--- title: "Drag on a cylinder and shedding frequency" subject: "Applied Computational Fluid Dynamics" chapter: 6 key: "semester_7/applied_computational_fluid_dynamics:cylinder-drag-vortex-shedding-numerical" source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/cylinder-drag-vortex-shedding-numerical site: BCT Engineering --- # Drag on a cylinder and shedding frequency Applied Computational Fluid Dynamics · Chapter 6: CFD Applications - Asked in: year not given, 5 marks - Video: [How to determine the total drag force and vortex shedding frequency](https://www.youtube.com/watch?v=GzQ0_LS--6c) — Study Tips - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/cylinder-drag-vortex-shedding-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-6 ## Question Air ($\rho = 1.2$ kg/m³, $\mu = 1.8\times10^{-5}$ Pa·s) flows at 15 m/s across a long circular cylinder of diameter 50 mm and length 2 m. (a) Find the Reynolds number. (b) Taking $C_D = 1.2$, find the drag force and power. (c) Estimate the vortex shedding frequency for $St = 0.2$. (d) Explain why a bluff body has a larger drag than a streamlined body and how CFD predicts it. ## Short answer Re = 5×10⁴. Drag force F_D =16.2 N, required power =243 W. Vortex shedding frequency f_s =60 Hz (St =0.2). Bluff bodies have large pressure drag from separated wakes; streamlined bodies have mainly skin friction. CFD predicts drag by integrating pressure and shear stress over the surface. ## Answer #### (a) Reynolds number $$ Re=\frac{\rho UD}{\mu}=\frac{1.2\times15\times0.05}{1.8\times10^{-5}}=5.0\times10^{4} $$ This is in the subcritical range (about $10^3$ to $2\times10^5$), where the laminar boundary layer separates near $80^\circ$ and $C_D\approx1.2$, so the given value is reasonable. #### (b) Drag force and power Reference area (projected) $A=DL=0.05\times2=0.10$ m². $$ F_D=\tfrac12\rho U^2C_DA=0.5\times1.2\times15^2\times1.2\times0.10=16.2\ \text{N} $$ Power to overcome drag: $$ P=F_DU=16.2\times15=243\ \text{W} $$ #### (c) Vortex shedding $$ f_s=\frac{St\,U}{D}=\frac{0.2\times15}{0.05}=60\ \text{Hz} $$ Alternating vortices (von Karman street) produce an oscillating lift force at this frequency, which can excite vibration and noise. #### (d) Bluff versus streamlined bodies ```text Bluff body Streamlined body ____ ______ / \ large wake ---/ \___ thin wake | O |::::::::::::: ---\______/ \____/ low pressure attached flow ``` - On a bluff body the boundary layer separates early because of the strong adverse pressure gradient. A broad low-pressure wake forms behind it, so pressure is much lower on the rear than the front: large **pressure (form) drag** dominates (friction drag is small). - On a streamlined body the flow stays attached up to the tail, the wake is thin, and drag is mainly **skin friction**, which is much smaller. CFD prediction: a 2D/3D domain extending about 10D upstream and 20D downstream is meshed with prism layers; RANS (k-$\omega$ SST) or unsteady URANS/LES is used; $C_D$ is computed by integrating surface pressure and shear stress over the body. Wake resolution and separation prediction are the main sources of error, so validation with experimental $C_D$ is needed. **Answer:** $Re=5\times10^4$; $F_D=16.2$ N; $P=243$ W; $f_s=60$ Hz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For air at 15 m/s over a 50 mm cylinder, what is Re? A. 5 x 10^4 B. 5 x 10^3 C. 5 x 10^5 D. 1.5 x 10^4 Answer: A. Re = 1.2 x 15 x 0.05/1.8e-5 = 5 x 10^4. (Key point: Reynolds number) 2. With A = D L = 0.10 m^2 and CD = 1.2, what are the drag force and power? A. F = 16.2 N; P = 243 W B. F = 32.4 N; P = 486 W C. F = 16.2 N; P = 16.2 W D. F = 243 N; P = 16.2 W Answer: A. F = 0.5 x 1.2 x 225 x 1.2 x 0.1 = 16.2 N, and P = F U = 16.2 x 15 = 243 W. (Key point: Drag force and power) 3. What is the shedding frequency for St = 0.2? A. f = St D/U = 0.00067 Hz B. f = St U/D = 60 Hz C. f = U/(St D) = 300 Hz D. f = St U D = 0.15 Hz Answer: B. f = St U/D = 0.2 x 15/0.05 = 60 Hz, producing an oscillating lift that can excite vibration. (Key point: Shedding frequency) 4. Why does a bluff body have higher drag than a streamlined body? A. Higher skin friction B. Early separation creates a broad low-pressure wake, so pressure (form) drag dominates C. It has a thinner wake D. Its flow stays attached to the tail Answer: B. Early separation from the adverse pressure gradient gives a big wake; streamlined bodies keep the flow attached and have mainly small skin friction. 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