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Chapter 6: Problem 73

A liquid of constant density \(\rho\) and constant viscosity \(\mu\) flows down a wide, long inclined flat plate. The plate makes an angle \(\theta\) with the horizontal. The velocity components do not change in the direction of the plate, and the fluid depth, \(h,\) normal to the plate is constant. There is negligible shear stress by the air on the fluid. Find the velocity profile \(u(y),\) where \(u\) is the velocity parallel to the plate and \(y\) is measured perpendicular to the plate. Write an expression for the volume flow rate per unit width of the plate.

Short Answer

Expert verified
The velocity profile of the fluid is \(u(y) = \frac{\rho g \sin(\theta)}{2\mu} y(2h - y)\) and the volume flow rate per unit width of the plate is \[Q = \frac{\rho g \sin(\theta)}{3\mu} h^3 \].

Step by step solution

01

Applying Navier-Stokes Equation

For flow along an inclined plate, the pressure is constant along the flow direction (x-direction). This results in the absence of pressure gradient in that direction. Then, the Navier-Stokes equation simplifies to \[ \rho g \sin(\theta) = \mu \frac{\partial^2 u}{\partial y^2}. \] Differential equations should be used to solve for \(u(y)\). This is the equation of motion for the fluid particle where the left side represents the force due to gravity and the right side represents viscous forces.
02

Solving the Differential Equation

Integrate the equation twice to solve for \(u\). The boundary conditions are \(u=0\) at \(y=0\) the no-slip condition at the plate surface and \(\frac{\partial u}{\partial y}= 0\) at \(y=h\), which represents zero shear stress at the fluid surface. Solving these gives the velocity profile \(u(y) = \frac{\rho g \sin(\theta)}{2\mu} y(2h - y). \]
03

Calculating Volume Flow Rate

The volume flow rate per unit width (\(Q\)) can be obtained by integrating the velocity profile across the fluid depth. Integrating gives \[Q = \frac{\rho g \sin(\theta)}{3\mu} h^3 \] which is the volume flow rate per unit width of the plate.

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Most popular questions from this chapter

A viscous fluid is contained between two infinitely long, vertical, concentric cylinders. The outer cylinder has a radius \(r_{o}\) and rotates with an angular velocity \(\omega\). The inner cylinder is fixed and has a radius \(r_{i} .\) Make use of the Navier-Stokes equations to obtain an exact solution for the velocity distribution in the gap. Assume that the flow in the gap is axisymmetric (neither velocity nor pressure are functions of angular position \(\theta\) within the gap) and that there are no velocity components other than the tangential component. The only body force is the weight.

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