256winter2001_modprob_0121

256winter2001_modprob_0121 - cellar We assume that the...

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Mth 256 W 2001 0121 Model Problems - Bent Petersen Practice problems. Do not turn in. Newton ’s law of cooling states if an object at temperature T is in contact with an object at temperature A then, ignoring all other eFects, the rate of change of the temperature T is proportional to the diFerence in the temperatures. Thus dT dt = k ( T A ) , or dT dt + kT = kA, where k is a constant. Problem 0121 – 1 . A cup of hot coFee initially at temperature 125 . 0 ±i sb rough tin t oaroomo f temperature A (the ambient). The coFee begins to cool down and, of course the room warms up a bit. However, the heat capacity of the room is so large compared to the cup of coFee, that we may assume A remains constant. After 2 minutes the coFee is observed to have the temperature 107 . 0 K. Another minute
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Unformatted text preview: cellar. We assume that the ambient temperature varies sinusoidally about a certain mean, and we wish to compute the interior temperature. Problem 0121 – 2 . Assume we have a sealed root cellar and that the ambient temperature varies sinu-soidally about a mean A with the amplitude of the fluctuation given by A 1 . Then A ( t ) = A + A 1 sin( ωt ) Compute the resulting temperature T in the root cellar (in terms of A , A 1 , k and a parameter determined by initial conditions). Identify the steady state component of T ( t ) and compute the phase shift and the attenuation factor in the case A = 18 ◦ C, A 1 = 4 ◦ C, ω = 5 π , k = 3. You may assume T (0) = 0 if you wish – it makes no diFerence. The graph below show the ambient A ( t ) and the response T ( t ). You can use it as a rough check on your calculations. 5 10 15 20 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 t...
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