Also prove that $U_{max}=\frac{3}{2} u_{avg} $

Solution :

Considering, Plates are at a distance (B) apart.

Consider a fluid element as shown with sides (dx, dy, dz).

The flow is a steady and uniform. There is no acceleration.

Sum of all forces in the direction of motion is zero.

Pressure Force + Shear Forces = 0.

$[Pdydz – (P + (\frac{∂p}{∂x})dx)dydz] +[(τ+(\frac{∂τ}{∂y})dy) dxdz – τ dxdz] = 0 $

Simplifying,

$\begin {aligned}–(\frac{∂p}{∂x})dxdydz + (\frac{∂τ}{∂y})dxdydz = 0 \end{aligned}$

Dividing by dxdydz, The volume of the parallel piped,

$$\begin {aligned} (\frac{∂p}{∂x}) = (\frac{∂τ}{∂y}) \end{aligned}$$

According to the Newton's law of viscosity,

$$\begin {aligned} τ = µ \frac {du}{dy} \end{aligned} $$

$$\begin {aligned} \therefore \frac{∂p}{∂x} &= µ \frac{∂^2u}{∂y^2} \\ \therefore \frac{∂^2u}{∂y^2} &= \frac{1}{\mu} \frac{∂p}{∂x} \end{aligned}$$

Since $ \frac{∂p}{∂x}$ is independent of y , integrating the above equation, we get

$$\begin {aligned} \frac{∂u}{∂y} = \frac{1}{\mu} \frac{∂p}{∂x} y + c_1 \end{aligned}$$

Integrating again,

$$\begin {aligned} u = \frac{1}{2µ} (\frac{∂p}{∂x})y^2 + c_1y + c_2 \end{aligned} $$

$c_1$ and $ c_2 $ are constants of integration.

At $y = 0, u = 0$

$\therefore c_2 = 0$ ; At $y=B, u=0 $

$$\begin {aligned} 0 = \frac{1}{2µ} (\frac{∂p}{∂x})B^2 + c_1B \\ c_1 =\frac{1}{2µ} (\frac{-∂p}{∂x}) B \end{aligned}$$

Substituting,

$$\begin {aligned} u = \frac{1}{2µ} (\frac{-∂p}{∂x}) [By – y^2] \end{aligned}$$

This equation shows that the velocity distribution for steady laminar flow between fixed parallel plates is parabolic.

$(\frac{∂p}{∂x}) $– Pressure decreases in the direction of flow and $(\frac{-∂p}{∂x}) $ is a positive quantity.

Max. Velocity occurs mid-way between the plates and can be obtained using

$y=\frac{B}{2}$

$$\begin {aligned} U_{max}=\frac{ B^2}{8µ} (\frac{-∂p}{∂x}) \end{aligned}$$

Discharge through the strip per unit width

$dq =$ Area × Velocity

$$\begin {aligned} dq= dy × 1× \frac{1}{2µ} (\frac{-∂p}{∂x}) [By – y^2] \end{aligned}$$

Discharge (q) per unit width of plate

$$\begin {aligned} q =\int_0^B \frac{1}{2µ} (\frac{-∂p}{∂x}) [By – y^2] dy \end{aligned} $$

Integrating above equation, we get

$$\begin {aligned} q =\frac{1}{2µ} (\frac{-∂p}{∂x}) [\frac{B^3}{2} – \frac{B^3}{3}] \end{aligned}$$

$$\begin {aligned} q= \frac{B^3}{12µ} (\frac{–∂p}{∂x}) \end{aligned} $$

Average velocity = $u_{avg }$= $\frac{q}{Area} $

Where, Area = B × 1;

$$\begin {aligned} u_{avg} =\frac{B^2}{12µ} (\frac{–∂p}{∂x}) \end{aligned}$$

Since $\begin {aligned} U_{max}=\frac{ B^2}{8µ} (\frac{-∂p}{∂x})\end{aligned} $

$$\begin {aligned} u_{avg}= \frac{2}{3} U_{max} \end{aligned} $$

In the case of steady laminar flow between two fixed parallel plates, the average velocity is equal to (2/3) maximum velocity