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Lets say the following is limited to 2.5 kW only and the thermostat and flow switch failed during use. After water is boiled away and bit of steam released. What would happen to the heating element inside? Can it just stay inside getting hot like a lamp? How do you compute the heat of the 2.5 kW heating element?enter image description here

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  • $\begingroup$ Power times time... or do you really mean final or highest temperature before failure as that will happen. $\endgroup$
    – Solar Mike
    Sep 18 '20 at 5:52
  • $\begingroup$ Yup highest temperature to know what will happen to the cooper tank, can it melt it? $\endgroup$
    – Jtl
    Sep 18 '20 at 6:50
  • $\begingroup$ Unlikely that the tank will melt- the element fails first. $\endgroup$
    – Solar Mike
    Sep 18 '20 at 7:03
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    $\begingroup$ How does the heating element fail? It melts inside? $\endgroup$
    – Jtl
    Sep 18 '20 at 7:10
  • $\begingroup$ Buy a cheap 1 or 2kW kettle element and see. $\endgroup$
    – Solar Mike
    Sep 18 '20 at 8:40
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Regarding the question title,

How to compute heat produced by 2.5 kW heating element.

That's easy. The heat generated will be 2.5 kW. For temperature generated see below.

After water is boiled away and bit of steam released.

Steam has a volume of 1760 times the water volume. If that's a 1 L tank then that's 1.7 m3 of steam.

What would happen to the heating element inside? Can it just stay inside getting hot like a lamp? How do you compute the heat of the 2.5 kW heating element?

The temperature will stabilise when power lost to the environment = electrical power in. There is no eay way to calculate this.

In general either the element will burn out or the insulation would break down and short the supply to the earthed element casing. The protective circuit breaker would then trip.

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Does the copper melt? Possibly, we can't be sure. Assuming 0.25m² of surface for the vessel, this formula for radiative heat transfer and an emissivity of 0.03, at 1068°C (melting point of copper) the vessel radiates about 1.4 kW, less then the 2.5kW input. So ignoring convective heat transfer, it would reach a high enough temperature.

We don't know enough to estimate convective heat transfer, as the vessel is in an enclosure. Where it open I would guesstimate that ~1kW convective heat transfer is entirely possible, but not guaranteed, in an enclosure all bets are off IMO.

Actually, things will probably go as per Transistors answer. Don't try this at home.

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    $\begingroup$ What about the heat transfer from the heating element? will surely fail before the canister... $\endgroup$
    – Solar Mike
    Sep 18 '20 at 11:01
  • $\begingroup$ Depends on what it's made of? But you are likely right. $\endgroup$
    – mart
    Sep 18 '20 at 11:14
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One very simplistic calculation is the following, with the following assumptions:

  • There is no heat insulation of the tank (no double wall etc)
  • the heat tank has infinite strength (will not blow up)
  • The heat tank temperature and its contents is uniform (e.g. the water and the walls of the tank do not have more that a few tenths of degrees K)

then you can assume that the boiler once it gets to thermal stability will "lose" 2.5[kW] of power. The losses will be a) convective, b) radiative and c) conductive.

Convective losses

$$\dot{Q}_{conv} = k_{conv} A_{conv} \delta T $$

where:

  • $ k_{conv}$: the convective constant typically between 0.1 and 1000[W/m^2K]
  • $A_{conv}$: the external surface of the boiler
  • $\delta T $: the temperature difference of the boiler surface with the environment ($T_{boiler}- T_{ambient}$)

Radiative losses

$$\dot{Q}_{r} = \sigma T^4 A_{conv}$$

where:

  • $ \sigma$: $5.6703\cdot 10^{-8} (W/m^2 K^4 )$ - The Stefan-Boltzmann Constant
  • $A_{conv}$: the external surface of the boiler (the same as
  • $T $: the temperature in K of the boiler

In general this could be neglected.

conductive losses

$$\dot{Q}_{r} = U A \delta T $$

where:

  • $ U $: coefficient of heat transfer.
  • $A_{cond}$: bracket crosssection. (Very small compared to surface
  • $\delta T $: the temperature difference of the boiler surface with the environment

In general this could be neglected, because the surface area of the boiler is much greater than the brackets.

Boiler temperature after ignoring conductive and radiative losses.

$$\dot{Q}_{conv} = k_{conv} A_{conv} \delta T $$ $$T_{boiler} = T_{ambient} + \frac{\dot{Q}_{conv}}{ k_{conv} A_{conv} }$$

Typical $k_{conv}$ values for non forced convection are in the order of a 2.5 to 20$[W/m^2K]$

You can substitute for your problem and obtain a ball park figure.

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