Question

Two large vertical parallel plates separated by a gap of $d = 1mm$ have a highly viscous liquid of density $800kg/{m^3}$ and viscosity coefficient $\eta = 5$$poise$ flowing steadily under gravity in between the gap. Find the velocity gradient of flow near plates surface in ${s^{ - 1}}$. $\left( {1poise = 0.1Ns/{m^2}} \right)$.

Answer 1

Hint Using the formula of tangential stress between two layers of a liquid, we can calculate the velocity gradient of flow by substituting the values given.
Formula used
$f \propto \dfrac{{dv}}{{dx}}$ where $\dfrac{{dv}}{{dx}}$ is the velocity gradient.
$f = \eta \dfrac{{dv}}{{dx}}$ where $\eta $ is known as the coefficient of viscosity.
$F = A\eta \dfrac{{dv}}{{dx}}$ where $F$ is the tangential force and $A$ is the area over which it acts.

Complete step by step solution
When a liquid flows slowly over a fixed horizontal surface, its layer in contact with the fixed surface is stationary and the velocity of each layer increases with its distance fixed from the surface. The difference in velocities between the adjacent layers gives rise to internal friction. This internal friction is known as the viscosity of the liquid.
If $f$ represents the tangential stress between two layers of a liquid then by Newton’s law we have,
$f \propto \dfrac{{dv}}{{dx}}$ where $\dfrac{{dv}}{{dx}}$ is the velocity gradient.
Then, $f = \eta \dfrac{{dv}}{{dx}}$ where $\eta $ is known as the coefficient of viscosity.
Now $f = \dfrac{F}{A}$where $F$is the tangential force and $A$is the area over which it acts.
Therefore, $F = A\eta \dfrac{{dv}}{{dx}}$
Now, since here we have two vertical plates so we can write $F = mg$where $g$ is the acceleration due to gravity.
Substituting this value we get,
$
  mg = \eta A\dfrac{{dv}}{{dx}} \\
   \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{mg}}{{\eta A}} \\
 $
The distance between the plates is $1mm = {10^{ - 3}}m$
In place of mass we can write $m = \rho \times A \times d$ as $\rho $ is defined as mass per unit volume.
Then we get,
 $
  \dfrac{{dv}}{{dx}} = \dfrac{{\rho Adg}}{{\eta A}} \\
   \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{\rho dg}}{\eta } \\
$
$
   \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{800 \times {{10}^{ - 3}} \times 9.8}}{{5 \times 0.1}} \\
   \Rightarrow \dfrac{{dv}}{{dx}} = 15.68{s^{ - 1}} \\
$

Therefore, the velocity gradient of flow near plate surface is $15.68{s^{ - 1}}$

Note When a liquid is at rest, we do not observe any shape rigidity or elasticity in it, but when the liquid is in orderly motion, it has been found that there comes into play a tangential stress called the viscosity.