Mass Conservation And Momentum Conservation

Lecture 23, March 3, 2004 • Reminder Quiz 2 Tuesday o Covers from the last quiz to end of Ch7 We saw last day how to reduce our mass conservation and momentum conservation (zero pressure gradient boundary layer equation) to single third order ordinary differential equation. Relatively speaking, this is very easy to solve.

Where

and

We need to set the boundary conditions on order to solve this equation, and we will need three of them since it is a third order equation. The no slip condition at the wall gives us two, u(0) = v(0) = 0. The final boundary condition comes from the edge of the boundary layer, where u = U∞. Now, we need to convert these boundary conditions into functions of f

and η, in order to solve the above. We need to use the equations developed last day in order to do this. Since, then,

The v velocity turned out to be,

Therefor, Finally,

so, This last boundary condition is a little problematic for most ‘solvers’ which require initial conditions (i.e. all the boundary conditions at η=0). We basically want to iterate on a boundary condition for f’’(0) until we see that we indeed get f’’(∞)=1. We can basically iterate on this condition until we see the physical solution that we want. This is reflected in the value of f’, which is the non-dimensional velocity

profile and which must approach a value of 1 as the free stream is reached. First let us try f’’(0) = 0.

This is not too interesting. Our velocity profile is simply all zero’s! Next let us try f ’’(0) = 1.0.

This is starting at least to look like a reasonable velocity profile, but the final value is 2.0, which is clearly too large. Lets try f ’’(0) = 0.5.

This is getting better – we are starting to get close. We find that f ’’(0) = 0.332 gives us exactly asymptotic behaviour we are looking for – f ’’(∞) = 1.0.

The next challenge is to define the edge of the boundary layer, and hence the boundary layer thickness. This is typically defined as the location where . We can explore our solution to see where this occurs. We cab see that this occurs very near η = 5.0. Others have investigated this in much more detail, and it is generally accepted that the edge of the momentum boundary layer in this flow occurs at η = 5.0. Note though that there is some arbitrariness in this! From the definition of η then,

Τhe edge of the boundary layer, δ, is y when η =5.0 We can simplify this, by introducing the Reynolds number,

What about the shear stress?

We found last day that therefore,

and f’’(0) is the boundary condition that we just figured out to be 0.332. We shall make this non-dimensional,

We saw earlier that the average value, when the function is of x^-1/2, is twice the local value. We therefore know the average friction coefficient as well.

We know need the energy equation, in order to find the functional form of the Nusselt number variation in order to characterize the heat transfer from the plate.

To begin, we need to non-dimensionalize our temperature.

And manipulate the energy equation as above to obtain,

An ordinary differential equation for T, which we can also easily solve.