1. Chilton and Colburn J-factor analogy
Recall: The equation for heat transfer in the turbulent regime
Sieder-Tate Equation
𝑁𝑢=0.023𝑅 𝑒 0.8 𝑃𝑟 1/3 𝜙 𝑣
𝜙 𝑣 = 𝜇 𝜇 𝑤
(for forced convection/ turbulent,
Re > & 0.5 < Pr < 100)
The most successful and most widely used analogy.
This analogy is based on experimental data for gases and liquids in both the laminar and turbulent regions
If we divide this by 𝑁 𝑅𝑒 𝑁 𝑃𝑟
𝑁 𝑁𝑢 𝑁 𝑅𝑒 𝑁 𝑃𝑟 = 𝑁 𝑅𝑒 𝑁 𝑃𝑟 𝜇 𝜇 𝑁 𝑅𝑒 𝑁 𝑃𝑟

2. Dimensionless Groups Dim. Group Ratio Equation Prandtl, Pr Schmidt, Sc
molecular diffusivity of momentum / molecular diffusivity of heat
𝑐 𝑃 𝜇 𝑘
Schmidt, Sc
momentum diffusivity/ mass diffusivity
ν 𝐷 𝐴𝐵
Lewis, Le
thermal diffusivity/ mass diffusivity
𝛼 𝐷 𝐴𝐵
Stanton, St
heat transferred/ thermal capacity
ℎ 𝑐 𝑝 𝜌 𝑣
Nusselt, Nu
convective / conductive heat transfer
across the boundary
hL 𝑘

3. Chilton and Colburn J-factor analogy
This can be rearranged as
𝑁 𝑁𝑢 𝑁 𝑅𝑒 𝑁 𝑃𝑟 = 𝑁 𝑅𝑒 𝑁 𝑃𝑟 𝜇 𝜇 𝑁 𝑅𝑒 𝑁 𝑃𝑟
𝑁 𝑆𝑡 𝑁 𝑃𝑟 𝜇 𝜇 1 −0.14 =0.023 𝑁 𝑅𝑒 −0.2
The most successful and most widely used analogy.
This analogy is based on experimental data for gases and liquids in both the laminar and turbulent regions
For the turbulent flow region, an empirical equation relating f and Re
𝑓 2 =0.023 𝑁 𝑅𝑒 −0.2

4. Chilton and Colburn J-factor analogy
𝑓 2 = 𝑁 𝑆𝑡 𝑁 𝑃𝑟 𝜇 𝜇 =0.023 𝑁 𝑅𝑒 −0.2
"𝑱 𝑯 "
This is called as the J-factor for heat transfer
JH- J- factor for heat transfer

5. Chilton and Colburn J-factor analogy
In a similar manner,
we can relate the mass transfer and momentum transfer using
the equation for mass transfer of all liquids and gases
𝑘 𝑐 ′ 𝐷 𝐷 𝑎𝑏 = 𝑁 𝑅𝑒 𝑁 𝑆𝑐
Sherwood number ata yung nasa left side of the equation.
If we divide this by 𝑁 𝑅𝑒 𝑁 𝑆𝑐
𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 𝑁 𝑅𝑒 =0.023 𝑁 𝑅𝑒 −0.2

6. Chilton and Colburn J-factor analogy
𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 𝑁 𝑅𝑒 =0.023 𝑁 𝑅𝑒 −0.2
Taking 𝑁 𝑅𝑒 0.03 =1
𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 =0.023 𝑁 𝑅𝑒 −0.2
𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 = 𝑓 2

7. Chilton and Colburn J-factor analogy
𝑓 2 = 𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 =0.023 𝑁 𝑅𝑒 −0.2
"𝑱 𝑫 "
This is called as the J-factor for mass transfer
JD- J- factor for mass transfer

8. Chilton and Colburn J-factor analogy
Extends the Reynolds analogy to liquids
f 2 = h c p 𝜌 𝑣 = 𝑘 𝑐 ′ 𝑣
f 2 = h c p 𝜌 𝑣 ( 𝑁 𝑃𝑟 ) 𝜇 𝜇 = 𝑘 𝑐 ′ 𝑣 ( 𝑁 𝑆𝑐 )
The most successful and most widely used analogy.
This analogy is based on experimental data for gases and liquids in both the laminar and turbulent regions

9. Chilton and Colburn J-factor analogy
If we let
𝜇 𝜇 =1
f 2 = h c p 𝜌 𝑣 ( 𝑁 𝑃𝑟 )= 𝑘 𝑐 ′ 𝑣 ( 𝑁 𝑆𝑐 )
"𝑱 𝑯 "
"𝑱 𝑫 "
Validity range covers almost all fluids of practical importance except liquid metals.
Applies to the following ranges:
For heat transfer:10,000 < Re < 300,000
0.6 < Pr < 100
For mass transfer: 2,000 < Re < 300,000
0.6 < Sc < 2,500
𝑓 2 = 𝐽 𝐻 = J D

10. Martinelli Analogy Reynolds Analogy Chilton-Colburn J-factor Analogy
 demonstrates similarity of mechanism
(the gradients are assumed equal)
 Pr = 1 and Sc = 1
Chilton-Colburn J-factor Analogy
 demonstrates numerical similarity
(implies that the correlation equations are not faithful statements of the mechanism, but useful in predicting numerical values of coefficients
 wider range of Pr and Sc

11. Martinelli Analogy Martinelli Analogy (heat and momentum transfer)
 applicable to the entire range of Pr number
Assumptions:
The T driving forces between the wall and the fluid is small enough so that μ/μ1 = 1
Well-developed turbulent flow exists within the test section
Heat flux across the tube wall is constant along the test section
Both stress and heat flux are zero at the center of the tube and increases linearly with radius to a maximum at the wall
At any point εq = ετ

12. Martinelli Analogy Assumptions: 6. The velocity profile
distribution given by
Figure 12.5 is valid

13. For cylindrical geometry
Martinelli Analogy
𝜏 𝑦 𝑟 𝑟 1 =− 𝜇 𝜌 + 𝜀 𝑡 𝑑 𝑣𝜌 𝑑𝑟
𝑞 𝐴 𝑟 𝑟 1 =− 𝛼+ 𝛼 𝑡 𝑑 𝜌 𝑐 𝑝 𝑇 𝑑𝑟
Both equal to zero;
For cylindrical geometry

14. For cylindrical geometry
Martinelli Analogy
𝜏 𝑦 𝑟 𝑟 1 =− 𝜇 𝜌 + 𝜀 𝑡 𝑑 𝑣𝜌 𝑑𝑟
Integrated and expressed
as function of position
Converted in the form
𝑵 𝑵𝒖 =𝒇( 𝑵 𝑹𝒆 , 𝑵 𝑷𝒓 , 𝒇)
𝑞 𝐴 𝑟 𝑟 1 =− 𝛼+ 𝛼 𝑡 𝑑 𝜌 𝑐 𝑝 𝑇 𝑑𝑟
Gives a complicated equation that can be presented as a plot of Nu vs Re at values of Pr varying betwee 0 to infinity (see next slide)
Both equal to zero;
For cylindrical geometry

15. Martinelli Analogy
Gives a complicated equation that can be presented as a plot of Nu vs Re at values of Pr varying betwee 0 to infinity (see next slide)

16. Martinelli Analogy Martinelli Analogy (heat and momentum transfer)
 applicable to the entire range of Pr number
 predicts Nu for liquid metals
 contributes to understanding of the mechanism of heat and momentum transfer
Gives a complicated equation that can be presented as a plot of Nu vs Re at values of Pr varying betwee 0 to infinity (see next slide)

17. Martinelli Analogy Martinelli Analogy (heat and momentum transfer)
 applicable to the entire range of Pr number
 predicts Nu for liquid metals
 contributes to understanding of the mechanism of heat and momentum transfer
Gives a complicated equation that can be presented as a plot of Nu vs Re at values of Pr varying betwee 0 to infinity (see next slide)

18. Analogies
EXAMPLE
Compare the value of the Nusselt number, given by the appropriate empirical equation, to that predicted by the Reynolds, Colburn and Martinelli analogies for each of the following substances at Re= 100,000 and f = Consider all substances at 1000F, subject to heating with the tube wall at 1500F.
Gives a complicated equation that can be presented as a plot of Nu vs Re at values of Pr varying betwee 0 to infinity (see next slide)

19. Example Sample Calculation For air, 𝑁 𝑁𝑢 =202 (𝑚𝑜𝑠𝑡 𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒)
𝑁 𝑁𝑢 = 𝑁 𝑅𝑒 𝑁 𝑃𝑟 𝜇 𝜇
𝑁 𝑁𝑢 = ,
Gives a complicated equation that can be presented as a plot of Nu vs Re at values of Pr varying betwee 0 to infinity (see next slide)
𝑁 𝑁𝑢 =202 (𝑚𝑜𝑠𝑡 𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒)

20. Example Sample Calculation For air, by Reynolds analogy
𝑁 𝑁𝑢 =163.3

21. Example Sample Calculation For air, by Colburn analogy 𝑁 𝑁𝑢 =202
𝑁 𝑆𝑡 = 𝑁 𝑁𝑢 𝑁 𝑅𝑒 𝑁 𝑃𝑟
𝑁 𝑁𝑢 = 𝑁 𝑅𝑒 𝑁 𝑃𝑟 𝑓 𝜇 𝜇
𝑁 𝑁𝑢 =
𝑁 𝑁𝑢 =202

22. Example
Sample Calculation For air, by Martinelli analogy
𝑁 𝑁𝑢 =170

23. FIN