Chapter 2, Problem 132P

Oil of viscosity $\mu =0.0357\text{Pa}\cdot \text{s}$ and density $\rho =0.796{\text{kg/m}}^{\text{3}}$ is sandwiched in the small gap between two very large parallel flat plates. A third plate of surface area $A=20.0\text{cm}\times 20.2\text{cm}$ (on one side) is dragged through the oil steady velocity $V=1.00\text{m/s}$ to the right as sketched.
The top plate is stationary, but the bottom plate is moving at velocity $V=0.300\text{m/s}$ to the left as sketched. The heights are $h_1=1.00\text{mm}$ and $h_2=165\text{mm}$ . The force required to pull the plate through the oil is F. (a) Sketch the velocity profiles and calculate the distance $y_A$ where the velocity is zero. Hint: Since the gaps are small and the oil is very viscous, the velocity profiles are linear in both gaps. Use the no-slip conditions at the walls the determine the velocity profile in each gap. (b) Calculate force F in newtons (N) required to keep the middle plate moving at constant speed.

To determine

(a)

Velocity profiles between the plates and the distance where velocity is zero.

Answer to Problem 132P

The below figure shows the velocity profile.

The velocity is zero from bottom at the height of $2.6\text{mm}$.

Explanation of Solution

Given information:

Dynamic viscosity is $0.0357\text{Pa}\cdot \text{s}$, density of the fluid is $0.796\text{kg}/{\text{m}}^{\text{3}}$, surface area of plate is $20.0\text{cm}\times 20.0\text{cm}$, steady velocity of middle plate is $1.00\text{m}/\text{s}$, velocity of top plate is $0,$ and velocity of bottom plate is $0.300\text{m}/\text{s}$.

The velocities will be zero at the boundaries of the plate due to no slip condition.

Write the expression for the velocity.

  $u=-\frac{1}{2\mu }\left(\frac{dP}{dh}\right)\left(h{y}_A-h^2\right)$....... (I)

Here, the pressure gradient is $\left(\frac{dP}{dh}\right)$, the distance between plates is $h,$ dynamic viscosity of the oil is $\mu ,$ and the distance from bottom where velocity is zero is $y_A$

Calculation:

Substitute $0$ for the $u$ in Equation (I).

  $\begin{array}{l}0=-\frac{1}{2\mu }\left(\frac{dP}{dh}\right)\left(h{y}_A-h^2\right)\\ \left(h{y}_A-h^2\right)=0\\ y_A=h\\ y_A=2.6\text{mm}\end{array}$

The below figure shows the velocity profile.

Figure-(1)

Conclusion:

The velocity is zero from bottom at the height of $2.6\text{mm}$.

To determine

(b)

The force required for moving the middle plate at constant speed.

Answer to Problem 132P

The force required for moving the middle plate at constant speed is $2.552\text{N}$.

Explanation of Solution

Write the expression for the force required.

  $F=\tau \cdot A$....... (II)

Here, shear stress is $\tau$, area is $A,$ and force is $F$.

Write the expression for the shear stress.

  $\tau =\mu \left(\frac{du}{dy}\right)$....... (III)

Here, shear stress is $\tau$, dynamic viscosity is $\mu ,$ and velocity gradient is $\left(du/dy\right)$.

Substitute $\mu \left(\frac{du}{dy}\right)$ for $\tau$ in Equation (II).

  $F=\mu \left(\frac{du}{dy}\right)A$....... (IV)

Write the expression for the force required between first and second plate.

  $F_{\left(1-2\right)}=\mu {\left(\frac{du}{dy}\right)}_{\left(1-2\right)}A$....... (V)

Here, velocity gradient between plate one and two is ${\left(\frac{du}{dy}\right)}_{\left(1-2\right)}$.

Write the expression for the force required between the second and third plate.

  $F_{\left(2-3\right)}=\mu {\left(\frac{du}{dy}\right)}_{\left(2-3\right)}A$....... (VI)

Here, velocity gradient between plate two and three is ${\left(\frac{du}{dy}\right)}_{\left(2-3\right)}$.

Write the expression for the total force.

  $F=\mu \left({\left(\frac{du}{dy}\right)}_{\left(1-2\right)}+{\left(\frac{du}{dy}\right)}_{\left(2-3\right)}\right)A$....... (VII)

Here, total force is $F$.

Write the expression for velocity gradient between plate one and two.

  ${\left(\frac{du}{dy}\right)}_{\left(1-2\right)}=\left(\frac{V-V_1}{h_1}\right)$....... (VIII)

Here, steady velocity of plate two is $V$, velocity of top plate is $V_1,$ and the gap between plate one and two is $h_1$.

Write the expression for velocity gradient between plate two and three.

  ${\left(\frac{du}{dy}\right)}_{\left(2-3\right)}=\left(\frac{V-V_2}{h_2}\right)$....... (IX)

Here, steady velocity of plate two is $V$, velocity of bottom plate is $V_2,$ and the gap between plate two and three is $h_2$.

Substitute $\left(\frac{V-V_1}{h_1}\right)$ for ${\left(\frac{du}{dy}\right)}_{\left(1-2\right)}$ and $\left(\frac{V-V_2}{h_2}\right)$ for ${\left(\frac{du}{dy}\right)}_{\left(2-3\right)}$ in Equation (VII).

  $F=\mu \left(\left(\frac{V-V_1}{h_1}\right)+\left(\frac{V-V_2}{h_2}\right)\right)A$....... (X)

Calculation:

Substitute $0.0357\text{Pa}\cdot \text{s}$ for $\mu$, $400{\text{cm}}^2$ for $A$, $0$ for $V_1$, $-0.300\text{m}/\text{s}$ for $V_2$, $1.00\text{mm}$ for $h_1$ and $1.65\text{mm}$ for $h_2$ in Equation (X).

  $\begin{array}{c}F=\left(0.0357\text{Pa}\cdot \text{s}\right)\left[\begin{array}{l}\left(\frac{1\text{m}/\text{s}-0}{1.0\text{mm}}\right)\\ +\left(\frac{1\text{m}/\text{s}-\left(-0.300\text{m}/\text{s}\right)}{1.65\text{mm}}\right)\end{array}\right]\left(20\text{cm}\times \text{20}\text{cm}\right)\\ =\left(0.0357\text{Pa}\cdot \text{s}\right)\left[\begin{array}{l}\left(\frac{1\text{m}/\text{s}-0}{1.0\text{mm}\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)}\right)\\ +\left(\frac{1\text{m}/\text{s}-\left(-0.300\text{m}/\text{s}\right)}{1.65\text{mm}\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)}\right)\end{array}\right]\left(400{\text{cm}}^2\right)\\ =0.0357\text{Pa}\cdot \text{s}\left(\frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)\left[\left({10}^3+0.7878\times {10}^3\right){\text{s}}^{-1}\right]\left(400{\text{cm}}^2\left(\frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}\right)\right)\\ =1.428\times {10}^{-3}\left(1787.8\text{N}\right)\end{array}$

  $F=2.552\text{N}$

Conclusion:

The force required for moving the middle plate at constant speed is $2.552\text{N}$.