rotational and inviscid

July 7, 2000, 13:46

Does anyone please tell me if inviscid flow can be rotational or not? What is the nature of rotational flow?

July 7, 2000, 14:09

Yes. Inviscid flow can be rotational. Rotational flow may be characterized by vorticity, which is the curl of velocity. A simple example is a vortex ring (like a cigarette smoke puff but without the diffusion process)

Check out fundamental fluid dynamic books.

It is important NOT to confuse potential flow with inviscid flow (which may be the source of your confusion). While potential flow satisfies the equation of motion of inviscid flow dynamics, the reverse is not necessarily true. That is, an inviscid flow is not necessarily potential (the former may contain vorticity, which induces rotational and not potential flow)

Adrin Gharakhani

July 7, 2000, 15:16

Inviscid flow can be rotational. As pointed out in an earlier reply, inviscid flows with vortices are good examples.

It might help to know that vorticity is conserved in inviscid flows, i.e., Euler equations are incapable of vorticity production. So in an inviscid flow computation, vorticity (in the form of point vortices or vortex blobs or continuous fields) can only come in as a part of the inflow and can not be produced or destroyed by the Euler equations.

Note also that the vortex rings can never be formed without viscosity though they can exist in a purely inviscid fluid.

July 7, 2000, 15:26

> Euler equations are incapable of vorticity production. So in an inviscid flow computation, vorticity (in the form of point vortices or vortex blobs or continuous fields) can only come in as a part of the inflow and can not be produced or destroyed by the Euler equations.

This is true _only_ for the incompressible (and constant density) case. Example: baroclinic vorticity generation (without solid boundaries).

Adrin Gharakhani

Shayor · July 7, 2000, 20:33

(1). It is not a good idea to mix these two terms at the same time. (2). At the begining, you have this equation called the Navier-Stokes equations, which represent the conservation of mass, momentum and energy. It contains viscosity terms. (3). Now, if you take the curl operation on the momentum equations, you will get this so-called "vorticity" equation. Remember that, the vorticity equation is derived from the momentum equation, so it has viscosity terms in it also. (4). If you set the viscosity terms to zero, you are going to get this mathematical equation, called "inviscid" equations, whether it is the inviscid momentum equations or the inviscid vorticity equation. (5). The inviscid momentum equation is generally called "Euler Equation" in contrast to the Navier-Stokes equations which has viscosity terms. (6). In the vorticity equation(derived from the Navier-Stokes equations), in addition to the viscosity terms, there are convection terms, stretching terms, production terms due to expansion, non-uniform density, and general body force. (7). So, even though you drop the viscosity terms, there are still many other terms in the vorticity equation, including the production terms. (8). So far, we have touched only the "inviscid" vs "viscous" forms of equation. (9). The rest of that is easy. The only time when the flow is "irrotational" is when the vorticity is everywhere zero. If there are still many terms in the vorticity equation (after dropping the viscosity terms), then you know that vorticity is normally non-zero. So, most of the time, it is rotational even without viscosity terms. (10). In old days, the inviscid equation was still very difficult to solve. So, the Euler equation was further simplified (second time) by setting the vorticity to zero (it is called "irrotational"). As a result, one can express the velocity in terms of the velocity potential and derive the so-called "potential flow"equation. (11). So, "potential flow" is "irrotational flow". And for the incompressible flow, the equation is the Laplace equation. The Laplace equation was the main focus in the 19th century. (12). Remember that, Euler equation and Laplace equation are simplified equations derived from the Navier-Stokes equations.(by setting the viscosity to zero, and then by setting the vorticity to zero, respectively)

July 14, 2000, 15:31

curved shocks can be a source of vorticity in an inviscid flow. i think there may be other potential sources but I can't remember what the others might be.

July 20, 2000, 12:39

where the velocity varies, where there exist vortex; where the entropy increases, where creates vortex.

Nick R · April 20, 2011, 02:25

Many thanks guys you explained things that confused me for awhile. Especial thanks to Adrian, John and Kalyan

silkworm · September 22, 2022, 05:37

I dont agree with you.
After a normal shock, velocity changes while no vortex exists.
In laminar boundary layer, there entropy increases but no vortex generates.

ichamail · November 2, 2023, 19:22

The vorticity equation derived from momentum equation for an inviscid incompressible flow is pressented below

$\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta} \cdot \nabla \right) \vec{V}$ , where $\vec{\zeta} = \nabla \times \vec{V}$

I read in several textbooks that, from this equation we come to the conclusion that if $\vec{\zeta}(t_0) = 0$ , then $\frac{d \vec{\zeta} }{dt} = 0$ for

Also in many of theses textbooks is written that this equation agrees with Kelvin's circulation theorem, which states that, if an inviscid fluid is initially irrotational then it remains irrotational at all subsequent times.

I can't understand, at least from a mathematical point of view, how that conclusion is made

LuckyTran · November 2, 2023, 19:30

Mathematically... starting with

$\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta} \cdot \nabla \right) \vec{V}$

put $\vec{\zeta}= 0$ into the RHS

Then you have $\frac{d \vec{\zeta} }{dt} = 0$

Hence if the vorticity was zero, it remains zero.

That's all there is to it

ichamail · November 2, 2023, 19:34

Quote:

Originally Posted by LuckyTran

Mathematically... starting with

$\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta} \cdot \nabla \right) \vec{V}$

put $\vec{\zeta}= 0$ into the RHS

Then you have $\frac{d \vec{\zeta} }{dt} = 0$

That's all there is to it

Now I feel dumb
Thank you

FMDenaro · November 3, 2023, 01:14

Note that d/dt is the lagrangian derivative, formally that means the vorticity along the path-line.

ichamail · November 3, 2023, 06:20

Quote:

Originally Posted by LuckyTran

Mathematically... starting with

$\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta} \cdot \nabla \right) \vec{V}$

put $\vec{\zeta}= 0$ into the RHS

Then you have $\frac{d \vec{\zeta} }{dt} = 0$

Hence if the vorticity was zero, it remains zero.

That's all there is to it

Revisiting your answer, wouldn't setting $\vec{\zeta}= 0$
lead to $\frac{d \vec{0} }{dt} = \left( \vec{0} \cdot \nabla \right) \vec{V} \implies 0 = 0$ ?

I mean, I would understand if the above equation was formulated as
$\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta_0} \cdot \nabla \right) \vec{V}$
where $\vec{\zeta_0}= \vec{\zeta}(x, y, z, t_0)$

that setting $\vec{\zeta_0}= \vec{\zeta}(x, y, z, t_0) = 0$ would lead to $\frac{d \vec{\zeta} }{dt} = 0$

But it's not formulated like that. Is there something too obvious that I'm missing?

LuckyTran · November 3, 2023, 08:31

Setting x=0 does not automatically imply anything about dx/dt

For example, if x=1, it does not imply that d(1)/dt is anything

Knowing f(a) doesn't tell you anything about f'(a)

ichamail · November 3, 2023, 10:01

Quote:

Originally Posted by LuckyTran

Setting x=0 does not automatically imply anything about dx/dt

For example, if x=1, it does not imply that d(1)/dt is anything

Knowing f(a) doesn't tell you anything about f'(a)

It actually does. that if x=1 then $\frac{dx}{dt} = \frac{d1}{dt} = 0$ . The only thing I can think of is that if $\frac{dx}{dt} = x$ is an ordinary diff equation of position then x = 0 is a solution to this ode and so it satisfies the differential equation $\frac{dx}{dt} - x = \frac{d0}{dt} - 0 = 0$ .
This type of solution means that if the system is at rest then it stays at rest.

Can this be an analogy to the pde equation of vorticity?

agd · November 3, 2023, 12:28

You have an initial value problem, with zero as the initial condition. The solution then is a constant, and that constant is the initial value zero. It really is that simple.

FMDenaro · November 3, 2023, 12:30

Quote:

Originally Posted by ichamail

It actually does. that if x=1 then $\frac{dx}{dt} = \frac{d1}{dt} = 0$ . The only thing I can think of is that if $\frac{dx}{dt} = x$ is an ordinary diff equation of position then x = 0 is a solution to this ode and so it satisfies the differential equation $\frac{dx}{dt} - x = \frac{d0}{dt} - 0 = 0$ .
This type of solution means that if the system is at rest then it stays at rest.

Can this be an analogy to the pde equation of vorticity?

You have to consider t=0, that is the equation writes

dz/dt|t=0 = (z.grad v)|t=0

LuckyTran · November 3, 2023, 14:39

When I say put vorticity=0 or x=1, I mean at one instance, not for all time.

For example, a new born child is born at a height (in US) of 20 inches. Eventually the baby grows into an adult and becomes much taller than 20 inches.

d(baby height)/dt is not zero (for all time).

Again.... f(a)=0 does not imply that f(t)=0 for all t. It simply means that f at t=a is 0. And it also does not imply anything about f'(t) = anything.

So yes, you are missing something very obvious, again and again about general mathematical notation.

Then you have the vorticity transport equation:

$\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta} \cdot \nabla \right) \vec{V}$

This equation is a general equation which relates the change of vorticity to velocity/vorticity field. Only from this equation can you evaluate what is $\frac{d \vec{\zeta} }{dt}$
Without this equation, you know nothing about the change in vorticity.

ichamail · November 3, 2023, 15:03

Quote:

Originally Posted by LuckyTran

When I say put vorticity=0 or x=1, I mean at one instance, not for all time.

For example, a new born child is born at a height (in US) of 20 inches. Eventually the baby grows into an adult and becomes much taller than 20 inches.

d(baby height)/dt is not zero (for all time).

Again.... f(a)=0 does not imply that f(t)=0 for all t. It simply means that f at t=a is 0. And it also does not imply anything about f'(t) = anything.

So yes, you are missing something very obvious, again and again about general mathematical notation.

Then you have the vorticity transport equation:

$\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta} \cdot \nabla \right) \vec{V}$

This equation is a general equation which relates the change of vorticity to velocity/vorticity field. Only from this equation can you evaluate what is $\frac{d \vec{\zeta} }{dt}$
Without this equation, you know nothing about the change in vorticity.

Let me use some different notation just in case we implie the same thing but we "lost in translation"

$\frac{d \vec{\zeta} }{dt}(t) = \left( \vec{\zeta}(t) \cdot \nabla \right) \vec{V}(t)$

If we know that $\vec{\zeta}(t_0) = 0$ then, all I can say is

$\frac{d \vec{\zeta} }{dt}(t_0) = \left( \vec{\zeta}(t_0) \cdot \nabla \right) \vec{V}(t_0) = 0$

This is all I can understand. If you are telling me that there is something more than that, I'm sorry but maybe I need to go back to the basics, and the come back again to ask for help

July 7, 2000, 13:46	rotational and inviscid	#1
Mike Guest Posts: n/a	Does anyone please tell me if inviscid flow can be rotational or not? What is the nature of rotational flow?

July 7, 2000, 14:09	Re: rotational and inviscid	#2
Adrin Gharakhani Guest Posts: n/a	Yes. Inviscid flow can be rotational. Rotational flow may be characterized by vorticity, which is the curl of velocity. A simple example is a vortex ring (like a cigarette smoke puff but without the diffusion process) Check out fundamental fluid dynamic books. It is important NOT to confuse potential flow with inviscid flow (which may be the source of your confusion). While potential flow satisfies the equation of motion of inviscid flow dynamics, the reverse is not necessarily true. That is, an inviscid flow is not necessarily potential (the former may contain vorticity, which induces rotational and not potential flow) Adrin Gharakhani

July 7, 2000, 15:16	Re: rotational and inviscid	#3
Kalyan Guest Posts: n/a	Inviscid flow can be rotational. As pointed out in an earlier reply, inviscid flows with vortices are good examples. It might help to know that vorticity is conserved in inviscid flows, i.e., Euler equations are incapable of vorticity production. So in an inviscid flow computation, vorticity (in the form of point vortices or vortex blobs or continuous fields) can only come in as a part of the inflow and can not be produced or destroyed by the Euler equations. Note also that the vortex rings can never be formed without viscosity though they can exist in a purely inviscid fluid.

July 7, 2000, 15:26	Re: rotational and inviscid	#4
Adrin Gharakhani Guest Posts: n/a	> Euler equations are incapable of vorticity production. So in an inviscid flow computation, vorticity (in the form of point vortices or vortex blobs or continuous fields) can only come in as a part of the inflow and can not be produced or destroyed by the Euler equations. This is true _only_ for the incompressible (and constant density) case. Example: baroclinic vorticity generation (without solid boundaries). Adrin Gharakhani

July 7, 2000, 20:33	Re: rotational and inviscid	#5
John C. Chien Guest Posts: n/a	(1). It is not a good idea to mix these two terms at the same time. (2). At the begining, you have this equation called the Navier-Stokes equations, which represent the conservation of mass, momentum and energy. It contains viscosity terms. (3). Now, if you take the curl operation on the momentum equations, you will get this so-called "vorticity" equation. Remember that, the vorticity equation is derived from the momentum equation, so it has viscosity terms in it also. (4). If you set the viscosity terms to zero, you are going to get this mathematical equation, called "inviscid" equations, whether it is the inviscid momentum equations or the inviscid vorticity equation. (5). The inviscid momentum equation is generally called "Euler Equation" in contrast to the Navier-Stokes equations which has viscosity terms. (6). In the vorticity equation(derived from the Navier-Stokes equations), in addition to the viscosity terms, there are convection terms, stretching terms, production terms due to expansion, non-uniform density, and general body force. (7). So, even though you drop the viscosity terms, there are still many other terms in the vorticity equation, including the production terms. (8). So far, we have touched only the "inviscid" vs "viscous" forms of equation. (9). The rest of that is easy. The only time when the flow is "irrotational" is when the vorticity is everywhere zero. If there are still many terms in the vorticity equation (after dropping the viscosity terms), then you know that vorticity is normally non-zero. So, most of the time, it is rotational even without viscosity terms. (10). In old days, the inviscid equation was still very difficult to solve. So, the Euler equation was further simplified (second time) by setting the vorticity to zero (it is called "irrotational"). As a result, one can express the velocity in terms of the velocity potential and derive the so-called "potential flow"equation. (11). So, "potential flow" is "irrotational flow". And for the incompressible flow, the equation is the Laplace equation. The Laplace equation was the main focus in the 19th century. (12). Remember that, Euler equation and Laplace equation are simplified equations derived from the Navier-Stokes equations.(by setting the viscosity to zero, and then by setting the vorticity to zero, respectively) Shayor likes this.

July 14, 2000, 15:31	Re: rotational and inviscid	#6
clifford bradford Guest Posts: n/a	curved shocks can be a source of vorticity in an inviscid flow. i think there may be other potential sources but I can't remember what the others might be.

July 20, 2000, 12:39	Re: rotational and inviscid	#7
Hua Zhou Guest Posts: n/a	where the velocity varies, where there exist vortex; where the entropy increases, where creates vortex.

April 20, 2011, 02:25		#8
Nick R Senior Member Nick Join Date: Nov 2010 Posts: 126 Rep Power: 15	Many thanks guys you explained things that confused me for awhile. Especial thanks to Adrian, John and Kalyan

September 22, 2022, 05:37		#9
silkworm New Member Wencan Wu Join Date: May 2018 Posts: 2 Rep Power: 0	I dont agree with you. After a normal shock, velocity changes while no vortex exists. In laminar boundary layer, there entropy increases but no vortex generates.

November 2, 2023, 19:22	stretching term of vorticity equation	#10
ichamail New Member Join Date: Jul 2022 Posts: 20 Rep Power: 3	The vorticity equation derived from momentum equation for an inviscid incompressible flow is pressented below $\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta} \cdot \nabla \right) \vec{V}$ , where $\vec{\zeta} = \nabla \times \vec{V}$ I read in several textbooks that, from this equation we come to the conclusion that if $\vec{\zeta}(t_0) = 0$ , then $\frac{d \vec{\zeta} }{dt} = 0$ for Also in many of theses textbooks is written that this equation agrees with Kelvin's circulation theorem, which states that, if an inviscid fluid is initially irrotational then it remains irrotational at all subsequent times. I can't understand, at least from a mathematical point of view, how that conclusion is made

November 2, 2023, 19:30		#11
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	Mathematically... starting with $\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta} \cdot \nabla \right) \vec{V}$ put $\vec{\zeta}= 0$ into the RHS Then you have $\frac{d \vec{\zeta} }{dt} = 0$ Hence if the vorticity was zero, it remains zero. That's all there is to it

November 3, 2023, 01:14		#13
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	Note that d/dt is the lagrangian derivative, formally that means the vorticity along the path-line. ichamail likes this.

November 3, 2023, 08:31		#15
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	Setting x=0 does not automatically imply anything about dx/dt For example, if x=1, it does not imply that d(1)/dt is anything Knowing f(a) doesn't tell you anything about f'(a)

November 3, 2023, 12:28		#17
agd Senior Member Join Date: Jul 2009 Posts: 353 Rep Power: 18	You have an initial value problem, with zero as the initial condition. The solution then is a constant, and that constant is the initial value zero. It really is that simple.

November 3, 2023, 14:39		#19
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,674 Rep Power: 65	When I say put vorticity=0 or x=1, I mean at one instance, not for all time. For example, a new born child is born at a height (in US) of 20 inches. Eventually the baby grows into an adult and becomes much taller than 20 inches. d(baby height)/dt is not zero (for all time). Again.... f(a)=0 does not imply that f(t)=0 for all t. It simply means that f at t=a is 0. And it also does not imply anything about f'(t) = anything. So yes, you are missing something very obvious, again and again about general mathematical notation. Then you have the vorticity transport equation: $\frac{d \vec{\zeta} }{dt} = \left( \vec{\zeta} \cdot \nabla \right) \vec{V}$ This equation is a general equation which relates the change of vorticity to velocity/vorticity field. Only from this equation can you evaluate what is $\frac{d \vec{\zeta} }{dt}$ Without this equation, you know nothing about the change in vorticity.

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