MrsDrPoe: Conservation of Momentum, Part IV

Thesis Tuesday is upon us yet again, ladies and gents! As promised, today we will be continuing our discussion of the Conservation of Momentum Equation with an example of how to solve the Navier Stokes equations.

We'll start with fluid flowing between two parallel plates, separated by H meters and infinite in the z-direction. (Note: infinite plates do not exist, but this consideration allows us to approximate the flow as 2D instead of 3D). We will consider the pressure driven flow case where both plates are stationary (I), as well as the Couette Flow case where the upper plate moves in the +x-direction with some velocity U m/s (II).

Given: gap height = H meters
Find: the flow field description (u, v, w, and p)
Assumptions: steady, laminar, incompressible, infinite plates (w = 0, d/dz(...) = 0), constant properties, ignore edge effects, Newtonian fluid, fully developed (d/dx(...) = 0)
Solution:

Step 1: We must first apply the continuity equation.

d/dt(rho) + d/dx(rho*u) + d/dy(rho*v) + d/dz(rho*w) = 0

Since the flow is steady (or because it's incompressible), the first term is equal to zero. For fully developed, laminar flow, d/dx(rho*u) = 0; for infinite plates, d/dz(rho*w) = 0. Because the fluid is incompressible, rho can be pulled out of the derivative in the remaining term. Thus, we are left with:

rho*d/dy(v) = 0 or d/dy(v) = 0

Since the derivative of v is equal to zero, that means that v must be constant (found by integrating both sides of the above equation):

v = const

But we know that because we are dealing with a viscous fluid, the no-slip condition holds. This tells us that the fluid adjacent to either plate has the same velocity as the plate; therefore, since the bottom plate is fixed in both cases, u = v = w = 0 at this location. Putting this boundary information with the result from the continuity equation, v = 0 throughout the domain.

Step 2: Next, we simplify the Navier Stokes equations.

x-momentum:

rho*[d/dt(u) + u*d/dx(u) + v*d/dy(u) + w*d/dz(u)] = -d/dx(p) + rho*gx + mu*[d/dx(d/dx(u)) + d/dy(d/dy(u)) + d/dz(d/dz(u))]

d/dt(u) = 0 since the flow is steady; d/dx(u) = 0 since the flow is fully developed; d/dy(u) = d/dz(u) = 0 because the flow is laminar; thus, the entire left side of the equation equals zero for this case. On the right-hand side, rho*gx = 0 because gravity only acts in the downward (-y) direction; d/dx(d/dx(u)) = 0 also because the flow is fully developed; d/dz(d/dz(u)) = 0 since we are dealing with infinite plates. Rearranging the two terms that are left, we have:

d/dx(p) = mu*d/dy(d/dy(u))

y-momentum:

rho*[d/dt(v) + u*d/dx(v) + v*d/dy(v) + w*d/dz(v)] = -d/dy(p) + rho*gy + mu*[d/dx(d/dx(v)) + d/dy(d/dy(v)) + d/dz(d/dz(v))]

From the result of our continuity equation examination, we know that all the terms containing a v must be equal to zero. We can quickly see that we are only left with two terms:

d/dy(p) = rho*gy

When both sides of this equation are integrated, we obtain the hydrostatic pressure distribution.

z-momentum:

rho*[d/dt(w) + u*d/dx(w) + v*d/dy(w) + w*d/dz(w)] = -d/dz(p) + rho*gz + mu*[d/dx(d/dx(w)) + d/dy(d/dy(w)) + d/dz(d/dz(w))]

Because our flow is laminar, all terms containing a w or d/dz(...) are equal to zero. Also, we know that the term rho*gz = 0, since gravity is only acting in the downward (-y) direction. This equation quickly reduces to:

0 = 0

Step 3: Then we integrate the reduced Navier-Stokes equations and solve for the unknowns.

z-momentum:

0 = 0 tells us nothing

y-momentum:

p = -rho*gy*y + fp(x)
essentially the hydrostatic pressure distribution, the integration constant in this case, fp(x), is constant in the y direction, but may be a function of x

x-momentum:

(1/mu)*d/dx(p) = d/dy(d/dy(u)) rearranging

(1/mu)*d/dx(p)*y + const1 = d/dy(u) integrating once

(0.5/mu)*d/dx(p)*y*y + const2 = u integrating twice

We now have an expression for each of the variables necessary for a full flow field description (v = 0, w = 0, p = ..., u = ...). To completely finish the problem, we must examine our given boundary conditions to determine

Step 4 (I): We need two boundary conditions since there are two constants of integration that need to be determined. For this case, both plates are stationary; therefore, the no-slip condition tells us that u = 0 at y = 0 and u = 0 at y = H (if we consider the bottom plate to be located at y = 0). Applying the first yields:

const2 = 0

Applying the second gives:

const1 = -(0.5/mu)*d/dx(p)*H

So: u = (0.5/mu)*d/dx(p)*(y*y - H*y)

Step 4 (II): We again need two boundary conditions since there are two constants of integration that need to be determined. For this case, the bottom plate is stationary and the top plate is moving with constant velocity, U; therefore, the no-slip condition tells us that u = 0 at y = 0 and u = U at y = H (if we consider the bottom plate to be located at y = 0). Applying the first yields:

const2 = 0

Applying the second gives:

const1 = -(0.5/mu)*d/dx(p)*H + U/H

So: u = (0.5/mu)*d/dx(p)*(y*y - H*y) + (U*y)/H


We can plot the velocity for both of these cases verses y for specified values of d/dx(p), U, mu, and H; this will give us the velocity distribution of the flow between the plates. Here you can see an animation of velocity profiles for various values of U and d/dx(p). If you'd like to play with this program yourself, you can download it (for free I think) from Wolfram Mathematics.

And that's the Navier Stokes equations in a nutshell! Like I mentioned last week, a lot of assumptions need to be made in order to obtain a version of these equations that we can work with; however, once these have been made, the resulting forms of the equations can be solved rather simply.

If you're feeling extra adventurous, try to go through the procedure again by yourself, examining what the velocity profile looks like for the case where the top plate is moving and bottom one is stationary or one plate is moving in the +x-direction and one is moving in the -x-direction or both plates are moving in the +x-direction at the same speed and at different speeds.

MrsDrPoe: Conservation of Momentum, Part III

Once again Thesis Tuesday is upon us, so we'll be continuing our discussion of the Conservation of Momentum equation. Today we'll change gears a bit, moving toward examination of the differential form of the momentum equation. When I taught this introductory fluid mechanics class, this form of the equations was the most loved and most hated at the same time by my students- loved because it was much simpler to work with and hated because I always made them start from the full form of the (inviscid or viscous) equation and tell me why they ended up with the final result.

Recall that the linear momentum equation is actually a vector equation that is composed of 3 scalar equations (for a 3D problem). The differential (scalar) form of the linear momentum equation is:

d/dt(rho*u) + u*d/dx(rho*u) + v*d/dy(rho*u) + w*d/dz(rho*u) = rho*gx + d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) + sum(Fx)

d/dt(rho*v) + u*d/dx(rho*v) + v*d/dy(rho*v) + w*d/dz(rho*v) = rho*gy + d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) + sum(Fy)

d/dt(rho*w) + u*d/dx(rho*w) + v*d/dy(rho*w) + w*d/dz(rho*w) = rho*gz + d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) + sum(Fz)

We've discussed each of the components in these equations before except for the S and T terms. These terms are ones from the stress tensor/matrix- S represents the normal stresses on a fluid due primarily to pressure forces; T represents the shear stresses on a fluid due primarily to the viscosity of the fluid. These stress terms reduce to different forms depending on whether we are dealing with a viscous or an inviscid fluid. Today we'll examine the viscous fluid equations.

The viscous form of the momentum equations is achieved by applying Stokes' postulates:

1) Each stress component is a linear function of strain rate (Newtonian)
2) The properties of a fluid are isotropic (not varying with direction)
3) when all strain rates are equal to zero, hydrostatic pressure is the only stress present

and Stokes' assumption:

lambda = -(2/3)*mu, where lambda is the bulk viscosity (related to the volumetric dilatation of the fluid) and mu is the viscosity

These allow the stress terms in the equations above to be rewritten as:

d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) = -d/dx(p) + mu*(d/dx(d/dx(u))+d/dy(d/dy(u)) + d/dz(d/dz(u)))

d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) = -d/dy(p) + mu*(d/dx(d/dx(v))+d/dy(d/dy(v)) + d/dz(d/dz(v)))

d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) = -d/dz(p) + mu*(d/dx(d/dx(w))+d/dy(d/dy(w)) + d/dz(d/dz(w)))

Placing these expressions in the initial differential form of the momentum equation gives us the Navier Stokes Equations. What these scalar equations physically represents is:

Left side- "mass times acceleration" of a fluid particle (per unit volume)

Right side-
First term- force of gravity of a fluid particle per unit volume; if positive, this terms acts to accelerate the particle in the +x-direction

Second term- pressure difference across the fluid particle; if negative, than this term acts to accelerate the particle in the +x-direction

Third-fifth terms- viscous stresses acting on the fluid particle; if sum is positive, than this group acts to accelerate the particle in the +x-direction

Sixth term- any other forces (support, etc.) per unit volume acting on the fluid; this term is often neglected

By combining these equations with the continuity equation, we can provide a complete mathematical description of the flow of an incompressible, Newtonian fluid: 4 equations allows us to solve for 4 possible unknowns (p, u, v, and w). If we have exactly four unknowns, our problem is well-defined. Fewer and our problem is over-defined; more (unknown external forces, etc.) and our problem is ill-posed and we cannot solve it unless more information is obtained.

These equations may seem a bit intimidating, but remember, we can always employ simplifying assumptions in order to obtain a more manageable set. In fact, that's what we'll look at next week- application of the Navier Stokes Equations. So until next time, happy studying!