Chapter 4, Problem 103P

To determine

The seven streak lines generated from a dye rake that introduce dye streaks at $\text{x}=\text{0}$and $y=0.2,0.4,0.6,0.8,1.0,1.2$and $1.4\text{mm}\text{.}$

Answer to Problem 103P

The seven streak lines generated from the dye rake are:

$x_{0.2}=0.4931\text{m}$

$x_{0.4}=0.8453\text{m}$

$x_{0.6}=1.0566\text{m}$

$x_{0.8}=1.1271\text{m}$

$x_{1.0}=1.0566\text{m}$

$x_{1.2}=0.8453\text{m}$

$x_{1.4}=0.4931\text{m}$

Explanation of Solution

Write the expression for the two dimensional Poiseuille flow.

$u=\frac{1}{2\mu }\frac{dP}{dx}\left(y^2-hy\right)$.

Here, the distance between the plates is $h$, the pressure gradient is $\frac{dP}{dx}$and the viscosity of the fluid is $\mu$.

Write the expression for the streak line generation equation.

$x={\displaystyle \int udt}$

Substitute $\frac{1}{2\mu }\frac{dP}{dx}\left(y^2-hy\right)$for u in the streak line generation equation.

$\begin{array}{l}x={\displaystyle \underset{t_o}{\overset{t}{\int }}\frac{1}{2\mu }\frac{dP}{dx}\left(y^2-hy\right)dt}\\ =\frac{1}{2\mu }\frac{dP}{dx}\left(y^2-hy\right){\left[t\right]}_{t_1}^{t_2}\\ =\frac{1}{2\mu }\frac{dP}{dx}\left(y^2-hy\right)\left[t_2-t_1\right]\end{array}$...... (I)

Conclusion:

Substitute 0 for $t_1$, 10 for $t_2$, $-230\text{N}/{\text{m}}^{\text{3}}$for $\frac{dP}{dx}$, $6.53\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}$for $\mu$and $1.6\times {10}^{-3}\text{m}$for $h$in Equation (I).

$\begin{array}{c}x=\left[\begin{array}{l}\frac{1}{2\times 6.53\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\times \\ \left(-230\text{N}/{\text{m}}^{\text{3}}\right)\left(y^2-\left(1.6\times {10}^{-3}\text{m}\right)y\right)\left(10\text{s}-0\text{s}\right)\end{array}\right]\\ \text{=}\left[\begin{array}{l}\frac{1}{2\times 6.53\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\times \\ \left(-230\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}{\text{m}}^{\text{3}}\right)\left(y^2-\left(1.6\times {10}^{-3}\text{m}\right)y\right)10\text{s}\end{array}\right]\\ \text{=}\left[\begin{array}{l}\frac{1}{2\times 6.53\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\times \\ \left(-230\text{kg}/{\text{m}}^2{\text{.s}}^{\text{2}}\right)\left(y^2-\left(1.6\times {10}^{-3}\text{m}\right)y\right)10\text{s}\end{array}\right]\end{array}$..... (II)

Substitute $y=0.2\times {10}^{-3}\text{m}$in Equation (II).

$\begin{array}{c}{x}_1=\left[\begin{array}{l}\frac{1}{2\times 6.53\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\left(-230\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\\ \times \left({\left(0.2\times {10}^{-3}\text{m}\right)}^2-\left(1.6\times {10}^{-3}\text{m}\right)0.2\times {10}^{-3}\text{m}\right)10\text{s}\end{array}\right]\\ =\left(765.6\text{m}\cdot \text{s}/\text{kg}\right)\times \left(-230\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\times \left(-0.28\times {10}^{-6}{\text{m}}^{\text{2}}\right)10\text{s}\\ =0.4931\text{m}\end{array}$

Substitute $y=0.4\times {10}^{-3}\text{m}$in Equation (II).

$\begin{array}{c}{x}_2=\left[\begin{array}{l}\frac{1}{2\times 6.53\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\left(-230\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\\ \times \left({\left(0.4\times {10}^{-3}\text{m}\right)}^2-\left(1.6\times {10}^{-3}\text{m}\right)0.4\times {10}^{-3}\text{m}\right)10\end{array}\right]\\ =\left(765.6\text{m}\cdot \text{s}/\text{kg}\right)\times \left(-230\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\times \left(4.8\times {10}^{-5}{\text{m}}^2\cdot \text{s}\right)\\ =0.8453\text{m}\end{array}$

Substitute $y=0.6\times {10}^{-3}\text{m}$in Equation (II).

$\begin{array}{c}{x}_3=\left[\begin{array}{l}\frac{1}{2\times 6.53\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\left(-230\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\\ \times \left({\left(0.6\times {10}^{-3}\text{m}\right)}^2-\left(1.6\times {10}^{-3}\text{m}\right)0.6\times {10}^{-3}\text{m}\right)10\text{s}\end{array}\right]\\ =\left(765.6\text{m}\cdot \text{s}/\text{kg}\right)\times \left(-230\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\times \left(-0.0077{\text{m}}^2\right)10\text{s}\\ =1.0566\text{m}\end{array}$

Substitute the values of $y$in Equation (II) and construct a table for the values of $x$.

$y_z$	$x_z=\frac{1}{2\times 6.53\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\left(-230\text{N}/{\text{m}}^{\text{3}}\right)\left(y^2-\left(1.6\times {10}^{-3}\text{m}\right)y\right)10$
$y_{0.2}=0.2\times {10}^{-3}\text{m}$	$x_{0.2}=0.4931\text{m}$
$y_{0.4}=0.4\times {10}^{-3}\text{m}$	$x_{0.4}=0.8453\text{m}$
$y_{0.6}=0.6\times {10}^{-3}\text{m}$	$x_{0.6}=1.0566\text{m}$
$y_{0.8}=0.8\times {10}^{-3}\text{m}$	$x_{0.8}=1.1271\text{m}$
$y_{1.0}=1.0\times {10}^{-3}\text{m}$	$x_{1.0}=1.0566\text{m}$
$y_{1.2}=1.2\times {10}^{-3}\text{m}$	$x_{1.2}=0.8453\text{m}$
$y_{1.4}=1.4\times {10}^{-3}\text{m}$	$x_{1.4}=0.4931\text{m}$

Table-(1)

Plot the values of $x$and $y$from Table-(1) to obtain the streak lines generated from the dye rake.

Figure-(1)

The figure (1) represents the graph generated from a dye rake.