04_thermal

025 m and surface area 12 m2 now have three

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Unformatted text preview: t’ add insulation instead of replacing the wood with insulation • Two-layer insulation – must compute temperature at interface – – – – – • Conduction plus Convection – skin temperature must be solved • Conduction plus Radiation – skin temperature must be solved each still has thickness 0.025 m; and surface area 12 m2 Now have three temperatures: Tin = 20°, Tmid, Tout = 1 0° Flow through first is: P 1 = 1·( Tin Tmid) ·A1/t1 Flow through second is: P 2 = 2·( Tmid Tout)·A2/t2 In thermal equilibrium, must have P 1 = P2 • else energy is building up or coming from nowhere • The whole enchilada – We know everything but Tmid, which we easily solve for: Tmid( 1A1/t1 + 2A 2/t2 ) = 2A 2T in/t2 + 2A2Tout/t2 – find Tmid = 9.412 or Tmid = 19.412 depending on which is interior or exterior – heat flow is 282 W (compare to 288 W before: wood hardly matters) – conduction, convection, radiation Winter 2008 13 Winter 2008 14 UCSD: Physics 121; 2008 UCSD: Physics 121; 2008 Conduction plus Convection Convection plus Radiation • Let’s take our hut with just wood, but considering Let’ convection • How warm should a room be to stand comfortably with no clothes? – The skin won’t necessarily be at Tout – Again, thermal equilibrium demands that power conducted through wall equals power wafted away in air – P = h·(Tskin Tout) ·A = ·( Tin Tskin) ·A/t – for which we find T skin = ( Tin/t + hTout)/(h + /t) = 16.7° C – so the skin is hot – P = ( 5)(26.7)(12) 1600 W – So a space heater actually could handle this (no radiation) – lesson: air could not carry heat away fast enough, so skin warms up until it can...
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This note was uploaded on 01/30/2014 for the course PHYS 121 taught by Professor Staff during the Winter '08 term at UCSD.

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