Can someone explain this equation for Reynolds' number?

[Jehannum]

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?

 

[_PJ_]

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.

 

[Andy Resnick]

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.

 

[Chestermiller]

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.

 

[Jehannum]

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.

 

[Chestermiller]

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.