1 D conduction

# 1 D conduction - CHAPTER 3 STEADY STATE 1-D CONDUCTION We...

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C HAPTER 3: S TEADY S TATE 1-D C ONDUCTION We already mentioned the parallels between heat flow and electrical current flow and the idea of k and h being measures of the resistance to heat transfer through a medium For electricity, resistance is related to the flow (of electrons) by Ohm’s Law: In parallel, for heat flow: R T q Δ = We will generate expressions for thermal resistance R in different cases Voltage – Driving Force R V I = Electrical Resistance to Flow Temperature Difference - Driving Force Thermal Resistance to Flow Flow (of electrons) Flow (of heat)

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Conduction Through Plane Walls In general: = + + + q z T k z y T k y x T k x & t T c p ρ For a slab or wall at steady state with no internal heat generation (applying the semi-infinite assumption) 0 = x T k x barb2right 0 2 2 = x T if k =constant Integrating this expression: T ( x ) = C 1 x + C 2 barb4right y = mx + b linear T profile inside slabs with no heat generation Since x T kA q = , therefore 1 kAC q = where C 1 =slope of temp. profile
If C 1 is the slope, x T x x T T C Δ Δ = = 1 2 1 2 1 So, R T x T kA q Δ = Δ Δ = Thus, thermal resistance is defined as kA x R Δ = (slab, 1-D, steady state) We could also get this result by applying our boundary conditions: In general, T ( x ) = C 1 x + C 2 B.C. #1 – Dirichlet - T=T 1 at x= 0 barb2right T 1 = C 1 (0)+ C 2 barb2right T 1 = C 2 B.C. #2 – Dirichlet - T=T 2 at x=L barb2right T 2 = C 1 L+C 2 = C 1 L+T 1 x T L T T C Δ Δ = = 1 2 1 x T 1 T 2 q x L

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Conduction Through Cylinders For a tube or hollow cylinder , it is harder to give an equation since the semi-infinite assumption may not hold i.e. the axis over which the heat flow is occurring may be different Radial flow is most common: 0 1 = r T kr r r (no heat generation) if k is constant 0 = r T r r Integrating, 1 C r T r = , r C r T 1 = Therefore, the temperature profile is: 2 1 ln C r C T + = Thus, in a cylinder, the temperature profile is logarithmic over the radius.
Applying boundary conditions: T = T i @ r = r i T = T o @ r = r o Solve for C 1 , C 2 via substitution: = o i o i r r T T C ln 1 ( ) ( ) = o i i o i o i i r r r T T r r T C ln ln ln 2 Substituting into 2 1 ln C r C T + = : o o i o o i T r r r r T T T + = ln ln ) ( or = o i o o i o r r r r T T T T ln ln and = o i o i r r T T r r T ln 1 (since d/dx(ln ax)=1/x) r o r i T o T i

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For a cylinder, the rate of heat flow is: r T r kA q = ) ( A(r)=radial cross-sectional area= 2 π rL Therefore, ( ) r T rL k q = π 2
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