AME$60634$$ Int.$HeatTrans.$ External Convection: …sst/teaching/AME60634/lectures/AME... ·...

13
AME 60634 Int. Heat Trans. D. B. Go External Convection: Laminar Flat Plate For a constant property, laminar flow a similarity solution exists for the flow field u(y) δ x = 5 x Re x 1 2 C f , x = 0.664 Re x 1 2 local boundary layer thickness local skin friction coefficient C f , x = τ s, x 1 2 ρu 2 τ s, x = 1 x τ s, x dx 0 x average skin friction coefficient Major flow parameters: C f , x = 1.328Re x 1 2

Transcript of AME$60634$$ Int.$HeatTrans.$ External Convection: …sst/teaching/AME60634/lectures/AME... ·...

Page 1: AME$60634$$ Int.$HeatTrans.$ External Convection: …sst/teaching/AME60634/lectures/AME... · External Convection: Laminar Flat Plate ... • The flow and heat transfer are affected

AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Laminar Flat Plate For a constant property, laminar flow a similarity solution exists for the flow field u(y)

δx =5xRex

12

Cf ,x = 0.664Rex− 12

local boundary layer thickness

local skin friction coefficient

Cf ,x =τ s,x12ρu∞

2⇒ τ s,x =

1x

τ s,x dx0

x

∫average skin friction coefficient

Major flow parameters:

⇒Cf ,x =1.328Rex−12

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Laminar Flat Plate For a constant property, laminar flow a similarity solution exists for the flow field u(y)

Nux = 0.332Rex12 Pr

13local Nusselt number

(Pr > 0.6)

local thermal boundary layer thickness δx δt,x = Pr13

average Nusselt number Nux =hxxk⇒ hx =

1x

hx dx0

x

∫ ⇒ Nux = 0.664Rex12 Pr

13

uniform surface temperature, Ts

Nux = 0.453Rex12 Pr

13

Ts −T∞( ) =##qsLNuL

=##qL

xkNux

dx0

L

∫ ⇒ NuL = 0.680ReL12 Pr

13

uniform surface heat flux, q”s

Major heat transfer parameters:

local Nusselt number (Pr > 0.6)

average Nusselt number

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Turbulent Flat Plate

Cf ,x = 0.0592Rex− 15local skin friction coefficient

Nux = 0.0296Rex45 Pr

13

local Nusselt number (Pr > 0.6)

average skin friction coefficient

average Nusselt number

NuL = 0.037ReL45 Pr

13

For xc= 0 or L >> xc (Rex,L >> Rex,c)

C f ,L = 0.074ReL− 15

Nux = 0.0308Rex45 Pr

13

uniform surface temperature, Ts

uniform surface heat flux, q”s

average skin friction coefficient

average Nusselt number

NuL = 0.037ReL45− 871#

$ % &

' ( Pr

13

assuming xc for Rex,c = 5×105

C f ,L = 0.074ReL− 15−1742ReL

−1

uniform surface temperature, Ts

uniform surface temperature, Ts

For turbulent flow, only empirical relations exist

τ s,L =1L

τ s,x,lamdx0

xc

∫ + τ s,x,turbdxxc

L

∫$

% & &

'

( ) )

h L =1L

hx, lamdx0

xc

∫ + hx,turbdxxc

L

∫#

$ % %

&

' ( (

Average parameters

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Starting Length •  The effect of an unheated starting length (USL) can be represented

on the local Nusselt number as:

•  Parameters a, b, C, & m depend on –  thermal boundary condition: uniform surface temperature (UST) or

uniform heat flux (UHF) –  flow conditions: laminar or turbulent

Nux =Nux ξ = 0

1− ξx

$ % & '

( ) a*

+ ,

-

. /

b where

Nux ξ = 0 = CRexm Pr

13 for Pr > 0.6

LAMINAR   TURBULENT  

a 3/4   3/4   9/10   9/10  

b 1/3   1/3   1/9   1/9  

C 0.332   0.453   0.0296   0.0308  

m 1/2     1/2   4/5   4/5  

" " q s

Ts

" " q s

Ts

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Starting Length •  Uniform Surface Temperature (UST)

•  Uniform Heat Flux (UHF)

•  The Nusselt number (and heat transfer coefficient) are functions of the fluid properties (ν,ρ,α,cp,k) –  the effect of variable properties may be considered by evaluating all

properties at the film temperature

–  most accurate solutions often require iteration on the film properties

Tf =Ts + T∞2

NuL = NuLξ = 0

LL −ξ

1− ξ L( ) 2p+1( ) 2p+2( )[ ]2p( ) 2p+1( ) p = 1 (laminar throughout)

p = 4 (turbulent throughout)

h L =1L

hx, lamdxξ

xc

∫ + hx,turbdxxc

L

∫$

% & &

'

( ) )

numerical integration for laminar/turbulent flow

" " q x = hx Ts −T∞( )⇒ q = h L As Ts −T∞( )

Ts x( ) = T∞ +# # q s

hx

⇒ q = # # q sAs

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Cylinder in Cross Flow •  As with flat plate flow, flow conditions determine heat transfer •  Flow conditions depend on special features of boundary layer

development, including onset at stagnation point, separation, and onset of turbulence

•  Stagnation point: location of zero velocity and maximum pressure –  boundary layer development under a favorable pressure gradient è

acceleration of the free stream flow

•  There is a minimum in the pressure distribution p(x) and toward the rear of the cylinder, the pressure increases. –  boundary layer development under an adverse pressure gradient €

dpdx < 0→ du∞

dx > 0

dpdx > 0

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Cylinder in Cross Flow •  Separation occurs when the momentum of the free stream flow is

insufficient to overcome the adverse pressure gradient –  the velocity gradient reduces to zero –  flow reversal occurs accompanied by a downstream wake

•  Location of separation depends on boundary layer transition

ReD =VDν

note:

V ≠ u∞

dudy y= 0

= 0

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Cylinder in Cross Flow •  Force (FD) imposed by the flow on the cylinder is composed of

two phenomena –  friction è boundary layer shear stress –  form drag (pressure drag) è pressure differential due to wake

CD ≡FD

Af ρV2 /2( )

drag coefficient

Af is the area projected perpendicular to free stream

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Cylinder in Cross Flow •  Thermal considerations: uniform surface temperature, Ts

–  heat transfer a function of the angel of separation θ –  empirical correlations are used to determine average Nusselt

numbers

•  Hilpert correlation: Pr ≥ 0.6 –  also suitable for non-circular cylinders

•  Churchill and Bernstein: ReDPr > 0.2

NuD =h Dk

NuD = CReDm Pr

13

ReD C m 0.4-­‐4   0.989   0.330  

4-­‐40   0.911   0.385  

40-­‐4000   0.683   0.466  

4000-­‐40,000   0.193   0.618  

40,000-­‐400,000   0.027   0.805  

NuD = 0.3+0.62ReD

1 2 Pr1 3

1+ 0.4 Pr( )2 3[ ]1 4 1+

ReD282,000"

# $

%

& ' 5 8(

) * *

+

, - -

4 5

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Sphere in Cross Flow •  Similar flow issues as cylinder in cross flow arise

•  Thermal considerations: uniform surface temperature, Ts –  heat transfer again defined by empirical correlations

•  Whitaker correlation: –  0.71 < Pr < 380 –  3.5 < ReD < 7.6×104

NuD = 2 + 0.4ReD1 2+ 0.06ReD

2 3( )Pr0.4 µµs

"

# $

%

& '

1 4

evaluate fluid properties at T∞ except for µs which is evaluated at Ts

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Impinging Jet •  Impinging jet consists of a high speed flow impacting a flat surface

–  generates large convection coefficients •  The flow and heat transfer are affected by a number of factors

–  shape/size of jet, velocity of jet, distance from plate, … •  Significant hydrodynamic features:

–  mixing and velocity profile development in the free jet –  stagnation point and zone –  velocity profile development in the wall jet

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AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Impinging Jet •  Local Nusselt number distribution:

•  Average Nusselt number based on empirical correlations for single nozzles and arrays of nozzles –  function of Reynolds number, Pr, distance along wall (r or x), height of

jet (H)

Nu = f ReDh,Pr, r or x( ) Dh ,H Dh( )

Page 13: AME$60634$$ Int.$HeatTrans.$ External Convection: …sst/teaching/AME60634/lectures/AME... · External Convection: Laminar Flat Plate ... • The flow and heat transfer are affected

AME  60634    Int.  Heat  Trans.  

D.  B.  Go  

External Convection: Impinging Jet •  Martin correlation: uniform surface temperature, Ts

–  single round nozzle

•  Martin correlation: uniform surface temperature, Ts –  single slot nozzle

NuPr0.42

=G r D,H D( )F1(ReD )

F1(ReD ) = 2ReD1 2 1+ 0.005ReD

0.55( )1 2⇒ ReD =

VexitDν

G =D r( ) 1−1.1 D r( )( )

1+ 0.1 H D− 6( ) D r( )

2000 ≤ ReD ≤ 400,0002 ≤ H D ≤122.5 ≤ r D ≤ 7.5

valid for

NuPr0.42

=3.06

x W + H W +2.78(ReDh

m )

⇒ ReDh=VexitDh

ν=Vexit (2W )

ν

m = 0.695 − x2W%

& '

(

) * +

H2W%

& '

(

) * 1.33

+ 3.06+

, -

.

/ 0

−1

3000 ≤ ReDh≤ 90,000

2 ≤ H W ≤104 ≤ x W ≤ 20

valid for