# The Rectangular Channel Steven A. Jones BIEN 501 The Rectangular Channel Steven A. Jones BIEN 501 Friday, April 4th, 2008 Louisiana Tech University Ruston, LA 71272 Slide 1 The Rectangular Channel Major Learning Objectives: 1. Deduce boundary conditions for a 2dimensional 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 2 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 3 Rectangular Channel y y h 2 z w 2 x z 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 4 Boundary Conditions y y h 2 z w 2 x z We can write down 6 boundary conditions, (but we only need 4). vx h / 2, z 0 vx y, w / 2 0 vx / y 0, z 0 vx / dz y, 0 0 Louisiana Tech University Ruston, LA 71272 Slide 5

Navier-Stokes Equations Look at v v Term for Fully Developed Flow vx vx vx vx vy vz x momentum x y z v y v y v y vx vy vz y momentum x y z vz vz vz vx vy vz z momentum

x y z With vx = vy = 0 and no z gradients, which terms go to zero? Louisiana Tech University Ruston, LA 71272 Slide 6 Navier-Stokes Equations vx vx vx vx vy vz x momentum x y z v y v y v y vx vy vz x momentum x y z vz

vz vz vx vy vz x momentum x y z All of them! Louisiana Tech University Ruston, LA 71272 Slide 7 Continuity vx v y vz 0 x y z 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 8 y-momentum

Before we look at x-momentum, look at y-momentum. v y t vx v y x vy v y y vz v y z 2 2 2 v v

vy 1 p y y 2 2 2 x y y z g y With no y and z velocities, this equation tells us that the y pressure gradient is cancelled by gravity. Louisiana Tech University Ruston, LA 71272 Slide 9 Constant Pressure Gradient 2 vx 2 vx 1 p 2 2 y z x

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, p / x cannot depend on x, y or z and must be constant. Louisiana Tech University Ruston, LA 71272 Slide 10 Particular Solution The equation: 2 2 vx vx 1 p 2 2 y z x 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. 1 p 4 y 2 It is easy to show that the solution: vx

1 2 x h Satisfies the equation (hint: substitute this function back into the equation). Louisiana Tech University Ruston, LA 71272 Slide 11 Particular Solution The function: 1 p 4 y 2 vx Vx 1 2 x h Also satisfies the boundary conditions at y h / 2 . It does not satisfy the boundary conditions for z w / 2 so our work is not quite done yet. However, we have made progress. Louisiana Tech University Ruston, LA 71272 Slide 12 Comments on the Particular Solution 1 p 4 y 2 Instead of: vx

1 2 x h 1 p 4 z 2 We could have used: vx 1 2 x w 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 13 Comments on Particular Solution When you have an equation like: 3vx 3vx 2 vx 2 vx vx 3 2 2 C0 3 y z y z y 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 vx y C0 x Louisiana Tech University Ruston, LA 71272 Slide 14 Exercise Find particular solutions to the following: 4 vx 4 vx 4 C0 4 y z 3 v x 3 v x 2 vx 2 vx 3 2 2 C0 3 y z y z

5 v x 3 v x vx C0 5 2 y zy 3 v x 3 v x 2 vx 2 C0 3 2 y zy y Louisiana Tech University Ruston, LA 71272 Slide 15 Exercise Answers 4 vx 4 vx C0 4 C0 4 4 C0 vx y or z 4 y z

24 24 3 v x 3 v x 2 vx 2 vx C0 2 C0 2 3 2 2 C0 vx y or z 3 y z y z 2 2 5 v x 3 v x vx C0 vx C0 5 2 y zy 3 v x 3 v x 2 vx C0 2 2 C0 vx

y 3 2 y zy y 2 Louisiana Tech University Ruston, LA 71272 Slide 16 Complete Solution The complete solution will be of the form: 1 p 4 y 2 vx Vx y y, z 1 2 y, z x h 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. Louisiana Tech University Ruston, LA 71272 Slide 17 Reduce the Equation

Plug: 1 p 4 y 2 vx 1 2 y, z x h 2 Into: To get: 2 vx vx 1 p 2 2 y z x These two terms cancel 2 2 1 p 4 y 2 y, z 1 2 2 y x h

y 2 2 2 1 p 4 y 2 y, z 1 p 2 1 2 2 z x h z x Louisiana Tech University Ruston, LA 71272 Slide 18 Reduce the Equation (continued) We are left with Laplaces equation: 2 y, z y But remember 2 2 y, z z

2 0 vx y, z Vx y y, z so y, z vx y, z Vx y Louisiana Tech University Ruston, LA 71272 Slide 19 Reduce the Equation (continued) If x, y vx x, y Vx y And if vx must satisfy the boundary conditions: vx h / 2, z 0 vx y, w / 2 0 vx / y 0, z 0 vx / dz y, 0 0 Then must satisfy the boundary conditions:

h / 2, z Vx h / 2 0 y, w / 2 Vx y 0 / y 0, z Vx 0 / y 0 / dz y, 0 Vx y z 0 Louisiana Tech University Ruston, LA 71272 Slide 20 Exercise Why is each of the indicated terms below zero? h / 2, z Vx h / 2 0 y, w / 2 Vx y 0 / y 0, z Vx 0 / y 0 / dz y, 0 Vx y z 0 Vx h / 2 0 Vx 0 / y 0 Vx y z 0 Louisiana Tech University Ruston, LA 71272

Slide 21 Exercise Answers Why is each of the indicated terms below zero? h / 2, z Vx h / 2 0 y, w / 2 Vx y 0 / y 0, z Vx 0 / y 0 / dz y, 0 Vx y z 0 Vx h / 2 0 From Couette flow (plug h/2 into Vx(y)) Vx 0 / y 0 From symmetry of Couette flow Vx y z 0 Because Vx(y) does not depend on z. Louisiana Tech University Ruston, LA 71272 Slide 22 Summary of Equations We must therefore solve Laplaces equation: 2 y, z y 2

2 y, z z 2 0 Subject to the following boundary conditions: h / 2, z 0 / y 0, z 0 Louisiana Tech University Ruston, LA 71272 1 p 4 y 2 y, w / 2 Vx y 1 2 0 x h / dz y, 0 0 Slide 23 Visual Vz y Louisiana Tech University

Ruston, LA 71272 Slide 24 Separable Solution to Homogeneous Equation y, z Y y Z z 2YZ 2YZ 0 2 2 z y 2 2 Y Z Z 2 Y 2 0 y z 2 2 2 2 1Y

1Z 1 d Z 2 2 1 d Y 2 , Y y 2 Z z 2 Z dz 2 Y dy 2 Louisiana Tech University Ruston, LA 71272 Slide 25 Solution to ODEs 2 d 2Y d Z 2 2

Y 0, Z 0 2 2 dy dz Z z C cosh z D sinh z Y y A cos y B sin y It may help to remember that the sin and sinh (cos and cosh) functions can be written as: eia e i ei e i cos , sin 2 2i ea e e e cosh , sinh 2 2 Louisiana Tech University Ruston, LA 71272

Slide 26 Solution to ODEs If Z z C cosh z D sinh z Y y A cos y B sin y And y, z Y y Z z Then anything that has this form: y, z C cosh z D sinh z A cos y B sin y Satisfies the homogeneous equation. Louisiana Tech University Ruston, LA 71272 Slide 27 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: y, z A C cosh z D sinh z A cos y B sin y all

Louisiana Tech University Ruston, LA 71272 Slide 28 Boundary Conditions in y y, z C cosh z D sinh z A cos y B sin y all First consider the boundary condition along the centerline where y = 0. 0, z y 0 A sin 0 B cos 0 0 B 0 Next, the boundary condition at y = h/2 requires: h / 2, z C cosh z D sinh z A cos h / 2 0 all This equation is true for all values of z only if: A cos h / 2 0 i.e. h / 2 n 12 2n 1 (The book uses n=2n+1) h Louisiana Tech University Ruston, LA 71272 Slide 29

Boundary Conditions in z We now have the following: 2n 1 2n 1 y, z Cn cosh z Dn sinh h h n=0 2n 1 z An cos y h 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 30

Boundary Conditions in z y, z C cosh z D sinh z A cos y n=0 Consider the boundary condition along the centerline where z = 0. y, 0 z 0 C sinh 0 D cosh 0 0 D 0 Next, the boundary condition at z = w/2 requires: 1 p 4 y 2 y, w / 2 A cosh w / 2 cos y 1 2 x h all This boundary condition is the one that requires the most work. Louisiana Tech University Ruston, LA 71272 Slide 31 Boundary Conditions in z Notice that this equation is a function of y only. 1 p 4 y 2

y, w / 2 A cosh w / 2 cos y 1 2 x h all 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). 2n 1 y p 4 y 2 1 2 n cos x h n 0 h We use the standard approach that was used to derive Fourier series. Louisiana Tech University Ruston, LA 71272 Slide 32 Orthogonal Expansion Multiply both sides of the equation by cos (m y ). cos m y p 4 y 2 cos m y A cosh n w / 2 cos n y 1 2

x h all n cos m y p 4 y 2 A cosh n w / 2 cos n y cos m y 1 2 x h all n Integrate from h/2 to h/2 cos m y p 4 y 2 h / 2 alln A cosh n w / 2 cos n y cos m y dy h / 2 x 1 h2 dy h/2 Louisiana Tech University Ruston, LA 71272 h/2 Slide 33 Orthogonality h/2 h/2

all n cos m y p 4 y 2 A cosh n w / 2 cos n y cos m y dy 1 2 dy h/2 x h h/2 The left hand side will be zero for all m except n = m so: cos m y p 4 y 2 h / 2 Am cosh m w / 2 cos m y dy h / 2 x 1 h2 dy h/2 2 h/2 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 34 Orthogonality

2 cos y p 4 y m 2 h / 2 Am cosh m w / 2 cos m y dy h / 2 x 1 h2 dy h/2 h/2 But h/2 h/2 cos 2 n y dy h / 2 So An h

cos n y p 4 y 2 cosh n w / 2 1 2 dy h / 2 2 x h Or 2 Am h cosh m w / 2 Louisiana Tech University Ruston, LA 71272 h/2 cos m y p 4 y 2 h / 2 x 1 h2 dy h/2 Slide 35 Integrating Cosine Squared cos (with 1/2)

cos squared (with 1/2) cos (with 3/2) cos squared (with 3/2) zero line 1.5 cosines 1 0.5 0 -0.5 -1 -1.5 -25 -20 -15 -10 -5 0 5 10

15 20 25 y The area of the rectangle is h. The area under cos2 is h/2. Louisiana Tech University Ruston, LA 71272 Slide 36 Orthogonality 2 Am h cosh m w / 2 cos m y p 4 y 2 h / 2 x 1 h2 dy h/2 So 2 p Am sin m y h cosh m w / 2 x

h 2 y h 2 4 2 h h/2 2 y cos y dy m h / 2 The final integral can be obtained with integration by parts (twice). Louisiana Tech University Ruston, LA 71272 Slide 37

Derived Information The final form of the solution is: 1 dp 4 y 2 vx y, z 1 2 A cosh z / 2 cos y dx h all We can obtain the shear stress from the stress tensor. Louisiana Tech University Ruston, LA 71272 Slide 38 The Stress Tensor For fluids: u1 x1 1 u u 2 2 1 2 x1 x2 1 u3 u1 2 x x 3

1 1 u1 u2 2 x2 x1 u2 x2 1 u3 u2 2 x2 x3 1 u1 u3 2 x3 x1 1 u2 u3 2 x3 x2 u3 x3

The shear stress has 4 non-zero components. Louisiana Tech University Ruston, LA 71272 Slide 39 Shear Stress, Bottom Surface 0 1 v 2 x 2 y 1 vx 2 z 1 vx 2 y 1 vx 2 z

0 0 0 0 Along the bottom surface, we are concerned only with xy. vx xy y Louisiana Tech University Ruston, LA 71272 Slide 40 Shear Stress, Bottom Surface y, z A cosh z / 2 cos y all

xy vx y xy A cosh z / 2 sin y all Louisiana Tech University Ruston, LA 71272 Slide 41 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. w/ 2 Q h/2 w/ 2 h/ 2 vx y, z dy dz 1 dp 4 y 2

1 2 A cosh z / 2 cos y dydz w/ 2 h/ 2 dx h all w/ 2 h/2 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 42 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/cm2 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? Louisiana Tech University Ruston, LA 71272

Slide 43