Lecture 7 annotated

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4364

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Oct 30, 2023

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9/12/23 1 MECE 4364 Heat Transfer Dr. Dong Liu Department of Mechanical Engineering University of Houston 1 Lecture 7 1 Last Lecture ¨ Boundary conditions 2 −" ! # $% ! $& " ! = −" # # $% # $& " ! % ! & $ = % # & $ 2

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9/12/23 2 1D Steady State Conduction (Chapter 3) 3 3 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 4 ( % % (& % = 0 % & = 0 = % & % & = * = % % ! " = ! ! − ! " % " + ! " ’ # = −() *! *" = () ! " − ! ! % = () ∆! % ’′′ # = ’ ) = ( ∆! % Linear temperature profile BCs 4 oT= T - Tz I Ex Ti -> Tz · o wall G = kA - 8 x = E -

9/12/23 3 Thermal Resistance ¨ In an electrical circuit ¤ Knowing the electrical resistance and voltage drop, the current is ¨ Can we draw the analogy for the conduction problem we just solved? ¤ Heat transfer through the wall is a current flow driven by a temperature drop ¤ Define thermal resistance R t – equivalent to electrical resistance R e 5 ’ = ∆* + = ,-./012/3- 56-037. 8169 ,-./012/3- 1.:2:03;/. < " = => ∆? @ = ∆? @ => = ?.A9.130B1. 8169 ?ℎ.1A3- 1.:2:03;/. = ∆? + # + # ≡ ∆? < = @ => (Unit: ℃//) + $ ≡ ∆* ’ = @ = $ > (Unit: Ω) 5 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 ¨ Physical meaning: resistance to heat flow ¨ Why? ¤ The wall is too thick ¤ The material is not conducting enough ¤ The surface area is too small ¨ Regardless, the heat flow for a prescribed ∆" 6 2 $ ≡ % () ’ # = ∆! 2 $ = ∆! % () 6 : of ① 0 & O R = & & Ke L ot => RekeA RI= 10 Re = 0 . 1 - =- L - 2 A T k - Rt * => Ex = t R = Conduction G = KA * = -

9/12/23 4 Thermal Resistance ¨ Now consider convection heat transfer from a solid surface ¤ Newton’s law of cooling ¤ Convection resistance 7 ’ = ℎ) % ! % − ! & ’ = ! % − ! & 1 ℎ) % 2 '()* = ! % − ! & ’ = 1 ℎ) % (Unit: ℃//) Physical meaning: resistance to heat flow due to convection Why? h is too small and/or surface area is too small … 7 Thermal Circuit (Thermal Network) 8 ¨ 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? Representative Circuit ( resistance in series ) The driving potential The current (heat flow) The resistance ∆! = ! &," − ! &,! 2 $($,- = 1 ℎ " ) + % () + 1 ℎ ! ) ’ = ∆! 2 $($,- = ! &," − ! &,! 1 ℎ " ) + % () + 1 ℎ ! ) R 1 R 2 R 3 Procedures for defining thermal circuit 1) Draw down important temperature nodes 2) Identify heat transfer modes between adjacent nodes 3) Connect the nodes with the corresponding resistances 4) Identify the current flow into/out of the circuit (or at a given node) 8 oT= Ts - To g = hA oT =n =8 e Entpp & T = To . 1 - To , 2 ⑧ ⑧ -> * Ettas-to . 2 0 ⑧ I M--wo A * A Ri R2 R3 Rtotal = Rs w ⑧ T - Rtotel= - & =- - - . - -

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9/12/23 5 ¨ How to find T s,1 and T s,2 ? ¤ Examine the first segment of thermal circuit and realize it is the same heat flow goes through the entire circuit Thermal Circuit (Thermal Network) 9 R 1 R 2 R 3 ’ = ! &," − ! %," 2 '()*," = ! &," − ! %," 1 ℎ " ) ! %," = ! &," − ’ ℎ " ) ! %,! = ! &,! + ’ ℎ " ) Similarly 9 Extension 1: Composite Plane Wall Total thermal resistance Heat transfer rate 10 ¨ 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? % & % % " & " % * & * % 2 $ = % " ( " ) + % ! ( ! ) ’ = ∆! 2 $ = ! " − ! ! % " ( " ) + % ! ( ! ) % & % % % &/% * & " & + * % " % + , , Heat flux ’′′ = ’ ) = ! " − ! ! % " ( " + % ! ( ! 10 · S 5 -> 0 · - Ex = ⑨ Ts ? Ta - --> X

9/12/23 6 Extension 1: Composite Plane Wall What is the temperature at the 1-2 contact T 1/2 ? It is the same heat flow that goes through the thermal circuit 11 % & % &/% % % " & " % * & % ! ’ = ! " − ! ! % " ( " ) + % ! ( ! ) = ! " − ! "/! % " ( " ) % & % % % &/% * & " & + * % " % + ! "/! = ! " − % " ( " ) % " ( " ) + % ! ( ! ) ! " − ! ! = % ! ( ! ! " + % " ( " ! ! % " ( " + % ! ( ! 11 Extension 1: Composite Plane Wall How do you compare " & and " % ? Recall the boundary condition at the interface 12 % & % &/% % % " & " % * & * % ( " 7 ) *! *" " = ( ! ) 7 *! *" ! % & % % % &/% * & " & + * % " % + ’ " = ’ ! Since 8 /0 /# " < 8 /0 /# ! , we know ( " > ( ! 12 +8 f Ri = # * ⑰ Rand=- KA - - ⊥ & know = x N ⑧ · illit :

9/12/23 7 Extension 2: Composite Plane Wall 13 13 Extension 3: Composite Plane Wall Total thermal resistance ( serials ) 14 ¨ 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. 2 $ = " 1 % 2 + 3 & 4 & 2 + 3 ' 4 ' 2 + 3 ( 4 ( 2 + " 1 ) 2 Heat transfer rate ’ = ∆! 2 $ = ! &," − ! &,! 1 ℎ " ) + % 2 ( 2 ) + % 5 ( 5 ) + % 6 ( 6 ) + 1 ℎ ! ) Heat flux ’′′ = ’ ) = ! &," − ! &,! 1 ℎ " + % 2 ( 2 + % 5 ( 5 + % 6 ( 6 + 1 ℎ ! How do you find T 1 , T 2 , T 3 and T 4 ? 14 0 - - 8 , = 8 a 0 - I = => Th III -> "

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9/12/23 8 Extension 4: Composite Plane Wall 15 ¨ 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 15 T 1 T 2 Extension 4: Composite Plane Wall 16 ¨ Thermal circuit ( parallel resistances ) T 1 T 2 q F q G , " = , F + , G ’ 7 = ! " − ! ! % 7 ( 7 )/2 ’ 8 = ! " − ! ! % 8 ( 8 )/2 ’ # = 1 % 7 ( 7 )/2 + 1 % 8 ( 8 )/2 ! " − ! ! 1 2 $($,- = 1 % 7 ( 7 )/2 + 1 % 8 ( 8 )/2 = 1 2 " + 1 2 ! ’ # = ! " − ! ! 2 $($,- 16 * # Rond= EF - fut = i + 2 ~ w Ts- T2 Ex = - Rtotel O 0 e -

9/12/23 9 Example 1 17 17 Example 1 18 18 Gx = Ex" : - - - Tqx" I I " - I * O . Ex -En i -- = 1 4 D - T S t ⊥ in is it = - Ton 0 T is it To , 0 oth of to 0 ⑧ ⑧ - ⊥ s Li LS I - Fi Es ho & - hi KS 1 , Y -

9/12/23 10 Example 1 19 19 Example 2 20 20 - -506 e -

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9/12/23 11 21 Example 2 21 22 Example 2 22 & Es To , i Ei To , Tr -> 11 -> 3 G , -okomorosf , " M -> ⑥ o o - ⑳ 0 ⑲ ⑤ G , " + 8.

9/12/23 12 Thermal Circuit (Thermal Network) 23 ¨ 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 ’ = ∆! 2 $($,- ’ = <)∆! <) = 1 2 $($,- ’ = ! &," − ! &,! 2 $($,- = ∆! 2 $($,- 23 Thermal Resistance ¨ Radiation resistance 24 (Unit: ℃//) ’ 9,/ = =>) % ! % : − ! %;9 : = ℎ 9,/ ) % ! % − ! %;9 2 9,/ = ! % − ! %;9 ’ 9,/ = 1 ℎ 9,/ ) % ℎ 9,/ = => ! % + ! %;9 ! % ! + ! %;9 ! 24

9/12/23 13 ¨ If % H ≈ % IJK ¨ 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 25 ’ = ’ '()* + ’ 9,/ = ℎ) % ! % − ! & + ℎ 9,/ ) % ! % − ! %;9 2 $($,- = ’ ! % − ! & = 1 ℎ + ℎ 9,/ ) % ’ = ℎ + ℎ 9,/ ) % ! % − ! & 1 2 $($,- = 1 2 '()* + 1 2 9,/ 2 $($,- = 1 2 '()* + 1 2 9,/ <" = 1 1 ℎ) % <" + 1 ℎ 9 ) % <" 25

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