Chapter 4, Problem 23P

A steady, incompressible, two-dimensional (in the xy-plane) velocity field is given by $\stackrel{\to }{V}=\left(0.523-1.88x+3.94y\right)\stackrel{\to }{i}+\left(-2.44+1.26x+1.88y\right)\stackrel{\to }{j}$ Calculate the acceleration at the point (x.y) = (-1.55, 2.07).

To determine

The acceleration at the point $\left(-1.55,2.07\right)$.

Answer to Problem 23P

The acceleration of the flow is $-23.7699\text{i+13}\text{.664j}$ with a magnitude of $27.417$.

Explanation of Solution

Given information:

Write the expression for the velocity field.

  $\overrightarrow{V}=\left(0.523-1.88x+3.94y\right)\overrightarrow{i}+\left(-2.44+1.26x+1.88y\right)\overrightarrow{j}$   ...... (I)

Here, the variables are $x$, $y$, the velocity field is $\overrightarrow{V},$ and the unit vectors are $\overrightarrow{i}$ and $\overrightarrow{j}$.

Write the expression for the velocity component along x direction.

  $u=0.523+1.88x+3.94y$   ...... (II)

Here, the variables are $x,y$ and the velocity component along x direction is $u$.

Write the expression for the velocity component in y direction.

  $v=-2.44+1.26x+1.88y$  ...... (III)

Here, the variables are $x,y$ and the velocity component along y direction is $v$.

Write the expression for the acceleration of a flow along horizontal axis at point $\left(-1.55,2.07\right)$.

  $a_x=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}$   ...... (IV)

Here, the velocity component in x direction is $u$, the velocity component in x direction is $v,$ and the acceleration is $a_x$.

Write the expression for the acceleration of a flow along vertical axis at point $\left(-1.55,2.07\right)$.

  $a_y=\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$   ...... (V)

Here, the velocity component in x direction is $u$, the velocity component in y direction is $v,$ and the acceleration is $a_y$.

Write the expression for the magnitude of the acceleration.

  $a=\sqrt{a_x^2+a_y^2}$   ...... (VI)

Here, the magnitude of the acceleration is $a$.

Write the expression for the vector notation of the acceleration.

  $a=a_x{\text{i+a}}_y\text{j}$   ...... (VII)

Calculation:

Differentiate Equation (II) with respect to $x$.

  $\begin{array}{l}\frac{\partial u}{\partial x}=\frac{\partial \left(0.523-1.88x+3.94y\right)}{\partial x}\\ \frac{\partial u}{\partial x}=-1.88\end{array}$

Differentiate Equation (II) with respect to $y$.

  $\begin{array}{l}\frac{\partial u}{\partial y}=\frac{\partial \left(0.523-1.88x+3.94y\right)}{\partial x}\\ \frac{\partial u}{\partial y}=3.94\end{array}$

Differentiate Equation (II) with respect to $t$.

  $\begin{array}{l}\frac{\partial u}{\partial t}=\frac{\partial \left(0.523-1.88x+3.94y\right)}{\partial x}\\ \frac{\partial u}{\partial t}=0\end{array}$

Differentiate Equation (III) with respect to $x$.

  $\begin{array}{l}\frac{\partial v}{\partial x}=\frac{\partial \left(-2.44+1.26x+1.88y\right)}{\partial x}\\ \frac{\partial v}{\partial x}=1.26\end{array}$

Differentiate Equation (III) with respect to $y$.

  $\begin{array}{l}\frac{\partial v}{\partial y}=\frac{\partial \left(-2.44+1.26x+1.88y\right)}{\partial x}\\ \frac{\partial u}{\partial y}=1.88\end{array}$

Differentiate Equation (III) with respect to $t$.

  $\begin{array}{l}\frac{\partial v}{\partial t}=\frac{\partial \left(-2.44+1.26x+1.88y\right)}{\partial x}\\ \frac{\partial u}{\partial t}=0\end{array}$

Substitute $-1.55$ for $x$ and $2.07$ for $y$ in Equation (II).

  $\begin{array}{c}u=0.523-\left(1.88\left(-1.55\right)\right)+\left(3.94\left(2.07\right)\right)\\ =0.523+2.914+8.1558\\ =11.5928\end{array}$

Substitute $-1.55$ for $x$ and $2.07$ for $y$ in Equation (III).

  $\begin{array}{c}v=-2.44+\left(1.26\left(-1.55\right)\right)+\left(1.88\times 2.07\right)\\ =-2.44-1.953+3.8916\\ =-0.5014\end{array}$

Substitute $11.5928$ for $y$, $-0.5014$ for $v$, $0$ for $\frac{\partial u}{\partial t}$, $-1.88$ for $\frac{\partial u}{\partial x}$ and $3.94$ for $\frac{\partial u}{\partial y}$ in Equation (IV).

  $\begin{array}{c}{a}_x=0+\left(\left(11.5928\right)\left(-1.88\right)\right)+\left(\left(-0.5014\right)\left(3.94\right)\right)\\ =-21.7944-1.9755\\ =-23.7699\end{array}$

Substitute $11.5928$ for $y$, $-0.5014$ for $v$, $0$ for $\frac{\partial v}{\partial t}$, $1.26$ for $\frac{\partial v}{\partial x}$ and $1.88$ for $\frac{\partial v}{\partial y}$ in Equation (V).

  $\begin{array}{c}{a}_y=0+\left(\left(11.5298\right)\left(1.26\right)\right)+\left(-0.5014\right)\left(1.88\right)\\ =14.527548-0.942632\\ =13.58\end{array}$

Substitute $-23.7699$ for $a_x$ and $13.664$ for $a_y$ in Equation (VI).

  $\begin{array}{c}a=\sqrt{{\left(-23.7699\right)}^2+{13.664}^2}\\ =\sqrt{565.008+186.704}\\ =27.417\end{array}$

Substitute $-23.7699$ for $a_x$ and $13.664$ for $a_y$ in Equation (VII).

  $a=-23.7699\text{i+13}\text{.664j}$

Conclusion:

The acceleration of the flow is $-23.7699\text{i+13}\text{.664j}$ with a magnitude of $27.417$.