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Please help thermodynamics

  1. Mar 31, 2009 #1
    O.D. Box 14" x 14" x 6"
    I.D. Box 12" x 12" x 6"

    Thermal conductivity of walls .194 Btu.in/(ft2.h.°F)

    Internal temp. -109 F
    block of ice (water) -109 F 2" x 12" x12"

    Outside temp 80 F

    No humidity,wind, leaks, perfect seal, in perfect conditions.

    How long till internal temp is 32 F (ice included).

    Please just explain in simple terms how to find the answer.

    Thank you
     
  2. jcsd
  3. Mar 31, 2009 #2

    Mapes

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    Hi timbdgr, welcome to PF. This model may be a good candidate for the "lumped capacitance" technique in heat transfer. This approach would take the ice to be a point mass; that is; all parts of the ice are assumed to be at the same temperature. This simplifies the math considerably because you only need to calculate the necessary change in energy based on the specific heat, and you don't worry about heat transfer through the ice.

    One would also assume that heat is diffusing through the box walls evenly. Is there a typo in the box description? It looks like the top of the box has zero thickness.

    The result of the lumped capacitance model is a simply exponential formula for temperature, which starts off rising quickly but would asymptotically approach 80°F because the driving force for heat flow drops as the contents reach the ambient temperature. Of course, in this case the ice doesn't make it past 32°F.

    On board so far?
     
  4. Mar 31, 2009 #3
    O.D. should be 8" sry.
    I understand so far
     
  5. Mar 31, 2009 #4

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    OK, the equation I propose applying is

    [tex]\frac{T-T_\infty}{T_i-T_\infty}=\exp\left(-\frac{kA}{\rho V L c}t\right)[/tex]

    where [itex]T[/itex] is the ice temperature, [itex]T_\infty[/itex] is the ambient temperature, [itex]T_i[/itex] is the initial ice temperature, [itex]k[/itex] is the wall thermal conductivity, [itex]A[/itex] is the effective surface area, [itex]\rho[/itex] is the ice density, [itex]V[/itex] is the ice volume, [itex]L[/itex] is the wall thickness, [itex]c[/itex] is the ice specific heat, and [itex]t[/itex] is time.

    There are a couple possible choices for [itex]A[/itex]: surface area of the ice, inner surface area of the box, outer surface area of the box, or a weighted combination of these. I'd use the inner surface area of the box as a first pass.

    Take a second to confirm that increasing or decreasing the controlling variables changes the temperature in the correct way (e.g., increasing the volume of the ice should slow the process, or predict a higher ice temperature at any given time). Also check that the units all work. Hopefully this equation will provide a good approximation for the time it takes for the ice to warm to 32°F.

    There a description of the technique http://www.nd.edu/~msen/Teaching/IntHT/Slides/05A-Chapter,%20Sec%205.1%20-%205.3%20Black-1D.ppt" [Broken] and at numerous places on the web. The illustrations in that presentation were taken from Incropera and DeWitt's Fundamentals of Heat and Mass Transfer, and that's the first place I'd suggest looking for more details; it's an outstanding and very readable textbook.
     
    Last edited by a moderator: May 4, 2017
  6. Mar 31, 2009 #5
    I'm sorry but, I got a couple of questions. T is the final temp? exp is for()^exp. Finally can you explain what this means ".194 Btu.in/(ft2.h.°F)" isnt k usually expressed as btu/Ft2.h.F

    thanks for you help
     
  7. Mar 31, 2009 #6

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    Yes, T is the final temperature (32°F in this case), [itex]\exp[/itex] is the exponential function. Your original units were correct, if a little unusual; thermal conductivity is measured in power per unit length per unit temperature.
     
  8. Mar 31, 2009 #7
    T=32 F
    Tinf=80 F
    T1=-109 F
    k=.194 Btu.in/(ft2.h.°F)
    A=864 in2
    p=.0331178 lb/in3
    V= 288 in3
    L= 1 in
    c=.485 BTU/lb

    32-80
    -109-80 =exp

    .194 Btu.in/(ft2.h.F)*864 in2
    .0331178 lb/in3 * 288 in3 * 1in * .485 btu/lb


    does this look right?

    or do i have to convert all the units to the same?
     
    Last edited: Mar 31, 2009
  9. Mar 31, 2009 #8

    Mapes

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    The units have to all cancel in the end. It may be easier to work in SI units, where thermal conductivity is measured simply in W m-1 °C -1. There are online converters to convert your BTU-based properties. I get a thermal conductivity of about 0.03 W m-1 °C -1, which I recognize as quite low. Is your box made of polystyrene foam or something similar? I'm only asking because I had to ship a frozen vial of cells to England recently and it was quite an effort. We ended up using 20 lbs dry ice along with a polystyrene container with 3" walls, which kept the cells frozen for a day or two.
     
  10. Mar 31, 2009 #9
    I realized that when my 1 st answer was rediculous. so the closest i have gotten i still have WJ/K is there a way to cancel them? It comes from c = 1719 J/KG C and k= W/mK or should i convert them to something else.

    It is actually Divinycell H45 made by the DIAB Group. The website actually lists .028 W/mK. It only comes in 4 x 8' sheets for around 170.
     
  11. Mar 31, 2009 #10

    Mapes

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    You can cancel °C and K because the increment size is the same. And don't forget 1 W = 1 J s-1!
     
  12. Mar 31, 2009 #11
    ok i got t= -19.46?
     
    Last edited: Mar 31, 2009
  13. Mar 31, 2009 #12

    Mapes

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    Yeah... no. :smile:

    We'd expect somewhere between an hour and a day.
     
  14. Mar 31, 2009 #13
    T= 273.15K
    Ti= 194.82K
    Tinf= 299.82K
    A= .371612 m2
    p= 916.7 Kg/m3
    c= 1719 J/Kg C
    K= .194 W/mK
    V= .03097 m3
    L= .0254 m


    -.0094389 ^.254
    1239.58 =1/t

    t= -19.9554

    where did i go wrong?
     
  15. Mar 31, 2009 #14

    Mapes

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    Hard to tell; can you format the equation in LaTeX? (Click on my equation above to see the format.)
     
  16. Mar 31, 2009 #15
    i can see the format.
     
  17. Mar 31, 2009 #16
    -.0094389 ^.254
    1239.58 =1/t
    [tex]
    {t^\ (-1)} = \left(-\frac{.0094389 }{ \ 1239.58}\right)^\ (.254)
    [/tex]


    t=-19.95546
     
    Last edited: Mar 31, 2009
  18. Mar 31, 2009 #17

    Mapes

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    Are you working from

    [tex]t=-\frac{\rho V L c}{kA}\ln\left(\frac{T-T_\infty}{T_i-T_\infty}\right)\quad\mathrm{?}[/tex]

    I can't tell.
     
  19. Mar 31, 2009 #18
    [tex]\frac{T-T_\infty}{T_i-T_\infty}=\exp\left(-\frac{kA}{\rho V L c}t\right)[/tex]
    this is the one you sent me I inputed the values divided by t and solved. what is ln?
     
    Last edited: Mar 31, 2009
  20. Mar 31, 2009 #19

    Mapes

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    It's the same equation, solved for t. [itex]\ln[/itex] is the natural logarithm (e.g., [itex]\ln(e)=1[/itex], [itex]\ln(1)=0[/itex], etc.).
     
  21. Mar 31, 2009 #20
    I didnt look b4 I signed up is this sight for students only? If it is im sorry im 29 and havent ben in school for 11 years. im trying to design a bag for my job could you possibly tell me just an approximate time frame ie with in an hour or 2.
     
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