Chapter 4, Problem 121P

A velocity field is given by $u=5y^2,v=3x,w=0$ . (Do not concern yourself with units in this problem.)
(a ) Is this flow steady or unsteady? Is it two- or three-dimensional?
(b ) At (x,y,z) = (3.2,-3), compute the velocity vector.
(c) At (x.y.z) = (3,2,-3), compute the heal (i.e., unsteady part) of the acceleration vector.
(d) At (x,y,z) = (3,2,-3), compute the convective (or advective) part of the acceleration vector.(e) At (x,y,z) = (3,2,-3), compute the (total) acceleration vector.

To determine

(a)

That the flow is steady or unsteady and is it two or three dimensional.

Answer to Problem 121P

The flow is steady and two-dimensional.

Explanation of Solution

Given information:

The velocity field is $u=5y^2$, $v=3x$, $w=0$.

Write the expression for the streamline for three- dimensional flow.

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

Here, the derivative of $u$ with respect to $x$ is $\frac{\partial u}{\partial x}$, the derivative of $v$ with respect to $y$ is $\frac{\partial v}{\partial y}$, and the derivative of $w$ with respect to $z$ is $\frac{\partial w}{\partial z}$.

Substitute $5y^2$ for $u$, $3x$ for $v$, and $0$ for $w$ in the Equation (I).

  $\begin{array}{l}\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}\\ =\frac{\partial \left(5y^2\right)}{\partial x}+\frac{\partial \left(3x\right)}{\partial y}+\frac{\partial \left(0\right)}{\partial z}\\ \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0\end{array}$

Conclusion:

The flow is steady and two-dimensional.

To determine

(b)

The velocity vector at $\left(x,y,z\right)=\left(3,2,-3\right)$.

Answer to Problem 121P

The velocity vector is $\left(20\right)\overrightarrow{i}+\left(9\right)\overrightarrow{j}$.

Explanation of Solution

Write the expression for the velocity vector.

  $\overrightarrow{V}=u\overrightarrow{i}+v\overrightarrow{j}+w\overrightarrow{k}$   ....... (II)

Here, $x$ -component of velocity is $u$, $y$ -component of velocity is $v$, $z$ -component of velocity is $w$.

Substitute $5y^2$ for $u$, $3x$ for $v$, and $0$ for $w$ in the Equation (II).

  $\begin{array}{l}\overrightarrow{V}=\left(5y^2\right)\overrightarrow{i}+\left(3x\right)\overrightarrow{j}+\left(0\right)\overrightarrow{k}\\ \overrightarrow{V}=\left(5y^2\right)\overrightarrow{i}+\left(3x\right)\overrightarrow{j}\end{array}$   ....... (III)

Here, the location points are x, and y.

Calculation:

Substitute $3$ for $x$, and $2$ for $y$ in the Equation (III).

  $\begin{array}{c}\overrightarrow{V}=\left(5{\left(2\right)}^2\right)\overrightarrow{i}+\left(3\left(3\right)\right)\overrightarrow{j}\\ \overrightarrow{V}=\left(20\right)\overrightarrow{i}+\left(9\right)\overrightarrow{j}\end{array}$

Conclusion:

The velocity vector is $\left(20\right)\overrightarrow{i}+\left(9\right)\overrightarrow{j}$.

To determine

(c)

The local (unsteady part) of the acceleration vector at $\left(x,y,z\right)=\left(3,2,-3\right)$.

Answer to Problem 121P

The local (unsteady part) of the acceleration is $0\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the expression for the local acceleration of the flow in $x$ -direction.

  $\frac{d{\overrightarrow{V}}_x}{dt}=\frac{\partial u}{\partial t}$   ....... (IV)

Here, the time derivative of $u$ is $\frac{\partial u}{\partial t}$, and the differential of velocity vector in $x$ -direction is ${\overrightarrow{V}}_x$.

Write the expression for the local acceleration of the flow in $y$ -direction.

  $\frac{d{\overrightarrow{V}}_y}{dt}=\frac{\partial v}{\partial t}$   ....... (V)

Here, the time derivative of $v$ is $\frac{\partial v}{\partial t}$, and the differential of velocity vector in $y$ -direction is ${\overrightarrow{V}}_y$.

Write the expression for the local acceleration of the flow in $z$ -direction.

  $\frac{d{\overrightarrow{V}}_z}{dt}=\frac{\partial w}{\partial t}$   ....... (VI)

Here, the time derivative of $w$ is $\frac{\partial w}{\partial t}$, and the differential of velocity vector in $z$ -direction is ${\overrightarrow{V}}_z$.

Substitute $5y^2$ for $u$ in the Equation (IV).

  $$$\begin{array}{c}\frac{d{\overrightarrow{V}}_x}{dt}=\frac{\partial \left(5y^2\right)}{\partial t}\\ =0\end{array}$

Substitute $3x$ for in the Equation (V).

  $\begin{array}{c}\frac{d{\overrightarrow{V}}_y}{dt}=\frac{\partial \left(3x\right)}{\partial t}\\ =0\end{array}$

Substitute $0$ for $w$ in the Equation (VI).

  $\begin{array}{c}\frac{d{\overrightarrow{V}}_z}{dt}=\frac{\partial \left(0\right)}{\partial t}\\ =0\end{array}$

Conclusion:

The local (unsteady part) of the acceleration is $0\text{m}/{\text{s}}^{\text{2}}$.

To determine

(d)

The convective (or advective) part of the acceleration vector at $\left(x,y,z\right)=\left(3,2,-3\right)$.

Answer to Problem 121P

The convective part of acceleration in $x$ -direction is $180\text{m}/{\text{s}}^{\text{2}}$, the convective part of acceleration in $y$ -direction is $60\text{m}/{\text{s}}^{\text{2}}$, and the convective part of acceleration in $z$ -direction is $0\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given information:

The velocity field is $u=5y^2$, $v=3x$, $w=0$.

Write the expression for the convective part of acceleration in $x$ -direction.

  $a_{xc}=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}$   ....... (VI)

Here, the velocity component in $x$ -direction is $u$, the derivative of the velocity component in $x$ -direction is $\frac{\partial u}{\partial x}$, the velocity component in $y$ -direction is $v$, the derivative of the velocity component in $y$ -direction is $\frac{\partial u}{\partial y}$, the velocity component in $z$ -direction is $w$, and the derivative of the velocity component in $z$ -direction is $\frac{\partial u}{\partial z}$.

Substitute $5y^2$ for $u$, $3x$ for $v$, and $0$ for $w$ in the Equation (VI).

  $\begin{array}{l}{a}_{xc}=\left(5y^2\right)\frac{\partial \left(5y^2\right)}{\partial x}+\left(3x\right)\frac{\partial \left(5y^2\right)}{\partial y}+\left(0\right)\frac{\partial \left(5y^2\right)}{\partial z}\\ a_{xc}=0+30xy+0\\ a_{xc}=30xy\end{array}$   ....... (VIII)

Here, the location points are $x$, and $y$

Write the expression for the convective part of acceleration in $y$ -direction.

  $a_{yc}=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}$   ....... (IX)

Here, the velocity component in $x$ -direction is $u$, the derivative of the velocity component in $x$ -direction is $\frac{\partial v}{\partial x}$, the velocity component in $y$ -direction is $v$, the derivative of the velocity component in $y$ -direction is $\frac{\partial v}{\partial y}$, the velocity component in $z$ -direction is $w$, and the derivative of the velocity component in $z$ -direction is $\frac{\partial v}{\partial z}$.

Substitute $5y^2$ for $u$, $3x$ for $v$, and $0$ for $w$ in the Equation (IX).

  $\begin{array}{l}{a}_{yc}=\left(5y^2\right)\frac{\partial \left(3x\right)}{\partial x}+\left(3x\right)\frac{\partial \left(3x\right)}{\partial y}+\left(0\right)\frac{\partial \left(3x\right)}{\partial z}\\ a_{yc}=15y^2+0+0\\ a_{yc}=15y^2\end{array}$   ....... (X)

Write the expression for the convective part of acceleration in $z$ -direction.

  $a_{zc}=u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}$   ....... (XI)

Here, the velocity component in $x$ -direction is $u$, the derivative of the velocity component in $x$ -direction is $\frac{\partial w}{\partial x}$, the velocity component in $y$ -direction is $v$, the derivative of the velocity component in $y$ -direction is $\frac{\partial w}{\partial y}$, the velocity component in $z$ -direction is $w$, and the derivative of the velocity component in $z$ -direction is $\frac{\partial w}{\partial z}$.

Substitute $5y^2$ for $u$, $3x$ for $v$, and $0$ for $w$ in the Equation (XI).

  $\begin{array}{l}{a}_{zc}=\left(5y^2\right)\frac{\partial \left(0\right)}{\partial x}+\left(3x\right)\frac{\partial \left(0\right)}{\partial y}+\left(0\right)\frac{\partial \left(0\right)}{\partial z}\\ a_{zc}=0\end{array}$

Calculation:

Substitute $3$ for $x$ and $2$ for $y$ in the Equation (VIII).

  $\begin{array}{l}{a}_{xc}=30\left(3\right)\left(2\right)\\ a_{xc}=180\text{unit}\end{array}$

Substitute $3$ for $x$ and $2$ for $y$ in the Equation (X).

  $\begin{array}{l}{a}_{yc}=15{\left(2\right)}^2\\ a_{yc}=60\text{unit}\end{array}$

To determine

(e)

The (total) acceleration vector at $\left(x,y,z\right)=\left(3,2,-3\right)$.

Answer to Problem 121P

The total acceleration is $189.73\text{m}/{\text{s}}^{\text{2}}.$

Explanation of Solution

Write the expression for the total acceleration.

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

Here, acceleration in $x$ -direction is $a_x$, acceleration in $y$ -direction is $a_y$, and acceleration in $z$ -direction is $a_z$.

Calculation:

Substitute $180\text{m}/{\text{s}}^{\text{2}}$ for $a_x$, $60\text{m}/{\text{s}}^{\text{2}}$ for $a_y$, and $0$ for $a_z$.

  $\begin{array}{l}a=\sqrt{{\left(180\text{m}/{\text{s}}^{\text{2}}\right)}^2+{\left(60\text{m}/{\text{s}}^{\text{2}}\right)}^2+{\left(0\right)}^2}\\ a=\sqrt{\left(32400{\left(\text{m}/{\text{s}}^{\text{2}}\right)}^2\right)+\left(3600{\left(\text{m}/{\text{s}}^{\text{2}}\right)}^2\right)}\\ a=\sqrt{36000{\left(\text{m}/{\text{s}}^{\text{2}}\right)}^2}\\ a=\left(189.73\text{m}/{\text{s}}^{\text{2}}\right)\end{array}$

Conclusion:

The total acceleration is $189.73\text{m}/{\text{s}}^{\text{2}}$