Chapter 15, Problem 15.1P

Consider the incompressible viscous flow of air between two infinitely long parallel plates separated by a distance h. The bottom plate is stationary. and the top plate is moving at the constant velocity u. in the direction of the plate. Assume that no pressure gradient exists in the flow direction.

a. Obtain an expression for the variation of velocity between the plates.

b. If $T=\text{constant}=320\text{K},u_e=30\text{m/s}$, and $h=0.01\text{m}$, calculate the shear stress on the top and bottom plates.

To determine

The expression for the variation of velocity between the plates.

Answer to Problem 15.1P

The expression for the variation of velocity between the plates is $u=\frac{u_{e}y}{h}$ .

Explanation of Solution

Given:

The distance between the plates is $h$ .

The velocity of the top plate is $u_e$ .

Formula used:

The expression for the pressure gradient is given as,

  $\frac{dp}{dx}=\frac{d}{dx}\left(\mu \frac{du}{dy}\right)$

Here, $\frac{dp}{dx}$ is the pressure gradient, $\mu$ is the viscosity of the fluid, and $\frac{du}{dy}$ is the velocity component in $y$ direction.

Calculation:

The figure (1) is showing the incompressible flow of air between two infinitely long plates,

  

Figure (1)

The given flow is in the x direction only and the no pressure gradient in the direction of flow,

  $\frac{dp}{dx}=0$

The expression for the pressure gradient is given as,

  $\frac{dp}{dx}=\frac{d}{dx}\left(\mu \frac{du}{dy}\right)$

Substitute the value of pressure gradient,

  $\frac{d}{dx}\left(\mu \frac{du}{dy}\right)=0$

Integrate the above expression,

  $\mu \frac{du}{dy}=C_1$

Again integrate the above expression,

  $\mu u=C_{1}y+C_2$ ........ (1)

Apply boundary condition at $y=0$ , $u=0$ ,

Substitute the values in equation (1)

  $\begin{array}{l}\mu \cdot 0=C_1\cdot 0+C_2\\ C_2=0\end{array}$

At $y=h$ , $u=u_e$

Substitute the values in equation (1),

  $\begin{array}{l}\mu u_e=C_{1}h+0\\ C_1=\frac{\mu u_e}{h}\end{array}$

Substitute the values of $C_1$ and $C_2$ in equation (1),

  $\begin{array}{l}\mu u=\frac{\mu u_e}{h}y+0\\ u=\frac{u_{e}y}{h}\end{array}$ ...... (2)

Conclusion:

Therefore, the expression for the variation of velocity between the plates is $u=\frac{u_{e}y}{h}$ .

To determine

The shear stress on the top and the bottom plates.

Answer to Problem 15.1P

The shear stress on the top and the bottom plates is $5.8\times {10}^{-2}\text{N}/{\text{m}}^{\text{2}}$ .

Explanation of Solution

Given:

The temperature of the fluid is $T=320\text{K}$ .

The velocity of the top plate is $u_e=30\text{m}/\text{s}$ .

The distance between the plates is $h=0.1\text{m}$ .

Formula used:

The expression to calculate the shear stress is given as,

  $\tau =\mu \frac{du}{dy}$

Here, $\tau$ is the stress, $\mu$ is the viscosity, and $\frac{du}{dy}$ is the velocity component in $y$ direction

The expression of the relation between the viscosity and the temperature is given as,

  $\frac{\mu }{{\mu }_0}={\left(\frac{T}{T_0}\right)}^{\frac{3}{2}}\left(\frac{T_0+110}{T+110}\right)$

Here, $\mu$ is the viscosity, $T_0$ is standard atmospheric temperature and ${\mu }_0$ is the viscosity constant.

Calculation:

It is known that the value of $T_0=288.16\text{K}$ and ${\mu }_0=1.7895\times {10}^{-5}\text{Pa}\cdot \text{s}$ .

The expression of the relation between the viscosity and the temperature is given as,

  $\frac{\mu }{{\mu }_0}={\left(\frac{T}{T_0}\right)}^{\frac{3}{2}}\left(\frac{T_0+110}{T+110}\right)$

Substitute the values in above expression

  $\begin{array}{c}\frac{\mu }{1.7895\times {10}^{-5}\text{Pa}\cdot \text{s}}={\left(\frac{320\text{K}}{288.16\text{K}}\right)}^{\frac{3}{2}}\left(\frac{288.16\text{K}+110}{320\text{K}+110}\right)\\ \mu =1.94\times {10}^{-5}\text{Pa}\cdot \text{s}\left(\frac{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{1\text{Pa}}\right)\\ \mu =1.94\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}\end{array}$

The expression to calculate the stress is given as,

  $\tau =\mu \frac{du}{dy}$

Substitute the values from equation (2),

  $\begin{array}{l}\tau =\mu \frac{d}{dy}\left(\frac{u_{e}y}{h}\right)\\ \tau =\frac{\mu u_e}{h}\end{array}$

Substitute the values in above expression,

  $\begin{array}{l}\tau =\frac{1.94\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}}{0.01\text{m}}\times 30\text{m}/\text{s}\\ \tau =5.82\times {10}^{-2}\text{m}/\text{s}\end{array}$

The value of shear stress is constant. Thus, the shear stress will be same on top and bottom of the plates.

Conclusion:

Therefore, the shear stress on the top and the bottom plates is $5.8\times {10}^{-2}\text{N}/{\text{m}}^{\text{2}}$ .