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ENPh257: Thermodynamics 00: Thermal waves

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1 ENPh257: Thermodynamics 00: Thermal waves
Β© Chris Waltham, UBC Physics & Astronomy, 2018

2 Thermal (diffusive) waves
The partial differential equation πœ•π‘‡ πœ•π‘‘ = π‘˜ π‘πœŒ πœ• 2 𝑇 πœ• π‘₯ 2 has solutions that are damped, diffusive, dispersive waves. Part of the lab task is to look for these waves. The steady-state form of the wave solutions is: 𝑇 π‘₯,𝑑 = 𝑇 0 exp βˆ’ π‘˜ πœ” π‘₯ cos πœ”π‘‘βˆ’ π‘˜ πœ” π‘₯ + 𝑇 Here, the frequency of power modulation (πœ”/2πœ‹) is chosen by you, 𝑇 0 is determined by the power amplitude, and 𝑇 is the mean temperature of the rod. The wave number π‘˜ πœ” = π‘πœŒπœ” 2π‘˜ and the wave (phase-) velocity is 𝑣=πœ”/π‘˜ πœ” = 2π‘˜πœ” π‘πœŒ (not constant). The damping distance 1/π‘˜ πœ” , which for brass, is 3 mm for 1 Hz, 3 cm for 0.01 Hz etc. The wavelength 2πœ‹/π‘˜ πœ” , which for brass, is 19 mm for 1 Hz, 19 cm for 0.01 Hz etc. Because the wavelength and damping distance are linked, observation is a trade-off between the two. Β© Chris Waltham, UBC Physics & Astronomy, 2018

3 The math This may not make sense until you have had more math courses. Postulate a wave solution to the thermal diffusion equation: 𝑇 π‘₯,𝑑 = 𝑇 0 exp {𝑗 πœ”π‘‘βˆ’ π‘˜ πœ” π‘₯ } + 𝑇 Test in the diffusion equation: π‘—πœ”= π‘˜ π‘πœŒ π‘˜ πœ” 2 𝑗 = Β±(1+𝑗)/ 2 , so π‘˜ πœ” has an oscillatory part and a exponential decay part (+ is non-physical). On a sunny day soils are hottest at noon on the surface, and hottest at midnight ~ 30 cm down. How deep would you find the soils warmest in mid-winter? Β© Chris Waltham, UBC Physics & Astronomy, 2018

4 The math On a sunny day soils are hottest at noon on the surface, and hottest at midnight ~ 30 cm down. How deep would you find the soils warmest in mid-winter? π‘—πœ”= π‘˜ π‘πœŒ π‘˜ πœ” 2 This is straightforward scaling problem: Frequency is the inverse of the period. Wavenumber is proportional to the inverse of the any characteristic length (e.g. wavelength). Multiply the period by 365, multiply all lengths by Answer: 5.7 m Temperature benefit will be the same as for diurnal changes; the x- axis of the graph on the right is just stretched, i.e. a few % of the surface amplitude. Β© Chris Waltham, UBC Physics & Astronomy, 2018


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