# Turbulent momentum and heat transfer over a flat plate

To obtain the boundary layer thickness for turbulent flow over a flat plate, von Kármán’s momentum integral can be employed. The integral momentum equation that was derived for the case of laminar flow is still valid except that the instantaneous velocity should be replaced by the time-averaged velocity. For the flat plate without bellowing or suction (vw = 0) and without pressure gradient, the equation is simplified to $\frac{\tau _{w}}{\rho }=\frac{d}{dx}\int_{0}^{\delta }{\bar{u}\left( U_{\infty }-\bar{u} \right)dy}$ (1)

While the velocity profile in the laminar boundary layer can be adequately described by a polynomial function, the velocity profile in a turbulent flow is too complicated to be described by a single function for the entire boundary layer. Equation (12) in Two-Layer Model states that the velocity profile in the turbulent boundary layer can be approximated as u + = 8.75(y + )1 / 7, which can be rewritten into the following dimensionless form: $\frac{{\bar{u}}}{U_{\infty }}=\left( \frac{y}{\delta } \right)^{1/7}$ (2)

Although eq. (2) can reasonably represent velocity profile in most parts of the boundary layer, the velocity gradients at both y = 0 and δ are incorrect: $(\partial \bar{u}/\partial y)_{y=0}\to \infty$ and $(\partial \bar{u}/\partial y)_{y=\delta }\ne 0$. To get the shear stress at the wall, let us substitute eq. (4) and (5) in Two-Layer Model into eq. (12) in Two-Layer Model: $\frac{U_{\infty }}{\sqrt{\tau _{w}/\rho }}=8.75\left( \frac{\delta \sqrt{\tau _{w}/\rho }}{\nu } \right)^{1/7}$ (3)

Solving for τw yields $\tau _{w}=0.0225\rho U_{\infty }^{2}\left( \frac{\delta U_{\infty }}{\nu } \right)^{-1/4}$ (4)

which is referred to as the Blasius relation and which is valid for $\operatorname{Re}_{x}<10^{7}$. Substituting eqs. (2) and (4) into eq. (1), one obtains $0.0225U_{\infty }^{2}\left( \frac{\delta U_{\infty }}{\nu } \right)^{-1/4}=\frac{d}{dx}\int_{0}^{\delta }{U_{\infty }^{2}\left[ \left( \frac{y}{\delta } \right)^{1/7}-\left( \frac{y}{\delta } \right)^{2/7} \right]dy}$ (5)

Performing the integration and differentiation on the right-hand side of eq. (5) yields the following differential equation for the boundary layer thickness: $0.0225U_{\infty }^{2}\left( \frac{\delta U_{\infty }}{\nu } \right)^{-1/4}=\frac{7}{72}\frac{d\delta }{dx}$ (6)

which can be integrated to obtain: $\left( \frac{\nu }{U_{\infty }} \right)^{1/4}x=3.45\delta ^{5/4}+C$ (7)

where C is the unspecified integration constant. If we assume that the turbulent boundary layer starts from the edge of the flat plate – which is not a good assumption – the integration constant C becomes zero, and eq. (7) becomes $\frac{\delta }{x}=\frac{0.376}{\operatorname{Re}_{x}^{1/5}}$ (8)

The local friction coefficient can be found as $c_{f}=\frac{\tau _{w}}{\rho U_{\infty }^{2}/2}=\frac{0.0576}{\operatorname{Re}_{x}^{1/5}}$ (9)

It should be pointed out that eqs. (8) and (9) are valid for the case that turbulent boundary layer starts from the leading edge of the flat plate and $\operatorname{Re}_{x}<10^{7}$ – beyond which the Blasius relation becomes invalid.

Figure to the right shows a comparison of the velocity profiles of laminar and turbulent boundary layers at $\operatorname{Re}_{x}=5\times 10^{6}$ . It can be seen that the turbulent boundary layer is much thicker than the laminar boundary layer at the same Reynolds number. The mean velocity in the turbulent boundary layer is much higher than that of the laminar boundary layer. The large mean velocity of the turbulent boundary layer allows for a much stronger momentum, heat and mass transfer. Another advantage of the turbulent boundary layer is that it can resist separation better than a laminar boundary layer. Due to its strong ability on momentum, heat and mass transfer and resisting separation, a turbulent boundary layer is desirable in many engineering applications.

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