Chapter 2, Problem 126P

(a)

To determine

The value for constant $a$.

The value for constant $b$.

The value for constant $c$.

Answer to Problem 126P

The value for constant $a$ is $9.5$.

The value for constant $b$ is $6$.

The value for constant $c$ is $7.5$.

Explanation of Solution

Given information:

The constant velocity of the liquid is $10\text{m}/\text{s}$, the thickness of the layer of the fluid is $0.5\text{m}$, the velocity profile for layer 1 is $V_1=6+ay-3y^2$, at $-0.5\le y\le 0$ and the velocity profile for layer 2 is $V_2=b+cy-9y^2$, at $0\le y\le -0.5$.

Write the expression for the velocity for layer 1.

   $V_1=6+ay-3y^2$    ...... (I)

Here, the velocity of the liquid 1 is $V_1$, the thickness of the film is $y,$ and the constant is $a$.

Write the expression for the velocity for layer 2.

   $V_2=b+cy-9y^2$    ...... (II)

Here, the constants are $b$ and $c$.

Since, both the liquids are in contact with each other, the velocity of liquids is equal at the interface.

   $\begin{array}{l}{V}_1=V_2\\ 6+ay-3y^2=b+cy-9y^2\end{array}$    ...... (III)

Calculation:

Substitute $0.5$ for $y$ and $10\text{m}/\text{s}$ for $V_1$ in Equation (I).

   $\begin{array}{l}10\text{m}/\text{s}=6+a\left(0.5\right)-3{\left(0.5\right)}^2\\ 10\text{m}/\text{s}=5.25+0.5a\\ a=\frac{4.75}{0.5}\\ a=9.5\end{array}$

Substitute $0.5$ for $y$ and $0$ for $V_2$ in Equation (II).

   $\begin{array}{l}0=b+c\left(0.5\right)-9{\left(0.5\right)}^2\\ c=\frac{b-2.25}{0.5}\end{array}$    ...... (IV)

Substitute $\frac{b-2.25}{0.5}$ for $b$ and $0$ for $y$ in Equation (III).

   $\begin{array}{l}6+\left(9.5\right)0-3{\left(0\right)}^2=b+\left(\frac{b-2.25}{0.5}\right)\left(0\right)-9{\left(0\right)}^2\\ 6=b+0\\ b=6\end{array}$

Substitute $6$ for $b$ in Equation (IV).

   $\begin{array}{c}c=\frac{6-2.25}{0.5}\\ =\frac{3.25}{0.5}\\ =7.5\end{array}$

Conclusion:

Thus, the value for constant $a$ is $9.5$.

Thus, the value for constant $b$ is $6$.

Thus, the value for constant $c$ is $7.5$.

(b)

To determine

The expression for the viscosity ratio.

Answer to Problem 126P

The expression for the viscosity ratio $\frac{{\mu }_1}{{\mu }_2}=0.79$.

Explanation of Solution

Write the expression for the shear stress in the liquid at the interface.

   $\begin{array}{l}{\tau }_1={\tau }_2\\ {\mu }_1\frac{d{V}_1}{dy}={\mu }_2\frac{d{V}_2}{dy}\\ \frac{{\mu }_1}{{\mu }_2}=\frac{\frac{d{V}_2}{dy}}{\frac{d{V}_1}{dy}}\end{array}$    ...... (V)

Here, the viscosity of the liquid 1 is ${\mu }_1$ and the viscosity of the liquid 2 is ${\mu }_2$.

Calculation:

Substitute $b+cy-9y^2$ for $V_2$ and $6+ay-3y^2$ for $V_1$ in Equation (V).

   $\begin{array}{c}\frac{{\mu }_1}{{\mu }_2}=\frac{\frac{d\left(b+cy-9y^2\right)}{dy}}{\frac{d\left(6+ay-3y^2\right)}{dy}}\\ =\frac{c-18y}{a-6y}\end{array}$    ...... (VI)

Substitute $0$ for $y$, $9.5$ for $a$ and $7.5$ for $c$ in Equation (VI).

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

Conclusion:

Thus, the expression for the viscosity ratio $\frac{{\mu }_1}{{\mu }_2}=0.79$.

To determine

(c)

The force and direction in the lower plate.

The force and direction in the upper plate.

Answer to Problem 126P

The force and direction in the lower plate is $0.08316\text{N}$ in right direction.

The force and direction in the upper plate is $0.05\text{N}$ in right direction.

Explanation of Solution

Given information:

The viscosity of the layer 1 is ${10}^{-3}\text{Pa}\cdot \text{s}$ and the surface area of the each plate is $4{\text{m}}^2$.

Write the expression for the force in the lower plate.

   $F_L=A\times {\mu }_2\left(c-18y\right)$    ...... (VII)

Here, the surface area of the plates is $A$.

Write the expression for the force in the upper plate.

   $F_u=A\times {\mu }_1\left(a-6y\right)$    ...... (VIII)

Write the expression for the viscosity of the liquid 2.

   ${\mu }_2=\frac{{\mu }_1}{0.79}$    ...... (IX)

Calculation:

Substitute ${10}^{-3}\text{Pa}\cdot \text{s}$ for ${\mu }_1$ in Equation (IX).

   $\begin{array}{c}{\mu }_2=\frac{\left({10}^{-3}\text{Pa}\cdot \text{s}\right)}{0.79}\\ =\frac{\left({10}^{-3}\text{Pa}\cdot \text{s}\right)}{0.79}\\ =1.26\times {10}^{-3}\text{Pa}\cdot \text{s}\end{array}$

Substitute $4{\text{m}}^2$ for $A$, $1.26\times {10}^{-3}\text{Pa}\cdot \text{s}$ for ${\mu }_2$, $-0.5$ for $y$, and $7.5$ for $c$ in Equation (VII).

   $\begin{array}{c}{F}_L=\left(4{\text{m}}^2\right)\left(1.26\times {10}^{-3}\text{Pa}\cdot \text{s}\right)\left(7.5-18\left(-0.5\right)\right)\\ =\left(4{\text{m}}^2\right)\left(1.26\times {10}^{-3}\text{Pa}\cdot \text{s}\right)\left(16.5\right)\\ =0.08316\text{Pa}\cdot \text{s}\cdot {\text{m}}^2\left(\frac{\text{N}\cdot \text{s}/{\text{m}}^2}{\text{Pa}\cdot \text{s}}\right)\\ =0.08316\text{N}\end{array}$

Thus, the force and direction in the lower plate is $0.08316\text{N}$ in right direction.

Substitute $4{\text{m}}^2$ for $A$, ${10}^{-3}\text{Pa}\cdot \text{s}$ for ${\mu }_1$, $-0.5$ for $y$, and $9.5$ for $a$ in Equation (VIII).

   $\begin{array}{c}{F}_u=\left(4{\text{m}}^2\right)\left({10}^{-3}\text{Pa}\cdot \text{s}\right)\left(9.5-6\left(-0.5\right)\right)\\ =\left(4{\text{m}}^2\right)\left({10}^{-3}\text{Pa}\cdot \text{s}\right)\left(12.5\right)\\ =0.05\text{Pa}\cdot \text{s}\cdot {\text{m}}^2\left(\frac{\text{N}\cdot \text{s}/{\text{m}}^2}{\text{Pa}\cdot \text{s}}\right)\\ =0.05\text{N}\end{array}$

Thus, the force and direction in the upper plate is $0.05\text{N}$ in right direction.

Conclusion:

Thus, the force and direction in the lower plate is $0.08316\text{N}$ in right direction.

Thus, he force and direction in the upper plate is $0.05\text{N}$ in right direction.