Lecture 6 annotated

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University of Houston *

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4364

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Mechanical Engineering

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Oct 30, 2023

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9/10/23 1 MECE 4364 Heat Transfer Dr. Dong Liu Department of Mechanical Engineering University of Houston 1 Lecture 6 1 Heat Conduction Equation (Chapter 2) 2 2
9/10/23 2 Last Lecture ¨ Before Lecture 4, we have ¤ Energy balance (conservation) equation ¤ Fourier’s law ¨ We applied these equations to a control volume and obtained ¤ If k is a constant 3 ! ! "" = −$ %& %’ ̇ ) #$ ̇ ) %&' + ̇ ) ( = ̇ ) )' ! * "" = −$ %& %+ !" !# $ ! ̇ $ "# ̇ & $ & $%&$ & ' & '%&' ! !+,! = ! ! + %! ! %’ ,’ % %’ $ %& %’ + % %+ $ %& %+ + ̇! = ./ - %& %0 % . & %’ . + % . & %+ . + ̇! $ = 1 2 %& %0 ! *+,* = ! * + %! * %+ ,+ 3 Boundary Conditions ¨ Consider the temperature variation along the wall of a brick house in winter ¨ The temperature at any point in the wall depends on the conditions at the surfaces of the wall and the initial condition ¤ Initial condition ( 1 st order in time ) ¤ Boundary conditions ( 2 nd order in space ) n Specified (constant) surface temperature 4 % . & %’ . = 1 2 %& %0 ! 0, $ = ! ! ! &, $ = ! " ! ’, $ = 0 = ! ’, 0 4
9/10/23 3 Example: Plane Wall ¨ A slab is at a steady state with dissimilar temperatures on either side and no internal heat generation. What are the temperature distribution and the heat flux through it? ¨ Step 1: T = T(x) for steady x-direction heat conduction ¨ Step 2: The steady 1-D heat equation without heat generation 5 ( " ! (’ " + ( " ! (* " + ( " ! (+ " + ̇- . = 0 0 " ! 0’ " = 0 Boundary conditions: T(x=0) = T s,1 and T(x=L) = T s,2 5 Plane Wall 6 6 = = 0 - - g = 0 T , - - - Tz - - - - D - x = 0 x = L - = c , = - - , B . Cs : x = 0 T = Ts - x L T = Tw #General Solution T(x = C , x + 2 #2 Apply B . Cs . x = r Tco) = Cr = T gi = - ke x = 2 Ti = G · L+ T1 #3 Specific solution = Tz I - T - T2 Tex = It x + T = kn L I of OT= Ti-Te =K-
9/10/23 4 Boundary Conditions ¨ Boundary conditions ¤ Specified (constant) surface heat flux ¤ Insulated surface (adiabatic) 7 A pan to cook beef stew on an electric oven 7 Boundary Conditions ¨ Boundary conditions ¤ Thermal symmetry n Some heat transfer problems possess thermal symmetry as a result of the symmetry in geometry and imposed thermal conditions n No heat flow across the center plane n The center plane can be viewed as an insulated surface n Note: For cylindrical or spherical bodies with thermal symmetry, the boundary condition requires that the first derivative of temperature with respect to r be zero at the centerline 8 3 %& %’ !/ 0 . = 0 3 %& %5 1/2 = 0 8 - + . -f + = Ent - - # = E = 0 = = - h k ite adiabatic di = L
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