Is the Newton's law of viscosity false in this condition???

abrahamgx · July 5, 2009, 04:04

We have learn that the relation of shear stress and shear strain in the fluid is:
t=miu(du/dy),
here miu is the dynamic viscosity of the fluid.
Now we assume that in 2D, the fluid is flow aroud a point, at a constant angular velocity Omega. That is to say, in different radius, the fluid have the same angular velocity.
Okey, in this condition, all the flows have a same angular velocity Omega, and it sames that there is no shear stress due to this.
But according to the Newton's law of viscosity, there is a velocity gratient along the radius direction.
So...is Newton's Law of Velocity false in this condition?

Dinocrack · July 5, 2009, 10:37

No. The Newton's law of viscosity still true in this condition.

The angular velocity omega is the same in all fluid, but the velocity is diferent along the radius, beucase w=v*r, this mean that exist a shear stress. You can work with radial velocity, and the shear stress is
approximately equal to the binary who gives the angular velocity. This will give something like:

t=miu(du/dy)=(3*B)/(2*pi*r^3),

where B is the binary and r is the radius.

This is the ideia used in rotational rheometry to measure the properties like viscosity, strain, etc, and this instrument is quite used in rheology because the use of a parallel reometry is not practical for big shear strain measurements.

abrahamgx · July 6, 2009, 02:59

Thank you Dinocrack!
Could you give a explanation of "Binary" in detail...I've never heard of that....

Dinocrack · July 6, 2009, 06:35

sorry i was thinking in my language.

what i mean with Binary is a Torque. This represents the magnitude of a force in a determinate distance, in a rotational system.

B=Fxd

where d is distance, F the force.

A practical case is the door. You have a point where the door rotate around. To open it near the oposite side of the rotation point, you need one amount of force. If you try to open it more closer to the rotation point you will need more force.

This is the basic concept and is a basic knowledge of mechnical engineering of physics. You can search more about it in wikipedia.

Gianluca · July 6, 2009, 16:48

You can define Newtonians fluids those for wich the dependence between the componens of viscous stress tensor and those of rate of strain is linear.

Dinocrack · July 6, 2009, 17:21

yes its true the upper laws is for newtonian fluids.

That law, can be the particular case from power law for non-newtonian fluid, when n=1

t=miu(du/dy)^n

abrahamgx · July 7, 2009, 00:50

Thank you Gianluca and Dinocrack.
But I think there are no relative movement in the fluid at different radius due to a same angular velocity, although the tangential velocity is different....
I am confused about the reateve stationary move and and the velocity gradient.....
The whole fluid can rotating as long as possible? Or it will stop as there are shear stresses....

sbaffini · July 7, 2009, 05:27

well, it's not that difficult to see that in a costant angular motion around a point the fluid has a linear velocity dependence with the cartesian coordinates so, following the definition of the stress tensor, it has a costant shear stress everywhere.

Ananda Himansu · July 7, 2009, 06:57

In 2-D "solid body" rotation of a fluid, every fluid particle displays the same angular velocity as every other. There is no shear strain rate in the fluid (except at or in the vicinity of the boundary of the container of a viscous fluid, if that boundary wall is not moving at the same angular velocity). The velocity gradient tensor (being a second rank tensor) can be uniquely written as the sum of a symmetric part and an antisymmetric part. For the velocity gradient tensor, the symmetric part is the strain rate tensor, and the antisymmetric part is 0.5 of the vorticity tensor. In this particular motion, the velocity gradient is nonzero. Its antisymmetric part is nonzero, and represents the angular velocity (0.5 vorticity) 0.5*(dv/dx-du/dy). Its symmetric part is zero, and represents the strain rate, of which the shear component is 0.5*(dv/dx+du/dy).

sbaffini · July 7, 2009, 09:59

Well...i was actually wrong, badly wrong.

Of course in this case there is no stress at all but just constant vorticity everywhere
and the fluid behaves in fact inviscidly.

However, to reach a prescribed angular velocity at the steady state, there will be an initial energy loss dependent of the velocity and the viscosity

By the way, i also appreciated the former personal opinion

Dinocrack · July 8, 2009, 06:34

The ideias that you introduce are new for me but seem very interesting.

But "solid body" rotation of a fluid is an ideal movement, isnt it?

July 5, 2009, 04:04	Is the Newton's law of viscosity false in this condition???	#1
abrahamgx New Member abrahamgx Join Date: Mar 2009 Posts: 11 Rep Power: 17	We have learn that the relation of shear stress and shear strain in the fluid is: t=miu(du/dy), here miu is the dynamic viscosity of the fluid. Now we assume that in 2D, the fluid is flow aroud a point, at a constant angular velocity Omega. That is to say, in different radius, the fluid have the same angular velocity. Okey, in this condition, all the flows have a same angular velocity Omega, and it sames that there is no shear stress due to this. But according to the Newton's law of viscosity, there is a velocity gratient along the radius direction. So...is Newton's Law of Velocity false in this condition?

July 7, 2009, 06:57		#9
Ananda Himansu New Member Ananda Himansu Join Date: Apr 2009 Location: Cleveland, Ohio, USA Posts: 17 Rep Power: 17	In 2-D "solid body" rotation of a fluid, every fluid particle displays the same angular velocity as every other. There is no shear strain rate in the fluid (except at or in the vicinity of the boundary of the container of a viscous fluid, if that boundary wall is not moving at the same angular velocity). The velocity gradient tensor (being a second rank tensor) can be uniquely written as the sum of a symmetric part and an antisymmetric part. For the velocity gradient tensor, the symmetric part is the strain rate tensor, and the antisymmetric part is 0.5 of the vorticity tensor. In this particular motion, the velocity gradient is nonzero. Its antisymmetric part is nonzero, and represents the angular velocity (0.5 vorticity) 0.5(dv/dx-du/dy). Its symmetric part is zero, and represents the strain rate, of which the shear component is 0.5(dv/dx+du/dy). Last edited by Ananda Himansu; July 7, 2009 at 07:02. Reason: removed personal opinion, changed stress to strain

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July 5, 2009, 10:37		#2
Dinocrack Member Nuno Gomes Join Date: May 2009 Location: Portugal Posts: 39 Rep Power: 17	No. The Newton's law of viscosity still true in this condition. The angular velocity omega is the same in all fluid, but the velocity is diferent along the radius, beucase w=vr, this mean that exist a shear stress. You can work with radial velocity, and the shear stress is approximately equal to the binary who gives the angular velocity. This will give something like: t=miu(du/dy)=(3B)/(2pir^3), where B is the binary and r is the radius. This is the ideia used in rotational rheometry to measure the properties like viscosity, strain, etc, and this instrument is quite used in rheology because the use of a parallel reometry is not practical for big shear strain measurements.

July 6, 2009, 02:59		#3
abrahamgx New Member abrahamgx Join Date: Mar 2009 Posts: 11 Rep Power: 17	Thank you Dinocrack! Could you give a explanation of "Binary" in detail...I've never heard of that....

July 6, 2009, 06:35		#4
Dinocrack Member Nuno Gomes Join Date: May 2009 Location: Portugal Posts: 39 Rep Power: 17	sorry i was thinking in my language. what i mean with Binary is a Torque. This represents the magnitude of a force in a determinate distance, in a rotational system. B=Fxd where d is distance, F the force. A practical case is the door. You have a point where the door rotate around. To open it near the oposite side of the rotation point, you need one amount of force. If you try to open it more closer to the rotation point you will need more force. This is the basic concept and is a basic knowledge of mechnical engineering of physics. You can search more about it in wikipedia.

July 6, 2009, 16:48		#5
Gianluca New Member Join Date: Jul 2009 Posts: 2 Rep Power: 0	You can define Newtonians fluids those for wich the dependence between the componens of viscous stress tensor and those of rate of strain is linear.

July 6, 2009, 17:21		#6
Dinocrack Member Nuno Gomes Join Date: May 2009 Location: Portugal Posts: 39 Rep Power: 17	yes its true the upper laws is for newtonian fluids. That law, can be the particular case from power law for non-newtonian fluid, when n=1 t=miu(du/dy)^n

July 7, 2009, 00:50		#7
abrahamgx New Member abrahamgx Join Date: Mar 2009 Posts: 11 Rep Power: 17	Thank you Gianluca and Dinocrack. But I think there are no relative movement in the fluid at different radius due to a same angular velocity, although the tangential velocity is different.... I am confused about the reateve stationary move and and the velocity gradient..... The whole fluid can rotating as long as possible? Or it will stop as there are shear stresses....

July 7, 2009, 05:27		#8
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,195 Blog Entries: 29 Rep Power: 39	well, it's not that difficult to see that in a costant angular motion around a point the fluid has a linear velocity dependence with the cartesian coordinates so, following the definition of the stress tensor, it has a costant shear stress everywhere.

July 7, 2009, 09:59		#10
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,195 Blog Entries: 29 Rep Power: 39	Well...i was actually wrong, badly wrong. Of course in this case there is no stress at all but just constant vorticity everywhere and the fluid behaves in fact inviscidly. However, to reach a prescribed angular velocity at the steady state, there will be an initial energy loss dependent of the velocity and the viscosity By the way, i also appreciated the former personal opinion

July 8, 2009, 06:34		#11
Dinocrack Member Nuno Gomes Join Date: May 2009 Location: Portugal Posts: 39 Rep Power: 17	The ideias that you introduce are new for me but seem very interesting. But "solid body" rotation of a fluid is an ideal movement, isnt it?