Thermodynamics HW Solutions 88

Thermodynamics HW Solutions 88 - Chapter 2 Heat Conduction...

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Chapter 2 Heat Conduction Equation 2-29 We consider a thin ring shaped volume element of width Δ z and thickness Δ r in a cylinder. The density of the cylinder is ρ and the specific heat is C. In general, an energy balance on this ring element during a small time interval Δ t can be expressed as t E Q Q Q Q z z z r r r Δ Δ = + Δ + Δ + element ) ( ) ( & & & & But the change in the energy content of the element can be expressed as Δ Δ Δ ΔΔ Δ E m C T T C r r z T T t tt t = EE tt element =− = ++ + () ( ) ρ π 2 Substituting, ( && )( )( ) QQ Cr rz TT t rr r zz z −+ −= + Δ Δ ρπ 2 Δ z r+ Δ r r Dividing the equation above by 2 rr z Δ Δ gives = 1 2 1 2 ππ + r z r r r z C t r z Δ Δ r 2 Noting that the heat transfer surface areas of the element for heat conduction in the r and z directions are Ar z == 2 Δ and , Δ respectively, and taking the limit as Δ Δ Δ t , and 0 yields t T C z T k z T k r r T kr r r ρ = +
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