**TI82** TxtView file generated by CalcText - Kouri  thermx  ˙df𝒑−𝒅𝒊𝒗(𝝋 ⃗ )=𝝆.𝑪𝒑.𝒅𝑻/𝒅𝒕 Local equation for conductive heat transfer in isotropic material is: 𝒑+𝒅𝒊𝒗(𝝀.(𝒈𝒓𝒂𝒅) ⃗(𝑻))=𝝆.𝑪𝒑.𝒅𝑻/𝒅𝒕 Be careful: p can vary with position, and with time 𝝀, 𝝆 and Cp can vary with T (so also with time), and with position 𝝀, 𝝆 and Cp only depend on T (and t) if material is homogeneous They can be considered as constants for low temperature variations. The solution is T(x,y,z,t) To get it, the previous differential equation has to be solved with the help of: Initial conditions for time T0(x, y, z) = T(x, y, z, t = 0) is known Boundary conditions for space either Temperature on an area S, TS(t) = T(S,t) is known either Heat flux density on an area S, (𝝋_𝑺 (𝒕)) ⃗=𝝋 ⃗(𝑺,𝒕) either Convective condition (h,Tref) on an area S such as 𝝋_𝑺 (𝒕)=±𝒉(𝑻_𝑺 (𝒕)−𝑻_𝒓𝒆𝒇) where 𝑻_𝑺 (𝒕) unknown At contact between 2 materials: Heat flux densities are equal Temperatures Are equal only in the case of an ideal contact Are not equal if the contact is not ideal (air, which is an insulation material is present) Under these conditions , electrical analogy can be used because it can be demonstrated that : ∅=(𝑻_𝟏−𝑻_𝟐)/𝑹_𝒕𝒉 As: I=(𝑉_1−𝑉_2)/𝑅_𝑒𝑙𝑒𝑐 Rth is a the thermal resistance between temperatures T1 and T2, expressed in K/W. Rth is not expressed in Ohm If the geometry is cartesian, resistance for conduction is expressed as : 𝑅_𝑐𝑑=𝐿/(𝜆.𝑆) Resistance for convection is expressed as : 𝑅_𝑐𝑣=1/(ℎ.𝑆) If the geometry is cylindrical, resistance for conduction is expressed as : 𝑅_𝑐𝑑=𝑙𝑛(𝑅2/𝑅1)/(𝜃.𝜆.𝐿) With L the length of the cylindrical surface With 𝜃 the angle of the cylindrical surface (2𝜋 for a cylinder Résistance aillette: 𝑅_𝑓𝑖𝑛=1/(ℎ(𝑆_𝑓𝑟𝑒𝑒+𝜂.𝑆_𝑓𝑖𝑛)) 𝜂=(𝑡ℎ(𝛼.𝐿))/(𝛼.𝐿) 𝛼= √((ℎ.𝑃)/(𝜆.𝑆)) ˙‘ö