James Prescott Joule (left) and Emil Khristianovich Lenz (right)
Electric heaters of various types have been used by mankind for centuries, due to the property electric current generate heat as it passes through a conductor. This phenomenon also has a negative factor - overheated electrical wiring due to too much current often became the cause of a short circuit and fires. The release of heat from the work of electric current was studied in the school physics course, but many have forgotten this knowledge.
For the first time, the dependence of heat release on the strength of the electric current was formulated and mathematically determined by James Joule in 1841, and a little later, in 1842, independently of him, by Emil Lenz. In honor of these physicists, the Joule-Lenz law was named, according to which the power of electric heaters and losses due to heat generation in power lines are calculated.
Definition of the Joule–Lenz law
In a verbal definition, according to the studies of Joule and Lenz, the law sounds like this:
The amount of heat released in a certain volume of the conductor during the flow of electric current is directly proportional to the multiplication of the density of the electric current and the magnitude of the electric field strength
In the form of a formula, this law looks like this:
Expression of the Joule-Lenz law Since the parameters described above are rarely used in everyday life, and given that almost all household calculations for the release of heat from the operation of electric current relate to thin conductors (cables, wires, filaments, power cords, conductive tracks on the board, etc.), use the Joule Lenz law with the formula presented in integral form:
Integral form of law In a verbal definition, Joule Lenz's law sounds like this:
Verbal definition of the Joule-Lenz law If we accept that the current strength and the resistance of the conductor do not change over time, then the Joule-Lenz law can be written in a simplified form:
Applying Ohm's law and algebraic transformations, we obtain the following equivalent formulas:
Equivalent expressions for heat according to Ohm's law Application and practical significance of the Joule–Lenz law
The studies of Joule and Lenz in the field of heat release from the work of electric current have significantly advanced the scientific understanding of physical processes, and the derived basic formulas have not changed and are still used in various branches of science and technology. In the field of electrical engineering, several technical tasks can be distinguished, where the amount of heat released during the flow of current is critical meaning when calculating the following parameters:
- heat losses in power lines;
- characteristics of wires of electrical wiring networks;
- thermal power (amount of heat) of electric heaters;
- operating temperature of circuit breakers;
- fuse melting point;
- heat dissipation of various electrical devices and elements of radio engineering.
Electrical appliances that use the thermal work of current The thermal effect of electric current in the wires of power lines (TL) is undesirable due to significant losses of electricity for heat generation.
According to various data, up to 40% of all electricity produced in the world is lost in power lines. To reduce losses during the transmission of electricity over long distances, they increase the voltage in power lines, making calculations according to the derivative formulas of the Joule-Lenz law.
Diagram of all possible losses of electricity, among which heat losses on overhead lines make up the lion's share (64%) Very simply, the thermal work of the current can be described as follows: electrons move between molecules, and from time to time collide with them, which makes their thermal vibrations become more intense. A visual demonstration of the thermal work of the current and associative explanations of the processes are shown in the video below:
Calculations of electricity losses in power lines
As an example, we can take a hypothetical section of a power line from a power plant to a transformer substation. Since the wires of the power line and the consumer of electricity (transformer substation) are connected successively, then the same current I flows through them. According to the Joule-Lenz law considered here, the amount of heat Q w (heat loss) released on the wires is calculated by the formula:
The power produced by electric current (Q c) in the load is calculated according to Ohm's law:
Thus, if the currents are equal, instead of I, the expression Q c / U c can be inserted into the first formula, since I = Q c / U c:
If we ignore the dependence of the resistance of conductors on temperature changes, then we can consider R w unchanged (constant). Thus, with a stable energy consumption of the consumer (transformer substation), the heat release in the wires of the power transmission line will be inversely voltage squared in end point lines. In other words, than more tension power transmission, the less power loss.
For power transmission high voltage large power pylons required The work of the Joule-Lenz law in everyday life
These calculations are also valid in everyday life when transmitting electricity over short distances - for example, from a wind generator to an inverter. With an autonomous power supply, every watt of energy generated by a low-voltage windmill is valued, and it may be more profitable to raise the voltage with a transformer right at the wind generator than to spend money on a large cable section in order to reduce power losses during transmission.
With a significant removal of a low-voltage wind generator alternating current to reduce power losses more profitable connection through a step-up transformer In household electrical wiring networks, the distances are extremely small in order to reduce heat losses by raising the voltage, therefore, when calculating the wiring, the thermal work of the current is taken into account, according to the Joule-Lenz law, when choosing the cross section of the wires so that they thermal heating did not lead to melting and ignition of insulation and surrounding materials. The choice of cable for power and wiring is carried out in accordance with the tables and regulatory documents of the PUE, and are described in detail on other pages of this resource.
Ratios of current strength and cross-section of conductors When calculating the heating temperature of radio engineering elements, a bimetallic circuit breaker or a fuse, the Joule-Lenz law is used in integral form, since the resistance of these materials changes with increasing temperature. These complex calculations also take into account heat transfer, heating from other heat sources, intrinsic heat capacity and many other factors.
Software simulation of heat dissipation of a semiconductor device Useful thermal work of electric current
The heat-generating work of an electric current is widely used in electric heaters, which use a series connection of conductors with different resistances. This principle works as follows: the same current flows in series-connected conductors, which means, according to the Joule-Lenz law, more heat will be released from the conductor material with high resistance.
The coil with increased resistance heats up, but the supply wires remain cold In this way, the power cord and lead wires of the hot plate remain relatively cold while the heating element heats up to the red glow temperature. As a material for conductors of heating elements, alloys with increased (relative to copper and aluminum electrical wiring) specific resistance are used - nichrome, constantan, tungsten and others.
The filament of an incandescent lamp is made from refractory tungsten alloys. When the conductors are connected in parallel, the heat dissipation will be more by heating element with less resistance, since when it decreases, the current of the relative neighboring component of the circuit increases. As an example, we can give an obvious example of the glow of two incandescent bulbs of different power - a more powerful lamp has more heat and luminous flux.
If you ring a light bulb with an ohmmeter, it turns out that a more powerful lamp has less resistance. In the video below, the author demonstrates a serial and parallel connection, but unfortunately, he made a mistake in the comment - the lamp will shine brighter with big resistance, and not vice versa.
The Joule-Lenz law is a law of physics that determines the quantitative measure of the thermal effect of an electric current. This law was formulated in 1841 by the English scientist D. Joule and completely separately from him in 1842 by the famous Russian physicist E. Lenz. Therefore, he received his double name - the Joule-Lenz law.
Law definition and formula
The verbal formulation is as follows: the power of heat released in the conductor when flowing through it is proportional to the product of the electric field density value and the strength value.
Mathematically, the Joule-Lenz law is expressed as follows:
ω = j E = ϭ E²,
where ω is the amount of heat released in units. volume;
E and j are the strength and density, respectively, of the electric fields;
σ is the conductivity of the medium. 
The physical meaning of the Joule-Lenz law
The law can be explained as follows: the current flowing through the conductor is a displacement electric charge under influence . Thus, the electric field does some work. This work is spent on heating the conductor.
In other words, energy transforms into its other quality - heat.
But excessive heating of conductors with current and electrical equipment must not be allowed, as this can lead to their damage. Severe overheating is dangerous with wires, when sufficiently large currents can flow through the conductors. 
In integral form for thin conductors Joule-Lenz law sounds like this: the amount of heat that is released per unit time in the section of the circuit under consideration is defined as the product of the square of the current strength and the resistance of the section.
Mathematically, this formulation is expressed as follows:
Q = ∫ k I² R t,
in this case, Q is the amount of released heat;
I is the current value;
R is the active resistance of the conductors;
t is the exposure time.
The value of the parameter k is usually called the thermal equivalent of work. The value of this parameter is determined depending on the digit capacity of the units in which the measurements of the values used in the formula are performed.
The Joule-Lenz law is quite general, since it does not depend on the nature of the forces that generate the current.
From practice, it can be argued that it is valid for both electrolytes and conductors and semiconductors.
Application area
Areas of application in everyday life of the law of Joule Lenz - great amount. For example, a tungsten filament in an incandescent lamp, an arc in electric welding, a heating filament in an electric heater, and more. etc. This is the most widely accepted physical law in everyday life.
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The amount of heat released per unit time in the section of the circuit under consideration is proportional to the product of the square of the current strength in this section and the resistance of the section
Joule Lenz's law in integral form in thin wires:
If the current strength changes with time, the conductor is stationary and there are no chemical transformations in it, then heat is released in the conductor.
- The power of heat released per unit volume of the medium during the flow of electric current is proportional to the product of the density of the electric current and the magnitude of the electric field
The conversion of electrical energy into thermal energy is widely used in electric furnaces and various electric heaters. The same effect in electrical machines and devices leads to involuntary energy costs (energy loss and reduced efficiency). Heat, by causing these devices to heat up, limits their load; In the event of an overload, an increase in temperature can damage the insulation or shorten the service life of the installation.
In the formula we used:
Quantity of heat
Current work
Conductor voltage
Current in the conductor
Time interval
Romanova_1 / kursachi / Romanov's kursovik / EXAMPLE / 13 Thermal calculation. Thermal power on the resistor formula
Task on the topic "Laws direct current". The task may be of interest to 10th grade students and graduates to prepare for the exam. By the way, this kind of task was on the exam in part 1 with a slightly different question (it was necessary to find the ratio of the amounts of heat released on the resistors).
On which of the resistors will the greatest (least) amount of heat be released? R1 = R4 = 4 ohms, R2 = 3 ohms, R3 = 2 ohms. Give a solution. To answer the question of the problem, it is necessary to compare the amount of heat released on each of their resistors. To do this, we use the formula of the Joule-Lenz law. That is, the main task will be to determine the current strength (or comparison) flowing through each resistor.
According to the laws of series connection, the current flowing through the resistors R1 and R2, and R3 and R4, is the same. 
To determine the current strength in the upper and lower branches, we use the law of parallel connection, according to which the voltage on these branches is the same. By writing the voltage on the lower and upper branches according to Ohm's law for a circuit section, we have: 
Substituting the numerical values of the resistances of the resistors, we obtain: That is, we obtain the ratio between the currents flowing in the upper and lower branches: Having determined the current strength through each of these resistors, we determine the amount of heat released on each of the resistors. 
Comparing the numerical coefficients, we conclude that maximum amount heat will be released on the fourth resistor, and the minimum amount of heat - on the second.
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Thermal power - calculation formula
Owners of private houses, apartments or any other objects have to deal with heat engineering calculations. This is the foundation of building design.
Understanding the essence of these calculations in official papers is not as difficult as it seems.
For yourself, you can also learn how to perform calculations in order to decide which insulation to use, how thick it should be, how much power the boiler should acquire, and whether there are enough existing radiators for a given area.
The answers to these and many other questions can be found if you understand what thermal power is. Formula, definition and scope - read the article.
What is thermal calculation?
Simply put, a thermal calculation helps you know exactly how much heat a building stores and loses, and how much energy heating needs to generate to keep a home comfortable.
When evaluating heat loss and the degree of heat supply, the following factors are taken into account:
- What kind of object is it: how many floors it has, the presence of corner rooms, whether it is residential or industrial, etc.
- How many people will "live" in the building.
- An important detail is the area of glazing. And the dimensions of the roof, walls, floor, doors, ceiling height, etc.
- What is the duration of the heating season, the climatic characteristics of the region.
- According to SNiPs, the temperature standards that should be in the premises are determined.
- The thickness of walls, ceilings, selected heat insulators and their properties.
Other conditions and features can be taken into account, for example, working days and weekends, the power and type of ventilation, the orientation of housing to the cardinal points, etc. are considered for production facilities.
What is a thermal calculation for?
How did the builders of the past manage to do without thermal calculations?
The surviving merchant houses show that everything was done simply with a margin: the windows are smaller, the walls are thicker. It turned out warm, but economically unprofitable.
Thermal engineering calculation allows you to build the most optimal. Materials are taken no more - no less, but exactly as much as needed. The dimensions of the building and the cost of its construction are reduced.
The calculation of the dew point allows you to build so that the materials do not deteriorate for as long as possible.
To determine the required power of the boiler, one cannot do without calculations. Its total power consists of energy costs for heating rooms, heating hot water for household needs, and the ability to block heat losses from ventilation and air conditioning. The power reserve is added for the period of peak cold weather.
When gasifying an object, coordination with the services is required. The annual gas consumption for heating and the total capacity of heat sources in gigacalories are calculated.
Calculations are needed when selecting elements of the heating system. The system of pipes and radiators is calculated - you can find out what their length, surface area should be. The power loss is taken into account when the pipeline turns, at the joints and the passage of valves.
When calculating the cost of thermal energy, knowledge of how to convert Gcal to kW and vice versa can be useful. The following article discusses this topic in detail with calculation examples.
A complete calculation of a warm water floor is given in this example.
Did you know that the number of sections of heating radiators is not taken "from the ceiling"? Too little of them will lead to the fact that the house will be cold, and too much of them will create heat and lead to excessive dryness of the air. The link http://microklimat.pro/sistemy-otopleniya/raschet-sistem-otopleniya/kolichestva-sekcij-radiatorov.html provides examples of the correct calculation of radiators.
Thermal power calculation: formula
Consider the formula and give examples of how to calculate for buildings with different coefficient scattering.
Vx(delta)TxK= kcal/h (heat output), where:
- The first indicator "V" is the volume of the calculated room;
- Delta "T" - the temperature difference - this is the value that shows how many degrees inside the room are warmer than outside;
- "K" is the dissipation coefficient (it is also called the "heat transmission coefficient"). The value is taken from the table. Usually the figure ranges from 4 to 0.6.
Approximate values of the dissipation factor for a simplified calculation
- If it is an uninsulated metal profile or a board, then “K” will be = 3 - 4 units.
- Single brickwork and minimal insulation - "K" \u003d from 2 to 3-ex.
- Two-brick wall, standard ceiling, windows and
- doors - "K" \u003d from 1 to 2.
- The warmest option. Double-glazed windows, brick walls with double insulation, etc. - "K" \u003d 0.6 - 0.9.
A more accurate calculation can be made by calculating the exact dimensions of the surfaces of the house that differ in properties in m2 (windows, doors, etc.), making a calculation for them separately and adding up the resulting indicators.
Example of heat output calculation
Let's take a certain room of 80 m2 with a ceiling height of 2.5 m and calculate how much boiler power we need to heat it.
First, we calculate the cubic capacity: 80 x 2.5 = 200 m3. Our house is insulated, but not enough - the dispersion coefficient is 1.2.
Frosts are up to -40 ° C, and in the room you want to have a comfortable +22 degrees, the temperature difference (delta "T") is 62 ° C.
We substitute the numbers in the formula for the power of heat losses and multiply:
200 x 62 x 1.2 \u003d 14880 kcal / h.
The resulting kilocalories are converted to kilowatts using the converter:
- 1 kW = 860 kcal;
- 14880 kcal = 17302.3 W.
We round up with a margin, and we understand that in the most severe frost of -40 degrees, we will need 18 kW of energy per hour.
Multiply the perimeter of the house by the height of the walls:
(8 + 10) x 2 x 2.5 = 90 m2 of wall surface + 80 m2 of ceiling = 170 m2 of surface in contact with cold. The heat losses calculated by us above amounted to 18 kW / h, dividing the surface of the house by the estimated energy consumed, we get that 1 m2 loses about 0.1 kW or 100 W hourly at an outdoor temperature of -40 ° C, and +22 ° C indoors .
These data can become the basis for calculating the required thickness of insulation on the walls.
Let's give another example of calculation, it is more complicated in some moments, but more accurate.
Formula:
Q = S x (delta)T / R:
- Q is the desired value of heat loss at home in W;
- S is the area of cooling surfaces in m2;
- T is the temperature difference in degrees Celsius;
- R is the thermal resistance of the material (m2 x K / W) (Square meters multiplied by Kelvin and divided by Watt).
So, to find "Q" of the same house as in the example above, let's calculate the area of its surfaces "S" (we will not count the floor and windows).
- "S" in our case = 170 m2, of which 80 m2 are ceilings and 90 m2 are walls;
- T = 62 °С;
- R is thermal resistance.
We are looking for "R" according to the table of thermal resistances or according to the formula. The formula for calculating the thermal conductivity coefficient is as follows:
R= H/ K.T. (H is the thickness of the material in meters, K.T. is the coefficient of thermal conductivity).
In this case, our house has walls in two bricks sheathed with foam plastic 10 cm thick. The ceiling is covered with sawdust 30 cm thick.
The heating system of a private house must be arranged taking into account the savings in energy costs. Calculation of the heating system of a private house, as well as recommendations for choosing boilers and radiators - read carefully.
How and how to insulate a wooden house from the inside, you will learn by reading this information. The choice of insulation and insulation technology.
From the table of thermal conductivity coefficients (measured by W / (m2 x K) Watt divided by the product of a square meter by Kelvin). We find the values for each material, they will be:
- brick - 0.67;
- polystyrene - 0.037;
- sawdust - 0.065.
- R (ceiling 30 cm thick) = 0.3 / 0.065 = 4.6 (m2 x K) / W;
- R (brick wall 50 cm) = 0.5 / 0.67 = 0.7 (m2 x K) / W;
- R (foam 10 cm) \u003d 0.1 / 0.037 \u003d 2.7 (m2 x K) / W;
- R (wall) \u003d R (brick) + R (foam) \u003d 0.7 + 2.7 \u003d 3.4 (m2 x K) / W.
Now we can proceed to the calculation of heat loss "Q":
- Q for the ceiling \u003d 80 x 62 / 4.6 \u003d 1078.2 watts.
- Q walls \u003d 90 x 62 / 3.4 \u003d 1641.1 watts.
- It remains to add 1078.2 + 1641.1 and convert to kW, it turns out (if rounded up immediately) 2.7 kW of energy in 1 hour.
It's all about the degree of fatigue of the houses (although, of course, the data could be different if we calculated the floor and windows).
Conclusion
The above formulas and examples show that in heat engineering calculations it is very important to take into account as many factors as possible that affect heat loss. This includes ventilation, and the area of the windows, the degree of their fatigue, etc.
And the approach, when 1 kW of boiler power is taken per 10 m2 of a house, is too approximate to seriously rely on it.
Related video
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13 Thermal calculation
10. Thermal calculation.
The design of the IC should be such that the heat released during its operation does not lead, under the most unfavorable operating conditions, to element failures as a result of overheating. The main fuel elements include, first of all, resistors, active elements and components. The power dissipated by capacitors and inductors is small. Film switching IC, due to the small electrical resistance and high thermal conductivity of metal films, contributes to the removal of heat from the most heated elements and equalizes the temperature of the GIS board and the semiconductor IC chip.
Rice. 10.1. Option for attaching the board to the case.
Thermal calculation of resistors.
The thermal resistance of the resistor is calculated by the formula (10.1)
p = 0.03 [W/cm °C] - coefficient of thermal conductivity of the substrate material;
δp = 0.06 cm is the board thickness.
RT=0.06/0.03=2 cm2∙°С/W
Calculate the temperature of film resistors using the formula
PR is the power dissipated in the resistor;
SR is the area occupied by the resistor on the board;
P0 is the total power allocated by all components of the microcircuit;
Sp is the area of the board.
PR = 0.43 mW is the power dissipated in the resistor;
SR = 0.426mm2 - the area occupied by the resistor;
Sn = 80 mm2 - board area;
RT = 2 cm2∙°С/W is the thermal resistance of the resistor;
Tacr.av = 40С – maximum ambient temperature;
T = 125°C = maximum allowable temperature of film resistors.
TR=(0.43∙10-3∙200)/0.426+(24.82∙10-3∙200)/80+40=40.26 С<125 С
The temperature of the remaining resistors is calculated similarly using the MathCad program. The calculation results are presented in Table 10.1
Table. 10.1
| Resistor | |||||
The table shows that for all film resistors the specified thermal regime is observed.
Thermal calculation for the hinged element.
Thermal resistance will be calculated by the formula:
k = 0.003 [W/cm °C] - thermal conductivity of the adhesive;
δc1 = 0.01 cm is the thickness of the adhesive.
Rt=(0.06/0.03)+(0.01/0.003)=5.33 cm2∙°С/W
Calculate the temperature of the hinged element using the formula:
Calculation of transistor KT202A, VT14
Pne \u003d 2.6 mW - power released on the transistor;
Sne = 0.49 mm2 - the area occupied by the transistor;
P0 = 24.82 mW – power dissipated by all board components;
Sn = 80 mm2 - board area;
Т0С = 40С – maximum ambient temperature;
T = 85°C = maximum allowable transistor temperature.
Tne=(2.6∙10-3∙533)/0.49+(24.82∙10-3∙533)/80+40=42.99С<85С
Therefore, the specified thermal regime is observed.
The temperature of the remaining transistors is calculated similarly using the MathCad program. The calculation results are presented in Table 10.2
Table 10.2
| Transistor | ||||||||
The table shows that for all transistors the specified thermal regime is observed. Consequently, the thermal conditions for the entire circuit are satisfied.
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Thermal power of electric current and its practical application
The reason for heating the conductor lies in the fact that the energy of the electrons moving in it (in other words, the energy of the current) is converted into a warm type of energy, or Q, in the sequential collision of particles with ions of the molecular lattice of a metal element, this is how the concept of "thermal power" is formed.
The work of the current is measured using the international system of units SI, applying joules (J) to it, the current power is defined as "watt" (W). Deviating from the system in practice, they can also use off-system units that measure the work of the current. Among them are watt-hour (W × h), kilowatt-hour (abbreviated kW × h). For example, 1 Wh denotes the work of a current with a specific power of 1 watt and a time duration of one hour. 
If electrons move along a fixed metal conductor, in this case all the useful work of the generated current is distributed to heat the metal structure, and, based on the provisions of the law of conservation of energy, this can be described by the formula Q=A=IUt=I2Rt=(U2/R)* t. Such ratios accurately express the well-known Joule-Lenz law. Historically, it was first determined empirically by the scientist D. Joule in the middle of the 19th century, and at the same time, independently of him, by another scientist - E. Lenz. Thermal power has found practical application in technical design since the invention in 1873 by the Russian engineer A. Ladygin of an ordinary incandescent lamp. 
The thermal power of the current is used in a number of electrical appliances and industrial installations, namely, in thermal measuring instruments, heating-type electric stoves, electric welding and inventory equipment, household appliances with an electric heating effect are very common - boilers, soldering irons, kettles, irons.
The thermal effect also finds itself in the food industry. With a high share of use, the possibility of electrocontact heating is used, which guarantees thermal power. It is due to the fact that the current and its thermal power, influencing the food product, which has a certain degree of resistance, causes uniform heating in it. We can give an example of how sausages are produced: through a special dispenser, minced meat enters metal molds, the walls of which simultaneously serve as electrodes. Here, a constant uniformity of heating is ensured over the entire area and volume of the product, the set temperature is maintained, the optimal biological value of the food product is maintained, together with these factors, the duration of technological work and energy consumption remain the smallest. 
The specific thermal power of the electric current (ω), in other words, the amount of heat that is released per unit volume for a certain unit of time, is calculated as follows. An elementary cylindrical volume of a conductor (dV), with a cross section of the conductor dS, a length dl parallel to the current direction, and a resistance make up the equations R=p(dl/dS), dV=dSdl.
According to the definitions of the Joule-Lenz law, for the allotted time (dt) in the volume taken by us, a heat level equal to dQ=I2Rdt=p(dl/dS)(jdS)2dt=pj2dVdt will be released. In this case, ω=(dQ)/(dVdt)=pj2 and, applying here Ohm's law to establish the current density j=γE and the relation p=1/γ, we immediately obtain the expression ω=jE= γE2. It in differential form gives the concept of the Joule-Lenz law.
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Embedder page » Thermal calculations
All electronic components generate heat, so the ability to calculate heatsinks so as not to fly by a couple of orders of magnitude is very useful for any electronics engineer.
Thermal calculations are very simple and have a lot in common with the calculations of electronic circuits. Here, look at a typical thermal design problem I just ran into.
A task
You need to choose a radiator for a 5-volt linear regulator, which is powered by 12 volts maximum and produces 0.5A. The maximum allocated power is (12-5) * 0.5 = 3.5W
Dive into theory
In order not to produce entities, people scratched a pumpkin and realized that heat is very similar to electric current, and for thermal calculations you can use the usual Ohm's law, only
Voltage (U) is replaced by temperature (T)
Current (I) replaced by power (P)
Resistance is replaced by thermal resistance. Conventional resistance is measured in Volts/Amps, while thermal resistance is measured in °C/Watt.
As a result, Ohm's law is replaced by its thermal counterpart:
A small note - in order to indicate that thermal (rather than electrical) resistance is meant, the letter theta is added to the letter R: 
I don’t have such a letter on the keyboard, and I’m too lazy to copy from the symbol table, so I’ll just use the letter R.
We continue
Heat is generated in the stabilizer crystal, and our goal is to prevent it from overheating (to prevent overheating of the crystal, not the case, this is important!).
To what temperature the crystal can be heated, it is written in the datasheet:
Usually, the limiting temperature of the crystal is called Tj (j = junction = junction - the temperature-sensitive insides of microcircuits mainly consist of pn junctions. We can assume that the temperature of the junctions is equal to the temperature of the crystal)
without heatsink
The thermal diagram looks very simple: 
Especially for cases of using a case without a radiator, the datasheets write the thermal resistance of the crystal-atmosphere (Rj-a) (you already know what j is, a = ambient = environment)
Note that the temperature of the "ground" is not zero, but equal to the temperature of the surrounding air (Ta). The air temperature depends on the conditions under which the radiator is located. If it is in the open air, then Ta = 40 ° C can be set, but if it is in a closed box, then the temperature can be much higher!
We write Ohm's thermal law: Tj = P*Rj-a + Ta. We substitute P = 3.5, Rj-a = 65, we get Tj = 227.5 + 40 = 267.5 °C. Too much, though!
We cling to the radiator
The thermal scheme of our example with a stabilizer on a radiator becomes like this:
- Rj-c is the resistance from die to case heatsink (c = case = case). Given in the datasheet. In our case - 5 °C / W - from the datasheet
- Rc-r - body-radiator resistance. It's not all that simple. This resistance depends on what is between the case and the heatsink. For example, a silicone gasket has a thermal conductivity coefficient of 1-2 W/(m*°C), and a KPT-8 paste has 0.75 W/(m*°C). Thermal resistance can be obtained from the thermal conductivity coefficient by the formula:
R = gasket thickness / (thermal conductivity * area of one side of the gasket)
Often Rc-r can be ignored altogether. For example, in our case (we use the TO220 case, with KPT-8 paste, the average depth of the paste taken from the ceiling is 0.05mm). Total, Rc-r = 0.5 °C/W. With a power of 3.5W, the temperature difference between the stabilizer body and the radiator is 1.75 degrees. That's not a lot. For our example, let's take Rc-r = 2 °C/W
Rr-a - thermal resistance between the radiator and the atmosphere. It is determined by the geometry of the radiator, the presence of airflow, and a bunch of other factors. This parameter is much easier to measure than to calculate (see at the end of the article). For example - Rr-c = 12.5 °C/W
Ta = 40°C - here we figured that the atmospheric temperature is rarely higher, you can take 50 degrees to be sure.
We substitute all these data into Ohm's law, and we get Tj = 3.5*(5+2+12.5) + 40 = 108.25 °C
This is significantly less than the limit of 150 °C. Such a radiator can be used. In this case, the radiator case will heat up to Tc = 3.5*12.5 + 40 = 83.75 °C. This temperature is already able to soften some plastics, so you need to be careful.
Measurement of radiator-atmosphere resistance.
Most likely, you already have a bunch of radiators lying around that can be used. Thermal resistance is very easy to measure. It needs resistance and power supply.
We sculpt the resistance on the radiator using thermal paste:
We connect the power source, and set the voltage so that some power is released on the resistance. It is better, of course, to heat the radiator with the power that it will dissipate in the final device (and in the position in which it will be, this is important!). I usually leave this design for half an hour so that it warms up well.
Once the temperature has been measured, the thermal resistance can be calculated.
Rr-a = (T-Ta)/P. For example, my radiator has warmed up to 81 degrees, and the air temperature is 31 degrees. thus Rr-a = 50/4 = 12.5 °C/W.
Estimated area of the radiator
In the ancient handbook of a radio amateur, a graph was given, according to which you can estimate the area of \u200b\u200bthe radiator. Here he is:
It is very easy to work with him. We select the overheating that we want to get and see what area corresponds to the required power with such overheating.
For example, with a power of 4W and overheating of 20 degrees, you will need 250 cm ^ 2 radiators. This graph gives an overestimation of the area, and does not take into account a bunch of factors such as forced airflow, fin geometry, etc.
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In 1841 and 1842, independently of each other, English and Russian physicists established the dependence of the amount of heat on the flow of current in a conductor. This relationship is called the "Joule-Lenz Law". The Englishman established dependence a year earlier than the Russian, but the law got its name from the names of both scientists, because their research was independent. The law is not theoretical in nature, but has great practical significance. And so let's briefly and clearly find out the definition of the Joule-Lenz law and where it is applied.
Wording
In a real conductor, when current flows through it, work is performed against friction forces. The electrons move through the wire and collide with other electrons, atoms and other particles. As a result, heat is released. The Joule-Lenz law describes the amount of heat generated when current flows through a conductor. It is directly proportional to the current strength, resistance and flow time.
In integral form, the Joule-Lenz Law looks like this:
The current strength is indicated by the letter I and is expressed in Amps, the Resistance is R in Ohms, and the time t is in seconds. The unit of measure for heat Q is the Joule, to convert to calories you need to multiply the result by 0.24. In this case, 1 calorie is equal to the amount of heat that must be brought to pure water in order to increase its temperature by 1 degree.
Such a formula is valid for a section of the circuit when the conductors are connected in series, when one current flows in them, but a different voltage drops at the ends. The product of current squared and resistance equals power. At the same time, power is directly proportional to the square of the voltage and inversely proportional to the resistance. Then for an electric circuit with a parallel connection, the Joule-Lenz law can be written as:
In differential form, it looks like this:
Where j is the current density A / cm 2, E is the electric field strength, sigma is the resistivity of the conductor.
It should be noted that for a homogeneous section of the circuit, the resistance of the elements will be the same. If there are conductors with different resistances in the circuit, a situation arises when the maximum amount of heat is released on the one that has the highest resistance, which can be concluded by analyzing the formula of the Joule-Lenz Law.
FAQ
How to find time? This refers to the period of current flow through the conductor, that is, when the circuit is closed.
How to find the resistance of a conductor? To determine the resistance, a formula is used, which is often called "rail", that is:
Here, the letter "Ro" denotes resistivity, it is measured in Ohm * m / cm2, l and S are the length and cross-sectional area. In calculations, square meters and centimeters are reduced and ohms remain.
Resistivity is a tabular value and it has its own for each metal. Copper has orders of magnitude less than high-resistance alloys such as tungsten or nichrome. For what it is used, we will consider below.
Let's move on to practice
The Joule-Lenz law is of great importance for electrical calculations. First of all, you can apply it when calculating heating devices. A conductor is most often used as a heating element, but not a simple one (like copper), but with a high resistance. Most often it is nichrome or kanthal, fechral.
They have high resistivity. You can also use copper, but then you will waste a lot of cable (sarcasm, copper is not used for this purpose). To calculate the heat power for a heating device, you need to determine which body and in what volumes you need to heat, take into account the amount of heat required and how long it takes to transfer it to the body. After calculations and transformations, you will get the resistance and current in this circuit. Based on the obtained data on resistivity, select the conductor material, its cross section and length.