Explain Chilton Colburn analogy.
Answers
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Chilton–Colburn J-factor analogy is a successful and widely used analogy between heat, momentum, and mass transfer. The basic mechanisms and mathematics of heat, mass, and momentum transport are essentially the same. Among many analogies (like Reynolds analogy, Prandtl–Taylor analogy) developed to directly relate heat transfer coefficients, mass transfer coefficients, and friction factors Chilton and Colburn J-factor analogy proved to be the most accurate.
It is written as follows,
{\displaystyle J_{M}={\frac {f}{2}}=J_{H}={\frac {h}{c_{p}\,G}}\,{Pr}^{\frac {2}{3}}=J_{D}={\frac {k'_{c}}{\overline {v}}}\cdot {Sc}^{\frac {2}{3}}} J_M=\frac{f}{2} = J_H = \frac{h}{c_p\, G}\,{Pr}^{\frac{2}{3}}= J_D = \frac{k'_c}{\overline{v}} \cdot {Sc}^{\frac{2}{3}}
This equation permits the prediction of an unknown transfer coefficient when one of the other coefficients is known. The analogy is valid for fully developed turbulent flow in conduits with Re > 10000, 0.7 < Pr < 160, and tubes where L/d > 60 (the same constraints as the Sieder–Tate correlation). The wider range of data can be correlated by Friend–Metzner analogy.
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Relationship between Heat and Mass;
{\displaystyle J_{M}={\frac {f}{2}}={\frac {Sh}{Re\,Sc^{\frac {1}{3}}}}=J_{H}={\frac {f}{2}}={\frac {Nu}{Re\,Pr^{\frac {1}{3}}}}} J_M = \frac{f}{2} = \frac{Sh}{Re\, Sc^{\frac{1}{3}}} = J_H = \frac{f}{2} = \frac{Nu}{Re\, Pr^{\frac{1}{3}}}