Question

Two large vertical parallel plates separated by
a gap of d have a highly viscous liquid of
density p and viscosity coefficient n, flowing
steadily under gravity in the gap. The velocity
gradient of flow near plates surface is​

Answer 1

Answer:

Two large vertical parallel plates separated by

a gap of d have a highly viscous liquid of

density p and viscosity coefficient n, flowing

steadily under gravity in the gap. The velocity

gradient of flow near plates surface

Answer 2

Given:

Separation of plates = d

Density of liquid = ρ

Coefficient of viscosity = η

Acceleration due to gravity = g

To find:

Velocity gradient at surface = $\frac{dv}{dz}$

Formula:

$F=\eta A \frac{dv}{dz}$

Solution:

Step 1 of 2

Velocity gradient is the rate of change in velocity with respect to the separation of plates.

In the formula, A is the cross section of the plates, F is the viscous force and dz is the separation of plates.

As the separation of plates is given as "d" in the question, so the formula will get modified as

$F=\eta A\frac{dv}{d}$

According to Newton's second law of motion, $F=ma$

Force is the product of mass with acceleration.

In this question, the liquid is flowing under the influence of gravity so, the acceleration will be equal to acceleration due to gravity.

$F=mg$

Step 2 of 2

With the above changes, the velocity gradient of the flow of liquid will be

$\frac{dv}{d}=\frac{F}{\eta A}\\\frac{dv}{d}=\frac{mg}{\eta A}$

The mass of the liquid can be determined by the product of density with volume as

$\rho=\frac{m}{V}\\m=\rho V$

So ,the velocity gradient will be

$\frac{dv}{d}=\frac{\rho Vg}{\eta A} \\\frac{dv}{d}=\frac{\rho 2Adg}{\eta A} \\\frac{dv}{d}=\frac{2\rho dg}{\eta }$

Thus, option (1) is the correct answer.

Final Answer:

Option (1) is the correct answer for velocity gradient of flow near plates surface.