Report

Heat Equations of Change II Outline 7. 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 Quiz – 2014.02.14 An electrically heated resistance wire has a diameter of 2 mm and a resistance of 0.10 ohm per foot of wire. The thermal conductivity of the wire is 20 W/m·K. At a current of 100 A, calculate the steady state temperature difference between the center and surface of the wire. TIME IS UP!!! From the previous lecture… 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? A Comparison Between Methods Solution! Energy Equation: = − ∙ − ∙ In spherical coordinates: For solids with generation = 2 + + We are left with: 1 2 0= 2 Because T depends on r only: 2 2 0= + + From the Energy Equation A Comparison Between Methods Solution! Another way!! Differential Equation of Heat Conduction: 2 1 + = In spherical coordinates: 1 1 1 2 1 2 + 2 sin + 2 2 + = 2 sin sin 2 We are left with: 1 2 0= 2 + Because T depends on r only: 2 From the Diff. Eqn 2 0= + of Heat Conduction A Comparison Between Methods Solution! Another way again!! The Shell: Overall Shell Heat Balance: Rate of Heat IN: 4 2 Rate of Heat OUT: 4 2 Generation: 4 2 ∆ +∆ Adding the terms and dividing 4∆: 2 2 − +∆ ∆ = 2 2 = 2 A Comparison Between Methods Solution! Another way again!! The Shell: Overall Shell Heat Balance: Rate of Heat IN: 4 2 Rate of Heat OUT: 4 2 Generation: 4 2 ∆ +∆ Inputting Fourier’s Law: 2 − = 2 2 = 2 A Comparison Between Methods Solution! Another way again!! The Shell: Overall Shell Heat Balance: Rate of Heat IN: 4 2 Rate of Heat OUT: 4 2 Generation: 4 2 ∆ +∆ Inputting Fourier’s Law: 2 − = 2 2 From the Overall 0= 2 + Shell Heat Balance A Comparison Between Methods Solution! Apparently… Using any method below, we can obtain the same ODE to solve!! Increasing range of applicability of the method 2 0= 2 + From the Energy Equation 2 From the Diff. Eqn 0= 2 + of Heat Conduction 2 From the Overall 0= 2 + Shell Heat Balance A Comparison Between Methods Solution! The next step is to solve the ODE and define boundary conditions. Integrating twice: 2 1 () = − − + 2 6 Boundary conditions: = 0, = , = ℎ − 2 0= 2 + From the Energy Equation 2 From the Diff. Eqn 0= 2 + of Heat Conduction 2 From the Overall 0= 2 + Shell Heat Balance A Comparison Between Methods Solution! Applying the boundary conditions: = 10 ℎ = 10 2 2 = + 1− 6 = 0.05 = 800 3 = 20° 2 + 3ℎ Final Answer: 21.37°C Outline 7. 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 1-D Transient Heat Conduction Example! Consider a flat slab of thickness L with an initial temperature of T0. It is later submerged in a large fluid with temperature T1 (> T0). Neglecting surface resistance (infinite h), determine the temperature at the center of the slab at any time t. 1-D Transient Heat Conduction Example! What do we expect to happen? • The surface temp. is held at T1. (because of infinite h) • Temperature profile at any time is symmetrical • Unidirectional (x only) heat flow only • The temp. at the center, T0, slowly approaches T1. T0 T1 1-D Transient Heat Conduction Example! Solution: Differential Equation of Heat Conduction: With the assumptions: 2 1 + = 2 = 2 Initial and Boundary conditions: Initial: Boundary: T0 T1 1-D Transient Heat Conduction Example! Solution: To facilitate solving, we will make this dimensionless! Implications: 2 = 2 Let: Y = dimensionless temp. 1 − = 1 − 0 *The choice of definition of Y is arbitrary. −1 = 1 − 0 −1 = 1 − 0 2 −1 2 = 2 1 − 0 2 2 − 1 − 0 = − 1 − 0 2 1-D Transient Heat Conduction Example! Solution: To facilitate solving, we will make this dimensionless! Implications: 1 = 2 = 2 Let: X = dimensionless length = *The choice of definition of X is arbitrary. 2 1 = 2 1 = 2 2 2 = 2 2 1-D Transient Heat Conduction Example! Solution: To facilitate solving, we will make this dimensionless! 2 = 2 2 Let: τ = dimensionless time = 2 *The choice of definition of τ is arbitrary. Implications: = 2 2 = 2 1-D Transient Heat Conduction Example! Solution: DIMENSIONLESS FORM: 2 = 2 Previous IC & BC: Easier to solve and no worries with units! New IC & BC: = 1, = 0, = 0, 1 − = 1 − 0 = = 0, = , = , = 2 = =0 =1 1-D Transient Heat Conduction A new dimensionless number… Dim. Group Fourier, Fo Ratio Rate of heat conduction/ Rate of heat storage Rate of heat conduction: 3 Rate of heat storage: 3 / 1 − = 1 − 0 = Equation 2 = 2 1-D Transient Heat Conduction Going back… DIMENSIONLESS FORM: 2 = 2 T0 New IC & BC: = 1, = 0, = 0, = 0, = , = , 1 − = 1 − 0 T1 = =0 =1 = = 2 1-D Transient Heat Conduction Going back… DIMENSIONLESS FORM: New IC & BC: = 1, = 0, = 0, 2 = 2 = 0, = , = , Solution: (A Fourier series) 4 (, ) = ∞ 2 2 − 2+1 =0 1 − = 1 − 0 2 + 1 sin 2 + 1 = = 2 = =0 =1 Gurney-Lurie Charts Gurney-Lurie Charts - plots of the dimensionless temperature Y against Fo with varying Bi and X for different geometries. - Each point in the curves are solutions to the PDE involving heat conduction + convection. 1 − = 1 − 0 = = 2 Gurney-Lurie Charts 1 − = 1 − 0 For flat plates with convection Geankoplis, Figure 5.3-6 = 2 Bi Gurney-Lurie Charts 1 − = 1 − 0 For long cylinders with convection Geankoplis, Figure 5.3-8 = 2 Bi Gurney-Lurie Charts 1 − = 1 − 0 For solid spheres with convection Geankoplis, Figure 5.3-10 = 2 Bi Gurney-Lurie Charts How do we compute the Biot Numbers for different geometries? Recall: Biot Number ℎ1 Characteristic length: = Volume/Surface Area From a while ago… Recall this 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/m2K. What is the temperature at the center of the sphere? From a while ago… Recall this exercise: A 10-cm diameter nickel-steel sphere has a thermal conductivity, k = 10 W/m-K. Within the sphere, 800 W/m3of heat is being generated.The sphere, initially at T0 = 30°C, is suddenly submerged into a fluid… The surrounding air is at ∞ = °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 across time? From a while ago… To solve the new problem: Differential Equation of Heat Conduction: 1 + = 2 In spherical coordinates: 1 1 1 2 1 2 + 2 sin + 2 2 + = 2 sin sin 2 Steps: 1. Turn the remaining PDE into dimensionless form. 2. Solve analytically for T(r, t). But there is an easier way! Lumped Systems Analysis Let’s assume that the sphere is too small for conduction to matter. The temperature distribution inside the sphere can, therefore, be assumed uniform! Heat Balance: ℎ ℎℎ = Initial Condition: = 0 , =0 ℎ ∞ − () = () ℎ = ∞ − Lumped Systems Analysis Initial Condition: = 0 , =0 () ℎ = ∞ − ℎ = ∞ − () Integrating and plugging the IC: ∞ − () ℎ − ln = ∞ − 0 Rearranging: () − ∞ −ℎ = exp 0 − ∞ Lumped Systems Analysis Temperature Profile at the center of the sphere across time: () − ∞ = − 0 − ∞ Rearranging: − −ℎ () − ∞ −ℎ = exp 0 − ∞ 2 Lumped Systems Analysis Temperature Profile at the center of the sphere across time: () − ∞ = − 0 − ∞ This Lumped System Analysis is sufficiently accurate only when Bi < 0.1. Checking: Given: = 10 ℎ = 10 2 2 0.05 = = 0.0167 < 0.1 3 10 10 = 0.05 0 = 30° ∞ = 20° Valid!! Lumped Systems Analysis For review: A person is found dead at 5 PM in a room where T = 20°C. The temperature of the body was measured at 25°C when found. The heat transfer coefficient is estimated to be 0.8 W/m2K. Estimate the time of death assuming: (1) The body can be modeled as a 30-cm diameter, 1.7-m long cylinder. (2) The thermal properties of the body and the heat transfer are constant. (3) The body temperature was 37°C at the time of death. (4) Since the human body has almost the same properties of water at the average temperature of 31°C: k = 0.617 W/mK, density = 996 kg/m3, and cp = 4178 J/kgK. (5) Radiation effects are negligible.