Calculation of a wire heater for an electric furnace.
This article reveals the biggest secrets in the design of electric furnaces - the secrets of heater calculations.
How are the volume, power and heating rate of a furnace related?
As stated elsewhere, there are no ordinary ovens. Likewise, there are no kilns for firing earthenware or toys, red clay or beads. Sometimes it’s just a stove (and here we’re talking exclusively about electric ovens) with some amount of usable space, made of some refractories. In this kiln you can place one large or small vase for firing, or you can put a whole stack of slabs on which thick fireclay tiles will lie. It is necessary to fire a vase or tiles, maybe at 1000 o C, and maybe at 1300 o C. For many production or household reasons, firing should take place in 5-6 hours or 10-12.
No one knows what you need from a stove better than you. Therefore, before you begin the calculation, you need to clarify all these issues for yourself. If you already have a stove, but you need to install heaters in it or replace old ones with new ones, there is no need for construction. If a furnace is being built from scratch, you need to start by finding out the dimensions of the chamber, that is, length, depth, width.
Let's assume you already know these values. Let's assume that you need a camera with a height of 490 mm, a width and a depth of 350 mm. Further in the text we will call an oven with such a chamber a 60-liter oven. At the same time, we will design a second furnace, larger, with height H=800 mm, width D=500 mm and depth L=500 mm. We will call this oven a 200-liter oven.
Oven 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 oven should be 60 liters, the volume of the second should really be 200! Don’t think that the author is being sarcastic: the most common errors in calculations are errors in dimensions!
Let's get started next question- What are the walls of the furnace made of? Modern furnaces are almost all made of lightweight refractories with low thermal conductivity and low heat capacity. Very old stoves are made of heavy fireclay. Such furnaces are easily recognized by their massive lining, the thickness of which is almost equal to the width of the chamber. If this is your 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 (MCRL-08, ShVP-350). Then only 50-80% of the energy will be spent on heating the walls.
The loading mass remains very uncertain. Although it is, as a rule, less than the mass of the refractories of the walls (plus the hearth and roof) of the furnace, this mass, of course, will contribute to the heating rate.
Now about the power. Power is how much heat the heater produces in 1 second. The unit of measurement for power is the watt (abbreviated W). A bright incandescent light bulb is 100 W, an electric kettle is 1000 W, or 1 kilowatt (abbreviated 1 kW). If you turn on a 1 kW heater, it will release heat every second, which, according to the law of conservation of energy, will go to heat the walls, products, and fly away with the air through the cracks. Theoretically, if there are no losses through cracks and walls, 1 kW is able to heat anything to an infinite temperature in an infinite time. In practice, real (approximate average) heat loss is known for furnaces, so there is the following rule-recommendation:
For a normal heating rate of a 10-50 liter oven, power is needed
100 W per liter of volume.
For a normal heating rate of a furnace of 100-500 liters, power is needed
50-70 W per liter of volume.
The value of the 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 loading. The larger the load mass, the larger the value you need to select. Otherwise, the oven will heat up, but it will take longer. Let’s choose a specific power of 100 W/l for our 60-liter tank, and 60 W/l for our 200-liter tank. Accordingly, we find that the power of the 60-liter heaters should be 60 x 100 = 6000 W = 6 kW, and the 200-liter heater should be 200 x 60 = 12000 W = 12 kW. Look how interesting it is: the volume has increased by more than 3 times, but the power has only increased by 2. Why? (Question for independent work).
It happens that there is no socket for 6 kW in the apartment, but there is only one for 4. But you need a 60-liter one! Well, you can calculate the heater at 4 kilowatts, but accept that the heating stage during firing will last 10-12 hours. It happens that, on the contrary, it is necessary to heat up a very massive load in 5-6 hours. Then you will have to invest 8 kW into a 60-liter stove and not pay attention to the red-hot wiring... For further discussion, we will limit ourselves to classic powers - 6 and 12 kW, respectively.
Power, amperes, volts, phases.
Knowing the power, we know the heat requirement for heating. According to the inexorable law of conservation of energy, we must take the same power from the electrical network. We remind you of the formula:
Heater Power (W) = Heater Voltage (V) x Current (A)
or P = U x I
There are two pitfalls in this formula. First: the voltage should be taken at the ends of the heater, and not generally at the outlet. Voltage is measured in volts (abbreviated as V). Second: we mean the current that flows specifically 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 is normal household outlet will be 220 V. Voltage in an industrial three-phase network between any phase and neutral wire is also 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 small - 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 law of the Great Ohm:
Ohm's law for a section of an electrical circuit:
Current (A) = Section Voltage (V) / Section Resistance (Ohm)
or I = U/R
In order to get 6 kW from a single-phase network, you need current I=P/U= 6000/220 = 27.3 amperes. This is a large, but real current of a good household network. For example, such a current flows in an electric stove with all the burners turned on at full power and the oven too. To get 12 kW in a single-phase network for a 200-liter tank, you will need twice as much current - 12000/220 = 54.5 amperes! This is not acceptable for any household network. It is better to use three phases, i.e. distribute power over three lines. Each phase will flow 12000/3/220 = 18.2 amperes.
Let's pay attention to the last calculation. On at the moment we DO NOT KNOW what kind of heaters will be in the oven, we DO NOT KNOW what voltage (220 or 380 V) will be supplied to the heaters. But we KNOW for sure that 12 kW must be taken from a three-phase network and the load distributed evenly, i.e. 4 kW in each phase of our network, i.e. 18.2A will flow through each phase wire of the input (general) furnace circuit breaker, 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 along the neutral wire due to the features of three-phase power supply. These features are ignored here, since we are only interested in the thermal work of current). If you have questions at this point in the presentation, read everything 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 with 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= 220 V / 27.3 A = 8.06 Ohm. Therefore, in the very general view The electrical circuit of the furnace will look like this:
A heater with a resistance of 8.06 Ohms should carry a current of 27.3 A
For a three-phase oven, three identical heating circuits will be required: the figure shows the most general electrical circuit of a 200-liter oven.
The power of a 200-liter stove must be evenly distributed over 3 circuits - A, B and C.
But each heater can be switched 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= 220 V / 18.2 A = 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, the current must be I=P/U= 4000/380 = 10.5 amperes, i.e. there must be resistance R=U/I= 380 V / 10.5 A = 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 be possible to simply change the power supply circuit from a star (12.08 Ohm heaters) to a triangle (36.19 Ohm heaters) - in each case you need your own heaters.
In a star circuit, each heating circuit
connected between phase and zero at a voltage of 220 Volts. Each heater with a resistance of 12.08 Ohms carries a current of 18.2 A. No current flows through wire N.
In a delta circuit, each heating circuit
connected between two phases at a voltage of 380 Volts. Each heater with a resistance of 36.19 Ohms carries a current of 10.5 A. The wire connecting point A1 to the power supply (point A) carries a current of 18.2 A, so 380 x 10.5 = 220 x 18.2 = 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 turned on in a triangle?
Maximum loads of wire heaters (Х23У5Т).
Complete victory! We know the heater resistance! All that remains is to unwind a piece of wire of the required length. Let's not get tired of calculations with resistivity - everything has long been calculated with sufficient accuracy for practical needs.
| Diameter, mm | Meters per 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 furnace you need 8.06 Ohms, let’s choose one and a half and we find that the required resistance will be provided by 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 has a resistance of 10 x 0.815 = 8.15 Ohms. The current at 220 volts will be 220 / 8.15 = 27 amperes. The power will be 220 x 27 = 5940 Watts = 5.9 kW. We wanted 6 kW. We haven’t made a mistake anywhere, the only alarming thing is that such stoves don’t exist...
A single hot heater in a 60-liter oven.
The heater is very small, or something. This is the feeling one gets when looking at the above picture. But we are engaged in calculations, not philosophy, so let’s move 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 (where else?) 5.9 kW is radiated into the volume of the furnace, i.e. per 1 sq. cm of area the emitted power is 12.5 watts. Omitting the details, we point out that the heater needs to heat up to a huge temperature before the temperature in the furnace increases 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 made from the domestic alloy X23Yu5T of any diameter (1.5-4 mm), you can use a value of 1.4-1.6 W/cm 2 for a temperature of 1200-1250 o C.
Physically, overheating can be associated with the difference in temperature on the surface of the wire and inside it. Heat is released throughout the entire volume, so the higher the surface load, the more these temperatures will differ. When the surface temperature is close to the operating limit temperature, the core temperature of the wire may approach the melting point.
If the furnace is designed for low temperatures, the surface load can be chosen larger, for example, 2 - 2.5 W/cm 2 for 1000 o C. Here we 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 type of kanthal is made by the Swedish company Kanthal.
Let's return to our 60-liter tube and select a thicker wire from the table - two. It is clear that twos will have to take 8.06 Ohm / 0.459 Ohm/m = 17.6 meters, and they will already weigh 440 grams. We calculate the surface load: p= 6000 W / (1760 x 0.2 x 3.14) cm 2 = 5.43 W/cm 2. Many. For a wire with a diameter of 2.5 mm, the result is 27.5 meters and p= 2.78. For three - 39 meters, 2.2 kilograms and p= 1.66. Finally.
Now we will have to wind 39 meters of the three-piece (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 Ohms, but twice as much. Therefore, for a double, you get two heaters of 17.6 x 2 = 35.2 m, each with 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 weight - 1.7 kg. We 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=2 W/cm 2 at 1000 o C | Current limit for p=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 |
An example of calculating a 200 liter stove.
Now that the basic principles are known, we will show how they are used in calculating a real 200 liter stove. 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 oven "in development". We seem to be looking at it from above, in the center - underneath, on the sides of the wall. We will calculate the area of all the walls so that we can then correctly organize the heat supply in proportion to the area.
"Unwrapping" of a 200-liter stove.
We already know that when connected by a star, a current of 18.2A should flow in each phase. From the above table on limiting currents it follows that for a wire with a diameter of 2.5 mm you can use one heating element (maximum current 20.7 A), and for a 2.0 mm wire you need to use two elements connected in parallel (since the maximum current is only 14.8A), in total there will be 3 x 2 = 6 in the oven.
Using Ohm's law, we calculate the required resistance of the heaters. For wire with a diameter of 2.5 mm R= 220 / 18.2 = 12.09 Ohms, or 12.09 / 0.294 = 41.1 meters. You will need 3 of these heaters, approximately 480 turns each, if wound on a 25 mm mandrel. Total weight 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 large - 24.18 Ohms. The length of each will be 24.18 / 0.459 = 52.7 meters. Each element will have 610 turns with the same winding. 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. For what? Firstly, 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 you to put a solid spiral into the oven. Then the door will need a separate spiral, and in the case of a diameter of 2.5 mm, we have only three of them...
We installed one phase of 2.5 mm wire. The heater was divided into 8 independent short spirals, all of them connected in series.
When we put all three phases in a similar way (see figure below), the following becomes clear. We forgot about the under! And it occupies 13.5% of the area. In addition, the spirals are in dangerous electrical proximity to each other. The proximity of the spirals on the left wall, where the voltage between them is 220 Volts (phase - zero - phase - zero...), is especially dangerous. If for some reason the adjacent spirals of the left wall touch each other, a large short circuit cannot be avoided. We suggest that you independently optimize the location and connection of the spirals.
All phases have been installed.
In case we decide to use a two, the diagram is shown below. Each element, 52.7 meters long, is divided into 4 successive spirals of 610 / 4 = 152 turns (winding on a 25 mm mandrel).
Option for the location of heaters in the case of 2.0 mm wire.
Features of winding, installation, operation.
The wire is convenient because it can be wound into a spiral, and the spiral can then be stretched as convenient. It is believed that the coiling diameter should be greater than 6-8 wire diameters. The optimal pitch between turns is 2-2.5 wire diameters. But you have to wind it turn by turn: stretching the spiral is very easy, compressing it is much more difficult.
Thick wire may break during winding. It’s especially frustrating if there are only 5 turns left to wind out of 200. It is ideal to wind on a lathe at a very slow rotation speed of the mandrel. Alloy Kh23Yu5T is produced tempered and untempered. The latter breaks especially often, so if you have a choice, be sure to purchase wire tempered for winding.
How many turns are needed? Despite the simplicity of the question, the answer is not obvious. Firstly, the diameter of the mandrel and, consequently, the diameter of one turn is not known exactly. Secondly, it is known for sure that the diameter of the wire varies slightly along its length, so the resistance of the spiral will also vary. Thirdly, the resistivity of a specific alloy may differ from the reference one. In practice, they wind the spiral 5-10 turns more than calculated, then measure its resistance - with a VERY ACCURATE device that you can trust, 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. The extra turns are bitten off.
It is best to place the spiral in a furnace on a mullite-silica tube (MSR). 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’s good if the stove is heated evenly on five sides (four walls + underneath). Significant power must be concentrated on the hearth, for example, 20 -25% of the total design power of the furnace. This compensates for the suction of cold air from outside.
Unfortunately, absolute heating uniformity still cannot be achieved. You can get closer to it using ventilation systems with BOTTOM air intake 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 remember 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 is simply lying on the bricks: iron oxides with aluminosilicates at high temperatures (the heater is one millimeter away!) 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 current strength I and voltage U at which the spiral will operate, and what material it is made of.
Instructions
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 ρ of the material 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 with a ruler on the spiral (for example: l=10cm=0.1m). Count the number of turns n coming to this length. Determine the spiral pitch H=l/n or measure it with a caliper.
Find how many turns N can be made from wire of length Lₒ: N= Lₒ/(πD+H).
Find the length of the spiral itself using the formula: L=Lₒ/N.
A spiral scarf is also called a boa scarf or a wave scarf. The main thing here is not the type of yarn, not the knitting pattern or the color of the finished product, but the technique of execution and the originality of the model. The spiral scarf represents festivity, splendor, and solemnity. It looks like an elegant lace frill, an exotic boa, and an ordinary, but very original scarf.
How to knit a spiral scarf with knitting needles
To knit a spiral scarf, cast on 24 stitches and knit the 1st row:
- 1 edge loop;
- 11 facial;
- 12 purl loops.
The quality and color of the yarn for this spiral scarf model is at your discretion.
1st row: first 1 edge loop, then 1 yarn over, then 1 knit stitch, then 1 yarn over and 8 knit loops. Slip one onto the right needle as a purl, and pull the yarn forward between the needles. Return the removed loop to the left knitting needle, pull the thread back between the knitting needles (in this case, the loop will turn out to be wrapped in thread). Turn the work and knit 12 purl stitches.
2nd row: first knit 1 edge loop, then 1 yarn over, then knit 3 knit stitches, 1 yarn over and 6 knit stitches. Slip one onto the right needle as a purl, and pull the yarn forward between the needles. Next, return the stitch to the left needle, pull the yarn back between the needles, then turn the work and purl 12 stitches.
3rd row: knit 1 edge loop, then 2 knit stitches together, then knit 1 stitch, then 2 knit stitches together and 4 knit stitches. Slip one onto the right needle as a purl, pull the yarn forward between the needles, return the stitch to the left needle, then pull the yarn back between the needles. After this, turn the work and knit 8 purl stitches.
4th row: knit 1 edge stitch, then knit 3 stitches together, then knit 4 stitches, *take out the wrapped loop from below and knit together with the next knit stitch, knit 1* (repeat * to * 3 times). Without turning the work over, knit the purl stitches.
In this way, knit a spiral scarf to the required length using 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 types: copper (silver)-containing devices and hormone-containing devices. Their size is 3X4 cm. The choice of contraceptive method and the device itself occurs at an appointment with a gynecologist. You shouldn't do this yourself. The intrauterine device is installed by a gynecologist during menstruation. It is small in size and resembles the letter T in shape.
The copper spiral is made from copper wire. Its peculiarity 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 hormonal IUD has a container that contains progestin. This hormone prevents the onset of ovulation. If a hormonal intrauterine device is used, sperm cannot fertilize an egg. As women note, when using such a spiral, menstruation becomes more scanty and less painful. However, this does not cause harm, because it is associated with the action of hormones located inside the spiral. Gynecologists recommend that women suffering from painful periods install a hormonal IUD.
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. This is influenced by many factors.
The Juno Bio spiral is quite popular among Russian women. It attracts, first of all, with its 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 accessible form of contraception.
This is a hormonal IUD. Its manufacturers promise that the Mirena IUD is less likely to move in the uterus or fall out. Namely, this leads to pregnancy, therefore patients are advised to regularly check the presence of an intrauterine contraceptive in the right place.
The standard voltage in the household electrical network is U=220V. The current strength is limited by fuses in the electrical panel and is usually equal to I = 16A.
Sources:
- Tables physical quantities, I.K. Kikoin, 1976
- spiral length formula
An electric soldering iron is a hand-held tool designed for fastening parts together using soft solders, by heating the solder to a liquid state and filling the gap between the parts being soldered with it.
Electric soldering irons are produced designed for mains voltages of 12, 24, 36, 42 and 220 V, and there are reasons for this. The main thing is human safety, the second is the network voltage in the place soldering work. In production, where all equipment is grounded and there is high humidity, it is allowed to use soldering irons with a voltage of no more than 36 V, and the body of the soldering iron must be grounded. The motorcycle's on-board network has voltage DC 6 V, a passenger car - 12 V, a truck - 24 V. In aviation, they use a network with a frequency of 400 Hz and a voltage of 27 V. There are also design limitations, for example, a 12 W soldering iron is difficult to make for a supply voltage of 220 V, since the spiral will need to be wound made of very thin wire and therefore winding many layers, the soldering iron will turn out to be large, not convenient for small work. Since the soldering iron winding is wound from nichrome wire, it can be powered with either alternating or direct voltage. The main thing is that the supply voltage matches the voltage for which the soldering iron is designed.
Electric soldering irons come in power ratings of 12, 20, 40, 60, 100 W and more. And this is also no coincidence. In order for the solder to spread well over the surfaces of the parts being soldered during soldering, they need to be heated to a temperature slightly higher than the melting point of the solder. Upon contact with a 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 heating element is small, then having given off heat, the tip will not be able to heat up to the set temperature, and soldering will be impossible. At best, the result will be loose and not strong soldering. 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 measuring 5 mm? True, there is a way out: several turns of copper wire with a diameter of 1 mm are wound around such a sting and the end of this wire is soldered. But the bulkiness of the soldering iron makes the work practically impossible. There is one more limitation. At high power, the soldering iron will quickly heat up the element, and many radio components do not allow heating above 70˚C and therefore the permissible soldering time is no more than 3 seconds. These are diodes, transistors, microcircuits.
Soldering iron device
The soldering iron is a red copper rod, which 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 from the soldering iron tip from the heating element. The end of the rod is wedge-shaped, is the working part of the soldering iron and is called the tip. The rod is inserted into a steel tube wrapped in mica or fiberglass. A nichrome wire is wound around the mica, which serves as a heating element.
A layer of mica or asbestos is wound over the nichrome, which serves to reduce heat loss and electrically insulate 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 electrical 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 with a metal plate; it is best to make the crimp from an aluminum plate, which has high thermal conductivity and will more effectively remove heat from the joint. For electrical insulation, tubes made of heat-resistant insulating material, fiberglass or mica are placed at the junction.
The copper rod and nichrome spiral are closed with a metal case consisting of two halves or a solid tube, as in the photo. The body of the soldering iron is fixed on the tube with cap rings. To protect a person’s hand from burns, a handle made of a material that does not transmit heat well, wood or heat-resistant plastic, is attached to the tube.
When you insert the soldering iron plug into a 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-sized radio components, thick wires and small parts. To solder large parts, for example, heat exchangers of a geyser, 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 heating temperature of the tip. 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, in order to maintain the optimal temperature of the soldering iron tip, it is necessary to connect it through thyristor power regulators with manual adjustment and automatic maintenance of the set temperature of the soldering iron tip.
Calculation and repair of the heating winding of a soldering iron
When repairing or making your own electric soldering iron or any other heating device, you have to wind the heating winding from nichrome wire. The initial data for calculating and selecting wire is the winding resistance of a soldering iron or heating device, which is determined based on its power and supply voltage. You can calculate what the winding resistance of a soldering iron or heating device should be using the table.
During repairs or self-production electric soldering iron or any other heating device, you have to wind a heating winding made of nichrome wire. The initial data for calculating and selecting wire is the winding resistance of a soldering iron or heating device, which is determined based on its power and supply voltage. You can calculate what the winding resistance of a soldering iron or heating device should be using the table.
Knowing the supply voltage and measuring resistance any heating electrical appliance, such as a soldering iron, or electric iron, you can find out the consumption of this household electrical appliance power b. For example, the resistance of a 1.5 kW electric kettle will be 32.2 Ohms.
| Table for determining the resistance of a nichrome spiral depending on power and supply voltage electrical appliances, Ohm | |||||
|---|---|---|---|---|---|
| Power consumption soldering iron, W | Soldering iron supply voltage, V | ||||
| 12 | 24 | 36 | 127 | 220 | |
| 12 | 12 | 48,0 | 108 | 1344 | 4033 |
| 24 | 6,0 | 24,0 | 54 | 672 | 2016 |
| 36 | 4,0 | 16,0 | 36 | 448 | 1344 |
| 42 | 3,4 | 13,7 | 31 | 384 | 1152 |
| 60 | 2,4 | 9,6 | 22 | 269 | 806 |
| 75 | 1.9 | 7.7 | 17 | 215 | 645 |
| 100 | 1,4 | 5,7 | 13 | 161 | 484 |
| 150 | 0,96 | 3,84 | 8,6 | 107 | 332 |
| 200 | 0,72 | 2,88 | 6,5 | 80,6 | 242 |
| 300 | 0,48 | 1,92 | 4,3 | 53,8 | 161 |
| 400 | 0,36 | 1,44 | 3,2 | 40,3 | 121 |
| 500 | 0,29 | 1,15 | 2,6 | 32,3 | 96,8 |
| 700 | 0,21 | 0,83 | 1,85 | 23,0 | 69,1 |
| 900 | 0,16 | 0,64 | 1,44 | 17,9 | 53,8 |
| 1000 | 0,14 | 0,57 | 1,30 | 16,1 | 48,4 |
| 1500 | 0,10 | 0,38 | 0,86 | 10,8 | 32,3 |
| 2000 | 0,07 | 0,29 | 0,65 | 8,06 | 24,2 |
| 2500 | 0,06 | 0,23 | 0,52 | 6,45 | 19,4 |
| 3000 | 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. In the leftmost column of the table, select 60 W. From the top 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 winding material, should be equal to 806 Ohms.
If you needed to make a soldering iron from a 60 W soldering iron, designed for a voltage of 220 V, for power supply from a 36 V network, then the resistance of the new winding should already be equal to 22 Ohms. You can independently calculate the winding resistance of any electric heating device using an online calculator.
After determining the required resistance value of the soldering iron winding, the appropriate diameter of the nichrome wire is selected from the table below, 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 X20N80. 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 a soldering iron spiral, the turns are laid close to each other. When heated red-hot, the 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 a second one is wound.
For electrical and thermal insulation of heating element windings 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 a soldering iron with wet asbestos, it is necessary to take into account that wet asbestos conducts electrical current well and it will be possible to turn on the soldering iron into the electrical network only after the asbestos has completely dried.
Winding a nichrome spiral for further heating is carried out mainly by trial and error. After winding, voltage is applied to the heating element and the required number of turns is determined by how the wire is heated.This process can take a long time. It is worth remembering that nichrome can lose when large quantities bends its characteristics. The wire will quickly burn out in areas of deformation. Ultimately, it may turn out that good material turn into scrap.
To correctly calculate a nichrome spiral, special tables are usually used, where the resistivity of nichrome wire = (Ohm mm2 / m). But, these tables display data for a voltage of 220V. To operate the heating element in industrial environment you will have to carry out the calculation yourself, substituting the available data.
Using the tabular data, you can accurately determine the winding length and the distance between the turns. Depending on the diameter of the wire and the diameter of the nichrome winding rod, it will not be difficult to recalculate the length of the spiral for operation at a different voltage. Here you need to use a simple mathematical proportion.
For example, if you need to calculate the length of the spiral for a voltage of 380 V using a wire with a diameter of Ø 0.6 mm and a winding rod Ø 6 mm. In the table you can see that the length of the spiral at a voltage of 220 V should be 30 cm. Next, we calculate using the following ratio:
220 V – 30 cm
380 V – X cm
Based on this data:
X= 380 30/220=52 cm
After the spiral has already been wound, it should be connected to the energy carrier and make sure that the winding is correct. In this case, the wound wire is not cut. For a spiral in a closed heater, the winding length should be 1/3 greater than the values given in the table.
Calculation of a heating element made of nichrome wire
The length of the wire is determined based on the indicators required power.As an example, we will carry out the following calculations based on the available indicators.
There are several types of tandoor heating. That's all today greater distribution receives electric method, since it does not require the purchase of fuel, does not emit combustion products, and makes it easier to use behind the stove.
Collapse
The device is heated by heating the spirals and subsequent uniform heat transfer. The article discusses in detail the features of the tandoor spiral. This information will help you select and install the heating element on the stove correctly.
What is a tandoor spiral?
The spiral is an important element of the tandoor; without it, the device will not work. Warms up quite quickly. Allows you to maintain the required temperature for a long time, which is especially important if you have to cook on the stove all day.
This is what a spiral looks like
The heating element is made from wire with high electrical resistance. The length of the wire is quite long, so it is twisted in turns for convenience. Spirals can have the shape of cylinders or flat coils and be equipped with contact leads. The heaters are attached to the furnace using ceramic or metal bases with special heat-resistant inserts or insulators.
Purpose of the spiral
The main function of a tandoor coil is incandescence and subsequent uniform distribution of heat. To do this, the element must have the following qualities:
- Heat resistance (does not collapse at high temperatures in tandoors).
- High current resistance (the heating rate, the resulting temperature, and the service life of the element depend on this).
- Constancy of properties (does not change depending on environmental conditions, duration of operation).
Species
The most practical materials For heating parts, nichrome and fechral compounds are used. Let's briefly consider their features.
Nichrome
Nichrome spirals are made from Cr+Ni. This alloy allows the device to heat up to 1200 degrees. It is characterized by creep resistance and oxidation resistance. Minus - smaller temperature regime in comparison with fechral alloys.
The price of nichrome products is affordable. For example, brand Х20Н80(20% chromium, 80% nickel), suitable for a standard voltage of 220 volts will cost 150-170 rubles. per meter
Fechral
Fechral is a combination chromium, iron, aluminum and titanium. The material is different good performance current resistance. It has increased heat resistance: the maximum melting temperature of spirals made of this material reaches 1500 degrees.
Fechral spiral
Types
When selecting a heating device, it is important to pay attention not only to the material, but also to the type of product: a 220 or 380 volt tandoor coil has some differences.
220 V is the standard voltage for home electrical networks (that is, for connecting to regular sockets in apartments and country cottages). Can also be used in small restaurants with low productivity. According to safety rules, spirals with a power of 3.5-7 kilowatts are connected to 220 volts.
A powerful tandoor is not connected to a standard consumer electrical network. This will cause the heater to burn out and short out. Requires connection to an industrial three-phase power supply of 380 volts. The power of each spiral in the tandoor in this case increases to 12 kilowatts. Special requirements for wires used in heating elements: they must have a cross-section of at least 4 mm.
How to choose the right spiral?
The dimensions of the wire used to create heaters are determined by the power of the tandoor, the voltage in the electrical network and the heat that should be produced by the stove. First, you need to determine the current strength using the formula: I = P:U
- P is the technical power of the furnace.
- U is the voltage in the electrical network.
For example, for an 800-watt stove and a mains voltage of 220 volts, the electric current will be 3.6 amperes. Afterwards, using the specified parameters (temperature and electric current), suitable wire dimensions are searched in a special table.
The length of the wire for the spiral is calculated by the formula l=RхS:ρ. For example, with a resistance of 61 Ohms, a cross-sectional size of 0.2 square meters. mm and a resistance of 1.1 requires a spiral made of wire 5.3 meters long.
Installation work
Specialists charge about 2300-3000 rubles for installing heating elements in a furnace. If you want to save money and install the spiral in the tandoor yourself, here are some important tips:
- You should not place the heating element vertically. The hot wire is soft and may bend due to gravity. It is better to lay it horizontally.
- It is not recommended to install the heater close to the insulating brick - the risk of overheating increases. A small “air cushion” is created between the walls of the furnace and the wire.
- When installing, you need to stretch the spiral so that all the turns are at a small distance from each other (experts recommend the distance between the rings is 1.5-2 times greater than the diameter of the wire).
An alternative option: a heating element (tubular electric heater with a wire spiral inside) is installed at the bottom of the tandoor. This is a convenient and safe option. But as practice shows, heating from the heating element will slower than in the case of an open spiral.
The photographs below show several types of spiral installation:
Example of spiral installation
Another way
heating element instead of a spiral
Conclusion
Correct and safe work tandoor depends on this important element like a spiral. When buying a ready-made stove or making a device with your own hands, it is important to choose suitable material, type, size of heaters. If you do not have confidence in your abilities and knowledge, it is better to entrust the selection and installation of foam spirals to specialists.
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Calculation of a nichrome spiral. We are ready to make a nichrome spiral for you
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 after heating the nichrome wire, the threads select the required number of turns.
Typically, such a procedure takes a lot of time, and nichrome loses its characteristics with multiple bends, which leads to rapid burning in places of deformation. In the worst case, business nichrome turns into nichrome scrap.
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 necessary 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 at a voltage of 220 V will be equal to 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 made of nichrome wire
The length of the nichrome wire for making a spiral is determined based on the required power.
Example: Determine the length of nichrome wire for a tile heating element with a power of P = 600 W at Unetwork = 220 V.
1) I = P/U = 600/220 = 2.72 A
2) R = U/I = 220/2.72 = 81 Ohm
3) Based on these data (see Table 1), we select d=0.45; S=0.159
then the length of nichrome
l = SR / ρ = 0.159 81 /1.1 = 11.6 m
where l is wire length (m)
S - wire cross-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 customer’s specifications and sketches
It is profitable to buy a nichrome spiral from PARTAL
Nichrome for spiral high quality only Russian production. Strict compliance with quality and brand
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Calculation of nichrome spiral | Useful
Calculating a nichrome spiral is, in fact, a very important process. Very often, in plants, factories, factories, this is neglected and calculations are made “by eye”, after which they connect the spiral to the network, and then select the required number of turns depending on the heating of the nichrome wire. This procedure may be very simple, but it takes a long period of 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. From the calculation that the resistivity of nichrome is equal to (Ohm mm2 / m)C, you can quickly calculate the length of winding turn to turn depending on the diameter of the rod on which the nichrome thread is wound, and directly on the thickness of the nichrome wire itself. 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 the winding rod is 4 mm in diameter. Looking at the table, you can see that the length of this spiral at a voltage of 220 Volts will be 22 cm. We make a simple ratio: 220 V - 22 cm 127 V - X cm then: X = 127 22 / 220 = 12.7 cm
Having wound the nichrome spiral, carefully connect it, without cutting it, to the voltage source and make sure of your calculations, or rather, the calculations of the correctness of 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, calculation of nichrome weight, use of nichrome
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We produce electrical spirals from NICHROMA according to the customer’s specifications and sketches
Nichrome spiral
Everyone knows what a nichrome spiral is. This is a heating element in the form of a wire, rolled 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 nichrome – 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 increase the machinability of the wire.
Based on these alloys, modifications with higher survivability and resistance to oxidation at high temperature.
These are brands X20N80-N (-N-VI) and X15N60 (-N-VI). They are used for heating elements in contact with air. Recommended maximum operating temperature – from 1100 to 1220 °C
Application of nichrome wire
The main quality of nichrome is high resistance electric current. It determines the applications of the alloy.
Nichrome spiral is used in two qualities - as a heating element or as a material for electrical resistances electrical diagrams.
For heaters, an electric spiral made of X20N80-N and X15N60-N alloys is used.
Application examples:
- household thermoreflectors and fan heaters;
- Heating elements for household heating devices and electric heating;
- heaters for industrial furnaces and thermal equipment.
Alloys Kh15N60-N-VI and Kh20N80-N-VI, produced in vacuum induction furnaces, used in industrial equipment increased reliability.
A spiral made of nichrome brands Х15Н60, Х20Н80, Х20Н80-VI, N80ХУД-VI is distinguished by the fact that its electrical resistance changes little with temperature changes.
Resistors and connectors are made from it electronic circuits, critical parts of vacuum devices.
How to wind a spiral from nichrome
A resistive or heating coil can be made at home. To do this, you need nichrome wire of a suitable grade 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 power, you need to take into account the maximum permissible current at which the spiral 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 to 900 °C.
There are reference tables for calculating temperature and current. But you still need to take into account 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 be reduced by 10-50%.
The intensity of heat removal, and therefore the temperature of the heater, is affected by the pitch of the spiral winding.
Densely spaced coils generate more heat, while larger pitches increase cooling.
It should be taken into account that all tabular calculations are given for a heater located horizontally. When the angle to the horizon changes, the heat removal conditions worsen.
Calculation of the resistance of a nichrome spiral and its length
Having decided on the power, we proceed to calculate the required resistance.
If the determining parameter is power, then first we find the required current using the formula I=P/U.
Having the current strength, we determine the required resistance. To do this, we use Ohm's law: R=U/I.
The notations here are generally accepted:
- P – allocated power;
- U is the voltage at the ends of the spiral;
- R – spiral resistance;
- 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 – required length;
- R – wire resistance;
- d – wire diameter;
- ρ – resistivity of nichrome;
- π – constant 3.14.
But it’s easier to take 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 changes in resistance when heated.
In this case, the calculation will look like this: L=R/ρld, where ρld is the resistance of one meter of wire having a diameter d.
Spiral winding
Now we will make a geometric calculation of the nichrome spiral. We have selected the wire diameter d, the required length L has been determined, and we have a rod with diameter D for winding. How many turns do you need to make? The length of one turn is: π(D+d/2). Number of turns – N=L/(π(D+d/2)). The calculation is completed.
In practice, rarely does anyone wind their own wire for a resistor or heater.
It’s 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 spirals 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 technical specifications customer.
partalstalina.ru
Calculation and repair of the heating winding of a soldering iron
When repairing or making your own electric soldering iron or any other heating device, you have to wind a heating winding made of nichrome wire. The initial data for calculating and selecting wire is the winding resistance of a soldering iron or heating device, which is determined based on its power and supply voltage. You can calculate what the winding resistance of a soldering iron or heating device should be using the table.
Knowing the supply voltage and measuring the resistance of any heating electrical appliance, such as a soldering iron, electric kettle, electric heater or electric iron, you can find out the power consumed by this household electrical 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. In the leftmost column of the table, select 60 W. From the top 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 winding material, should be equal to 806 Ohms.
If you needed to make a soldering iron from a 60 W soldering iron, designed for a voltage of 220 V, for power supply from a 36 V network, then the resistance of the new winding should already be equal to 22 Ohms. You can independently calculate the winding resistance of any electric heating device using an online calculator.
After determining the required resistance value of the soldering iron winding, the appropriate diameter of the nichrome wire is selected from the table below, 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 X20N80. 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 a soldering iron spiral, the turns are laid close to each other. When heated red-hot, the 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 a second one is wound.
For electrical and thermal insulation of the heating element winding, the best materials are 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 a soldering iron with wet asbestos, it is necessary to take into account that wet asbestos conducts electrical current well and it will be possible to turn on the soldering iron into the electrical network only after the asbestos has completely dried.
felstar.mypage.ru
HOW TO CALCULATE A NICHROME SPIRAL?
Post written by admin at 01/18/2015 23:23Categories: 3. Home electrical, Electrical workshop
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Winding a nichrome spiral for heating devices is often done “by eye”, and then, including the spiral in the network, the required number of turns is selected based on the heating of the nichrome wire. Usually this procedure takes a lot of time, and nichrome is wasted.
When using a 220 V spiral, you can use the data given in the table, based on the calculation that the resistivity of nichrome ρ=(Ohm mm2/m). Using this formula, you can quickly determine the length of 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 you need to determine the length of a spiral for a voltage of 127 V from nichrome wire with a thickness of 0.3 mm, the winding rod dia. 4 mm. The table shows that the length of such a spiral at a voltage of 220 V will be equal to 22 cm.
Let's make a simple ratio:
220 V - 22 cm
X = 127 * 22 / 220 = 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.
Legend in the table: D - diameter of the rod, mm; L - length of the spiral, cm.
| dia. nichrome0.2 mm | dia. nichrome0.3 mm | dia. nichrome0.4 mm | dia. nichrome0.5 mm | dia. nichrome0.6 mm | dia. nichrome0.7 mm | dia. nichrome0.8 mm | dia. nichrome0.9 mm | dia. nichrome1.0 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 | ||||||||||||||||
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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, and the grade of the alloy. The general formula for active resistance is: R = ρ l / S R - active electrical resistance (Ohm), ρ - specific electrical resistance (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, based on the calculation that the resistivity of nichrome = (Ohm mm2 / m)C. With its help, you can quickly determine the length of 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, you need 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 at a voltage of 220 V will be equal to 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 the 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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| Calculation of the heating element Calculation example. Given: U=220V,t=700°C, type X20H80, d=0.5mm-----------L,P-?Solution: From Table 1 we find that the diameter d=0.5mm corresponds to S = 0.196 mm², and the current at 700 ° C I = 5.2 A. The type of alloy X20N80 is nichrome, the resistivity 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. Determine the power of the heating element: P = U I = 220 5.2 = 1.15 kW .When winding the spiral, adhere to the following relationship: 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 a closed version of the heater, the current should be reduced by 1.2-1.5 times. A smaller coefficient is taken for thicker wire, a larger one for thinner wire. For the first case, the coefficient is chosen exactly the opposite. Let me make a reservation: we are talking about a simplified calculation of the heating element. Perhaps someone will need a table of electrical resistance values for 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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