Chilton and Colburn J-factor analogy
Recall: The equation for heat transfer in the turbulent regime
Sieder-Tate Equation
𝑁𝑢 = 0.023𝑅𝑒 0.8 𝑃𝑟1/3 𝜙𝑣
(for forced convection/ turbulent,
Re > 10000 & 0.5 < Pr < 100)
𝜇
𝜙𝑣 =
𝜇𝑤
If we divide this by 𝑁𝑅𝑒 𝑁𝑃𝑟
𝑁𝑁𝑢
= 0.023
𝑁𝑅𝑒 𝑁𝑃𝑟
1
0.8
𝑁𝑅𝑒
𝑁𝑃𝑟 3
𝑁𝑅𝑒 𝑁𝑃𝑟
𝜇
𝜇1
0.14
0.14
Dimensionless Groups
Dim. Group
Ratio
Equation
Prandtl, Pr
molecular diffusivity of momentum /
molecular diffusivity of heat
Schmidt, Sc
momentum diffusivity/ mass diffusivity
Lewis, Le
thermal diffusivity/ mass diffusivity
Stanton, St
heat transferred/ thermal capacity
𝑐𝑃 𝜇
𝑘
ν
𝐷𝐴𝐵
𝛼
𝐷𝐴𝐵
ℎ
𝑐𝑝 𝜌𝑣
Nusselt, Nu
convective / conductive heat transfer
across the boundary
hL
𝑘
Chilton and Colburn J-factor analogy
This can be rearranged as
𝑁𝑁𝑢
= 0.023
𝑁𝑅𝑒 𝑁𝑃𝑟
2
3
𝑁𝑆𝑡 𝑁𝑃𝑟
𝜇
𝜇1
1
𝑁𝑅𝑒 0.8 𝑁𝑃𝑟 3
𝑁𝑅𝑒 𝑁𝑃𝑟
𝜇
𝜇1
0.14
−0.14
−0.2
= 0.023𝑁𝑅𝑒
For the turbulent flow region, an empirical equation relating f and Re
𝑓
−0.2
= 0.023𝑁𝑅𝑒
2
Chilton and Colburn J-factor analogy
2
𝑓
𝜇
3
= 𝑁𝑆𝑡 𝑁𝑃𝑟
2
𝜇1
0.14
−0.2
= 0.023𝑁𝑅𝑒
"𝑱𝑯 "
This is called as the J-factor for heat transfer
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
𝑘𝑐′ 𝐷
= 0.023 𝑁𝑅𝑒
𝐷𝑎𝑏
0.83
𝑁𝑆𝑐
0.33
If we divide this by 𝑁𝑅𝑒 𝑁𝑆𝑐
2
𝑘𝑐′
3
𝑁𝑆𝑐
𝑣
𝑁𝑅𝑒
0.03
−0.2
= 0.023𝑁𝑅𝑒
Chilton and Colburn J-factor analogy
2
𝑘𝑐′
3
𝑁𝑆𝑐
𝑣
𝑁𝑅𝑒
0.03
−0.2
= 0.023𝑁𝑅𝑒
0.03
Taking 𝑁𝑅𝑒
=1
2
𝑘𝑐′
3
−0.2
𝑁𝑆𝑐
= 0.023𝑁𝑅𝑒
𝑣
2
𝑘𝑐′
𝑓
3
𝑁𝑆𝑐 =
𝑣
2
Chilton and Colburn J-factor analogy
2
𝑓 𝑘𝑐′
3
−0.2
=
𝑁𝑆𝑐
= 0.023𝑁𝑅𝑒
2
𝑣
"𝑱𝑫 "
This is called as the J-factor for mass transfer
Chilton and Colburn J-factor analogy
Extends the Reynolds analogy to liquids
f
h
𝑘𝑐′
=
=
2 cp 𝜌 𝑣
𝑣
2
f
h
𝜇
3
=
(𝑁𝑃𝑟 )
2 cp 𝜌 𝑣
𝜇1
0.14
𝑘𝑐′ 23
= (𝑁𝑆𝑐 )
𝑣
Chilton and Colburn J-factor analogy
If we let
𝜇
𝜇1
0.14
=1
2
2
′
f
h
𝑘
𝑐
3
3
=
(𝑁𝑃𝑟 ) = (𝑁𝑆𝑐
)
2 cp 𝜌 𝑣
𝑣
"𝑱𝑯 "
𝑓
= 𝐽𝐻 = JD
2
"𝑱𝑫 "
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
Martinelli Analogy
Reynolds 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
Martinelli Analogy
Martinelli Analogy (heat and momentum transfer)
 applicable to the entire range of Pr number
Assumptions:
1. The T driving forces between the wall and the fluid is small
enough so that μ/μ1 = 1
2. Well-developed turbulent flow exists within the test section
3. Heat flux across the tube wall is constant along the test
section
4. Both stress and heat flux are zero at the center of the tube
and increases linearly with radius to a maximum at the wall
5. At any point εq = ετ
Martinelli Analogy
Assumptions:
6. The velocity profile
distribution given by
Figure 12.5 is valid
Martinelli Analogy
𝜏𝑦
𝑟
𝜇
=−
+ 𝜀𝑡
𝑟1
𝜌
𝑞 𝑟
= − 𝛼 + 𝛼𝑡
𝐴 𝑟1
𝑑 𝑣𝜌
𝑑𝑟
𝑑 𝜌𝑐𝑝 𝑇
𝑑𝑟
Both equal to zero;
For cylindrical geometry
Martinelli Analogy
𝜏𝑦
𝑟
𝜇
=−
+ 𝜀𝑡
𝑟1
𝜌
𝑑 𝑣𝜌
𝑑𝑟
Integrated and expressed
as function of position
Converted in the form
𝑞 𝑟
= − 𝛼 + 𝛼𝑡
𝐴 𝑟1
𝑑 𝜌𝑐𝑝 𝑇
𝑑𝑟
Both equal to zero;
For cylindrical geometry
𝑵𝑵𝒖 = 𝒇(𝑵𝑹𝒆 , 𝑵𝑷𝒓 , 𝒇)
Martinelli Analogy
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
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
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 = 0.0046. Consider
all substances at 1000F, subject to heating with the tube wall
at 1500F.
Example
Sample Calculation
For air,
𝑁𝑁𝑢 = 0.023
𝜇
𝜇1
1
0.8
𝑁𝑅𝑒
𝑁𝑃𝑟 3
𝑁𝑁𝑢 = 0.023 100,000
0.8
0.71
1
3
0.14
0.018
0.02
𝑁𝑁𝑢 = 202 (𝑚𝑜𝑠𝑡 𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒)
0.14
Example
Sample Calculation
For air, by Reynolds analogy
𝑁𝑁𝑢
f
𝑁𝑆𝑡 =
=
𝑁𝑅𝑒 𝑁𝑃𝑟 2
𝑁𝑁𝑢
f
0.0046
= 𝑁𝑅𝑒 𝑁𝑃𝑟 =
2
2
𝑁𝑁𝑢 = 163.3
105 (0.71)
Example
Sample Calculation
For air, by Colburn analogy
2
f
𝜇
3
= 𝑁𝑆𝑡 (𝑁𝑃𝑟 )
2
𝜇1
𝑁𝑁𝑢 =
𝑁𝑁𝑢 = 105 0.71
1
3
0.14
𝑁𝑁𝑢
𝑁𝑆𝑡 =
𝑁𝑅𝑒 𝑁𝑃𝑟
1
𝑁𝑅𝑒 𝑁𝑃𝑟 3
0.0046
2
0.018
0.02
𝑓
2
𝜇
𝜇1
0.14
0.14
𝑁𝑁𝑢 = 202
Example
Sample Calculation
For air, by Martinelli analogy
𝑁𝑁𝑢 = 170
FIN