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Heating material

  1. Aug 15, 2011 #1
    If I had an input of 4KW with
    Material S/S 304
    Density(0.29 lbs/in3)
    Specific Heat(0.12 BTU/lb/°F)

    What temperature would the S/S level off at when the 4KW is inputted??

    If I am missing any info to help solve this let me know...
     
  2. jcsd
  3. Aug 15, 2011 #2
    There are a few things we need to know to solve this:

    a) The dimensions of the SS304.
    b) The environment of the sample. For example, is it outside in the sun? In an oven? In front of a fan? Indoors?
    c) Is the 4 kW uniformally distributed on one or two sides, or concentrated at some location?
     
  4. Aug 15, 2011 #3

    a) 12" width x 12" length x12" height
    b) It is indoors but if you could expand this to a fan in front blowing on it that is something I would like to know how to include this if I know the CFM.
    c)The input is uniformally distributed on both sides.
     
  5. Aug 15, 2011 #4
    Just to clarify: this is a solid cube of steel or is it hollow on the inside?
     
  6. Aug 16, 2011 #5
    Update for givens:

    1.Material S/S 304
    2.Density(0.29 lbs/in3)
    3.Specific Heat(0.12 BTU/lb/°F)
    4. 12" width x 12" length x12" height(solid)
    5. It is indoors but if you could expand this to a fan in front blowing on it that is something I would like to know how to include this if I know the CFM.
    6. The input is uniformally distributed on both sides.
    7. Emssivity: 0.075
     
    Last edited: Aug 16, 2011
  7. Aug 16, 2011 #6

    RonL

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    I'm not sure exactly what you are trying to set up, or how to get the answer to what you ask, but some thoughts in my mind for heating in a uniform method,....I would convert AC to DC and control amps, 208volts@19.230amps = 4,000 watts. A proper sized hole of some depth, to receive a (firm) press fit conductor wire, in the center of each surface (6).
    The block of SS will weigh just about 501 pounds, a minimum shaft diameter, three point support, with (almost) pin point tips, to minimize any heat transfer. Height above floor????

    http://en.wikipedia.org/wiki/U-factor

    A fan blowing air past the block, will produce a thermal transfer from the block, that will require difficult calculations (beyond my ability or time).

    Someone else might step in, but as I see in my mind,....if the fan is blowing straight at the block,
    1. front surface, heat will radiate away from and toward the fan, the air flow will appear to be flowing over an umbrella, with little motion of air in the central area. Heat will move outward from surface and fold into the air flow, as it moves along the four sides.
    2. thermal transfer from the sides, will have a different rate due to the preheat in the air flow, from the frontal heat discharge, the air flow will continue to increase in temperature as it moves along the four sides.
    3. as the air flows past the four sides and some distance beyond the rear block surface, there should be a dome like area of thermal heated air (normal low pressure draft area) I'm not clear in my thoughts if heat from the block will make this a positive pressure area in relation to the air flow as it passes the sides, or how far down stream a swirling, mixing effect will be taking affect.

    Just a few thoughts, if i'm not too for wrong, this might give you a bit of insight of what might make u-factors hard to calculate.
    The other thing I noticed in the link, was thermal radiation factor had 12.11~45.0 ? you gave 304 as a grade, maybe it has a set rate, I did not look.

    Your block will for sure not have a uniform temperature. I think.

    Ron
     
  8. Aug 16, 2011 #7
    I am giving up on finding an exact number for now. Lets start with the basics..

    I know that 1 Watt = 1 Joule/Second

    That means if I input 4000 Kw into a piece of steel of any sizefor 60 seconds that will produce 240,000 Joules or 240KJ and will give off a temperature hotter than it was initial with no Kw input. Next I input the 4000Kw for 120 Seconds and then the amount of joules produced is at 480,000 Joules or 480kJ making the temperature hotter than ran for 60 second. Will this just go on forever where over time the more I keep the 4 Kw input running over time it will keep on adding more heat until it melts and even when it melts because the 4Kw over time it will just keep on adding more and more joules to the material or am I missing something.

    I just want to see how if I supply Kw over time will the temp rise to infinity(in a vacuum) or taper off at one point due to actual conditions around it..???
     
  9. Aug 16, 2011 #8

    RonL

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    In a word, No.
    Take for example an arc-welder, the rod makes contact with the metal that is in good contact with the ground lead. If the rod sticks to the metal, a very large amp flow will start to heat the rod, and if left in this shorted condition, it will reach a yellow to white color as it melts (assuming) the welder coils did not burn out first.
    If the rod is lifted the correct amount, a plasma arc is created, metal between the rod and ground lead is melted, the plasma field is positioned by the location of the rod tip and will melt a portion of the base metal, along with the piece being joined.

    If the rod is moved away from the molten metal (but not so far that conduction is broken), the electric generated plasma arc will increase in temperature and (I think) electromagnetic pressure due to liberated electrons, the molten metal will be blasted away, which creates a hole, or, as some people actually cut sheet or other forms of metal objects, using this method.

    The metal is in some portion vaporized, into a gas, while most is hardened by the atmosphere, into metal BB's.

    Any material will cease to hold increased temperature beyond the point of melting, you might look into vapor deposition of materials.

    Hope this is some help, if not, someone else will have to answer.

    Ron
     
    Last edited: Aug 16, 2011
  10. Aug 16, 2011 #9
    It will taper off.

    To get started, you can to a back-of-the-envelope estimate as follows:

    Q = hA (Ts - Ambient) + eps * sigma * A * (Ts^4 - Tambient^4)

    where,

    Q = heater input = 4 kW
    h = natural convection heat transfer coefficient
    A = surface area ~ area of 4 sides of cube (assume two heated)
    Ts = surface temperature of cube
    Tambient = surrounding temperature ~ 25C for indoors
    eps = emissivity of material
    sigma = Stefan–Boltzmann constant

    Solve for Ts.

    The idea is that the surface will get hot enough to move the 4 kW into the surroundings.
     
  11. Aug 16, 2011 #10
    ok from what I read...

    As the 4kW is being supplied to the material it will heat up forever and at the same time the supplied 4 Kw will be removed at the same time due to free convection, radiation, to level the temeperature at one point.

    Now this is where I am at....
    Q=M*Cp*DeltaT/ T
    <http://www.engineeringtoolbox.com/heat-up-energy-d_1055.html> [Broken]

    Q=4.0Kw, supplied to material
    Cp=.5 for s/s
    M=.= 4.776 Kg
    DeltaT=? Change in Temperature
    T=Time to heat up(seconds)<-----this is where i am confused. Am I supposed to know this already where it heats up to the max temp or is this over time. 1 second to heat up is different then 2 seconds and so forth.

    I did this stuff in the textbooks but now I am trying to relate it to before I test sutff and something does not add up because the values are way to off.
     
    Last edited by a moderator: May 5, 2017
  12. Aug 16, 2011 #11

    QuantumPion

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    If you want a time-dependent solution you will have to solve http://en.wikipedia.org/wiki/Heat_equation" [Broken]. Maybe there's some empirical charts out there somewhere. :uhh:

    You can't calculate this analytically. However you can do it empirically (with a thermometer and a stopwatch). Once you experimentally determine the steel's peak temperature and how long it takes to get there with your known input power, you can solve for the rest (total energy input, radiated power, etc).
     
    Last edited by a moderator: May 5, 2017
  13. Aug 16, 2011 #12
    math111, the simple formula I posted was for the "steady state" behavior of the solid. In other words, the temperature has stopped changing with time.

    If you would like the time dependent solution, as Post #11, discussed, you can tackle the heat equation.

    My old heat transfer text book has a nice treatment of this, and I programmed (in MathCAD) the model of a 2D finite cylinder a few months ago to solve a "skunk work" problem at work.

    This book is Incropera & DeWitt, 4th Edition, Chapter 5, Section 5.8. It has a solution for "rectangular parallelepiped". I think most good heat transfer texts have this (usually just after the "lumped capacity" discussion).
     
  14. Aug 17, 2011 #13
     
    Last edited by a moderator: May 5, 2017
  15. Aug 17, 2011 #14
    I apalogize for all the confusion with this question but maybe I am getting confused with this How power relates to temperature!!!!!

    Does using power at a constant value raise the heat and in turn raise the temperature or does it stay at a constant value over time?

    Like you use in your house a toaster. This toaster plugs in to power it and using ? amount of power and then turn it off. I know this toaster does not get hotter and hotter and hotter until it is unplugged.
     
  16. Aug 17, 2011 #15

    QuantumPion

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    Both. It will heat up until it reaches an equilibrium temperature where the power input is balanced by heat output, at which point its temperature will remain constant as long as the power is applied.

    It is exactly analogous to a light bulb or an oven element. A light bulb filament has a large surface area but small mass, so it heats up to its equilibrium temperature very quickly.

    You maybe be able to calculate an approximation for your example if you assume your steel bar is perfectly insulated except for radiation. You can calculate the peak temperature using [(q / ε σ A) + Tc^4)]^-4 = Th. Then with Th known, you can calculate the time, t = m cp (Th - Tc) / q. Accuracy not guaranteed. :)
     
  17. Aug 17, 2011 #16
    Now help me if this makes sense

    Q=MCpDeltaT
    Q=660Watts supplied for 1 min---> 1 min = .0167 Hrs---> Power Used =660W*.0167Hr = 11Watt-Hr-->Convert to BTU(37.533 Btu)
    M=.0265 Lb<--very small s/s fin plate 2"x1.375"x.035"
    Cp of stainless steel(.12 Btu/Lb-F)

    37.533 = .0265 X .12 X DeltaT
    DeltaT = 11,802.83F!!!

    How could the temp rise be this crazy unless it is right or my units are wrong?

    PS. I like using american units
     
  18. Aug 17, 2011 #17

    QuantumPion

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    You can't make an adiabatic approximation when your situation is clearly not adiabatic. The amount of power you are adding is large compared to the size of the object. 660 watts through 0.02 lb of steel is like a stadium floodlight. You can't assume that you can put that much energy in without any coming out.
     
  19. Aug 17, 2011 #18
    Does the math make sense at least. The only way I could round this down is I am using the .02 lb of material for a heating fin which 33 of them are used attached to a s/s tube so I could multiply the 600Watts for 33 heating elements instead of using 660 for 1 which you told me would be for a flood light of a stadium!! Comparing this to something real does help me have a better feel for how much power is being used in such small materials to give much larger rise easily. I am trying to just instead of testing this come up with a theoretical temp before I start modifying the different Kw versions..
     
  20. Aug 17, 2011 #19

    QuantumPion

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    Well a regular tungsten lightbulb runs at ~5000 F so you're only off by a factor of 2. :tongue2:

    You need to use the radiation power equation mentioned previously to estimate the peak temperature for a given object and power.
     
  21. Aug 17, 2011 #20
    The final word is theoretically/analytically whatever you want to call it before it is really made if in ideal conditions with what I mentioned for that S/S plate only of that size with those properties and tested/supplied with 660watts for 1 min mark( will we have a temp difference of 11,802F with a small difference when physcial tested or is that number seem crazy except you said it is off by a factor of two so is that bad or good?

    I really want to walk away with this knowing that
    Q=McpDeltaT is a reliable equation if I know most of the givens. How do people design stuff otherwise. Build and then test but if thats the case how do all these text books help us then what we used in college if their used for not ideal situations? I did take heater transfer and work for a place that does electrical heating eq. with from 500w up to 6kW for finned tubing heaters and want to be able to find temepratures but at certain points is this just overwheling to find this...

    I did take mesaurements of the 4Kw heater with a thermocouple/probe...

    1. With the forced Convection(2-3in from the exit): 160F
    2. With the forced Convection(placed on the heating fins):220F
    3. Without Forced convection: got up to 900F and stopped it at that temp due to smoke

    So how do I find these values analytically or close without designing a 6Kw...??
     
    Last edited: Aug 17, 2011
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