Chapter 4, Problem 4.3P

To determine

(a)

The acceleration $a_x$.

The acceleration $a_y$.

Answer to Problem 4.3P

The acceleration $a_x$ is $18\text{units}$.

The acceleration $a_y$ is $26\text{units}$.

Explanation of Solution

Given information:

The two dimensional velocity field is,

$V=\left(x^2-y^2+x\right)i-\left(2xy+y\right)j$

Here, the variables are $x$, $y$ and $z$.

The given point is $\left(x,y\right)=\left(1,2\right)$.

Write the general expression for velocity.

$V=ui+vj+wk$

Here, the velocity function for the $i$ coordinate is $u$, the velocity function for the $j$ coordinate is $v$, the velocity function for $k$ coordinate is $w$.

Write the expression for the acceleration in x-direction.

$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}$ ..... (I)

Here, the velocity gradient for $u$ in x-coordinate is $\frac{\partial u}{\partial x}$, the velocity gradient for $u$ in y-coordinate is $\frac{\partial u}{\partial y}$ and the velocity gradient for $u$ in z-coordinate is $\frac{\partial u}{\partial z}$.

Write the expression for the acceleration in y-direction.

$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}$ ..... (II)

Here, the velocity gradient for $v$ in x-coordinate is $\frac{\partial v}{\partial x}$, the velocity gradient for $v$ in y-coordinate is $\frac{\partial v}{\partial y}$ and the velocity gradient for $v$ in z-coordinate is $\frac{\partial v}{\partial z}$.

Substitute $\left(x^2-y^2+x\right)$ for $u$, $-\left(2xy+y\right)$ for $v$ and $0$ for $w$ in Equation (I).

$\begin{array}{c}{a}_x=\left[\begin{array}{l}\left(x^2-y^2+x\right)\frac{\partial \left(x^2-y^2+x\right)}{\partial x}+\left[-\left(2xy+y\right)\right]\frac{\partial \left[x^2-y^2+x\right]}{\partial y}\\ +\left(0\right)\frac{\partial \left(x^2-y^2+x\right)}{\partial z}\end{array}\right]\\ =\left(x^2-y^2+x\right)\left(2x+1\right)+\left[-\left(2xy+y\right)\right]\left(-2y\right)+0\\ =\left(x^2-y^2+x\right)\left(2x+1\right)+\left[\left(2xy+y\right)\right]\left(2y\right)\end{array}$ ..... (III)

Substitute $\left(x^2-y^2+x\right)$ for $u$, $-\left(2xy+y\right)$ for $v$ and $0$ for $w$ in Equation (II).

$\begin{array}{c}{a}_y=\left[\begin{array}{l}\left(x^2-y^2+x\right)\frac{\partial \left[-\left(2xy+y\right)\right]}{\partial x}+\left[-\left(2xy+y\right)\right]\frac{\partial \left[-\left(2xy+y\right)\right]}{\partial y}\\ +\left(0\right)\frac{\partial \left[-\left(2xy+y\right)\right]}{\partial z}\end{array}\right]\\ =\left(x^2-y^2+x\right)\left(-2y\right)+\left[-\left(2xy+y\right)\right]\left(-2x-1\right)\\ =\left(x^2-y^2+x\right)\left(-2y\right)+\left[\left(2xy+y\right)\left(2x+1\right)\right]\end{array}$ ..... (IV)

Calculation:

Substitute $1$ for $x$ and $2$ for $y$ in Equation (III).

$\begin{array}{c}{a}_x=\left[{\left(1\right)}^2-2^2+1\right]\left[2\left(1\right)+1\right]+\left[\left(2\left(1\right)\left(2\right)+2\right)\right]\left(2\times 2\right)\\ =-2\times 3+24\\ =24-6\\ =18\text{units}\end{array}$

Substitute $1$ for $x$ and $2$ for $y$ in Equation (IV).

$\begin{array}{c}{a}_y=\left(1^2-2^2+1\right)\left(-2\left(2\right)\right)+\left[\left(2\left(1\right)\left(2\right)+2\right)\left(2\left(1\right)+1\right)\right]\\ =\left(-2\right)\left(-4\right)+18\\ =26\text{units}\end{array}$

Conclusion:

Thus, the acceleration $a_x$ is $18\text{units}$.

Thus, the acceleration $a_y$ is $26\text{units}$.

To determine

(b)

The velocity component in the direction $\theta =40°\text{C}$.

Answer to Problem 4.3P

The velocity component in the direction $\theta =40°$ is $5.3882\text{units}$.

Explanation of Solution

Given information:

The angle for the velocity component is $\theta =40°$.

Write the expression for two-dimensional velocity field.

$V=\left(x^2-y^2+x\right)i-\left(2xy+y\right)j$ ..... (V)

Write the expression for the normal vector at any angle.

$n=\cos \theta i+\sin \theta j$ ..... (VI)

Here, the angle with the x-axis is $\theta$.

Write the expression for the magnitude of velocity in direction of $n$.

$V=V\cdot n$ ..... (VII)

Calculation:

Substitute $1$ for $x$ and $2$ for $y$ in Equation (V).

$\begin{array}{c}V=\left(1^2-2^2+1\right)i-\left(2\left(1\right)\left(2\right)+2\right)j\\ =-2i-6j\end{array}$

Substitute $40°$ for $\theta$ in Equation (VI).

$\begin{array}{c}n=\cos \left(40°\right)i+\sin \left(40°\right)j\\ =0.766i+0.6427j\end{array}$

Substitute $-2i-6j$ for $V$ and $0.766i+0.6427j$ for $n$ in Equation (VII).

$\begin{array}{c}V=\left(-2i-6j\right)\cdot \left(0.766i+0.6427j\right)\\ =\left(-2\right)\left(0.766\right)+\left(-6\right)\left(0.6427\right)\\ =-1.532-3.8562\\ =5.3882\text{units}\end{array}$

Conclusion:

Thus, the velocity component in the direction $\theta =40°$ is $5.3882\text{units}$.

To determine

(c)

The direction of maximum velocity.

Answer to Problem 4.3P

The direction of maximum velocity is $55.3°$.

Explanation of Solution

Given data The direction of maximum velocity will be equal to the direction of maximum acceleration.

Write the expression for direction of maximum acceleration.

$\begin{array}{l}\tan \theta =\left(\frac{a_y}{a_x}\right)\\ \theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)\end{array}$ ..... (VIII)

Calculation:

Substitute $26$ for $a_y$ and $18$ for $a_x$ in Equation (VIII).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{26}{18}\right)\\ ={\tan }^{-1}\left(1.444\right)\\ =55.3°\end{array}$

Conclusion:

Thus, the direction of maximum velocity is $55.3°$.

To determine

(d)

The direction of maximum acceleration.

Answer to Problem 4.3P

The direction of maximum acceleration is $55.3°$.

Explanation of Solution

Write the expression for direction of maximum acceleration.

$\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$ ..... (IX)

Calculation:

Substitute $26$ for $a_y$ and $18$ for $a_x$ in Equation (IX).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{26}{18}\right)\\ ={\tan }^{-1}\left(1.444\right)\\ =55.3°\end{array}$

Conclusion:

Thus, the direction of maximum velocity is $55.3°$.