A mathematical model for natural cooling of a cup of tea/coffee/ etc.

Apr 2015
13
0
hello

I have some problem modeling an approximate model of natural cooling of a cup of tea/coffee left in a closed and controlled room.

I attached a detailed PDF document (called 01.10.19) that describes all the details regarding the questions I am going to ask below.

1st questions: when I tried to take in account all the heat transfer mechanisms besides radiation (convection from the free upper surface + conduction and afterwards convection through the cup peripheral wall) I got an expression in which I have an extra variable (Tw) that makes my equation become partial differential equation rather than ordinary differential equation. The equation appears in an attached photo -called Capture1.

I tried to check what happens when assuming that conduction and afterwards convection through the cup peripheral wall are negligible, but got a horrible typical time constant, 186.6 days.
The solution I got appears in an attached photo -called Capture2.

Thanks a lot!
 

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Oct 2017
578
297
Glasgow
You forgot to include the convection heat lost from the surface of the cup to the external air:

\(\displaystyle Q_{conv} = hA_s (T_s - T_w)\)

Neglecting radiative losses to the environment, conservation of energy transfer across the boundary of the control volume at the surface of the coffee cup means that the convective heat losses have to equal the conduction heat transfer across the cup (due to Fourier's law). That should allow you to eliminate \(\displaystyle T_w\).
 
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Apr 2015
13
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hey benit13 and thanks for replying.
I originally included convection heat lost from the surface of the cup to the external air.
It appears in "capture1" as h*A2*(Tw-Ts)=(1/R2)*(Tw-Ts).

Can you help me and write in detail the relation between Ts and Tw ?
 
Oct 2017
578
297
Glasgow
Ahhh, I see...

In your thermal system you have set the total heat leaving the cup to be the sum of

1. the conduction across the surface of the cup
2. the convection across the surface of the cup
3. the convection across the surface of the coffee

but the ability for the convection to occur on the surface of the cup is dependent on the conduction across the cup... the heat that conducts across the cup doesn't leave the system immediately; it reaches the surface of the cup and then convects from it.

You should have:

\(\displaystyle Q_{out} = Q_{conv,lid} + Q_{conv,cup}\)

and

\(\displaystyle Q_{conv,cup} = Q_{cond}\)

so, in other words:

\(\displaystyle hA_2 (T_W - T_S) = k \frac{T_L - T_W}{\Delta x}\)

where \(\displaystyle \Delta x\) is the width of the cup. Rearrange that equation for \(\displaystyle T_S\) and substitute into the next one:

\(\displaystyle Q_{out} = hA_3 (T_L - T_S) + hA_2 (T_W - T_S)\)

to get an equation that only depends on \(\displaystyle T_L\) and \(\displaystyle T_W\). All this assumes a steady-state in thermal equilibrium
 
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Apr 2017
525
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No one has yet brought up evaporation loss ... this must surely be the biggest way heat is lost ...

If your model is correct it should show the coffee is below room temperature after about 4 hours ... air humidity should figure in the calculation
 
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Oct 2017
578
297
Glasgow
No one has yet brought up evaporation loss ... this must surely be the biggest way heat is lost ...

If your model is correct it should show the coffee is below room temperature after about 4 hours ... air humidity should figure in the calculation
Good point.

The OP is approximating the evaporation losses using a convection term. I just assumed the OP is okay with that assumption.

Modelling the loss of enthalpy from the coffee to the air using psychrometrics will give you the most accurate result, but it's more complex. Using an effective convective heat transfer coefficient should be a reasonable approximation for a coffee cup.
 

ChipB

PHF Helper
Jun 2010
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Morristown, NJ USA
This is slightly off topic, but this post reminds me of a question a VP of Engineering at Polaroid (remember them?) asked me during a job interview when I was a senior in college studying mechanical engineering. I suppose this was to see how well I could think in real time under a bit of pressure. No fair using pencil and paper, writing equations, etc.:

Suppose you are at work and decide to walk to the coffee machine to get a hot cup of coffee. Your intent is to get the coffee, carry it back to your desk, and drink it there. You like cream in your coffee, and you want the hottest possible mixture when you start to drink the coffee at your desk. It's a 3 minute walk from the coffee machine to your desk. Assume that the cream comes in a small container, and you have the choice of either adding the cream to the coffee when it's dispensed from the machine, or carry the coffee and creamer separately back to your desk and mix them there. Which approach gives you the warmest coffee to drink at your desk?

By the way, the job I was interviewing for was for an industrial engineer position in Polaroid's new factory manufacturing instant movie film. This was before VHS became the popular standard for home video use, but even then I was wise enough to pass on the opportunity. Polaroid went bankrupt a few years later.
 
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Apr 2017
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Suppose you are at work and decide to walk to the coffee machine to get a hot cup of coffee. Your intent is to get the coffee, carry it back to your desk, and drink it there. You like cream in your coffee, and you want the hottest possible mixture when you start to drink the coffee at your desk. It's a 3 minute walk from the coffee machine to your desk. Assume that the cream comes in a small container, and you have the choice of either adding the cream to the coffee when it's dispensed from the machine, or carry the coffee and creamer separately back to your desk and mix them there. Which approach gives you the warmest coffee to drink at your desk?
Definitely adding the cream first .... I'm always thinking about these sort of questions ....

I haven't much use for mathematical modeling ... I think this tread illustrates the limitations ... an extraordinarily difficult problem to model , the way the humid air would rise from the fluid surface is impossible to model...

The way forward is experiment ... two hot cups of water with thermocouples inside to measure temp. One cup has a drop of oil added , this will spread over the surface and stop heat loss by evaporation .... I suspect the temperature difference between the two as they cool will be dramatic.
 
Oct 2017
578
297
Glasgow
Definitely adding the cream first .... I'm always thinking about these sort of questions ....

I haven't much use for mathematical modeling ... I think this tread illustrates the limitations ... an extraordinarily difficult problem to model , the way the humid air would rise from the fluid surface is impossible to model...
It's not particularly difficult, but it requires knowledge of psychrometrics, which is usually taught in engineering rather than in physics. It's definitely not a one-line formula you can ad-hoc include into an existing model thermal model. If you hit the engineering references, there's plenty of mathematical models available that are not too crazy to implement.

I've been researching existing models of evaporative cooling. Cooling towers in particular use evaporation for the cooling of warm water for cooling systems (e.g. typically power stations and HVAC systems). There's several calculation methods. The two main ones are:

1. Merkel theory method
2. Effectiveness-NTU method

There's others too, but they're mainly to take into account the particulars of cooling tower engineering nuances rather than just evaporation in general.

I'll put some details down in another thread perhaps for anyone that's interested. However, I think it's safe to say that approximating thermal losses due to evaporation using a convection term is a reasonable assumption.