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Boundary Layer Similarity

Mathematic

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Boundary Layer Similarity

  • B. L. Equations:






≈0


≈0

For low-speed, forced convection flows :

RHS = diffusion terms

LHS = convection terms

The last three equations are very similar

B.L. Similarity Parameters: Re, Pr, Sc

L - is some characteristic length

U - is the free stream velocity

T - is the free stream temperature

TS - is the surface temperature

CA,S – is the concentration of species A at the …… surface

CA,∞ - is the concentration of species A …… free stream

Using the above dimensionless independent variables, the B.L. equations become:

≈0

Reynolds #:

Prandtl #: Pr≈1 for gases

Pr>>1 for oils

Pr<<1 for liquid metane

Schmidth #:

Lewis #:

Other dimensionless groups:

Biot #:

Nusselt #:

Stanton #:

Sherwood #:

Mass Transfer

Stanton #:

Functional Form of the solutions for the B.L. Equations

Eq. 2 :

≈ on the geometry of the surface

universally applicable to different fluids and over a wide range of values for U and L

Eq. 3 :

Eq. 4:

The similarity parameters reduce the independent variables and make the solutions universally applicable

Boundary Layer Analogies

  • provide relations between Cf, Nu and Sh

The Heat and Mass Transfer Analogy



Eq. (3) and (4), as well as the associated boundary conditions are analogous → T* and C*A must be of the same functional form

└→ heat and mass transfer relations for a particular geometry are interchangeable

Hence, h can be obtained from knowledge of hm and vice-versa.

The same relation can be applied to and , and it may be used in turbulent, as well as laminar flow.

Example

T=200C

V=100 m/s TS=800C

P=1 at


L=1m

  • The Reynolds Analog

If and Sc=Pr => Eqs. (2)-(4) are of precisely the same form

u*,T* and CA* are equivalent as solutions.

“Reynolds analogy”

It has been shown that, with certain corrections, the analogy may be applied over a wide range of Pr and Sc.


“modified Reynolds” or “Chilton-Colburn analogies”

!!! also valid for averages:

End Note:

  • T* and CA* - fields may be obtained analytically only for a few simple cases: h→hm
  • The more practical approach frequently involves calculating h and hm from empirical equations; the empirical equations are obtained by correlating measured convection and mass transfer results in terms of appropriate dimensionless groups

PROBLEM 1.

KNOWN: Air flow conditions and drag force associated with a heater of prescribed surface temperature and area,

FIND: Required heater power.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Reynolds analogy is applicable, (3) Bottom surface is adiabatic.

PROPERTIES: Air (Tf=350K, 1 at): ρ=0.995 kg/m3, cp=1009 J/Kg K, Pr=0.7

ANALYSIS: The average shear stress and friction coefficient are

From the Reynolds analogy, modified:

Solving for and substituting numerical values, find

=0.995 kg/m3(15 m/s) 1009 J/kg K (8.93x10-3/2)(0.7)-2/3 ;

=85 W/m2K

Hence, the heat rate is :

q=A(TS-T )=85W/m2K(0.25m2)(140-15)oC;

q=2.66kW

COMMENTS: Due to bottom heat losses, which have been assumed negligible, the actual power requirement would exceed 2.66kW.







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