4. Reynolds Analogy

Reynolds assumed that the entire flow field consisted of a single zone of highly turbulent region. That is, he neglected the presence of the viscous sublayer and the buffer layer. In such a turbulent core, \[\nu \ll \varepsilon_m \qquad \text{ and } \alpha \ll \varepsilon_h\] In addition, he assumed that the turbulent diffusivities are equal. \[\varepsilon_m = \varepsilon_m = \varepsilon\] With the above assumptions, \[\begin{aligned} \frac{\tau}{\rho} &= \varepsilon \frac{dv}{dy} \\ \frac{q}{\rho C_P} &= -\varepsilon \frac{dT}{dy} \end{aligned}\]

Combining the above two equations, we get \[dT = -\frac{q}{\tau C_P}dv\] Integration limits: \[\begin{aligned} {3} \ & \text{Wall conditions: } & \quad T&=T_w \quad \text{ and }\quad v=0 \\ \ & \text{Bulk stream conditions: } & T&= T_m \quad \text{ and } \quad v = v_m \end{aligned}\] Assuming \(q/\tau\) remains constant, \[\begin{align*} \int_{T_w}^{T_m}dT &= -\frac{q}{\tau C_P}\int_{0}^{v_m}dv \\ T_w-T_m &= \frac{qv_m}{\tau C_P} \tag*{(1)} \end{align*}\]

By definition, \[\begin{aligned} h &= \frac{q}{T_w-T_m} {and} f &= \frac{\tau}{\rho v_m^2/2} \end{aligned}\] Therefore, Eqn.(1) becomes, \[\boxed{\frac{h}{\rho C_P v_m} =\text{ St }= \frac{f}{2}}\] This result is known as Reynolds analogy for momentum and heat transfer in fully developed turbulent flow in a pipe. It is valid for \(\text{Pr} \approx 1\), and negligible pressure gradient (\(dp/dx \approx 0\)).