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Boundary Layer Similarity


_{}
≈0
_{}
_{}
≈0
_{}
For lowspeed, 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
T_{S}  is the surface temperature
_{} C_{A,S} – is the concentration of species A at the …… surface
C_{A,∞}  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 #: _{}
Sherwood #: _{}
Mass Transfer
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
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 h_{m} and viceversa.
The same relation can be applied to _{} and _{}, and it may be used in turbulent, as well as laminar flow.
Example
T_{∞}=20^{0}C
V=100 m/s T_{S}=80^{0}C
P=1 at
L=1m
_{}
If _{} and Sc=Pr => Eqs. (2)(4) are of precisely the same form
u^{*},T^{*} and C_{A}^{*} 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 “ChiltonColburn analogies”
!!! also valid for averages: _{}
End Note:
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) Steadystate conditions, (2) Reynolds analogy is applicable, (3) Bottom surface is adiabatic.
PROPERTIES: Air (T_{f}=350K, 1 at): ρ=0.995 kg/m^{3}, c_{p}=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/m^{3}(15 m/s) 1009 J/kg K (8.93x10^{3}/2)(0.7)^{2/3 };
_{}=85 W/m^{2}K
Hence, the heat rate is :
q=_{}A(T_{S}T_{ })=85W/m^{2}K(0.25m^{2})(14015)^{o}C;
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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