# What rate is needed for turbulent flow? [Challenge Solved]

Given a tubular membrane 4 inches in diameter what flow rate is needed
for turbulent flow?

Facts:

• A desalinization plant has upgraded to a new micro filtration system as part of its pretreatment of sea water.
• The new system uses a tubular membrane, four inches in diameter.
• Turbulent flow conditions must exist within the tube for best performance.

Assuming well developed flow, and that sea water has essentially the same properties as pure water, what flow rate is needed to achieve fully turbulent
conditions?

It is past time to look at this month's Chenected challenge problem, so lets get started!

This month's problem asked what flow rate of water would be needed to have fully developed turbulent flow in a circular tube. The Reynolds number is a dimensionless term used to determine if a fluid low is laminar or turbulent. The Reynolds number is defined as

Re = density (kg/m3) * fluid velocity (m/s) * diameter (m) / viscosity (N*s/m2)

(Reynolds number is some times defined in terms of kinematic viscosity, which is density over viscosity)

To put the equation above in terms of a mass flow rate we can use the relationship

Density (kg/m3) * fluid velocity (m/s) * cross sectional area (m2) = mass flow rate (kg/s)

If we combine those equations and solve for mass flow rate we get the following equation (keep in mind for our circular pipe cross sectional area = pi * D2 / 4).

Mass flow rate = Re * pi * D * viscosity / 4

According to "Fundamentals of Heat and Mass Transfer" by Incropera Dewitt Bergman and Lavine (sixth edition) fully developed turbulent flow occurs in circular pipes at a Reynolds number of 10000 (the laminar region ends at Re = 2500, flow is considered turbulent at Re ~ 5000 and fully developed at Re = 10000). The viscosity of water at 22 C is 9.59 e-4 N s / m2. Using those assumptions

(1e4 * 3.14 * ( 4 in = 0.1016 m) * 9.59 e-4 N s /m2 ) / 4 = 0.765 kg/s

The minimum flow rate of water to achieve fully turbulent flow in this tubular system is 0.765 kg/s or about 12 GPM.

Thanks to everyone who submitted an answer, including Trent Benanti and Bob McGurk! Be sure to participate in the next challenge. Stay tuned, we'll have it up soon. If you have an Idea for a ChEnected challenge problem we would love to hear it! Have a challenge of your own for the community? Let us know.