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!
