Chilton and Colburn J-factor analogy

Recall: The equation for heat transfer in the turbulent regime

Sieder-Tate Equation 𝑁𝑢=0.023𝑅𝑒0.8 𝑃𝑟 1 /3𝜙𝑣

𝜙𝑣=( 𝜇𝜇𝑤 )0.14

(for forced convection/ turbulent, Re > 10000 & 0.5 < Pr < 100)

If we divide this by

𝑁𝑁𝑢

𝑁𝑅𝑒𝑁 𝑃𝑟=0.023

(𝑁𝑅𝑒 )0.8 (𝑁 𝑃𝑟 )13 ( 𝜇𝜇1 )

0.14

𝑁𝑅𝑒𝑁 𝑃𝑟

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

𝑁𝑁𝑢

𝑁𝑅𝑒𝑁 𝑃𝑟=0.023

(𝑁𝑅𝑒 )0.8 (𝑁 𝑃𝑟 )13 ( 𝜇𝜇1 )

0.14

𝑁𝑅𝑒𝑁 𝑃𝑟

𝑁 𝑆𝑡𝑁𝑃𝑟

23 ( 𝜇𝜇1 )

− 0.14

=0.023𝑁𝑅𝑒− 0.2

𝑓2 =0.023𝑁𝑅𝑒

−0.2

For the turbulent flow region, an empirical equation relating f and Re

Chilton and Colburn J-factor analogy

𝑓2

=𝑁 𝑆𝑡𝑁 𝑃𝑟

23 ( 𝜇𝜇1 )

0.14

=0.023𝑁𝑅𝑒−0.2

} rsub { } 𝑱 𝑯 ¿

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

𝑘𝑐′ 𝐷𝐷𝑎𝑏

=0.023 (𝑁𝑅𝑒 )0.83 (𝑁𝑆𝑐 )0.33

the equation for mass transfer of all liquids and gases

If we divide this by

𝑘𝑐′

𝑣(𝑁 𝑆𝑐

23 ) (𝑁𝑅𝑒 )0.03=0.023𝑁𝑅𝑒

− 0.2

Chilton and Colburn J-factor analogy

T

𝑘𝑐′

𝑣(𝑁 𝑆𝑐

23 )=0.023 𝑁𝑅𝑒

−0.2

𝑘𝑐′

𝑣(𝑁 𝑆𝑐

23 ) (𝑁𝑅𝑒 )0.03=0.023𝑁𝑅𝑒

− 0.2

𝑘𝑐′

𝑣(𝑁 𝑆𝑐

23 )= 𝑓

2

Chilton and Colburn J-factor analogy

𝑓2

=𝑘𝑐′

𝑣(𝑁 𝑆𝑐

23 )=0.023𝑁𝑅𝑒

−0.2

This is called as the J-factor for mass transfer

} rsub { } 𝑱 𝑫 ¿

Chilton and Colburn J-factor analogy

Extends the Reynolds analogy to liquids

f2= hc p𝜌 𝑣

=𝑘𝑐′

𝑣

f2= hc p𝜌 𝑣

(𝑁 𝑃𝑟

23 )( 𝜇𝜇1 )

0.14

=𝑘𝑐′

𝑣 (𝑁 𝑆𝑐

23 )

Chilton and Colburn J-factor analogy

If we let

f2= hc p𝜌 𝑣

(𝑁 𝑃𝑟

23 )=

𝑘𝑐′

𝑣 (𝑁 𝑆𝑐

23 )

} rsub { } 𝑱 𝑯 ¿ } rsub { } 𝑱 𝑫 ¿

𝑓2 = 𝐽 𝐻=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

( 𝜇𝜇1 )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 μ/μ1 = 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 εq = ετ

Martinelli Analogy

Assumptions:

6. The velocity profile distribution given by Figure 12.5 is valid

Martinelli Analogy

𝑞𝐴 ( 𝑟𝑟1 )=− (𝛼+𝛼 𝑡 ) (𝑑 (𝜌𝑐𝑝𝑇 )

𝑑𝑟 )

𝜏 𝑦( 𝑟𝑟1 )=−(𝜇𝜌+𝜀𝑡)( 𝑑 (𝑣 𝜌 )𝑑𝑟 )

Both equal to zero;For cylindrical geometry

Martinelli Analogy

𝑞𝐴 ( 𝑟𝑟1 )=− (𝛼+𝛼 𝑡 ) (𝑑 (𝜌𝑐𝑝𝑇 )

𝑑𝑟 )

𝜏 𝑦( 𝑟𝑟1 )=−(𝜇𝜌+𝜀𝑡)( 𝑑 (𝑣 𝜌 )𝑑𝑟 )

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,

𝑁𝑁𝑢=0.023 (𝑁𝑅𝑒 )0.8 (𝑁 𝑃𝑟 )13 ( 𝜇𝜇1 )

0.14

𝑁𝑁𝑢=0.023 (100,000 )0.8 (0.71 )13 ( 0.018

0.02 )0.14

𝑁𝑁𝑢=20 2(𝑚𝑜𝑠𝑡 𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒𝑣𝑎𝑙𝑢𝑒)

Example

Sample CalculationFor air, by Reynolds analogy

𝑁 𝑆𝑡=𝑁 𝑁𝑢

𝑁𝑅𝑒𝑁 𝑃𝑟= f

2

𝑁𝑁𝑢=f2𝑁𝑅𝑒𝑁 𝑃𝑟=( 0.0046

2 ) (105 )(0.71)

𝑁𝑁𝑢=16 3.3

Example

Sample CalculationFor air, by Colburn analogy

𝑁 𝑆𝑡=𝑁 𝑁𝑢

𝑁𝑅𝑒𝑁 𝑃𝑟

f2=𝑁 𝑆𝑡 (𝑁 𝑃𝑟

23 )( 𝜇𝜇1 )

0.14

𝑁𝑁𝑢=𝑁𝑅𝑒 (𝑁 𝑃𝑟 )13 ( 𝑓2 )( 𝜇𝜇1 )

0.14

𝑁𝑁𝑢=105 (0.71 )13 ( 0.0046

2 )( 0.0180.02 )

0.14

𝑁𝑁𝑢=202

Example

Sample CalculationFor air, by Martinelli analogy

𝑁𝑁𝑢=170

FIN