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Chapter 2 Introduction to Conduction. Conduction Rate Equation Cartesian Cylindrical Spherical Isotherm: The direction of heat flow will always be normal.

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Presentation on theme: "Chapter 2 Introduction to Conduction. Conduction Rate Equation Cartesian Cylindrical Spherical Isotherm: The direction of heat flow will always be normal."— Presentation transcript:

1 Chapter 2 Introduction to Conduction

2 Conduction Rate Equation Cartesian Cylindrical Spherical Isotherm: The direction of heat flow will always be normal to a surface of constant temperature.

3 Thermal Conductivity (W/mK) 300°K400°K500°K AgSilver429425419 AlAluminum237240236 AuGold317311304 CrChromium949186 CuCopper401393386 MoMolybdenum138134130 NiNickel918072 PbLead363534 PtPlatinum727272 SnTin676260 TiTitanium232120 WTungsten174159146 See Incropera Appendix A for full listing

4

5 Heat Diffusion Equation y x z q x+dx qxqx qzqz q z+dz q y+dy qyqy dz dy dx

6 Where Energy Balance

7 Recall so

8 From Fourier’s law Divide by

9 If k is constant or Whereis the thermal diffusivity which has units of m 2 s -1

10 Heat Diffusion Equation Another Approach y x z q x+Δx qxqx qzqz q z+Δz q y+Δy qyqy ΔzΔz ΔyΔy ΔxΔx

11 Energy Balance Recall So dividing the energy balance equation by and taking the limits to zero yields for the x direction

12 So again we find

13 Heat Diffusion Equation Cylindrical Coordinates Heat Diffusion Equation Spherical Coordinates

14 Boundary conditions for heat diffusion equation at surface x =0 unsteady State Constant Surface Temperature Constant finite surface heat flux Insulated Surface Convection x x x x

15 Temperature Distribution T1T1 T2T2 x1x1 x2x2 L One dimensional wall system No source or sink of energy Steady state Constant conductivity

16 integratingby separation of variables Integrating again by separation of variables

17 Finally we determine the constants using the boundary conditions and so and Therefore Applying Fourier’s law to determine the heat transfer rate

18 Temperature Distribution Similarly for a cylinder with no source or sink of energy, at steady state and with a constant conductivity Cold air Hot fluid L r2r2 r1r1 T2T2 T1T1

19 Integrating twice gives From the boundary conditions and we have Applying Fourier’s law to determine the heat transfer rate


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