4 內流場之熱對流 (Internal Flow Heat Convection)

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1 4 內流場之熱對流 (Intenal Flow Heat Convection)

2 Pipe cicula coss section. Duct noncicula coss section. Tubes small-diamete pipes.

3 4.1 Aveage Velocity (V avg ) Mass flowate and aveage fluid velocity in a cicula tube: m = 質量流率 = ρv A = ρ u() da avg C C A c V avg ρ u () da C o () A c = ρ u πd 平均流速 = ρa = C ρπ = o o o u() d o

4 4. Aveage Fluid Tempeatue (T m ) ( ) δ ρ () () E = mc T = c T m = c T u da fluid p m p p c m A c = 流體能量 T m = 平均溫度 c () () () p T δ m cp T ρ u πd m A c = = = mc ρv c V p ( π ) avg o p avg o o T () u () d The mean tempeatue T m o T b of a fluid changes duing heating: Idealized( 一維 ) Actual

5 4.3 Local Convective Heat Tansfe Coefficient (h) and Nusselt Numbe (Nu) - 局部熱對流係數與紐塞數 Definition: h x qs ( x) k ( T / ) = = T T T T f = o s b s b local convective heat tansfe coefficient whee T b = T m = aveage tempeatue, T = T(,x) Nu x h x D h / k f = Nusselt numbe = 紐塞數流體 x T b (x) T s (x) o q s (x)

6 4.4 One-Dimensional Analysis

7 A Flow inside a Tube with Constan Suface Tempeatue 流體通過等溫管壁之內部時溫度之變化管壁等溫 (T s =constant) Tube inne diamete = D Flow T i X 方向 T o L T dx T s (constant) dq T o T i dt X = L x

8 Q = ( mc ) ( T T) = total heat tansfe ate p c o i m = mass flowate (kg/s) ; c = specific heat (J/kg-K) (i) Enegy Balance (cold fluid) : dq = ( mc ) dt (ii) Heat Tansfe : dq = h( pdx)( T T ) s p p c Combined, (i) and (ii) T o dt hp To Ts hpl = [ ] dx; so, ln( ) = ( T T ) ( mc ) T T ( mc ) T i L s p c i s To T s hpl Following, = exp Ti Ts ( mc p ) c hpl Hence, To = Ts ( Ts Ti) exp ( mc p ) c =? Note: h value needs to be detemined by expeimental measuement o by two/thee dimensional analysis p c

9 4.5 Bounday Laye and Reynolds Numbe - 內流場流體之邊界層與雷諾數

10 Lamina and Tubulent Flow in Tubes Reynolds numbe is defined as: Re = ρv avg µ D h = V avg ν D h hydaulic diamete D h is defined as: D h = 4A c P 圓管 : D h = D (1) () lamina flow: Re 3 tansition: 3 < Re < 1 fully tubulent flow: Re 1,

12 邊界層 (Bounday Laye) Iotational flow Bounday laye The thickness of this bounday laye inceases in the flow diection until it eaches the pipe cente.

13 4.6 Flow Fields fo Lamina Flow and Tubulent Flow - 管道內之層流與紊流之流場

14 Hydodynamic Entance Region The velocity pofile in the fully developed egion is paabolic in lamina flow, and somewhat flatte o fulle in tubulent flow. Lamina flow: (x h /D).5 Re D Tubulent flow: (x h /D) 1 X h = hydodynamic entance length

15 Themal Entance Region Themally fully developed egion the egion beyond the themal entance egion in which the dimensionless tempeatue pofile expessed as (T s -T)/(T s -T m ) emains unchanged. Lamina flow: (x t /D).5 Re D P Tubulent flow: (x t /D) 1 x t = themal entance length

16 x T(x,) T s T m T s Ts T In the themally fully-developed egion: ( ) = x T T s m

17 In hydodynamic fully-developed egion: Invaiant velocity pofile. In the themal fully-developed egion: h value does not vay. Ts T Ts T ( ) = i.e. is not a function of x! x T T T T s m s m Ts T Then, it must be = a function of only T T s m Ts T (T/ ) ( ) = a constant = T T T T s m = s m o = o T Fouie's Law: qs = k f( ) = and Newton's Cooling Law: q o s = h(tm T s) T h (T/ ) = o k f( ) = = h(t o m T s) = = 常數 k T T f s m If k is a constant, h will also be a constant! f

18 T T x T T s ( ) s m = 1 T T T T T T = s s s m Hence, ( ) ( ) Ts Tm x x (Ts T m) x x T dt T T dt T T dt = + x dx T T dx T T dx s s s s Afte aangement, ( ) ( ) s m s m (i) Fo constant suface heat flux case: qs = h(tm T s) = constant dtm dts (Tm T s) must also be a constant. So, =, substitute it back, dx dx T dts it yields =. x dx dts (ii) Fo constant suface tempeatue case: = dx T Ts T dtm It yields, = ( ) x T T dx s m m

19 4.7 Continuity, Momentum and Enegy Equations fo Lamina Flow in Cicula Tube

20 Unifom V avg, T in φ x 圓柱體座標 : 方向之速度 : v x 方向之速度 : u φ 方向之速度 : v φ If Lamina Flow, 速度分佈最終變為拋物線 (Paabola) (velocity is fullydeveloped!) Continuity (incompessible flow 不可壓縮流體之質量不滅 ): vφ v v 1 u = φ x symmety

21 Momentum Equations (i) - 方向 Momentum Equation ( 動量 ): Steady flow v v vφ v vφ v ρ( + v + + u ) t φ x p v 1 v v 1 v v vφ v = B + µ( ) φ φ φ x Symmety (ii) φ - 方向 Momentum Equation ( 動量 ): v v v v v v v φ φ φ φ φ φ ρ( + v + + u ) t φ x 1 p vφ 1 vφ vφ 1 vφ v vφ = B φ + µ( ) φ φ φ x Usually can be neglected! (iii) x - 方向 Momentum Equation ( 動量 ): u u vφ u u p u 1 u 1 u u ρ( + v + + u ) = B x + µ( ) t φ x x φ x steady flow

22 簡化後, (i) di. Momentum Equation ( 動量 ): v v p v 1 v v + = + + ρ(v u ) µ( ) x (iii) x - di. Momentum Equation ( 動量 ): u u p u 1 u + = + + x ρ(v u ) µ( ) x 需要之條件 : (i) at x =, u = V avg, v = and p = p (constant) u v p (iii) at =, v = u = (no slip condition) (ii) at =, = = = (symmety) o

23 Enegy Equation ( 能量平衡 ): 忽略 dissipation effect! T T v T T 1 T 1 T T ( v φ u ) = α ( ) + + t φ x φ x 需要三個條件 : 通常亦可略 (usually negligible)! Unless, it is a liquid metal. (i) at x =, T = Tin T (ii) at =, = (symmety) (iii) at =, T = T (if wall is isothemal) o s

24 Simplified Govening Equations (i) Continuity: v v u x + + = (ii) x - Momentum Equation: u u dp u 1 u ρ(v + u ) = + µ ( + ) x dx Solving u, v, p (iii) x diection bulk mass balance: = o ρv avg[( π o ) ( )(π o )] = u da ρ = = u ( d) o = ρ π = T solved sepaately (iv) Enegy Equation ( 能量平衡 ): T T 1 T (v + u ) = α [ ( )] x

25 4.8 Dacy 摩擦係數 --- fiction facto Definition of Dacy fiction facto (f) f 之定義 : dynamic pessue L ρv PL = f D fiction facto avg

26 Velocity Pofile of Lamina Flow in Cicula Tube (Fully-Developed Region) x - Momentum Equation: dp ρ(v u ) µ( ) u u u 1 u + = + + x dx A constant In the hydodynamically fully developed egion: u v = = x

27 µ d du dp = d d dx Reaanging and integating it twice to give 1 dp u( ) = + C ln+ C 4µ dx 1 Bounday Conditions: symmety about the centeline : u/ = at =, no-slip condition: u = at = o. with the bounday conditions ( ) o dp u = 1 4µ dx o

28 Pefoming the integation gives the aveage velocity o o o dp o dp Vavg = u ( ) d = 1 d = o o 4µ dx o 8µ dx Combining the last two equations, the velocity pofile is u ( ) = V 1 ; u = V R avg max avg 拋物線 Integating fom x=x 1 whee the pessue is P 1 to x=x =L whee the pessue is P gives ( 速度場完全發展區域 ) dp dx = constant = P P L 1 Substitute back, 8µ LV 3µ LV P= P P = = avg 1 o D avg

29 Cicula tube, lamina, fully-developed flow: f 64µ 64 = = ρdv Re avg L D P ( 壓力降或壓差 )

30 Moody Diagam

31 Diffeence between Dacy fiction facto and Fanning fiction facto P f = Dacy fiction facto (L/D)( ρv C C f,x f τ ( ρv L 1 L ( ρv nomal foce P = = coss sectional aea Substitute w avg τ w avg τ = Fanning fiction facto / ) / ) w dx τ w = ( ρv back, C avg f / ) A wτ A τ w ( ρv avg c w ( πdl) τ w = πd /4 = / ) avg ( ρv / ) avg 4L = ( ) τ D w P / )(4L / D) Hence, f = 4 C f 流體 x L D P ( 壓力降或壓差 )

32 4.9 Velocity Pofile of Lamina Flow in Cicula Tube (Hydo-dynamically Developing Region) (i) Continuity: v v u x + + = (ii) x - Momentum Equation: u u dp u 1 u ρ(v + u ) = + µ( + ) x dx Solving u, v, p: 由式 (ii) & (iii) 計算一個 x 位置之 u 分佈與 dp/dx, 再由式 (i) 計算此位置之 v 分佈 (iii) x diection mass balance: = o ρv avg[( π o ) ( )(π o )] = u da ρ = = u ( d) o = ρ π =

33 無因次化統御方程式 : Define : u v p, ( )Re, ρ u * v * p* V V V / ρvd x x / D Re, x * =, * µ ( )Re Re ( ie.. = ) o s s 1 u* v* v* (i) Continuity: ( ) + + = x* * * 1 u* u* 1 dp* u* (ii) Momentum: u* ( ) + v * ( ) = ( ) + [ ( * )] x* * 4 dx* * * * * * = 1 * * * * = (iii) x-di. Mass Balance: ρπ π [ ( )] = ρu (π d ) s 由式 (ii) & (iii) 計算一個 x* 位置之 u* 分佈與 dp*/dx*, 再由式 (i) 計算此位置 v* 之分佈, 接著計算下一個 x* 位置之 u* 分佈與 dp*/dx*, 再由式 (i) 計算此位置 v* 之分佈, 如此整個流場之 u* 與 v* 可以獲得

34 fully-developed egion i.e. (x/d) >.5 Re cente-line at the wall Fig. 7.6 Velocity Distibution

35 appaent mean fiction coefficient ( c ), mean fiction coefficient ( c ) and local fiction coefficient ( c f f app ) - Based on Fanning's Definition fm 包含 momentum 改變造成之結果由 τ w 計算之結果 x ρv p = 4 c ( )( ) f,app D As Re/(x/D), Rec = 16, that means, f = 64/Re f,app

36 4.1 Nusselt Numbe fo Fully-Developed Flow in a Cicula Tube

37 Two Limiting Bounday Conditions (1) Constant suface tempeatue 等溫管壁 (T s = constant) () Constant suface heat flux 管壁等熱通量 (q s = constant)

38 Vaiations of Heat Tansfe coefficient and Fiction facto Developing egion Fully developed egion L h 與 L t 何者較長?

39 Fluid 流體 (i) Constant Suface Heat Flux Lamina Fully Developed Flow T w (x) x T b (x) o Velocity Pofile q s = constant Enegy balance u( ) = V 1 avg o q x ρv c T = s = avg p o constant Enegy Equation u T α T = x d 其中 k α ρc p Odinay Diffeential Equation 常微分方程式 4qs 1 d dt 1 = ko o d d

40 Sepaating the vaiables and integating twice q 4 s T= + C1+ C ko 4o Bounday conditions Symmety at = : ( ) T = = C 1 = At = o : T( = o ) = T s C T q = T s o s 4 k 4 o 4o

41 T = 平均流體溫度 = T( ) u( ) d m Vavgo o 11 q s o Tm = Ts 4 k q ht T ( ) s s m 4 k 48 k k h = = = D D Constant heat flux (cicula tube, lamina) hd N u = = 4.36 = 常數 k o

42 (ii) Constant Suface Tempeatue Lamina Fully Developed Flow α T T Enegy eqaution: = u d x V Fo hydaudynamically fully-developed flow: u = ( 1 ) α o Fo themally fully-developed flow with constant suface tempeatue: x T T = So, T T(x,) function of "" only s s ( ) Ts Tm Ts T m(x) T Ts T dt = ( ) m x Ts Tm dx 1 T V Ts T dtm Hence, ( ) = (1 )( ) (1) α o Ts Tm dx Ts T(x,) n n Assume = a n ( ) = ao + a 1( ) + a ( ) a n( ) Ts T m ( x) n= o o o o =

43 T(x, ) = T (T T )[a + a ( ) + a ( ) a ( ) ] () n s s m o 1 n o o o T a a 3a na Diffeentiation, = (T T )[ ] (3) T = (T T )[a ( ) + a ( ) + 3a ( s m 1 3 o o n n s m 3 n o o o o o 3 n n o n 1 1 s m 3 3 n n o o o o n 1 s m 3 3 n n o o o o s m n n= 1 n ) na ( ) ] T a ( ) = (T T )[ + 4a ( ) + 9a ( ) n a ( )] 1 T a 1 ( ) = (T T )[ + 4a ( ) + 9a ( ) n a ( )] = (T T ) n a ( Fom bounday condition (1): at = (wall), T n o ) o = left side = T = a + a + a a Fom bounday condition (): at = (centeline), a = 1 o 1 n s T = (symmety)

44 能量平衡 : mc dt = h(t T )( πd)dx p m s m dt m πdh 4h πd = (T T ) = (T T ) whee m =ρv( ) dx mc VDc 4 p s m s m ρ p T s Contol volume T m m T m + dt m dx

45 代回式之右側, V s ight side = (1 )( ) α o s m V n 4h n s m k o n= o ρvdcp p T T dt T T dx = (1 ) [ a ( ) ] [ (T T )] ( ) ρc m 4h Nu = ( )(1 ) [ a ( ) ](T T ) = ( )(1 ) [ a ( ) k n n s m n o o n= o o o n= o n ](T T ) s m ight side = left side Nu (T T ) = ( n n s m na( n ) ( )(1 )[ n a( n )] Ts Tm n= 1 o o o n= o a 1 [ 4a ( ) 9a ( ) 16a ( ) 5a ( )... n a ( )] 3 n n n o o o o o o Nu = ( )(1 )[ a o o o + a ( ) + a ( ) a ( ) ] n 1 n o o o )

46 [4a 9a ( ) 16a ( ) 5a ( )... n a ( )] n n n o o o o = ( Nu)[a + (a a )( ) + (a )( ) + (a a )( ) + (a )( ) Nu a = ( )a o o o o o o o n... (a n a n )( )...] o 比較項, 得知右邊並無項故必定 n ( ) ( ), a 3 =! o o a = a = a = a = a = ( 下標奇數者均為 ) Nu ( )( ) = Nu 同時, 4a 4 = ( )(a a o)... (n + ) Nu n+ = n n a ( )(a a ) 4 a a o

47 a 1 4a a a geneal fom: = ( )( ) a (n ) a a a n+ n n o + o o o fo n = 偶數, 且 a 1 a a = = a 4 a a 4 (i) n, ( )( )( 1) (4) (ii) n = 4, o o o a6 1 a a4 a = ( )( )( ) (5) a 9 a a a o o o o a 1 a a a = = a 16 a a a (iii) n 6, ( )( )( ) (6) o o o o... (iv) n n, a 1 a a a = ( )( ) [ ] 4 n+ n n = ao (n + ) ao ao ao

48 Numeical Method: 已知 Fom bounday condition (1) = a + a + a + a + a a o n = a + a + a + a a o 4 6 n a a4 a6 a8 an = ( ) (7) a a a a a o o o o o a a4 首先假設一個 ( ) 值, 代入式 (4) 中, 計算一個 ( ) 值, a a o a 再代入式中計算一個將所得到之 6 (5) ( ),..., ao a a 所有之值再代入式中, 即可以得到一個新的值, n ( ) (7) ( ) ao ao 重複以上之過程直至答案收斂! 所選擇之 n值為 16, 即可得到準確之答案 (Nu = 3.657)! o

49 Lamina Flow in Fully-Developed Region Nu = hd h / k f = aveage Nusselt numbe whee h = aveage convective heat tansfe coefficient

50 4.11 Nusselt Numbe fo Developing Lamina Flow in the Entance Region

51 (1) Hagen-Poiseuille Poblem - Themally Developing, Hydodynamically Developed Lamina Flow X= Stat to heat the tube wall T s = a constant

52 Developing Lamina Flow in the Entance Region Fo a cicula tube of length L subjected to constant suface tempeatue, the aveage Nusselt numbe: Hausen equation [1943]: (themally developing and hydodynamically developed flow).668 Gz = (Gz) Nu /3 Gz = Gaetz numbe = Re D P (D/L)

53 X= Stat to heat the tube wall " q s = a constant

54 Nusselt Numbe fo Themally Developing, Hydodynamically Developed Lamina Flow

55 Themally Developing, Hydodynamically Developed Lamina Flow (1) Enegy Equation ( 能量平衡 ): U, T o Fully-developed velocity pofile x T x Stat to heat the plate T s =a constant T 1 T u T T + = α x x Note: v = (fully-developed velocity pofile).

56 Ts T u + + define: θ u Ts T o U x/ x/ xα x k ReP (U D /α) U U ρc Dimensionless fom: + o o x = = = h o o 通常可省略 + θ 1 T u θ 1 θ + = x (ReP) x whee ReP Peclet numbe Bounday conditions: + (i) at x =, T = T at x =, θ = 1 (ii) at = o, T = T + s at = 1, θ = + + (iii) at =, T / = at =, θ / =

57 P aabolic 速度分佈 : u ( ) = V 1- V = = U 平均速度 o + + Dimensionless fom: u = (1 ) T s q s = k( ) = = k ( )( ) + o + = 1 o T T θ This deivative needs to be known! () Solution of Enegy Equation: + θ 1 T (1 ) θ + = x Use method of sepaation of vaiables: θ(x, ) = R( ) X(x ) Substitute back, 1 " ' + ' R X + R X = (1 )RX, + " ' R 1 R [ + ] X 1 X + ' Hence R R = λ (a nega + function of + function of x + tive value)

58 λ is the eigenvalue (chaacteistic value)! It yields, 1 " ' + (i) R + R λ (1 )R = + a Stum-Louville type equation ' (ii) X X +λ = The solution is X(x ) = ce + λ x + The Stum-Louville type equation needs to be solved using the powe-seies method. Assuming the solution is a powe seies as follows: R( ) = 1 + a + a ( ) + a ( ) + a ( ) + a ( ) + a ( ) a ( ) a ( ) i 7 i

59 + Diffeentiation with espect to, R ( ) = a + a + 3a ( ) + 4a ( ) + 5a ( ) + 6a ( ) ' a ( ) + 8a ( ) ia ( ) i i R ( ) = a + 6a + 1a ( " a ( ) + 7a ( ) ) + a ( ) + 3a ( ) + 4a ( ) i(i 1)a ( ) i 7 8 i Substituting back, a + (4a +λ ) + (9a +λ a ) + (16a +λ a λ ) + (5a + λ a λ a )( ) + (36a +λ a λ a )( ) ( ) (49a + λ a λ a )( ) + (64a + λ a λ a )( ) (81a + λ a λ a )( ) + (1a + λ a λ a )( ) (11a + λ a λa )( ) + (144a + λ a λ a )( )

60 It must be, a = (o fom bounday condition (iii)) 4a a / 4 9a a a 16a a a ( /16)(a 1), Note : a 5a a a a 36a a a a ( / 36)(a a ) 49a + a a a 64a + a a a ( / 64)(a a ) 81a + a a 1a + a a 1 +λ = = λ 3+λ 1 = 3 = 4 +λ λ = 4 = λ 5 +λ 3 λ 1 = 5 = 6 +λ 4 λ = 6 = λ 4 7 λ 5 λ 3 = 7 = 8 λ 6 λ 4 = 8 = λ λ 7 λ 5 = 1 λ 8 λ 6 = etc. = 1 In summay, (i) when i is an odd numbe : ai = (ii) when i is an even numbe : a i = λ ( / i )(ai a i 4) (iii) a can be evaluated using a and a. i i i 4 Hence, R( ) = 1 + a ( ) + a + + ( ) + a ( ) + a ( ) a ( ) i i

61 + Fom bounday condition (ii) : at = 1, θ (1, x) = R(1) = It yields, R(1) = 1 + a + a + a + a a = i κλ = 1 λ / 4 + ( λ /16)( λ / 4 + 1) + a + a a = Let ( ) 1 + a + a 4 + a 6 + a ai 6 8 i Thee ae infinite numbe of oots ( λn, n = 1, ) fo κ( λ ) =. Using numeical scheme and selecting appopiate ai tems (i value), the fist few oots of the above equation can be easilly found. Fo evey oot ( λn ), thee is a coesponding solution (R ) fo the Stum-Louville type equation. The solution can be expessed as: n + R n ( ) = 1 + ( λ / 4)( ) + ( λ /16)( λ / 4 + 1)( ) +... κ( λ ) n n n 1. λ 1 λ λ 3 λ 4 λ 5 λ

62 若 R 為以上 Stum-Louville 公式之之解答, 則無因次化溫度之解答 n 可以表示為以下之多項式 (Polynomial), Cn R n( ) exp λn + n = + θ(x, ) = Substituting bounday condition (i): at x =, θ = 1 Hence, = C R ( ) n= n m n + + ( x ) Multiplying both sides by R [(1 ) ] and integating the equation fom = to =1, it obtains R m[(1 ) ] d = R m[(1 ) ][ CnRnd ] d n= side = R m[(1 ) ][ CnR nd ] d n= m m m m The ight + = (othogonality) = R [(1 ) ]C R d + R [(1 ) ]C R d +... [(1 ) ]C R d +...

63 n m m m So, R [(1 ) ] d = C [(1 ) ] R d and o C n R m[(1 ) ] d C m = m =, 1,, 3,... 1 [(1 ) ] R d = m R [(1 ) ] d n [(1 ) ] R d n =, 1,, 3,... whee R ( ) = 1 + ( λ / 4)( ) + ( λ /16)( λ / 4 + 1)( ) n n n n + Using λ, the R ( ) can be assued and the C can be evaluated. Similaly, the C 1, C, C 3... can be evaluated using λ, λ, λ... espectively. 1 3

64 (3) Evaluation of Local Nusselt Numbe (Nu x ): hd x o s Local Nusselt numbe Nu x = = [ ] k k Tm Ts whee T T T θ = s q (x) = k( ) k( )( ) s = + o + = 1 o Ts T ' + k( ) CnR n(1)exp( nx ) o n= Ts T + n n o n= = λ q (x) k( )[ G exp( λ x )] whee ' n n n G (C /)R (1) θ T T s m Define m, Ts T q (x) θ Hence, Nu = [ ] = ( ) o s x + + = 1 k θm(ts T ) θm

65 θ In ode to obtain the Nu x, θm and ( ) + must be known. + = 1 Diffeentiating θ with espect to, n = + θ ' + ( ) + = C + 1 n R n(1) exp( nx ) = λ ' n n n Define G (C / ) R (1) Hence, θ Nu = ( ) θ x + + = 1 m 4 + = ( ) Gnexp( λnx ) θ m n = Note : G can be ' n n n known fom the infomation of C and R (1)!

66 mass flowate = m mean fluid tempeatue Contol volume unifom fluid tempeatue = T = T m x = q(x) s x tube Enegy balance fo the contol volume: x o s m o (π ) q (x) dx = mc(t T ) whee m=u ( π ) ρ Afte aangement, π = θ 1) x o [ ] q s(x) dx mc( (T -T s) π x o Hence, θ m = 1 [ ] q s(x) dx = 1 mc(t -T s) Gn + Gn = ( )exp( λ nx ) 8 n= λn n= λn m x + 8 G exp( λ x ) dx n= n n + +

67 Gn whee 8 = 8[ ] = 1 n= λn It yields, G + θ m = 8 ( )exp( λnx ) λ n= n n Substitute back to the expession fo Nu, Fo x + θ 4 + Nu x = ( ) + = ( ) G + = 1 n exp( λnx ) θ θ = m n = n= n n n n G exp( λ x ) m G + ( )exp( λnx ) λ + n = >.1, only the fist tems in the two polynomials ae significant: + λ λ G exp( x ) Nu x = G + ( )exp( λ x ) λ

68 + λ Fo x >.1, Nu x = This is the Nusselt numbe in the themally and hydodynamically fully developed egion. Hence, the themal enty length must be at x =.1. + (x / D) x =.1 Re P x That is, ( ) fully developed.5rep D + Example : If Re = 5 x (i) Fo ai : P.7, ( ) fully developed 17.5 D x (ii) Fo oil: if P 1, ( ) fully developed 5 D Hence, fo heat exchanges applied to oil, it is necessay to check whethe the Nu value (3.658) in the fully developed egion is appopiate to be used in design.

69 Nu x x + =.1 x +

70 (4) Evaluation of Aveage Nusselt Numbe (Nu m ): Aveage Nusselt numbe (Nu m ): hmd Nu m k 1 x whee hm = h(x) dx aveage convective heat tansfe coefficient x = 1 x [ h(x) dx] D 1 x x 1 x + Hence, Num = Nu x dx = Nu x dx + k x x T dt m T m (fluid) T s (tube wall) dx x

71 Enegy balance fo the contol volume (dx): mc dt m = h x(πodx)(tm T s) Afte aangement, integate the equation fom x = to x = x, d(t T) πh dx It yields, Tm x m s o x = T T m T s mc T T π x m s o ln( ) = ( ) h xdx T T mc π x o x Hence, θ m = exp[ ( ) h dx] mc + + x x + = exp[ Nu x dx ] = exp[ x ( Nu x dx )] + x + = exp[ x Nu m] So, Nu m = ( )ln( ) = ( )ln[ ] + + x θm x Gn + 8 ( )exp( λ nx ) λ n= n

72 () Simplified Govening Equations fo Hydodynamically and Themally Developing Flow Cicula Tube (i) Continuity: v v u x + + = (ii) x - Momentum Equation: u u dp u 1 u ρ(v + u ) = + µ( + ) x dx (iii) x diection bulk mass balance: = o ρv avg[( π o ) ( )(π o)] = u da ρ = = u ( d) o = ρ π = Solving u, v, p: 由式 (ii) & (iii) 計算一個 x 位置之 u 分佈與 dp/dx, 再由式 (i) 計算此位置之 v 分佈 T solved sepaately (iv) Enegy Equation ( 能量平衡 ): T T 1 T (v + u ) = α [ ( )] x

73 無因次化統御方程式 ( V = V ) : Define : avg u v p / u *, v * ( )Re, p* V V ρ V ρvd x x / D Re, x * =, * µ ( )Re Re 1 u v v * * * (i) Continuity: ( ) + + = * * * s x 1 u* u* 1 dp* u* (ii) Momentum: u* ( ) + v * ( ) = ( ) + [ ( * )] x* * 4 dx* * * * = 1 ρπ π * * = ρ * π * * * = T* 4 T* 1 T* T* * * * * * * * (iii) x-di. Mass Balance: [ ( )] u ( d ) (iv) Enegy: u ( ) = [ + ( )] v ( ) x P s 由式 (ii) & (iii) 計算一個 x* 位置之 u* 分佈與 dp*/dx*, 再由式 (i) 計算此位置 v* 之分佈, 當整個流場之 u* 與 v* 獲得後, 再由式 (iv) 計算 T*

74 Siede and Tate equation [1936]: (themally and hydodynamically developing flow) = Re P µ L/D µ D 1/3 Nu 1.86( ) ( ) w Re P µ > < < L/D µ w µ.44 < < 9.75 µ w ReDP Gaetz numbe L/D D 1/3 whee [( ) ( )] ;.48 P 167 ;

75 x= Stat to heat the tube wall T s = a constant

76 x= Stat to heat the tube wall q s = a constant

77 Mean Nusselt Numbe fo Themally and Hydodynamically Developing Lamina Flow Constant Wall Tempeatue Cicula Tube

78 (3) Flow between Two Paallel Mean Nusselt Numbe fo Themally and Hydodynamically Developing Lamina Flow Constant Wall Tempeatue Paallel Plate

79 4.1 Tubulent Flow in Tubes - Expeimentally Detemined

80 Fo smooth tubes, the Dacy fiction facto can be detemined by Petukhov equation [197]: ( ) 6 f =.79ln Re < Re < 5 1 (1) Dittus Boelte equation [193]:.8 n Re > 1, n=.4 heating Nu =.3Re P.7 P 16 n =.3 cooling () Seide and Tate equation [1936]: Nu = 4/5 1/3 µ.7red P ( ) whee.7 < P < 167 ; ReD > 1 µ (3) Gnielinski equation [1976]: w 流體被加熱 Nu = (f/8)(re 1)P D D /3 [1+ ( ) ] whee.6 < P < ; Re 1/ /3 D (f/8) (P 1) L > 3

81 Example 1 Cicula Tube - 等溫管壁管壁等溫 (T s = 5 C) T i = C X 方向 T o L=.3 m D=.8 m (i) Ai at U = 3 m/s πd π(.8) 3 m = ρuac = ρu( ) =.73(kg/m ) 3(m/s) [ ](m ) = (kg/s) 入口處之體積流率 (volumetic flowate at the inlet) = = = 4 3 UAC (m /s) 9.4(L/min) - cp 1 (J/kg-K) ; p = πd =.51 1 (m) ; P.7 ;

82 3 ρud 4m.75(kg/m ) 3(m/s).8(m) Re = = = = µ π Dµ.36 1 (kg/m s) -1 Gz = Invese of Gaetz numbe = (L/D)/[Re P] (.3/.8) = =.7, Fom Figue: Num = 4.8 k.34(w/m K) h = Nu m = (4.8) = 19.(W/m K) D.8(m) hpl = T o T s (Ts T) i exp (mc p ) c Lamina flow ( 層流 ) - 19.(W/m K).51 1 (m).3(m) = 5 (5 ) exp (kg/s) 114 (J/kg-K) = 91 C

83 壓力降 (pessue dop) Re = 763 (Lamina Flow) fom Moody Diagam o by diect calculation f = 64/Re =.839 (assume in fully-developed egion 完全發展區域 ) dynamic pessue fiction facto L ρv 平均值 avg PL = f D 3 = (.839) (.3/.8) [(1/)(.84 kg/ m )(3 m/s) ] = 11.9 (N/ m) =11. 9 (pa) 以上之估算不計入出口處之影響, 所有管道亦均假設於 fully developed egion, 實際數值會比此數值高! X h = hydodynamic entance length =.5 D Re D =.35 (m) Flow is still in the developing egion!

84 Pessue Dop fiction plus flow acceleation x ρv p = 4 c f,app ( )( ); V = V D avg

85 壓力降 (pessue dop) Re = 763 (Lamina Flow) Re/(x/D) = 763/(.3/.8) =.3 查前圖 Re (c ) 3 f,app c.31 f,app x ρv p = 4 c ( )( ) f,app D.3 (.84)(3) = 4 (.31)( )[ ].8 = 17.1 (pa) Incease almost 44%!

86 4.13 Heat Tansfe Enhancement ( 管內之熱傳增強 )

87 Tube with Intenal Fins and Tube with Intenal Roughened Suface

88 Tubes Inseted with Twisted Tape and Coiled Sping

89 Seconday Flow Induced by Coiled Sping

90 熱傳增強之原理 (Pinciple of Heat Tansfe Enhancement) 增加熱傳面積 (Incease of heat tansfe aea) 增加流體之紊流度 (Incease of tubulence) 增加流體之比熱與導熱性 (Incease of specific heat and themal conductivity of fluid) 技術之實用性 (Pacticality) 便宜 (cheap pice)? 耐久 (enduance)? 無污染 (pollution)?

= lim(x + 1) lim x 1 x 1 (x 2 + 1) 2 (for the latter let y = x2 + 1) lim

= lim(x + 1) lim x 1 x 1 (x 2 + 1) 2 (for the latter let y = x2 + 1) lim 1061 微乙 01-05 班期中考解答和評分標準 1. (10%) (x + 1)( (a) 求 x+1 9). x 1 x 1 tan (π(x )) (b) 求. x (x ) x (a) (5 points) Method without L Hospital rule: (x + 1)( x+1 9) = (x + 1) x+1 x 1 x 1 x 1 x 1 (x + 1) (for the