Heat Diffusion EquationApply this equation to a solid undergoing conduction heat transfer:E=mcpT=(ρV)cpT=ρ(dxdydz)cpTdydxEnergy balance equation: dEdtdEdtg=+−+−−qq EEdWdtin out!!12qxqx+dxxyqKATxKdydzTxq q dq qqxdxxxxxdx x x xx=−∂∂=−∂∂=+ =+∂∂+()All go to zeroHeat Diffusion Equation (2)()Energy Storage = Energy Generation + Net Heat Transfert()()()()()pxxdxxpxxpc T dxdydz qdxdydz q qqTc dxdydz qdxdydz q q dxtxTqdxdydz k dxdydzxxTTTTcqk k ktxxyyzzρρρ+∂=+−∂∂∂=+−+∂∂∂∂=−−∂∂∂∂∂∂∂∂∂=+ + +∂∂∂∂∂∂∂!!!!Generalized to three-dimensionalNote: partial differential operatoris used since T=T(x,y,z,t)Heat Diffusion Equation (3)ρρcTtqxkTxykTyzkTzcTtkTxTyTzqkTqwherexyzTTxTyTzpp∂∂=+∂∂∂∂+∂∂∂∂+∂∂∂∂∂∂=∂∂+∂∂+∂∂+=∇ +∇=∂∂+∂∂+∂∂∂∂=∇=∂∂+∂∂+∂∂=!() () ()!()!!,!,Special case1: no generation q = 0Special case 2: constant thermal conductivity k = constant is the Laplacian operatorSpecial case 3: t and q = 0 The famous Laplace'22222222222222222222200 s equation1-D, Steady Heat TransferAssume steady and no generation, 1- D Laplace' s equationfunction of x coordinate aloneNote: ordinary differential operator is used since T = T(x) onlyThe general solution of this equation can be determined by integting twice:First integration leads to dTdx Integrate again T(x) = CSecond order differential equation: need two boundary conditions todetermine the two constants C and C112dTdxTxyxt Txcons t CxC22120====+,(,,,) (),tan ..T(x=0)=100°C=C2T(x=1 m)=20°C=C1+C2, C1=-80°CT(x)=100-80x (°C)10020Tx1-D Heat Transfer (cont.)Recall Fourier' s Law:q=-kAdTdx If the temperature gradient is a constantq = constent (heat transfer rate is a constant)dTdxwhere ,,(),()=−== ==TTLTx TTx L T21120T1T2xLqkAdTdxkATTLTTLkAqTTRwhere RLkA=− =−=−=−=12 1212(/ ), :thermal resistanceq (I)T1 (V1)T2 (V2)R (R)Electric circuit analogyI = (V1-V2)/RComposite Wall Heat TransferT1T2L1L2k1k2T2T1R1=L1/(k1A)R2=L2/(k2A)T12 12 121212121111111TAlso, q= ,TT TT TTqRRRLLkA kATLTTqRTqRkA−− −== =++−=− =−Note: In the US, insulation materials are often specified in terms of their thermal resistance in (hr ft2°F)/Btu ----> 1 Btu=1055 J.R-value = L/k, R-11 for wall, R-19 to R-31 for ceiling.TR valueThe thermal resistance of insulation material can be characterized by its R-value. R is defined as the temperature difference across the insulation by the heat flux going through it:RTqTkTxxk== =∆∆∆∆∆"The typical space inside the residential frame wall is 3.5 in. Find the R-value if the wall cavity is filled with fiberglass batt. (k=0.046 W/m.K=0.027 Btu/h.ft.R)RxkftBtu h ft RRft h Btu R== = ≈−∆0 2920 02710 8 112../...( . . /
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