating devices is a resistor with high resistivity. And one of the priority materials is chromium-nickel alloy. Since the resistance of nichrome wire is high, this material takes the lead as a raw material for various kinds electrothermal installations. The calculation of the heater from nichrome wire is carried out in order to determine the dimensions of the heating element.
Basic concepts
In general, it is necessary to calculate the heating element from nichrome according to four calculations: hydraulic, mechanical, thermal and electrical. But usually calculations are carried out only in two stages: according to thermal and electrical indicators.
Thermal characteristics include:
- thermal insulation;
- coefficient useful action by warmth;
- required heat transfer surface.
The main purpose of calculating nichrome is to determine the geometric dimensions of the heating resistance.
To the electrical parameters of heaters are:
- supply voltage;
- power control method;
- power factor and electrical efficiency.
When choosing a supply voltage for heating devices, preference is given to that which poses a minimum threat to animals and service personnel. Mains voltage in installations Agriculture is 380/200 volts with a current frequency of 50 Hertz. In the case of electrical installations in particularly damp rooms, with increased electrical hazard, the voltage should be reduced. Its value should not exceed 12, 24, 36 volts.
Adjust the temperature and power of the heater can be done in two ways:
- changing voltage;
- change in resistance value.
The most common way to change power is to turn on a certain number of sections of a three-phase installation. In modern heating installations, the power is changed by adjusting the voltage using thyristors.
The calculation for the operating current is based on a tabular relationship that relates the current load on a nichrome conductor, its cross-sectional area and temperature.
The tabular data was compiled for nichrome wire, which was stretched in air without taking into account oscillations and vibrations at a temperature of 20 ° C.
In order to move to real conditions, it is necessary to use correction factors in the calculations.
The calculation of a nichrome spiral should be carried out in stages, using the initial information about the heater: the required power and brand of nichrome.
Power of one section:
P - plant power, W;
m - number of phases, for single-phase m = 1;
n - number of sections in one phase, for installations with a capacity of about 1 kW n = 1.
Operating current of one heater section:
U - mains voltage, for single-phase installations U = 220 V
Estimated wire temperature:
θр = θd/(Km Ks)
θd - allowable operating temperature, selected from table 1 depending on the material, °C.
Table 1- Material options for electric heaters.
Km - installation factor, selected from table 2 depending on the design.
table 2- Mounting factor for some types of heater designs in still air flow.
The role of the installation coefficient is that it makes it possible to take into account the increase in the temperature of the heater in real conditions in comparison with the data of the reference table.
Kc - coefficient environment, is determined from Table 3.
Table 3- Correction factor for some environmental conditions.
The environmental factor corrects for improved heat transfer due to ambient conditions. Therefore, the actual calculation results will differ slightly from the table values.
Diameter d, mm and area cross section S, mm 2 is selected according to the operating current and design temperature from table 4
Table 4- Permissible load on nichrome wire at 20 °C, suspended horizontally in still air.
Wire length of one section:
L \u003d (U f 2 S * 10 -6) / (ρ 20 Rs x10 3)
ρ 20 - resistivity at a temperature of 20 ° C, selected from table 1;
α - temperature coefficient of resistance, determined from the corresponding column in table 1.
Spiral diameter:
D = (6…10) d, mm.
Determine the pitch of the spiral:
h = (2…4) d, mm
The pitch of the helix affects the productivity of the work. At higher values, the heat transfer increases.
Number of turns of the spiral
W = (lx10 3)/ (√h 2 +(πD) 2)
Spiral length:
If the purpose of the wire heater is to increase the temperature of the liquid, the operating current is increased by 1.5 times the calculated value. In the case of calculating a heater with a closed type, it is recommended to reduce the operating current by 1.2 times.
Classification of heaters by temperature
Heaters according to the maximum allowable temperature are divided into five classes:
Troubleshooting Options
The greatest probability of failure of electric heaters due to oxidation of the surface of the heating resistance.
Factors that affect the rate of destruction of the heater:
Due to the fact that electric heating installations operate in excess of the permissible values of these parameters, the most frequent breakdowns occur: burning of contacts, violation of the mechanical strength of nichrome wire.
Repair of a heating element made of nichrome is carried out by soldering or twisting.