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3. Steady-State Conduction – One Dimension.pptx
- 2. Steady-State andOne Dimension HeatTransfer Heat loss through a wall Isothermal Heat conduction in a medium can be; one-dimensional multi-dimensional (two dimensional or three dimensional) Heat transfer problems are often classified as being: steady (also called steady state) 𝑄𝑐𝑜𝑛𝑑,𝑤𝑎𝑙𝑙 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 transient (also called unsteady).
- 3. The Plane Wall Heat flow in a normal direction Isothermal temperature at all points of the wall Heat transfer steady and one dimensional 𝑇1 and 𝑇2 are constant – steady flow rate 𝑄𝑐𝑜𝑛𝑑,𝑤𝑎𝑙𝑙 = −𝑘𝐴 𝑑𝑇 𝑑𝑥 (𝑊) 𝑄𝑐𝑜𝑛𝑑,𝑤𝑎𝑙𝑙 = 𝑘𝐴 𝑇1−𝑇2 𝐿 (𝑊) Wall parameters: k, A and L
- 4. TheThermal Resistance Concept Heat conduction through a plane wall 𝑄𝑐𝑜𝑛𝑑, 𝑤𝑎𝑙𝑙 = 𝑇1−𝑇2 𝑅𝑤𝑎𝑙𝑙 (W) 𝑅𝑤𝑎𝑙𝑙 = 𝐿 𝑘𝐴 - thermal resistance of the wall against heat conduction or conduction resistance of the wall Newton’s law of cooling for convection heat transfer rate 𝑄𝑐𝑜𝑛𝑣 = ℎ𝐴𝑠 𝑇𝑠 − 𝑇∞ 𝑄𝑐𝑜𝑛𝑣 = 𝑇1−𝑇2 𝑅𝑐𝑜𝑛𝑣 𝑅𝑐𝑜𝑛𝑣 = 1 ℎ𝐴𝑠 -thermal resistance of the surface against heat convection or convection resistance of the surface ℎ → ∞; 𝑅𝑐𝑜𝑛𝑣 = 0 For radiation 𝑄𝑟𝑎𝑑 = 𝜀𝜎𝐴𝑠 𝑇𝑠 4 − 𝑇𝑠𝑢𝑟𝑟 4 = ℎ𝑟𝑎𝑑𝐴𝑠 𝑇𝑠 − 𝑇𝑠𝑢𝑟𝑟 = 𝑇𝑠−𝑇𝑠𝑢𝑟𝑟 𝑅𝑟𝑎𝑑 𝑅𝑟𝑎𝑑 = 1 ℎ𝑟𝑎𝑑𝐴𝑠 - thermal resistance of the surface against radiation ℎ𝑟𝑎𝑑 = 𝑄𝑟𝑎𝑑 𝐴𝑠 𝑇𝑠−𝑇𝑠𝑢𝑟𝑟 = 𝜀𝜎 𝑇𝑠 2 − 𝑇𝑠𝑢𝑟𝑟 2 𝑇𝑠 − 𝑇𝑠𝑢𝑟𝑟 (𝑊 𝑚2 . 𝐾) – radiation heat transfer coefficient ℎ𝑐𝑜𝑚𝑏𝑖𝑛𝑒𝑑 = ℎ𝑐𝑜𝑛𝑣 + ℎ𝑟𝑎𝑑 - combined heat transfer coefficient (conv + rad)
- 5. Thermal Resistance Network Therefore, the rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.
- 6. TheOverall Heat-Transfer Coefficient 𝑄 = 𝑇∞1−𝑇∞2 𝑅𝑡𝑜𝑡𝑎𝑙 𝑄 = ∆𝑇 𝑅 the larger the resistance, the larger the temp drop. Expressing heat transfer through a medium as Newton’s law of cooling 𝑄 = 𝑈𝐴∆𝑇 where U is the overall heat transfer coefficient. 𝑈𝐴 = 𝑄 ∆𝑇 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑄 ∆𝑇 = 1 𝑅𝑡𝑜𝑡𝑎𝑙 ∴ 𝑈𝐴 = 1 𝑅𝑡𝑜𝑡𝑎𝑙 Therefore, for a unit area, the overall heat transfer coefficient is equal to the inverse of the total thermal resistance. No need to know the surface temperatures of walls to calculate the rate of steady heat transfer.
- 7. Multilayer PlaneWalls Total thermal resistance Different thermal conductivities Determining temp of any surface Temp drop across a layer: ∆𝑇 = 𝑄𝑅 𝑄 - constant, steady heat transfer with no heat generation
- 8. Thermal Contact Resistance Thermal contact resistance, 𝑅𝑐 is the resistance to heat transfer per unit interface area that arises when surfaces are pressed against each other. Air gaps act as insulators because of the low thermal conductivity of air. 𝑅𝑐, determined experimentally Heat transfer through the interface: 𝑄 = 𝑄𝑐𝑜𝑛𝑡𝑎𝑐𝑡 + 𝑄𝑔𝑎𝑝 OR 𝑄 = ℎ𝑐𝐴∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 where ℎ𝑐 - thermal contact conductance ℎ𝑐 = 𝑄 𝐴 ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 𝑊 𝑚2 .𝑂 𝐶 : 𝑅𝑐 = 1 ℎ𝑐 = ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 𝑄 𝐴 𝑚2 .𝑂 𝐶/𝑊) The thermal contact resistance (𝑅𝑐) for the entire interface is obtained by dividing 𝑅𝑐 by the apparent interface area A. Factors affecting thermal contact resistance: - surface roughness - material properties – temperature and pressure at the interface - type of fluid trapped at the interface - *bolts, screws, rivets Good conductors vs insulators for thermal contact resistance Minimizing thermal contact resistance: - Using thermally conducting liquid (thermal grease) – replace the air at the interface with better conducting gas (helium/hydrogen) – insert a soft metallic foil (tin/silver/nickel/aluminum) between the two surfaces
- 9. Generalized Thermal Resistance Networks Parallel layers or combined series-parallel arrangements. Results are approximate (isothermal, heat transfer btn 1+2) Assumptions in solving multidimensional heat transfer problems: any plane wall normal to the x-axis is isothermal any plane parallel to the x-axis is adiabatic
- 10. Heat Conduction in Cylinders and Spheres Fourier’s law of heat conduction for heat transfer through the cylindrical layer can be expressed as: 𝑄𝑐𝑜𝑛𝑑, 𝑐𝑦𝑙 = −𝑘𝐴 𝑑𝑇 𝑑𝑟 (𝑊) where 𝐴 = 2𝜋𝑟𝐿 𝑄𝑐𝑜𝑛𝑑, 𝑐𝑦𝑙 = 2𝜋𝐿𝑘 𝑇1−𝑇2 ln(𝑟2 𝑟1) 𝑄𝑐𝑜𝑛𝑑,𝑐𝑦𝑙 = 𝑇1−𝑇2 𝑅cy𝑙 (𝑊) 𝑅cy𝑙 = ln(𝑟2 𝑟1) 2𝜋𝐿𝑘 = ln(𝑜𝑢𝑡𝑒𝑟 𝑟𝑎𝑑𝑖𝑢𝑠 𝑖𝑛𝑛𝑒𝑟 𝑟𝑎𝑑𝑖𝑢𝑠) 2𝜋×𝐿𝑒𝑛𝑔𝑡ℎ×𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦 - is the thermal resistance of the cylindrical layer against heat conduction, or conduction resistance of the cylinder layer. Fourier’s law of heat conduction for heat transfer through a spherical layer can be expressed as: 𝑄𝑐𝑜𝑛𝑑, 𝑠𝑝ℎ = −𝑘𝐴 𝑑𝑇 𝑑𝑟 (𝑊) where 𝐴 = 4𝜋𝑟2 𝑄𝑐𝑜𝑛𝑑, 𝑠𝑝ℎ = 4𝜋𝑟2𝑟1𝑘 𝑇1−𝑇2 𝑟2−𝑟1 𝑄𝑐𝑜𝑛𝑑, 𝑠𝑝ℎ = 𝑇1−𝑇2 𝑅cy𝑙 (𝑊) 𝑅sph = 𝑟2−𝑟1 4𝜋𝑟2𝑟1𝑘 = 𝑜𝑢𝑡𝑒𝑟 𝑟𝑎𝑑𝑖𝑢𝑠 −𝑖𝑛𝑛𝑒𝑟 𝑟𝑎𝑑𝑖𝑢 4𝜋×𝑜𝑢𝑡𝑒𝑟 𝑟𝑎𝑑𝑖𝑢𝑠×𝑖𝑛𝑛𝑒𝑟 𝑟𝑎𝑑𝑖𝑢𝑠×𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦 - is the thermal resistance of the spherical layer against heat conduction, or conduction resistance of the spherical layer.
- 12. Critical Radius of Insulation 𝑄𝑐𝑜𝑛𝑣 = ℎ𝐴𝑠 𝑇𝑠 − 𝑇∞ : 𝑄𝑐𝑜𝑛𝑑 = −𝑘𝐴 𝑑𝑇 𝑑𝑥 Insulation, heat transfer area, thermal resistance Increase in conduction resistance, decrease in convection resistance Critical radius of insulation of a cylindrical body, 𝑟𝑐𝑟,𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝑘 ℎ (𝑚) Critical radius of insulation of a spherical body, 𝑟𝑐𝑟,𝑠𝑝ℎ𝑒𝑟𝑒 = 2𝑘 ℎ (𝑚) Critical radius less in forced convection Radius of electric wires - safety
- 13. HeatTransfer From Finned Surfaces 𝑄𝑐𝑜𝑛𝑣 = ℎ𝐴𝑠 𝑇𝑠 − 𝑇∞ 𝑄𝑐𝑜𝑛𝑑 = −𝑘𝐴 𝑑𝑇 𝑑𝑥 • Fins – highly conductive material like aluminum • Extruding, welding, or wrapping a thin metal sheet on a surface • Assumptions: - steady operation – no heat generation – thermal conductivity of material is constant – heat transfer coefficient, h is constant and uniform over fin surface • Many fins may decrease the overall heat transfer
- 14. Ref: Yunus A. Cengel, (2000). HeatTransfer,A PracticalApproach, Second Edition.