# Video: Onset of Turbulence in Fluid Flow

At what flow rate might turbulence begin to develop in a water main with a 0.200 m diameter? Use a value of 8.94 Γ 10β»β΄ Paβs for the viscosity of water and assume that a Reynolds number of 2000 corresponds to the onset of turbulence.

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### 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 weβve 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 donβt 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 πΏ, all divided by its viscosity π.

When we write out the Reynolds number equation for our scenario, the characteristic length πΏ 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 itβs 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, weβre ready to plug in and solve for π. Just before we do though, we notice that a factor of the pipeβs diameter π· cancels out. Now when weβre 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. Thatβs the flow rate at which turbulence begins to develop in this water main.