The velocity distribution for flow of a thin viscous film...

The velocity distribution for flow of a thin viscous film down an inclined plane surface was developed in Example 5.9. Consider a film $7 \mathrm{mm}$ thick, of liquid with $\mathrm{SG}=1.2$ and dynamic viscosity of $1.60 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}$. Derive an expression for the shear stress distribution within the film. Calculate the maximum shear stress within the film and indicate its direction. Evaluate the volume flow rate in the film, in $\mathrm{mm}^{3} / \mathrm{s}$ per millimeter of surface width. Calculate the film Reynolds number based on average velocity.

Key Concepts

Newton's Law of Viscosity

This principle states that the shear stress in a fluid is directly proportional to the rate of change of velocity with respect to distance perpendicular to the flow. In mathematical terms for a Newtonian fluid, ? = ?(dv/dy), where ? is the shear stress, ? is the dynamic viscosity, and dv/dy is the velocity gradient. This relationship underpins the derivation of shear stress distributions in viscous flows.

Thin Film Flow (Lubrication Approximation)

Thin film flow refers to fluid flow where one dimension (typically the film thickness) is much smaller than the others, allowing simplifications in the governing Navier-Stokes equations. Known as the lubrication approximation, this approach significantly simplifies the analysis by assuming that variations in the flow direction dominate over those in the perpendicular direction, making it easier to derive expressions for velocity profiles and shear stress distributions.

Shear Stress Distribution

In the context of viscous film flow down an inclined plane, the shear stress distribution is derived from the balance of forces in the fluid. Due to the no-slip condition at the solid boundary and the free surface condition (usually zero shear at the free surface), the shear stress typically varies linearly across the film thickness, reaching a maximum at the interface with the surface.

Volume Flow Rate Calculation

The volume flow rate per unit width in a thin film is obtained by integrating the velocity profile over the film thickness. This integration sums the contributions of all infinitesimal layers within the film and results in an expression that relates the flow rate to the physical properties of the fluid (such as viscosity and density), the film thickness, and driving forces like gravity.

Reynolds Number

The Reynolds number is a dimensionless parameter defined as the ratio of inertial forces to viscous forces in the flow. In the case of thin film flow, it is typically calculated using the average velocity and a characteristic length scale, such as the film thickness. This number helps predict the transition between laminar and turbulent flow regimes, with lower values indicating laminar behavior.

Transcript

00:01 Here in this problem consider the equation of velocity distribution for the flow of a thin viscous film down and inclined plane.

00:08 So we will write the velocity profile uy, uy will be equal to row g side theta, rg sine theta divided by mu multiplied by hy minus y squared by 2, y square divided by 2.

00:26 Let us assume this equation be equation number 1.

00:28 Now, now the density is given as specific gravity multiplied by the density of water.

00:36 So it will be equals to 1 ,200 kg per meter cube.

00:42 Now we will calculate the shear stress.

00:44 C .r stress is given as star is equals to mu, d u divided by d .y.

00:50 Let this equation be equation number 2.

00:52 So from equation 1 we can write uy, uy is equal to row g, r g s s theta divided by mu multiplied by h y minus y square so y minus y squared by two or we can further write d u divided by d y d u divided by d y is equal to row g sine theta r g sine theta divided by mu multiplied by h minus y h minus y so let this equation be equation number three now we'll substitute the value of the the value of the u by an equation in the equation two we will find tau is equals to mu multiplied by the d u by d y we have already found as rog sine tita divided by mu multiplied by h minus y or upon further solve the right tau row g sine theta multiplied by h minus y so let's put the values here row is 200 multiplied by g is 9 .81 multiplied by sine 3rd 5 x x the value of h h is 7m so 7 multiplied by 10 to 5 minus 3 meter minus y is 0 so upon solving we get this will be taumax at h is equal to at y is equal to zero we have taum x so taum x is equal to so taum x from here we get 21 .3 pascal so this is the maxim c a stress in x direction 21 .3 pascal now we will calculate the flow rate using equation q is equal to integration u b a so it will be equal to integration from 0 to h multiplied by u y multiplied by a is w multiplied by d y so substitute the value of u in this expression we get q is equal to w multiplied by 0 to h 0 to h rog r g sine theta divided by mu multiplied by multiplied by hy minus y squared by 2, dy, or upon further solving, we get q is equals to rho g, w, sine theta, divided by mu, divided one new multiplied by hq divided by 2 minus h q divided by 6, bracket close.

03:54 So finally, we can write discharge is equal to row g, w rog, sine theta h cube divided by mu, so we will put the values here, w is 1 .2 multiplied by 10 to the power of 3 3 this is the value of this is the value of and the value of g is 9 .81 9 .81 and the theta is 15 degrees so sign 15 degree multiplied by the value of h is 7 multiplied by 10 to the power of minus 3 this will be in cube divided by 3 multiplied by 3 multiplied the viscosity is given as 1 .6 1 .6.

05:02 Let's write it is q divided by w...

Please give Ace some feedback