This screencasts uses this the idea of boundary layers and boundary layer thickness. As well
as flow regime, laminar or turbulent. Where the flow turns turbulent. So what we have
here. Is we have a flat plate, and on that flat plate. We have water that flows by it.
The flat plate is oriented to the flow, and a boundary layer is created. So we want to
know, at a distance of 0.5 m down the plate. So something like this. Were the boundary
layer thickness is 6 mm. Is the flow laminar or turbulent. So in order to find that out.
We need to calculate the Re, because if it is less then 5 times ten to the 5. Then it
is laminar. If it is greater then 5 times ten to the 5 it is turbulent. So our Re is
U*x divided by the kinematic viscosity of water, and we notice we have our x, and we
can certainly look up the viscosity, but we are missing our U. So we go back to what we
are given. Which is the boundary layer thickness. We can use the relationship, that the boundary
layer thickness is 5 times the square root of the kinematic viscosity times x over big
U or our free stream velocity. We have everything except for our. So we can solve for it. So
we rewrite this at U equals kinematic viscosity times x times 5 over del squared, and when
we put in our numbers. Here is our kinematic viscosity 1.12 times ten to the minus 6 m^/s,
and our x, which is 0.5 m, and our 5 over 6 times 10^-3 meters. We come out with a velocity
of about 0.39 m/s. So now we can check Re, 0.39 times 0.5 m divided by 1.12*10^-6 m^2/s,
and we come out with a Re of 1.74 times 10^5, which is less then 5*10^5. So our flow is
laminar. So lets see where the flow becomes turbulent, and that means that our Re has
to be 5*10^5. So we can rewrite this as our x equals our critical Re times our kinematic
viscosity divided by our free stream velocity, and we got this by rearranging our Re number
such that x equals Re times the kinematic viscosity divided by U. So when we calculate
our x critical. We find that it is 1.43 m down the plate. What is the boundary layer
thickness there. Again we are going to use this equation, but now our x is going to be
1.43 m. Everything else is going to stay the same. So this is going to be 5 times the square
root of 1.12*10^-6 times 1.43 meters divided by in this case is all under the square root,
0.39 m/s, and we find out that the boundary layer thickness there is 0.01 m. One thing
though that you have to remember is that we calculated this boundary layer thickness assuming
at this point that the flow was still laminar.