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Chilton and Colburn J-factor analogy
Recall: The equation for heat transfer in the turbulent regime
Sieder-Tate Equation ππ’=0.023π π0.8 ππ 1 /3ππ£
ππ£=( πππ€)
0.14(for forced convection/ turbulent, Re > 10000 & 0.5 < Pr < 100)
If we divide this by
πππ’
ππ ππ ππ
=0.023
(ππ π )0.8 (π ππ )13 ( ππ1
)0.14
ππ ππ ππ
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
Nusselt, Nu convective / conductive heat transfer across the boundary
Chilton and Colburn J-factor analogy
This can be rearranged as
πππ’
ππ ππ ππ
=0.023
(ππ π )0.8 (π ππ )13 ( ππ1
)0.14
ππ ππ ππ
π ππ‘πππ
23 ( ππ1
)β 0.14
=0.023ππ πβ 0.2
π2
=0.023ππ πβ0.2
For the turbulent flow region, an empirical equation relating f and Re
Chilton and Colburn J-factor analogy
π2
=π ππ‘π ππ
23 ( ππ1
)0.14
=0.023ππ πβ0.2
} rsub { } π± π― ΒΏ
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
ππβ² π·π·ππ
=0.023 (ππ π )0.83 (πππ )0.33
the equation for mass transfer of all liquids and gases
If we divide this by
ππβ²
π£(π ππ
23 ) (ππ π )0.03=0.023ππ π
β 0.2
Chilton and Colburn J-factor analogy
T
ππβ²
π£(π ππ
23 )=0.023 ππ π
β0.2
ππβ²
π£(π ππ
23 ) (ππ π )0.03=0.023ππ π
β 0.2
ππβ²
π£(π ππ
23 )= π
2
Chilton and Colburn J-factor analogy
π2
=ππβ²
π£(π ππ
23 )=0.023ππ π
β0.2
This is called as the J-factor for mass transfer
} rsub { } π± π« ΒΏ
Chilton and Colburn J-factor analogy
Extends the Reynolds analogy to liquids
f2= hc pπ π£
=ππβ²
π£
f2= hc pπ π£
(π ππ
23 )( ππ1
)0.14
=ππβ²
π£(π ππ
23 )
Chilton and Colburn J-factor analogy
If we let
f2= hc pπ π£
(π ππ
23 )=
ππβ²
π£(π ππ
23 )
} rsub { } π± π― ΒΏ } rsub { } π± π« ΒΏ
π2
= π½ π»=J D
Applies to the following ranges:For heat transfer:10,000 < Re < 300,0000.6 < Pr < 100For mass transfer: 2,000 < Re < 300,0000.6 < Sc < 2,500
( ππ1)
0.14
=1
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 ΞΌ/ΞΌ1 = 12. Well-developed turbulent flow exists within the test section3. Heat flux across the tube wall is constant along the test
section4. Both stress and heat flux are zero at the center of the tube
and increases linearly with radius to a maximum at the wall5. At any point Ξ΅q = Ξ΅Ο
Martinelli Analogy
Assumptions:
6. The velocity profile distribution given by Figure 12.5 is valid
Martinelli Analogy
ππ΄ ( ππ1
)=β (πΌ+πΌ π‘ ) (π (ππππ )ππ )
π π¦( ππ1)=β(ππ+ππ‘)( π (π£ π )
ππ )
Both equal to zero;For cylindrical geometry
Martinelli Analogy
ππ΄ ( ππ1
)=β (πΌ+πΌ π‘ ) (π (ππππ )ππ )
π π¦( ππ1)=β(ππ+ππ‘)( π (π£ π )
ππ )
Both equal to zero;For cylindrical geometry
Integrated and expressed as function of position
Converted in the form
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
EXAMPLECompare 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 CalculationFor air,
πππ’=0.023 (ππ π )0.8 (π ππ )13 ( ππ1
)0.14
πππ’=0.023 (100,000 )0.8 (0.71 )13 ( 0.018
0.02 )0.14
πππ’=202(πππ π‘ππππ’πππ‘ππ£πππ’π)
Example
Sample CalculationFor air, by Reynolds analogy
π ππ‘=π ππ’
ππ ππ ππ
= f2
πππ’=f2ππ ππ ππ=( 0.0046
2 ) (105 )(0.71)
πππ’=16 3.3
Example
Sample CalculationFor air, by Colburn analogy
π ππ‘=π ππ’
ππ ππ ππ
f2=π ππ‘ (π ππ
23 )( ππ1
)0.14
πππ’=ππ π (π ππ )13 ( π2 )( ππ1
)0.14
πππ’=105 (0.71 )13 ( 0.0046
2 )( 0.0180.02 )
0.14
πππ’=202
Example
Sample CalculationFor air, by Martinelli analogy
πππ’=170
FIN
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