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Unformatted text preview: External Convective Heat External Transfer1
The Thermal Boundary Layer Newton’s Cooling Law Overview
• Boundary layers
– Velocity boundary layer – Thermal boundary layer – Mechanism of convection • Newton’s Cooling Law
– The convective heat transfer coefficient – The Nusselt number – Comparison with the Biot number • Three heat transfer mechanisms Boundary layers... Boundary Boundary layers and heat transfer
• Convection occurs in liquids and gases. • Energy is carried with fluid motion when convection occurs. An alternative term is “advection.”
y x z Fluid flow, V = (u,v,w), T(x,y,z) Wall: Tw or qw The velocity boundary layer The
δ
THE REGION NEAR THE WALL IS THE VELOCITY “BOUNDARY LAYER” LAYER” THE LARGEST CHANGES IN VELOCITY OCCUR AT THE WALL. THE HIGHEST VALUES OF SHEAR OCCUR AT THE WALL. The temperature boundary layer
T∞ , U ∞ U∞
δ(x) δT(x) Tw − T∞ Heated Wall at Tw > T∞ Thermal Boundary Layer Thermal boundary layers are also thin. The velocity and thermal boundary layers are not generally equal. Wall shear stress
U∞ τW ∂u =−µ ∂y y =0 Force of wall on fluid via viscosity, τw Shear force on wall by the fluid, τw Shear Heat transfer at the wall
Temperature distribution in the fluid.
y Tw
x q ′′ = − k fluid w
Ts ∂T ∂y y =0 Temperature distribution in the wall. Newton’s cooling law... Newton Newton’s Cooling Law q ∝ Tw − Ta (23.1a) (23.1b) q = hA (Tw − Ta ) •The quantity h is the convective heat transfer coefficient. coefficient •It is dependent on the type of fluid flowing past the wall and the velocity distribution. •Thus, h is not a thermophysical property. Newton’s Cooling Law Newton
Q = q ′′ = h (TW − T ∞ ) A ⇒ h= q ′′ w (TW − T∞ ) h= ∂T −kf ∂y (TW − T∞ ) y =0 (23.2) The Biot number & Newton’s Cooling Law
Also, evaluating the heat flux in the solid, h= ∂T − kw ∂y (Tw − T∞ ) y =0,w (23.3) The Nusselt number The
Consider the fluid side and the definition of h. Let η= y/L, where L is a length scale, e.g., the width of the plate. Then, −k h= f ∂T ∂y −
y=0 (T w − T∞ ) ⇒ hL η =0 = kf (T w − T ∞ )
(23.4) ∂T ∂η Typical values of h
Convective Process Free convection Gases Liquids Forced convection Gases Liquids Convective phase change h (W/m2K) 2  25 501000 25250 50  20 ,000 2,500  200,000 The Nusselt Number, hL/kf The
The group hL/kf is a nondimensional group that is is defined as the Nusselt number. Nu = hL kf = convective flux at the surface conductive flux in the fluid (23.5) Convective mechanisms...
Single phase flow and heat transfer Convective mechanisms • Single phase fluids (gases and liquids)
• Forced convection • Free convection, or natural convection • Mixed convection (forced plus free) Forced convection
Fluid forced past surface by a fan or blower.
y U ∞ , T∞
dP/dx < 0
x Tw > T∞ = 0 (steady state) y U ∞ T∞ ,
x δ ( x)
δ T ( x)
T w > T∞ ∂ ∂t ∫∫∫ ρedV =
V − ∫∫ p u ⋅ ndΩ − ∫∫ (τ ⋅ u )⋅ ndΩ
Ω Ω ∫∫ q ′′ ⋅ ndΩ − ∫∫ e(ρ u ) ⋅ ndΩ
Ω Ω (23.6) y U ∞ T∞ ,
x δ ( x)
δ T ( x)
T w > T∞ FS + ∫∫∫
V ∂ BρdV = ∂t + ∫∫∫ ρ udV
V = 0 (steady state) << momentum ∫∫ (ρu )u ⋅ ndΩ
Ω
(23.7) y U ∞ T∞ ,
x δ ( x)
δ T ( x)
T w > T∞ = 0 (steady state) FS = (ρu )u ⋅ ndΩ ∫∫
Ω (23.8) Surface forces include the pressure forces and shear forces. Velocity boundary layers & Cfx... Velocity boundary layer Velocity
Reynolds number = Re
x = ρU ∞ x U ∞ x = µ ν
(25.1)
δ(x) Re x < 5 x10 5 xcr x The boundary layer thickness, δ(x) δ ( x)
x 4 . 641 = Re 1x/ 2
δ (x)
x = (25.2) This says that the
Note that the exact solution is (x) ~ x1/2
5 Re 1/ 2 x The local skin friction coefficient, Cfx The
C fx = τw
1 2 ρU ∞ 2 − µ ⎛ ∂u ⎞ ⎜ ∂y ⎟ ⎝ ⎠ y =0 = 1 2 ρU ∞ 2 (25.3) (25.4) C fx 0 .646 = Re 1x/ 2 Average skin friction coefficient C fx 0.646 = 1/ 2 2 Re L (25.5) The average viscous drag over a finite length is twice the local value at that length for a laminar boundary layer. Laminar boundary layer Laminar heat transfer: the relation between δT(x) and h(x)... The case when the heated length is the entire plate, x > 0, and (0) = T(0) = 0. U ∞ , T∞
y
δΤ(x) δ(x)
Thermal boundary layer Heated wall, Tw > T∞ x The case when the heated The length begins at x0 > 0, and 0, T(x0) = 0. U ∞ , T∞ y
δ(x) δΤ(x) Heated wall, Tw > T∞ x0 x The convective heat transfer mechanism
• The thermal boundary layer is where conduction effects become significant. The thermal boundary layer can be approximated as a “conduction region.”
Heat transfer is dominated by conduction in the thermal boundary layer. U ∞ , T∞ Tw > T∞ The convective heat transfer mechanism The
• At the wall, the heat transfer on the fluid side can be evaluated via Fourier’s Law at y = 0. U∞ , T∞
∂T ∂y T(x,y) Tw > T∞
q w′ = − k f ′
y =0 The convective heat transfer mechanism
• The local heat transfer coefficient is defined as the local heat flux divided by the local temperature difference. q ′′ w = hx = T w − T∞ −kf ∂T ∂y y =0 T w − T∞ (25.5) Approximate analysis for h(x) Approximate
Recall that we assumed that the thermal boundary layer transfered all of its heat via conduction, not too bad an assumption for a rough approximation and analysis. Use the definition of h(x) and an approximation for Eq. (25.5) based on the Fourier Law applied to the entire thermal boundary layer: q′′ = hx ∆T w ∆T ≈ kf δ T ( x)
(25.6) hx ≈ kf δ T ( x)
(25.7) The xdependence of h(x) with heating from x = 0 kf hx ~ δ T (x) (25.7) x The heat transfer coefficient, h(x), with The heating starting from x0
h(x) ∼δ(x)−1 x0 x Nondimensional local heat transfer coefficient kf hx ~ δ T ( x) (25.7) δ T (x ) ≈ kf hx ⇒ δT
x ≈ kf hx x
(25.8) Nondimensional local Non heat transfer coefficient, h(x) δT k f ≈ (25.8) x hx x δT hx x 1
x ≈ Nux , Nux ≡ kf (25.9) Equation (25.9) relates the local Nusselt number to the thickness of the thermal boundary layer. Overall heat transfer Overall coefficient... Average heat transfer coefficient Average
kf hx ~ δ T (x)
(25.7) 1L hL = hx ( x)dx L0 k f L dx = L 0 δ T ( x) ∫ ∫ (25.10) Skin friction and heat transfer… Integral analysis of the Integral thermal boundary layer... Integral analysis of the thermal boundary layer
U ∞ , T∞ y
δ(x) Heated wall,
δΤ(x) Tw > T∞ x0 Assumptions: (a) Constant velocity (pressure) free stream. (b) Constant thermophysical properties (c) Wall at y = 0 is at a constant temperature x Integral analysis of the Integral thermal boundary layer
y
δ(x) U ∞ , T∞ Control Volume
For Integral Analysis δΤ(x) Heated wall, Tw > T∞
x x0 Momentum and Heat Flow In Control volume boundary Momentum and Heat Flow In Momentum and Heat Flow Out Heat Flow In at the Wall Approximate velocity and Approximate temperature profiles... Approximate velocity and temperature profiles for laminar flow.
y y=H A A T(x,y) U∞ T∞ u(x,y)
⎛ ∂T ⎞ dq′′ = −k f ⎜ ⎟ w ⎜ ∂y ⎟ ⎝ ⎠w 1 x
dx 2 Tw > T∞ The approximate profiles The for laminar flow Velocity u(x,y):
u 3⎛ y⎞ 1⎛ y⎞ = ⎜ ⎟− ⎜ ⎟ U∞ 2 ⎝ δ ⎠ 2 ⎝ δ ⎠
3 δT T(x,y) (24.1) Temperature T(x,y):
Compatibility and physical conditions on the profile T = Tw , y = 0 ∂ T = 0, y = δ T ∂y T = T∞ , y = δT ∂ 2T ∂y 2 = 0, y=0 (24.2) The approximate temperature profile T ( x , y ) = a + by + cy 2 + dy 3
From the conditions on T(x,y): (24.3) θ 3⎛ y ⎞ 1⎛ y = ⎜ ⎟− ⎜ θw 2 ⎜δT ⎟ 2 ⎜δT ⎝ ⎝⎠ θ ( x, y ) = T ( x, y ) − Tw ⎞ ⎟ ⎟ ⎠ 3 (24.4) The energy balance A A 1 2 + Energy convected in across 1A and AA + Heat transfer at the wall + Viscous work within the CV Energy convected = out across 2A
(24.5) Energy convected in across 1A and A A d⎛ H ⎞ = ρCT ∞ ⎜ 0udy ⎟ dx dx ⎝ ⎠ ∫ (24.6) Viscous work within the CV =µ ∫ H 0 ⎛ ∂u ⎞ ⎜ ⎟ dy ⎜ ∂y ⎟ ⎝ ⎠ (24.7) 2 Note: Viscous work has to be done owing to simultaneous existence of shear and rate of strain. Heat transfer at the wall Energy convected out across 2A ⎛ ∂T ⎞ = −k f ⎜ ⎜ ∂y ⎟ dx ⎟ ⎝ ⎠0
(24.8) H d⎛ = ρC uTdy + ρC uTdy ⎞ dx ⎜ ⎟ 0 0 dx ⎝ ⎠ ∫ H ∫ (24.9) The energy balance
d⎡ dx ⎢ ⎣ (T − T ∞ )udy ⎤ + µ ∫0 ⎥ ρC ⎦
H ⎡ ⎢ ⎢ ⎣ ∫ H 0 2 ⎤ ⎛ ∂u ⎞ ⎡ ∂T ⎤ ⎜ ⎟ dy ⎥ = α ⎢ ⎥ ⎜ ∂y ⎟ ⎥ ⎝ ⎠ ⎣ ∂y ⎦ o ⎦ To solve, for δ T(x), assume: (1) δ T(x) < δ (x) (x) (2) Viscous work is negligible (24.10) Thus all integrations are for 0 < y < δ T. The relation of the The thermal to the velocity boundary layer thickness, δΤ(x)/δ(x) in laminar flow… The thermal boundary layer, δT(x)
Substituting the approximate velocity and temperature profiles in to the Eq. (24.10) gives:
4 ⎡ ⎛ 3 δ T2 3 δT d − θ ∞U ∞ ⎢δ ⎜ 2 280 δ 4 dx ⎢ ⎜ 20 δ ⎣⎝ ⎞⎤ 3 αθ ∞ ⎟⎥ = ⎟⎥ 2 δ T ⎠⎦
(24.11) Neglect this term because δ T / δ < 1 and this term is much smaller than the first term inside the brackets. Let ζ = δT(x)/δ(x); Eq. (24.11) can be written: 3 3 αθ ∞ d θ ∞U ∞ δζ 2 = 20 2 δζ dx ( ) 1 dζ dδ ⎞ α ⎛ +ζ 2 U ∞ ⎜ 2δζ ⎟= 10 dx dx ⎠ δζ ⎝ dδ 140 ν 280 νx = with δ , and δ 2 = dx 13 U ∞ 13 U ∞ dζ 13 α ζ 3 + 4 xζ 2 = dx 14 ν (24.12) Solution for δT(x) / δ(x) = ζ ζ 3 + 4 xζ 2
3 dζ 13 α = dx 14 ν
−3 / 4 (24.12) 13 α + ζ = Cx 14 ν x = x0 ζ =0 (24.13) ζ 3 = Cx −3 / 4 + ζ =0 13 α 14 ν x = x0 (24.13) ⎡ ⎛ x0 ⎞ 3 / 4 ⎤ δT 1 −1 / 3 = ζ= Pr ⎢1 − ⎜ ⎟ ⎥ δ 1.026 ⎢ ⎝x⎠ ⎥ ⎣ ⎦ 1/ 3 where Pr = ν α (24.14) Role of the Prandtl number
When the plate is heated over its entire length, x0 = 0. ζ= 1 δT = Pr −1 / 3 1 .026 δ (24.15) The Prandtl number thus indicates the relative thickness of the viscous and thermal boundary layers. For large Pr, the viscous boundary layer is much thicker than the thermal boundary layer. The role of the Prandtl number The
Note that in Eq. (24.15 ), if we assume that 13/14 ~ 1, which is not a bad assumption, we get the result that δT −1 / 3 ≅ Pr δ (24.16) Large and small Prandtl number cases: δ δT δT δ Pr >> 1 δ >> δT Pr << 1 δ << δT The local and average The Nusselt numbers... The heat transfer coefficient, h(x)
− k f ⎛ ∂T ⎞ ⎜ ∂y ⎟ 3k f ⎝ ⎠w 3 k f = = hx = Tw − T∞ 2 δ T 2ζδ hx = 0.332 Pr
1/ 3 (24.17) ⎛U∞ ⎞ ⎜ ⎟ ⎝ νx ⎠ 1/ 2 ⎡ ⎛ x0 ⎞ ⎢1 − ⎜ ⎟ ⎢ ⎝x⎠ ⎣ 3/ 4 ⎤ ⎥ ⎥ ⎦ −1 / 3 (24.18) The heat transfer coefficient, h(x) The
h(x) ~x1/2 x0 x The Local Nusselt number
Define the local Nusselt number as the dimensionless heat transfer coefficient. Nu x = hx x kf (24.19)
−1 / 3 ⎡ ⎛ x0 ⎞ 3 / 4 ⎤ 1/ 3 1/ 2 Nu x = 0.332 Pr Re x ⎢1 − ⎜ ⎟ ⎥ ⎢ ⎝x⎠ ⎥ ⎦ ⎣ (24.20) Average Nusselt number for x0 = 0 Nu x = 0 . 332 Pr 1 / 3 Re 1x / 2
(24.21) Integrating from x = 0 to L, ∫ h ( x ) dx h=
L 0 Nu L L = 2 Nu x = L = 2hL (24.22) (24.23) The average Nusselt number Nu L ,avg = 0.664 Pr 1 / 3 Re1 / 2 L
(24.24) (Heating from the leading edge, x = 0) Special Problem Special
23h.1 Consider the physical and compatibility conditions that apply to the temperature distribution in the thermal boundary layer. Assume that Eq. (24.3) holds. Derive Eq. (24.3). θ 3⎛ y ⎞ 1⎛ y = ⎜ ⎟− ⎜ θw 2 ⎜δT ⎟ 2 ⎜δT ⎝ ⎝⎠ θ ( x, y ) = T ( x, y ) − Tw ⎞ ⎟ ⎟ ⎠ 3 (24.4) Special problem
Numerically demonstrate using any equation solver or spread sheet program that the average heat transfer coefficient is twice its local value for a given location on the flat plate. ⎡ ⎛ x0 ⎞ 3 / 4 ⎤ Nu x = 0.323 Pr 1 / 3 Re1 / 2 ⎢1 − ⎜ ⎟ ⎥ x ⎢ ⎝x⎠ ⎥ ⎣ ⎦ −1 / 3 (23h.20) The Reynolds analogy The for laminar flow... The role of the Prandtl number The
Note that in Eq. (24.15 ), if we assume that 13/14 ~ 1, which is not a bad assumption, we get the result that δT −1 / 3 ≅ Pr δ (26.1) Skin friction and heat transfer
• A relation between the heat transfer coefficient and the shear stress at the wall is suggested by Eq. (25.1). • Physically this is expected because pumping, or fan power, is expected to be related to the Reynolds number. • The Reynolds number appears in the expressions for the Nusselt number (heat transfer coefficient) and the skin friction coefficient. Skin friction and heat transfer Skin
The average Nusselt number is given by Eq. (24.24) as Nu L , avg = 0 .663 Pr 1 / 3 Re 1 / 2 L
(26.2) The average skin friction coefficient is given by as C fx , avg 2 0 .664 = Re 1 / 2 L (26.3) Reynolds analogy for laminar flow
Combine Eqs. (26.1) and (26.2). Nu L ,avg Re L ⎛1 ⎞ = ⎜ C fx ,avg ⎟ Pr −1 / 3 (26.4a) ⎝2 ⎠ Similarly for local values, Nu x ⎛ 1 ⎞ = ⎜ C fx ⎟ Pr −1 / 3 Re L ⎝ 2 ⎠ (26.4b) The Reynolds analogy The Nux ⎛ 1 ⎞ −1 / 3 = ⎜ C fx ⎟ Pr Re L ⎝ 2 ⎠ (26.4b) Eq. (26.4b) expresses the relation between the heat transfer coefficient and the wall shear. Note that the Reynolds number and Prandtl number remain formally a part of the relation. Reynolds analogy for laminar flow
The grouping Nu/RePr is defined as the Stanton number. Thus we have for average and local values: ⎛1 ⎞ StL,avg = ⎜ C fx,avg ⎟ Pr−2 / 3 , ⎝2 ⎠ ⎛1 ⎞ Stx = ⎜ C fx ⎟ Pr−2 / 3 ⎝2 ⎠
(26.5) Eqs. (26.4a,b) or (26.5) expresses the Reynolds analogy for laminar flow. The Stanton number The
hx x k f Nu x St x = = Re x Pr ⎛ Ux ⎞⎛ ν ⎞ ⎜ ⎟⎜ ⎟ ⎝ ν ⎠⎝ α ⎠ hx = ρUC p = hx (Tw − T∞ ) = ρUC p (Tw − T∞ ) Energy convected with the flow. Heat transfer at the wall due to convection Comments on the Reynolds analogy
Eq. (26.5) is termed the “analogy between skin friction and heat transfer.” A similarity between the momentum and thermal boundary layers is implied. The similarity between the thermal and momentum problems holds for: (a) Similar boundary conditions at the wall and the free stream for velocity and temperature. (b) The following relation between the boundary layer thickness: δ = Pr 1 / 3 δT (26.6) Example problem... Example The boundary layer thickness
Air at 27o C and 1 atm flows over a flat plate at a speed of 2 m/s. Calculate the boundary layer thickness at distances of 20 cm and 40 cm from the leading of the plate. Calculate the mass flow that enters the boundary layer between x = 20 and x = 40 cm. The viscosity of air at 27o C is 0.00198 kg/ms. Assume a unit depth in the zdirection. U = 2 m/s y T = 27o C xcr
x Solution:
ρ=
1.0132 x 10 5 P = = 1.177 RT (287 )( 300 ) ( kg / m 3 ) [0.073 lbm / ft 3 )] At x = 20 cm and 40 cm: Re Re x = 0 .20 x = 0 .40 = (1.177 )( 2.0)(0.2)
−5 1.98 x 10 = 45, 540 = 23, 770 Both Reynolds numbers are less than the critical value. Thus, we are in the laminar boundary layer region, and: δ
δ (0.2)
0.2 =
= 4.64 ≈ x Re 1x/ 2
(m) 4.64 4.64 = 0.030, δ ≈ 0.00602 1/ 2 = Re x 23770
4.64 1/ 2 = Re x 4.64 = 0.021, δ ≈ 0.00851 47540 δ (0.4)
0.4 ( m) δ ~ 0.0060 m (0.24 in) δ ~ 0.0085 m (0.34 in.) 0.2 0.4 x At any xlocation, the mass flow in the boundary layer is mδ = ρudy = ρU
0 ∫ δ ∫ δ 0 ⎡ 3 ⎛ y ⎞ 1 ⎛ y ⎞3 ⎤ 5 ⎢ ⎜ ⎟ − ⎜ ⎟ ⎥dy = ρUδ 8 ⎢2 ⎝δ ⎠ 2 ⎝δ ⎠ ⎥ ⎦ ⎣ δ ~ 0.006 m (0.24 in) δ ~ 0.009 m (0.34 in.) 0.2 0.4 x The mass addition is thus the difference between mδ at the two xlocations: x ∆mδ = 5 5 ρ U (δ (0 .4 ) − δ ( 0 .2 ) ) = (1 .177 )( 2 )( 0 .0085 − 0 .0060 ) 8 8 −3 = 3 .68 x10 ( kg / s ) (8 .1x10 − 3 lbm / s ) Consider the previous plate as being heated over its entire length at 60o C. Determine the heat transferred over the first 20 cm and over the first 40 cm of the plate. T∞ = 27 o C
Tw = 60 o
0.2 0.4 U = 2 m/s T = 27o C δ ~ 0.006 m (0.24 in) δ ~ 0.009 m (0.34 in.) Tw = 600 C
x To evaluate properties, use the standard definition of the film temperature: w ∞ f T= T +T 2 δT
x δ Solution:
60 + 27 = 43 . 5 deg C = 316 . 5 2 m2 /s ν = 17 . 36 x10 − 6 k = 0 . 02749 W /m − K Tf = C p = 1 . 006 Pr = 0 . 7
At x = 0.2 m K (110 . 3 o F ) kJ / kg − K Re x = ( 2 )( 0 . 4 ) = 23 ,041 ν 17 .36 x10 − 6 Nu x = 0 .332 Re 1x/ 2 Pr 1 / 3 = ( 0 .332 ) 23041 ( 0 .7 ) 1 / 3 = 44 . 74 = U∞x k (44.74)(0.02749) = = 6.15 x 0.2 h = 2hx = 12.3 W / m 2 − K hx = Nu x W / m2 − K (1.083 BTU .h − ft 2 − o ) Q = h A(Tw − T∞ ) = 12.3(0.2)(60 − 27) = 81.8 W (277BTU / h) Similarly for x = 0.4 m, h = 8.698 W / m2 − K (1.53BTU/ fts − h −o F) W (392 / h) BTU Q = (8.698 0.4)(60− 27) =1148 )( . Determine the drag on the plate on the first 40 cm using the analogy between fluid friction and heat transfer. Solution: St avg Pr 2/ 3 1 = C f , avg 2 T f = 316 .5 K kg / m 3 ρ= P = 1.115 RT St avg = 1 2 8.698 h = = 3.88 x10 − 3 ρC P U (1.115)(1006 )(2) C f , avg = 3.06 x10 − 3 τ W = C f , avg ρU 2 2 D = ( 0.0136 )(0.4) = 5.46 = 0.0136 N / m2 N/m (1.23 x10 − 3 lb f / ft ) Laminar external flow Laminar on a flat plate...
Heating from the leading edge U ∞ , T∞
y
δΤ(x) δ(x) Heated wall, Tw > T∞ x Nu x = 0 . 332 Pr 1/ 3 Re (27.1) 1/ 2 x Average Nusselt number for x0 = 0 Nu x = 0 . 332 Pr 1 / 3 Re 1x / 2
(27.1) Integrating from x = 0 to L, h= Nu
L ∫ h ( x ) dx
L 0 L = 2 Nu = 2hL (27.2) (27.3) x=L The average Nusselt number The Nu L ,avg = 0.664 Pr 1/ 3 Re 1/ 2 L (27.4) (Heating from the leading edge, x = 0) The Prandtl number in convection... Approximate values of Pr 102 Liquid Metals 1 Gases 10 Water 102 104 Viscous Oils Pr = υ α Prandtl number ranges
Type of fluid Liquid Metals Gases Water Viscous Oils Earth Mantle Rock Typical Prandtl Number < 102 ~ 0.7 ~3  15 > 100 ~1020 Heat transfer for in laminar boundary layer Heat flow  Small Prandtl numbers
U ∞ , T∞
y
δΤ(x) δ(x) Heated wall, Tw > T∞ x Heat transfer for in laminar boundary layer flow  Small Prandtl numbers
δΤ(x) U ∞ , T∞
y u(x,y)
δ(x) T(x,y) Heated wall, Tw > T∞ x Small Prandtl numbers Small
Performing the integral analysis with gives:
1/ 2 u ≈ U∞ Nu x = 0.565 (Re x Pr ) , Pr ≤ 0.5, Re x Pr ≥ 100
Note: RePr is termed the Peclet number, Pe (27.5) A generalized correlation... Generalized correlation Generalized The ChurchillOzoe correlation:
Nu x = [1 + (0.0468 / Pr ) ] 0.3387 Re 1 / 2 Pr 1 / 3 x
2 / 3 1/ 4 , Re x Pr ≥ 100
(27.6) Restrictions and Extensions: (1) All Prandtl numbers (2) Average value is twice local value Nusselt number for a constant wall heat flux... Nusselt number for constant heat Nusselt flux at the wall
Using the same methods as in the constant wall temperature case, we have for a constant wall heat flux: heat hx Nu x = = 0.453 Pr 1 / 3 Re 1x/ 2 kf
(27.7) Average temperature for the constant flux case
Nu
x = q ′′ x w k f (T w − T ∞ ) (T w − T∞ )avg = 1 L ∫ L 0 (T w − T ∞ ) dx = k 1 L ∫ L 0 q ′′ x w dx k f Nu x = q ′′ L w f 0 . 6795 Re 1L/ 2 Pr 1 / 3 (27.8) Practical Practical considerations for applications... Application methods & rules
Eq. (27.1) and (27.6) can be used with under the following restrictions:
(a) 0.6 < Pr < 50, which applies for most gases and fluids, except liquid metals (very small Pr). (b) For Pr ~ 0.7 or 1, T = (c) For very small Pr, a different kind of analysis is needed and is quite easily carried out. (d) Fluid properties are evaluated at a “film temperature” defined as Tf = Tw + T∞ 2 (27.9) Example problem... Example Heat transfer under constant flux conditions
A 1.0 kW heater is made in the shape of a 60 cm x 60 cm glass plate with an electrically conducting film that produces a constant heat flux. The air stream is 27o C, 1 atm, with U = 5 m/s. Calculate the average plate temperature difference along the plate and the temperature difference at the trailing edge.
y Tw = Tw ( x ) δ δΤ Qw = 1 kW x Initially evaluate the properties at 27 deg C because we Initially cannot predict the plate temperature at the outset. T ∞ = 300 K ν = 16 .84 x 10 − 6 m2 / s Pr = 0.708 k = 0.0264 W /m−K UL Re L = = 1.78 x 10 5 ν Use the following relations for the average wall temperature: Nu x = hx x = 0.453 Pr 1 / 3 Re 1x/ 2 kf (Tw − T∞ )avg = q ′′ L w kf 0.6795 Re1 / 2 Pr 1 / 3 L From laminar boundary analysis, we have (Tw − T∞ )avg = (1000 / 0.6 x 0.6 )(0.6)0.0264 q ′′ L / k w = 1/ 2 1/ 3 0.6795 Re L Pr 0.6795 (1.78 x10 −5 ) 1 / 2 (0.708 )1 / 3 = 248 .6 0 C Evaluate the film temperature using the above result. This sets up an interative type of calculation process. Tf = (248.6 + 27) / 2 = 137.8C = 410.8K ν = 27.18m2 / s, Pr = 0.687, k = 0.0344 / m − K W Re L = 1.10 x10 5 (Tw − T∞ )avg = 243.6 o C
Here we stop the calculation, but we could correct it one more time to see how much the average wall temperature would vary upon successive approximations owing to improved values of the film temperature. At x = 0.6 m, we use the NusseltReynolds number relation for constant flux to get: Nu L = 0.453 Re1 / 2 Pr1 / 3 , (Tw − T∞ ) x= L = L (Tw − T∞ ) L=0.6 = 365.4 o C qw x kNuL External Forced External Convection  7
Turbulent Boundary Layer Heat Transfer Integral analysis of turbulent boundary layer flow and heat transfer on a constant temperature plate Overview
• Review: The laminar boundary layer • Integral analysis of the turbulent boundary layer
– Empirical velocity profile – The Blasius equation for wall shear stress – The boundary layer thickness, δ(x), using Prandtl’s approxmation – Laminar sublayer model Overview Overview • Approximate models for the laminarturbulent thermal boundary layer • Integral analysis for heat transfer The laminar boundary layer... The laminar boundary layer The
y x
Transition region
Fully turbulent boundary layer Laminar boundary layer δT − 1/ 3 = Pr δ C fx
2 (29.1a) = 0.332 Re 1 / 2 x (29.1b) (29.1c) ⎛1 ⎞ St x = ⎜ C fx ⎟ Pr − 2 / 3 ⎝2 ⎠ The turbulent boundary layer... The overall fluid mechanics The
δ ~ x4/5
y δ ~ x1/2 x
Laminar boundary layer Transition region
Fully turbulent boundary layer, The turbulent boundary layer
y x
Transition region
Fully turbulent boundary layer Laminar boundary layer C fx = 0.0592 Re −1 / 5 x δ = 0 . 38 Re x (29.2a)
−1 / 5 x (29.2b) (29.2c) ⎛1 ⎞ StL = ⎜ CfL ⎟ Pr−2/ 3 ⎝2 ⎠ Comments on Eqs. Comments (29.2a,b,c)... C fx = 0.0592 Re −1 / 5 x (29.2a)
This is an empirical result based on a compilation of measurements from a large number of studies. It applies to flows for which 5x105 < ReL < 107. For larger Reynolds numbers from 107 < ReL < 109, the following formula is recommended. C fx = 0.370 [log 10 Re L ] −2.584
(29.3) Additional useful results for viscous drag on a flat plate Additional in uniform flow: Cf = (log10 Re L )2.584
5x105 1742 0.455 − A , Re L < 10 9 Re L
(29.4) 106 3340 3x106 8940 where
Recrit 3x103 A 1055 An approximate expression is, Cf = 0.074 A − , Re L < 10 7 0.20 Re L Re L (29.5) δ
x = 0.37Re −1/ 5 x (29.2b) This is is an approximate result based on an integral analysis using Eq. (29.2a) and the 1/7th velocity profile within the turbulent boundary layer. Eq. (29.2a) gives the wall shear, and the velocity profile represents the turbulent core, i.e., u ⎛ y⎞ =⎜ ⎟ U∞ ⎝ δ ⎠ 1/ 7 (29.6) ⎛ 1 ⎞ −2/3 StL =⎜ CfL ⎟Pr ⎝ 2 ⎠ (29.2c)
Eq. (29.2c) applies to the entire laminarturbulent boundary laminarlayer on a plate of length L. Local values are given by the following formulas based empirical skin friction data. Stx Pr2 / 3 = 0.0296Re1 5 , 5 ×105 < Rex < 107 x Stx Pr2 3 = 0.185(log10 Rex )
−2.584 , 107 < Rex < 109
(29.7) Approximate models for the heat transfer coefficient for the laminarturbulent boundary layer... Analogybased model Analogy
From Eq. (29.5) and the Reynolds analogy, Eq. (29.2c), St L Pr2 3 = 0.037Re4 5 − 871 , L Recrit = 5 ×105 ReL < 107
(29.8a) N u L = 0 . 037 Re 4 / 5 − 871 Pr 1 / 3 L
(29.8b) [ Average heat transfer coefficient 1⎡ hL = L⎢ ⎣ ∫ xcr h( x ) dx + 0 ∫ h( x ) dx ⎤ ⎥ xcr ⎦
L
(29.9) Laminar portion of the flow. Turbulent portion of the flow for which the Reynolds analogy for local values of the Stanton number. Note that the choice of the transition Reynolds number is important and determines the form of the local skin friction coefficient. Integral analysis using Integral the laminar sublayer model... Integral analysis
T∞ ,U ∞
δ(x) x=0 Xcr Assume that the boundary layer begins at the leading edge and is fully turbulent. (Prandtl’s approximation.) From tube flow measurements due to Blasius, we have for shear stress at the wall: 1/ 4
2⎡ ν ⎤ − τ w = 0.0235 (ρ U ∞ ) ⎢U δ ⎥ ⎣∞⎦ (29.10) Integral analysis Integral
Also from measurement, have for the velocity profile in the turbulent boundary layer: u ⎛ y⎞ = U∞ ⎝ δ ⎠
The profile holds in the core of the turbulent boundary layer. 1/ 7 (29.11) U∞ Integral analysis
Shear stress at the wall from Eq. (29.11) is predicted to be infinite. ∂u τ w = −µ ∂y ∂u 1 U ∞ = ∂y 7 δ 1/ 7 y 6/ 7 (29.12) U∞
Laminar sublayer ub In the laminar sublayer, turbulence is observed to In disappear, that is shear is dominated by viscous transport. The velocity at the edge of the laminar sublayer is donoted by ub. Integral analysis  Prandtl’s model
The Prandtl model is to use Eq. (29.5) for smooth surfaces with the velocity profile Eq. (29.6). ρ U (U ∞ − u )dy = − τ w
0 1/ 7 1/ 7 ⎡ ⎛y⎞ ⎤ ⎢1 − ⎜ ⎟ ⎥ dy = 0 . 0235 ⎝δ ⎠ ⎥ ⎢ ⎦ ⎣ ∫ δ ∫ δ 0 ⎛y⎞ ⎜⎟ ⎝δ ⎠ ⎛ν ⎜ ⎜U δ ⎝∞ ⎞ ⎟ ⎟ ⎠ 1/ 4 (29.13) Integral analysis  Prandtl’s model Integral
⎛ν⎞ 7 dδ ⎟ = 0.0235 ⎜ 72 dx ⎝ U ∞δ ⎠ ⎛ν⎞ ⎟ δ 1 / 4 d δ = 0.242 ⎜ ⎝U∞ ⎠
1/ 4 1/ 4 dx x + Const δ 5/4 ⎛ν⎞ ⎟ = 0.303 ⎜ ⎝U∞ ⎠
1/ 5 1/ 4 ⎛ν⎞ ⎟ δ = 0.384 ⎜ ⎝U∞ ⎠ x 4 /5 (29.14) Integral analysis  Prandtl’s model
T∞ ,U ∞
δ(x) x=0 δ 0.384 = x Re 1/ 5 x (29.15) This is an approximation. The actual velocity boundary layer is thicker. Integral analysis Laminar sublayer Integral model
T∞ ,U ∞
δ(x) x=0 T∞ ,U ∞ δ(x)
δturb δb x=0 xcr Integral analysis  Laminar sublayer model
δ(x) U∞
δturb Approximate laminar sublayer velocity profile. δb
Laminar sublayer ub Integral analysis  Laminar sublayer model Integral
In the laminar sublayer, the shear stress is constant at the wall value. Thus, −τ w = µ u y 2 ρU∞ ⎛ ν ⎞ ⎜ ⎟ u = 0 .0235 µ ⎝ U ∞δ ⎠ 1/4 y
3/4 ⎛ν⎞ u δb 1 ⎜ ⎟ =b δ U ∞ 0 .0235 ⎝ U ∞ δ ⎠ (29.16) Integral analysis  Laminar sublayer model δ b ⎡ ub ⎤ =⎢ ⎥ δ ⎣ U∞ ⎦ 7 δ(x) U∞
ub δturb
(29.17) δb Combining Eqs. (29.15) and (29.16) gives: ub − = 1.878 Re δ 1/ 8 U∞ (29.18) Integral analysis  Laminar sublayer model Integral
Using the laminar sublayer model and the above results for the velocity profile, we have 0.0296 2 −τ w = µ = ρU∞ Re 0.2 δb x (29.19)
This applies to the turbulent boundary layer. The total resistive force on the wall is determined using the laminarturbulent model as before. ub Integral analysis  Laminar sublayer model δ b ⎡ ub ⎤ =⎢ ⎥ δ ⎣ U∞ ⎦ 7 194 δ b ⎡ ub ⎤ = = 0.7 Re x δ ⎢U∞ ⎥ ⎣⎦ 7 (29.20) Integral analysis  Laminar sublayer Integral model
Prandtl proposed that the critical Reynolds number is 485,000. Thus the average skin friction coefficient over the plate is C f ,L 0 . 074 1700 = − 1/ 5 Re L Re L (29.21) The result agrees well with experiments. Integral analysis  laminar sublayer Integral model C f ,L 0.074 1700 = − 1/ 5 Re L Re L (29.21) The heat transfer coefficient is obtained with the Reynolds analogy to obtain values of average Stanton numbers. Key terms and concepts Key
Laminar starting length Laminar sublayer Prandtl model for the turbulent boundary layer  This model employs use of empirical wall shear stress expression obtained from tube flow and an empirically obtained 1/7th velocity profile for the core region to determine the relation between boundary layer thickness and distance from the leading edge, δ(x). The major assumption is that the boundary layer is turbulent from the leading edge, i.e., the laminar starting length and transition region are assumed not to exist. Transition region Turbulent boundary layer Turbulent core region ...
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This note was uploaded on 12/23/2010 for the course MECH 412 taught by Professor Drnesreenghaddar during the Spring '09 term at American University of Beirut.
 Spring '09
 DrNesreenGhaddar
 Heat Transfer

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