Chapter 7, Problem 136P

To determine

(a)

The six appropriate boundary conditions on both velocity and pressure.

Answer to Problem 136P

The first boundary condition is $u_1=0$.

The second boundary condition is $u_2=h_1+h_2$.

The third boundary condition is $u_2=u_1$.

The fourth boundary condition is ${\mu }_1\frac{d{u}_1}{dz}={\mu }_2\frac{d{u}_2}{dz}$.

The fifth boundary condition is $P=P_0$.

The sixth boundary condition

is $P_1=P_2$.

Explanation of Solution

Given information:

The following figure shows that two parallel flat plates.

  

  Figure-( 1)

Assume, at the point of wall and fluid, the velocity of the fluid is equal to zero.

Write the expression for velocity of the fluid 1,

  $u_1=0$   ....... (I)

Here, the velocity of fluid 1 is $u_1$.

Assume, the velocity of the fluid 2 at the free surface of the wall is equal to the velocity of the moving plates.

Write the expressions for the velocity of fluid 2.

  $\begin{array}{c}{u}_2=V\\ =h_1+h_2\end{array}$   ....... (II)

Here, the velocity of fluid 2 is $u_2$, the velocity of the flow is $V$, the height of the fluid 1 from the wall is $h_1$ and the height of the fluid 2 from the interface is $h_2$.

Write the expression for velocity at interface.

  $u_2=u_1$   ....... (III)

Write the expression for rate of shear stress.

  $\tau =\mu \frac{du}{dz}$   ....... (IV)

Here, the kinematic coefficient of fluid is $\mu$ and the rate of shear stress is $\tau$.

Write the expression for the shear stress acting on fluid 1.

  ${\tau }_1={\mu }_1\frac{d{u}_1}{dz}$   ...... (V)

Here, the kinematic coefficient of fluid 1 is ${\mu }_1$ and the shear stress acting on the fluid 1 is ${\tau }_1$.

Write the expression for the shear stress acting on fluid 2.

  ${\tau }_2={\mu }_2\frac{d{u}_2}{dz}$   ....... (VI)

Here, the kinematic coefficient of fluid 2 is ${\mu }_2$ and the shear stress acting on the fluid 2 is ${\tau }_2$.

Write the expression for the rate of shear stress at interface.

  ${\mu }_1\frac{d{u}_1}{dz}={\mu }_2\frac{d{u}_2}{dz}$   ...... (VII)

Write the expression for pressure at the bottom of the flow,

  $P=P_0$   ...... (VIII)

Here, the pressure is $P$ and the pressure at the bottom of the flow is $P_0$.

Write the expression for the pressure at the interface of fluid 1.

  $P=P_1$   ...... (IX)

Here, the pressure at the fluid 1 is $P_1$.

Write the expression for the pressure at the interface of fluid 2.

  $P=P_2$   ....... (X)

Here, the pressure at the fluid 1 is $P_2$.

Assume, at the interface of the fluid the pressure cannot have discontinuity and the surface is ignored.

Write the expression for the pressure at interface of fluid.

  $P_1=P_2$   ...... (XI)

Conclusion:

The first boundary condition is $u_1=0$.

The second boundary condition is $u_2=h_1+h_2$.

The third boundary condition is $u_2=u_1$.

The fourth boundary conditions is ${\mu }_1\frac{d{u}_1}{dz}={\mu }_2\frac{d{u}_2}{dz}$.

The fifth boundary condition is $P=P_0$.

The sixth boundary condition

is $P_1=P_2$.

To determine

(b)

The expressions for the velocity of fluid 1 and 2.

Answer to Problem 136P

The expression for the velocity of fluid 1 is $\left(\frac{{\mu }_2{C}_3}{{\mu }_1}\right)z$.

The expression for the velocity of fluid 2 is $V+\frac{u_1^2}{{\mu }_2{h}_1}\left(z+\left(h_1+h_2\right)\right)$.

Explanation of Solution

Write the expression for $x$ momentum equation of flow for fluid 1.

  $\frac{d^2{u}_1}{d{z}^2}=0$   .......(XII)

Here, the velocity of flow for fluid 1 is $u_1$.

Write the expression for $x$ momentum equation of flow for fluid 2.

  $\frac{d^2{u}_2}{d{z}^2}=0$   ....... (XIII)

Here, the velocity of flow for fluid 2 is $u_2$.

Calculation:

Integrate Equation (XIII) with respect to $z$.

  $\begin{array}{c}{\displaystyle \int \frac{d^2{u}_1}{d{z}^2}}={\displaystyle \int 0}\\ \frac{d{u}_1}{dz}=C_1\end{array}$   ...... (XIV)

Here, the constant is $C_1$.

Integrate Equation (XIII) with respect to $z$.

  $\begin{array}{c}{\displaystyle \int \frac{d{u}_1}{dz}}={\displaystyle \int C_1}\\ u_1=C_1{z}^{0+1}+C_2\\ u_1=C_{1}z+C_2\end{array}$   ....... (XV)

Here, the constant is $C_2$ and the distance is $z$.

Integrate Equation (XIV) with respect to $z$.

  $\begin{array}{c}{\displaystyle \int \frac{d^2{u}_2}{d{z}^2}}={\displaystyle \int 0}\\ \frac{d{u}_2}{dz}=C_3\end{array}$   ...... (XVI)

Here, the constant is $C_3$.

Integrate Equation (XIV) with respect to $z$.

  $\begin{array}{c}{\displaystyle \int \frac{d{u}_2}{dz}}={\displaystyle \int C_3}\\ u_2=C_3{z}^{0+1}+C_4\\ u_2=C_{3}z+C_4\end{array}$   ....... (XVII)

Here, the constant is $C_4$.

Substitute $0$ for $z$ and $0$ for $u_1$ in Equation (XV).

  $\begin{array}{l}0=C_1\left(0\right)+C_2\\ 0=0+C_2\\ C_2=0\end{array}$

Substitute $0$ for $C_2$ and $h_1$ for $z$ in Equation (XV).

  $\begin{array}{l}{u}_1=C_1{h}_1+0\\ u_1=C_1{h}_1\end{array}$...  ....... (XVIII)

Substitute $h_1+h_2$ for $z$ and $V$ for $u_2$ in Equation (XVII).

  $V=C_3\left(h_1+h_2\right)+C_4$  .......(XIX)

Substitute $h_1$ for $z$ and $u_1$ for $u_2$ in Equation (XVII).

  $u_1=C_3{h}_1+C_4$   ....... (XX)

Substitute $C_{3}z$ for $u_2$ and $C_{1}z$ for $u_1$ in Equation (VII).

  ${\mu }_1\frac{d\left(C_{1}z\right)}{dz}={\mu }_2\frac{d\left(C_{3}z\right)}{dz}$   ...... (XXI)

Differentiate Equation (XXI) with respect to $z$.

  $\begin{array}{l}{\mu }_1{C}_1{z}^{1-1}={\mu }_2{C}_3{z}^{1-1}\\ {\mu }_1{C}_1={\mu }_2{C}_3\\ C_1=\frac{{\mu }_2{C}_3}{{\mu }_1}\end{array}$   ....... (XXII)

Substitute $\frac{{\mu }_2{C}_3}{{\mu }_1}$ for $C_1$ in Equation (XVIII).

  $\begin{array}{l}{u}_1=\frac{{\mu }_2{C}_3}{{\mu }_1}\times h_1\\ \frac{u_1^2}{{\mu }_2{h}_1}=C_3\\ C_3=\frac{u_1^2}{{\mu }_2{h}_1}\end{array}$   ....... (XXIII)

Substitute $\frac{u_1^2}{{\mu }_2{h}_1}$ for $C_3$ in Equation (XIX).

  $\begin{array}{l}V=\frac{u_1^2}{{\mu }_2{h}_1}\left(h_1+h_2\right)+C_4\\ C_4=V-\frac{u_1^2}{{\mu }_2{h}_1}\left(h_1+h_2\right)\end{array}$   ....... (XXIV)

Substitute $V-\frac{u_1^2}{{\mu }_2{h}_1}\left(h_1+h_2\right)$ for $C_4$ and $\frac{u_1^2}{{\mu }_2{h}_1}$ for $C_3$ in Equation (XVII).

  $\begin{array}{l}{u}_2=\frac{u_1^2}{{\mu }_2{h}_1}z+V-\frac{u_1^2}{{\mu }_2{h}_1}\left(h_1+h_2\right)\\ u_2=V+\frac{u_1^2}{{\mu }_2{h}_1}\left(z+\left(h_1+h_2\right)\right)\end{array}$

Substitute $\frac{{\mu }_2{C}_3}{{\mu }_1}$ for $C_1$ and $0$ for $C_2$ in Equation (XV).

  $\begin{array}{l}{u}_1=\left(\frac{{\mu }_2{C}_3}{{\mu }_1}\right)z+0\\ u_1=\left(\frac{{\mu }_2{C}_3}{{\mu }_1}\right)z\end{array}$

Conclusion:

The expression for the velocity of fluid 1 is $\left(\frac{{\mu }_2{C}_3}{{\mu }_1}\right)z$.

The expression for the velocity of fluid 2 is $V+\frac{u_1^2}{{\mu }_2{h}_1}\left(z+\left(h_1+h_2\right)\right)$.

To determine

(c)

The expressions for pressure of fluid 1 and 2.

Answer to Problem 136P

The expression for the pressure of fluid 1 is $-{\rho }_{1}gz+P_o$.

The expression for the pressure of fluid 2 is $P_o+g{\rho }_2{h}_1-gz\left({\rho }_1+{\rho }_2\right)$.

Explanation of Solution

Write the expression for $z$ momentum equation of flow of fluid 1.

  $\frac{d{P}_1}{dz}=-{\rho }_{1}g$   ...... (XXV)

Here, the density of the fluid 1 is ${\rho }_1$ and the acceleration due to gravity is $g$.

Write the expression for $z$ momentum equation of flow of fluid 2.

  $\frac{d{P}_2}{dz}=-{\rho }_{2}g$   ...... (XXVI)

Here, the density of the fluid 2 is ${\rho }_2$.

Calculation:

Integrate Equation (XXV) with respect to $z$.

  $\begin{array}{l}{\displaystyle \int \frac{d{P}_1}{dz}}={\displaystyle \int -{\rho }_{1}g}\\ P_1=-{\rho }_{1}gz+C_5\end{array}$   ...... (XXVII)

Here, the constant is $C_5$.

Substitute $P_o$ for $P_1$ and $0$ for $z$ in Equation (XXVII).

  $\begin{array}{l}{P}_o=-{\rho }_{1}g\left(0\right)+C_5\\ C_5=P_o\end{array}$   ...... (XXVIII)

Substitute $P_o$ for $C_5$ in Equation (XXVII).

  $P_1=-{\rho }_{1}gz+P_o$   ....... (XXIX)

Integrate Equation (XXVI) with respect to $z$.

  $\begin{array}{l}{\displaystyle \int \frac{d{P}_2}{dz}}={\displaystyle \int -{\rho }_{2}g}\\ P_2=-{\rho }_{2}gz+C_6\end{array}$   ....... (XXX)

Here, the constant is $C_6$.

Substitute $P_1$ for $P_2$ and $h_1$ for $z$ in Equation (XXX).

  $P_1=-{\rho }_{2}g{h}_1+C_6$   ....... (XXXI)

Substitute $-{\rho }_{1}gz+P_o$ for $P_1$ in Equation (XXXI).

  $\begin{array}{l}-{\rho }_{1}gz+P_o=-{\rho }_{2}g{h}_1+C_6\\ C_6=-{\rho }_{1}gz+P_o+{\rho }_{2}g{h}_1\\ C_6=P_o+g\left({\rho }_2{h}_1-{\rho }_{1}z\right)\end{array}$   ....... (XXXII)

Substitute $P_o+g\left({\rho }_2{h}_1-{\rho }_{1}z\right)$ for $C_6$ in Equation (XXXI).

  $\begin{array}{c}{P}_2=-{\rho }_{2}gz+P_o+g\left({\rho }_2{h}_1-{\rho }_{1}z\right)\\ =P_o+g{\rho }_2{h}_1-g{\rho }_{1}z-{\rho }_{2}gz\\ =P_o+g{\rho }_2{h}_1-gz\left({\rho }_1+{\rho }_2\right)\end{array}$

Conclusion:

The expression for the pressure of fluid 1 is $-{\rho }_{1}gz+P_o$.

The expression for the pressure of fluid 2 is $P_o+g{\rho }_2{h}_1-gz\left({\rho }_1+{\rho }_2\right)$.

To determine

(d)

The plot $u$ as a function of $z$ across the entire channel.

Answer to Problem 136P

The following Figure-(2) represents the velocities of fluid 1 and 2.

  

Explanation of Solution

Given information:

The fluid 1 be water and the fluid 2 be unused engine oil, at $80°C$. The height of fluid 1 is $5\text{mm}$ and the height of fluid 2 is $8\text{mm}$. The velocity of flow is $10\text{m}/\text{s}$.

The following figure shows that two parallel flat plates.

Write the expression for the velocity of fluid 1.

  $u_1=\left(\frac{{\mu }_{2}V}{{\mu }_2{h}_1+{\mu }_1{h}_2}\right)z$   ....... (XXXIII)

Here, the distance is $z$, the velocity of fluid 1 is $u_1$, the velocity of fluid 2 is $u_2$, the velocity of the flow is $V$, the height of the fluid 1 from the wall is $h_1$ and the height of the fluid 2 from the interface is $h_2$.

Write the expression for the velocity of fluid 2.

  $u_2=\left(\frac{V}{{\mu }_2{h}_1+{\mu }_1{h}_2}\right)\left[\mu \left(z-h_2\right)+{\mu }_2{h}_1\right]$  ...... (XXXIV)

Calculation:

Refer the Table-A-3E, "Properties of saturated water", to obtain the value of dynamic viscosity of water is $\text{0}\text{.2958}\text{kg}/\text{m}\cdot \text{s}$ corresponding to the temperature $80°C$.

Refer the Table-A-7E, "Properties of the atmosphere at high attitude", to obtain the value of dynamic viscosity of unused engine oil is $\text{0}\text{.355}\times {\text{10}}^{-3}\text{kg}/\text{m}\cdot \text{s}$ corresponding to the temperature $80°C$.

Substitute $\text{0}\text{.2958}\text{kg}/\text{m}\cdot \text{s}$ for ${\mu }_2$, $\text{0}\text{.355}\times {\text{10}}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for ${\mu }_1$, $10\text{m}/\text{s}$ for $V$, $5\text{mm}$ for $h_1$ and $8\text{mm}$ for $h_2$ in Equation (XXXIII).

Substitute $\text{0}\text{.2958}\text{kg}/\text{m}\cdot \text{s}$ for ${\mu }_2$, $\text{0}\text{.355}\times {\text{10}}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for ${\mu }_1$, $10\text{m}/\text{s}$ for $V$, $5\text{mm}$ for $h_1$ and $8\text{mm}$ for $h_2$ in Equation (XXXIV).

The following graph represents the velocities of fluid 1 and 2.

  

  Figure-(2)

In the fluid 1 the linear curve is increasing with respect to the velocity of flow and height of fluid 1. In the fluid 2 the curve is increasing with respect to the velocity of flow and height of fluid 2.

Conclusion:

The following Figure-(2) represents the velocities of fluid 1 and 2.