A decent k value for Newton's law of cooling for water?

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  • #1
Chernoobyl
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A decent "k" value for Newton's law of cooling for water?

Recently I've been trying to cool some water to a specific temperature from boiling. It doesn't have to be super accurate (within about 5° degrees Fahrenheit or 2° Celsius) but the only thermometer I have access to is just for ambient temperature, and I'm not about to dunk that thing in boiling water until the temperature drops ~30 degrees. So what I need is a k value I can use for water so I can be relatively close to my desired temperature.

What I'm assuming is that I have a magical floating cube of water where the only insulation is from the water its self. I'd also like to be able to make the constant based on various volumes of water so if that could be kept as a variable, V, that would make it much easier for me for when I plug this whole thing into excel.

It's been quite a while since I've had anything I could consider a thermodynamics class so forgive me if I'm a little slow on picking up what you're putting down (don't you love the dated phraseology?). Any and all help would be greatly appreciated.
 

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  • #2
Integral
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From the information you have, you will not be able to calculate a time or temperature. Your best bet it to do trials. To get temperature and time requires solution of a PDE, and you need good knowledge of the energy exchange. The environment is as critical in determining the final temperature as the properties of the water.
 
  • #3
Chernoobyl
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Well I can make more assumptions, tell me what I need to assume and I can give values.
 
  • #4
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What is the specific geometry you are interested in, and what are the thermal conditions applied at the boundary(ies)? For example, constant temperature surface, insulated surface, specified heat flux surface, etc. What are the physical dimensions? These are all considerations needed to arrive at your answer.
 
  • #5
Chernoobyl
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Well the water is a magical floating cube so that means the dimensions are ³√V for all sides and I'll assume it is a constant temperature surface at the ambient temperature of the room on all 6 sides... however, it doesn't seem that accurate so if it isn't too much of a bear to solve for let's make it so 5 of the 6 sides are perfectly insulated and the remaining side is the only one with the constant temperature surface (although both will probably get me close enough).
 
  • #6
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Well the water is a magical floating cube so that means the dimensions are ³√V for all sides and I'll assume it is a constant temperature surface at the ambient temperature of the room on all 6 sides... however, it doesn't seem that accurate so if it isn't too much of a bear to solve for let's make it so 5 of the 6 sides are perfectly insulated and the remaining side is the only one with the constant temperature surface (although both will probably get me close enough).
So let me get this straight. You have a glob of water which is somehow levitated in the air in the room (say glob and the room are in free fall). You are assuming that the glob is cubical, but I'm guessing a spherical shape would be just as good as a cube. Typically, the main resistance to heat transfer is going to be on the air side of the interface, so the temperature at the interface is going to be much closer to the average water temperature than to the bulk air temperature. For this type of situation, the heat transfer coefficient to a sphere is approximately
[tex]h=2k_a/D=k_a/R[/tex]
where ka is the thermal conductivity of air and D is the diameter of the sphere. This assumes that there is no natural convection, which would be the case if the sphere of water is levitated in the room since the water and the room are in free fall. Natural convection would enhance the rate of heat transfer, so using this equation for the heat transfer coefficient would predict the slowest rate of cooling. A heat balance on the sphere, assuming that the heat transfer resistance within the sphere is small compared to the heat transfer resistance outside the sphere gives:
[tex]\left(\frac{4}{3}\pi R^3\right)ρC_p\frac{dT}{dt}=-4\pi R^2h(T-T_a)[/tex]
I leave it up to you to substitute the heat transfer coefficient into this equation, and work the equation into the mathematical form of Newton's law of cooling.

Chet
 
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  • #7
Chernoobyl
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Many thanks Chestermiller. I feel nightmares from my thermodynamics classes flooding back. Really appreciate the work. Stay awesome.
 

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