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DQ/dt=kA(dT/dx) = Heat Transfer. What is Q(t) Solution ?

  1. Apr 9, 2015 #1

    morrobay

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    1. The problem statement, all variables and given/known data
    With the time rate of change for heat transfer:
    dQ/dt = kA(dT/dx)
    k = thermal conductivity constant
    A = boundary surface area
    dT/dx = temp gradient
    2. Relevant equations
    Can the solution for Q(t) be obtained from modifying Newtons Law of Cooling since dQ/dt is proportional to T0 - Tambient and with A , x and k as constants ?

    3. The attempt at a solution
    Q(t) = Ta + (Ta-T0)e-ktAx
     
  2. jcsd
  3. Apr 10, 2015 #2

    Simon Bridge

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    What sort of heat transfer does Newton's law of cooling work for?
    What sort of heat transfer is the equation above for?
     
  4. Apr 10, 2015 #3

    morrobay

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    Newtons Law of Cooling can apply for the cooling of a hot liquid in cooler surroundings.
    The equation above for dQ/dt applies to conductive/contact heat transfer.(edit)
    So why not similar solutions, T(t) for Newton solution and Q(t) for the above equation
     
    Last edited: Apr 10, 2015
  5. Apr 10, 2015 #4

    Simon Bridge

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    Newtonian cooling only applies where the temperature closest to the object being cooled stays the same.
    Is that the case for the other relation?
    Is the solution you proposed from considering the Newton case a solution to the DE for the contact-conduction case?
     
    Last edited: Apr 10, 2015
  6. Apr 10, 2015 #5

    morrobay

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    For Newtons DE, dT/dt = -k(T-Ta)
    with Tambient remaining the same. And dQ/dt = -kA(dT/dx) there is heat transfer with temp change. for two objects. But in both cases does this apply ? : y(t) = y0 e-kt and therefore my proposed solution.with additional constants A,and x.
    If not then what function is Q(t) ?
     
    Last edited: Apr 10, 2015
  7. Apr 10, 2015 #6

    Simon Bridge

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    That's what you have to figure out.
    Generally "it depends".
     
  8. Apr 12, 2015 #7

    morrobay

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    Ok, this is a more complete description ; Two 1kg cubes of copper one at 100 C (930 calories) the other at 400C (3720 calories) are in contact with a .5 meter conduction bar with area .0025 m2 area. Area of one face of copper cube, other faces are insulated. This is not steady state sinceT2 temp decreasing and T1 temp increasing. So dQ/dt = kA dT/dx may not apply. I assume the DE is related to dQ/dt = -k(Q2-Q1) And if the solution is related to y(t) = y0 e-kt then therefore proposed solution : Q(t) = 930 +2790e-ktAx
    Note: I could not find anything in Halliday Resnick text for this.
     
  9. Apr 13, 2015 #8

    morrobay

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    I plugged in all the values for k(t3)Ax and above is obviously not the solution for either Q(t) or T(t) for the copper cube that is 400 C at t0
     
    Last edited: Apr 13, 2015
  10. Apr 14, 2015 #9
    Is the thermal inertia of the conduction bar significant (i.e.,does it have significant heat capacity) or can its heat capacity be neglected?

    In this problem, you assume that the resistance to heat transfer within the copper cubes is negligible, so that the two copper cubes are each at a uniform temperature at all times. Is this assumption a problem for you?

    Chet
     
  11. Apr 14, 2015 #10

    morrobay

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    Yes , heat capacity of conduction bar is negligible. By uniform if you mean even heat distribution in cubes then yes
    The cubes at t0 are 100 and 400 with T1 absorbing heat and T2 transferring heat. I would appreciate the DE and Q(t) or T(t) solution. Thanks
     
    Last edited: Apr 14, 2015
  12. Apr 14, 2015 #11
    Then the temperature profile along the bar responds instantaneously to the changes in the temperature of the cubes at each of its ends, and the temperature profile along the bar is always linear. So, if TH is the temperature of the hotter cube at any time, and TC is the temperature of the colder cube at any time, then the rate of heat flow from the hotter cube to the colder cube at any time is equal to ##kA\frac{(T_H-T_C)}{L}##. Since this is the rate of heat loss of the hot cube, how is the rate of change of the hot cube temperature related to this rate of heat loss?

    Chet
     
  13. Apr 14, 2015 #12

    morrobay

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    So dQ/dt = kA/dT/dx applies to both steady( T1 and T2 fixed) and non steady state time rate of change heat transfer which of course decreases with heat loss of hotter cube. In steady state with thermal conductivity coefficient Q(t) for hotter cube can be obtained. But in non steady state situation dQ/dt rate is not linear. So how do you get Q(t) for hotter cube ? Thanks
     
    Last edited: Apr 14, 2015
  14. Apr 14, 2015 #13
    Maybe this will help:
    $$mC\frac{dT_H}{dt}=-kA\frac{(T_H-T_C)}{L}$$
    $$mC\frac{dT_C}{dt}=+kA\frac{(T_H-T_C)}{L}$$
    where m is the mass of each cube and C is the heat capacity of copper. I hope this makes sense to you, because it is correct.

    Chet
     
  15. Apr 15, 2015 #14

    morrobay

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    The question is not only about rate of heat transfer, dQ/dt, which is constant in steady state where T1 and T2 are fixed.
    And in a non steady state where T2 decreasing and T1 increasing and therefore dQ/dt is not constant. The question is for the function Q(t) , the quantity of calories in the hotter cube at any time (t) for non steady state. Since heat transfer rate dQ/dt in non steady state is not constant where both cubes have changing temperatures , I dont understand how equation above can obtain Q(t) ? For example in a non steady state at t0 T2 = 400 C. Then at suppose t8 where T2 is less than 40 0 how many calories are left in hotter cube or what is its temperature ? Even for the case in Newtons Cooling where only T2 is changing (cooling) T(t) is a DE solution. In the case above where T2 > T1 and both are changing it appears a DE and
    solution for Q(t) is involved.
     
    Last edited: Apr 15, 2015
  16. Apr 15, 2015 #15
    That's right. And all this become clearer to you once you solve those two differential equations I wrote down, subject to the initial conditions on TH and TC. Do you know how to solve those two simultaneous first order ordinary differential equations for TH and TC as functions of time? If so, let's see what you obtain.

    Chet
     
  17. Apr 15, 2015 #16
    If I subtract the cold equation from the hot equation, I get:
    $$mC\frac{d(T_H-T_C)}{dt}=-2kA\frac{(T_H-T_C)}{L}$$
    where m is the mass of each cube and C is the heat capacity of copper. The solution to this differential equation is:

    $$(T_H-T_C)=(T_{H0}-T_{C0})e^{-\frac{2kAt}{mCL}}$$

    So, Q(t) (as you call it) is given by:

    $$\frac{dQ}{dt}=\frac{kA}{L}(T_{H0}-T_{C0})e^{-\frac{2kAt}{mCL}}$$

    $$Q(t)=\frac{mC}{2}(T_{H0}-T_{C0})\left(1-e^{-\frac{2kAt}{mCL}}\right)$$

    Any questions?

    Chet[/QUOTE]
     
    Last edited: Apr 15, 2015
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