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The Reynolds analogy between the coefficients of heat transfer and wall shear can be suggested on the basis of the following simple analysis. (a)

  (a) Divide the expression for shear stress [ tau=rholeft(u+varepsilon_{m}right) frac{d u}{d y} ] by the expressionto obtain [ frac{tau C_{p}}{q}=frac{u+varepsilon_{m}}{alpha+varepsilon_{H}} frac{d u}{d T} ] (b) Assume that the m  

The Reynolds analogy between the coefficients of heat transfer and wall shear can be suggested on the basis of the following simple analysis. (a) Divide the expression for shear stress T = P(v + m) by the expression for heat flux du dy q = p ( + H) dT dy to obtain Tw CP (1 9w TCP 9 (b) Assume that the molecular and turbulent Prandtl numbers are unity so that the velocity and temperature profiles are nearly the same and du/Dt=const. Replace the left-hand side by wall values and then integrate to obtain h pCUv v + m du a + EH dT (Tmax - Twall) = (Umax = (Umax - Uwall) (c) Set T = (Cf/2)pU and qw=h(Tmixing Uav 2 Umax cup Tmixing cup - Twall) to obtain Tmax - Twall Twall Uwall Tmixing cup - - which, since the bracketed term must be nearly unity, can be rephrased as Nu/Re Pr = C/2 and constitutes Reynolds' analogy.

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