Chapter 6.1, Problem 1P

To determine

The local component of acceleration in $x$ direction.

The convective component of acceleration in $x$ direction.

The local component of acceleration in $y$ direction.

The convective component of acceleration in $y$ direction.

The magnitude and direction of velocity at $x=y=2\text{ft}$ and time $t=0$.

The magnitude of acceleration at $x=y=2\text{ft}$ and time $t=0$.

The direction of acceleration at $x=y=2\text{ft}$ and time $t=0$.

Answer to Problem 1P

The local component of acceleration in $x$ direction is $2x$.

The local component of acceleration in $y$ direction is $-2y$.

The convective component of acceleration in $x$ direction is $4x{t}^2$.

The convective component of acceleration in $y$ direction is $4y{t}^2$.

The magnitude of acceleration at $x=y=2\text{ft}$ and time $t=0$ is $5.66\text{ft}/\text{s}$.

The direction of acceleration at $x=y=2\text{ft}$ and time $t=0$ is $-45°$.

Explanation of Solution

Write the expression of velocity in a certain two dimensional flow field.

    $V=2xt\widehat{i}-2yt\widehat{j}$                                                                                           (I)

Here, the displacement in $x$ direction is $x$, the displacement in $y$ direction is $y$, the time is $t$, the unit vector in $x$ direction is $\widehat{i}$ and the unit vector in $y$ direction is $\widehat{j}$.

Write the expression of velocity.

    $V=u\widehat{i}+v\widehat{j}$                                                                                                (II)

Here, the $x$ component of velocity is $u$ and the $y$ component of the velocity is $v$.

Compare the term of unit vector from Equation (I) and (II).

    $u=2xt$                                                                                                      (III)

    $v=-2yt$                                                                                                    (IV)

Write the expression of local component of acceleration in $x$ direction.

    ${\left(a_x\right)}_{\text{Local}}=\frac{\partial u}{\partial t}$                                                                                         (V)

Write the expression of local component of acceleration in $y$ direction.

    ${\left(a_y\right)}_{\text{Local}}=\frac{\partial v}{\partial t}$                                                                                         (VI)

Write the expression of convective acceleration in $x$ direction.

    ${\left(a_x\right)}_{\text{Convective}}=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}$                                                                     (VII)

Write the expression of convective acceleration in $y$ direction.

    ${\left(a_y\right)}_{\text{Convective}}=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$                                                                     (VIII)

Write the expression of acceleration.

    $a=\left\{{\left(a_x\right)}_{\text{Local}}+{\left(a_x\right)}_{\text{Convection}}\right\}\widehat{i}+\left\{{\left(a_y\right)}_{\text{Local}}+{\left(a_y\right)}_{\text{Convection}}\right\}\widehat{j}$                       (IX)

Substitute $2x$ for ${\left(a_x\right)}_{\text{Local}}$, $-2y$ for ${\left(a_y\right)}_{\text{Local}}$, $4x{t}^2$ for ${\left(a_x\right)}_{\text{Convection}}$ and $4y{t}^2$ for ${\left(a_y\right)}_{\text{Convection}}$ in Equation (IX).

    $a=\left\{2x+4x{t}^2\right\}\widehat{i}+\left\{-2y+4y{t}^2\right\}\widehat{j}$                                                      (X)

Write the expression of acceleration.

    $a=a_x\widehat{i}+a_y\widehat{j}$                                                                                      (XI)

Here, the acceleration component in $x$ direction is $a_x$ and the acceleration component in $y$ direction is $a_y$.

Write the expression of direction of acceleration.

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

Write the expression of magnitude of the acceleration.

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

Conclusion:

Substitute $2xt$ for $u$ in Equation (V).

    $\begin{array}{c}{\left(a_x\right)}_{\text{Local}}=\frac{\partial }{\partial t}\left(2xt\right)\\ =2x\times 1\\ =2x\end{array}$

Thus, the local component of acceleration in $x$ direction is $2x$.

Substitute $2yt$ for $v$ in Equation (VI).

    $\begin{array}{c}{\left(a_y\right)}_{\text{Local}}=\frac{\partial }{\partial t}\left(-2yt\right)\\ =-2y\times 1\\ =-2y\end{array}$

Thus, the local component of acceleration in $y$ direction is $-2y$.

Substitute $2xt$ for $u$ in Equation (VII).

    $\begin{array}{c}{\left(a_x\right)}_{\text{Convective}}=\left(2xt\right)\frac{\partial }{\partial x}\left(2xt\right)+\left(-2yt\right)\frac{\partial }{\partial y}\left(2xt\right)\\ =\left(2xt\right)\times 2t\times 1-\left(2yt\right)\times 2t\times 0\\ =4x{t}^2\end{array}$

Thus, the convective component of acceleration in $x$ direction is $4x{t}^2$.

Substitute $-2yt$ for $v$ in Equation (VIII).

    $\begin{array}{c}{\left(a_x\right)}_{\text{Convective}}=\left(2xt\right)\frac{\partial }{\partial x}\left(-2yt\right)+\left(-2yt\right)\frac{\partial }{\partial y}\left(-2yt\right)\\ =\left(2xt\right)\times 2t\times 0-\left(2yt\right)\times \left(-2t\right)\times 1\\ =4y{t}^2\end{array}$

Thus, the convective component of acceleration in $y$ direction is $4y{t}^2$.

Substitute $2\text{ft}$ for $x$ and $0$ for $t$ in Equation (III).

    $\begin{array}{c}u=2\times \left(2\text{ft}\right)\times 0\\ =0\end{array}$

Substitute $2\text{ft}$ for $y$ and $0$ for $t$ in Equation (IV).

    $\begin{array}{c}y=2\times \left(2\text{ft}\right)\times 0\\ =0\end{array}$

Substitute $0$ for $u$ and $0$ for $v$ in Equation (V).

    $\begin{array}{c}V=0\widehat{i}+0\widehat{j}\\ =0\end{array}$

Thus, The magnitude and direction of velocity at $x=y=2\text{ft}$ and time $t=0$ is $0$.

Substitute $2\text{ft}$ for $x$, $2\text{ft}$ for $y$ and $0$ for $t$ in Equation (X).

    $\begin{array}{c}a=\left\{2\left(2\text{ft}\right)+4\left(2\text{ft}\right)\times 0\right\}\widehat{i}+\left\{-2\left(2\text{ft}\right)+4\left(2\text{ft}\right)\times 0\right\}\widehat{j}\\ =\left(4\text{ft}\right)\widehat{i}-\left(4\text{ft}\right)\widehat{j}\end{array}$                            (XIV)

Compare the unit vector of Equation (XI) and (XIV).

    $a_x=4\text{ft}/\text{s}$

    $a_y=-4\text{ft}/\text{s}$

Substitute $4\text{ft}/\text{s}$ for $a_x$ and $-4\text{ft}/\text{s}$ for $a_y$ in Equation (XIII).

    $\begin{array}{c}a=\sqrt{{\left(4\text{ft}/\text{s}\right)}^2+{\left(-4\text{ft}/\text{s}\right)}^2}\\ =\sqrt{\left(16{\text{ft}}^2/{\text{s}}^2\right)+\left(16{\text{ft}}^2/{\text{s}}^2\right)}\\ =\sqrt{\left(32{\text{ft}}^2/{\text{s}}^2\right)}\\ =5.66\text{ft}/\text{s}\end{array}$

Thus, the magnitude of acceleration at $x=y=2\text{ft}$ and time $t=0$ is $5.66\text{ft}/\text{s}$.

Substitute $4\text{ft}/\text{s}$ for $a_x$ and $-4\text{ft}/\text{s}$ for $a_y$ in Equation (XII).

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-4\text{ft}/\text{s}}{4\text{ft}/\text{s}}\right)\\ ={\tan }^{-1}\left(-1\right)\\ =-45°\end{array}$

Thus, the direction of acceleration at $x=y=2\text{ft}$ and time $t=0$ is $-45°$.