thick layer of oil of viscosity 0.25 Ns/m. Determine:

(i) the pressure gradient required to cause no shear stress at the belt surface

(ii) the average velocity and the discharge of oil to be maintained for the above.

Given:

w=0.5 m (Width of the flat belt)

U = 4 m/s (Relative velocity of the belt with respect to the lower surface

h = 6 cm (Gap between the belt conveyor and lower surface)

μ = $0.25 Ns/m ^2$(Viscosity of the oil)

To find:

"Pressure Gradient $(∆P/∆x) such that τ=0 at y=h$

"Average velocity of the oil $(U_avg)$

"Discharge of oil (Q ̇)

Sol: We know from the definition of the problem that the flow is a couette flow,

Now, Using the condition τ=0 at y=h

$τ_(y=h)=μ \frac{du}{dy}_{y=h}$=0

∴$[\frac{1}{2μ} \frac{∆P}{∆x} (2y-h)+(\frac{U}{h})]_{y=h}$=0

∴$[\frac{1}{2μ} \frac{∆P}{∆x} (2h-h)+(\frac{U}{h})]_{y=h}$=0

$\frac{∆P}{∆x}=-(\frac{2μU}{h}^2) $

Substituting in the above equation,

$\frac{∆P}{∆x}=-\frac{2(0.25)(4))}{0.06}^2$

∴$\frac{∆P}{∆x}=-555.5 ̅ N/m^3$

…this is the Pressure gradient required to cause no shear stress at the belt.

Now, the average velocity of the oil is given by,

∴$ U_avg=\frac{1}{h }∫_0^h u dy$

∴$ U_{avg}=\frac{1}{h} ∫_0^h{\frac{1}{2μ} \frac{∆P}{∆x} y(y-h)+(\frac{Uy}{h})} dy$

∴$U_avg=\frac{1}{h} \frac{[1}{2μ} (\frac{∆P}{∆x} (\frac{y^3}{3}-\frac{hy^2}{2})+(\frac{Uy^2}{2h})]_0^h$

∴$U_{avg}=\frac{1}{h} [\frac{1}{2μ} \frac{∆P}{∆x} (\frac{h^{3}}{3}-\frac{h^{3}}{2})+(\frac{Uh^2}{2h})]$

∴$U_{avg}=[\frac{1}{2μ} \frac{∆P}{∆x} (\frac{h^2}{3}-\frac{h^2}{2})+(\frac{U}{2})]$

Substituting in the above equation,

∴$U_{avg}=[\frac{1}{2μ}(-555.5 ̅)(\frac{0.06^{2}}{3}-\frac{0.06^{2}}{2})+(\frac{4}{2})]$

∴$ U_{avg}=2.6 ̅$

…this is the average velocity of the oil.

The discharge of the oil is given by,

Q ̇=$ U_{avg}∙h∙w$

Substituting in the above equation,

∴$ Q ̇=2.6667(0.06)(0.5)$

∴ $Q ̇= 0.08 m^3/s$

…this is the discharge required.