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

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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 ⑯ - 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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9/10/23 5 ¨ Boundary conditions ¤ Convection boundary condition – most common BC in heat transfer ¤ Surface energy balance Boundary Conditions 9 The interface does not store energy 9 Boundary Conditions ¨ Boundary conditions ¤ Radiation boundary condition 10 The interface does not store energy Surface energy balance where 6 3 and 6 . are the surface emissivities 7 = 5.67×10 45 W/m 2 · K 4 & )&11,3 and & )&11,. are the surface temperatures Note that the temperatures in radiation calculations must be expressed in K 10 T =- - - --- - "D * 0 B . C . & x-0 Text I S Ein-Fout+Ege= Es -> Ein = Ent ↓ => h , [Tr , -T10 - ] = -k / = - => - = = hu)Tao23 Tsuri - ② O I - -

9/10/23 6 Boundary Conditions ¨ Boundary conditions ¤ Interface boundary condition – multilayer materials 11 BC at an interface are based on two requirements ¨ two bodies in contact must have the same temperature at the area of contact ¨ an interface cannot store energy, and thus the heat flux on the two sides of an interface must be the same. 11 Example 12 12 I en - I I I en - - at =E - 2 - K- 0 A dxC - TA . ) = TBCs · ) t - -- Kin - Kis -> i x + bx + c Ia Eil - I = Es B - - - C - l - Es"- Gr "s , Set as t unit fit - sit · KB Ka I kB> Kc

9/10/23 7 Boundary Conditions 13 13 1D Steady State Conduction (Chapter 3) 14 14

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9/10/23 8 Plane Wall A slab is at a steady state with dissimilar temperatures on either side and no internal heat generation. We want to know the temperature distribution, the heat transfer rate and the heat flux through it. ¨ Assumptions ¤ Steady ¤ One-dimensional since the wall is large relative to its thickness ¤ Thermal conductivity k is constant ¤ No heat generation ¨ Analysis 15 0 " ! 0’ " = 0 ! ’ = 0 = ! ! ! ’ = & = ! " & ’ = & . − & 3 = ’ + & 3 ! ! = −$> ,& ,’ = $> & 3 − & . = = $> ∆& = !′′ ! = ! > = $ ∆& = Linear temperature profile BCs 15 Thermal Resistance ¨ In an electrical circuit ¨ Can we draw the analogy for the conduction problem we just solved? 16 16

9/10/23 9 Plane Wall A slab is at a steady state with dissimilar temperatures on either side and no internal heat generation. We want to know the temperature distribution, the heat transfer rate and the heat flux through it. ¨ Thermal resistance of the plane wall 17 A ' ≡ = $> 17 Thermal Resistance ¨ Now consider convection heat transfer from a solid surface ¤ Newton’s law of cooling 18 ! = ℎ> ) & ) − & 7 18

9/10/23 10 Thermal Circuit (Thermal Network) 19 ¨ A slab is at a steady state with dissimilar ambient temperatures and convection heat transfer coefficients on either side and no internal heat generation. What is the the heat transfer rate through it? 19 ¨ How to find T s,1 and T s,2 ? Thermal Circuit (Thermal Network) 20 R 1 R 2 R 3 20

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9/10/23 11 Extension 1: Composite Plane Wall 21 ¨ A composite slab is at a steady state with dissimilar ambient temperatures on either side and no internal heat generation. What is the heat transfer rate through it? ! ! . ! . " & ! & " - - 21 Extension 1: Composite Plane Wall What is the temperature at the 1-2 contact T 1/2 ? 22 ! ! ! !/" ! " . ! . " & ! = . ! ! ! " ! !/" & ! . ! 1 & " . " 1 22

9/10/23 12 Extension 1: Composite Plane Wall How do you compare . ! and . " ? 23 ! ! ! !/" ! " . ! . " & ! & " ! ! ! " ! !/" & ! . ! 1 & " . " 1 23 Extension 2: Composite Plane Wall 24 24

9/10/23 13 Extension 3: Composite Plane Wall 25 ¨ A composite slab is at a steady state with dissimilar ambient temperatures and convection heat transfer coefficients on either side and no internal heat generation. We want the heat transfer rate through it. 25 Extension 4: Composite Plane Wall 26 ¨ A composite slab is at a steady state with dissimilar temperatures on either side and no internal heat generation. What is the the heat transfer rate through it? T 1 T 2 26

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9/10/23 14 T 1 T 2 Extension 4: Composite Plane Wall 27 - $ = - % + - & 27 Example 1 28 28

9/10/23 15 Example 1 29 29 Example 1 30 30

9/10/23 16 Example 2 31 31 32 Example 2 32

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9/10/23 17 33 Example 2 33 Thermal Circuit (Thermal Network) 34 ¨ It is sometimes convenient to express heat transfer through a medium in the form of Newton’s law of cooling where U is the overall heat transfer coefficient (Unit: W/m 2 K) ¤ With UA, we do not need to know T s,1 and T s,2 in order to evaluate heat transfer through the wall ! = ∆& A '%'89 ! = D>∆& D> = 1 A '%'89 ! = & 7,3 − & 7,. A '%'89 = ∆& A '%'89 34

9/10/23 18 Thermal Resistance ¨ Radiation resistance 35 (Unit: ℃/G) ! 18, = 67> ) & ) : − & )&1 : = ℎ 18, > ) & ) − & )&1 A 18, = & ) − & )&1 ! 18, = 1 ℎ 18, > ) ℎ 18, = 67 & ) + & )&1 & ) . + & )&1 . 35 ¨ If ! ' ≈ ! ()* ¨ Consider thermal resistors in parallel Thermal Resistance ¨ A surface exposed to the surrounding air involves convection and radiation simultaneously ¨ The total heat transfer at the surface is determined by adding the radiation and convection components 36 ! = ! ;%$< + ! 18, = ℎ> ) & ) − & 7 + ℎ 18, > ) & ) − & )&1 A '%'89 = ! & ) − & 7 = 1 ℎ + ℎ 18, > ) ! = ℎ + ℎ 18, > ) & ) − & 7 1 A '%'89 = 1 A ;%$< + 1 A 18, A '%'89 = 1 A ;%$< + 1 A 18, 43 = 1 1 ℎ> ) 43 + 1 ℎ 1 > ) 43 36