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Heating oil from -20 degrees Celcius with an immersion heater

  1. Mar 25, 2017 #1
    Hi

    I am planning on building an immersion heater to heat oil when it is -20 C degrees outside. I need some help calculating the heat transfer. I want to find out how long it would take to heat around 10 litres of oil from -20 degrees to 10 degrees using an immersion heater with a constant surface temperature of 100 degrees C and a surface area of 0.1m^2. The oil is in a iron container with a dry weight around 10kg I am stuck when it comes to accounting for the heat loss due to the cold iron and the outside air. It would be really helpful if someone had an idea which equations to combine.

    Many thanks
     
  2. jcsd
  3. Mar 25, 2017 #2

    Baluncore

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    Welcome to PF.

    A complex situation like that will come down to equations, but it will need to be a very complicated set of equations if it takes into account all heat flows. It might be easier to do an experiment to find the power in watts needed to maintain the 100°C surface temperature of the heater. In general, heat flows will be proportional to temperature difference. Do you know the composition and thermal capacity of the oil? At what temperature does the oil become a jelly? What is the external surface area of the container?

    For each watt of power that leaves the immersion heater, that watt must enter the oil. That heat will spread into the oil by conduction and by convection circulation. Circulation will depend on viscosity and so on the temperature of different parts of the oil in the container. The same effects will occur where heat is being lost through the steel container external wall.

    Outside the container, air circulation will make a big difference. There will be a convective flow upwards past the container into a thermal plume, but if the wind is blowing on the container the cooling effect will be greater.

    You will need to insulate the oil container when the temperature difference across the container wall exceeds a certain value. Without that insulation, the immersion heater will be very inefficient and may never heat the oil sufficiently. An alternative would be to avoid the immersion heater by using a jacket for the container that has an electric heater layer like an electric blanket, surrounded by thermal insulation and wind proofing.
     
  4. Mar 27, 2017 #3
    Dear Baluncore,

    Thank you for taking time to answer.

    This is a project of mine in the early planning stage.

    I am thinking of building the immersion heater using a thin metal thread with high electrical resistivity rolled up into a spiral. The thread will be insulated with rubber so that it will not be in contact with the oil. The optimal solution would be find a metal thread with a size and resisitivity so that the thread reaches a final temperature of 100 degrees Celsius using around 500W. I think that a surface temperature around 100 degrees would be good so that there is no risk for the oil to become jelly closest to the heater. The geometry of the immersion heater is limited to a cylindrical volume with a diameter of 5 mm and a height of 500 mm.
    The oil in question is ordinary synthetic engine oil that is used in cars. I think that the thermal capacity for engine oil is around 2000 J/(kg*K).
    The steel container will be protected from wind but due to the application it is not possible to thermal insulate the steel container. Therefor the air around it will be -20 degrees celsius. The container has a surface area around 0.82 m^2.

    I can calculate how much energy is needed to heat the oil from -20 to 10 degrees using Q = m * c * (T2-T1). But this does not account for the heat loss. It will also take some time heat the oil due to the small size of the heater and it would be interesting to find out if it will take a couple of hours or a day. This will determine if it is worth building or not.

    Any ideas or help calculating the time needed to heat from -20 to 10 degrees or ideas of materials and size for the thread would be appreciated.

    This project might be a bit overkill but i like challenges and building things on my spare time :)

    Many thanks
     
  5. Mar 27, 2017 #4

    mfb

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    If the concept should work at all, heat loss at 10 degrees should be notably smaller than heating power - which means your first estimate for the heating time should not be too far away from the actual heating time.
    The steel container will have the same temperature as the oil, and outside convection will dominate, which is difficult to model.

    Rubber directly next to a metal wire that is heated? I'm not sure if that is a good idea. It is not a good thermal conductor either. Is there nothing else that provides some chemical separation and ideally some electrical isolation as well?
     
  6. Mar 27, 2017 #5

    Baluncore

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    Your question cannot be answered without knowing the temperature rise of the steel container above the –20°C environment when a heat flow of 500 watts occurs out through the wall.

    It is a risky assessment, but if we assume that the steel container remains at –20°C we can estimate an average temperature for the oil.

    1. Heat flow from the heating element to the oil is proportional to element surface area multiplied by the temperature difference. The element surface will remain close to 100°C.
    The 500mm x 5mm element will have an area of 0.5 * 0.005 * Pi = 0.00785 m2
    Relative thermal resistance = 1 / 0.00785 = 127.4

    2. Heat flow from the oil to the steel container is proportional to case surface area multiplied by the temperature difference. The steel container will remain close to –20°C.
    The steel container has a surface area of 0.82 m2.
    Relative thermal resistance = 1 / 0.82 = 1.22

    So heat will flow about 100 times, (127.4 / 1.22), faster through the steel container wall per °C than from the element.

    If we assume all the oil is at the same temperature between those two surfaces and that the steel container remains at –20°C the temperature of the oil can be calculated.
    heat flow = dT / (R1 + R2) = ( 100 + 20 ) / ( 127.4 + 1.22 ) = 0.933
    127.4 * 0.933 = 118.86°C across the element surface.
    1.22 * 0.933 = 1.14°C across the steel container wall surface.

    Average temp of oil will be –20 + 1.14 = –18.86°C
    So, I really do not like your chances of heating the oil by more than a couple of degrees.
     
  7. Mar 27, 2017 #6

    mfb

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    The steel container won't stay at -20°C. The difference between container wall and air outside will be significant.
    I don't see how you calculated the thermal resistance without taking oil flow or the thickness of the steel cylinder into account.
     
  8. Mar 27, 2017 #7
    Thickness of the steel container is probably insignificant since it most likely a thin shell and one can assume the inside temperature to be the same as the outside. The flow of heat from the container to the air and also from the oil to the steel should more than make up the resistance of the steel.

    The heat produced from the heater has to travel through the oil to the shell, so here the conductivity of the oil has to be taken into account.
    If we assume no mixing for the moment, the temperature degrades from the heater wall, through the oil, to the shell interface, to the ambient air. It is not inconceivable that the oil near the heater could have several degrees higher temperature than the oil near the wall.

    The surface temperature of the heater is an unknown. It may turn out that a 100 C surface temperature is what you get, but that would be by fluke more than anything else.

    The criteria mentioned is the change in temperature of the oil - 30 C - the and the surface area of the container - 0.82 m^2.
    From http://www.engineersedge.com/heat_transfer/convective_heat_transfer_coefficients__13378.htm
    there is a table of expected convective heat transfer coefficients for air over a surface.

    Just picking a value of say 10 W/m^/C for air, a difference of container temperature and air temperature halfway between the oil temp and air temp giving 15 C, and upping the surface area to 1 m^2 for ease of calculation, gives a very CRUDE ESTIMATE of 150 W.( which is probably a lowball figure )

    Right now, even if crude, it does seem in line with the wattage of block heaters for a car with most being around 500 W or less, and they give a what 30 degree F temperature rise in the water jacket after 4 hours.
     
  9. Mar 27, 2017 #8

    Baluncore

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    I did not. I only considered the area of the film of oil on the boundary layer of the two extreme isothermal exchange surfaces.

    I expect that to be the case, which is why the thermal gradient will be steepest at the heating element which has only 1% of the steel container's surface area. As heat flows outwards, the area of the isotherms increases progressively from the element to the steel shell. I was considering that tapered resistance to heat flow and trying to show that the heating element end of the path had insufficient area to keep the element temperature below 100°C at the start of the process.

    I would change the element design to a flat strip which could have a much greater surface area. Only then can we expect a 100°C heater element to become effective. But the heating element temperature and area has been constrained by the OP specification.

    I said it was a risky assessment. When there is insufficient data, all I can do is to optimise the known parameters.
     
  10. Mar 27, 2017 #9

    mfb

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    The heater temperature won't drop - if it is below 100°C at the end, it is lower than that initially as well.

    If we take the 10W/K estimate of 256bits, we need 300 W from the heating element.

    Let's take the worst case approach: No convection, only thermal conduction, a long cylindrical case. 10 liters with a length of 500 mm leads to a radius of 8 cm.
    Thermal conductivity: 0.13 W/(m*K)
    We have an area of 0.5m*2*pi*R, leading to a temperature gradient of 300W/(0.13W/(m*K)*0.5m*2*pi*R) = 730 K/R. Integrating from 2.5 mm to 8 cm leads to a temperature gradient of 2500 K. Obviously that doesn't work. We get a 70 K drop within the first 250 micrometers.

    What about convection? At up to 90 K temperature gradient, we have to cycle through ~2 gram per second. That is a tiny amount - the nearest 250 micrometers around the heater. We need good convection close to the heater to make that work.
     
  11. Apr 3, 2017 #10

    rbelli1

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    http://www.engineeringtoolbox.com/heat-loss-open-water-tanks-d_286.html

    According to the above site it is approximately 300W/m2 for 30 degrees rise in a bare steel tank. This is for water tanks so your temperature gradient will be much greater for an oil tank.

    Once you are at temperature you will have very approximately twice the heat that you need. The temperature will be at about a delta of nearly 45C at equilibrium. That is the surface temperature so the temperature near the heater will be higher and you need to calculate for a larger heat stored in the fluid than that temperature implies.

    These numbers are fairly close to what 256bits came up with.

    The viscosity of your oil will have a huge impact on the details.

    BoB
     
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