Turbulence Intensity vs Reynolds number

jason580315 · February 26, 2017, 01:09

Hi everyone
I don't have too much background in fluid dynamics. From ANSYS user manual, turbulence intensity (I) of fully developed pipe flow is defined as :

I=0.16 *Re_dh^-1/8.

where Re_dh is the Reynolds number. It seems that turbulence intensity is inverse proportional to Reynolds number. We know that Reynolds number of laminar flow is much smaller than that of turbulent flow. But according to this equation, does that mean turbulence intensity of laminar flow is higher than that of turbulent flow? This doesn't make any sense because laminar flow should shouldn't have turbulence intensity or very low turbulence intensity. Which part I didn't understand correctly? Thanks.

piu58 · February 26, 2017, 03:00

I don't think that this formula is thought to be applicated for creeping flow or Re just beyond it.

In turbulent cases you have an effect: With lower Re the eddies take more room, because there is more time for expansion. An example is a well devloped Kármán vortex street. At much higher values of Re the turbulent area gets much more norrw, but with vortices containing more energy. A missile does't give a wide vortex street.

Thats the way I explain for me what the formula contains.

jason580315 · March 1, 2017, 01:50

Thanks Uwe.

abdelaziz begag · March 1, 2017, 05:21

hello every one.
I have a question about the flow regime, according to my bibliographic research I found that for low numbers of Reynolds i.e: 600- 1400,
The k-e turbulence model is introduced into the calculation, but the regime is laminar, can you please answer me.

February 26, 2017, 01:09	Turbulence Intensity vs Reynolds number	#1
jason580315 New Member Jason Join Date: Oct 2016 Posts: 11 Rep Power: 9	Hi everyone I don't have too much background in fluid dynamics. From ANSYS user manual, turbulence intensity (I) of fully developed pipe flow is defined as : I=0.16 *Re_dh^-1/8. where Re_dh is the Reynolds number. It seems that turbulence intensity is inverse proportional to Reynolds number. We know that Reynolds number of laminar flow is much smaller than that of turbulent flow. But according to this equation, does that mean turbulence intensity of laminar flow is higher than that of turbulent flow? This doesn't make any sense because laminar flow should shouldn't have turbulence intensity or very low turbulence intensity. Which part I didn't understand correctly? Thanks. BlnPhoenix likes this.

February 26, 2017, 03:00		#2
piu58 Senior Member Uwe Pilz Join Date: Feb 2017 Location: Leipzig, Germany Posts: 744 Rep Power: 15	I don't think that this formula is thought to be applicated for creeping flow or Re just beyond it. In turbulent cases you have an effect: With lower Re the eddies take more room, because there is more time for expansion. An example is a well devloped Kármán vortex street. At much higher values of Re the turbulent area gets much more norrw, but with vortices containing more energy. A missile does't give a wide vortex street. Thats the way I explain for me what the formula contains. __________________ Uwe Pilz -- Die der Hauptbewegung überlagerte Schwankungsbewegung ist in ihren Einzelheiten so hoffnungslos kompliziert, daß ihre theoretische Berechnung aussichtslos erscheint. (Hermann Schlichting, 1950)

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March 1, 2017, 01:50		#3
jason580315 New Member Jason Join Date: Oct 2016 Posts: 11 Rep Power: 9	Thanks Uwe.

March 1, 2017, 05:21		#4
abdelaziz begag New Member aziz Join Date: Nov 2016 Posts: 1 Rep Power: 0	hello every one. I have a question about the flow regime, according to my bibliographic research I found that for low numbers of Reynolds i.e: 600- 1400, The k-e turbulence model is introduced into the calculation, but the regime is laminar, can you please answer me.