Chapter 4, Problem 4.1P

To determine

(a)

Whether the flow is steady or unsteady.

Whether the flow is two or three dimensional.

The acceleration vector of the velocity field.

Answer to Problem 4.1P

The given flow is an unsteady flow.

The flow is a three dimensional flow.

The acceleration vector of the velocity field is $a=-4\left[1+4t^2\right]i-4t\left[1-t^3\right]j$.

Explanation of Solution

Given information:

Write the expression for the idealized velocity field.

$V=4txi-2t^{2}yj+4xzk$

Here, the variables for position are $x$, $y$ and $z$, and the variable for time is $t$.

The given point on the velocity field is $\left(-1,1,0\right)$.

A unsteady flow is a flow that changes with respect to time.

The given velocity field has a component of time. Hence, the flow is an unsteady flow.

A two dimensional flow is a flow that has two components of velocity and a three dimensional flow has three components of velocity.

As the given flow field has three components of velocity hence, it is a three dimensional flow.

Write the general expression for the velocity field.

$V=ui+vj+wk$

Here, the velocity function in x-coordinate is $u$, the velocity function in y-coordinate is $v$ and the velocity component in z-coordinate is $w$.

Write the expression for the acceleration function along x-coordinate with respect to time.

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

Here, the velocity gradient of $u$ with respect to x coordinate is $\frac{\partial u}{\partial x}$, the velocity gradient of $u$ with respect to y-coordinate is $\frac{\partial u}{\partial y}$ and the velocity gradient of $u$ with respect to z-coordinate is $\frac{\partial u}{\partial z}$.

Write the expression for the acceleration function in y-coordinate with respect to time.

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

Here, the velocity gradient of $v$ with respect to x coordinate is $\frac{\partial v}{\partial x}$, the velocity gradient of $v$ with respect to y-coordinate is $\frac{\partial v}{\partial y}$ and the velocity gradient of $v$ with respect to z-coordinate is $\frac{\partial v}{\partial z}$.

Write the expression for the acceleration function in z-coordinate with respect to time.

$\frac{dw}{dt}=\frac{\partial w}{\partial t}+u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}$ ..... (III)

Here, the velocity gradient of $w$ with respect to x coordinate is $\frac{\partial w}{\partial x}$, the velocity gradient of $w$ with respect to y-coordinate is $\frac{\partial w}{\partial y}$ and the velocity gradient of $w$ with respect to z-coordinate is $\frac{\partial w}{\partial z}$.

Write the expression for acceleration.

$a=\left(\frac{du}{dt}\right)i+\left(\frac{dv}{dt}\right)j+\left(\frac{dw}{dt}\right)k$ ..... (IV)

Substitute $4xt$ for $u$, $-2t^{2}y$ for $v$ and $4xz$ for $w$ in Equation (I).

$\begin{array}{c}\frac{du}{dt}=\frac{\partial \left(4xt\right)}{\partial t}+\left(4xt\right)\frac{\partial \left(4xt\right)}{\partial x}+\left(-2t^{2}y\right)\frac{\partial \left(4xt\right)}{\partial y}+\left(4xz\right)\frac{\partial \left(4xt\right)}{\partial z}\\ =4x+16x{t}^2\end{array}$

Substitute $4xt$ for $u$, $-2t^{2}y$ for $v$ and $4xz$ for $w$ in Equation (II).

$\begin{array}{c}\frac{dv}{dt}=\frac{\partial \left(-2t^{2}y\right)}{\partial t}+\left(4xt\right)\frac{\partial \left(-2t^{2}y\right)}{\partial x}+\left(-2t^{2}y\right)\frac{\partial \left(-2t^{2}y\right)}{\partial y}+\left(4xz\right)\frac{\partial \left(-2t^{2}y\right)}{\partial z}\\ =-4ty+4t^{4}y\end{array}$

Substitute $4xt$ for $u$, $-2t^{2}y$ for $v$ and $4xz$ for $w$ in Equation (III).

$\begin{array}{c}\frac{dw}{dt}=\frac{\partial \left(4xz\right)}{\partial t}+\left(4xt\right)\frac{\partial \left(4xz\right)}{\partial x}+\left(-2t^{2}y\right)\frac{\partial \left(4xz\right)}{\partial y}+\left(4xz\right)\frac{\partial \left(4xz\right)}{\partial z}\\ =16xtz+16x^{2}z\end{array}$

Substitute $16xtz+16x^{2}z$ for $\frac{dw}{dt}$, $-4ty+4t^{4}y$ for $\frac{dv}{dt}$ and $4x+16x{t}^2$ for $\frac{du}{dt}$ in Equation (IV).

$a=\left[4x+16x{t}^2\right]i+\left[-4ty+4t^{4}y\right]j+\left[16xtz+16x^{2}z\right]k$ ..... (V)

Calculation:

Substitute $-1$ for $x$, $1$ for $y$ and $0$ for $z$ in Equation (V).

$\begin{array}{c}a=\left[4\left(-1\right)+16\left(-1\right)t^2\right]i+\left[-4t\left(1\right)+4t^4\left(1\right)\right]j+\left[16\left(-1\right)t\left(0\right)+16{\left(-1\right)}^2\left(0\right)\right]k\\ =\left[-4-16t^2\right]i+\left[-4t+4t^4\right]j\\ =-4\left[1+4t^2\right]i-4t\left[1-t^3\right]j\end{array}$

Conclusion:

Thus, the given flow is an unsteady flow.

Thus, the given flow is three-dimensional.

Thus, the acceleration vector of the velocity field is $a=-4\left[1+4t^2\right]i-4t\left[1-t^3\right]j$.

To determine

(b)

The unit vector normal to the acceleration.

Answer to Problem 4.1P

The unit vector normal to the acceleration is ${\widehat{a}}_n=\pm \frac{\left[t\left(t^3-1\right)\right]i+\left[\left(1+4t^2\right)\right]j}{\sqrt{{\left(1+4t^2\right)}^2+{\left[t\left(t^3-1\right)\right]}^2}}$.

Explanation of Solution

Write the general expression for the acceleration.

$a=a_{x}i+a_{y}j$

Here, the component of acceleration in x-coordinate is $a_x$ and the component of acceleration in y-coordinate is $a_y$.

Write the expression for unit vector normal to acceleration.

${\widehat{a}}_n=\pm \frac{a_{y}i-a_{x}j}{\sqrt{a_x^2+a_y^2}}$ ..... (VI)

Calculation:

Substitute $-4\left[1+4t^2\right]$ for $a_x$ and $-4t\left[1-t^3\right]$ for $a_y$ in Equation (VII).

$\begin{array}{c}{\widehat{a}}_n=\pm \frac{\left[-4t\left(1-t^3\right)\right]i-\left[-4\left(1+4t^2\right)\right]j}{\sqrt{\left[-4\left(1+4t^2\right)\right]+{\left[-4t\left(1-t^3\right)\right]}^2}}\\ =\pm \frac{\left[t\left(t^3-1\right)\right]i+\left[\left(1+4t^2\right)\right]j}{\sqrt{{\left(1+4t^2\right)}^2+{\left[t\left(t^3-1\right)\right]}^2}}\end{array}$

Conclusion:

Thus, the unit vector normal to the acceleration is ${\widehat{a}}_n=\pm \frac{\left[t\left(t^3-1\right)\right]i+\left[\left(1+4t^2\right)\right]j}{\sqrt{{\left(1+4t^2\right)}^2+{\left[t\left(t^3-1\right)\right]}^2}}$.