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Temperature Profile in Lake

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Lakes become stratified in temperature over the course of the year because of the way that heat transfer occurs. In the spring, the temperature of a lake as a function of depth is reasonably constant. During the summer, as heat enters the lake from solar radiation, the upper layers begin to warm. Since warm fluids tend to rise and the warmest water is already on the surface, there is not much mixing in the lake during this time. As a result, a temperature profile like that attached below tends to form.


    Fig 1 – Temperature profile in a stratified lake – From Thibodeaux, Environmental Chemodynamics, 2nd ed.

    Later in the year, the air becomes cooler than the water and the lake loses heat. Eventually, the surface waters cool enough that they sink, causing the lake to mix during the winter and setting up a new cycle for the following year.

    During the time of year that the lake is stratified it is of interest to know the depth at which the separation of layers occurs – because physical processes such as transport of pollutants or nutrients are very different in the different layers. The separation of layers is considered to occur at the thermocline, which is defined as the location of the steepest slope in the temperature gradient. Mathematically, this occurs at the inflection point – so the position of the thermocline can be found from the following criterion:

    (1)∂2T/∂y2=0

    where y is the depth (measured from the lake surface) and T is the temperature.

    A mathematical model for temperature as a function of depth y (in m) and time t (in days) is:

    (2)T(y,t)-T0/Tsurf(t)-T0=exp(-y2/4αt)


    where Tsurf(t) is the water temperature of the lake surface at time t, α is a property called the “eddy thermal diffusivity” and T0 is the lake temperature at time zero. Time zero must be chosen to be on a day when the lake temperature is more or less uniform.

    Here are the specific tasks:

    (1) Apply equation (1) to equation (2) and develop an expression for the location ytc of the thermocline as a function of time.

    (2) The speed at which the thermocline moves vtc is defined as

    vtc=∂ytc/∂t

    Use your results from (1) to obtain an expression for vtc as a function of time.


    2. Relevant equations

    [STRIKE]if you need to see the picture it is attached here: [/STRIKE]
    https://www.physicsforums.com/attachment.php?attachmentid=57866&d=1365975296

    3. The attempt at a solution
    First I expressed the function as T(y,t)=e^(-y2/4αt)(Tsurf(t)-T0)+T0....

    Then I got the second partial derivative with respect to y which gave me:
    ytc=1/4α2t2(Tsurf(t)-T0)e-y2/4αt=0.

    Im not sure if this is the way to go so I just wanted to check to see if I was in the right path. Also for number 2 im not sure how to differentiate with respect to t when I have Tsurf(t) as a function of time. Any help would be appreciated.
     
    Last edited by a moderator: Apr 17, 2013
  2. jcsd
  3. Apr 16, 2013 #2

    haruspex

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    That's wrong in several ways. Please post all your working to get to that point.
     
    Last edited by a moderator: Apr 17, 2013
  4. Apr 16, 2013 #3
    from T(y,t)=e^(-y2/4αt)(Tsurf(t)-T0)+T0 I know that only the exponential expression contains a y so everything else becomes a constant and the last T0 drops off...

    so for the first partial derivative this becomes (Tsurf(t)-T0)e^(-y2/4αt)(-2/4αt)

    then for the second partial its pretty much the same as the first one only now we also have a value of (-2/4αt) so this becomes: (1/4α2t2)(Tsurf(t)-T0)e^(-y2/4αt)
     
  5. Apr 16, 2013 #4

    haruspex

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    No, that would be the result of differentiating wrt y2. Use the chain rule.
     
  6. Apr 17, 2013 #5
    oh wow yeah...
    so the first derivative would come out to (Tsurf(t)-T0)e^(-y2/4αt)(-2y/4αt) which then simplifies to (-y/2αt)(Tsurf(t)-T0)e^(-y2/4αt), correct?

    And then for the second derivate I would need to product rule this all up...
     
  7. Apr 17, 2013 #6

    haruspex

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    Yes.
     
  8. Apr 17, 2013 #7
    thank you...
     
  9. Apr 17, 2013 #8
    so for the second partial I get:

    (-1/2αt)[Tsurf(t)-T0]e^(-y2/4αt)+(y2/4α2t2)[Tsurf(t)-T0]e^(-y2/4αt)

    and the second part asks me to get the partial of this with respect to "t"... when doing this does Tsurf(t) just becomes T'surf(t)?
     
  10. Apr 17, 2013 #9
    You first have to solve for the value of y that makes this expression for the 2nd derivative equal to zero.
     
  11. Apr 17, 2013 #10
    y=±sqrt(2αt)... so know I differentiate this wrt "t", correct?
     
  12. Apr 17, 2013 #11
    Yes, except that the negative value for y is located outside the lake water, so you drop that root.
     
  13. Apr 18, 2013 #12
    thank you, I really appreciate you guys help.
     
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