Chapter 9, Problem 112P

To determine

The velocity field and possibility of the emergence of the Froude number and the Reynolds number and the plotting of profile $w^{\ast }$ versus $x^{\ast }$.

Answer to Problem 112P

The velocity field is given by $w=\frac{\rho gx}{2\mu }\left(x-2h\right)+V$.

The emergence of Froude number and Reynolds number is given as $w^{\ast }=\frac{x^{\ast }}{2}\left(x^{\ast }-2\right)+\frac{{\text{Fr}}^2}{\mathrm{Re}}$.

The plot of $w^{\ast }$ versus $x^{\ast }$ is shown below:

  

Explanation of Solution

Given information:

The assumptions made are stated below:

• The flow is parallel, steady, and laminar.
• The gravity acts in the negative z direction.
• The wall is infinite in yz-plane.
• The constant pressure exits in the free surface.
• The incompressible fluid is Newtonian having constant properties.
• The field velocity is two dimensional and all partial derivatives with respect to $y$ are zero.

The boundary conditions are given below:

• No slip exists in the wall.
• Negligible shear exists at free surface which means at $x=h$, $\frac{\partial w}{\partial x}=0$.

Write the expression for the continuity equation in the Cartesian coordinates.

  $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$  ...... (I)

Here, the velocity component along the x-axis is $u$, the velocity component along the y-axis is $v$, the velocity component along the z-axis is $w$, the distance along the x-direction is $x$, the distance along y-direction is $y$ and the distance along z-direction is $z$.

Write the expression for the $z$ component for the Navier Stokes equation which is incompressible.

  $\rho \left(\frac{\partial w}{\partial t}+u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}\right)=\left(\begin{array}{l}-\frac{\partial P}{\partial z}+\rho g_z\\ +\mu \left(\frac{{\partial }^{2}w}{\partial x^2}+\frac{{\partial }^{2}w}{\partial y^2}+\frac{{\partial }^{2}w}{\partial z^2}\right)\end{array}\right)$  ...... (II)

Here, the pressure exerted by the fluid is $P$, the density of the fluid is $\rho$, the viscosity is $\mu$, the gravitational acceleration along z-direction is $g_z$ and the partial derivative of velocity component $\left(w\right)$ with respect to time $\left(t\right)$ is $\frac{\partial w}{\partial t}$.

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{\rho Vh}{\mu }$

Here, the velocity function is $V$.

Write the expression for Froude number.

  $\begin{array}{l}\text{Fr}=\frac{V}{\sqrt{gh}}\\ {\text{Fr}}^2=\frac{V^2}{gh}\end{array}$

Calculation:

There is no sip in the wall which means that the velocity components $u$ and $v$ are zero and the velocity component $w$ is equal to the velocity function $\left(V\right)$ when the value of $x$ is zero. The shear is negligible at the free surface which means that the partial derivative of $w$ with respect to $x$ is zero when the value of $x$ is $h$. Here, the random length along the x-direction is $h$.

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

  $\begin{array}{l}\frac{\partial \left(0\right)}{\partial x}+\frac{\partial \left(0\right)}{\partial y}+\frac{\partial w}{\partial z}=0\\ \frac{\partial w}{\partial z}=0\end{array}$

Hence, $w$ is not a function of $z$. Therefore, $w$ will be the function of $x$.

  $w=w\left(x\right)$.

Substitute $0$ for $u$, $0$ for $\frac{\partial w}{\partial t}$, $0$ for $v$, $0$ for $\frac{\partial w}{\partial z}$, $0$ for $\frac{\partial P}{\partial z}$, $0$ for $\frac{{\partial }^{2}w}{\partial y^2}$, $0$ for $\frac{{\partial }^{2}w}{\partial z^2}$ and $-g$ for $g_z$ in Equation (II).

  $\begin{array}{l}\rho \left(\begin{array}{l}0\\ +\left(0\right)\frac{\partial w}{\partial x}\\ +\left(0\right)\frac{\partial w}{\partial y}\\ +w\left(0\right)\end{array}\right)=\left(\begin{array}{l}-\left(0\right)+\rho \left(-g\right)\\ +\mu \left(\frac{{\partial }^{2}w}{\partial x^2}+\left(0\right)+\left(0\right)\right)\end{array}\right)\\ 0=-\rho g+\mu \frac{{\partial }^{2}w}{\partial x^2}\\ \mu \frac{{\partial }^{2}w}{\partial x^2}=\rho g\\ \frac{{\partial }^{2}w}{\partial x^2}=\frac{\rho g}{\mu }\end{array}$  ...... (III)

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

  $\begin{array}{l}{\displaystyle \int \frac{{\partial }^{2}w}{\partial x^2}}={\displaystyle \int \frac{\rho g}{\mu }}\\ \frac{\partial w}{\partial x}=\frac{\rho g}{\mu }x+C_1\end{array}$  ...... (IV)

Here, the constant is $C_1$.

Apply boundary condition that is at $x=h$, $\frac{\partial w}{\partial x}=0$ in Equation (IV).

Substitute $h$ for $x$ and $0$ for $\frac{\partial w}{\partial x}$ in Equation (IV).

  $\begin{array}{l}0=\frac{\rho g}{\mu }h+C_1\\ C_1=-\frac{\rho gh}{\mu }\end{array}$

Substitute $-\frac{\rho gh}{\mu }$ for $C_1$ in Equation (IV).

  $\frac{\partial w}{\partial x}=\frac{\rho g}{\mu }x+-\frac{\rho gh}{\mu }$  ...... (V)

Integrate Equation (V) with respect to $x$.

  $\begin{array}{l}{\displaystyle \int \frac{\partial w}{\partial x}}={\displaystyle \int \left(\frac{\rho g}{\mu }x+-\frac{\rho gh}{\mu }\right)}\\ {\displaystyle \int \frac{\partial w}{\partial x}}={\displaystyle \int \frac{\rho g}{\mu }x}+{\displaystyle \int -\frac{\rho gh}{\mu }}\\ w=\frac{\rho g}{\mu }\frac{x^2}{2}-\frac{\rho gh}{\mu }x+C_2\end{array}$  ...... (VI)

Here, the constant is $C_2$.

Apply boundary condition that is at $x=0$, $w=V$.

Substitute $0$ for $x$ and $V$ for $w$ in Equation (VI).

  $\begin{array}{l}V=\frac{\rho g}{\mu }\frac{{\left(0\right)}^2}{2}-\frac{\rho gh}{\mu }\left(0\right)+C_2\\ V=C_2\\ C_2=V\end{array}$

Substitute $V$ for $C_2$ in Equation (VI).

  $\begin{array}{l}w=\frac{\rho g}{\mu }\frac{x^2}{2}-\frac{\rho gh}{\mu }x+V\\ w=\frac{\rho gx}{2\mu }\left(x-2h\right)+V\end{array}$  ...... (VII)

Thus, the velocity field is given by $w=\frac{\rho gx}{2\mu }\left(x-2h\right)+V$.

Rearrange the terms in Equation (VII).

  $\begin{array}{l}w=\frac{\rho gx}{2\mu }\left(x-2h\right)+V\\ \frac{w\mu }{\rho g{h}^2}=\frac{x}{2h^2}\left(x-2h\right)+\frac{V\mu }{\rho g{h}^2}\\ \frac{w\mu }{\rho g{h}^2}=\frac{x}{2h}\left(\frac{x}{h}-2\right)+\frac{V\mu }{\rho g{h}^2}\end{array}$  ...... (VIII)

Assume $x^{\ast }$ for $\frac{x}{h}$ and $w^{\ast }$ for $\frac{w\mu }{\rho g{h}^2}$.

Substitute $x^{\ast }$ for $\frac{x}{h}$ and $w^{\ast }$ for $\frac{w\mu }{\rho g{h}^2}$ in Equation (VIII).

  $\begin{array}{c}{w}^{\ast }=\frac{x^{\ast }}{2}\left(x^{\ast }-2\right)+\frac{V\mu }{\rho g{h}^2}\\ w^{\ast }=\frac{x^{\ast }}{2}\left(x^{\ast }-2\right)+\frac{V^2}{gh}\cdot \frac{1}{\frac{\rho Vh}{\mu }}\end{array}$  ...... (IX)

Substitute ${\text{Fr}}^2$ for $\frac{V^2}{gh}$ and $\mathrm{Re}$ for $\frac{\rho Vh}{\mu }$ in Equation (IX).

  $w^{\ast }=\frac{x^{\ast }}{2}\left(x^{\ast }-2\right)+\frac{{\text{Fr}}^2}{\mathrm{Re}}$  ...... (X)

Substitute $0.5$ for $\text{Fr}$, $0.5$ for $\mathrm{Re}$ and $0$ for $x^{\ast }$ in Equation (X).

  $\begin{array}{c}{w}^{\ast }=\frac{0}{2}\left(0-2\right)+\frac{{\left(0.5\right)}^2}{0.5}\\ w^{\ast }=\frac{0}{2}\left(0-2\right)+0.5\\ =0.5\end{array}$

The table shown below shows the value of $w^{\ast }$, $x^{\ast }$ and the value $\text{Fr}$ for $\mathrm{Re}=0.5$.

S.NO.	$x^{\ast }$	$w^{\ast }$	$\text{Fr}$	$\mathrm{Re}$
$1$.	$0$	$0.5$	$0.5$	$0.5$
$2$.	$0.2$	$0.32$	$0.5$	$0.5$
$3$.	$0.4$	$0.18$	$0.5$	$0.5$
$4$.	$0.6$	$0.08$	$0.5$	$0.5$
$5$.	$0.8$	$-0.64$	$0.5$	$0.5$
$6$.	$1$	$0$	$0.5$	$0.5$

Substitute $0.5$ for $\text{F}r$, $1.0$ for $\mathrm{Re}$ and $0$ for $x^{\ast }$ in Equation (X).

  $\begin{array}{c}{w}^{\ast }=\frac{0}{2}\left(0-2\right)+\frac{{\left(0.5\right)}^2}{1}\\ w^{\ast }=0+0.25\\ =0.25\end{array}$

The table shown below shows the value of $w^{\ast }$, $x^{\ast }$ and the value $\text{Fr}$ for $\mathrm{Re}=1$.

S.NO.	$x^{\ast }$	$w^{\ast }$	$\text{Fr}$	$\mathrm{Re}$
$1$.	0	$0.25$	$0.5$	$1$
$2$.	0.2	$0.07$	$0.5$	$1$
$3$.	0.4	$-0.07$	$0.5$	$1$
$4$.	0.6	$-0.17$	$0.5$	$1$
$5$.	0.8	$-0.23$	$0.5$	$1$
$6$.	1	$-0.25$	$0.5$	$1$

Substitute $0.5$ for $\text{F}r$, $5$ for $\mathrm{Re}$ and $0$ for $x^{\ast }$ in Equation (X).

  $\begin{array}{c}{w}^{\ast }=\frac{0}{2}\left(0-2\right)+\frac{{\left(0.5\right)}^2}{5}\\ w^{\ast }=0+0.05\\ =0.05\end{array}$

The table shown below shows the value of $w^{\ast }$, $x^{\ast }$ and the value $\text{Fr}$ for $\mathrm{Re}=5$.

S.NO.	$x^{\ast }$	$w^{\ast }$	$\text{Fr}$	$\mathrm{Re}$
$1$.	$0$	$0.05$	$0.5$	$5$
$2$.	$0.2$	$-0.13$	$0.5$	$5$
$3$.	$0.4$	$-0.27$	$0.5$	$5$
$4$.	$0.6$	$-0.37$	$0.5$	$5$
$5$.	$0.8$	$-0.43$	$0.5$	$5$
$6$.	$1$	$-0.55$	$0.5$	$5$

Plot $w^{\ast }$ versus $x^{\ast }$ profile graph by using the tabular data.

  

Figure (1)

Conclusion:

The velocity field is given by $w=\frac{\rho gx}{2\mu }\left(x-2h\right)+V$.

The emergence of Froude number and Reynolds number is given as $w^{\ast }=\frac{x^{\ast }}{2}\left(x^{\ast }-2\right)+\frac{{\text{Fr}}^2}{\mathrm{Re}}$.

The plot of $w^{\ast }$ versus $x^{\ast }$ is show below.