Question

luids of viscosities 1  =0.1 N.s/ms and 2  =0.15 N.s/m2 are contained between two plates (each plate 1
m2 area). The thicknesses are h1=.5 mm and h2=0.3 mm, respectively. Find the force F to make the upper
plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids?

Answer 1

Answer:

a) 1 / 3 N

b) 5/3 m/s

Explanation:

For constant speed V at the interface of the two fluids the net force is zero.

Hence, F top = F bottom thus the shear stresses at top and bottom must be the same.

$t_{top} = t_{bottom}\\where\\t = u.\frac{dU}{dt} \\Hence\\(\frac{V - 1}{h_{2} })*u_{top} = (\frac{V}{h_{1} })*u_{bottom} \\\\0.3V = 0.5\\\\V = 5/3 m/s$

For force of top plate

$F_{top} = u_{top} * \frac{V-U}{h_{2} }*A=( 0.15)*(\frac{5/3 -1}{0.3 } )*(1)\\F_{top} = \frac{1}{3} N$

Answer 2

The force F to make the upper plate move at a speed of 1 m/s is;  143 N

The fluid velocity at the interface between the two fluids is; 0.714 m/s

Fluid Velocity

We are given fluid viscosities as;

₁ = 0.1 N.s/m²

₂ = 0.15 N.s/m²

Thicknesses of plates are;

h₁ = 0.5 mm and h₂ = 0.3 mm

The shear stress is the same throughout and as such;

τ = ₁(du₁/dy₁) = ₂(du₂/dy₂)

⇒ ₁(Ui/h₁) = ₂(U - Ui)/h₂

where Ui is the interface velocity and U is the speed at which the upper body moved, we have;

Ui = U/(1 + (₁/₂)*(h₂/h₁))

⇒ Ui = 1/(1 + (0.1/0.15)*(0.3/0.5))

⇒ Ui = 0.714 m/s

Force required is;

F = ₁ * (Ui/h₁) * (A)

We are given A = 1 m²

Thus;

F = 0.1 * (0.714/0.0005) * 1

F ≈ 143 N

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