--- title: "Velocity defect law for smooth and rough pipes" subject: "Hydraulics" chapter: 1 key: "semester_4/hydraulics:smooth-rough-pipe-velocity-defect-law" source: https://www.bctengineer.com/q/semester_4/hydraulics/smooth-rough-pipe-velocity-defect-law site: BCT Engineering --- # Velocity defect law for smooth and rough pipes Hydraulics · Chapter 1: Pipe flow - Asked in: 2070 Magh, 8 marks - Video: [Prandtl's Velocity Distribution In Pipes - Turbulent Flow - Fluid Mechanics 2](https://www.youtube.com/watch?v=Teqo-xCW2z8) — Ekeeda - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/smooth-rough-pipe-velocity-defect-law - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1 ## Question Show that in both smooth and rough pipes for turbulent flow $\frac{u-V}{v_*} = 5.75\log\left(\frac{y}{R}\right) + 3.75$, where V = mean velocity; u = point velocity at distance y from boundary; $v_*$ = shear velocity; R = radius of pipe. ## Short answer Subtract the mean velocity law from the point velocity law. For smooth pipes: (u/V*) - (V/V*) = 5.75 log(y/R) + (5.5 - 1.75) = 5.75 log(y/R) + 3.75. For rough pipes: (8.5 - 4.75) = 3.75. The wall constants cancel, yielding a universal defect law valid for both smooth and rough pipes. ## Answer **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 ($y = R$): $\dfrac{u_{max} - V}{v_*} = 3.75$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does subtracting the mean-velocity law from the point-velocity law give one equation for both smooth and rough pipes? A. Because smooth and rough pipes have the same u* B. Because the wall-dependent terms (nu or k) and the constants combine to cancel, leaving only y/R C. Because the mean velocity is zero at the wall D. Because roughness has no effect on turbulent flow Answer: B. The terms log(v*/nu) or log(1/k) appear in both u and V and cancel, so the difference depends only on y/R. (Key point: Idea of the derivation) 2. For the smooth pipe, how does the constant 3.75 come out? A. 8.5 - 4.75 B. 5.5 + 1.75 C. 5.5 - 1.75 D. 5.75 x 0.4343 Answer: C. u/v* has constant 5.5 and V/v* has constant 1.75, and 5.5 - 1.75 = 3.75. (Key point: Smooth-pipe constant) 3. For the rough pipe, which subtraction gives 3.75? A. 8.5 + 4.75 B. 8.5 - 3.75 C. 5.5 - 1.75 D. 8.5 - 4.75 Answer: D. The rough point law has 8.5 and the mean law has 4.75, so the difference is 3.75. (Key point: Rough-pipe constant) 4. What does the defect law give at the axis (y = R)? A. (u_max - V)/v* = 3.75 B. (u_max - V)/v* = 0 C. (u_max - V)/v* = 5.75 D. (u_max - V)/v* = 1.75 Answer: A. At y = R, log(y/R) = 0, leaving 3.75. 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