Question Video: Computing the Reynolds Number of Water Flow through a Pipe | Nagwa Question Video: Computing the Reynolds Number of Water Flow through a Pipe | Nagwa

Question Video: Computing the Reynolds Number of Water Flow through a Pipe Physics

Water flowing through a pipe has a dynamic viscosity of 8.9 × 10⁻⁴ Pa⋅s and a density of 1,000 kg/m³. The pipe has an internal diameter of 0.025 m and the water flows at an average speed of 0.15 m/s. What is the Reynolds number for the flow?

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Video Transcript

Water flowing through a pipe has a dynamic viscosity of 8.9 times 10 to the negative four pascal seconds and a density of 1000 kilograms per cubic meter. The pipe has an internal diameter of 0.025 meters and the water flows at an average speed of 0.15 meters per second. What is the Reynolds number for the flow?

Let’s say that this is our pipe with water flowing through it. The Reynolds number of this flow is a number that indicates how laminar or turbulent the flow of water is. The Reynolds number for the flow of a fluid, represented this way, is equal to the density 𝜌 of a fluid multiplied by that fluid speed times what’s called the characteristic dimension of the flow, we’ll call it capital 𝐿, all divided by the dynamic viscosity 𝜇 of the fluid. In our scenario, we’ll use the same general form of the Reynolds number equation where, for us, the characteristic dimension of our fluid flow, that’s the most important length scale of the flow, is the inner diameter capital 𝐷 of our pipe. This inner diameter determines the flow’s characteristics more than any other length.

So then, we’ll calculate the Reynolds number for this flow of water by 𝜌 times 𝑣 times 𝐷, the inner diameter of the pipe, divided by the dynamic viscosity 𝜇 of our water. Note that we’re given values for all four of these quantities. The density 𝜌 of water is 1000 kilograms per cubic meter. The water flows at an average speed of 0.15 meters per second. It flows through a pipe with an inner diameter of 0.025 meters. And it has a dynamic viscosity or just viscosity for short of 8.9 times 10 to the negative four pascal seconds.

If we look at the units in this expression, believe it or not, they will cancel out entirely from numerator and denominator. We can start to see this by recalling that a pascal is defined as a newton per meter squared and that a newton is a kilogram meter per second squared. Combining these equations, we can write that a pascal is a kilogram per second squared meter. If we then replace the unit of pascal in our denominator with kilograms per second squared meter, we see that in the denominator, one factor of seconds cancels out. And we get overall units in our denominator of kilograms per second meter.

Considering next the units in our numerator, we have kilograms per meter cubed multiplied by meters per second multiplied by meters. We see that in this product, two factors of meters cancel from numerator and denominator. We get final units of kilograms per second meter. If we collect all the units in our numerator to the far right, we see that these units cancel out entirely with the units in our denominator.

When we calculate our Reynolds number, we’ll get a pure number with no units. That number rounded so that it matches the level of precision of our given information is 4200. This is the Reynolds number for the flow of water. And in general, a Reynolds number of higher than 2000 tends to indicate turbulent fluid flow. We would say then that the water in this pipe experiences turbulent rather than laminar flow.

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