--- title: "Streamline and acceleration of a flow field" subject: "Applied Computational Fluid Dynamics" chapter: 1 key: "semester_7/applied_computational_fluid_dynamics:streamline-acceleration-numerical" source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/streamline-acceleration-numerical site: BCT Engineering --- # Streamline and acceleration of a flow field Applied Computational Fluid Dynamics · Chapter 1: Introduction to CFD and Fluid Mechanics Review - Asked in: year not given, 6 marks - Video: [Fluid Kinematics: The Acceleration field](https://www.youtube.com/watch?v=HsVM2y-QutQ) — Sonya Lopez - 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/streamline-acceleration-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-1 ## Question The velocity field of a steady, two-dimensional flow is $u = 2x$ and $v = -2y$ (in m/s, with $x, y$ in m). (a) Show that the flow is incompressible. (b) Find the equation of the streamline passing through the point (2, 1). (c) Determine the acceleration of a fluid particle at that point. (d) Is the flow irrotational? ## Short answer Divergence is 2-2=0, so continuity is satisfied. The streamline through (2,1) is xy=2. Acceleration components are ax=8, ay=4 m/s^2, magnitude 8.94 m/s^2. Vorticity is zero, so the flow is irrotational. ## Answer Given: $u=2x$, $v=-2y$, steady 2D flow. #### (a) Incompressibility $$ \nabla\cdot\vec V=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=2+(-2)=0 $$ Since the divergence is zero, the continuity equation for incompressible flow is satisfied. #### (b) Streamline through (2, 1) $$ \frac{dx}{u}=\frac{dy}{v}\ \Rightarrow\ \frac{dx}{2x}=\frac{dy}{-2y}\ \Rightarrow\ \frac{dx}{x}+\frac{dy}{y}=0 $$ Integrating gives $\ln x+\ln y=\ln C$, so $xy=C$. At (2, 1): $C=2\times1=2$. **Streamline: $xy = 2$** (a rectangular hyperbola). This is a stagnation-point flow: fluid approaches the origin along the y-axis and leaves along the x-axis. #### (c) Acceleration The flow is steady, so only the convective part remains: $$ a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=(2x)(2)+0=4x $$ $$ a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=0+(-2y)(-2)=4y $$ At (2, 1): $a_x=8\ \text{m/s}^2$, $a_y=4\ \text{m/s}^2$. $$ |a|=\sqrt{8^2+4^2}=8.94\ \text{m/s}^2 $$ Direction: $\tan^{-1}(4/8)=26.6^\circ$ above the x-axis. #### (d) Rotation $$ \omega_z=\frac12\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)=\frac12(0-0)=0 $$ The flow is irrotational. **Answer:** (a) $\nabla\cdot\vec V=0$; (b) $xy=2$; (c) $a=8.94\ \text{m/s}^2$ ($a_x=8$, $a_y=4$ m/s²); (d) irrotational. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For u = 2x, v = -2y, how is incompressibility shown? A. The divergence du/dx + dv/dy = 2 - 2 = 0 B. The curl dv/dx - du/dy = 0 C. u and v are both linear D. The acceleration is zero Answer: A. Zero divergence of velocity satisfies the incompressible continuity equation. (Key point: Incompressibility test) 2. The streamline through (2, 1) is: A. x/y = 2 B. x^2 + y^2 = 5 C. xy = 2 D. x + y = 3 Answer: C. dx/(2x) = dy/(-2y) integrates to xy = C, and C = 2 x 1 = 2. (Key point: Streamline equation) 3. Why is only the convective part of the acceleration used here? A. Because the flow is incompressible B. Because the flow is irrotational C. Because the viscosity is zero D. Because the flow is steady, so the local term du/dt is zero Answer: D. In steady flow the local acceleration vanishes, leaving a_x = u du/dx + v du/dy. (Key point: Acceleration in steady flow) 4. What is the acceleration at (2, 1)? A. a_x = 4, a_y = 8, |a| = 8.94 m/s^2 B. a_x = 8, a_y = 4, |a| = 8.94 m/s^2 C. a_x = 4, a_y = -4, |a| = 5.66 m/s^2 D. a_x = 0, a_y = 0 Answer: B. a_x = 4x = 8 and a_y = 4y = 4, so |a| = sqrt(64+16) = 8.94 m/s^2. 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