Chapter 10, Problem 56RQ

(a)

To determine

The values of constants a, b and c.

Explanation of Solution

Given:

The velocity profiles for the layers are as follows:

$\begin{array}{l}{V}_1=6+ay-3y^2,0\le y\le -0.5\\ V_2=b+cy-9y^2,-0.5\le y\le 0\end{array}$

Calculation:

The conditions to be satisfied by the velocity profiles are as follows,

  $\begin{array}{l}{V}_1(h)=10\\ V_2(-h)=0\end{array}$

  $\begin{array}{l}{V}_1(0)=V_2(0)\\ V_1(0)=6\text{m}/\text{s}\end{array}$

Determine the value of a.

  $\begin{array}{l}{V}_1=6+ay-3y^2\\ 10=6+a\times 0.5-3\times {(0.5)}^2\\ a=9.5\end{array}$

Thus, the value of constant a is $\underset{\_}{9.5}$.

Determine the value of b.

  $\begin{array}{l}{V}_2=b+cy-9y^2\\ V_2(0)=6=b+c\times 0-9{(0)}^2\\ b=6\end{array}$

Thus, the value of constant b is $\underset{\_}{6}$.

Determine the value of c.

  $\begin{array}{l}{V}_2=b+cy-9y^2\\ V_2(-h)=0\\ 0=6+c\times (-0.5)-9{(-0.5)}^2\\ c=7.5\end{array}$

Thus, the value of constant c is $\underset{\_}{7.5}$.

The expressions of velocity profiles are as follows:

  $\begin{array}{l}{V}_1=6+9.5y-3y^2,0\le y\le -0.5\\ V_2=6+7.5y-9y^2,-0.5\le y\le 0\end{array}$

(b)

To determine

The viscosity ratio.

Explanation of Solution

The shear stress at the interface is unique.

  $\begin{array}{l}{\left({\mu }_1\frac{d{V}_1}{dy}\right)}_{y=0}={\left({\mu }_2\frac{d{V}_2}{dy}\right)}_{y=0}\\ \frac{{\mu }_1}{{\mu }_2}=\frac{{\left(\frac{d{V}_2}{dy}\right)}_{y=0}}{{\left(\frac{d{V}_1}{dy}\right)}_{y=0}}\\ \frac{{\mu }_1}{{\mu }_2}={\left(\frac{7.5-18y}{9.5-6y}\right)}_{y=0}\\ \frac{{\mu }_1}{{\mu }_2}=0.79\end{array}$

Thus, the viscosity ratio is $\underset{\_}{0.79}$.

(c)

To determine

The forces and their directions exerted by the liquids.

Explanation of Solution

Determine the force in the lower plate.

  $\begin{array}{l}{F}_L={\left({\mu }_2\frac{d{V}_2}{dy}\right)}_{y=-h}\\ A=\left(\frac{{10}^{-3}\text{N}\cdot \text{s}/{\text{m}}^2}{0.79}\right)\times {[}_{7.5}\times \left(4{\text{ m}}^2\right)\\ A=0.0835\text{ N to the right }\end{array}$

Thus, the force in the lower plate is $\underset{\_}{0.0835\text{ N to the right }}$.

Determine the force in the upper plate.

  $\begin{array}{l}{F}_U={\left({\mu }_1\frac{d{V}_1}{dy}\right)}_{y=h}\\ A=\left({10}^{-3}\text{N}\cdot \text{s}/{\text{m}}^2\right)\times {[}_{9.5}\times \left(4{\text{ m}}^2\right)\\ A=0.05\text{ N to the right }\end{array}$

Thus, the force in the upper plate is $\underset{\_}{0.05\text{ N to the right }}$.