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add insulation
instead of replacing the wood with insulation • Twolayer 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.
 Winter '08
 staff
 Energy, Thermal Energy

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