MrsDrPoe: Momentum Integral Boundary Layer Equation

Happy Thesis Tuesday to all!  Today we'll be looking at another case of external flow- the momentum integral boundary layer equation.

Defining the boundary layer height as the height from the plate where the velocity reached 99% of the free stream velocity is appealing, but often this is difficult to physically measure.  Instead, we can measure the displacement thickness, delta*, which is how far the free stream streamlines are shifted (or displaced) due to the boundary layer.  We can derive an expression for the displacement thickness by examining conservation of mass in a control volume that has a top control surface along a displaced streamline:

d/dt(int(rho)dV) + int(rho*(V.n))dA = 0

We will assume that the flow is steady and 1D with the flow crossing through two areas of the control surface with unit width (y(1) and H(1)) such that:

int(rho*u)dy|0,y - int(rho*Uinfinity)dy|0,H = 0

Assuming the flow is also incompressible, we can divide the rho out of both integrals:

int(u)dy|0,y - int(Uinfinity)dy|0,H = 0

We know that the free stream velocity is a constant and that y can be written as H + delta* and u can be written as Uinfinity + u - Uinfinity:

Uinfinity*H = int(Uinfinity + u - Uinfinity)dy|0,y = int(Uinfinity)dy|0,H+delta* + int(u-Uinfinity)dy|0,y = Uinfinity*H + Uinfinity*delta* + int(u - Uinfinity)dy|0,y

Further simplification yields:

Uinfinity*delta* = int(Uinfinity - u)dy|0,y or delta* = int(1 -(u/Uinfinity))dy|0,y

As previously mentioned, this value is much easier to measure than delta.

Another boundary layer thickness definition commonly employed is the momentum thickness (Theta), which is related to the amount of momentum "lost" due to drag in the boundary layer.  We can derive an expression for this thickness by applying the x-momentum equation to the control volume previously defined:

d/dt(int(rho*V)dV) + int(rho*V*(V.n))dA = Sum(F)

becomes:

int(rho*V*u)dA - int(rho*V*Uinfinity)dA = Fdrag

Taking the x-momentum equation:

int(rho*u*u)dA - int(rho*Uinfinity*Uinfinity)dA = -Fdrag

and applying the flow through an area of unit width with Fdrag = rho*Uinfinity*Uinfinity*Theta, then:

-rho*Uinfinity*Uinfinity = int(rho*u*u)dy|0,y - int(rho*Uinfinity*Uinfinity)dy|0,H = int(rho*u*u)dy|0,y - Uinfinity*int(rho*Uinfinity)dy|0,H

Employing the continuity equation:

-rho*Uinfinity*Uinfinity*Theta = int(rho*u*u)dy|0,y - Uinfinity*int(rho*u)dy|0,y

We can divide both sides by -rho*Uinfinity*Uinfinity (assuming an incompressible fluid) to obtain:

Theta = -int((u*u)/(Uinfinity*Uinfinity))dy|0,y + int(u/Uinfinity)dy|0,y = int((u/Uinfinity) - (u*u)/(Uinfinity*Uinfinity))dy|0,y

or finally:

Theta = int((u/Uinfinity)*(1 - (u/Uinfinity)))dy|0,y

This momentum thickness is also more easily measured than delta.  For most cases, Theta < delta* < delta.  These integrals are valid for any y in the free stream.  It can also be noted that the shear stress on a flat plate can be written in terms of the momentum integral:

tauw = rho*U*U*d/dx(Theta)

Next week, we'll look at a quick example of these equations in action, but until then, happy studying!

MrsDrPoe: Blasius Boundary Layer Example

It's another Thesis Tuesday on the blog, and as promised, I'll be giving you an example of the Blasius Boundary Layer Solution.

Problem: A viscous fluid with a known density (rho = 1300 kg/m^3) flows past a flat plate such that the boundary layer thickness at a distance 1.3 m from the leading edge is 12 mm.  Determine the boundary layer thickenss, shear stress and friction coefficient at a distance of 0.2 m from the leading edge if the free stream velocity is 1.5 m/s.

Given: xk = 1.3 m, deltak = 12 mm, rho = 1300 kg/m^3, U = 1.5 m/s
Find: delta, tauw, and Cf at 0.2 m
Solution:

Rek = (5*xk/deltak)^2 = 2.934x10^5

nu = U*xk/Rek = 6.646x10^-6 m*m/s

mu = nu*rho = 8.64x10^-3 N*s/m*m

Rex = U*x/nu =  45,140

delta = 5*x/sqrt(Rex) =4.707 mm

tauw = 0.3332*U^(3/2)*sqrt(rho*mu/x) = 4.571 Pa

Cf = 0.664/sqrt(Rex) = 3.125x10^-3

Pretty simple, huh?  Until next time...keep studying!

MrsDrPoe: The Blasius Boundary Layer Solution

It's Thesis Tuesday on the blog, and today we'll be looking at the Blasius Boundary Layer solution.

Often when discussing boundary layer flow over a flat plate, the Blasius boundary layer solution is examined.  This solution is based on the fact that, even though the boundary layer profile changes along the plate, its general shape remains the same.  To capture this relationship, Blasius defined a similarity variable, eta = y*sqrt(Uinfinity/(2*nu*x)).  We can define a stream function for the flow based on the volumetric flow rate between streamlines: psi = sqrt(2*nu*Uinfinity*x)*f(eta).  Using our stream function/velocity relations and the product rule, we can see that:

u = d/dy(psi) = Uinfinity*f'(eta)

v = -d/dx(psi) = sqrt((nu*Uinfinity)/(4*x))*(eta*f'(eta) - f(eta)).

Since we now have our velocity components in terms of x and y, we can plug these expressions into the boundary layer momentum equation to obtain (with MUCH manipulation): 

f''' + f*f'' = 0.

The boundary conditions for this differential equation are f = f' = 0 at eta = 0 and f' -> 1 as eta -> infinity.  Check out the figure here.

From the computed solution, it can be seen that u = 0.99*Uinfinity when eta = 5.0.  This gives us: delta = 5*sqrt(nu*x/Uinfinity) = 5*x/sqrt(Rex), where Rex is the local Reynolds number at any x-location along the plate (Rex = Uinfinity*x/nu).  Furthermore, since we now know the velocity profile in the boundary layer, we can calculate the shear stress at any x-location along the plate by tauw = 0.332*Uinfinity^(3/2)*sqrt(rho*mu/x).  We can also determine the friction coefficient at a particular x-location Cf = 0.664/sqrt(Rex).

Next week we'll look at an example problem involving these equations.  Happy studying!

MrsDrPoe: Boundary Layer Equations

It's once again Thesis Tuesday on the blog, and today we'll look some more at external flow.

One of the most important considerations of external flow is this region around the body in the flow that is affected by viscous stresses.  This region, called a boundary layer, possesses a large velocity gradient as the velocity transitions from 0 at the wall (from the no-slip condition) to 0.99*Uinfinity at the top of the boundary layer.  Since tau = mu*d/dy(u), we know that it is because of this velocity gradient that viscous stresses are important!  Typically, the height (distance from the surface to the top of the boundary layer) is denoted delta.

Fluid flow in the boundary layer is governed by a special reduced form of the governing equations.  We can find these equations using simple scaling analysis.  We will begin with the continuity equation for flow over a flat plate:

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

Reducing for incompressible, 2D flow:

d/dx(u) + d/dy(v) = 0

We can say that du scales with Uinfinity (Uinfinity is a representative x-velocity), dx scales with L, dv scales with vs, and dy scales with delta.   For continuity to be satisfied, these two scaled terms must be of the same magnitude.  Thus, setting them equal, we find that vs = U*delta/L.

Next we will look at the Navier-Stokes y-momentum equation (so we've made assumptions of Newtonian Fluid, steady, constant properties, laminar, incompressible) for 2D flow:

rho*(u*d/dx(v) + v*d/dy(v)) = -d/dy(p) + v*(d/dx(d/dx(v)) + d/dy(d/dy(v)))

Applying our scaling arguments as before with the exception of p scaling with rho*U*U: rho*U*vs/L, rho*vs*vs/delta, rho*U*U/delta, v*vs/(L*L), v*vs/(delta*delta).  If we substitute in the value for vs that we found from the continuity equation:

rho*U*(1/L)*(U*delta/L), rho*(U*delta/L)*(1/delta)*(U*delta/L), rho*U*U/delta, v*(1/(L*L))*(U*delta/L), v*(1/delta)*(U*delta/L)

We can make these terms dimensionless by multiplying both sides by delta/(U*U*rho): (delta*deta)/(L*L), (delta*delta)/(L*L), 1, (delta*delta)/(L*L), 1/Re, 1/Re.  If we then take the limit as delta/L (the boundary layer is VERY thin) -> 0 and Re -> infinity, we see that the only term that does not go to zero is the pressure term, d/dy(p).  From these arguments, we can see that d/dy(p) = 0, or the pressure does not vary significantly in the direction normal to the wall.

Finally, we will apply scaling arguments to the Navier-Stokes x-momentum equation (again already reduced for 2D flow): rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*(d/dx(d/dx(u)) + d/dy(d/dy(u))).  The terms become: rho*U*U/L, rho*vs*U/delta, rho*U*U/L, v*U/L, v*U/delta.  Substituting our expression for vs: rho*U*U/L, rho*(U*delta/L)*U/delta, rho*U*U/L, v*U/L, v*U/delta and non-dimensionalizing by multiplying by L/(U*U*rho): 1, 1, 1, 1/Re, (L*L/delta*delta), 1/Re.  If we again take the limit as delta/L -> 0 and Re -> infinity, we can see that the fourth term goes to zero; however, it is unclear what the last term becomes.  We can examine three possibilities: a) it's less that 1, b) it's greater than 1, or c) it's equal to 1.  

For case (a), scaling allows us to eliminate this term (since it is insignificant compared to the terms that reduced to 1) leaving: rho*(u*d/dx(v) + v*d/dy(v)) = -d/dx(p).  This equation is true for potential flow (irrotational, no viscous stresses), but since we definitely have these stresses present in the boundary layer, this option cannot be true.  

For case (b), the fifth term is the most significant term in the momentum equation so it would become: 0 = v*d/dy(d/dy(u)).  While this equation accounts for viscous stresses, it is purely diffusive and incorrect (we can see that d/dy(u) changes along the plate in the x-direction and is not constant).  

So case (c) must be correct, which means our boundary layer x-momentum equation can be written: rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*d/dy(d/dy(u)).

A few miscellaneous notes:

Our full set of boundary layer equations consists of: 

d/dx(u) + d/dx(v) = 0

 

rho*(u*d/dx(u) + v*d/dy(u)) = -d/dx(p) + v*d/dy(d/dy(u))

 

0 = d/dy(p)

The boundary conditions for these equations are: u(x,0) = 0 (no-slip), v(x,0) = 0 (no-slip), u(x,delta) = Uinfinity, and u(x0,y) = uin(y) (starting velocity profile).

Incidentally, for our scaling: delta/L scales with 1/sqrt(Re).  Also, just outside the boundary layer, Bernoulli's equation applies.