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How to calculate necessary heat power

  1. Feb 9, 2009 #1

    I have a problem in my factory, and contracted a company to solve it, but unfortunately they were unable to do it.

    I have a tank with 1.000 liters of water mixed with sand, that is pumped with a high capacity pump, through a compressed air system, to clean some small metal parts.

    My problem is that I need the water mix to remain at 80ºC, and for that I have four 6.000 W electrical heaters, submerged in the water, for a total of 24.000 W of heating capacity. The temperature was my requirement, and 24.000 W was the solution the contracted company installed.

    After 2 hours and 30 minutes working, the temperature drops to 30ºC, with all the electrical heaters working.

    What I would like to know is the way to calculate the amount of heating power needed to maintain the water mix at 80ºC.
    Knowing that the 1.000 liters water mix, starting at 80ºC, with 24.000 W of heating power turned on, after 2 hours and 30 minutes drops to 30ºC, is it possible to calculate it?

    I really don't want to guess (like the contracted company did), and keep trying until I get enough power... I think that the correct way is to do the math.

    Can anyone help me out?


    Jose Marques
  2. jcsd
  3. Feb 10, 2009 #2


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    In theory the calculation is easy - that is, when you neglect practical factors like the heat capacity of your tank, heat loss to the environment, dependence on the mixture of water/sand, etc.
    Probably assuming that the dissipated power is constant in time, you can estimate an effective heat capacity of the "non-water" in the system and try to do the calculation. However, I think that there would still be dependences on the type of water (sweet / salt, clear / polluted), the amount of sand.

    However, I am a theoretician, not an engineer, so maybe someone can offer a practical solution :smile:
  4. Feb 10, 2009 #3
    Yes, it is possible to calculate this.

    If you have a drawing of the pipes, it would help greatly in figuring this one out.

    I need to know this:
    *How much water per hour you use.
    *The correct mix of water and sand(does sand go through your piping as well?)
    *Celsius or fahrenheit
    *Is this a closed system?
  5. Feb 10, 2009 #4


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    Add the following questions to the list:

    Do you stir your tank?
    How often do you replenish the water content and what temperature is that new water at?
  6. Feb 10, 2009 #5
    Ok, lets see:

    *How much water per hour you use.​

    The pump has a power of 28 cubic meters / hours of circulation.

    *The correct mix of water and sand(does sand go through your piping as well?)​

    The sand goes to the pipe also, and we have 100kg of sand in the water.

    *Celsius or fahrenheit​


    *Is this a closed system?
    *How often do you replenish the water content and what temperature is that new water at?
    This is a closed system, so no water comes into the system. We start with 1000 liters of water, at 80ºC, and the temperature keeps dropping.
  7. Feb 10, 2009 #6
    I missed one. I stir the tank, to keep the sand mixed.
  8. Feb 10, 2009 #7
    So, what you are asking is, how much more kW you need to have this stay at 80 C?
  9. Feb 10, 2009 #8


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    Sorry, but one more bit of confusion on my part...

    I am envisioning a wet type of sandblasting operation. Is this correct? The slurry mix is sprayed at parts at atmospheric pressure and then collected through a drain to be sent back to the reservoir. Is this what is really happening?
  10. Feb 10, 2009 #9
    That is exactly what I need to know... I am trying not to guess, or to test trial.
  11. Feb 10, 2009 #10
    That's it, wet sandblasting. I was having trouble finding the correct English word.

    You correctly described the operation.
  12. Feb 10, 2009 #11
    Did the temperature level out at 30C
    What is the environmental temperature?

    I will try my hand at the deduction.

    Assuming 10C for the environmental temperature.
    And assuming that the temperature did level out at 30C

    Power loss per deg C. = 24000/(30-10) = 1250W

    Power required = (80-10)*1250 = 87500W

    I of course could be wrong, and it would be easy to prove. You can use the math above to predict what would happen to the temperature when one of the heaters are turned off.

    I have designed and built quite a few specialty furnaces, but clearly do not have the experience that other members of this forum have so I would still seek a concurring opinion before using this advice.
  13. Feb 10, 2009 #12
    I might add to the previous post that the prediction assumes specific things, and one of them is total power input. One significant source of heat is the pump, and can change the prediction. So one would need to include actual power consumed by the pump, multiplied by the electrical efficiency of the pump (perhaps 75%).
  14. Feb 10, 2009 #13

    Ok, this is a very good idea. If I try with only 2 heaters turned on, with a total of 12000W, using the same math the temperature should stabilize at 19.6ºC.

    If I have to use around 90.000W, I must prove it first, because I cannot use electric heaters for that kind of power... and the investment in a furnace is much higher.

    Thanks, that was very helpful.
  15. Feb 11, 2009 #14


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    That's all well and good, but I see a major issue in this whole scenario; it seems that there is no real effort on anyone's part to really understand your heat losses. Here are some of the areas that would need to be pinned down to do a proper heat transfer analysis:

    - Size, material, insulation of the main tank
    - Piping run distances, pipe insulation to end use
    - Temperature of returning liquids and what rate (you are shooting this stuff through a high pressure nozzle, it will expand and drop in temperature among other losses).
    - There has to be make up water entering the system. You have to be experiencing some evaporative losses in this system.

    To me it looks like you will run into one of two results...
    1) The total power input is theoretically correct, but does not account for all of the losses in your system, so the heaters are undersized.
    2) You will grossly oversize the heaters to get it to work and will not be as efficient as you could be.

    If your contracting company was worth the money you spent, they should have spent some time characterizing your system and not only putting big immersion heaters in your tank, but also looking at ways to minimize heat loss throughout the system. It looks like they did half the job. I hope you paid them only half of the money.
  16. Feb 11, 2009 #15


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    I think the first thing you need to do is to understand the energy balance of your system, and to estimate the different parameters of it. You can draw imaginary control volume around your system and ask you self what cross the borders of this control volume. For example: X Watts are lost from the pipes by natural convection Y Watts enter your C.V from the heater Z Watts enter from the pumps etc. After you understand the energy balance of your system you can you solve the problem.
  17. Feb 12, 2009 #16


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    What's the exhaust rate (volume per unit time) from the system?

    Other thing --- 2 1/2 hrs X 24 kW ~ 200 MJ ~ 50 C X 4MJ/ton water K --- del Hvap ~ 2MJ/kg which implies you could be losing 100 liters water over the 2 1/2 hrs; you say 28 m3/hr, and what? 10 times that volume of air? If you're using a "dry" compressed air system you can easily lose that much water --- saturating it at 80 C before running it into the blasting cabinet is one approach you can take, you'll still need another 24 kW to do it.
  18. Feb 12, 2009 #17
    Whoa! I missed that one! Bystander you have quite the point. The compressed air must saturate after expansion, and that could take allot of water away. On top of that it also indicated that the system is not closed, and can carry allot of heat out through evaporation alone with out the aid of the compressed air flow.

    If all of this is correct. That will take allot of heat to maintain temperature!
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