Question

Two large plane surfaces are 25 mm apart and the space between them is filled with a liquid of viscosity μ = 0.958 Pa- s. Assuming the velocity gradient to be a straight line, what force is required to pull a very thin plate of 0.37 m2 area at a constant speed of 0.3 m/s if the plate is 8.4 mm from one of the surfaces.

Answer 1

Explanation:

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Answer 2

Answer:

The force required to pull a thin plate of area 0.37$m^2$ at a constant speed of 0.3m/s placed between two large plane surfaces that are 25mm apart and the gap filled with a liquid of viscosity 0.958 pa-s is 19.17N.

Explanation:

Viscosity:

• It is the resistance of a fluid to a change in shape or movement of neighbouring portions relative to one another.
• If a fluid is filled in between two plane surfaces of area A whose separation is y and the flow velocity u, then the force acting on the surface is proportional to its area A, the flow velocity u and inversely proportional to the separation y. Writing this proportionality as an equation we have,

                $F=\eta A\frac{u}{y}$

where $\eta$ is the coefficient of viscosity.

• If the velocity is varying with the separation then,

               $F=\eta A\frac{du}{dy}$

Step 1:

Given two large plane surfaces that are 25mm apart and the gap filled with a liquid of viscosity 0.958 pa-s. From one of the surfaces at a distance of 8.4mm, a thin plate of area 0.37$m^2$ is to be pulled at a constant velocity of 0.3m/s. We are required to find the force required to pull this plate.

              $y=25mm=25 \times 10^{-3}m\\\eta=0.958pa-s\\A=0.37 m^2\\u=0.3m/s$

Step 2:

Since the plate is not exactly at the centre of the gap, the force exerted by the liquid at the top and at the bottom varies depending on the distance from the surfaces, Let's consider the plate to be y= 8.4mm from the bottom plate and so, x=16.6mm from the top. Let $F_u$ be the force by the liquid on the top surface of the plate and $F_b$ be the force by the liquid on the bottom surface of the plate. Then,

                    $F_b=\eta A\frac{u}{y}\\\\ F_u=\eta A\frac{u}{x}$

Step 3:

                   $F_b=\eta A\frac{u}{y}\\F_b = 0.958 \times 0.37\times 0.3(\frac{1}{8.4 \times 10^{-3}})\\ F_b=12.76N$

Therefore, the force by the liquid on the bottom surface of the plate is 12.76N.

Step 3:

                   $F_u=\eta A\frac{u}{x}\\F_u = 0.958 \times 0.37\times 0.3(\frac{1}{16.6 \times 10^{-3}})\\ F_u=6.38N$

Therefore, the force by the liquid on the top surface of the plate is 6.38N.

Step 4:

Net force = $F_u+F_b=12.76+6.38=19.17N$

Adding the forces because both these forces oppose the motion of the plate.

Therefore, The force required to pull a thin plate is 19.17N.