Boundary Layer Similarity
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:
Prandtl #: Pr≈1 for gases
Pr>>1 for oils
Pr<<1 for liquid metane
Other dimensionless groups:
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 :
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 hm and vice-versa.
The same relation can be applied to and , and it may be used in turbulent, as well as laminar flow.
V=100 m/s TS=800C
If and Sc=Pr => Eqs. (2)-(4) are of precisely the same form
u*,T* and CA* are equivalent as solutions.
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:
KNOWN: Air flow conditions and drag force associated with a heater of prescribed surface temperature and area,
FIND: Required heater power.
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 ;
Hence, the heat rate is :
COMMENTS: Due to bottom heat losses, which have been assumed negligible, the actual power requirement would exceed 2.66kW.
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