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I'm looking to buy an induction hob to heat some water and I would like to know if its going to be powerful enough. If so, can it be done in, say, 1 hour? Additionally, what power rating could I go down to if I were to increase the time to boil by 50%?

The hob is rated at 2000W and I would like to know if it can reach, and maintain, a rolling boil of 11 litres of water in a container with diameter of 280mm. I have seen from wikipedia in a statement from the U.S DoE that the efficiency of induction hobs is approx. 84%

So far I have the following. (Apologies, I'm clueless as how to input maths into this and would appreciate an edit.)

Estimated power delivered to the water is approx. 2000x84% = 1680W / 1680J/s

So in one hour, theoretically it should be able to deliver 6.1MJ to the water.

Q(total)=Q(in)-Q(out)

Energy required to raise the water from tap water temperature to boiling is:-

Q=m.C(water).(T2-T1)

Q=(density).(volume).C(water).(T2-T1) Q=1000x0.011x4181x95 Q=4.37MJ

I need to include the energy lost to the environment through evaporation and heatloss at the top of the pot in general but I'm unsure how to do this beyond thinking it's to do with the specific latent heat, surface area exposed to the atmosphere, ambient temperature etc. etc.

So if,

Q(out) < 6.1MJ - 4.37MJ then I should be okay, correct?

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closed as off-topic by Kyle Oman, Kyle Kanos, ACuriousMind, yuggib, Qmechanic Jul 1 '15 at 13:08

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  • $\begingroup$ Important point: does the pot have a lid? And of what material is the pot made, and what is its mass? $\endgroup$ – tok3rat0r Jun 30 '15 at 13:20
  • $\begingroup$ The pot possibly has a lid - it hasn't arrived yet so I'm unsure and it wasn't clear in the description when buying! I'd like to err on the side of caution for this calculation if possible but would be very interested to see how it would differ if a lid was being used. The pot is made of ferritic stainless steel and weighs 3kg. Thanks! $\endgroup$ – Phizzy Jun 30 '15 at 13:23
  • $\begingroup$ I'm surprised (but not surprised :-) ) that nobody considers my answer"useful". Note that insulation (mentioned below) is entirely feasible to reduce thermal losses. Base heat rises until the temperature difference is sufficient to cause thermal transfer into the liquid - which is well within the range of many commonly available materials. I use towels around the outside of resistance heated slow cookers to improve their performance. These are thermostatically controlled but typical cheap Asian sourced ones have no thermal insulation and can struggle to reach thermostat temperature. $\endgroup$ – Russell McMahon Jul 1 '15 at 7:12
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    $\begingroup$ @Phizzy For interest I took the largest pot to hand & tried boiling it. This was a stainless steel saucepan with lid. Pot contained 4.5 l of water. Using a conventional electrical "radiant" element with 1.3 kW in (and higher losses than induction heater) I got a good boil with the lid off and an extremely enthusiastic one with the lid on (I took photos). An 11 l pot would not have vastly more losses. Putting the lid on very greatly increases the boiling rate. I'd estimate that say 1500 W would produce a good boil with lid on in 11l. $\endgroup$ – Russell McMahon Jul 1 '15 at 14:42
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    $\begingroup$ @Phizzy What is your application. If this is for water sterilisation you may be trying harder than needed. If that's your application please advise as I may be able to make some useful comment. $\endgroup$ – Russell McMahon Jul 1 '15 at 14:43
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Summary:

  • Based on provided data you will be able to comfortably bring the water to boiling point in 1 hour. The energy input required to heat the water alone is significantly less than the available energy and the difference is greater than typical thermal losses.
    ie Water_mass x delta_temperature x Water_specific_heat < energy input
    Details below.

    As an induction cooker produces temperatures in the base of the cooking vessel which are only slightly higher than the liquid contents (enough to cause thermal transfer) it is possible to insulate the cooking container to reduce thermal losses if desired.
    ie If desired and there are no other special factors the pot can be wrapped in a towel or other insulator to reduce thermal losses.
    (Remove towel before using pot or electric or gas stove :-) ).

    A rolling boil will easily be achieved once the water reaches 100 C as the energy required to increase temperature to boiling point is now all available to cause phase change (aka boiling) while thermal losses will be about the same as during prior heating.


Details:

You use a temperature delta of 95 degrees implying cold water at 5 degrees to start.
In many cases water will be somewhat warmer as it tends to be at in-ground temperature (depending on source) but a delta of 95 makes for a close to worst case example.

I'll assume that energy to heat water is ~= 4200 J/kg/K (you use essentially same 4181 )
and that water density is 1 kg/l. In practice both water density and specific heat vary with temperature across the temperature range concerned. A rounded figure of 4.2 kJ/kg/K is accurate enough for for practical purposes given the various other uncertainties such as the electronic grid power to thermal conversion efficiency and thermal losses.

Required: Heat 11 litres of water x 95 C in 1 hour.

Energy = 4200 x kg x delkta_t J = 4200 x 11 x 95 ~= 4.4 MJ
Same as your answer (no great surprise :-) ).

Required electrical input at 85%
= 4.4MJ / 0.84 = 5.2 MJ into heater.
Power to achieve this in one hour = 5.2E6 /(3600s x 1000W) kW
= 1.44 kW

ie if the container had no other losses apart from what was assumed by Wikipedia then a 2000 Watt input unit would achieve the result comfortably (notionally in about 1440/2000 x 60 ~= 43 minutes).

To boil in an hour you can lose (2000-1440)/2000 = 0.28
or ABOUT 25% of your actual heat energy to thermal losses from convection, radiation and conduction. This sounds very achievable, and with an induction cooker you can insulate the pot by eg wrapping it in a towel! - but NOT under the base. (Depending on pot material you can put a thin layer of insulator under the pot an still heat - noting that this is already done as the "cooking surface" is an insulator (probably glass).

Latent heat of vaporization or water (energy required to convert 1 kg from liquid to "steam" is 2260 kJ/kg = 2260 J/g = 2260 J/cc

If 1440 J/s (Watts) was available for the actual boiling process then rate of conversion to steam = Power/(latent heat of vaporisation)
= 1440/2260 = 0.636 cc/second
= 2.29 l/hour

That's a rather small percentage of total volume per second.
However, mentally visualising a 11 litre pot of water on a standard radiant element at full power (ay 1500 Watt) with the water at 100 degrees C - I'm essentially certain that it would boil very vigorously indeed - especially so if it ws possible to insulate the sides of the container, as it probably is in this case.


Additionally, what power rating could I go down to if I were to increase the time to boil by 50%?

As above, if 2000 Watt nominal input gives an ~= 45 minutes to boiling, and the target was < 1 hour, then you'd expect an increase in time to boiling by a factor of 1.5 to 90 minutes to require ABOUT 2000 x 45/90 = 1000 Watts electrical input if thermal losses are able to be scaled down proportionally. With "normal" cooking energy sources adding insulation is problematic but with induction heating adding another towel wrapper and a layer of padding on top is actually feasible. If losses cannot be scaled down then see above calculations for assumed losses and recalculate accordingly.

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  • $\begingroup$ Your point about wrapping a towel round the pot is a good one, but in the general case is there a 'rule of thumb' for estimating total power loss from an object (in air) with a given temperature difference between the object and the air? One could fairly easily work out the radiative part, but I'm guessing convection would dominate... $\endgroup$ – tok3rat0r Jul 1 '15 at 7:27
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    $\begingroup$ Note: C for ferritic steel in this temperature range is approx 500 J/kg/K, so total energy to increase temperature of pot from 5 degrees to 100 degrees is 142.5 kJ (with no losses). If this were to be achieved in the same 43 minutes as the estimated time for the water to boil, it would require a mere 55 W. So the energy required to heat the pot is negligible. $\endgroup$ – tok3rat0r Jul 1 '15 at 7:38

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