Best fit value of eddy thermal diffusivity

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Homework Help Overview

The discussion revolves around a mathematical model for temperature as a function of depth and time, specifically focusing on determining the best fit value of eddy thermal diffusivity (α) from temperature data collected on July 19th. The problem involves fitting an equation to the data and addressing challenges related to logarithmic calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods for fitting the temperature data to the model, including the suggestion of using non-linear least squares fitting. There are questions about how to handle values that lead to logarithms of negative numbers and whether to average the calculated values of α.

Discussion Status

Some participants have offered guidance on plotting data to analyze the relationship between temperature and depth, suggesting that a semi-log plot could help visualize the results. There is an ongoing exploration of how to derive α from the slope of the line obtained from the plot.

Contextual Notes

Participants note the importance of ensuring that temperature values used in calculations are greater than the baseline temperature (T0) to avoid undefined logarithmic results. The original poster also mentions specific constraints regarding the choice of time zero for the temperature measurements.

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Homework Statement



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

(T(y,t)-T0)/(Tsurf(t)-T0)=e^(-y2/4αt) (2)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.

Fit equation (2) to the data for July 19th to obtain the best fit value of α.

20-Dec 18-Apr 16-May 19-Jul
y (m) T(C) T(C) T(C) T(C)
0 10.8 19.1 22.2 28.4
1 10.7 18.7 21.8 27.9
2 10.5 18 21.4 28
3 10.5 17.4 21.2 27.9
4 10.5 17 21.1 27.4
5 10.5 16.4 20.7 26.2
6 10.5 16 19.3 23.6
7 10.5 15.2 17.1 21.4
8 10.5 14.7 15.6 19.3
9 10.5 13.7 14.6 17.9
10 10.5 12.9 14.1 16.8
11 10.5 12.1 13.2 15.9
12 10.5 11.6 12.7 15
13 10.5 11.1 12.1 14.1
14 10.5 10.7 11.6 13.2
15 10.4 10.4 11.3 12.4
20 10.3 9.3 9.9 10.6
25 10.3 8.9 9.4 9.8
30 10.1 8.7 9.1 9.3
35 10.3 8.7 8.8 9.1

Homework Equations


T0=10.5
t=211 since Dec. 20th is t=0

The Attempt at a Solution


I solved for α which gives the equation -y^2/[844ln(T - 10.5)/17.9]

so I started going down the list using the values from July 19th which gave me a different value for every value of T but once i got to T=9.8 I couldn't get any value for alpha because it would be the ln of a negative number... my question is how exactly should I look for the best fit value of α? is it the average value of the ones I was able to calculate? Or maybe I am missing something? Any help would greatly be appreciated.
 
Last edited:
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Sorry, here is a better look at the Temperatures from July 19th

19-Jul
T(C)
28.4
27.9
28
27.9
27.4
26.2
23.6
21.4
19.3
17.9
16.8
15.9
15
14.1
13.2
12.4
10.6
9.8
9.3
9.1
 
You probably want to do a non-linear least squares fit to the data, minimizing the error with respect to the eddy diffusivity.
 
Chestermiller said:
You probably want to do a non-linear least squares fit to the data, minimizing the error with respect to the eddy diffusivity.

Im not quite sure how I would go about doing that; but thanks for the reply.
 
Make a plot of T(y,t)-T(0) vs y2 on a semi-log plot, including only the points for which T(y,t) > T(0). The semi-log parameter should be T(y,t) - T(0). You should get something close to a straight line. The slope of this line should be -1/(4αt). Draw your best straight line in, and then calculate the slope. The points where T(y,t) < T(0) are omitted because, within experimental uncertainty, they are essentially equal to T(0).
 
so once I have the slope just solve for α?
 
Yes.
 

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