04_thermal

8 t 293 k 100 51 456 t t 105 so the room is

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Unformatted text preview: carry enough heat away—at the same time reducing T across wood – h may tend higher due to self-induced airflow with large T – also, a breeze/wind would help cool it off Winter 2008 Lecture 4 15 – assume you can put out P = 100 W metabolic power • 2000 kcal/day = 8,368,000 J in 86400 sec – – – – – – – 100 W P = h·(Tskin Tout) ·A + A (Tskin4 Tout4) (hA + 4 A T3) T with emissivity = 0.8, T = 293 K 100 = ((5)(1) + 4.56) T T = 10.5° so the room is about 310 10.5 = 299.5 K = 26.5° C = 80° F iterating (using T = 299.5); 4.56 4.87; T 10.1° assumes skin is full internal body temperature • some conduction in skin reduces skin temperature • so could tolerate slightly cooler Winter 2008 16 4 Thermal Considerations 01/17/2008 UCSD: Physics 121; 2008 UCSD: Physics 121; 2008 The whole enchilada The enchilada calculation • Let’s take a cubic box with a heat source inside and Let’ consider all heat transfers • power generated = power conducted = power convected plus power radiated away – – – – – – – P = 1 W internal source inside length = 10 cm thickness = 2.5 cm R-value = 5 so 5.67 t/ = 5 = 0.028 W/m/K effective conductive area is 12.5 cm cube A c = 0.09375 m2 external (radiative, convective) area is 15 cm cube A ext = 0.135 m2 – assume h = 5 W/K/m2, = 0.8, Text = 293 K – assume the air inside is thoroughly mixed (perhaps 1 W source is a fan!) Winter 2008 17 P = ·(Tin Tskin) ·Ac/t = ( hAext + 4 A ext T3)·( Tskin Text) – first get Tskin from convective/radiative piece – Tskin = Text + P/ (hAext + 4 A ext T3) = 20° + 1.0/(0.675+0.617) – Tskin = 20.8° (...
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