Velocity Profiles for Flow Between Parallel Plates. In Example 8.3-2, a fluid is flowing between vertical parallel plates with one plate moving. Do as follows: 1. Determine the average velocity and the maximum velocity. 2. Make a sketch of the velocity profile for three cases where the surface is moving upward, moving downward, and stationary.

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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Velocity Profiles for Flow Between Parallel Plates. In Example 8.3-2, a fluid is flowing between vertical parallel plates with one plate moving. Do as follows:

1. Determine the average velocity and the maximum velocity.

2. Make a sketch of the velocity profile for three cases where the surface is moving upward, moving downward, and stationary.

**Example 8.3-2: Laminar Flow Between Vertical Plates with One Plate Moving**

A Newtonian fluid is confined between two parallel and vertical plates as shown in Fig. 8.3-2 (W6). The surface on the left is stationary, and the other is moving vertically at a constant velocity \( v_o \). Assuming that the flow is laminar, solve for the velocity profile.

**Solution:** 

The equation to use is the Navier–Stokes equation for the y coordinate, Eq. (8.2-37):

\[
\rho \left( \frac{\partial v_y}{\partial t} + v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} + v_z \frac{\partial v_y}{\partial z} \right) = \mu \left( \frac{\partial^2 v_y}{\partial x^2} + \frac{\partial^2 v_y}{\partial y^2} + \frac{\partial^2 v_y}{\partial z^2} \right) - \frac{\partial p}{\partial y} + \rho g_y
\]

At steady-state, \(\partial v_y/\partial t = 0\). The velocities \( v_x \) and \( v_z = 0 \). Also, \(\partial v_y/\partial y = 0\) from the continuity equation, \(\partial v_y/\partial z = 0\), and \(\rho g_y = -pg\). The partial derivatives become derivatives and Eq. (8.2-37) becomes

\[
\mu \frac{d^2 v_y}{dx^2} - pg = 0 \quad (8.3-10)
\]

This is similar to Eq. (8.3-2) in Example 8.3-1. The pressure gradient \(-dp/dy\) is constant. Integrating Eq. (8.3-10) once yields
Transcribed Image Text:**Example 8.3-2: Laminar Flow Between Vertical Plates with One Plate Moving** A Newtonian fluid is confined between two parallel and vertical plates as shown in Fig. 8.3-2 (W6). The surface on the left is stationary, and the other is moving vertically at a constant velocity \( v_o \). Assuming that the flow is laminar, solve for the velocity profile. **Solution:** The equation to use is the Navier–Stokes equation for the y coordinate, Eq. (8.2-37): \[ \rho \left( \frac{\partial v_y}{\partial t} + v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} + v_z \frac{\partial v_y}{\partial z} \right) = \mu \left( \frac{\partial^2 v_y}{\partial x^2} + \frac{\partial^2 v_y}{\partial y^2} + \frac{\partial^2 v_y}{\partial z^2} \right) - \frac{\partial p}{\partial y} + \rho g_y \] At steady-state, \(\partial v_y/\partial t = 0\). The velocities \( v_x \) and \( v_z = 0 \). Also, \(\partial v_y/\partial y = 0\) from the continuity equation, \(\partial v_y/\partial z = 0\), and \(\rho g_y = -pg\). The partial derivatives become derivatives and Eq. (8.2-37) becomes \[ \mu \frac{d^2 v_y}{dx^2} - pg = 0 \quad (8.3-10) \] This is similar to Eq. (8.3-2) in Example 8.3-1. The pressure gradient \(-dp/dy\) is constant. Integrating Eq. (8.3-10) once yields
The image presents a mathematical derivation related to fluid mechanics or a similar field, showing steps to solve a differential equation. Here's the transcribed content for an educational purpose:

1. The differential equation is given as:

   \[
   \frac{dv_y}{dx} - \frac{x}{\mu} \left( \frac{dp}{dy} + \rho g \right) = C_1
   \]

   (Equation 8.3-11)

2. Integrating the equation again provides:

   \[
   v_y - \frac{x^2}{2\mu} \left( \frac{dp}{dy} + \rho g \right) = C_1 x + C_2
   \]

   (Equation 8.3-12)

3. The boundary conditions are specified:

   - At \( x = 0 \), \( v_y = 0 \)
   - At \( x = H \), \( v_y = v_0 \)

   Solving for the constants gives:

   - \( C_1 = v_0 / H - (H / 2\mu) (dp/dy + \rho g) \)
   - \( C_2 = 0 \)

   Consequently, Equation 8.3-12 simplifies to:

   \[
   v_y = -\frac{1}{2\mu} \left( \frac{dp}{dy} + \rho g \right) (Hx - x^2) + v_0 \frac{x}{H}
   \]

   (Equation 8.3-13)

This derivation involves integrating a differential equation with respect to specified variables under given boundary conditions to express the velocity component \( v_y \) in terms of position \( x \), pressure gradient \( \frac{dp}{dy} \), and other parameters such as fluid viscosity \( \mu \) and gravitational effect \( \rho g \).
Transcribed Image Text:The image presents a mathematical derivation related to fluid mechanics or a similar field, showing steps to solve a differential equation. Here's the transcribed content for an educational purpose: 1. The differential equation is given as: \[ \frac{dv_y}{dx} - \frac{x}{\mu} \left( \frac{dp}{dy} + \rho g \right) = C_1 \] (Equation 8.3-11) 2. Integrating the equation again provides: \[ v_y - \frac{x^2}{2\mu} \left( \frac{dp}{dy} + \rho g \right) = C_1 x + C_2 \] (Equation 8.3-12) 3. The boundary conditions are specified: - At \( x = 0 \), \( v_y = 0 \) - At \( x = H \), \( v_y = v_0 \) Solving for the constants gives: - \( C_1 = v_0 / H - (H / 2\mu) (dp/dy + \rho g) \) - \( C_2 = 0 \) Consequently, Equation 8.3-12 simplifies to: \[ v_y = -\frac{1}{2\mu} \left( \frac{dp}{dy} + \rho g \right) (Hx - x^2) + v_0 \frac{x}{H} \] (Equation 8.3-13) This derivation involves integrating a differential equation with respect to specified variables under given boundary conditions to express the velocity component \( v_y \) in terms of position \( x \), pressure gradient \( \frac{dp}{dy} \), and other parameters such as fluid viscosity \( \mu \) and gravitational effect \( \rho g \).
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