--- title: "Derive laminar velocity in inclined pipe" subject: "Hydraulics" chapter: 1 key: "semester_4/hydraulics:laminar-inclined-pipe-velocity-derivation" source: https://www.bctengineer.com/q/semester_4/hydraulics/laminar-inclined-pipe-velocity-derivation site: BCT Engineering --- # Derive laminar velocity in inclined pipe Hydraulics · Chapter 1: Pipe flow - Asked in: 2078 Chaitra, 6 marks - Video: [Flow through an inclined pipe Lecture 21](https://www.youtube.com/watch?v=m1I5aRGUz6I) — Dr. Madeeha Tahir - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/laminar-inclined-pipe-velocity-derivation - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1 ## Question Consider steady, incompressible, laminar flow of a Newtonian fluid in an infinitely long round pipe of diameter D or radius R = D/2 inclined at angle $\alpha$ as shown in the figure. The fluid flows down the pipe due to gravity alone. Consider the coordinate system shown, with X down the axis of pipe. Derive an expression for the X-component of velocity u as a function of radius r and the other parameters of the problem. [Figure: inclined pipe wall at angle $\alpha$ to the horizontal, fluid of density $\rho$ and viscosity $\mu$, radial coordinate r, x measured down the pipe axis, gravity g vertical] ## Short answer For steady gravity-driven laminar flow, the x-momentum reduces to μ (1/r) d/dr(r du/dr) + ρg sinα = 0. Integrating with u(R)=0 yields u(r) = (ρg sinα/(4μ)) (R² - r²), a parabolic profile. ## Answer **Set-up.** Pipe of radius $R$ inclined at angle $\alpha$ below the horizontal; the fluid flows down the pipe under gravity alone, so there is no externally imposed pressure gradient: $\dfrac{\partial p}{\partial x} = 0$ (the pressure along the pipe is atmospheric/constant). Flow is steady, laminar, fully developed, so $u = u(r)$ only, and $v_r = v_\theta = 0$. ```text x (down the pipe) ___ \ / \ alpha / pipe \ /_________\ ____ horizontal gravity component along x: g sin(alpha) ``` **Momentum equation along $x$** (Navier-Stokes in cylindrical coordinates, reduced for this case): $$ 0 = \rho g\sin\alpha + \mu\,\frac{1}{r}\frac{d}{dr}\left(r\frac{du}{dr}\right) $$ (Equivalent force balance on a cylinder of radius $r$: weight component $\rho g \sin\alpha\,\pi r^2 L$ is balanced by the shear force $-\mu\dfrac{du}{dr}\,2\pi r L$.) **Integrate.** $$ \frac{d}{dr}\left(r\frac{du}{dr}\right) = -\frac{\rho g\sin\alpha}{\mu}\,r $$ $$ r\frac{du}{dr} = -\frac{\rho g\sin\alpha}{2\mu}\,r^2 + C_1 $$ $C_1 = 0$, since $du/dr$ must be finite at $r = 0$. Then $$ \frac{du}{dr} = -\frac{\rho g\sin\alpha}{2\mu}\,r \quad\Rightarrow\quad u = -\frac{\rho g\sin\alpha}{4\mu}\,r^2 + C_2 $$ **Boundary condition.** No slip at the wall: $u = 0$ at $r = R$, so $C_2 = \dfrac{\rho g\sin\alpha}{4\mu}R^2$. $$ \boxed{u(r) = \frac{\rho g\sin\alpha}{4\mu}\left(R^2 - r^2\right)} $$ The profile is parabolic with $u_{max} = \dfrac{\rho g\sin\alpha}{4\mu}R^2$ at the axis. With $R = D/2$, $u = \dfrac{\rho g\sin\alpha}{16\mu}\left(D^2 - 4r^2\right)$. The mean velocity is $V = u_{max}/2 = \dfrac{\rho g D^2\sin\alpha}{32\mu}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What drives the flow in this problem? A. An imposed pressure gradient B. The component of gravity along the pipe, rho g sin(alpha) C. The pipe rotation D. Surface tension Answer: B. dp/dx = 0, so the weight component along x balances viscous shear. (Key point: Driving force) 2. Which two conditions fix the constants? A. u = 0 at r = 0 and du/dr = 0 at r = R B. du/dr finite at r = 0 (C1 = 0) and no slip u = 0 at r = R C. u = u_max at r = R only D. du/dr = 0 at the wall Answer: B. Finite velocity gradient at the axis removes the 1/r singularity and no slip fixes C2. (Key point: Boundary conditions) 3. What is the velocity profile? A. u = (rho g sin(alpha)/4mu)(R^2 - r^2) B. u = (rho g cos(alpha)/4mu)(R^2 - r^2) C. u = (rho g sin(alpha)/2mu)(R - r) D. u = (rho g sin(alpha)/mu)(R^2 + r^2) Answer: A. Integrating twice gives a parabola with the sin(alpha) component of gravity and maximum at the axis. 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