1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Heat flow rate with non-constant k through plane wall

  1. Oct 9, 2014 #1

    Maylis

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Consider a large plane wall with thickness L whose thermal conductivity varies in a specified temperature range as
    [tex] k(T) = k_{0}(1 + \beta T) [/tex]
    where ##k_{0}## and ##\beta## are two specified constants. The wall surface at ##x=0## is maintained at a constant temperature of ##T_{1}##, while the surface at ##x = L## is maintained at ##T_{2}##. Assuming steady one dimensional heat transfer by conduction through the wall only, obtain a relation for the heat transfer rate through the wall.

    Sketch temperature and heat flux profiles for zero, positive and negative coefficients. Explain briefly why gases usually have a positive coefficient a, but liquids and solids show a negative coefficient. Water is an exceptional liquid that shows a positive coefficient β, explain.


    2. Relevant equations


    3. The attempt at a solution
    I tried this two different ways, and am stuck a little bit. The first approach I did a general energy balance
    FIRST METHOD
    [tex] \frac {dE}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x[/tex]
    [tex] \frac {d(\rho \hat c_{p} A \Delta x T)}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x[/tex]
    [tex]\rho \hat c_{p} A \Delta x \frac {dT}{dt} = \dot Q_{x} - \dot Q_{x + \Delta x} + \dot e_{gen}A \Delta x[/tex]
    divide through by ##A \Delta x##
    [tex]\lim_{\Delta x \to 0} \rho \hat c_{p} \frac {dT}{dt} = \frac {1}{A \Delta x} (\dot Q_{x} - \dot Q_{x + \Delta x}) + \dot e_{gen}[/tex]

    Assuming that no heat is generated and using Fourier's law, ##\dot Q = -kA \frac {dT}{dx}##, this reduces to

    [tex]\rho \hat c_{p} \frac {dT}{dt} = \frac {-1}{A} \frac {d \dot Q}{dx}[/tex]
    [tex]\rho \hat c_{p} \frac {dT}{dt} = \frac {-1}{A} \frac {d}{dx} (-kA \frac {dT}{dx})[/tex]
    The Area terms cancel out as well as the negative terms on the RHS,
    [tex]\rho \hat c_{p} \frac {dT}{dt} = \frac {d}{dx} (k \frac {dT}{dx})[/tex]
    With the assuming of steady state and the problem statement stating that ##k = k_{0}(1 + \beta T)##, we get

    ##\frac {d}{dx}[k_{0}(1 + \beta T) \frac {dT}{dx}] = 0##

    Now I am hesitant to try to integrate this thing, with the ##\frac {d}{dx}## not combining to make ##k_{0}(1+ \beta T) \frac {d^{2}T}{dx^2}##

    SECOND METHOD
    For this, I just went straight away with Fourier's Law
    [tex] \dot Q = -kA \frac {dT}{dx}[/tex]
    [tex] \dot Q = -k_{0}(1 + \beta T)A \frac {dT}{dx} [/tex]
    [tex] \frac {\dot Q}{-k_{0}A} dx = (1 + \beta T) dT [/tex]
    [tex] \frac {\dot Q}{-k_{0}A} \int dx = \int (1 + \beta T) dT [/tex]
    This integrates to be
    [tex] \frac {\dot Q}{-k_{0}A}x = T + \frac {\beta}{2}T^{2} + C [/tex]
    Using the boundary condition of ##T = T_{1}## at ##x = 0##, this will be
    [tex] 0 = T_{1} + \frac {\beta}{2}T_{1}^{2} + C [/tex]
    [tex] C = -T_{1} - \frac {\beta}{2}T_{1}^{2}[/tex]
    Putting back into the original equation
    [tex]\frac {\dot Q}{-k_{0}A}x = T + \frac {\beta}{2}T^{2} -T_{1} - \frac {\beta}{2}T_{1}^{2}[/tex]
    [tex] \dot Q = \frac {-k_{0}A}{x} [T + \frac {\beta}{2} T^{2} - T_{1} - \frac {\beta}{2}T_{1}^{2}][/tex]
    So I do get ##\dot Q = f(T)##, but it seems like I completely ignored the second boundary condition. It seems like this should have two constants of integration, which would be closer to what I saw in the first method. So I am both not sure how to integrate the first method, and here unsure of my boundary conditions for the second method.
     
    Last edited: Oct 9, 2014
  2. jcsd
  3. Oct 9, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    I think that in your second method you can make both integrals definite, incorporating your boundary conditions.
    $$\frac {\dot Q}{-k_{0}A} \int_0^L dx = \int_{T_1}^{T_2} (1 + \beta T) dT $$
     
    Last edited: Oct 9, 2014
  4. Oct 9, 2014 #3
    Both methods are correct, and give the same answer.
    This equation is correct. The term in brackets must be a constant if its derivative is zero. That constant is minus the heat flux.
    This result is correct so far. Just substitute the other boundary condition at x = L, and you will get the value of Q dot (which is constant, and independent of both x and T).

    You were soooo close to getting the right answer with both methods.

    Chet
     
  5. Oct 9, 2014 #4

    Maylis

    User Avatar
    Gold Member

    So then, that means

    [tex] \dot Q = \frac {-k_{0}A}{L} [T_{2} + \frac {\beta}{2} T_{2}^{2} - T_{1} + \frac {\beta}{2} T_{1}^{2}] [/tex]

    I didn't know you can just substitute two boundary conditions into one equation in sequence, that seems strange...unless there were two constants of integration
     
    Last edited: Oct 9, 2014
  6. Oct 9, 2014 #5
    In method 1 there actually are two constants of integration. In method 2, you already recognized that the heat flux is constant, and, to find out what that constant is, you needed to apply the condition on the other boundary.

    Chet
     
  7. Oct 10, 2014 #6

    Maylis

    User Avatar
    Gold Member

    I'm now trying to sketch the temperature and heat flux profiles for ##\beta = 0##, ##\beta < 0##, and ##\beta > 0##. I think for ##\beta = 0##, the temperature profile will be a straight negative slope, and heat flux is just a straight horizontal line.

    However, I am not sure about the other two

    If the thermal conductivity of the wall is changing, how can the heat flux be constant conceptually? I think the conductivity will tend to go down as temperature goes up. At the beginning the temperature T1 > T2, the thermal conductivity will be at its lowest value, but will start decreasing as it goes along. If the material isn't uniformly conducting heat, how can the heat flow rate be a constant?

    I was doing another problem, and I came across something peculiar. When thermal conductivity is a small number, it seems like the temperature profile has a very steep slope. How can this be? When you have a small thermal conductivity, you shouldn't conduct heat very well, so the temperature should not change very much.

    Also, then next part of this question
    Consider a 1.5-m-high and 0.6-m-wide plate whose thickness is 0.15 m. One side of the plate is
    maintained at a temperature of 500 K while the other side is maintained at 350 K. The thermal
    conductivity of the plate can be assumed to vary linearly in that temperature range as
    ##k(T) = k_{0}(1 + \beta T)##, where ##k_{0}## =18 W/m.K and ##\beta## = 8.7 x 10^-4 K^-1. Disregarding the edge effects and assuming steady one-dimensional heat transfer, determine the rate of heat conduction through the plate.

    I just used the equation for ##\dot Q## that I derived earlier, and got -1300 J/s. How can this be physically interpreted? If I used a convention that positive ##\dot Q## is from ##T_{1}## to ##T_{2}##, where ##T_{1} > T_{2}##, doesn't this mean that the heat is flowing backwards, from the cold side of the wall to the hot side of the wall??
     
    Last edited: Oct 11, 2014
  8. Oct 11, 2014 #7

    gneill

    User Avatar

    Staff: Mentor

    With a constant heat flux (or heat current), the temperature change per unit distance through the wall will vary with the local conductivity. So long as all the temperature changes add up to the total temperature difference between one side and the other of the wall nature is satisfied.

    It’s very much like a series electrical circuit where there’s a potential difference across a string of resistors. The current only depends upon the total resistance and total potential difference. While the resistors may have different values along the way, the current is the same through each resistor. Each resistor will have its own potential change depending upon the current and the resistor’s value.
    When you wrote Fourier’s Law the ##Q## represented the heat on the “source” side of the wall and ##\dot Q## the change in heat on the source side. So a negative value represents heat leaving the source. By the way, when I run your numbers through the formula I don’t get your –1300 J/s. I get something quite a bit larger. So it might be worthwhile checking your calculation there.
     
  9. Oct 11, 2014 #8

    Maylis

    User Avatar
    Gold Member

    So then the heat flux profile for all 3 situations will just be a straight horizontal line. Only the temperature profile will change with ##\beta##

    Also,
    ##\dot Q = - \frac {(18)(0.9)}{0.15}[350 + \frac {8.7 \times 10^{-4}}{2} (350)^2 - 500 + \frac {8.7 \times 10^{-4}}{2} (500)^2] = -1300 W##

    How do you calculate it different??
     
  10. Oct 11, 2014 #9

    gneill

    User Avatar

    Staff: Mentor

    Shouldn’t the second T2 term be negative?
     
  11. Oct 11, 2014 #10

    Maylis

    User Avatar
    Gold Member

    Woops, I was reading my post #4, I changed that sign accidentally. I get about 22,190 J/s. However, from what you said, that means heat is coming into the source, which means that heat is flowing into the hot part of the wall??
     
    Last edited: Oct 11, 2014
  12. Oct 11, 2014 #11
    Can't even been to comment on your equations, but its my understanding that as heat goes up resistance goes down.
     
  13. Oct 11, 2014 #12

    gneill

    User Avatar

    Staff: Mentor

    I retract my interpretation of sign for the direction of Q flow. Sorry about that bit of fuzzy thinking:oops: Dunno where my brain went to for that one. Your original interpretation is correct.
     
  14. Oct 11, 2014 #13
    This presents what Gneill said in a slightly different way.
    The temperature gradient decreases in such a way, that, when multiplied by the thermal conductivity, their product (the heat flux) is constant. The heat has nowhere else to go, so the local heat flux through the wall must be constant (as a function of position x).
    If the temperature difference is fixed, then, for smaller thermal conductivity, the heat flow will decrease. If the heat flow is fixed, then for smaller thermal conductivity, the temperature gradient will have to be steeper. Again, the product of the temperature gradient and the thermal conductivity is equal to the heat flux.
     
  15. Oct 11, 2014 #14

    Maylis

    User Avatar
    Gold Member

    Ok, well both can't be constant then, right? But the temperature difference is fixed in this problem as well, but I think the heat flux is as well. For ##\beta = 0##, then if the heat flux is a constant (q vs. x is a straight horizontal line), then

    ## \dot q = \frac {-k_{0}}{L} [T_{2} + \frac {\beta}{2} T_{2}^{2} - T_{1} - \frac {\beta}{2} T_{1}^{2}]##, which for a fixed ##T_{1}## and ##T_{2}## is a constant.

    going back to ##\dot q = -k_{0} \frac {dT}{dx}##, if ##\dot q## is known to be constant, then integration yields
    [tex] \frac {\dot q}{-k_{0}} \int dx = \int dT[/tex]
    [tex] \frac {\dot q}{-k_{0}} x = T + C [/tex]
    from the boundary condition at ##x = 0##, ##T = T_{1}##
    [tex] 0 = T_{1} + C [/tex]
    [tex] C = -T_{1} [/tex]
    so then
    [tex] T = \frac {\dot q}{-k_{0}} x + T_{1} [/tex]
    of course I just ignored the second boundary condition, so something still isn't write here, but it suggests the temperature gradient will be linear with a slope of ##\frac {\dot q}{-k_{0}} ##

    I suppose if I wrote it the other way and ignored the boundary condition at ##x=0##, I would get
    [tex] \frac {\dot q}{-k_{0}} \int dx = \int dT[/tex]
    [tex] \frac {\dot q}{-k_{0}} x = T + C [/tex]
    [tex] \frac {\dot q}{-k_{0}}L = T_{2} + C [/tex]
    [tex] C = \frac {\dot q}{-k_{0}}L - T_{2} [/tex]
    hence,
    [tex] T = \frac {\dot q}{-k_{0}}(x - L) + T_{2} [/tex]

    I just have a feeling these equations are wrong. At ##x = L##, then ##T = T_{2}##, which is fine. However, at ##x = 0##, ##T = \frac {\dot q L}{k_{0}} + T_{2}##, and I calculate ##T = 555.45 K##, which is not correct, it should be 500 K. The other boundary conditions yields the same, giving the correct ##T_{1}## at ##x = 0##, but seems to fall apart at ##x = L##. By the way this is for when ##\beta## is that number given in the problem statement, 8.7*10^-4
     
    Last edited: Oct 11, 2014
  16. Oct 11, 2014 #15

    gneill

    User Avatar

    Staff: Mentor

    At steady state the heat flux and temperature difference are both constant.

    The heat flux is the result of the fixed temperature difference over the net conductivity once the system settles out and the temperature profile (and thus conductivity profile) has settled into its steady state.
     
  17. Oct 11, 2014 #16
    What you're missing is that ##\dot{q}=-k\frac{(T_2-T_1)}{L}## (for the case of constant conductivity). This satisfies both your equations.

    Chet
     
  18. Oct 11, 2014 #17

    Maylis

    User Avatar
    Gold Member

    How did I not get that when I did the integration? Where did I go wrong...plus I am trying to find equation for T, not ##\dot q##. I want the temperature profile. By the way, is the heat flux rate constant no matter the sign of beta, or will the heat flux rate not always be constant.
     
  19. Oct 11, 2014 #18

    The heat flux will be constant, irrespective of beta.
    Go back to your final equation using Method 1:
    [tex]\frac{d}{dx}\left(k_0(1+\beta T)\frac{dT}{dx}\right)=0[/tex]
    Integrating once gives:
    [tex](k_0(1+\beta T)\frac{dT}{dx}=C_1[/tex]
    Integrating again gives:
    [tex]k_0(T+\beta \frac{T^2}{2})=C_1x+C_2[/tex]
    Use the boundary conditions to determine C1 and C2

    Chet
     
  20. Oct 11, 2014 #19

    Maylis

    User Avatar
    Gold Member

    Darn. Why the heck does method 2 not produce the same results? Or do I have to do some handwavy math to make it that way..
     
  21. Oct 11, 2014 #20
    Method 2 does produce the same results. When you yourself did the integration for method 2, you implicitly assumed that Q dot is constant(check the math). And this assumption was correct. If you evaluate Q dot using both boundary conditions, and then eliminate Q dot from the equation for the temperature, you get the exact same result as with method 1.

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Heat flow rate with non-constant k through plane wall
  1. (Rate of Heat Flow) (Replies: 1)

Loading...