Finding Surface area using Newton's Law

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SUMMARY

This discussion focuses on applying Newton's Law of Cooling to determine the surface area of an object. The formula dQ/dt = -kA(T - Ts) is central to this analysis, where k represents the heat conduction coefficient, A is the surface area, T is the object's temperature, and Ts is the surrounding temperature. By manipulating the differential equation and understanding the convection process, one can derive the surface area from known values of thermal power exchanged (q), convection coefficient (h), and temperature differences. The discussion emphasizes the complexity of accurately calculating the convection coefficient, h, which often requires experimental correlations or advanced fluid dynamics techniques.

PREREQUISITES
  • Understanding of Newton's Law of Cooling
  • Familiarity with differential equations
  • Basic knowledge of heat transfer principles
  • Introduction to fluid dynamics concepts
NEXT STEPS
  • Study the derivation of Newton's Law of Cooling and its applications
  • Learn about calculating the convection coefficient (h) using empirical correlations
  • Explore the Navier-Stokes equations for fluid dynamics analysis
  • Investigate practical examples of heat transfer in engineering contexts
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Students studying thermodynamics, engineers involved in heat transfer analysis, and researchers exploring practical applications of Newton's Law of Cooling.

maccaman
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As part of a homework assignment, I am required to research Newton's Law of Cooling and evaluate it with three practical examples. One must be time of death. So far i have created a time of death example, and used an example to cool food. However the last one i heard someone say is that you could use Newton's Law of cooling to find the surface area of the object that is under question. I have also read that this has something to do with the value of the constant k. If anyone could help me it would be greatly appreciated.
 
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I don't know that much about this subject and I do not have a book at hand for formulas, but maybe this will help:

Newtons law of cooling is something like: dQ/dt=-kA(T-Ts) with k the "heat conduction coefficiënt" or something, A the surface area, T the temperature of the body in question and Ts the temperature of the surrounding.

If you use: dQ=CdT (C the heat capacity) you get: dT/dt=(-kA/C)(T-Ts) with the (only initially! correct) solution: T(t)=Ts+(T0-Ts)e^-(kA/C)t by applying the appropriate boundary conditions.

So the 'k' in your other question (https://www.physicsforums.com/showthread.php?t=38534) is something like kA/C. So by studying the cooling of an object you can get information about it's surface area by using the formulas above

I know the solution of the differential equation above is only an approximate solution. But maybe the equation in your other post itself isn't much more than that. Anybody care to help some more...?!
 
maccaman said:
However the last one i heard someone say is that you could use Newton's Law of cooling to find the surface area of the object that is under question. I have also read that this has something to do with the value of the constant k. If anyone could help me it would be greatly appreciated.

Of course. Do you ever have heard to talk about the use of fins like heat dissipators?. If q=A*h*(T-Ts) where:

A=surface area of convection;
h=convection coefficient ( I advise you to name it as I have done).
T=surface temperature;
Ts=surroundings temperature
q=thermal power exchanged,

Thus, if q, T, Ts and h are known, A could be calculated. You should know Newton's Law talk to us of another mechanism of heat transfer: the convection. Heat Convection is a function of the area of exchange. To completely understand the convection mechanism you must know something about Fluid Dynamics. The constant of convection h is a very hard variable of computing. If you advance in the future in this topic you will fight a lot in order to obtain an accurate value of h. Usually engineers calculate it with correlations that emphazises in one experiment in concrete. But other brave folks calculate h solving the Navier Stokes equations over the external flow.
 

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