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Syed Hameed Anwar Sahib V
B A S I S
Based on experimental data for gases and liquids in
both the laminar and turbulent flow regions
I M P ORTA N C E
Lies in heat transfer characterization for flows having
Prandtl Number not equal to unity
U S UA L F O R M A N D I T S U S E
jH = jD = 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
A S S U M PTION
Pr ≠ Sc ≠1 ; No form drag exit condition
where,
Pr = Prandtl Number = Momentum Diffusivity/Thermal Diffusivity = Cpμ/k
Sc = Schmidt Number = Momentum Diffusivity/Mass Diffusivity = μ/ρDAB
D E R I VAT I O N
From Sieder Tate Equation for turbulent flow,
Nu = 0.023 Re0.8 Pr⅓ (μ/μw)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 = Dvρ/μ
Pr = Prandtl Number = Momentum Diffusivity/Thermal Diffusivity = Cpμ/k
Dividing equation (1) by RePr,
Nu/RePr = 0.023 (Re0.8 Pr⅓ (μ/μw)0.14/RePr)
Rearranging,
St Pr⅔ (μ/μw)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 Pr⅔ (μ/μw)0.14 = 0.023 Re-0.2 jH = j-factor for Heat Transfer
D E R I VAT I O N
Similarly, Relating Mass and Momentum Transfer using Mass Transfer Equation,
kc
’D/DAB = 0.023 Re0.8 Sc⅓
Dividing equation (2) by ReSc,
kc
’/v Sc⅔ Re0.03 = 0.023 Re-0.2 jD = j-factor for Diffusivity
Simplified Form of Chilton Colburn Analogy
This analogy is also known as Modified Reynolds Analogy
where,
StH = Nu/RePr = Heat/Thermal Capacity and StM = Sh/ReSc = Mass/Density*Velocity
(2)
f/2 = StH Pr⅔ = jH = StM Sc⅔ = jD | 0.6 < Pr < 60 , 0.6 < Sc < 3000
VA L I D I T Y
 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
C H I LTON CO L B URN J - FAC TORS
F O R H E AT A N D M A S S T R A N S F ER
Heat Transfer quantities
(Pure Fluids)
Binary Mass Transfer
quantities
(Isothermal Fluids)
MOLAR UNITS
Binary Mass Transfer
quantities
(Isothermal Fluids)
MASS UNITS
jH = Nu Re-1 Pr-⅓
= h/ρCpv (Cpμ/k)2/3
jD = Sh Re-1 Sc-⅓
= k/cv (μ/ρDAB)2/3
jD = Sh Re-1 Sc-⅓
= k/ρv (μ/ρDAB)2/3
C H I LTO N CO L B U R N A N A LO G Y
F O R F LO W S W I T H A N D W I T H O U T
F O R M D R AG
 For flow past a flat plate or a pipe where no form drag is
present,
f/2 = jH = jD
 For flow in packed beds or other blunt objects where
form drag is present,
f/2 > jH or jD
and jH = jD
D I M E NS IONLESS G R O U P S
F O R H E AT A N D M A S S T R A N S F ER
Heat Transfer quantities
(Pure Fluids)
Binary Mass Transfer quantities
(Isothermal Fluids)
MOLAR UNITS
Binary Mass Transfer quantities
(Isothermal Fluids)
MASS UNITS
Re = Lvρ/μ
Fr = v²/gL
Nu = hL/k
Pr = Cpμ/k
Gr = L³ρ²gβ∆T/μ²
Pé = LvCp/k
Sh = kL/cDAB
Sc = μ/ρDAB
Gr = L³ρ²gε∆x/μ²
Pé = Lv/DAB
Sh = kL/ρDAB
Sc = μ/ρDAB
Gr = L³ρ²gε∆w/μ²
Pé = Lv/DAB
C H I LTON CO L B URN A N A LO G Y
F O R LO N G S M O OTH T U B E S
In highly turbulent range (Re > 10000), the heat transfer ordinate
agrees approximately with f/2 for the long smooth pipes under
consideration
f/2 = jH
where,
jH = Nu Re-1 Pr-⅓ = h/ρCpv (Cpμ/k)2/3 = hS/wCp (Cpμ/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
A D VA N TA G E S
Useful to find each term for highly turbulent flow as f equation is known
f = 0.058 Re-1/5
L I M I TAT I O N S
1. Only applicable when no form drag is present in flows
2. 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)
A P P L I C AT I O N S
1. Evaluation of heat transfer in internal forced flow
2. Heat Exchanger Design
3. Reactor Design
Chilton Colburn Analogy - Overall Concept

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Chilton Colburn Analogy - Overall Concept

  • 2. B A S I S Based on experimental data for gases and liquids in both the laminar and turbulent flow regions I M P ORTA N C E Lies in heat transfer characterization for flows having Prandtl Number not equal to unity
  • 3. U S UA L F O R M A N D I T S U S E jH = jD = 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 A S S U M PTION Pr ≠ Sc ≠1 ; No form drag exit condition where, Pr = Prandtl Number = Momentum Diffusivity/Thermal Diffusivity = Cpμ/k Sc = Schmidt Number = Momentum Diffusivity/Mass Diffusivity = μ/ρDAB
  • 4. D E R I VAT I O N From Sieder Tate Equation for turbulent flow, Nu = 0.023 Re0.8 Pr⅓ (μ/μw)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 = Dvρ/μ Pr = Prandtl Number = Momentum Diffusivity/Thermal Diffusivity = Cpμ/k Dividing equation (1) by RePr, Nu/RePr = 0.023 (Re0.8 Pr⅓ (μ/μw)0.14/RePr) Rearranging, St Pr⅔ (μ/μw)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 Pr⅔ (μ/μw)0.14 = 0.023 Re-0.2 jH = j-factor for Heat Transfer
  • 5. D E R I VAT I O N Similarly, Relating Mass and Momentum Transfer using Mass Transfer Equation, kc ’D/DAB = 0.023 Re0.8 Sc⅓ Dividing equation (2) by ReSc, kc ’/v Sc⅔ Re0.03 = 0.023 Re-0.2 jD = j-factor for Diffusivity Simplified Form of Chilton Colburn Analogy This analogy is also known as Modified Reynolds Analogy where, StH = Nu/RePr = Heat/Thermal Capacity and StM = Sh/ReSc = Mass/Density*Velocity (2) f/2 = StH Pr⅔ = jH = StM Sc⅔ = jD | 0.6 < Pr < 60 , 0.6 < Sc < 3000
  • 6. VA L I D I T Y  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
  • 7. C H I LTON CO L B URN J - FAC TORS F O R H E AT A N D M A S S T R A N S F ER Heat Transfer quantities (Pure Fluids) Binary Mass Transfer quantities (Isothermal Fluids) MOLAR UNITS Binary Mass Transfer quantities (Isothermal Fluids) MASS UNITS jH = Nu Re-1 Pr-⅓ = h/ρCpv (Cpμ/k)2/3 jD = Sh Re-1 Sc-⅓ = k/cv (μ/ρDAB)2/3 jD = Sh Re-1 Sc-⅓ = k/ρv (μ/ρDAB)2/3
  • 8. C H I LTO N CO L B U R N A N A LO G Y F O R F LO W S W I T H A N D W I T H O U T F O R M D R AG  For flow past a flat plate or a pipe where no form drag is present, f/2 = jH = jD  For flow in packed beds or other blunt objects where form drag is present, f/2 > jH or jD and jH = jD
  • 9. D I M E NS IONLESS G R O U P S F O R H E AT A N D M A S S T R A N S F ER Heat Transfer quantities (Pure Fluids) Binary Mass Transfer quantities (Isothermal Fluids) MOLAR UNITS Binary Mass Transfer quantities (Isothermal Fluids) MASS UNITS Re = Lvρ/μ Fr = v²/gL Nu = hL/k Pr = Cpμ/k Gr = L³ρ²gβ∆T/μ² Pé = LvCp/k Sh = kL/cDAB Sc = μ/ρDAB Gr = L³ρ²gε∆x/μ² Pé = Lv/DAB Sh = kL/ρDAB Sc = μ/ρDAB Gr = L³ρ²gε∆w/μ² Pé = Lv/DAB
  • 10. C H I LTON CO L B URN A N A LO G Y F O R LO N G S M O OTH T U B E S In highly turbulent range (Re > 10000), the heat transfer ordinate agrees approximately with f/2 for the long smooth pipes under consideration f/2 = jH where, jH = Nu Re-1 Pr-⅓ = h/ρCpv (Cpμ/k)2/3 = hS/wCp (Cpμ/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
  • 11. A D VA N TA G E S Useful to find each term for highly turbulent flow as f equation is known f = 0.058 Re-1/5 L I M I TAT I O N S 1. Only applicable when no form drag is present in flows 2. 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) A P P L I C AT I O N S 1. Evaluation of heat transfer in internal forced flow 2. Heat Exchanger Design 3. Reactor Design