written-lesson fluid mechanics
Explanation and Analysis: A couple things to review: Basic Equations: I’m going to skip ahead and hope that you’ve covered this in class, but a momentum balance can be represented by the Cauchy momentum equation. The general form of this equation looks like this:
where is the density of the continuum, is the stress tensor, and contains all of the body forces per unit volume (often simply density times gravity). is the velocity vector field, which depends on time and space. (http://en.wikipedia.org/wiki/Cauchy_momentum_equation)
Assuming constant viscosity and fluid velocity, we can then derive the Navier Stokes equation, which looks like this:
where v is the flow velocity, ρ is the fluid density, p is the pressure, is the (deviatoric) component of the total stress tensor, and f represents body forces (per unit volume) acting on the fluid and ∇ is the del operator. (http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equation) Recall also, that we can always do a mass balance on a small piece of fluid. This can be represented by the Continuity of Mass Equation:
where ∇• is divergence, · t is time, · σ is the generation of q per unit volume per unit time. Your textbook should probably have some information about how these are derived, so I didn’t going to go into that. These governing equations are fairly well-established and so usually you don’t need to re-derive them. However, you do need to know how to transform these equations into a form that you can actually solve. Velocity Profile: Whenever you do a problem like this, you should have a picture that looks something like this in your head:
or
or something similar!
Where longer arrows means the fluid is flowing faster, and shorter arrows means the fluid is flowing slower. This is what part c and d of this problem are asking for.
Solving the Problem Now that we have some of the basics covered and have a general idea of what we’re trying to do, we can begin solving the problem. a) Set up a momentum balance on a fluid shell under the conveyor belt. This part should be easy. We already said that the Cauchy momentum Equations describe the momentum balance; we also know that viscosity r and density m are constant so we can use the Navier Stokes equations. However, in order to actually use those equations, we need to choose a coordinate system that makes sense. Some things to think about: Which coordinate system would you use? In this case, because we are looking at a flat rectangular conveyor belt, I’ve decided to use a Cartesian coordinate system. I’ve also decided to position my coordinate system x axis parallel to the slope. Here, u, v, and w represent the flowrates
\
Z axis goes into the page y x
b) Simplify the equations in a by stating reasons and write boundary conditions This is my favorite part: crossing out everything that we don’t need! Some things to think about: 1) What does “steady volumetric flow” tell you? 2) Does the flow change in all directions (x, y, AND z) as you move up the slope? 3) What do we know about pressure in the x, y, and z directions? For the boundary conditions, think about what happens at the wall and at the interface between liquid and the conveyor belt. 4) And just, intuitively, what do you think the flow should look like? Where will flow be fastest? Slowest?
Here’s what I think: 1) First of all, “steady volumetric flow” suggests to me that the velocity profile will not change with time. So whether you look at the pipe now, or 10 minutes from now, we can assume the place where the flow is fastest will still look the fastest. Because there is no time-dependence, we can cross out all of the terms with dt in them. 2) Remember that u,v, and w are components of the total velocity vector V. While I imagine that the flow might change in the x and y directions because there are forces acting on it, I don’t think there is any flow in the z direction because there are no forces acting in the z direction. Thus, we can immediately cross out all of the velocity derivative terms with dz, z subscript, w, or dw in them. 3) I am also going to assume that there is no fluid flow in the y direction, i.e. that all of the fluid runs parallel to the conveyor belt. Thus I can eliminate all of the v terms. 4) Finally, I am going to make an assumption that the velocity profile of the fluid any point in the pipe will not change depending on how far it is into the pipe, i.e. “fully developed flow”. Thus, I will cross out all of the velocity derivative terms with dx in them. 5) In the x direction, the pressure doesn’t change as you go up the slope in the x direction because we know that the pressure is at Patm at both ends, so dp/dx is equal to 0. (Note that we exclude the pressure exerted due to gravity because it is accounted for in the NS equation already).
Crossing everything out, we get (lets call this EQUATION 1 and 2) 0 = -dp/dx +mu* (d2u/dy2) + rho*gx 0 = -dp/dy + rho*gy Much better. Finally, boundary conditions: We usually assume the “no-slip” condition, which says that the fluid flow velocity at an interface with a solid is the same as the velocity of that solid. Thus, for V(x,y) according to the axes I drew earlier, V = 0 at the wall y =0, and V = v at y =B, where little v is the velocity of the conveyor belt. For the pressure, we know that P(0,0) = Patm+ rho*g*sinβ∗Β
c) Solve equations in (b) and derive an expression for velocity profile. First we need an expression for gx and gy Breaking gravity acceleration into x and y components, we get gx = g*sinβ, gy = g*cosβ gx = - g*sin(β)
gy = - g*cos(β)
Now we can solve EQUATION 1 AND 2 From Equation 2 (the question does NOT ask for this, but for reference) 0 = -dp/dy + ρgy and P(0,0) = Patm+ ρgB*cosβ Py (y) = -ρgy*cosβ + Patm+ ρgB*sinβ From Equation 1: 0 = mu* (d2u/dy2) + ρ*gx d2u/dy2 = - ρ g*cosβ/ µ Integrating once, du/dy = -ρ yg/ µ + a Integrating again, u = -ρ y2g*cosβ/(2 µ) + a*y + b where a and b are constants
(eq*)
Using boundary condition: u=v when y=B and u= 0 when y =0 to plug into (eq*) above, we get: b=0 a = v + ρ Bg*cosβ/(2 µ) Therefore, V = u = -ρ y g*cosβ/(2 µ) + y*v/B+ y*ρBg*cosβ/(2 µ)
where is the density of the continuum, is the stress tensor, and contains all of the body forces per unit volume (often simply density times gravity). is the velocity vector field, which depends on time and space. (http://en.wikipedia.org/wiki/Cauchy_momentum_equation)
Assuming constant viscosity and fluid velocity, we can then derive the Navier Stokes equation, which looks like this:
where v is the flow velocity, ρ is the fluid density, p is the pressure, is the (deviatoric) component of the total stress tensor, and f represents body forces (per unit volume) acting on the fluid and ∇ is the del operator. (http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equation) Recall also, that we can always do a mass balance on a small piece of fluid. This can be represented by the Continuity of Mass Equation:
where ∇• is divergence, · t is time, · σ is the generation of q per unit volume per unit time. Your textbook should probably have some information about how these are derived, so I didn’t going to go into that. These governing equations are fairly well-established and so usually you don’t need to re-derive them. However, you do need to know how to transform these equations into a form that you can actually solve. Velocity Profile: Whenever you do a problem like this, you should have a picture that looks something like this in your head:
or
or something similar!
Where longer arrows means the fluid is flowing faster, and shorter arrows means the fluid is flowing slower. This is what part c and d of this problem are asking for.
Solving the Problem Now that we have some of the basics covered and have a general idea of what we’re trying to do, we can begin solving the problem. a) Set up a momentum balance on a fluid shell under the conveyor belt. This part should be easy. We already said that the Cauchy momentum Equations describe the momentum balance; we also know that viscosity r and density m are constant so we can use the Navier Stokes equations. However, in order to actually use those equations, we need to choose a coordinate system that makes sense. Some things to think about: Which coordinate system would you use? In this case, because we are looking at a flat rectangular conveyor belt, I’ve decided to use a Cartesian coordinate system. I’ve also decided to position my coordinate system x axis parallel to the slope. Here, u, v, and w represent the flowrates
\
Z axis goes into the page y x
b) Simplify the equations in a by stating reasons and write boundary conditions This is my favorite part: crossing out everything that we don’t need! Some things to think about: 1) What does “steady volumetric flow” tell you? 2) Does the flow change in all directions (x, y, AND z) as you move up the slope? 3) What do we know about pressure in the x, y, and z directions? For the boundary conditions, think about what happens at the wall and at the interface between liquid and the conveyor belt. 4) And just, intuitively, what do you think the flow should look like? Where will flow be fastest? Slowest?
Here’s what I think: 1) First of all, “steady volumetric flow” suggests to me that the velocity profile will not change with time. So whether you look at the pipe now, or 10 minutes from now, we can assume the place where the flow is fastest will still look the fastest. Because there is no time-dependence, we can cross out all of the terms with dt in them. 2) Remember that u,v, and w are components of the total velocity vector V. While I imagine that the flow might change in the x and y directions because there are forces acting on it, I don’t think there is any flow in the z direction because there are no forces acting in the z direction. Thus, we can immediately cross out all of the velocity derivative terms with dz, z subscript, w, or dw in them. 3) I am also going to assume that there is no fluid flow in the y direction, i.e. that all of the fluid runs parallel to the conveyor belt. Thus I can eliminate all of the v terms. 4) Finally, I am going to make an assumption that the velocity profile of the fluid any point in the pipe will not change depending on how far it is into the pipe, i.e. “fully developed flow”. Thus, I will cross out all of the velocity derivative terms with dx in them. 5) In the x direction, the pressure doesn’t change as you go up the slope in the x direction because we know that the pressure is at Patm at both ends, so dp/dx is equal to 0. (Note that we exclude the pressure exerted due to gravity because it is accounted for in the NS equation already).
Crossing everything out, we get (lets call this EQUATION 1 and 2) 0 = -dp/dx +mu* (d2u/dy2) + rho*gx 0 = -dp/dy + rho*gy Much better. Finally, boundary conditions: We usually assume the “no-slip” condition, which says that the fluid flow velocity at an interface with a solid is the same as the velocity of that solid. Thus, for V(x,y) according to the axes I drew earlier, V = 0 at the wall y =0, and V = v at y =B, where little v is the velocity of the conveyor belt. For the pressure, we know that P(0,0) = Patm+ rho*g*sinβ∗Β
c) Solve equations in (b) and derive an expression for velocity profile. First we need an expression for gx and gy Breaking gravity acceleration into x and y components, we get gx = g*sinβ, gy = g*cosβ gx = - g*sin(β)
gy = - g*cos(β)
Now we can solve EQUATION 1 AND 2 From Equation 2 (the question does NOT ask for this, but for reference) 0 = -dp/dy + ρgy and P(0,0) = Patm+ ρgB*cosβ Py (y) = -ρgy*cosβ + Patm+ ρgB*sinβ From Equation 1: 0 = mu* (d2u/dy2) + ρ*gx d2u/dy2 = - ρ g*cosβ/ µ Integrating once, du/dy = -ρ yg/ µ + a Integrating again, u = -ρ y2g*cosβ/(2 µ) + a*y + b where a and b are constants
(eq*)
Using boundary condition: u=v when y=B and u= 0 when y =0 to plug into (eq*) above, we get: b=0 a = v + ρ Bg*cosβ/(2 µ) Therefore, V = u = -ρ y g*cosβ/(2 µ) + y*v/B+ y*ρBg*cosβ/(2 µ)
