# Louisiana Tech University Ruston, LA 71272 Slide 1 The Rectangular Channel Steven A. Jones BIEN 501 Friday, April 4th, 2008.

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Louisiana Tech University Ruston, LA 71272 Slide 1 The Rectangular Channel Steven A. Jones BIEN 501 Friday, April 4th, 2008

Louisiana Tech University Ruston, LA 71272 Slide 2 The Rectangular Channel Major Learning Objectives: 1.Deduce boundary conditions for a 2- dimensional laminar internal flow problem. 2.Reduce continuity and momentum for the problem. 3.Divide the momentum equation into its homogeneous and non-homogeneous components. 4.Transform the boundary conditions for the new momentum equation.

Louisiana Tech University Ruston, LA 71272 Slide 3 The Rectangular Channel Major Learning Objectives (continued): Use separation of variables to deduce the form of the solutions. Apply boundary conditions along three boundaries. Use superposition to deduce the series form of the complete solution. Use orthogonality (from Sturm-Liouville) to deduce the coefficients in the infinite series.

Louisiana Tech University Ruston, LA 71272 Slide 4 Rectangular Channel Look at the half-channel. z=0 is the midline of the channel. z ranges from –w/2 to +w/2 y ranges from –h/2 to +h/2 Fully Developed Flow No-Slip Boundary Conditions

Louisiana Tech University Ruston, LA 71272 Slide 5 Boundary Conditions We can write down 6 boundary conditions, (but we only need 4).

Louisiana Tech University Ruston, LA 71272 Slide 6 Navier-Stokes Equations Look at Term for Fully Developed Flow With v x = v y = 0 and no z gradients, which terms go to zero?

Louisiana Tech University Ruston, LA 71272 Slide 7 Navier-Stokes Equations All of them!

Louisiana Tech University Ruston, LA 71272 Slide 8 Continuity All terms in the continuity equation are zero. This result tells us that our assumptions are consistent with continuity.

Louisiana Tech University Ruston, LA 71272 Slide 9 y-momentum With no y and z velocities, this equation tells us that the y pressure gradient is cancelled by gravity. Before we look at x -momentum, look at y-momentum.

Louisiana Tech University Ruston, LA 71272 Slide 10 Constant Pressure Gradient We learned from the y and z momentum that pressure did not depend on y and z. So the right hand side of the above equation can only depend on x, but the left hand of the equation cannot depend on x because of the fully developed flow assumption. Therefore, cannot depend on x, y or z and must be constant.

Louisiana Tech University Ruston, LA 71272 Slide 11 Particular Solution Is non-homogeneous, meaning that the right hand side is not zero. A standard method for solving this type of equation is to first find a “particular solution” and subtract that solution from the equation to find a new homogeneous equation. It is easy to show that the solution: Satisfies the equation (hint: substitute this function back into the equation). The equation:

Louisiana Tech University Ruston, LA 71272 Slide 12 Particular Solution Also satisfies the boundary conditions at. It does not satisfy the boundary conditions for so our work is not quite done yet. However, we have made progress. The function:

Louisiana Tech University Ruston, LA 71272 Slide 13 Comments on the Particular Solution We could have used: Instead of: This solution would have reversed the roles of y and z, but the procedure would otherwise be the same.

Louisiana Tech University Ruston, LA 71272 Slide 14 Comments on Particular Solution Or something more complicated, it is generally easy to find a particular solution. Choose one of the variables, say y, and ask if there is a function f(y) that will yield a constant when differentiated an amount of times equal to the lowest order differential. Since it is not a function of z, the derivatives in z do not contribute, nor do the higher order derivatives in y (because the derivative of a constant is zero). For example, a particular solution to the above equation is When you have an equation like:

Louisiana Tech University Ruston, LA 71272 Slide 15 Exercise Find particular solutions to the following:

Louisiana Tech University Ruston, LA 71272 Slide 16 Exercise Answers

Louisiana Tech University Ruston, LA 71272 Slide 17 Complete Solution Where the particular solution part will handle the nonhomogeneity in the partial differential equation and the second part,  ( y, z ), will satisfy the homogeneous equation and satisfy the boundary conditions. The complete solution will be of the form:

Louisiana Tech University Ruston, LA 71272 Slide 18 Reduce the Equation Into: Plug: To get: These two terms cancel

Louisiana Tech University Ruston, LA 71272 Slide 19 Reduce the Equation (continued) We are left with Laplace’s equation: But remember so

Louisiana Tech University Ruston, LA 71272 Slide 20 Reduce the Equation (continued) If And if v x must satisfy the boundary conditions: Then  must satisfy the boundary conditions:

Louisiana Tech University Ruston, LA 71272 Slide 21 Exercise Why is each of the indicated terms below zero?

Louisiana Tech University Ruston, LA 71272 Slide 22 Exercise Answers Why is each of the indicated terms below zero? From Couette flow (plug h/2 into V x ( y )) From symmetry of Couette flow Because V x (y) does not depend on z.

Louisiana Tech University Ruston, LA 71272 Slide 23 Summary of Equations We must therefore solve Laplace’s equation: Subject to the following boundary conditions:

Louisiana Tech University Ruston, LA 71272 Slide 24 Visual

Louisiana Tech University Ruston, LA 71272 Slide 25 Separable Solution to Homogeneous Equation

Louisiana Tech University Ruston, LA 71272 Slide 26 Solution to ODEs It may help to remember that the sin and sinh (cos and cosh) functions can be written as:

Louisiana Tech University Ruston, LA 71272 Slide 27 Solution to ODEs And If Then anything that has this form: Satisfies the homogeneous equation.

Louisiana Tech University Ruston, LA 71272 Slide 28 Superposition Since there may be multiple values of l that work, a complete solution must consider all possible such solutions. We also note that the equations are linear, so that we can add solutions and still have a solution. Thus, we can write:

Louisiana Tech University Ruston, LA 71272 Slide 29 Boundary Conditions in y First consider the boundary condition along the centerline where y = 0. Next, the boundary condition at y = h/2 requires: This equation is true for all values of z only if: (The book uses n’=2n+1 )

Louisiana Tech University Ruston, LA 71272 Slide 30 Boundary Conditions in z We now have the following: But it is a lot to write, so we will continue to write it in terms of for now.

Louisiana Tech University Ruston, LA 71272 Slide 31 Boundary Conditions in z Consider the boundary condition along the centerline where z = 0. Next, the boundary condition at z = w/2 requires: This boundary condition is the one that requires the most work.

Louisiana Tech University Ruston, LA 71272 Slide 32 Boundary Conditions in z Notice that this equation is a function of y only. It can be interpreted to mean that the right hand side is being expanded as a Fourier cosine series within the interval of interest (i.e. –h/2 < y < h/2). We use the standard approach that was used to derive Fourier series.

Louisiana Tech University Ruston, LA 71272 Slide 33 Orthogonal Expansion Multiply both sides of the equation by cos ( m y ). Integrate from –h/2 to h/2

Louisiana Tech University Ruston, LA 71272 Slide 34 Orthogonality The left hand side will be zero for all m except n = m so: Note that the sum disappeared because only the value of n that is equal to m is needed.

Louisiana Tech University Ruston, LA 71272 Slide 35 Orthogonality But So Or

Louisiana Tech University Ruston, LA 71272 Slide 36 Integrating Cosine Squared The area of the rectangle is h. The area under cos 2 is h/2.

Louisiana Tech University Ruston, LA 71272 Slide 37 Orthogonality So The final integral can be obtained with integration by parts (twice).

Louisiana Tech University Ruston, LA 71272 Slide 38 Derived Information The final form of the solution is: We can obtain the shear stress from the stress tensor.

Louisiana Tech University Ruston, LA 71272 Slide 39 The Stress Tensor For fluids: The shear stress has 4 non-zero components.

Louisiana Tech University Ruston, LA 71272 Slide 40 Shear Stress, Bottom Surface Along the bottom surface, we are concerned only with  xy.

Louisiana Tech University Ruston, LA 71272 Slide 41 Shear Stress, Bottom Surface

Louisiana Tech University Ruston, LA 71272 Slide 42 Relationship of Flow Rate to Pressure Gradient To obtain flow rate in terms of pressure gradient, we must integrate the velocity over the cross-section. This relationship could be used, for example, to determine how much pressure is required to drive blood through a microchannel device at a given flow rate.

Louisiana Tech University Ruston, LA 71272 Slide 43 Example You are interested in designing a microdevice that samples blood from a vein and causes it to flow with a shear rate of 15 dynes/cm 2 over a microchannel that is coated with fibrinogen. The pressure difference driving the flow is the venous pressure. If you use a vacuum container at the downstream end of the device, can you obtain the required shear stress, and if so, what should be the dimensions of the channel?

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