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Chapter 8: Problem 28

Consider steady, fully developed laminar flow of a viscous liquid down an inclined surface. The liquid layer is of constant thickness, \(h\), Use a suitably chosen differential control volume to obtain the velocity profile. Develop an expression for the volume flow rate.

Short Answer

Expert verified
The velocity profile is parabolic given by \(v_x = \frac{gh\sin(\beta)}{2\mu}(2y - y^2)\) and the volume flow rate per unit width is \(Q = \frac{gh^3\sin(\beta)}{3\mu}\).

Step by step solution

01

Defining the Differential Control Volume

Since the flow of the liquid down an inclined surface is steady and fully developed, one may assume that the velocity profile is only a function of y-coordinate (taking direction perpendicular to surface as y-direction). The suitable differential control volume is thus a differential slice of liquid layer of width \(dx\), height \(dy\) and with unit cross sectional area perpendicular to the plane.
02

Applying Newton鈥檚 second law in the direction of flow

Summing up forces in the x-direction (direction of the flow) due to pressure and shear stress and setting it equal to mass times acceleration (due to Newton鈥檚 law) gives the momentum equation. For the steady flow, the net pressure force is zero and the net force due to shear stress is \(\mu \frac{d^2v_x}{dy^2}dx\), where \(v_x\) is the x-component velocity and \(\mu\) is the liquid viscosity. As there's no acceleration, balance gives \(\mu \frac{d^2v_x}{dy^2} = 0\). The solution of this differential equation gives a linear velocity profile: \(v_x = C_1y + C_2 \).
03

Applying Boundary Conditions

To find \(C_1\) and \(C_2\), we use the no-slip condition at the liquid-solid interface (velocity is zero at y=0 and y=h). This gives the velocity distribution as: \(v_x = \frac{gh\sin(\beta)}{2\mu}(2y - y^2)\), where \(g\) is the gravitational constant, and \(\beta\) is the inclination angle of the surface.
04

Calculating the Volume Flow Rate

The volume flow rate per unit width can be calculated by integrating the velocity profile from y=0 to h, which gives: \(Q = \frac{gh^3\sin(\beta)}{3\mu}\).

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Most popular questions from this chapter

Gasoline flows in a long. underground pipeline at a constant temperature of \(15^{\circ} \mathrm{C}\). Two pumping stations at the same elevation are located \(13 \mathrm{km}\) apart. The pressure drop between the stations is \(1.4 \mathrm{MPa}\). The pipeline is made from 0.6-m-diameter pipe. Although the pipe is made from commercial steel, age and corrosion have raised the pipe roughness to approximately that for galvanized iron. Compute the volume flow rate.

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