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Heat Equations of Change I Outline So far… 1. Heat Transfer Mechanisms 2. Conduction Heat Transfer 3. Convection Heat Transfer 4. Combined Heat Transfer 5. Overall Shell Heat Balances 6. Heat Equations of Change Outline 6. Heat Equations of Change 7.1. Derivation of Basic Equations 7.1.1. Differential Equation for Heat Conduction 7.1.2. Energy Equation 7.1.3. Buckingham Pi Method 7.2. Unsteady-state Conduction 7.2.1. Gurney-Lurie Charts 7.2.2. Lumped Systems Analysis Differential Equation for Heat Conduction Consider a differential element balance: Assumptions: 1. Solid conduction thermal resistance only. 2. Constant density, thermal conductivity and specific heat. Differential Equation for Heat Conduction Consider a differential element balance: Rate of HEAT in - out: βπ¦βπ§ ππ₯ π₯ − βπ¦βπ§ ππ₯ π₯+βπ₯ Rate of HEAT generation: π βπ₯βπ¦βπ§ Rate of HEAT accumulation: In Chem 16, this is mcPdT ππ πππ βπ₯βπ¦βπ§ ππ‘ Differential Equation for Heat Conduction Consider a differential element balance: Heat Balance: ππ βπ¦βπ§ ππ₯ − βπ¦βπ§ ππ₯ + π βπ₯βπ¦βπ§ = πππ βπ₯βπ¦βπ§ ππ‘ π₯ π₯+βπ₯ Dividing by βπ₯βπ¦βπ§ : ππ₯ π₯ − ππ₯ βπ₯ π₯+βπ₯ ππ + π = πππ ππ‘ Taking the limit βπ₯ → 0: πππ₯ ππ − + π = πππ ππ₯ ππ‘ Differential Equation for Heat Conduction Consider a differential element balance: πππ₯ ππ − + π = πππ ππ₯ ππ‘ Substituting Fourier’s Law: Noting that k is constant: Extending to 3D space: π ππ ππ − −π + π = πππ ππ₯ ππ₯ ππ‘ π2π ππ π + π = πππ 2 ππ₯ ππ‘ π π» 2π ππ + π = πππ ππ‘ Differential Equation for Heat Conduction Consider a differential element balance: π Recall the definition of thermal diffusivity: π» 2π ππ + π = πππ ππ‘ Measure of how quickly a material can carry heat away from a source. π πΌ= πππ Dividing everything by k: Differential Equation for Heat Conduction π»2π π 1 ππ + = π πΌ ππ‘ Differential Equation for Heat Conduction Differential Equation for Heat Conduction π» 2π π 1 ππ + = π πΌ ππ‘ Simplifications of the equation: 1) No heat generation: 2) Steady-state: 3) Steady-state & no heat generation: 1 ππ π» π= πΌ ππ‘ π 2 π» π+ =0 π 2 π»2π = 0 Fourier’s Second Law of Conduction Poisson’s Equation Laplace’s Equation Differential Equation for Heat Conduction Differential Equation for Heat Conduction π» 2π π 1 ππ + = π πΌ ππ‘ The equation in different coordinate systems: 1) Rectangular: 2) Cylindrical: 3) Spherical: π 2 π π 2 π π 2 π π 1 ππ + 2+ 2+ = 2 ππ₯ ππ¦ ππ§ π πΌ ππ‘ 1 π ππ 1 π 2 π π 2 π π 1 ππ π + 2 2+ 2+ = π ππ ππ π ππ ππ§ π πΌ ππ‘ 1 π ππ 1 π ππ 1 π 2 π π 1 ππ 2 π + 2 sin π + 2 2 + = π 2 ππ ππ π sin π ππ ππ π sin π ππ 2 π πΌ ππ‘ Differential Equation for Heat Conduction Example! Determine the steady-state temperature distribution and the heat flux in a slab in the region 0 ≤ x ≤ L for thermal conductivity k and a uniform heat generation in the medium at a rate of g0 when the boundary surface at x = 0 is kept at a uniform temperature T0 and the boundary surface at x = L dissipates heat by convection into an environment at a constant temperature T∞ with a heattransfer coefficient h. Differential Equation for Heat Conduction Example! Assumptions (or given): 1. Steady-state 2. Unidirectional heat flow (x only) 3. Constant k, ρ, cP, and h. Differential Equation for Heat Conduction: π 1 ππ π» π+ = π πΌ ππ‘ 2 π 2 π π0 + =0 2 ππ₯ π Differential Equation for Heat Conduction Example! π2 π π0 + =0 2 ππ₯ π After 1st and 2nd integration: ππ π0 = − π₯ + πΆ1 ππ₯ π 1 π0 2 π π₯ =− π₯ + πΆ1 π₯ + πΆ2 2π Differential Equation for Heat Conduction Example! ππ π0 = − π₯ + πΆ1 ππ₯ π 1 π0 2 π π₯ =− π₯ + πΆ1 π₯ + πΆ2 2π Boundary conditions: ππ‘ π₯ = 0, ππ‘ π₯ = πΏ, π(0) = π0 ππ π = β π∞ − π(πΏ) ππ₯ *The second B.C. denotes that the heat leaving by conduction is equal to the heat entering by convection. Differential Equation for Heat Conduction Example! ππ π0 = − π₯ + πΆ1 ππ₯ π 1 π0 2 π π₯ =− π₯ + πΆ1 π₯ + πΆ2 2π Applying B.C. 1: C2 = T0 Applying B.C. 2: π − π0 π0 2 πΏ + πΆ1 = β π∞ + πΏ − πΆ1 πΏ − π0 π 2π πΆ1 βπΏ + π = β π∞ − π0 − π0 πΏ βπΏ + 2π 2π Differential Equation for Heat Conduction Example! Applying B.C. 1: C2 = T0 Applying B.C. 2: π0 π0 2 π − πΏ + πΆ1 = β π∞ + πΏ − πΆ1 πΏ − π0 π 2π π0 πΏ πΆ1 βπΏ + π = β π∞ − π0 − βπΏ + 2π 2π β π∞ − π0 π0 πΏ βπΏ + 2π πΆ1 = − βπΏ + π 2π βπΏ + π πΆ2 = π0 1 π0 2 π π₯ =− π₯ + πΆ1 π₯ + πΆ2 2π After substitution… Differential Equation for Heat Conduction Example! 1 π0 2 π π₯ =− π₯ + πΆ1 π₯ + πΆ2 2π After substitution… 1 π0 2 β π∞ − π0 π0 π₯πΏ βπΏ + 2π π π₯ =− π₯ + π₯− + π0 2π βπΏ + π 2π βπΏ + π Further manipulation into a desired form: 1 π0 2 π∞ − π0 π π₯ − π0 = − π₯ + π 2π 1+ βπΏ 2π π₯ π0 π₯πΏ 1 + βπΏ − πΏ 2π 1 + π βπΏ Differential Equation for Heat Conduction Example! 1 π0 2 π∞ − π0 π π₯ − π0 = − π₯ + π 2π 1+ βπΏ 2π π₯ π0 π₯πΏ 1 + βπΏ − πΏ 2π 1 + π βπΏ Manipulating into a desired form even further: πΏ2 π∞ − π0 π₯ π0 π π₯ − π0 = − π πΏ 2π 1+ βπΏ 2π 1+ βπΏ π 1+ βπΏ Now, we introduce a new dimensionless number… π₯ π₯ − πΏ πΏ 2 Differential Equation for Heat Conduction Dim. Group Biot, Bi Ratio convection at body’s surface/ conduction within the body Equation βπΏ π Manipulating into a desired form even further: πΏ2 π∞ − π0 π₯ π0 π π₯ − π0 = − π πΏ 2π 1+ βπΏ Finally: 2π 1+ βπΏ π 1+ βπΏ π∞ − π0 π₯ π0 πΏ2 π π₯ − π0 = − 1 + 1 π΅π πΏ 2π π₯ π₯ − πΏ πΏ 1 + 2 π΅π 1 + 1 π΅π 2 π₯ π₯ − πΏ πΏ 2 Differential Equation for Heat Conduction Example! π∞ − π0 π₯ π0 πΏ2 π π₯ − π0 = − 1 + 1 π΅π πΏ 2π 1 + 2 π΅π 1 + 1 π΅π Special cases of the problem: I. The Biot Number approaches infinity. In this case, the boundary conditions should have been: ππ‘ π₯ = 0, π(0) = π0 ππ‘ π₯ = πΏ, π πΏ = π∞ *When Bi approaches infinity, then the heat transfer coefficient, h, approaches infinity also. π₯ π₯ − πΏ πΏ 2 Differential Equation for Heat Conduction Example! π∞ − π0 π₯ π0 πΏ2 π π₯ − π0 = − 1 + 1 π΅π πΏ 2π 1 + 2 π΅π 1 + 1 π΅π Special cases of the problem: I. The Biot Number approaches infinity. The resulting equation when π΅π → ∞ is: π∞ − π0 π0 πΏ2 π₯ π₯ π π₯ − π0 = π₯+ − πΏ 2π πΏ πΏ Recall the result when g0 is zero! 2 π₯ π₯ − πΏ πΏ 2 Differential Equation for Heat Conduction Example! π∞ − π0 π₯ π0 πΏ2 π π₯ − π0 = − 1 + 1 π΅π πΏ 2π 1 + 2 π΅π 1 + 1 π΅π Special cases of the problem: II. The Biot Number approaches zero. In this case, the boundary conditions should have been: ππ‘ π₯ = 0, π(0) = π0 ππ ππ‘ π₯ = πΏ, =0 ππ₯ *When Bi approaches zero, then the heat transfer coefficient, h, approaches zero also. π₯ π₯ − πΏ πΏ 2 Differential Equation for Heat Conduction Example! π∞ − π0 π₯ π0 πΏ2 π π₯ − π0 = − 1 + 1 π΅π πΏ 2π 1 + 2 π΅π 1 + 1 π΅π Special cases of the problem: II. The Biot Number approaches zero. The resulting equation when π΅π → 0 is: π0 πΏ2 π₯ π₯ π π₯ − π0 = 2 − 2π πΏ πΏ Q: What does dT/dx = 0 imply? 2 π₯ π₯ − πΏ πΏ 2 Differential Equation for Heat Conduction Exercise! A 10-cm diameter nickel-steel sphere has a thermal conductivity, k = 10 W/m-K. Within the sphere, 800 W/m3 of heat is being generated. The surrounding air is at 20°C and the heat transfer coefficient from the surroundings to the surface of the sphere is 10 W/m2-K. What is the temperature at the center of the sphere? Energy Equation Consider a differential volume element: Recall: Combined Energy Flux 1 2 π= ππ£ + ππ π + π β π + π 2 Recall: First Law of Thermodynamics Energy Equation Consider a differential volume element: Rate of Increase in KE and Internal Energy: (Accumulation) Rate of Energy IN – OUT: Rate of Work Done by External Forces, g: Energy Equation Consider a differential volume element: Combining them: Expanding the combined energy flux term… Energy Equation Consider a differential volume element: THE ENERGY EQUATION Energy Equation Consider a differential volume element: The complete form of the Energy Equation Energy Equation Consider a differential volume element: If we subtract the mechanical energy balance from the energy equation: THE EQUATION OF CHANGE FOR INTERNAL ENERGY Energy Equation Consider a differential volume element: If we subtract the mechanical energy balance from the energy equation: THE EQUATION OF CHANGE FOR INTERNAL ENERGY π π·π ππ + π» β πππ = π ππ‘ π·π‘ Energy Equation Consider a differential volume element: Putting the internal energy in substantial derivative form: By absorbing the pressure force term, U becomes H. Since Dπ» π·π‘ = π·π ππΆπ π·π‘ then at constant pressure: π·π ππΆπ = − π» β π − π β π»π π·π‘ Convenient Form! Energy Equation Special Cases of the Energy Equation: π·π ππΆπ = − π» β π − π β π»π π·π‘ 1. Fluid at constant pressure and small velocity gradients. R: C: S: π·π πππ = ππ» 2 π π·π‘ Energy Equation Special Cases of the Energy Equation: π·π ππΆπ = − π» β π − π β π»π π·π‘ 2. For solids ππ ππ = ππ» 2 π π ππ‘ Fourier’s Second Law of Conduction 3. With Heat Generation (simply added) π·π πππ = ππ» 2 π + π π·π‘ Energy Equation Example! π·π ππΆπ = − π» β π − π β π»π π·π‘ A solid cylinder in which heat generation is occurring uniformly as g W/m3 is insulated on the ends. The temperature of the surface of the cylinder is held constant at Tw K. The radius of the cylinder is r = R m. Heat flows only in the radial direction. Using the Energy Equation only, derive the temperature profile at steadystate if the solid has a constant k. Energy Equation Example! π·π ππΆπ = − π» β π − π β π»π π·π‘ Using the solids special case with cylindrical coordinates: This can be rewritten as: Energy Equation Example! π·π ππΆπ = − π» β π − π β π»π π·π‘ From here on, the solution is just the same as with the electrical wire:
