Nichrome was invented in 1905 by Albert Marsh, who combined nickel (80%) and chromium (20%). Today there are about ten modifications of alloys of various brands. Aluminum, manganese, iron, silicon, titanium, molybdenum, etc. are added as additional alloying impurities. Due to its outstanding qualities, this metal has become widely used in the production of electrical equipment.

Basic qualities of nichrome

Nichrome is different:

• high heat resistance. At high temperatures, its mechanical properties do not change;
• plasticity, which allows you to make nichrome spirals, wires, tapes, threads from the alloy;
• ease of processing. Products made from nichrome are well welded and stamped;
• high resistance to corrosion in various environments.
• Nichrome resistance is high.

Basic properties

• Density is 8200-8500 kg/m3.
• The melting point of nichrome is 1400 C.
• The maximum operating temperature is 1100°C.
• Strength - 650-700 MPa.
• The resistivity of nichrome is 1.05-1.4 Ohm.

Marking nichrome wire

Nichrome wire is an excellent material for various electric heating elements, which are used in almost all industries. Almost every household heating device has elements made of nichrome.

Letter marking of wire:

• “H” - used, as a rule, in heating elements.
• “C” - used in resistance elements.
• “TEN” - intended for tubular electric heaters.

According to domestic standards, there are several main brands:

• Double wire X20N80. The alloy composition includes: nickel - 74%, chromium - 23%, as well as 1% each of iron, silicon and manganese.
• Triple X15N60. The alloy consists of 60% nickel and 15% chromium. The third component is iron (25%). Saturation of the alloy with iron makes it possible to significantly reduce the cost of nichrome, the price of which is quite high, and at the same time maintain its heat resistance. In addition, its machinability increases.
• The cheapest option for nichrome is X25N20. It is an iron-rich alloy in which mechanical properties are maintained, but the operating temperature is limited to 900°C.

Application of nichrome

Thanks to their high-quality and unique characteristics, nichrome products can be used where reliability, strength, and resistance to chemically aggressive environments and very high temperatures are needed.

Nichrome spirals and wire are an integral part of almost all types of heating devices. Nichrome is present in toasters, bakeries, heaters, and ovens. The alloy has also found application in resistors and rheostats operating under high heat. Nichrome is also found in electric lamps and soldering irons. Nichrome spirals have heat resistance and significant resistance, which allows them to be used in high-temperature drying and firing ovens.

Nichrome scrap is also used. It is melted down, and the material is used again. An alloy of nickel and chromium is used in chemical laboratories. This composition does not react with most alkalis and acids. Deformed nichrome heating coils are used in electronic cigarettes.

Compared to iron previously used for these purposes, nichrome products are safer, do not spark, do not rust, and do not have melted areas.

The melting point of nichrome is 1400°C, so no foreign odors or fumes are felt when cooking.

Engineers are still exploring the unique properties of this material, constantly expanding the scope of its application.

At home, nichrome wire is used to make homemade equipment, jigsaws and cutters, such as, for example, a foam or wood cutting machine, a soldering iron, a wood burning device, welding machines, household heaters, etc.

The most popular wires are X20H80 and X15H60.

Where can I buy nichrome wire?

This product is sold in rolls (coils, spools) or in the form of tape. The cross-section of nichrome wire can be in the form of an oval, circle, square, or trapezoid; the diameter ranges from 0.1 to 1 millimeter.

Where can I get or buy nichrome products? We suggest considering the most common and possible options:

1. First of all, you can contact the organization that manufactures these products and place an order. You can find out the exact address of such enterprises in special information desks on goods and services, which are available in almost all major cities and towns. The operator will be able to tell you where to buy it and give you a phone number. In addition, information about the range of such products can be found on the official websites of manufacturers.
2. You can buy nichrome products in specialized stores, for example, those selling radio components, materials for craftsmen like “Skillful Hands,” etc.


3. Buy from private individuals selling radio components, spare parts and other metal products.
4. At any hardware store.
5. On the market you can buy some old device, for example a laboratory rheostat, and take nichrome.
6. Nichrome wire can also be found at home. For example, it is from this that the spiral of an electric stove is made.

If you need to place a large order, then the first option is most suitable. If you need a small amount of nichrome wire, in this case you can consider all the other items on the list. When purchasing, be sure to pay attention to the labeling.

Nichrome spiral winding

Today, the nichrome spiral is one of the main elements of many heating devices. After cooling, nichrome is able to retain its plasticity, thanks to which a spiral made of such material can be easily removed, changed its shape, or, if necessary, adjusted to a suitable size. Winding of the spiral in industrial conditions is carried out automatically. At home, you can also do manual winding. Let's take a closer look at how to do this.

If the parameters of the finished nichrome spiral in its working state are not too important, during winding you can make the calculation, so to speak, “by eye.” To do this, you should select the required number of turns depending on the heating of the nichrome wire, while periodically including the spiral in the network and decreasing or increasing the number of turns. This winding procedure is very simple, but it can take quite a lot of time, and some of the nichrome is wasted.

To increase the simplicity and accuracy of spiral winding calculations, you can use a special online calculator.

Having calculated the required number of turns, you can begin winding it on the rod. Without cutting the wire, you should carefully connect the nichrome spiral to the voltage source. Then check the correctness of the calculations for winding the spiral. It is important to consider that for closed-type spirals, the winding length should be increased by a third of the value obtained in the calculation.

To ensure the same distance between adjacent turns, you need to wind two wires: one - nichrome, the second - any copper or aluminum, with a diameter that is equal to the required gap. When winding is completed, the auxiliary wire should be carefully wound.

Cost of nichrome

The only drawback that nichrome has is the price. Thus, a two-component alloy when purchased at retail is estimated at approximately 1000 rubles per kilogram. The cost of nichrome stamps with a ligature is about 500-600 rubles.

Conclusion

When choosing nichrome products, it is necessary to take into account data on the chemical composition of the product of interest, its electrical conductivity and resistance, physical characteristics of diameter, cross-section, length, etc. It is also important to inquire about compliance documentation. In addition, you need to be able to visually distinguish the alloy from its, so to speak, “competitors”. The correct choice of material is the key to the reliability of electrical engineering.

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EXAMPLES OF TASKS

Part 1

1. The current strength in the conductor was increased by 2 times. How will the amount of heat released in it per unit time change if the resistance of the conductor remains constant?

1) will increase 4 times
2) will decrease by 2 times
3) will increase 2 times
4) will decrease by 4 times

2. The length of the electric stove spiral was reduced by 2 times. How will the amount of heat released in the spiral per unit time change at a constant network voltage?

1) will increase 4 times
2) will decrease by 2 times
3) will increase 2 times
4) will decrease by 4 times

3. The resistance of the resistor ​(R_1)​ is four times less than the resistance of the resistor ​(R_2)​. Current work in resistor 2

1) 4 times more than in resistor 1
2) 16 times more than in resistor 1
3) 4 times less than in resistor 1
4) 16 times less than in resistor 1

4. The resistance of the resistor ​(R_1)​ is 3 times greater than the resistance of the resistor ​(R_2)​. The amount of heat that will be released in resistor 1

1) 3 times more than in resistor 2
2) 9 times more than in resistor 2
3) 3 times less than in resistor 2
4) 9 times less than in resistor 2

5. The circuit is assembled from a current source, a light bulb and a thin iron wire connected in series. The light bulb will glow brighter if

1) replace the wire with a thinner iron one
2) reduce the length of the wire
3) swap the wire and the light bulb
4) replace the iron wire with nichrome

6. The figure shows a bar graph. It shows the voltage values ​​at the ends of two conductors (1) and (2) of the same resistance. Compare the values ​​of current work ​(A_1)​ and ​(A_2)​ in these conductors for the same time.

1) ​(A_1=A_2)​
2) (A_1=3A_2)
3) (9A_1=A_2)
4) (3A_1=A_2)

7. The figure shows a bar graph. It shows the current values ​​in two conductors (1) and (2) of the same resistance. Compare the values ​​of current work (A_1)​ and ​(A_2) in these conductors for the same time.

1) ​(A_1=A_2)​
2) (A_1=3A_2)
3) (9A_1=A_2)
4) (3A_1=A_2)

8. If you use lamps with a power of 60 and 100 W in a chandelier to illuminate the room, then

A. A large current will be in a 100 W lamp.
B. A 60 W lamp has greater resistance.

The following statement(s) are true:

1) only A
2) only B
3) both A and B
4) neither A nor B

9. An electric stove connected to a direct current source consumes 108 kJ of energy in 120 s. What is the current strength in the tile spiral if its resistance is 25 Ohms?

1) 36 A
2) 6 A
3) 2.16 A
4) 1.5 A

10. An electric stove with a current of 5 A consumes 1000 kJ of energy. What is the time it takes for the current to pass through the spiral of the tile if its resistance is 20 Ohms?

1) 10000 s
2) 2000 s
3) 10 s
4) 2 s

11. The nickeline spiral of the electric stove was replaced with a nichrome one of the same length and cross-sectional area. Establish a correspondence between physical quantities and their possible changes when the tile is connected to the electrical network. Write down the selected numbers in the table under the corresponding letters. The numbers in the answer may be repeated.

PHYSICAL QUANTITY
A) electrical resistance of the spiral
B) the strength of the electric current in the spiral
B) electrical power consumed by the tile

NATURE OF CHANGE
1) increased
2) decreased
3) has not changed

12. Establish a correspondence between physical quantities and the formulas by which these quantities are determined. Write down the selected numbers in the table under the corresponding letters.

PHYSICAL QUANTITIES
A) current work
B) current strength
B) current power

FORMULAS
1) ​(frac(q)(t))​
2) ​(qU)​
3) (frac(RS)(L))​
4) ​(UI)​
5) (frac(U)(I))​

Part 2

13. The heater is connected in series with a rheostat with a resistance of 7.5 Ohms in a network with a voltage of 220 V. What is the resistance of the heater if the power of the electric current in the rheostat is 480 W?

Very often, if you want to make or repair heater When making electric furnaces with your own hands, a person has many questions. For example, what diameter should the wire be taken, what should its length be, or what power can be obtained using a wire or tape with the given parameters, etc. With the right approach to solving this issue, it is necessary to take into account quite a lot of parameters, for example, the strength of the current passing through heater, operating temperature, type of electrical network and others.

This article provides background information on the materials most common in the manufacture of heaters electric ovens, as well as methods and examples of their calculation (calculation of heaters for electric furnaces).

Heaters. Materials for the manufacture of heaters

Directly heater- one of the most important elements of the furnace, it is it that carries out heating, has the highest temperature and determines the performance of the heating installation as a whole. Therefore, heaters must meet a number of requirements, which are given below.

Requirements for heaters

Basic requirements for heaters (heater materials):

• Heaters must have sufficient heat resistance (scale resistance) and heat resistance. Heat resistance - mechanical strength at high temperatures. Heat resistance - resistance of metals and alloys to gas corrosion at high temperatures (the properties of heat resistance and heat resistance are described in more detail on the page).
• Heater in an electric furnace must be made of a material with high electrical resistivity. In simple terms, the higher the electrical resistance of a material, the more it heats up. Therefore, if you take a material with lower resistance, you will need a heater of greater length and with a smaller cross-sectional area. It is not always possible to place a sufficiently long heater in the oven. It is also worth considering that, the larger the diameter of the wire from which the heater is made, the longer its service life . Examples of materials with high electrical resistance are chromium-nickel alloy, iron-chromium-aluminum alloy, which are precision alloys with high electrical resistance.
• A low temperature coefficient of resistance is an essential factor when choosing a material for a heater. This means that when the temperature changes, the electrical resistance of the material heater doesn't change much. If the temperature coefficient of electrical resistance is high, to turn on the furnace in a cold state it is necessary to use transformers that initially provide a reduced voltage.
• The physical properties of heater materials must be constant. Some materials, for example carborundum, which is a non-metallic heater, can change their physical properties over time, in particular electrical resistance, which complicates their operating conditions. To stabilize electrical resistance, transformers with a large number of steps and a voltage range are used.
• Metal materials must have good technological properties, namely ductility and weldability, so that they can be used to make wire, tape, and from the tape - heating elements of complex configuration. Also heaters can be made from non-metals. Non-metallic heaters are pressed or molded into a finished product.

Materials for the manufacture of heaters

The most suitable and most used in the production of heaters for electric furnaces are precision alloys with high electrical resistance. These include alloys based on chromium and nickel ( chromium-nickel), iron, chromium and aluminum ( iron-chromium-aluminum). The grades and properties of these alloys are discussed in “Precision alloys. Stamps". Representatives of chromium-nickel alloys are grades X20N80, X20N80-N (950-1200 °C), X15N60, X15N60-N (900-1125 °C), iron-chromium-aluminum alloys - grades X23Yu5T (950-1400 °C), X27Yu5T (950-1350 °C ), X23Yu5 (950-1200 °C), X15YU5 (750-1000 °C). There are also iron-chromium-nickel alloys - Kh15N60Yu3, Kh27N70YUZ.

The alloys listed above have good heat resistance and heat resistance properties, so they can operate at high temperatures. good heat resistance provides a protective film of chromium oxide that forms on the surface of the material. The melting point of the film is higher than the melting point of the alloy itself; it does not crack when heated and cooled.

Let us give a comparative description of nichrome and fechral.
Advantages of nichrome:

• good mechanical properties at both low and high temperatures;
• the alloy is creep-resistant;
• has good technological properties - ductility and weldability;
• well processed;
• does not age, non-magnetic.

Disadvantages of nichrome:

• high cost of nickel - one of the main components of the alloy;
• lower operating temperatures compared to fechral.

Advantages of fehrali:

• a cheaper alloy compared to nichrome, because does not contain ;
• has better heat resistance compared to nichrome, for example, fechral X23Yu5T can operate at temperatures up to 1400 °C (1400 °C is the maximum operating temperature for a heater made of wire Ø 6.0 mm or more; Ø 3.0 - 1350 °C; Ø 1.0 - 1225 °C; Ø 0.2 - 950 °C).

Disadvantages of fehrali:

• a brittle and weak alloy, these negative properties are especially pronounced after the alloy has been at temperatures above 1000 °C;
• because Since fechral contains iron, this alloy is magnetic and can rust in a humid atmosphere at normal temperatures;
• has low creep resistance;
• interacts with fireclay lining and iron oxides;
• During operation, fechral heaters elongate significantly.

Also comparison of alloys fechral And nichrome produced in the article.

Recently, alloys of the Kh15N60Yu3 and Kh27N70YUZ types have been developed, i.e. with the addition of 3% aluminum, which significantly improved the heat resistance of the alloys, and the presence of nickel practically eliminated the disadvantages of iron-chromium-aluminum alloys. Alloys Kh15N60YUZ, Kh27N60YUZ do not interact with fireclay and iron oxides, are fairly well processed, mechanically strong, and non-fragile. The maximum operating temperature of the X15N60YUZ alloy is 1200 °C.

In addition to the above-mentioned alloys based on nickel, chromium, iron, and aluminum, other materials are used for the manufacture of heaters: refractory metals, as well as non-metals.

Among non-metals for the manufacture of heaters, carborundum, molybdenum disilicide, coal, and graphite are used. Carborundum and molybdenum disilicide heaters are used in high-temperature furnaces. In furnaces with a protective atmosphere, coal and graphite heaters are used.

Among the refractory materials, tantalum and niobium can be used as heaters. In high-temperature vacuum furnaces and furnaces with a protective atmosphere, they are used molybdenum heaters And tungsten. Molybdenum heaters can operate up to temperatures of 1700 °C in a vacuum and up to 2200 °C in a protective atmosphere. This temperature difference is due to the evaporation of molybdenum at temperatures above 1700 °C in a vacuum. Tungsten heaters can operate up to 3000 °C. In special cases, heaters made of tantalum and niobium are used.

Calculation of electric furnace heaters

Typically, the initial data for this are the power that the heaters must provide, the maximum temperature that is required to carry out the corresponding technological process (tempering, hardening, sintering, etc.) and the dimensions of the working space of the electric furnace. If the furnace power is not specified, it can be determined using a rule of thumb. When calculating heaters, it is necessary to obtain the diameter and length (for wire) or cross-sectional area and length (for tape), which are necessary for manufacturing of heaters.

It is also necessary to determine the material from which to make heaters(this point is not discussed in the article). In this article, a chromium-nickel precision alloy with high electrical resistance, which is one of the most popular in the manufacture of heating elements, is considered as a material for heaters.

Determination of the diameter and length of the heater (nichrome wire) for a given furnace power (simple calculation)

Perhaps the simplest option heater calculations from nichrome is the choice of diameter and length for a given heater power, supply voltage, as well as the temperature that the heater will have. Despite the simplicity of the calculation, it has one feature, which we will pay attention to below.

An example of calculating the diameter and length of a heating element

Initial data:
Device power P = 800 W; mains voltage U = 220 V; heater temperature 800 °C. Nichrome wire X20N80 is used as a heating element.

1. First you need to determine the current strength that will pass through the heating element:
I=P/U = 800 / 220 = 3.63 A.

2. Now you need to find the heater resistance:
R=U/I = 220 / 3.63 = 61 Ohm;

3. Based on the value of the current strength obtained in step 1 passing through nichrome heater, you need to select the wire diameter. And this point is important. If, for example, with a current of 6 A you use nichrome wire with a diameter of 0.4 mm, it will burn. Therefore, having calculated the current strength, it is necessary to select the appropriate wire diameter value from the table. In our case, for a current of 3.63 A and a heater temperature of 800 °C, we select nichrome wire with a diameter d = 0.35 mm and cross-sectional area S = 0.096 mm 2.

General rule for choosing wire diameter can be formulated as follows: it is necessary to select a wire whose permissible current strength is not less than the calculated current strength passing through the heater. In order to save heater material, you should choose a wire with the nearest higher (than calculated) permissible current strength.

Table 1

Permissible current passing through a nichrome wire heater corresponding to certain heating temperatures of the wire suspended horizontally in calm air at normal temperature

Diameter, mm	Cross-sectional area of ​​nichrome wire, mm 2	Nichrome wire heating temperature, °C
		200	400	600	700	800	900	1000
		Maximum permissible current, A
5	19,6	52	83	105	124	146	173	206
4	12,6	37,0	60,0	80,0	93,0	110,0	129,0	151,0
3	7,07	22,3	37,5	54,5	64,0	77,0	88,0	102,0
2,5	4,91	16,6	27,5	40,0	46,6	57,5	66,5	73,0
2	3,14	11,7	19,6	28,7	33,8	39,5	47,0	51,0
1,8	2,54	10,0	16,9	24,9	29,0	33,1	39,0	43,2
1,6	2,01	8,6	14,4	21,0	24,5	28,0	32,9	36,0
1,5	1,77	7,9	13,2	19,2	22,4	25,7	30,0	33,0
1,4	1,54	7,25	12,0	17,4	20,0	23,3	27,0	30,0
1,3	1,33	6,6	10,9	15,6	17,8	21,0	24,4	27,0
1,2	1,13	6,0	9,8	14,0	15,8	18,7	21,6	24,3
1,1	0,95	5,4	8,7	12,4	13,9	16,5	19,1	21,5
1,0	0,785	4,85	7,7	10,8	12,1	14,3	16,8	19,2
0,9	0,636	4,25	6,7	9,35	10,45	12,3	14,5	16,5
0,8	0,503	3,7	5,7	8,15	9,15	10,8	12,3	14,0
0,75	0,442	3,4	5,3	7,55	8,4	9,95	11,25	12,85
0,7	0,385	3,1	4,8	6,95	7,8	9,1	10,3	11,8
0,65	0,342	2,82	4,4	6,3	7,15	8,25	9,3	10,75
0,6	0,283	2,52	4	5,7	6,5	7,5	8,5	9,7
0,55	0,238	2,25	3,55	5,1	5,8	6,75	7,6	8,7
0,5	0,196	2	3,15	4,5	5,2	5,9	6,75	7,7
0,45	0,159	1,74	2,75	3,9	4,45	5,2	5,85	6,75
0,4	0,126	1,5	2,34	3,3	3,85	4,4	5,0	5,7
0,35	0,096	1,27	1,95	2,76	3,3	3,75	4,15	4,75
0,3	0,085	1,05	1,63	2,27	2,7	3,05	3,4	3,85
0,25	0,049	0,84	1,33	1,83	2,15	2,4	2,7	3,1
0,2	0,0314	0,65	1,03	1,4	1,65	1,82	2,0	2,3
0,15	0,0177	0,46	0,74	0,99	1,15	1,28	1,4	1,62
0,1	0,00785	0,1	0,47	0,63	0,72	0,8	0,9	1,0

Note :

• if the heaters are located inside the heated liquid, then the load (permissible current) can be increased by 1.1 - 1.5 times;
• with a closed arrangement of heaters (for example, in chamber electric furnaces), it is necessary to reduce the load by 1.2 - 1.5 times (a smaller coefficient is taken for thicker wire, a larger one for thinner wire).

4. Next, determine the length of the nichrome wire.
R = ρ l/S ,
Where R - electrical resistance of the conductor (heater) [Ohm], ρ - specific electrical resistance of the heater material [Ohm mm 2 / m], l - length of conductor (heater) [mm], S - cross-sectional area of ​​the conductor (heater) [mm 2 ].

Thus, we obtain the length of the heater:
l = R S / ρ = 61 · 0.096 / 1.11 = 5.3 m.

In this example, nichrome wire Ø 0.35 mm is used as a heater. In accordance with "Wire made of precision alloys with high electrical resistance. Technical specifications" the nominal value of the electrical resistivity of nichrome wire grade X20N80 is 1.1 Ohm mm 2 / m ( ρ = 1.1 Ohm mm 2 / m), see table. 2.

The result of the calculations is the required length of nichrome wire, which is 5.3 m, diameter - 0.35 mm.

table 2

Determination of the diameter and length of the heater (nichrome wire) for a given furnace (detailed calculation)

The calculation presented in this paragraph is more complex than the one above. Here we will take into account the additional parameters of the heaters and try to understand the options for connecting the heaters to a three-phase current network. We will calculate the heater using an electric furnace as an example. Let the initial data be the internal dimensions of the furnace.

1. The first thing you need to do is calculate the volume of the chamber inside the oven. In this case let's take h = 490 mm, d = 350 mm and l = 350 mm (height, width and depth respectively). Thus, we get the volume V = h d l = 490 · 350 · 350 = 60 · 10 6 mm 3 = 60 l (measure of volume).

2. Next, you need to determine the power that the oven should produce. Power is measured in Watts (W) and is determined by rule of thumb: for an electric oven with a volume of 10 - 50 liters, the specific power is 100 W/l (Watt per liter of volume), for a volume of 100 - 500 liters - 50 - 70 W/l. Let us take the specific power of 100 W/l for the furnace in question. Thus, the heater power of the electric furnace should be P = 100 · 60 = 6000 W = 6 kW.

It is worth noting that with a power of 5-10 kW heaters are usually manufactured single-phase. At high powers, to ensure even loading of the network, the heaters are made three-phase.

3. Then you need to find the current passing through the heater I=P/U , Where P - heater power, U - voltage across the heater (between its ends), and heater resistance R=U/I .

There may be two options for connecting to the electrical network:

• to a single-phase household network - then U = 220 V;
• to an industrial three-phase current network - U = 220 V (between neutral wire and phase) or U = 380 V (between any two phases).

Further calculations will be carried out separately for single-phase and three-phase connections.

I=P/U = 6000 / 220 = 27.3 A - current passing through the heater.
Next, you need to determine the resistance of the furnace heater.
R=U/I = 220 / 27.3 = 8.06 Ohm.

Figure 1 Wire heater in a single-phase current network

The required values ​​of the wire diameter and its length will be determined in paragraph 5 of this paragraph.

With this type of connection, the load is distributed evenly across three phases, i.e. 6 / 3 = 2 kW per phase. So we need 3 heaters. Next, you need to select a method for connecting the heaters (load) directly. There can be 2 ways: “STAR” or “TRIANGLE”.

It is worth noting that in this article the formulas for calculating the current strength ( I ) and resistance ( R ) for a three-phase network are not written in classical form. This is done in order not to complicate the presentation of the material on calculating heaters with electrical terms and definitions (for example, phase and linear voltages and currents and the relationships between them are not mentioned). The classical approach and formulas for calculating three-phase circuits can be found in specialized literature. In this article, some mathematical transformations carried out on classical formulas are hidden from the reader, and this does not have any effect on the final result.

When connecting “STAR” type the heater is connected between phase and zero (see Fig. 2). Accordingly, the voltage at the ends of the heater will be U = 220 V.
I=P/U = 2000 / 220 = 9.10 A.
R=U/I = 220 / 9.10 = 24.2 Ohms.

Figure 2 Wire heater in a three-phase current network. STAR connection

When connecting the “TRIANGLE” type the heater is connected between two phases (see Fig. 3). Accordingly, the voltage at the ends of the heater will be U = 380 V.
Current passing through the heater -
I=P/U = 2000 / 380 = 5.26 A.
Resistance of one heater -
R=U/I = 380/ 5.26 = 72.2 Ohm.

Figure 3 Wire heater in a three-phase current network. Connection according to the "TRIANGLE" scheme

4. After determining the resistance of the heater with appropriate connection to the electrical network it is necessary to select the diameter and length of the wire.

When determining the above parameters, it is necessary to analyze specific surface power of the heater, i.e. power that is released per unit area. The surface power of the heater depends on the temperature of the material being heated and on the design of the heaters.

Example
From the previous calculation points (see paragraph 3 of this paragraph), we know the resistance of the heater. For a 60 liter stove with a single-phase connection it is R = 8.06 Ohm. Let's take 1mm diameter as an example. Then, to obtain the required resistance, it is necessary l = R / ρ = 8.06 / 1.4 = 5.7 m nichrome wire, where ρ - nominal value of electrical resistance of 1 m of wire, [Ohm/m]. The mass of this piece of nichrome wire will be m = l μ = 5.7 · 0.007 = 0.0399 kg = 40 g, where μ - mass of 1 m of wire. Now you need to determine the surface area of ​​a piece of wire 5.7 m long. S = l π d = 570 · 3.14 · 0.1 = 179 cm 2, where l – wire length [cm], d – wire diameter [cm]. Thus, 6 kW should be released from an area of ​​179 cm2. Solving a simple proportion, we find that power is released from 1 cm 2 β = P/S = 6000 / 179 = 33.5 W, where β - surface power of the heater.

The resulting surface power is too high. Heater will melt if heated to a temperature that would provide the resulting surface power value. This temperature will be higher than the melting point of the heater material.

The given example is a demonstration of the incorrect choice of the diameter of the wire that will be used to make the heater. In paragraph 5 of this paragraph an example will be given with the correct selection of diameter.

For each material, depending on the required heating temperature, the permissible value of surface power is determined. It can be determined using special tables or graphs. These calculations use tables.

For high temperature furnaces(at temperatures above 700 - 800 °C) permissible surface power, W/m2, is equal to β additional = β eff · α , Where β eff – surface power of heaters depending on the temperature of the heat-receiving medium [W/m2], α – radiation efficiency coefficient. β eff selected according to table 3, α - according to table 4.

If low temperature oven(temperature less than 200 - 300 °C), then the permissible surface power can be considered equal to (4 - 6) · 10 4 W/m2.

Table 3

Effective specific surface power of heaters depending on the temperature of the heat-receiving medium

Temperature of the heat-receiving surface, °C	β eff, W/cm 2 at heater temperature, °C
	800	850	900	950	1000	1050	1100	1150	1200	1250	1300	1350
100	6,1	7,3	8,7	10,3	12,5	14,15	16,4	19,0	21,8	24,9	28,4	36,3
200	5,9	7,15	8,55	10,15	12,0	14,0	16,25	18,85	21,65	24,75	28,2	36,1
300	5,65	6,85	8,3	9,9	11,7	13,75	16,0	18,6	21,35	24,5	27,9	35,8
400	5,2	6,45	7,85	9,45	11,25	13,3	15,55	18,1	20,9	24,0	27,45	35,4
500	4,5	5,7	7,15	8,8	10,55	12,6	14,85	17,4	20,2	23,3	26,8	34,6
600	3,5	4,7	6,1	7,7	9,5	11,5	13,8	16,4	19,3	22,3	25,7	33,7
700	2	3,2	4,6	6,25	8,05	10,0	12,4	14,9	17,7	20,8	24,3	32,2
800	-	1,25	2,65	4,2	6,05	8,1	10,4	12,9	15,7	18,8	22,3	30,2
850	-	-	1,4	3,0	4,8	6,85	9,1	11,7	14,5	17,6	21,0	29,0
900	-	-	-	1,55	3,4	5,45	7,75	10,3	13	16,2	19,6	27,6
950	-	-	-	-	1,8	3,85	6,15	8,65	11,5	14,5	18,1	26,0
1000	-	-	-	-	-	2,05	4,3	6,85	9,7	12,75	16,25	24,2
1050	-	-	-	-	-	-	2,3	4,8	7,65	10,75	14,25	22,2
1100	-	-	-	-	-	-	-	2,55	5,35	8,5	12,0	19,8
1150	-	-	-	-	-	-	-	-	2,85	5,95	9,4	17,55
1200	-	-	-	-	-	-	-	-	-	3,15	6,55	14,55
1300	-	-	-	-	-	-	-	-	-	-	-	7,95

Table 4

Wire spirals, semi-closed in lining grooves

Wire spirals on shelves in tubes

Wire zigzag (rod) heaters

Let's assume that the heater temperature is 1000 °C, and we want to heat the workpiece to a temperature of 700 °C. Then, according to Table 3, we select β eff = 8.05 W/cm2, α = 0,2, β additional = β eff · α = 8.05 · 0.2 = 1.61 W/cm2 = 1.61 · 10 4 W/m2.

5. After determining the permissible surface power of the heater, it is necessary find its diameter(for wire heaters) or width and thickness(for tape heaters), as well as length.

The wire diameter can be determined using the following formula: d - wire diameter, [m]; P - heater power, [W]; U - voltage at the ends of the heater, [V]; β additional - permissible surface power of the heater, [W/m 2 ]; ρ t - specific resistance of the heater material at a given temperature, [Ohm m].
ρ t = ρ 20 k , Where ρ 20 - specific electrical resistance of the heater material at 20 °C, [Ohm m] k - correction factor for calculating changes in electrical resistance depending on temperature (by ).

The length of the wire can be determined using the following formula:
l - wire length, [m].

Select the diameter and length of the wire from nichrome X20N80. The specific electrical resistance of the heater material is
ρ t = ρ 20 k = 1.13 · 10 -6 · 1.025 = 1.15 · 10 -6 Ohm m.

Single-phase household network
For a 60 liter stove connected to a single-phase household network, it is known from the previous stages of calculation that the stove power is P = 6000 W, voltage at the ends of the heater - U = 220 V, permissible surface heater power β additional = 1.6 · 10 4 W/m2. Then we get

The resulting size must be rounded to the nearest larger standard. Standard sizes for nichrome and fechral wire can be found in, Appendix 2, Table 8. In this case, the nearest larger standard size is Ø 2.8 mm. Heater diameter d = 2.8 mm.

Heater length l = 43 m.

It is also sometimes necessary to determine the mass of the required amount of wire.
m = l μ , Where m - weight of a piece of wire, [kg]; l - wire length, [m]; μ - specific gravity (weight of 1 meter of wire), [kg/m].

In our case, the heater mass m = l μ = 43 · 0.052 = 2.3 kg.

This calculation gives the minimum wire diameter at which it can be used as a heater under given conditions. From the point of view of saving material, this calculation is optimal. In this case, wire of a larger diameter can also be used, but then its quantity will increase.

Examination
Calculation results can be checked in the following way. A wire diameter of 2.8 mm was obtained. Then the length we need will be
l = R / (ρ k) = 8.06 / (0.179 1.025) = 43 m, where l - wire length, [m]; R - heater resistance, [Ohm]; ρ - nominal value of electrical resistance of 1 m of wire, [Ohm/m]; k - correction factor for calculating changes in electrical resistance depending on temperature.
This value is the same as the value obtained from another calculation.

Now we need to check whether the surface power of the heater we have chosen will not exceed the permissible surface power, which was found in step 4. β = P/S = 6000 / (3.14 · 4300 · 0.28) = 1.59 W/cm2. Received value β = 1.59 W/cm 2 does not exceed β additional = 1.6 W/cm2.

Results
Thus, the heater will require 43 meters of X20N80 nichrome wire with a diameter of 2.8 mm, which is 2.3 kg.

Three-phase industrial network
You can also find the diameter and length of the wire required for the manufacture of furnace heaters connected to a three-phase current network.

As described in paragraph 3, each of the three heaters accounts for 2 kW of power. Let's find the diameter, length and mass of one heater.

STAR connection(see Fig. 2)

In this case, the nearest larger standard size is Ø 1.4 mm. Heater diameter d = 1.4 mm.

Single heater length l = 30 m.
Weight of one heater m = l μ = 30 · 0.013 = 0.39 kg.

Examination
A wire diameter of 1.4 mm was obtained. Then the length we need will be
l = R / (ρ k) = 24.2 / (0.714 · 1.025) = 33 m.

β = P/S = 2000 / (3.14 · 3000 · 0.14) = 1.52 W/cm2, it does not exceed the permissible limit.

Results
For three heaters connected in a “STAR” configuration, you will need
l = 3 30 = 90 m of wire, which is
m = 3 · 0.39 = 1.2 kg.

TRIANGLE connection(see Fig. 3)

In this case, the nearest larger standard size is Ø 0.95 mm. Heater diameter d = 0.95 mm.

Single heater length l = 43 m.
Weight of one heater m = l μ = 43 · 0.006 = 0.258 kg.

Examination
A wire diameter of 0.95 mm was obtained. Then the length we need will be
l = R / (ρ k) = 72.2 / (1.55 · 1.025) = 45 m.

This value practically coincides with the value obtained as a result of another calculation.

The surface thickness will be β = P/S = 2000 / (3.14 · 4300 · 0.095) = 1.56 W/cm2, it does not exceed the permissible limit.

Results
For three heaters connected in a “TRIANGLE” pattern, you will need
l = 3 43 = 129 m of wire, which is
m = 3 · 0.258 = 0.8 kg.

If you compare the 2 options for connecting heaters to a three-phase current network discussed above, you will notice that “STAR” requires a larger diameter wire than “TRIANGLE” (1.4 mm versus 0.95 mm) to ensure a given oven power of 6 kW. Wherein the required length of nichrome wire when connecting according to the “STAR” scheme is less than the length of the wire when connecting according to the “TRIANGLE” type(90 m versus 129 m), and the required mass, on the contrary, is greater (1.2 kg vs. 0.8 kg).

Spiral calculation

During operation, the main task is to place the heater of the calculated length in the limited space of the furnace. Nichrome and fechral wire are wound in the form of spirals or bent in the form of zigzags, the tape is bent in the form of zigzags, which allows you to fit a larger amount of material (along the length) into the working chamber. The most common option is the spiral.

The relationship between the pitch of the spiral and its diameter and the diameter of the wire is chosen in such a way as to facilitate the placement of heaters in the furnace, ensure their sufficient rigidity, eliminate local overheating of the turns of the spiral itself to the maximum extent possible, and at the same time do not impede the heat transfer from them to the products.

The larger the diameter of the spiral and the smaller its pitch, the easier it is to place heaters in the furnace, but as the diameter increases, the strength of the spiral decreases and the tendency of its turns to lie on top of each other increases. On the other hand, with an increase in the winding frequency, the shielding effect of the part of its turns facing the products on the rest increases and, consequently, the use of its surface deteriorates, and local overheating may also occur.

Practice has established well-defined, recommended relationships between the diameter of the wire ( d ), step ( t ) and the diameter of the spiral ( D ) for wire Ø from 3 to 7 mm. These ratios are as follows: t ≥ 2d And D = (7÷10) d for nichrome and D = (4÷6) d - for less durable iron-chromium-aluminum alloys, such as fechral, ​​etc. For thinner wires the ratio D And d , and t usually take more.

Conclusion

The article discussed various aspects related to calculation of electric furnace heaters- materials, calculation examples with the necessary reference data, links to standards, illustrations.

In the examples, calculation methods were considered only wire heaters. In addition to wire made of precision alloys, tape can also be used to make heaters.

The calculation of heaters is not limited to the choice of their sizes. Also it is necessary to determine the material from which the heater should be made, the type of heater (wire or tape), the type of location of the heaters and other features. If the heater is made in the form of a spiral, then it is necessary to determine the number of turns and the pitch between them.

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Bibliography

• Dyakov V.I. "Typical calculations for electrical equipment".
• Zhukov L.L., Plemyannikova I.M., Mironova M.N., Barkaya D.S., Shumkov Yu.V. "Alloys for heaters".
• Sokunov B.A., Grobova L.S. "Electrothermal installations (electric resistance furnaces)".
• Feldman I.A., Gutman M.B., Rubin G.K., Shadrich N.I. "Calculation and design of resistance electric furnace heaters".
• http://www.horss.ru/h6.php?p=45
• http://www.electromonter.info/advice/nichrom.html

When completing this task you should:

2. Analyze the left column and realize what the given quantities characterize (body property, interaction, state, change of state, etc.). In this example, the given values ​​characterize the state of the body and their change is associated with a change in state.

3. Analyze the process described in the condition and compare the nature of their change in this process with physical quantities.

4. Write down the numbers of the selected elements in the right column in the table.

Tasks for independent work

147. The lead ball is cooled in the refrigerator. How does the internal energy of the ball, its mass and the density of the substance of the ball change?

For each physical quantity, determine the corresponding nature of change.

1) increased

2) decreased

3) has not changed

Write down the selected numbers in the table under the corresponding letters.

The numbers in the answer may be repeated.

INTERNAL ENERGY	DENSITY OF MATTER