--- title: "Acceleration and streamline curvature, u = Axy" subject: "Fluid Mechanics" chapter: 4 key: "semester_3/fluid_mechanics:acceleration-radius-curvature-streamline" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/acceleration-radius-curvature-streamline site: BCT Engineering --- # Acceleration and streamline curvature, u = Axy Fluid Mechanics · Chapter 4: Hydrokinematics - Asked in: 2078 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/acceleration-radius-curvature-streamline - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-4 ## Question The x component of velocity in a two-dimensional, incompressible flow field is given by u = Axy; the coordinates are measured in meters and A = 2 m⁻¹s⁻¹. There is no velocity component or variation in the z direction. Calculate the acceleration of a fluid particle at point (x, y) = (2, 1). Estimate the radius of curvature of the streamline passing through this point. Plot the streamline and show both the velocity vector and the acceleration vector on the plot. ## Short answer At (2,1), a = (4 i + 2 j) m/s², |a|=4.47 m/s². Radius of curvature R=5.84 m. Streamline: xy²=2 m³. ## Answer **Given:** $u = Axy$, $A = 2\ \text{m}^{-1}\text{s}^{-1}$, 2-D incompressible, steady, no $w$. **Step 1: find v from continuity** $$ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0\;\Rightarrow\;\frac{\partial v}{\partial y} = -Ay\;\Rightarrow\; v = -\frac{Ay^2}{2} $$ (taking $v = 0$ at $y = 0$). At $(2, 1)$: $$ u = 2\times 2\times 1 = 4\ \text{m/s},\qquad v = -\frac{2\times 1}{2} = -1\ \text{m/s} $$ **Step 2: acceleration** (steady flow, so only the convective part) $$ \begin{aligned} a_x &= u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = u(Ay) + v(Ax)\\ &= 4(2) + (-1)(4) = 4\ \text{m/s}^2\\ a_y &= u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = 0 + (-1)(-Ay) = (-1)(-2) = 2\ \text{m/s}^2 \end{aligned} $$ $$ \vec a = 4\hat i + 2\hat j,\qquad |a| = 4.472\ \text{m/s}^2\ \text{at } 26.6^\circ \text{ above +x} $$ **Step 3: radius of curvature.** The normal component of acceleration is $a_n = V^2/R$ and $a_n = |\vec V\times\vec a|/|V|$. $$ \begin{aligned} |V| &= \sqrt{4^2 + 1^2} = 4.123\ \text{m/s}\\ |\vec V\times\vec a| &= |u a_y - v a_x| = |4(2) - (-1)(4)| = 12\\ R &= \frac{|V|^3}{|\vec V\times\vec a|} = \frac{4.123^3}{12} = 5.84\ \text{m} \end{aligned} $$ **Step 4: streamline.** $$ \frac{dy}{dx} = \frac{v}{u} = \frac{-Ay^2/2}{Axy} = -\frac{y}{2x}\;\Rightarrow\; \frac{dy}{y} = -\frac{dx}{2x}\;\Rightarrow\; xy^2 = C $$ Through $(2,1)$: $C = 2$, so $xy^2 = 2\ \text{m}^3$. | y (m) | 0.5 | 0.75 | 1 | 1.5 | 2 | 3 | |---|---|---|---|---|---|---| | x (m) | 8.0 | 3.56 | 2.0 | 0.89 | 0.5 | 0.22 | ```text y 3 |. 2 | . 1.5| . 1 | .--->V(4,-1) at (2,1) | ^ . | a(4,2) . . 0.5| . . . +----------------------------- x 0 1 2 3 ... 8 ``` At (2,1) the velocity vector is directed down-right (slope $-0.25$, tangent to the streamline); the acceleration vector $(4, 2)$ points up-right, and its normal component points towards the centre of curvature (the concave side of the streamline). **Answer: $\vec a = (4\hat i + 2\hat j)$ m/s², $|a| = 4.47$ m/s²; $R = 5.84$ m; streamline $xy^2 = 2$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is v obtained from u = Axy for 2-D incompressible flow? A. From continuity: dv/dy = -du/dx = -Ay, so v = -Ay^2/2 B. By differentiating u with respect to y C. By setting v = u D. From the Bernoulli equation Answer: A. Continuity du/dx + dv/dy = 0 gives dv/dy = -Ay; integrating with v=0 at y=0 gives v = -Ay^2/2. (Key point: Finding v) 2. Why does the acceleration at (2,1) contain only convective terms? A. The flow is incompressible, so convective terms vanish B. The flow is steady, so du/dt = 0 C. Because there is no z component D. Because velocity is zero at the point Answer: B. The velocity field has no time dependence, so local acceleration is zero and a = u du/dx + v du/dy etc. giving a = (4, 2) m/s^2. (Key point: Acceleration) 3. Which expression gives the radius of curvature of the streamline? A. R = |V| / |a| B. R = |a|^2 / |V| C. R = |V|^3 / |V x a| D. R = |V| x |a| Answer: C. a_n = V^2/R and a_n = |V x a|/|V|, so R = |V|^3/|V x a| = 4.123^3/12 = 5.84 m. (Key point: Radius of curvature) 4. What is the streamline through (2,1)? A. x y = 2 B. x^2 y = 4 C. y = x/2 D. x y^2 = 2 Answer: D. dy/dx = v/u = -y/(2x) integrates to x y^2 = C, and C = 2*1 = 2. 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