Chapter 2, Problem 81P

A thin $30-\text{cm}\times 30-\text{cm}$ flat plate is pulled at 3 m/s horizontally through a 3.6-mm-thick oil layer sandwiched between two plates, one stationary and the other moving at a constant velocity of 0.3 m/s, as shown in Fig. P2-77. The dynamic viscosity of the oil is $0.027\text{Pa}\cdot \text{s}$ . Assuming the velocity in each oil layer to vary linearly, (a) plot the velocity profile and find the location where the oil velocity is zero and (b) determine the force that needs to be applied on the plate to maintain this motion. Fixed wall

FIGURE P2-77

To determine

(a)

The location where the oil velocity is zero.

Answer to Problem 81P

The location where the oil velocity is zero is $2.364\text{m}$.

Explanation of Solution

Given information:

The cross section of the plate is $30\text{cm}\times 30\text{cm}$, the velocity of the plate is $3\text{m}/\text{s}$, thickness of oil layer is $3.6\text{mm}$, velocity of other plate $0.3\text{m}/\text{s},$ and the dynamic viscosity of the oil is $0.027\text{Pa}\cdot \text{s}$.

The following figure gives the velocity profile of the plate:

Figure-(1)

Write the expression for the viscous force.

  $\begin{array}{c}{F}_2=\eta A\frac{d{V}_2}{d{z}_2}\\ =\eta A\frac{\left(V-V_w\right)}{d{z}_2}\end{array}$   ..... (I)

Here, the viscous force is $F_2$, the co-efficient of viscosity is $\eta$, the velocity of the plate is $V$, the velocity of the oil is $V_w$, the thickness of the oil layer is $d{z}_2$.

Write the expression for the thickness of the oil film.

  $dz=\eta A\frac{\left(V-V_w\right)}{F_2}$   ..... (II)

Calculation:

Substitute $0.027\text{Pa}\cdot \text{s}$ for $\eta$, $3\text{cm}\times \text{3}\text{cm}$ for $A$ and $3\text{m}/\text{s}$ for $V$, $-0.3\text{m}/\text{s}$ for $V_w$ and $3.6\text{mm}$ for $dz$ in Equation (I)

  $\begin{array}{c}{F}_2=\left(0.027\text{Pa}\cdot \text{s}\right)\left(30\text{cm}\times \text{30}\text{cm}\right)\frac{\left(\left(3\text{m}/\text{s}\right)-\left(-0.3\text{m}/\text{s}\right)\right)}{3.6\text{mm}}\\ =\left(0.027\text{Pa}\cdot \text{s}\right)\left(30\text{cm}\times \text{30}\text{cm}\right){\left(\frac{1\text{m}}{100\text{cm}}\right)}^2\frac{\left(\left(3\text{m}/\text{s}\right)-\left(-0.3\text{m}/\text{s}\right)\right)}{\left(3.6\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =\left(3.084\text{Pa}\cdot {\text{m}}^2\right)\left(\frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)\\ =3.084\text{N}\end{array}$

Substitute $0.027\text{Pa}\cdot \text{s}$ for $\eta$, $3\text{cm}\times \text{3}\text{cm}$ for $A$ and $3\text{m}/\text{s}$ for $V$, $0\text{m}/\text{s}$ for $V_w$ and $3.084\text{N}$ for $F_2$ in Equation (II)

  $\begin{array}{c}dz=\left(0.027\text{Pa}\cdot \text{s}\right)\left(30\text{cm}\times 30\text{cm}\right)\frac{\left(3\text{m}/\text{s}\right)}{\left(3.084\text{N}\right)}\\ =\left(0.027\text{Pa}\cdot \text{s}\right)\left(\frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)\left(30\text{cm}\times 30\text{cm}\right){\left(\frac{1\text{m}}{100\text{cm}}\right)}^2\frac{\left(3\text{m}/\text{s}\right)}{\left(3.084\text{N}\right)}\\ =\left(0.002364\text{m}\right)\left(\frac{1000\text{mm}}{1\text{m}}\right)\\ =2.364\text{m}\end{array}$

Conclusion:

The location where the oil velocity is zero is $2.364\text{m}$.

To determine

(b)

The force required to maintain the motion.

Answer to Problem 81P

The force required to maintain the motion is $10.374\text{N}$.

Explanation of Solution

Write the expression for the viscous force.

  $F_1=\eta A\frac{d{V}_1}{d{z}_1}$   ..... (III)

Here, the viscous force is $F_1$, the co-efficient of viscosity is $\eta$, the change in velocity of upper portion of the plate is $d{V}_1$, the thickness of the oil layer is $d{z}_2$.

Write the expression for the total applied force.

  $F=F_1+F_2$   ..... (IV)

Here, the total force is $F$.

Calculation:

Substitute $0.027\text{Pa}\cdot \text{s}$ for $\eta$, $3\text{cm}\times \text{3}\text{cm}$ for $A$ and $3\text{m}/\text{s}$ for $d{V}_1$, and $1\text{mm}$ for $dz$ in Equation (III)

  $\begin{array}{c}{F}_1=\left(0.027\text{Pa}\cdot \text{s}\right)\left(30\text{cm}\times \text{30}\text{cm}\right)\frac{\left(3\text{m}/\text{s}\right)}{1\text{mm}}\\ =\left(0.027\text{Pa}\cdot \text{s}\right)\left(30\text{cm}\times \text{30}\text{cm}\right){\left(\frac{1\text{m}}{100\text{cm}}\right)}^2\frac{\left(3\text{m}/\text{s}\right)}{\left(1\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =\left(7.29\text{Pa}\cdot {\text{m}}^2\right)\left(\frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)\\ =7.29\text{N}\end{array}$

Substitute $7.29\text{N}$ for $F_1$ and $3.084\text{N}$ for $F_2$.

  $\begin{array}{c}F=\left(7.29\text{N}\right)\text{+}\left(\text{3}\text{.084}\text{N}\right)\\ =10.374\text{N}\end{array}$

Conclusion:

The force required to maintain the motion is $10.374\text{N}$.