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Calculating time taken to boil water

  1. Jan 21, 2014 #1

    Need help please... my very limited knowledge of physics is very rusty lol...Im trying to work out the following:

    Basically if i have 50litres of water in a metal box how long will it take to boil off when heated to 800°C?

    And is there an equation for?

    Thanks very much in advance
  2. jcsd
  3. Jan 21, 2014 #2


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    Welcome to PF!

    As asked, the question doesn't work: water boils at 100C so under normal conditions it can't be heated to 800C. We need to know a lot more about the setup and known and unknown parameters. It mat be a simple matter of knowing the starting conditions and heat flow rate and assuming everything else, but you will have to tell us.
  4. Jan 21, 2014 #3
    I'm guessing that you have a water boiler with a burner that's 800C. So you would need to know the initial temperature of the water, the heat transfer function, and the ambient pressure. You need to bring the water temp to 100C (approximately) and then overcome the latent heat of evaporation.
  5. Jan 21, 2014 #4
    sorry Im struggling to use the correct terms.

    so I would like to put a data logger inside a water tank which will then go inside a kiln at 800°C

    (water, water tank and logger will start at room temp)

    Im wondering if my water tank was 50litres capacity for example, how long before all the water has been evaporated (protection for logger gone).

    hope that makes sense?

  6. Jan 21, 2014 #5
    Is there a lid on the tank of water, or is the tank open to the surroundings. If there is no lid, is the kiln sealed, or is it able to allow gas to escape?
  7. Jan 21, 2014 #6
    yes tank is sealed other than two baffled pipes for steam to escape (also used to fill with water)
  8. Jan 21, 2014 #7
    OK. So the tank is not sealed, and steam can escape the tank. How big is the kiln, and can gas escape from the kiln?
  9. Jan 21, 2014 #8
    the kiln is a typical pottery size kiln and i don't think the steam can escape.
  10. Jan 21, 2014 #9
    In other words, no matter high the pressure gets within the kiln, no vapor will escape?

    Let me say back what is happening so we understand. The kiln is at 800 C (inside), and the inside surfaces of the kiln are held at that temperature. The water is placed in the kiln at room temperature. There are 50 liters of water, which is about 15 gallons, which is about 2 cubic ft. Most of the pottery kilns on the internet are less than 1 cubic ft. So, I ask again, what is the volume of your kiln?

    The metal box sits on the base of the base of the kiln. How thick is the wall of the box. What is the design geometry of the base of the kiln, including the location of the heating elements, wall thickness, insulation, etc?

    This is not going to be an easy problem to model.
  11. Jan 21, 2014 #10


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    It strikes me that all that water is going to cool down the kiln, if it's placed in with the kiln hot. Kiln heaters are not designed to deliver that much heat quickly. I'd imagine they are deliberately made low power so that the pottery can warm up slowly and acclimatise without cracking.. If the kiln is cold when the water is placed in, the power of the heater is needed in order to do any calculations. The surface temperature of the tank will be near 100C until the water has all boiled away - how long, will depend upon the power of the element.
    This has not been described in enough detail yet, to allow any sensible answers. Please tell us more.
  12. Jan 22, 2014 #11
    sorry all the factors are examples (clearly bad ones)
    • The heat in the kiln is from all sides.
    • It is a 3 phase kiln
    • water tank size will be as neccesary (1, 10 or 50 litres...unknown yet)
    • kiln will be at 800°C when water tank is inserted.

    so im after... how long will a certain amount of water take to evaporate at 800°C? - is there an equation for this?

    I read on another post here that water heats up 16°C every minute!?

    does this depend on amount? - if not then it will take 6+mins to boil? - is there a similar rule for water evaporating or am i well off the mark?

    lets say for this purpose
    • i have a water tank 500mm wide, 500mm long & 200mm high giving me 50ltrs.
    • the tank skin is 2mm thick 304 stainless steel.

    sorry if this is still not clearer
  13. Jan 22, 2014 #12
    Is there a fan in the kiln?
  14. Jan 22, 2014 #13


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    The basic answer, clearly from all of the responses, is that no, there is no single, simple equation to describe this. It is a very complicated process.

    You can, however, do just a little bit of testing and create an accurate model of what is happening...
  15. Jan 22, 2014 #14


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    It would help if we had some context. You have yet to ask the right question, I think. Once you have found the right question, the right answer will not take long to emerge.
    What's it all about and what do you want to achieve with your kiln full of water? Cooking rice?
  16. Jan 22, 2014 #15


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    You could very well damage the kiln by doing this, since it will very rapidly (and unevenly) cool down the section of the kiln where you place the water. If the kiln material isn't designed for this kind of rapid temperature change and strong thermal gradient, it could crack or fail.

    Why do you want to put a pot of water in a kiln anyways?
  17. Jan 22, 2014 #16
    The most basic thing needed here is the thermal power of the kiln. If nothing else is known or too difficult to model, that should give you a ballpark idea.
  18. Jan 23, 2014 #17
    thanks for you input so far

    ok the kiln im using is:
    • 700mm wide, 700mm high & 900mm long (internal/usable space)
    • 3 phase 239amps @400 volts & 38kw
    • it does have an extraction vent

    my water tank made from 2mm thick stainless steel is 504.25mm wide, 500mm long & 200mm high (50.425litres) - minus the datalogger cavity 25mm high, 100mm wide & 230mm long (minus 0.575litres)

    so total water to evaporate off is 50litres

    FYI - the kiln is designed to go to 1400°C and handle steam.

    Also 'Heat transfer seventh edition' by J.P. Holman has the following equation:

    Elp = (3.212+0.02224u)(Ps - Pw)0.88

    Elp = land-pan evaporation, mm/day
    u = daily wind movement measured 152mm above the pan rim, km/day
    Ps = saturation vapor pressure
    Pw = actual vapor pressure

    now im pretty sure im well off here but this is the only thing that points to evaporation BUT "daily wind movement" is irrelevant and its a tank not a pan.

    i am stabbing in the dark here and hopefully you guys can point me in the right direction now ive found the kiln power specs

    thanks very much in advance
  19. Jan 23, 2014 #18


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    You really need to forget about the idea of a kiln here.
    You have crudely a half meter cube space that you're putting a twentieth meter cube of water into.
    Water has enormous heat capacity, so your kiln will chill down to 100C while the water is getting boiled off.
    The relevant parameters are that the kiln is fed by a 38kW power supply and that you have 50 liters of water.
    It takes about 0.75 kWhrs of power to boil away 1 liter of water, so your 38kW power supply will be able to boil away the 50 liters in about an hour. Until that water is gone, the water tank acts like an ice cube in the kiln, it chills everything because it is so much colder.
  20. Jan 23, 2014 #19
    A clearcut picture is beginning to emerge now. If you calculate the maximum amount of water vapor that a kiln of this volume can hold in the gas phase at one atmosphere and 800 C, it comes out to be on the order of about 100 gm = 0.1 liters. The vent on the kiln is enough to allow the air to be purged from the kiln by the steam, and for maintaining the pressure within the kiln at 1 atm. So, after only a tiny amount of water vapor has boiled off from the tank, virtually all the air will have been purged, and the gas phase will consist of pure water vapor. It is reasonable to assume that this state is reached in no time. The water vapor in the gas phase will be at 800 C, and most of it will not be in equilibrium with the liquid water, which will be at 100 C. There will be a thermal boundary layer around the tank in which the gas temperature varies from 100C at the internal surface of the tank to 800 C in the bulk of the gas phase. The rate of boiling is thus going to be heat transfer rate limited. The temperature driving force for heat transfer to the liquid water will be 800-100 = 100C. As pointed out in Post #3 by .Scott, to calculate the heat transfer rate, we are going to need to know the heat transfer coefficient (equivalent to the thermal boundary layer thickness) and heat transfer area surrounding the liquid water. The heat transfer rate will be equal to the heat transfer coefficient h times the heat transfer area times thetemperature driving force: hA (800 - 700). This will be equal to the heat of vaporization at 100 C times the evaporation rate. So, to get the evaporation rate, we need to be able to estimate the heat transfer coefficient and the heat transfer area. The bottom of the tank is in contact with the bottom of the kiln, so, aside from interface resistance between the kiln and tank, the heat transfer coefficient can be probably be approximated by conduction through the tank wall. Of course, if the bottom of the tank is acting as a heat sink for the kiln wall in the bottom, we may also have to consider conduction within the walls of the kiln, and this will reduce the effective heat transfer coefficient. The heat transfer on the sides of the tank is going to be controlled by the thermal boundary layer in the gas phase near the tank surface. This will probably be pretty low compared to the heat transfer through the base. The first thing to do is to neglect the heat transfer between the side walls of the tank and assume that all the heat transfer occurs between the kiln wall and the base of the tank. The heat transfer rate can be bounded on the high side by assuming that it is equal to that obtain by pure conduction through the base of the tank, and that the temperature driving force is 700 C. This will give the maximum evaporation rate possible. All other cases will give lower evaporation rate. To do a more accurate job, one would need to model heat transfer that includes temperature gradients within the kiln wall, from 800 C at lateral locations away from the tank to closer to 100 C underneath the tank. This can be modeled most easily with finite element heat transfer software. But, maybe the upper bound to the heat transfer rate obtained using the simple method above would be sufficient to answer many questions.

  21. Jan 23, 2014 #20


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    Surely what you need is a slow supply of water, fed into the kiln, which will vaporise at a steady rate, purging the air but not affecting the internal temperature of the kiln (as you seem to have plenty of Power to produce superheated steam in the quantity you need it.

    You see, this is very difficult for us because we are having to read between the lines to find out what it is your really need - rather than what you think you need.
  22. Jan 23, 2014 #21
    I've done some quick calculations to see what the heat transfer rate to the water would be if the base of the water tank could be kept at 800C by the kiln. The thermal conductivity of stainless steel is 16 W/(mK), and the wall thickness is 0.002 m, so the heat transfer coefficient of the base would be 2000 W/(m2K). The area of the base is 0.25 m2, and the temperature driving force is 700 C. So the heat flow rate would be 1400 kW. That compares with the 38 kW that the kiln is actually capable of delivering. So the conclusion here is that the base of the kiln underneath the tank could not be maintained anywhere close to 800 C. This confirms etudient's and sophiecentaur's observations that the water tank is going to act as a huge sink for heat and cause cooling of the kiln wall, certainly in the region where the water tank is in contact with the base. So, in order to estimate the heat transfer rate accurately enough, some serious heat conduction calculations are going to have to be carried out to quantify the conduction of heat laterally within the kiln walls to the region underneath the water tank.

  23. Jan 23, 2014 #22


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    Why do you need to heat all the water at once?
  24. Jan 23, 2014 #23


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    You still haven't told us precisely what you are trying to achieve - except to boil away gallons and gallons of water.
  25. Jan 24, 2014 #24
    sorry guys.. in my mind i have told you what im after but im clearly struggling to get it down on paper

    so...in a nut shell all i want is to know how long (hrs/mins) will an amount of water take to evaporate at 800°C or any temp for that matter

    The reason im after this is to work out how much protection (hrs/mins) the logger will have so i can repeat/modify the application/calculation when using in a furnace or large industrial paint oven etc.

    The logger cant see much more above 100°C so it is important i know when to pull the tank out of the application before the logger is toast.

    hope this helps and please forgive my lack of experience
  26. Jan 24, 2014 #25


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    You have just repeated the "evaporate at 800C" statement and I thought it had been pointed out that it's a bit of a nonsense to ask that.
    You are still not saying what the whole thing is about and that is preventing you from getting a useful answer. If you want to measure a temperature of 800C then a logger that works at 100C is not the right instrument - without a lot of additional experimenting.
    You mention your lack of experience - so let PF use its combined experience by telling us what you 'really' want to find out. Are you interested in the thermal properties of Water, the Kiln, the heater?? After 24 posts, it is still a guessing game.
    People will lose interest soon and you won't get anything useful out of the thread.
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