Can someone explain this equation for Reynolds' number?

In summary, the British Standard 6891:2015 states that in the UK, the Reynolds number is equivalent to 25 043 x Q/d for natural gas and 83 955 x Q/d for LPG, where Q is the flow rate in cubic meters per hour and d is the pipe diameter in millimeters. While many sources state that the Reynolds number is proportional to the pipe diameter, this formula shows that it is actually inversely proportional. This is due to the fact that when expressed in terms of flow rate, the Reynolds number is inversely proportional to diameter, while it is directly proportional when expressed in terms of velocity. This demonstrates the importance of understanding the relationships between variables in equations.
  • #1
Jehannum
102
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The following is quoted from British Standard 6891:2015, the standard for the installation of domestic gas pipework:

"In the UK, the Reynolds number is taken to be equivalent to:
25 043 x Q/d for natural gas; and
83 955 x Q/d for LPG."

[Q = flow rate, cubic metre per hour, d = pipe diameter, mm]

In every internet source I've looked at, Reynolds' number is proportional to pipe diameter; in this BS 6891 formula it's inversely proportional. How can this be?
 
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  • #2
I am not versed at all in fluid dynamics, but a brief scour of a few reliable (I believe( locations seemed to support that indeed the Re is proportional to the inverse cross-section area. This also makes sense, since less viscous material would flow 'better' in a wider pipe.
 
  • #3
Jehannum said:
The following is quoted from British Standard 6891:2015, the standard for the installation of domestic gas pipework:

"In the UK, the Reynolds number is taken to be equivalent to:
25 043 x Q/d for natural gas; and
83 955 x Q/d for LPG."

[Q = flow rate, cubic metre per hour, d = pipe diameter, mm]

In every internet source I've looked at, Reynolds' number is proportional to pipe diameter; in this BS 6891 formula it's inversely proportional. How can this be?

Reynolds number Re = uL/ν can be re-written by substituting in an expression for the velocity u (say, Poiseulle flow) and integrating to obtain an expression in terms of the volume flow instead of the pressure gradient. When all is done, Re = 4Q/πνd for cylindrical pipes.
 
  • #4
Andy Resnick said:
Reynolds number Re = uL/ν can be re-written by substituting in an expression for the velocity u (say, Poiseulle flow) and integrating to obtain an expression in terms of the volume flow instead of the pressure gradient. When all is done, Re = 4Q/πνd for cylindrical pipes.
This equation is correct, but no integration is required.
 
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  • #5
I solved this problem and learned something in the solving.

Re = density . v . d / viscosity [eqn. 1]

Q = v . cross-section = velocity . pi . (d / 2)^2

Transposing: v = 4 Q / (pi . d^2) [eqn. 2]

Substituting eqn. 2 into eqn. 1: Re = 4 . Q . density . d / (pi . d^2 . viscosity)

Therefore Re is inversely proportional to diameter when the relation is expressed in terms of flow rate, and Re is directly proportional to diameter when the relation is expressed in terms of velocity.

What I learned was the surface-level proportionalities you see in equations can be misleading when variables are not independent.
 
  • #6
Jehannum said:
I solved this problem and learned something in the solving.

Re = density . v . d / viscosity [eqn. 1]

Q = v . cross-section = velocity . pi . (d / 2)^2

Transposing: v = 4 Q / (pi . d^2) [eqn. 2]

Substituting eqn. 2 into eqn. 1: Re = 4 . Q . density . d / (pi . d^2 . viscosity)

Therefore Re is inversely proportional to diameter when the relation is expressed in terms of flow rate, and Re is directly proportional to diameter when the relation is expressed in terms of velocity.

What I learned was the surface-level proportionalities you see in equations can be misleading when variables are not independent.
That should not be a d^2 in the denominator of your last equation.
 

1. What is Reynolds' number and why is it important in fluid mechanics?

Reynolds' number is a dimensionless quantity used to predict the type of fluid flow (laminar or turbulent) in a specific situation. It is calculated by dividing the product of velocity, length, and fluid density by the fluid viscosity. This number helps determine the behavior of fluids in different systems, making it a crucial tool in fluid mechanics.

2. How does Reynolds' number relate to the flow regime of a fluid?

Reynolds' number is used to classify the flow regime of a fluid. If the number is below 2300, the flow is considered laminar, meaning the fluid particles move in smooth, parallel layers. If the number is above 4000, the flow is turbulent, characterized by chaotic movement and mixing of fluid particles. Numbers between 2300 and 4000 indicate transitional flow, which can exhibit elements of both laminar and turbulent flow.

3. What is the significance of the parameters involved in the calculation of Reynolds' number?

The parameters used to calculate Reynolds' number represent various physical properties of the fluid and the system. The velocity represents the rate of flow, the length represents the characteristic size of the system, and the fluid density and viscosity represent the resistance to flow. These parameters help determine the type of flow and provide valuable information for designing and analyzing fluid systems.

4. Can Reynolds' number be used for all types of fluids and flow situations?

Reynolds' number can be used for any type of fluid, including gases and liquids, as long as the fluid properties and flow conditions are known. However, it is most commonly used for incompressible fluids with a constant viscosity and density. For compressible fluids, a modified version of Reynolds' number is used, taking into account the changes in density and viscosity with pressure and temperature.

5. How is Reynolds' number applied in practical engineering and research?

Reynolds' number is widely used in various fields, such as aerodynamics, hydraulics, and chemical engineering. It is used to predict the flow behavior of fluids in pipes, pumps, turbines, and other systems. It also helps in the design and optimization of these systems, as well as in research to better understand fluid dynamics and improve fluid-related technologies.

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