The energy equation for this problem must take into account several approximations. We will neglect viscous...

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The energy equation for this problem must take into account several approximations. We will neglect viscous dissipation since the rod is solid. The only velocity is in the z-direction and that velocity will be constant. There will be no temperature gradient in the angular or radial directions. With that in mind, the energy equation looks like: * 3²T - pc,(v. 37) k Əz² The boundary conditions are: z=0 T = T₁ b z = Z, Solving gives: T=T T = T₁+ T=-Ca C. ( =*) EXP(-1 + ²) + C₁ exp -z 0 m exp T-T b m VZ 00 α *exp(-x + =]] The energy equation for this problem must take into account several approximations. We will neglect viscous dissipation since the rod is solid. The only velocity is in the z-direction and that velocity will be constant. There will be no temperature gradient in the angular or radial directions. With that in mind, the energy equation looks like: * 3²T - pc,(v. 37) k Əz² The boundary conditions are: z=0 T = T₁ b z = Z, Solving gives: T=T T = T₁+ T=-Ca C. ( =*) EXP(-1 + ²) + C₁ exp -z 0 m exp T-T b m VZ 00 α *exp(-x + =]]

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The equation k 0 represents the heat conduction equation in one dimension in the zdi... View the full answer