# Thermal Physics: Cooling by both Conduction & Convection

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1. Jun 1, 2015

### Ying_Thes

Hi all, I am not sure if the is the right place to post the question being new to the forum, but I am looking for some help with a heat transfer experiment that I ran for my honours thesis.

Essentially the aim is to determine if warmed saline fluid bags taped to 3 different types of tree barks will cool at the same rate. We had 4 fluid bags in total all starting at the same temperature, 3 of which were taped to a rough, smooth and paper bark tree respectively and the 4th was hung in the shade and had no contact to any surfaces.

I initially thought that I would just measure the rate of heat loss per area using the following:

Heat loss C/sec/cm2 = {[k * (Ti - Tf)] / As} / t

Where k = Thermal conductivity of tree bark or air
Ti = initial temperature
Tf = final temperature
As = Surface area exposed to either wood or air
t = time

Then I read up more and decided that a time constant formula should be used instead. The problem I have is that I am not sure which formula/equation/law to use and what else I should include to compare between the cooling rates of the 4 bags. After reading multiple journals and texts, there was still no solution to this so I thought of posting it on the forum, the major issue is how do I incorporate both convective cooling by air and conductive cooling by heat loss to the cooler tree trunks into the a thermal time constant formula?

I tried using the Logarithmic Method on this webpage : http://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/SysDyn/SysDyn3TCBasic.htm#Measuring to generate a slope and get an equation for the slope (using the trendline in excel), and for one of the bags it was "y = -0.0156x + 7.4019", which didn't make any sense to me, because time constant would be -0.0156?

I also read up on lumped system analysis but it just made me very confused. Then there was interface boundary conditions where 2 materials of different thermal conductivities are in imperfect thermal contact and share a common boundary which I am not sure what to use that for.

Things to note:
• We only measured the surface temperatures of the bags and the tree trunks above and below where the bags were sitting. The data seems to indicate that the two surfaces (bag and tree trunk) never came to equilibrium after 2 hours of data recording.
• The ambient temperature and humidity fluctuates as it was not in an environmentally controlled room, but wind speed was not an issue as there was not much wind around. As all the bags were in shaded areas, I don't think radiation is of much concern.
• Each fluid bag seemed to have a temperature gradient within it, with the top of the bags warmer than the bottom despite us having mixed the fluid withing the bags prior to taping them onto the trees to ensure a uniform temperature throughout.
• All tree trunks were cooler than the saline bags and the ambient temperature was also consistently cooler than the bags.
I am looking for somebody to point me in the right direction of what kind of formulas I should be using and what kind of variables to include, if there are any texts, websites or articles I should read that would really help!

Thanks so much~

2. Jun 2, 2015

### Simon Bridge

Tree bark is primarily a thermal insulator - I would expect bags in contact with the bark to cool slightly more slowly than bags exposed on all sides to air.
It makes a difference whether the air is still or if there is a breeze - still air by a warm object gets warmer, which slows the rate of cooling.
In general the rate of cooling is proportional to the temperature difference - all other things remaining equal.

You should be using your data to reach the stated aims, not some theoretical model.
Your aim is to discover differences in the rate of cooling in the three situations compared with the control.
You do not need a mathematical formula to do this. Did you plot the data (with error bars?) Which bag cooled the fastest?

What is the context this experiment is being conducted in? (i.e. school level, work type, hobby...)

3. Jun 2, 2015

### Ying_Thes

Hi Simon,

From the rate of cooling which I initially calculated it showed that the bags against the tree trunks were cooling faster than the control which was hanging off a branch.

This experiment is part of my undergrad honours project and we were just testing the potential of koalas hugging cooler tree trunks in an attempt to keep their bodies cool during warm weather (put forward by another study). The bags, being a koala model...we realise that it's a poor model but it really is to just see whether there is a potential with the tree bark being an insulator, the hypothesis was that as the tree barks are insulators + time to reach thermal equilibrium, it is possible that the tree trunks do not act as an effective heat sink.

4. Jun 2, 2015

### Simon Bridge

Oh I see, the tree-bark is cold. That would make a difference... it would depend on how cold.
But it looks like you were anticipating a similar result to me - the Koalas body warms the tree-bark right under it and after that gains no additional cooling benefit without moving. Your data does not to support the hypothesis - aim satisfied. Well done.

5. Jun 2, 2015

### Ying_Thes

I was thinking the same in regards that I already established the results, but my supervisors seems to think that the thermal time constant plays a big role and wants me to include a thermal time constant model if possible, is there any references sources you would refer me to for that?

6. Jun 2, 2015

### Simon Bridge

How were you defining "rate of cooling" before?
Did you just take the change in temperature over the total time period?
That would be the average rate of cooling - the samples did not cool at the same rate all through the experiment though did they?

Did you take lots of data? Did you graph it?

7. Jun 2, 2015

### Ying_Thes

As I mentioned in the original post, I calculated heat loss for the bags with this formula

Heat loss C/sec/cm2 = {[k * (Ti - Tf)] / As} / t

Where k = Thermal conductivity of tree bark or air
Ti = initial temperature
Tf = final temperature
As = Surface area exposed to either tree bark or air
t = time

I measured the surface area of the bag that was in contact with the tree trunk (cooling by conduction) and subtracted that from the total surface area to get surface area of cooling by convection, then plugged in the amount for each bag with the corresponding thermal conductivities.

Yes, lots of data, lots of graphs, the temperatures were logged automatically every 15secs for around 2 hours.

8. Jun 2, 2015

### Simon Bridge

OK - the formula you used assumes that the cooling rate was a constant throughout the experiment - if that was the case, then the graph of temperature vs time would be a straight line.

[Aside: You got lots of graphs and didn't do anything with them? You got 480 data points and only used 2 in your analysis? What is the point of doing all that work if you don't use the results you get? This is basically why your supervisors want you to try harder.]
Have you ever completed any teaching laboratories as part of your more junior undergrad work?

Look at your graphs - do you have one for temperature vs time? If yes, then, is it a straight line?

9. Jun 2, 2015

### Ying_Thes

Just wondering, what do you mean by straight line? Horizontally straight? -No, it's not, Vertically straight? - No, it's not.
But it is a straight downwards slope for the duration of the experiment.

10. Jun 2, 2015

### Simon Bridge

A straight line graph follows the form of equation $y=mx+c$ ... just like you learned in secondary school.
A graph is a good candidate for a straight line if you can draw one through all the error bars such that the data points are roughly evenly distributed on either side.
If this is the case for your data, then the time constant will be the (absolute value of the) slope of the graph.

It is possible to graphs to look like they may be straight when they actually follow a slight curve so be careful.
Most cooling follows an exponential curve of form $y=A\exp[-t/T]$ where T is the time constant.

11. Jun 2, 2015

### Ying_Thes

Oh~ Thanks alot, I tend to take things too literally with mathematical language, kinda explains my grades back then.

I will do that the error distribution plot.

So if I get y = -0.0156x + 7.4019 do I solve it to get the time constant? Because I read somewhere that m (slope), in this case -0.0156, is the time constant but that is not right...is it?

I will be careful with the graph make sure to see if it is a curve or line~ Thanks again!

12. Jun 2, 2015

### Simon Bridge

if y is the temperature and x is the time, and the line is y=mx+c then the time coefficient is |m|.
That equation does not, properly, have a time constant in it. The time constant belongs to an exponential equation.

do you have a plot of log-temperature vs time?
Is that a better or worse straight line than the graph of temperature vs time?

13. Jun 3, 2015

### Ying_Thes

I have plotted a graph of log-temperature vs time by transforming the temperature data using LOG10 (data), is this correct for getting the log-temperature? The line mimics that of the temperature vs time plot very closely.

If the data does not produce a straight line plot, is there another way to calculate time constant? I ask because we had 2 temperature loggers on each fluid bag one on the outside and the other on the contact side where the bag is in contact with the tree trunk and whilst the contact side temperature is constantly decreasing in a nice straight line, the temperature on the outside is decreasing but has quite a lot of fluctuations causing wavy lines.

14. Jun 3, 2015

### Simon Bridge

You need to use the natural logarithm (log base e) rather than the log10, which is log base 10.

Temperatures cooling usually follow a curve that looks like y=Yexp(-t/T) and T is called the "time constant".
If you take the natural logarithm of both sides you get:

ln(y) = ln(Y) - t/T

The slope of the ln(y) vs t graph will be m=-1/T so T=-1/m

Fluctuations don't matter so much as long as the fluctuations are within the uncertainty limits of the experiment.
You would take the log-lin plot as before and find the best fit line.

If the fluctuations look periodic rather than random, the you may have another influence on the experiment you have not accounted for.
Remember - straight lines don't actually have a time constant - they have a slope of a time-coefficient.
Time constants are also called decay constants - they are characteristic of exp functions.

Is this a senior college or senior secondary school project?

15. Jun 4, 2015

### Ying_Thes

Hi Simon,

The plot of lnTemp vs time and Temp vs Time are virtually identical, and they both show a slight curve but the lnTemp vs Time plot does fit the straight line slightly better.

So if I get the equation for the fitted line in the format of y=mx + c how would I be able to get the Time Constant from the time-coefficient?

This is a undergraduate honours project, so I think it would be college on your side of the world.

16. Jun 4, 2015

### Simon Bridge

All right then - you will have to argue that the temp vs time plot is linear - with a slope = ... whatever slope you get.
You'll have to do this by providing the different plots and asserting that the lin-lin plot is the best fit to a line.
Use a least-squares fit to get the slope and the uncertainty on the slope.

You cannot get the time constant from the time coefficient - the graph is linear. You can cal it "the time constant" if you like but it is unlikely that this is what your supervisor is thinking of.

I am puzzled that you have not learned to do this as a 1st year tertiary student.
You should review your previous years experimental work.

17. Jun 4, 2015

### Ying_Thes

Hi Simon,
I was a Veterinary Technology major, and am currently doing Animal Sciences, not physics which would probably explain why I haven't learned about the time constant etc and physics in general since we didn't have any classes for those. This is my first time actually doing thing kind of experiments.

Thanks for all the help.