### Video Transcript

At what flow rate might turbulence
begin to develop in a water main with a 0.200-meter diameter? Use a value of 8.94 times 10 to the
negative fourth pascal seconds for the viscosity of water. And assume that a Reynolds number
of 2000 corresponds to the onset of turbulence.

In this exercise, we want to solve
for a flow rate we can call lowercase π. In this example, we have a water
main of diameter weβve called capital π· of 0.200 meters, allowing water with
viscosity 8.94 times 10 to the negative fourth pascal seconds to flow through
it. The flow rate π we want to solve
for corresponds to the onset of turbulence in the pipe. This means that the speed at which
the water flows increases until the Reynolds number of that flow is equal to
2000.

We know that, in general, flow rate
is equal to the area through which a fluid is flowing multiplied by the speed of
that fluid. If we call the cross-sectional area
of our pipe capital π΄, we can solve for that area based on the diameter of the pipe
π·. But we still donβt know the rate of
fluid flow π£. To solve for that speed π£, we can
recall that the Reynolds number describing fluid flow is equal to the density of
that fluid times its speed multiplied by a characteristic length πΏ, all divided by
its viscosity π.

When we write out the Reynolds
number equation for our scenario, the characteristic length πΏ is the diameter π· of
the pipe. We notice that fluid speed π£ is in
this equation for Reynolds number. So when we rearrange to solve for
π£, we find itβs equal to the Reynolds number multiplied by fluid viscosity divided
by fluid density times pipe diameter π·. If we substitute this expression in
for π£ in our equation for flow rate π and then replace the cross-sectional area π΄
with the area of the cross section in terms of diameter π·. We now have an expression for the
flow rate π in terms of known values, except for the density π of the water. The density of pure water is very
nearly 1000 kilograms per cubic meter.

So with that, weβre ready to plug
in and solve for π. Just before we do though, we notice
that a factor of the pipeβs diameter π· cancels out. Now when weβre entering the values
of these variables, when we plug in for the diameter π·, the Reynolds number of
2000, the viscosity of water, and its density π. When we enter these values on our
calculator, we find, to three significant figures, 2.81 times 10 to the negative
fourth cubic meters per second. Thatβs the flow rate at which
turbulence begins to develop in this water main.