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L33 - Fall 2003 Math 308/501502 3 Mathematical Models 3.3...

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Fall 2003 Math 308/501–502 3 Mathematical Models 3.3 Heating and Cooling [of Buildings] Wed, 17/Sep c 2003, Art Belmonte Summary Newton’s Law of Cooling/Heating The rate of change d Q / dt of the temperature of an object is proportional to the difference between the temperature M ( t ) of the surrounding medium and the temperature Q (t) of the object itself; i.e., d Q / dt = K ( M - Q ) , where K is a proportionality constant. In applications of this law, M is often constant. Heating and Cooling of Buildings A general model is d Q dt = K ( M ( t ) - Q ( t )) + H ( t ) + U ( t ) . M ( t ) is the temperature of the surrounding medium. Q ( t ) is the temperature inside the building. H ( t ) is the rate of increase in inside temperature due to people, lights, machines, etc. U ( t ) is the rate of increase/decrease in inside temperature due to heating/air conditioning, respectively. The reciprocal 1 / K is known as the time constant for the building. Note on nomenclature MATLAB on all platforms is case insensitive. That is, it internally regards T and t as different variables. The TI-89, like DOS or Windows, does not. Accordingly, I have chosen to use Q for T above, so as to pick something that will work on all platforms. (Of course, this is q on the ‘89.) Hand/MATLAB Examples 107/4 A red wine is brought up from the wine cellar, which is a cool 10 C, and then left to breathe in a room of temperature 23 C. If it takes 10 minutes for the wine to reach 15 C, when will the temperature of the wine reach 18 C and be ready to drink? Solution Let Q ( t ) be the temperature of the wine t minutes after being brought up from the cellar. Newton’s Law gives d Q dt = K ( 23 - Q ), Q ( 0 ) = 10 .
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