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 0.14 (for forced convection/ turbulent, Re > 10000 & 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 =0.023 𝑁 𝑅𝑒 0.8 𝑁 𝑃𝑟 1 3 𝜇 𝜇 1 0.14 𝑁 𝑅𝑒 𝑁 𝑃𝑟

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

Chilton and Colburn J-factor analogy This can be rearranged as 𝑁 𝑁𝑢 𝑁 𝑅𝑒 𝑁 𝑃𝑟 =0.023 𝑁 𝑅𝑒 0.8 𝑁 𝑃𝑟 1 3 𝜇 𝜇 1 0.14 𝑁 𝑅𝑒 𝑁 𝑃𝑟 𝑁 𝑆𝑡 𝑁 𝑃𝑟 2 3 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

Chilton and Colburn J-factor analogy 𝑓 2 = 𝑁 𝑆𝑡 𝑁 𝑃𝑟 2 3 𝜇 𝜇 1 0.14 =0.023 𝑁 𝑅𝑒 −0.2 "𝑱 𝑯 " This is called as the J-factor for heat transfer JH- 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 𝑘 𝑐 ′ 𝐷 𝐷 𝑎𝑏 =0.023 𝑁 𝑅𝑒 0.83 𝑁 𝑆𝑐 0.33 Sherwood number ata yung nasa left side of the equation. If we divide this by 𝑁 𝑅𝑒 𝑁 𝑆𝑐 𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 2 3 𝑁 𝑅𝑒 0.03 =0.023 𝑁 𝑅𝑒 −0.2

Chilton and Colburn J-factor analogy 𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 2 3 𝑁 𝑅𝑒 0.03 =0.023 𝑁 𝑅𝑒 −0.2 Taking 𝑁 𝑅𝑒 0.03 =1 𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 2 3 =0.023 𝑁 𝑅𝑒 −0.2 𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 2 3 = 𝑓 2

Chilton and Colburn J-factor analogy 𝑓 2 = 𝑘 𝑐 ′ 𝑣 𝑁 𝑆𝑐 2 3 =0.023 𝑁 𝑅𝑒 −0.2 "𝑱 𝑫 " This is called as the J-factor for mass transfer JD- J- factor for mass transfer

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 2 3 ) 𝜇 𝜇 1 0.14 = 𝑘 𝑐 ′ 𝑣 ( 𝑁 𝑆𝑐 2 3 ) 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

Chilton and Colburn J-factor analogy If we let 𝜇 𝜇 1 0.14 =1 f 2 = h c p 𝜌 𝑣 ( 𝑁 𝑃𝑟 2 3 )= 𝑘 𝑐 ′ 𝑣 ( 𝑁 𝑆𝑐 2 3 ) "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

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

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

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

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

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

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)

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)

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)

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. 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)

Example Sample Calculation For air, 𝑁 𝑁𝑢 =202 (𝑚𝑜𝑠𝑡 𝑎𝑐𝑐𝑢𝑟𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒) 𝑁 𝑁𝑢 =0.023 𝑁 𝑅𝑒 0.8 𝑁 𝑃𝑟 1 3 𝜇 𝜇 1 0.14 𝑁 𝑁𝑢 =0.023 100,000 0.8 0.71 1 3 0.018 0.02 0.14 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)

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

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 1 3 𝑓 2 𝜇 𝜇 1 0.14 𝑁 𝑁𝑢 = 10 5 0.71 1 3 0.0046 2 0.018 0.02 0.14 𝑁 𝑁𝑢 =202

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

FIN