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Heat transfer time

  1. May 22, 2017 #1
    How to calculate cooling time for a aluminum block in a steel crate placed in a room...both at 90°c ....how much time aluminum block will tak take to cool down to 25°c?
     
  2. jcsd
  3. May 22, 2017 #2
    What is the exact geometry of the block and the crate, the thickness of the crate wall, the heat transfer coefficient between the crate and the outside air, the thermal conductivity of the crate wall, and the geometric arrangement of the block and crate relative to one another?
     
  4. May 22, 2017 #3

    russ_watters

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    Staff: Mentor

    ...and the ambient temperature of the room?
     
  5. May 22, 2017 #4
    Aluminum block is of brick shape with dimensions 10x 10 x 10 mm.

    The steel crate wall is 5 mm.

    The heat transfer coefficient is 25 w/m2 k

    The thermal conductivity of crate wall is 40 w/m k

    The block and crate are arranged in such a way that there is no gap between both . The crate covers the block from all side except from the top face which is exposed to environment/room

    The ambient temperature is 25 deg c
     
  6. May 22, 2017 #5
    The ambient temperature is 25 deg c
     
  7. May 23, 2017 #6

    Nidum

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    Science Advisor
    Gold Member

    About 20 minutes if you sit the box on a larger steel block in the open air .

    About 40 minutes if you sit the box on a brick and put it in a cupboard .

    About 60 seconds if you drop the box in water .
     
    Last edited: May 23, 2017
  8. May 23, 2017 #7
    Can you please share the formulas u used sequence wise
     
  9. May 23, 2017 #8
    From the data you provided, it looks like the dominant resistance to heat transfer is on the outside of the crate, assuming that the heat transfer resistance between the crate and the block is negligible. If the crate are assumed to be levitated in the air (and neglecting the internal heat transfer resistance of the block and crate), the equation for heat transfer would be:
    $$(M_BC_B+M_CC_C)\frac{dT}{dt}=-hA(T-T_a)$$where the M's are the masses of the block and crate, the C's are the heat capacities of of the block and crate, h is the heat transfer coefficient, A is the total outside surface area of the block and crate, and ##T_a## is the ambient temperature
     
  10. May 23, 2017 #9
    was viewing it on cellphone so it wasnt showing the post properly.Now its clear on the mac
     
  11. May 23, 2017 #10
    $$\ln{\frac{(T-T_a)}{(T_0-T_a)}}=-\frac{hAt}{(M_BC_B+M_CC_C)}$$
     
    Last edited: May 23, 2017
  12. May 28, 2017 #11
    hello again chester.....Sir this discussion we had about the heat transfer we considered the our objects as lumped system......what if the mass is quite large and we cant use lumped approach as there are temperature difference over along the length itself. For example i have an aluminum pipe with ID=0.3m diameter and OD=1 m diameter and length=6m.........how would i find out time to cool it down to 30 c........from my understanding now due to large size of object,there will be conduction inside the pipe and then convection from the surface of the pipe exposed to environment......how can i find cooling time for this pipe......initial temperature that the pipe was heated to is 90 c and then it was left to cool down to a final temperature that is 30 c and ambient is 25 c....please help
     
  13. May 29, 2017 #12
    You are correct in what you are saying. To take the conduction within the solid into consideration, you need to learn about transient solutions to the heat conduction equation. This is not the proper format for a complete tutorial on transient heat conduction within solids. But, if I were solving the specific problem you have posed, as an experienced heat transfer guy, this is how I would proceed:

    I would focus on trying to determine the maximum possible time that the cooling would require, by bounding the answer. (I assume that you want to find out the maximum possible time you would have to wait for the object to cool.) The approach I would use would deliver not only the maximum, but also a close approximation to the actual time. I would start out by neglecting the heat transfer at the inner diameter (because of the small amount of surface there) and at the ends of the cylinder (because of the relatively large aspect ratio). So the heat transfer would then be purely radial. I would also mathematically fill in the hollow part with metal, so that I now have a solid cylinder rather than an annulus. So now I have a problem of transient radial heat conduction in an (essentially) infinite cylinder.

    From transient solutions to the heat conduction equation in a cylinder, I would know that the heat transfer coefficient on the solid side of the interface (based on the temperature difference between the interface and the bulk mean temperature) starts out very high at short times, and then decreases monotonically to a lower limiting asymptotic value at very long times. This asymptotic value is given by:$$h=8\frac{k_m}{D}$$where ##k_m## is the thermal conductivity of the solid and D is the cylinder diameter. So the heat flux at the cylinder boundary is given by:
    $$q=8\frac{k_m}{D}(T_i-\bar{T})$$where ##\bar{T}## is the average temperature of the cylinder at time t.

    Beyond this, you just solve the problem as you did for the lumped model, of course based on the overall heat transfer coefficient from the combination of conduction and external convection.

    You can see from all this that being able to solve a problem like this requires considerable experience with solution to the transient heat conduction equation, including (a) recognizing the kinds of approximations that you can make and (b) familiarity with the asymptotic limiting behavior of the solutions (for different kinds of boundary conditions). If you wanted to be even more accurate, you could use graphs in McAdams (Heat Transfer) showing the dimensionless temperature change as a function of the dimensionless time (Fourier Number) and the dimensionless external heat transfer coefficient (Biot number), or solve the transient heat conduction equation yourself. But, for what you are asking, I think that the bounding approach would be adequate.
     
  14. May 29, 2017 #13
    thakyou so much chester. after finding the conductive heat flux......how can i combine the two (bounded + lumped approah)for finding time
     
  15. May 29, 2017 #14
    The equation is now: $$\ln{\frac{(T-T_a)}{(T_0-T_a)}}=-\frac{UAt}{(MC)}$$where $$\frac{1}{U}=\frac{1}{h_c}+\frac{D}{8k_m}$$where T is the average temperature of the solid and ##h_c## is the convective heat transfer coefficient.
     
  16. May 29, 2017 #15
    Thank you chester
     
  17. May 31, 2017 #16
     

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  18. May 31, 2017 #17
    Please see the hollow cylinder.The red colored sides are insulated and the only exposed surface is the hollow pipe surfaces
     
  19. May 31, 2017 #18
    hello chester. I have posted a picture of the cylinder i was discussing with you. Please see the cylinder is insulated from the sides with red color (the bottom side isn't visible in the pic but that is also insulated similar to top side) and the yellow color surfaces in the hollow pipe area is exposed to room conditions. The length of the steel cylinder is 6 m, OD is 1m,ID is 0.3 m,To is 333k ,Ta is 298k, T is 299k , specific heat is 480 j/kg k, Thermal conductivity is 0.4 w/m k........the insulation is of NBR with specific heat is 1430 j/kg k, Thermal conductivity is 0.4 w/m k...can you help me find the cooling time.......I m getting too many days which seems wrong
     
  20. May 31, 2017 #19
    correction: thermal conductivity of steel cylinder 51.9 w/m k
     
  21. May 31, 2017 #20
    OK. What do you calculate for the overall heat transfer coefficient U?
     
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