Chilton-Colburn J-factor Analogy: Heat & Mass Transfer
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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 π 0.8 ππ1/3 ππ£ (for forced convection/ turbulent, Re > 10000 & 0.5 < Pr < 100) π ππ£ = ππ€ If we divide this by ππ π πππ πππ’ = 0.023 ππ π πππ 1 0.8 ππ π πππ 3 ππ π πππ π π1 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 ππ π π ν π·π΄π΅ πΌ π·π΄π΅ β ππ ππ£ Nusselt, Nu convective / conductive heat transfer across the boundary hL π Chilton and Colburn J-factor analogy This can be rearranged as πππ’ = 0.023 ππ π πππ 2 3 πππ‘ πππ π π1 1 ππ π 0.8 πππ 3 ππ π πππ π π1 0.14 −0.14 −0.2 = 0.023ππ π For the turbulent flow region, an empirical equation relating f and Re π −0.2 = 0.023ππ π 2 Chilton and Colburn J-factor analogy 2 π π 3 = πππ‘ πππ 2 π1 0.14 −0.2 = 0.023ππ π "π±π― " 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 ππ′ π· = 0.023 ππ π π·ππ 0.83 πππ 0.33 If we divide this by ππ π πππ 2 ππ′ 3 πππ π£ ππ π 0.03 −0.2 = 0.023ππ π Chilton and Colburn J-factor analogy 2 ππ′ 3 πππ π£ ππ π 0.03 −0.2 = 0.023ππ π 0.03 Taking ππ π =1 2 ππ′ 3 −0.2 πππ = 0.023ππ π π£ 2 ππ′ π 3 πππ = π£ 2 Chilton and Colburn J-factor analogy 2 π ππ′ 3 −0.2 = πππ = 0.023ππ π 2 π£ "π±π« " This is called as the J-factor for mass transfer Chilton and Colburn J-factor analogy Extends the Reynolds analogy to liquids f h ππ′ = = 2 cp π π£ π£ 2 f h π 3 = (πππ ) 2 cp π π£ π1 0.14 ππ′ 23 = (πππ ) π£ Chilton and Colburn J-factor analogy If we let π π1 0.14 =1 2 2 ′ f h π π 3 3 = (πππ ) = (πππ ) 2 cp π π£ π£ "π±π― " π = π½π» = JD 2 "π±π« " 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 ο demonstrates similarity of mechanism (the gradients are assumed equal) ο Pr = 1 and Sc = 1 Chilton-Colburn J-factor Analogy ο demonstrates numerical similarity (implies that the correlation equations are not faithful statements of the mechanism, but useful in predicting numerical values of coefficients ο wider range of Pr and Sc Martinelli Analogy Martinelli Analogy (heat and momentum transfer) ο 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 ππ¦ π π =− + ππ‘ π1 π π π = − πΌ + πΌπ‘ π΄ π1 π π£π ππ π πππ π ππ Both equal to zero; For cylindrical geometry Martinelli Analogy ππ¦ π π =− + ππ‘ π1 π π π£π ππ Integrated and expressed as function of position Converted in the form π π = − πΌ + πΌπ‘ π΄ π1 π πππ π ππ Both equal to zero; For cylindrical geometry π΅π΅π = π(π΅πΉπ , π΅π·π , π) Martinelli Analogy Martinelli Analogy Martinelli Analogy (heat and momentum transfer) ο applicable to the entire range of Pr number ο predicts Nu for liquid metals ο contributes to understanding of the mechanism of heat and momentum transfer Martinelli Analogy Martinelli Analogy (heat and momentum transfer) ο applicable to the entire range of Pr number ο predicts Nu for liquid metals ο 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, πππ’ = 0.023 π π1 1 0.8 ππ π πππ 3 πππ’ = 0.023 100,000 0.8 0.71 1 3 0.14 0.018 0.02 πππ’ = 202 (πππ π‘ ππππ’πππ‘π π£πππ’π) 0.14 Example Sample Calculation For air, by Reynolds analogy πππ’ f πππ‘ = = ππ π πππ 2 πππ’ f 0.0046 = ππ π πππ = 2 2 πππ’ = 163.3 105 (0.71) Example Sample Calculation For air, by Colburn analogy 2 f π 3 = πππ‘ (πππ ) 2 π1 πππ’ = πππ’ = 105 0.71 1 3 0.14 πππ’ πππ‘ = ππ π πππ 1 ππ π πππ 3 0.0046 2 0.018 0.02 π 2 π π1 0.14 0.14 πππ’ = 202 Example Sample Calculation For air, by Martinelli analogy πππ’ = 170 FIN
