Joule's law describes how electrical energy is converted into heat when current flows through a resistance. The heat generated, Q = I² · R · t, grows with the square of the current I, with the resistance R and with the time t, and the corresponding heating power is P = I² · R (Wikipedia, Joule heating). James Prescott Joule published the relationship in 1840 and had demonstrated the proportionality to I² by 1842, and Heinrich Lenz found it independently, which is why the law is also called the Joule-Lenz law or, in German, the Stromwärmegesetz. This article derives the formula, explains the symbols and units, works through a house cable, a heating wire and a motor winding with real material data, and shows why the resistive heat of the winding determines the thermal class of the insulation materials that GOBA supplies for it.
From electrical work to resistive heat: the derivation in three lines
If a current I flows through a resistance R at a voltage U for a time t, the current performs the electrical work W = U · I · t. In an ohmic resistance all of that work ends up as heat, because there is no motor to turn it into motion and no battery to turn it into chemistry. Ohm's law U = R · I replaces the voltage, and U · I · t becomes I² · R · t. Replacing the current instead gives the second form U² / R · t. Both describe the same amount of heat: the first works with the current through the component, the second with the voltage across it.
The factor I² is the core of the law. The current appears twice in the formula: once as the number of charges passing through the conductor every second, and once through the voltage drop R · I that each of those charges has to overcome. Double the current therefore means four times the heat, half the current a quarter. Microscopically, the accelerated electrons collide with the atomic cores of the lattice and hand over their kinetic energy as lattice vibration. In a superconductor with R = 0 no resistive heat is generated at all (Wikipedia, Stromwärmegesetz).
The terminology follows the purpose. Where the heat is wanted, as in a heating wire, it is called Joule heat or resistive heat. Where it is unwanted, as in a cable, a transformer or a motor winding, the same heat is called resistive loss, ohmic loss or copper loss. Joule heating in the narrow sense is the temperature rise this heat produces in a component with heat capacity C: Δϑ = Q / C, as long as no heat flows away. Losses that arise inside the insulating material itself through polarisation are a different matter and are covered in the article on dielectric losses.
Symbols and units: joule, watt and kilowatt hour
The unit of the heat Q is the joule, named after the same physicist. One joule is one watt second, the energy that a power of one watt converts in one second, and at the same time one newton metre. For energy bills the kilowatt hour is more practical: 1 kWh is 1,000 W for 3,600 s, which is 3.6 million joules. The table summarises the formulas that are enough for any calculation of resistive heat.
| Quantity | Symbol | Unit | Formula |
|---|---|---|---|
| Heat, energy | Q, W | J = W · s, 1 kWh = 3.6 MJ | Q = I² · R · t |
| Heating power | P | W | P = I² · R = U · I = U² / R |
| Current | I | A | I = U / R |
| Resistance of a conductor | R | Ω | R = ρ · l / A |
| Resistivity | ρ | Ω · mm²/m | copper 0.0175, aluminium 0.0278 (HTW Dresden) |
| Resistance at temperature ϑ | R(ϑ) | Ω | R(ϑ) = R20 · (1 + α · (ϑ − 20 °C)) |
| Temperature rise without heat loss | Δϑ | K | Δϑ = Q / (m · c) |
The formula for the conductor resistance turns the law into a tool: length l in metres, cross-section A in square millimetres and the resistivity ρ of the material give R, and with the current the heat is fixed. For copper, the Deutsches Kupferinstitut data sheet for Cu-ETP gives 0.018 Ω · mm²/m in the annealed state and 0.017 to 0.018 cold worked, while university lab scripts use 0.0175. How the reciprocal of this quantity is stated in data sheets is explained in the article on electrical conductivity.
Worked examples with real numbers: cable, heating wire, motor winding
Example 1: house cable with 2.5 mm² and 16 A
A copper socket circuit with 2.5 mm² cross-section runs 25 m from the distribution board to the load. The current flows out and back, so the conductor length is 50 m, and the resistance is R = 0.0175 · 50 / 2.5 = 0.35 Ω. At 16 A, the rated current of the circuit breaker, the cable generates P = 16² · 0.35 = 89.6 W of resistive heat, spread along its whole length. Over eight hours at full load that is 0.72 kWh nobody uses. With 4 mm² the loss drops to 56 W, with 10 A instead of 16 A to 35 W. Cross-section acts linearly, current quadratically. That is why DIN VDE 0298-4 limits the current rating through the conductor temperature: 2.5 mm² with PVC insulation may carry around 21 to 27 A depending on the installation method, so that the conductor stays below 70 °C (Elekrechner, ampacity table).
Example 2: heating coil of a 2,000 W kettle
Here the heat is the product. At 230 V a 2,000 W kettle draws I = 2,000 / 230 = 8.7 A, so the heating coil needs R = U² / P = 230² / 2,000 = 26.45 Ω. Made of copper that would no longer be a wire but a reel: at 0.5 mm diameter (0.196 mm²) copper has 0.089 Ω per metre, so 26.45 Ω would take 297 m. Heating conductors are therefore alloys with high resistivity: constantan 0.49, nickel-chromium 80/20 1.1 and Kanthal A-1 1.45 Ω · mm²/m (Wikipedia, Heizleiterlegierung). With Kanthal A-1 at 0.5 mm diameter, 3.58 m are enough, and the alloy withstands up to 1,400 °C, where copper would have melted long before. The 2,000 W heat one litre of water with c = 4,190 J/(kg · K) (el-gor.at) from 20 to 100 °C with Q = 1 · 4,190 · 80 = 335 kJ, which takes 168 s if no heat escapes to the surroundings.
Example 3: stator winding of a three-phase motor
In a three-phase winding the losses of the three phases add up: P = 3 · R · I² with the DC resistance of one phase. The German metrology institute PTB measured this on a frame size 132 induction motor: the stator winding with 3.74 kg of copper dissipates 135.5 W at 50 Hz and 20 Nm, the calculation gave a winding temperature rise of 36.1 K and the measurement 35.3 K. At 36.2 Nm the losses rose to 422.7 W and the measured temperature rise to 86.2 K (Lehrmann, Bulletin SEV/VSE 2006). The torque increased by 80 %, the winding losses by 210 %, which is what the square looks like in practice.
The heat capacity shows how little time there is. 3.74 kg of copper at 0.386 J/(g · K) (Kupferinstitut) store 1,440 J per kelvin. At 422.7 W and without any heat removal the temperature would rise by 0.29 K per second, and 90 K would be reached after five minutes. In the real motor most of the heat flows through the slot liner and the lamination stack into the housing, which is why the winding settles in continuous duty. The PTB puts the thermal resistance between winding and lamination stack at 0.059 K/W: every watt generated in the winding lifts it 0.059 K above the stack. How the winding in which this heat arises is built is described in the article on the electric motor winding.
Resistance rises with temperature, and the heat rises with it
Joule's law has a feedback loop that encyclopaedia entries leave out. The resistance of metals increases with temperature. For Cu-ETP the Kupferinstitut gives 0.018 Ω · mm²/m at 20 °C, 0.023 at 100 °C and 0.03 at 200 °C, and the temperature coefficient α is 0.00393 per kelvin, valid from −100 to 200 °C. As a rule of thumb the resistance of copper and aluminium grows by 0.4 % per kelvin (Wikipedia, Temperaturkoeffizient). A winding with 150 W of loss when cold dissipates 150 · (1 + 0.00393 · 95) = 206 W at a winding temperature of 115 °C, 37 % more at the same current. More heat, higher resistance, still more heat: without adequate cooling this loop runs into overheating.
The same dependence is also the measuring instrument. Measure the resistance of a copper winding cold and again warm after switching off, and you get its mean temperature without a single sensor: Δϑ = (Rwarm − Rcold) / Rcold · (234.5 + ϑcold). The copper constant of 234.5 K is nothing other than 1/α − 20 °C, and for aluminium it is 228.1 K (IEEE 399, cited by Industrial Monitor Direct). IEC 60034-1 prescribes this resistance method for the temperature rise of windings. Heating conductors go the opposite way: constantan has a temperature coefficient of only ±0.00004 per kelvin, its resistance rises by just 4 % from 20 to 600 °C (Elektropraktiker, resistance materials), so the heating power stays stable.
Why high voltage reduces the losses
For a fixed power P = U · I the current falls in the same proportion as the voltage rises, and the losses I² · R fall with the square. An example: 100 kW are to be sent over a line with 0.5 Ω of resistance. At 400 V the current is 250 A and the loss is 250² · 0.5 = 31.25 kW, almost a third of the power. At 20 kV medium voltage the current is 5 A and the loss is 12.5 W. The voltage ratio is 50, the loss factor 50² = 2,500. That is why every grid steps the voltage up for transport and back down only at the consumer, and the transformers themselves cost 0.5 to 2 % copper losses at rated load for it (Wikipedia, Kupferverluste).
The measured grid losses confirm the calculation. In Germany around 5.7 % of the electrical energy fed into the grid is lost, and a 380 kV line loses a good 1 % per 100 km at full load (Wikipedia, Übertragungsverlust). The Munich utility SWM reports the losses per grid level (cited by Proteus Solutions):
| Grid level | Voltage | Loss |
|---|---|---|
| Extra-high voltage | 220 and 380 kV | 0.44 % |
| High voltage | 110 kV | 0.29 % |
| Transformation high to medium voltage | transformer | 0.37 % |
| Medium voltage | 10 to 30 kV | 0.60 % |
| Transformation medium to low voltage | transformer | 1.13 % |
| Low voltage | 230 and 400 V | 2.42 % |
The low-voltage level loses five times more than the extra-high-voltage level, although its lines are short. The level with the smallest U carries the largest current per kilowatt, and that enters squared.
The resistive heat of the winding determines the thermal class of the insulation
In an electric motor the resistive heat of the stator winding is the largest loss item: the trade journal Elektrotechnik puts it at 34 % of the total motor losses. This heat arises in the copper, but it ages the insulation that surrounds the copper: wire enamel, slot liner, phase separator, slot closure. IEC 60034-1 therefore limits the temperature rise of the winding above 40 °C coolant according to the thermal class of the insulation system: 80 K for class B, 105 K for class F, 125 K for class H, measured by the resistance method (IEC Motores). The difference to the limit temperature of the class is the allowance for the hottest spot of the winding, which the average from the resistance measurement does not show.
| Thermal class | Temperature rise to IEC 60034-1 | Limit temperature | Flexible insulation materials at GOBA |
|---|---|---|---|
| B | 80 K | 130 °C | polyester film, DMD laminate, polyester fleece |
| F | 105 K | 155 °C | NMN laminate, impregnated mica products |
| H | 125 K | 180 °C | Nomex 410, NKN laminate |
| N and R | values in IEC 60034-1 | 200 and 220 °C | Kapton polyimide film, micanite |
The design follows from this in two steps. First, the resistive heat of the winding at rated current determines the temperature rise, and temperature rise plus 40 °C plus allowance gives the minimum class. Second, anyone who wants a reserve picks the class above. The usual choice is class F insulation with class B utilisation: the insulation tolerates 155 °C, the machine is operated up to 130 °C. According to the 10 kelvin rule the 25 K reserve extends the life of the insulation roughly by a factor of 5, because every sustained 10 K above the limit temperature halves it. The full class table from Y to C is in the article on insulation classes, and the materials above 130 °C with continuous and short-term values are in the article on high-temperature insulation.
Heat removal through thin insulation layers: where GOBA comes in
The resistive heat has to get out of the winding, and its path leads through the insulation into the lamination stack. Electrical insulating materials conduct heat poorly: polyester film 0.2, Nomex 0.14, Kapton 0.12, mica and micanite 0.3 to 0.7 W/(m · K), against 394 for copper (Kupferinstitut). What counts for heat transfer is therefore the thermal conductivity divided by the thickness. A slot liner of 0.25 mm polyester film passes 800 W per square metre and kelvin, the same film at 0.5 mm only 400 W. Doubling the thickness doubles the temperature difference between winding and core. How λ is measured and which values the individual materials have is covered in the article on thermal conductivity.
Our recommendation follows from this, and it contradicts a common reflex: a thicker insulation is rarely the safer one. It withstands the voltage better, but it also keeps the heat in the winding, and every 10 K more halves the life of that very insulation. We size the thickness to the electrical requirement, with a reserve for test voltage and edge quality, without a habitual surcharge. GOBA slits insulating materials from 0.023 mm thickness for this, as slit rolls from 5 to 1,600 mm wide with plus or minus 0.1 mm, and die-cuts slot liners, phase separators and slot closures up to 1,000 x 2,000 mm. In a hairpin winding the flat conductor lies against the slot wall with its whole face, so the insulation layer decides the heat path even more, which is why for drives in electromobility we match material and thickness to the current density together with the customer. Send us the thermal class, the rated current and your drawing, and we will recommend the build-up from our range of insulation and molded parts.
