Chilton Colburn Analogy - Overall Concept

6,133 views 12 slides Oct 09, 2019
Slide 1
Slide 1 of 12
Slide 1
1
Slide 2
2
Slide 3
3
Slide 4
4
Slide 5
5
Slide 6
6
Slide 7
7
Slide 8
8
Slide 9
9
Slide 10
10
Slide 11
11
Slide 12
12

About This Presentation

This presentation is made to provide the overall conceptual knowledge on Chilton Colburn Analogy. It includes basis, importance, assumption, advantages, limitations and applications in addition to the derivation. Make It Useful!


Slide Content

CHILTON COLBURN ANALOGY Syed Hameed Anwar Sahib V Submitted By

BASIS Based on experimental data for gases and liquids in both the laminar and turbulent flow regions IMPORTANCE Lies in heat transfer characterization for flows having Prandtl Number not equal to unity

USUAL FORM AND ITS USE j H = j D = a function of Re, Geometry and Boundary Conditions has proven to be useful for transverse flow around cylinders , flow through packed beds and flow in tubes at high Reynolds Numbers ASSUMPTION Pr ≠ Sc ≠1 ; No form drag exit condition where , Pr = Prandtl Number = Momentum Diffusivity/Thermal Diffusivity = C p μ/k Sc = Schmidt Number = Momentum Diffusivity/Mass Diffusivity = μ/ ρ D AB

DERIVATION From Sieder Tate Equation for turbulent flow, Nu = 0.023 Re 0.8 Pr ⅓ ( μ/μ w ) 0.14 , 0.7 ≤ Pr ≤ 16700 , Re > 10000 (1) where, Nu = Nusselt Number = Resistance to Conduction/Resistance to Convection = hL/k Re = Reynolds Number = Inertial Force/Viscous Force = Dv ρ / μ Pr = Prandtl Number = Momentum Diffusivity/Thermal Diffusivity = C p μ/k Dividing equation (1) by RePr, Nu/RePr = 0.023 (Re 0.8 Pr ⅓ (μ/μ w ) 0.14 /RePr) Rearranging, St Pr ⅔ ( μ/μ w ) 0.14 = 0.023 Re -0.2 For turbulent flow region, an empirical relation of f and Re, f/2 = 0.023 Re -0.2 f/2 = St Pr ⅔ (μ/μ w ) 0.14 = 0.023 Re -0.2 j H = j-factor for Heat Transfer

DERIVATION Similarly, Relating Mass and Momentum Transfer using Mass Transfer Equation, k c ’ D/D AB = 0.023 Re 0.8 Sc ⅓ Dividing equation (2) by ReSc, k c ’ /v Sc ⅔ Re 0.03 = 0.023 Re -0.2 j D = j-factor for Diffusivity Simplified Form of Chilton Colburn Analogy This analogy is also known as Modified Reynolds Analogy where, St H = Nu/RePr = Heat/Thermal Capacity and St M = Sh/ReSc = Mass/Density*Velocity (2) f/2 = St H Pr ⅔ = j H = St M Sc ⅔ = j D | 0.6 < Pr < 60 , 0.6 < Sc < 3000

VALIDITY Validity of Chilton Colburn Analogy depends on property of the fluid and length of the plate This analogy is valid for flow around spheres only when Nu and Sh are replaced by (Nu-2) and (Sh-2) This analogy is not valid below Re = 10000

CHILTON COLBURN j-Factors for heat and mass transfer Heat Transfer quantities (Pure Fluids) Binary Mass Transfer quantities (Isothermal Fluids) MOLAR UNITS Binary Mass Transfer quantities (Isothermal Fluids) MASS UNITS j H = Nu Re -1 Pr - ⅓ = h/ρC p v (C p μ/k) 2/3   j D = Sh Re -1 Sc - ⅓ = k/cv (μ/ρ D AB ) 2/3 j D = Sh Re -1 Sc - ⅓ = k/ρv (μ/ρ D AB ) 2/3

Chilton colburn analogy for flows with and without form drag For flow past a flat plate or a pipe where no form drag is present , f/2 = j H = j D For flow in packed beds or other blunt objects where form drag is present , f/2 > j H or j D and j H = j D

Dimensionless groups for heat and mass transfer Heat Transfer quantities (Pure Fluids) Binary Mass Transfer quantities (Isothermal Fluids) MOLAR UNITS Binary Mass Transfer quantities (Isothermal Fluids) MASS UNITS Re = Lv ρ / μ Fr = v²/gL Nu = hL/k Pr = C p μ/k Gr = L³ ρ² g β∆ T/ μ ² Pé = Lv C p /k Sh = kL/cD AB Sc = μ/ ρ D AB Gr = L³ ρ² g ε∆ x/ μ ² Pé = Lv /D AB Sh = kL/ ρ D AB Sc = μ/ ρ D AB Gr = L³ ρ² g ε∆ w/ μ ² Pé = Lv /D AB

Chilton colburn analogy for long smooth tubes In highly turbulent range (Re > 10000), the heat transfer ordinate agrees approximately with f/2 for the long smooth pipes under consideration f/2 = j H where , j H = Nu Re -1 Pr - ⅓ = h/ρC p v ( C p μ/k) 2/3 = hS/wC p (C p μ/k) 2/3 where , S = Area of the tube cross section w = Mass flowrate through the tube f/2 is obtainable using Re = Dw/Sμ

Advantages Useful to find each term for highly turbulent flow as f equation is known f = 0.058 Re -1/5 limitations Only applicable when no form drag is present in flows Only applicable to conditions 0.6 < Pr < 60 , 0.6 < Sc < 3000 3. Not applicable to heavy oils (Pr>60) and liquid metals (Pr<0.6) applications Evaluation of heat transfer in internal forced flow Heat Exchanger Design 3. Reactor Design