Velocity defect law for smooth and rough pipes

Idea. Write the log law for the point velocity $u$ and for the mean velocity $V$, then subtract. The wall constants (and the roughness $k$ or $\nu$) cancel, so one law describes both smooth and rough pipes. Let $v_*$ be the shear velocity and $y$ the distance from the wall.

Point velocity (from Prandtl's mixing length theory and experiments)

• Smooth pipe: $\dfrac{u}{v_*} = 5.75\log_{10}\dfrac{v_*y}{\nu} + 5.5$ ... (1)
• Rough pipe: $\dfrac{u}{v_*} = 5.75\log_{10}\dfrac{y}{k} + 8.5$ ... (2)

Mean velocity

$$V = \frac{1}{\pi R^2}\int_0^R u\,2\pi r\,dr = \frac{2}{R^2}\int_0^R u\,(R - y)\,dy$$

Write (1) as $\dfrac{u}{v_*} = 5.75\log_{10}\dfrac{v_*R}{\nu} + 5.5 + 5.75\log_{10}\dfrac{y}{R}$. The first two terms are constant. For the last term, with $\eta = y/R$ and $\log_{10}\eta = 0.4343\ln\eta$:

$$2\int_0^1 (1-\eta)\ln\eta\,d\eta = 2\left[-1 + \frac14\right] = -\frac32$$

so its average is $5.75(0.4343)(-1.5) = -3.75$. Hence

• Smooth: $\dfrac{V}{v_*} = 5.75\log_{10}\dfrac{v_*R}{\nu} + 1.75$ ... (3)
• Rough (same steps with (2)): $\dfrac{V}{v_*} = 5.75\log_{10}\dfrac{R}{k} + 4.75$ ... (4)

Subtract to get the velocity defect law

Smooth pipe, (1) $-$ (3):

$$\frac{u - V}{v_*} = 5.75\left[\log_{10}\frac{v_*y}{\nu} - \log_{10}\frac{v_*R}{\nu}\right] + 5.5 - 1.75 = 5.75\log_{10}\frac{y}{R} + 3.75$$

Rough pipe, (2) $-$ (4):

$$\frac{u - V}{v_*} = 5.75\left[\log_{10}\frac{y}{k} - \log_{10}\frac{R}{k}\right] + 8.5 - 4.75 = 5.75\log_{10}\frac{y}{R} + 3.75$$

Both give the same result:

$$\boxed{\frac{u - V}{v_*} = 5.75\log_{10}\left(\frac{y}{R}\right) + 3.75}$$

Because the viscosity and roughness have dropped out, the shape of the velocity profile, measured as the defect from the mean velocity, is the same in smooth and rough pipes. At the axis ($y = R$): $\dfrac{u_{max} - V}{v_*} = 3.75$.