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how do a calculate the maximum temperature that a certain wattage can get on a wire? I know how to convert wattage to "Celsius heat units (IT) per hour", but eventually the wire has to stop heating up, right?

EXAMPLE: if i have 144 watts, converting it to Celsius heat units (IT) per hour would be 272 degrees Celsius. But i still don't know, if that is the hottest it will get.

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    \$\begingroup\$ (1) What does (IT) mean? (2) Where did you get this conversion formula? \$\endgroup\$
    – Transistor
    Commented Jul 18, 2020 at 22:58
  • \$\begingroup\$ here: conversion-website.com/power/watt-to-Celsius-heat-unit-IT-per-hour.html \$\endgroup\$
    – user256116
    Commented Jul 18, 2020 at 23:00
  • \$\begingroup\$ Samuel, the Celsius heat unit is energy or work. Work divided by time is power. But none of this tells you the equilibrium temperature of a wire dissipating some number of watts. Yes, the ratio of watts to your crazy unit of Celsius heat units per hour is 273.154. But this ratio is unitless and doesn't carry Celsius as its units and you cannot use that ratio as a way to figure out the equilibrium temperature of the wire. \$\endgroup\$
    – jonk
    Commented Jul 19, 2020 at 1:35
  • \$\begingroup\$ Don't you just love units derived from combining metric and English systems! \$\endgroup\$ Commented Jul 19, 2020 at 13:15

4 Answers 4

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Probably about as many people calculate the maximum temperature of wire as use units like CHU per hour or furlongs per fortnight. Most people find tables in electrical codes for building wiring or published by wire manufacturers and others for equipment wiring. You need to know the wire material, the ambient temperature, the current that the wire carries and whether the wire is mostly surrounded by air or bundled with other wires. Tables give ampacity for various size wires and conditions. Some conditions are handled by the use of adjustment factors.

Here are some examples:

IEWC Global Solutions

Cooner Wire

Belden

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I found a definition of the "Celsius heat unit" but I have no idea if it is true or correct:

The Celsius heat unit is the energy required to increase the temperature of one pound of water by one degree Celsius (or 1.8 degrees Fahrenheit) at a constant pressure of one atmosphere. One Celsius heat unit is equal to 1.8 British thermal units. Source: Conversion Website

The definition seems to be a confused mess of metric and imperial units and you will find it far better to work in SI (metric) units.

  • The Celsius scale measures temperature.
  • Use watts (W) to measure power.
  • Use joules (J) to measure energy.

if i have 144 watts, converting it to Celsius heat units (IT) per hour would be 272 degrees Celsius. But i still don't know, if that is the hottest it will get.

The temperature will stabilise when the heat lost to the surroundings of the wire is equal to the electrical power input to the wire. To calculate this you would need to know the length, surface area and emissivity of the wire to calculate radiated energy and the temperature and flow rate of the surrounding fluid (e.g., air) for convective losses. You might be able to ignore conduction losses depending on the geometry of the setup.

Forget about "Celsius heat units (IT) per hour". It's not used in general or electrical engineering.

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    \$\begingroup\$ Thanks! I knew something didn't look right, but it was late at night. Fixed. \$\endgroup\$
    – Transistor
    Commented Jul 19, 2020 at 9:28
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You'll need to consider the thermal resistance from the wire to ambient surroundings, call it \$\theta_{wa}\$ (units of degrees Kelvin per Watt). This will need to take into account any forced airflow, convection, contact with supports/substrate, and direct radiation as applicable. This may be hard to calculate and thus might require experimental techniques to determine.

The wire will reach a temperature of \$P \cdot \theta_{wa} + T_{\text{ambient}}\$ under the assumption that the power is constant as temperature changes. This is another approximation that may not hold; as the wire heats up its resistance typically increases, which may lead to an increase or decrease in dissipated heat depending on how the wire is connected to the circuit. If driven with a nearly constant voltage, power will decrease as the wire heats up; if driven with a nearly constant current, the dissipated power will increase.

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  • \$\begingroup\$ Ok. What is the P in the equation? \$\endgroup\$
    – user256116
    Commented Jul 19, 2020 at 21:24
  • \$\begingroup\$ @SamuelWalker P is the dissipated power in the wire, i.e. I^2R \$\endgroup\$
    – nanofarad
    Commented Jul 19, 2020 at 21:24
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According to this, the "celsius heat unit" is an actual unit of energy, roughly equal to the amount of energy required to raise the temperature of one pound of water by 1 C.

I have never previously heard of this unit. It is not used regularly in academic physics or in electrical engineering. Perhaps there is some industry where it is used or has been used historically; refrigeration and building heating or air conditioning seem likely, given the scale of the unit.

But given that the CHU is a measure of energy, it is not a measure of temperature. You cannot compare CHU and degrees celsius directly.

As the other answers have said, the temperature achieved by a wire (or any other object) being heated by a constant power depends entirely on how that wire is cooled (by conduction, convection, and radiation) and not at all on the conversion factor between watts and CHU per hour.

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