--- title: "Wall friction force for developing pipe flow" subject: "Hydraulics" chapter: 1 key: "semester_4/hydraulics:friction-force-parabolic-velocity-pipe" source: https://www.bctengineer.com/q/semester_4/hydraulics/friction-force-parabolic-velocity-pipe site: BCT Engineering --- # Wall friction force for developing pipe flow Hydraulics · Chapter 1: Pipe flow - Asked in: 2076 Bhadra, 6 marks - Video: [Fluid Mechanics Lesson 08C: Fully Developed Pipe Flow](https://www.youtube.com/watch?v=0rPMBtY4zwM) — John Cimbala - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/friction-force-parabolic-velocity-pipe - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1 ## Question A fluid of constant density $\rho$ enters a horizontal pipe of radius R with uniform velocity V and pressure $p_1$. At a downstream section the pressure is $p_2$ and the velocity varies with radius r according to the equation $u = 2V\{1 - (r^2/R^2)\}$. Show that the friction force at the pipe walls from the inlet to the section considered is given by $\pi R^2\left(p_1 - p_2 - \frac{\rho V^2}{3}\right)$. ## Short answer Using a horizontal control volume between inlet and the parabolic profile section, the increase in momentum flux is ρπR²V²/3. Applying the momentum equation gives the wall friction force as πR²(p₁ − p₂ − ρV²/3). ## Answer **Control volume.** Take the fluid in the pipe between the inlet section 1 (uniform velocity $V$, pressure $p_1$) and the section 2 (velocity $u = 2V(1 - r^2/R^2)$, pressure $p_2$). The pipe is horizontal. Let $F$ be the friction force exerted by the wall on the fluid (equal and opposite to the friction force of the fluid on the wall). **Continuity check.** Mean velocity at section 2: $$ \bar u = \frac{1}{\pi R^2}\int_0^R 2V\left(1 - \frac{r^2}{R^2}\right)2\pi r\,dr = \frac{4V}{R^2}\left[\frac{R^2}{2} - \frac{R^2}{4}\right] = V $$ So the discharge is the same at both sections. **Momentum flux at section 2.** The momentum flux is $\rho\int u^2\,dA$, since $u$ is not uniform: $$ \int_0^R u^2\,2\pi r\,dr = 4V^2\cdot 2\pi\int_0^R\left(1 - \frac{r^2}{R^2}\right)^2 r\,dr = 8\pi V^2\left[\frac{R^2}{2} - \frac{R^2}{2} + \frac{R^2}{6}\right] = \frac{4}{3}\pi R^2V^2 $$ (using $\int_0^R (r - 2r^3/R^2 + r^5/R^4)\,dr = R^2/2 - R^2/2 + R^2/6 = R^2/6$). Momentum flux at section 1 (uniform): $\rho V^2\,\pi R^2$. **Momentum equation (flow direction).** Net force = rate of change of momentum flux: $$ p_1\pi R^2 - p_2\pi R^2 - F = \rho\left(\frac{4}{3}\pi R^2V^2 - \pi R^2V^2\right) = \frac{\rho \pi R^2V^2}{3} $$ $$ \boxed{F = \pi R^2\left(p_1 - p_2 - \frac{\rho V^2}{3}\right)} $$ This is the friction force of the wall on the fluid, and by Newton's third law it is also the friction drag on the pipe walls from the inlet to the section. The term $\rho V^2/3$ is the pressure drop used to build up the parabolic profile (increase in momentum flux), not lost by friction. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why must momentum flux at section 2 be computed as rho * integral of u^2 dA? A. Because the velocity is not uniform across the section B. Because density varies C. Because the pipe is inclined D. Because pressure is uniform Answer: A. With a non-uniform profile, the momentum flux is not rho Q V. (Key point: Momentum flux) 2. What is the momentum flux at section 2 for u = 2V(1 - r^2/R^2)? A. rho pi R^2 V^2 B. (4/3) rho pi R^2 V^2 C. 2 rho pi R^2 V^2 D. (1/3) rho pi R^2 V^2 Answer: B. The integral of u^2 dA is (4/3) pi R^2 V^2. (Key point: Flux value) 3. What does the rho V^2/3 term in the answer represent? A. Pressure lost to friction B. Pressure drop that builds up the parabolic profile (momentum increase), not friction C. Elevation change D. Exit loss Answer: B. Part of p1 - p2 raises momentum flux from rho V^2 to 4/3 rho V^2; friction is the remainder. 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