Chapter 6, Problem 1P

An incompressible frictionless flow field is given by $\stackrel{\to }{V}=(Ax+By)\hat{i}+(Bx-Ay)\hat{j}$, where A = 2 s⁻¹ and B = 2 s⁻¹, and the coordinates are measured in meters. Find the magnitude and direction of the acceleration of a fluid particle at point (x, y) = (2, 2). Find the pressure gradient at the same point, if $\stackrel{\to }{g}=-g\hat{j}$ and the fluid is water.

To determine

The magnitude of the acceleration of a fluid particle at point $\left(2,2\right)$.

The direction of the acceleration of a fluid particle at point $\left(2,2\right)$.

The pressure gradient of fluid at point $\left(2,2\right)$.

Explanation of Solution

Given:

The flow field $\left(\overrightarrow{V}\right)$ is $\left(Ax+By\right)\widehat{i}+\left(Bx-Ay\right)\widehat{j}$.

The constants $A$ and $B$ are $2{\text{s}}^{-1}$ and $2{\text{s}}^{-1}$.

Fluid particle point $\left(x,y\right)$ is $\left(2,2\right)\text{m}$.

Consider the density of water $\left(\rho \right)$ is $999\text{kg}/{\text{m}}^3$.

Calculations:

From the flow field $\left(\overrightarrow{V}\right)$, the components are:

  $\begin{array}{c}u=Ax+By\\ v=Bx-Ay\end{array}$

Calculate the acceleration of the particle along x direction $\left(a_x\right)$.

  $\begin{array}{c}{a}_x=u\frac{\partial }{\partial x}u+v\frac{\partial }{\partial y}u\\ =\left(Ax+By\right)\frac{\partial }{\partial x}\left(Ax+By\right)+\left(Bx-Ay\right)\frac{\partial }{\partial y}\left(Ax+By\right)\\ =A\left(Ax+By\right)+\left(Bx-Ay\right)B\\ =\left(A^2+B^2\right)x\end{array}$

  $\begin{array}{c}{a}_x=\left[{\left(2{\text{s}}^{-1}\right)}^2+{\left(2{\text{s}}^{-1}\right)}^2\right]\left(2\text{m}\right)\\ =16\text{m}/{\text{s}}^2\end{array}$

Calculate the acceleration of the particle along y direction $\left(a_y\right)$.

  $\begin{array}{c}{a}_y=u\frac{\partial }{\partial x}v+v\frac{\partial }{\partial y}v\\ =\left(Ax+By\right)\frac{\partial }{\partial x}\left(Bx-Ay\right)+\left(Bx-Ay\right)\frac{\partial }{\partial y}\left(Bx-Ay\right)\\ =B\left(Ax+By\right)-\left(Bx-Ay\right)A\end{array}$

  $\begin{array}{c}{a}_y=\left(A^2+B^2\right)y\\ =\left[{\left(2{\text{s}}^{-1}\right)}^2+{\left(2{\text{s}}^{-1}\right)}^2\right]\left(2\text{m}\right)\\ =16\text{m}/{\text{s}}^2\end{array}$

Calculate the resultant acceleration of a fluid particle at point.

  $\begin{array}{c}a=\sqrt{{\left(a_x\right)}^2+{\left(a_y\right)}^2}\\ =\sqrt{{\left(16\text{m}/{\text{s}}^2\right)}^2+{\left(16\text{m}/{\text{s}}^2\right)}^2}\\ =22.627\text{m}/{\text{s}}^2\end{array}$

Thus, the magnitude of the acceleration of a fluid particle at point $\left(2,2\right)$ is $\underset{\_}{22.627\text{m}/{\text{s}}^2}$.

Calculate the direction of the acceleration of fluid particle $\left(\theta \right)$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)\\ ={\tan }^{-1}\left(\frac{16\text{m}/{\text{s}}^2}{16\text{m}/{\text{s}}^2}\right)\\ =45°\end{array}$

Thus, the direction of the acceleration of a fluid particle at point $\left(2,2\right)$ is $\underset{\_}{45°}$.

Calculate the pressure gradient in x direction.

  $\begin{array}{c}\frac{\partial }{\partial x}p=-\rho a_x\\ =-\left(999\text{kg}/{\text{m}}^3\right)\times 16\text{m}/{\text{s}}^2\times \frac{\text{N}\cdot {\text{s}}^2}{\text{kg}\cdot \text{m}}\times \frac{1\text{kPa}}{1000\text{N}/{\text{m}}^2}\\ =-15.984\text{kPa}/\text{m}\end{array}$

Thus, the pressure gradient of fluid at point $\left(2,2\right)$ in x direction is $\underset{\_}{-15.984\text{kPa}/\text{m}}$.

Calculate the pressure gradient in y direction.

  $\begin{array}{c}\frac{\partial }{\partial y}p=-\rho g_y-\rho a_y\\ =-\left(999\text{kg}/{\text{m}}^3\right)\times \left(9.81\text{m}/{\text{s}}^2+16\text{m}/{\text{s}}^2\right)\times \frac{\text{N}\cdot {\text{s}}^2}{\text{kg}\cdot \text{m}}\times \frac{1\text{kPa}}{1000\text{N}/{\text{m}}^2}\\ =-25.784\text{kPa}/\text{m}\end{array}$

Thus, the pressure gradient of fluid at point $\left(2,2\right)$ in y direction is $\underset{\_}{-25.784\text{kPa}/\text{m}}$.