--- title: "Continuity in cylindrical coordinates, radial flow" subject: "Fluid Mechanics" chapter: 4 key: "semester_4/fluid_mechanics_mechanical:continuity-cylindrical-radial-flow" source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/continuity-cylindrical-radial-flow site: BCT Engineering --- # Continuity in cylindrical coordinates, radial flow Fluid Mechanics · Chapter 4: Basic Equations for Fluid Flow - Asked in: year not given, 3+3 marks - Video: [Derivation of the Continuity Equation in Cylindrical Coordinates](https://www.youtube.com/watch?v=95jHgmcLzoM) — Fluid Matters - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/continuity-cylindrical-radial-flow - Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-4 ## Question (a) Write the continuity equation for steady incompressible flow in cylindrical coordinates $(r, \theta, z)$ and explain each term. (b) Water enters at the centre of two parallel circular discs 20 mm apart and flows radially outward. The discharge is 0.04 m$^3$/s. Show that the flow satisfies continuity and find the radial velocity at radii 0.10 m and 0.25 m. ## Short answer The steady incompressible continuity equation in cylindrical coordinates is (1/r)∂(rv_r)/∂r + (1/r)∂v_θ/∂θ + ∂v_z/∂z = 0. For radial flow between parallel discs, v_r = Q/(2πrb). With Q=0.04 m³/s and b=0.02 m, v_r = 3.18 m/s at r=0.10 m and 1.27 m/s at r=0.25 m. ## Answer #### (a) Continuity equation in cylindrical coordinates With velocity components $v_r$ (radial), $v_\theta$ (tangential) and $v_z$ (axial): $$ \frac{1}{r}\frac{\partial (r v_r)}{\partial r} + \frac{1}{r}\frac{\partial v_\theta}{\partial \theta} + \frac{\partial v_z}{\partial z} = 0 $$ - First term: net radial outflow per unit volume. The factor $r$ appears because the flow area $r\,d\theta\,dz$ grows with $r$. - Second term: net tangential outflow per unit volume. The factor $1/r$ appears because the arc length of the element is $r\,d\theta$. - Third term: net axial outflow, the same as in the Cartesian form. For compressible unsteady flow the term $\partial\rho/\partial t$ is added and $\rho$ is included inside the derivatives. For axisymmetric flow, $\partial/\partial\theta = 0$. #### (b) Radial flow between discs Given $b = 0.02\ \text{m}$, $Q = 0.04\ \text{m}^3/\text{s}$. The flow is radial and symmetric, so $v_\theta = 0$, $v_z = 0$, and the flow area at radius $r$ is $2\pi r b$. From continuity (a volume balance), $v_r = \dfrac{Q}{2\pi r b}$, so $r v_r = \dfrac{Q}{2\pi b} = \text{constant}$. Hence $$ \frac{1}{r}\frac{\partial (r v_r)}{\partial r} = \frac{1}{r}\frac{\partial}{\partial r}\left(\frac{Q}{2\pi b}\right) = 0 $$ and the cylindrical continuity equation is satisfied. Velocities: $$ \begin{aligned} v_r(0.10) &= \frac{0.04}{2\pi \times 0.10 \times 0.02} = 3.18\ \text{m/s} \\ v_r(0.25) &= \frac{0.04}{2\pi \times 0.25 \times 0.02} = 1.27\ \text{m/s} \end{aligned} $$ The velocity falls inversely with radius, as the flow area increases. **Answer: $v_r = 3.18\ \text{m/s}$ at $r = 0.10\ \text{m}$ and $1.27\ \text{m/s}$ at $r = 0.25\ \text{m}$** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For steady incompressible flow, the continuity equation in cylindrical coordinates is: A. d(v_r)/dr + dv_theta/dtheta + dv_z/dz = 0 B. (1/r) d(r v_r)/dr + (1/r) dv_theta/dtheta + dv_z/dz = 0 C. (1/r) d(v_r)/dr + dv_theta/dtheta + dv_z/dz = 0 D. d(r v_r)/dr + d(r v_theta)/dtheta + d(r v_z)/dz = 0 Answer: B. The factor r inside the radial term arises because the flow area r dtheta dz grows with r; the tangential term has 1/r from arc length. (Key point: Cylindrical continuity equation) 2. For radial flow between discs of gap b, the radial velocity at radius r is: A. Q / (pi r^2) B. Q / (2 pi r b) C. Q b / (2 pi r) D. Q / (2 pi b) Answer: B. The flow area at radius r is 2 pi r b, so v_r = Q/(2 pi r b), giving 3.18 m/s at r = 0.10 m. (Key point: Radial velocity between discs) 3. Why is continuity satisfied by v_r = Q/(2 pi b r)? A. v_r is constant with r B. r v_r is constant, so d(r v_r)/dr = 0 C. v_theta is large D. Q varies with r Answer: B. r v_r = Q/(2 pi b) is constant, making (1/r) d(r v_r)/dr zero; velocity falls as 1/r. 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