Incompressible fluid definition
$begingroup$
In a fluid mechanics course I found that an incompressible fluid flow means literally:
$$rho = text{constant} quad forall vec r,, forall t$$
Where $vec r = (x, y, z)$
In my understanding, this means literally that the fluid density is uniform? (Am I wrong?)
In the other hand we can find also that an incompressible fluid means:
$$dfrac{Drho}{Dt} = 0$$ which does not necessarily mean that the fluid density is uniform.
What's wrong here?
fluid-dynamics
New contributor
$endgroup$
add a comment |
$begingroup$
In a fluid mechanics course I found that an incompressible fluid flow means literally:
$$rho = text{constant} quad forall vec r,, forall t$$
Where $vec r = (x, y, z)$
In my understanding, this means literally that the fluid density is uniform? (Am I wrong?)
In the other hand we can find also that an incompressible fluid means:
$$dfrac{Drho}{Dt} = 0$$ which does not necessarily mean that the fluid density is uniform.
What's wrong here?
fluid-dynamics
New contributor
$endgroup$
$begingroup$
When just looking at the math: Say that $rho$ is constant. If the density is constant for all positions and all time then you have a blob in space (filling all the space where it it's position is defined) not moving. It is not changing shape, there are however certain types of ways the matter can flow. The derivative being $0$ means that you could have a $rho(vec{r})$ not constant, that just doesn't change wrt time. So it doesn't move but certain flow inside is still allowed. The density can be a time independent non constant function of position.
$endgroup$
– ty.
6 hours ago
add a comment |
$begingroup$
In a fluid mechanics course I found that an incompressible fluid flow means literally:
$$rho = text{constant} quad forall vec r,, forall t$$
Where $vec r = (x, y, z)$
In my understanding, this means literally that the fluid density is uniform? (Am I wrong?)
In the other hand we can find also that an incompressible fluid means:
$$dfrac{Drho}{Dt} = 0$$ which does not necessarily mean that the fluid density is uniform.
What's wrong here?
fluid-dynamics
New contributor
$endgroup$
In a fluid mechanics course I found that an incompressible fluid flow means literally:
$$rho = text{constant} quad forall vec r,, forall t$$
Where $vec r = (x, y, z)$
In my understanding, this means literally that the fluid density is uniform? (Am I wrong?)
In the other hand we can find also that an incompressible fluid means:
$$dfrac{Drho}{Dt} = 0$$ which does not necessarily mean that the fluid density is uniform.
What's wrong here?
fluid-dynamics
fluid-dynamics
New contributor
New contributor
New contributor
asked 6 hours ago
IamNotaMathematicianIamNotaMathematician
375
375
New contributor
New contributor
$begingroup$
When just looking at the math: Say that $rho$ is constant. If the density is constant for all positions and all time then you have a blob in space (filling all the space where it it's position is defined) not moving. It is not changing shape, there are however certain types of ways the matter can flow. The derivative being $0$ means that you could have a $rho(vec{r})$ not constant, that just doesn't change wrt time. So it doesn't move but certain flow inside is still allowed. The density can be a time independent non constant function of position.
$endgroup$
– ty.
6 hours ago
add a comment |
$begingroup$
When just looking at the math: Say that $rho$ is constant. If the density is constant for all positions and all time then you have a blob in space (filling all the space where it it's position is defined) not moving. It is not changing shape, there are however certain types of ways the matter can flow. The derivative being $0$ means that you could have a $rho(vec{r})$ not constant, that just doesn't change wrt time. So it doesn't move but certain flow inside is still allowed. The density can be a time independent non constant function of position.
$endgroup$
– ty.
6 hours ago
$begingroup$
When just looking at the math: Say that $rho$ is constant. If the density is constant for all positions and all time then you have a blob in space (filling all the space where it it's position is defined) not moving. It is not changing shape, there are however certain types of ways the matter can flow. The derivative being $0$ means that you could have a $rho(vec{r})$ not constant, that just doesn't change wrt time. So it doesn't move but certain flow inside is still allowed. The density can be a time independent non constant function of position.
$endgroup$
– ty.
6 hours ago
$begingroup$
When just looking at the math: Say that $rho$ is constant. If the density is constant for all positions and all time then you have a blob in space (filling all the space where it it's position is defined) not moving. It is not changing shape, there are however certain types of ways the matter can flow. The derivative being $0$ means that you could have a $rho(vec{r})$ not constant, that just doesn't change wrt time. So it doesn't move but certain flow inside is still allowed. The density can be a time independent non constant function of position.
$endgroup$
– ty.
6 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The definition of incompressible is often unclear and changes depending on which community uses it. So let's look at some common definitions:
Constant density
This means the density is constant everywhere in space and time. So:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla{rho} = 0
$$
Because density is constant everywhere in space and time, the temporal derivative is zero, and the spatial gradient is zero.
Low Mach Number
This shows up when the flow velocity is relatively low and so all pressure changes are hydrodynamic (due to velocity motion) rather than thermodynamic. The effect of this is that $partial rho / partial p = 0$. In other words, the small changes in pressure due to flow velocity changes do not change the density. This has a secondary effect -- the speed of sound in the fluid is $partial p/partial rho = infty$ in this instance. So there is an infinite speed of sound, which makes the equations elliptic in nature.
Although we assume density is independent of pressure, it is possible for density to change due to changes in temperature or composition if the flow is chemically reacting. This means:
$$
frac{Drho}{Dt} neq 0
$$
because $rho$ is a function of temperature and composition. If, however, the flow is not reacting or multi-component, you will also get the same equation as the constant density case:
$$
frac{Drho}{Dt} = 0
$$
Therefore, incompressible can mean constant density, or it can mean low Mach number, depending on the community and the application. I prefer to be explicit in the difference because I work in the reacting flow world where it matters. But many in the non-reacting flow communities just use incompressible to mean constant density.
Example of non-constant density
Since it was asked for an example where the material derivative is zero but density is not constant, here goes:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla rho = 0
$$
Rearrange this:
$$
frac{partial rho}{partial t} = -vec{u}cdotnablarho
$$
gives a flow where $rho neq text{const.}$ yet $Drho/Dt = 0$. It has to be an unsteady flow.
Is there another example of steady flow? In steady flow, the time derivative is zero, so you have:
$$
vec{u}cdotnablarho = 0
$$
If velocity is not zero, $vec{u} neq 0$, then we have $nabla rho = 0$ and so any moving, steady flow without body forces (gravity) or temperature/composition differences must have constant density.
If velocity is zero, you can have a gradient in density without any issues. Think of a column of the atmosphere for example -- density is higher at the bottom than the top due to gravity, and there is no velocity. So again, $Drho/Dt = 0$ but density is not constant everywhere. The challenge here of course is that the continuity equation is not sufficient to describe the situation since it becomes $0 = 0$. You would have to include the momentum equation to incorporate the gravity forces.
$endgroup$
$begingroup$
But without speaking about compressibility effects. Why $dfrac{Drho}{Dt} =0$ does not imply $rho = text{cst}$ everywhere at any time
$endgroup$
– IamNotaMathematician
6 hours ago
$begingroup$
Could you please give me an example when $dfrac{Drho}{Dt} = 0$ and the desnity is not constant everywhere?
$endgroup$
– IamNotaMathematician
6 hours ago
1
$begingroup$
@IamNotaMathematician Done. These examples come from inspecting the equation. I encourage you to understand what the terms mean in the equations you write out and to think about what happens under various situations (like steady, or no velocity, or spatial gradients exist, etc). Everything will fall out of the equations.
$endgroup$
– tpg2114
6 hours ago
$begingroup$
In the last example, I imagine that as follow: $vec u$ is a vector and $nabla rho$ (gradient) is a vector.The scalar product = 0 implies that they are perpendicular that the direction of density change is perpendicular to flow velocity.
$endgroup$
– IamNotaMathematician
5 hours ago
$begingroup$
@IamNotaMathematician I'd have to think about whether such a flow could exist and be stable. Remember, this is just the continuity equation, a real flow will have to satisfy the momentum equation also. A flow with velocity perpendicular to density gradient will end up generating baroclinic torque which will try to align the velocity and density gradients. So I suppose mathematically looking at this one equation, yes, flow perpendicular to density gradient is a possible solution. But I don't think it would hold for a real flow.
$endgroup$
– tpg2114
5 hours ago
add a comment |
$begingroup$
The general definition of an incompressible flow is $frac{Drho }{Dt}=0$ : the density of a fluid particle does not change along its path.
For example, if $overrightarrow{v}=v(x)overrightarrow{{{e}_{x}}}$ and $rho =rho (y)$ : the path lines are horizontal lines and on such a line, the density does not change.
The condition $rho =cst$ is a particular case ("incompressible fluid" rather than "incompressible flow").
But frequently, one mean $rho =cst$ when one speak of incompressible fluid !
Sorry for my poor english !
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
IamNotaMathematician is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f464208%2fincompressible-fluid-definition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The definition of incompressible is often unclear and changes depending on which community uses it. So let's look at some common definitions:
Constant density
This means the density is constant everywhere in space and time. So:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla{rho} = 0
$$
Because density is constant everywhere in space and time, the temporal derivative is zero, and the spatial gradient is zero.
Low Mach Number
This shows up when the flow velocity is relatively low and so all pressure changes are hydrodynamic (due to velocity motion) rather than thermodynamic. The effect of this is that $partial rho / partial p = 0$. In other words, the small changes in pressure due to flow velocity changes do not change the density. This has a secondary effect -- the speed of sound in the fluid is $partial p/partial rho = infty$ in this instance. So there is an infinite speed of sound, which makes the equations elliptic in nature.
Although we assume density is independent of pressure, it is possible for density to change due to changes in temperature or composition if the flow is chemically reacting. This means:
$$
frac{Drho}{Dt} neq 0
$$
because $rho$ is a function of temperature and composition. If, however, the flow is not reacting or multi-component, you will also get the same equation as the constant density case:
$$
frac{Drho}{Dt} = 0
$$
Therefore, incompressible can mean constant density, or it can mean low Mach number, depending on the community and the application. I prefer to be explicit in the difference because I work in the reacting flow world where it matters. But many in the non-reacting flow communities just use incompressible to mean constant density.
Example of non-constant density
Since it was asked for an example where the material derivative is zero but density is not constant, here goes:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla rho = 0
$$
Rearrange this:
$$
frac{partial rho}{partial t} = -vec{u}cdotnablarho
$$
gives a flow where $rho neq text{const.}$ yet $Drho/Dt = 0$. It has to be an unsteady flow.
Is there another example of steady flow? In steady flow, the time derivative is zero, so you have:
$$
vec{u}cdotnablarho = 0
$$
If velocity is not zero, $vec{u} neq 0$, then we have $nabla rho = 0$ and so any moving, steady flow without body forces (gravity) or temperature/composition differences must have constant density.
If velocity is zero, you can have a gradient in density without any issues. Think of a column of the atmosphere for example -- density is higher at the bottom than the top due to gravity, and there is no velocity. So again, $Drho/Dt = 0$ but density is not constant everywhere. The challenge here of course is that the continuity equation is not sufficient to describe the situation since it becomes $0 = 0$. You would have to include the momentum equation to incorporate the gravity forces.
$endgroup$
$begingroup$
But without speaking about compressibility effects. Why $dfrac{Drho}{Dt} =0$ does not imply $rho = text{cst}$ everywhere at any time
$endgroup$
– IamNotaMathematician
6 hours ago
$begingroup$
Could you please give me an example when $dfrac{Drho}{Dt} = 0$ and the desnity is not constant everywhere?
$endgroup$
– IamNotaMathematician
6 hours ago
1
$begingroup$
@IamNotaMathematician Done. These examples come from inspecting the equation. I encourage you to understand what the terms mean in the equations you write out and to think about what happens under various situations (like steady, or no velocity, or spatial gradients exist, etc). Everything will fall out of the equations.
$endgroup$
– tpg2114
6 hours ago
$begingroup$
In the last example, I imagine that as follow: $vec u$ is a vector and $nabla rho$ (gradient) is a vector.The scalar product = 0 implies that they are perpendicular that the direction of density change is perpendicular to flow velocity.
$endgroup$
– IamNotaMathematician
5 hours ago
$begingroup$
@IamNotaMathematician I'd have to think about whether such a flow could exist and be stable. Remember, this is just the continuity equation, a real flow will have to satisfy the momentum equation also. A flow with velocity perpendicular to density gradient will end up generating baroclinic torque which will try to align the velocity and density gradients. So I suppose mathematically looking at this one equation, yes, flow perpendicular to density gradient is a possible solution. But I don't think it would hold for a real flow.
$endgroup$
– tpg2114
5 hours ago
add a comment |
$begingroup$
The definition of incompressible is often unclear and changes depending on which community uses it. So let's look at some common definitions:
Constant density
This means the density is constant everywhere in space and time. So:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla{rho} = 0
$$
Because density is constant everywhere in space and time, the temporal derivative is zero, and the spatial gradient is zero.
Low Mach Number
This shows up when the flow velocity is relatively low and so all pressure changes are hydrodynamic (due to velocity motion) rather than thermodynamic. The effect of this is that $partial rho / partial p = 0$. In other words, the small changes in pressure due to flow velocity changes do not change the density. This has a secondary effect -- the speed of sound in the fluid is $partial p/partial rho = infty$ in this instance. So there is an infinite speed of sound, which makes the equations elliptic in nature.
Although we assume density is independent of pressure, it is possible for density to change due to changes in temperature or composition if the flow is chemically reacting. This means:
$$
frac{Drho}{Dt} neq 0
$$
because $rho$ is a function of temperature and composition. If, however, the flow is not reacting or multi-component, you will also get the same equation as the constant density case:
$$
frac{Drho}{Dt} = 0
$$
Therefore, incompressible can mean constant density, or it can mean low Mach number, depending on the community and the application. I prefer to be explicit in the difference because I work in the reacting flow world where it matters. But many in the non-reacting flow communities just use incompressible to mean constant density.
Example of non-constant density
Since it was asked for an example where the material derivative is zero but density is not constant, here goes:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla rho = 0
$$
Rearrange this:
$$
frac{partial rho}{partial t} = -vec{u}cdotnablarho
$$
gives a flow where $rho neq text{const.}$ yet $Drho/Dt = 0$. It has to be an unsteady flow.
Is there another example of steady flow? In steady flow, the time derivative is zero, so you have:
$$
vec{u}cdotnablarho = 0
$$
If velocity is not zero, $vec{u} neq 0$, then we have $nabla rho = 0$ and so any moving, steady flow without body forces (gravity) or temperature/composition differences must have constant density.
If velocity is zero, you can have a gradient in density without any issues. Think of a column of the atmosphere for example -- density is higher at the bottom than the top due to gravity, and there is no velocity. So again, $Drho/Dt = 0$ but density is not constant everywhere. The challenge here of course is that the continuity equation is not sufficient to describe the situation since it becomes $0 = 0$. You would have to include the momentum equation to incorporate the gravity forces.
$endgroup$
$begingroup$
But without speaking about compressibility effects. Why $dfrac{Drho}{Dt} =0$ does not imply $rho = text{cst}$ everywhere at any time
$endgroup$
– IamNotaMathematician
6 hours ago
$begingroup$
Could you please give me an example when $dfrac{Drho}{Dt} = 0$ and the desnity is not constant everywhere?
$endgroup$
– IamNotaMathematician
6 hours ago
1
$begingroup$
@IamNotaMathematician Done. These examples come from inspecting the equation. I encourage you to understand what the terms mean in the equations you write out and to think about what happens under various situations (like steady, or no velocity, or spatial gradients exist, etc). Everything will fall out of the equations.
$endgroup$
– tpg2114
6 hours ago
$begingroup$
In the last example, I imagine that as follow: $vec u$ is a vector and $nabla rho$ (gradient) is a vector.The scalar product = 0 implies that they are perpendicular that the direction of density change is perpendicular to flow velocity.
$endgroup$
– IamNotaMathematician
5 hours ago
$begingroup$
@IamNotaMathematician I'd have to think about whether such a flow could exist and be stable. Remember, this is just the continuity equation, a real flow will have to satisfy the momentum equation also. A flow with velocity perpendicular to density gradient will end up generating baroclinic torque which will try to align the velocity and density gradients. So I suppose mathematically looking at this one equation, yes, flow perpendicular to density gradient is a possible solution. But I don't think it would hold for a real flow.
$endgroup$
– tpg2114
5 hours ago
add a comment |
$begingroup$
The definition of incompressible is often unclear and changes depending on which community uses it. So let's look at some common definitions:
Constant density
This means the density is constant everywhere in space and time. So:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla{rho} = 0
$$
Because density is constant everywhere in space and time, the temporal derivative is zero, and the spatial gradient is zero.
Low Mach Number
This shows up when the flow velocity is relatively low and so all pressure changes are hydrodynamic (due to velocity motion) rather than thermodynamic. The effect of this is that $partial rho / partial p = 0$. In other words, the small changes in pressure due to flow velocity changes do not change the density. This has a secondary effect -- the speed of sound in the fluid is $partial p/partial rho = infty$ in this instance. So there is an infinite speed of sound, which makes the equations elliptic in nature.
Although we assume density is independent of pressure, it is possible for density to change due to changes in temperature or composition if the flow is chemically reacting. This means:
$$
frac{Drho}{Dt} neq 0
$$
because $rho$ is a function of temperature and composition. If, however, the flow is not reacting or multi-component, you will also get the same equation as the constant density case:
$$
frac{Drho}{Dt} = 0
$$
Therefore, incompressible can mean constant density, or it can mean low Mach number, depending on the community and the application. I prefer to be explicit in the difference because I work in the reacting flow world where it matters. But many in the non-reacting flow communities just use incompressible to mean constant density.
Example of non-constant density
Since it was asked for an example where the material derivative is zero but density is not constant, here goes:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla rho = 0
$$
Rearrange this:
$$
frac{partial rho}{partial t} = -vec{u}cdotnablarho
$$
gives a flow where $rho neq text{const.}$ yet $Drho/Dt = 0$. It has to be an unsteady flow.
Is there another example of steady flow? In steady flow, the time derivative is zero, so you have:
$$
vec{u}cdotnablarho = 0
$$
If velocity is not zero, $vec{u} neq 0$, then we have $nabla rho = 0$ and so any moving, steady flow without body forces (gravity) or temperature/composition differences must have constant density.
If velocity is zero, you can have a gradient in density without any issues. Think of a column of the atmosphere for example -- density is higher at the bottom than the top due to gravity, and there is no velocity. So again, $Drho/Dt = 0$ but density is not constant everywhere. The challenge here of course is that the continuity equation is not sufficient to describe the situation since it becomes $0 = 0$. You would have to include the momentum equation to incorporate the gravity forces.
$endgroup$
The definition of incompressible is often unclear and changes depending on which community uses it. So let's look at some common definitions:
Constant density
This means the density is constant everywhere in space and time. So:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla{rho} = 0
$$
Because density is constant everywhere in space and time, the temporal derivative is zero, and the spatial gradient is zero.
Low Mach Number
This shows up when the flow velocity is relatively low and so all pressure changes are hydrodynamic (due to velocity motion) rather than thermodynamic. The effect of this is that $partial rho / partial p = 0$. In other words, the small changes in pressure due to flow velocity changes do not change the density. This has a secondary effect -- the speed of sound in the fluid is $partial p/partial rho = infty$ in this instance. So there is an infinite speed of sound, which makes the equations elliptic in nature.
Although we assume density is independent of pressure, it is possible for density to change due to changes in temperature or composition if the flow is chemically reacting. This means:
$$
frac{Drho}{Dt} neq 0
$$
because $rho$ is a function of temperature and composition. If, however, the flow is not reacting or multi-component, you will also get the same equation as the constant density case:
$$
frac{Drho}{Dt} = 0
$$
Therefore, incompressible can mean constant density, or it can mean low Mach number, depending on the community and the application. I prefer to be explicit in the difference because I work in the reacting flow world where it matters. But many in the non-reacting flow communities just use incompressible to mean constant density.
Example of non-constant density
Since it was asked for an example where the material derivative is zero but density is not constant, here goes:
$$
frac{Drho}{Dt} = frac{partial rho}{partial t} + vec{u}cdotnabla rho = 0
$$
Rearrange this:
$$
frac{partial rho}{partial t} = -vec{u}cdotnablarho
$$
gives a flow where $rho neq text{const.}$ yet $Drho/Dt = 0$. It has to be an unsteady flow.
Is there another example of steady flow? In steady flow, the time derivative is zero, so you have:
$$
vec{u}cdotnablarho = 0
$$
If velocity is not zero, $vec{u} neq 0$, then we have $nabla rho = 0$ and so any moving, steady flow without body forces (gravity) or temperature/composition differences must have constant density.
If velocity is zero, you can have a gradient in density without any issues. Think of a column of the atmosphere for example -- density is higher at the bottom than the top due to gravity, and there is no velocity. So again, $Drho/Dt = 0$ but density is not constant everywhere. The challenge here of course is that the continuity equation is not sufficient to describe the situation since it becomes $0 = 0$. You would have to include the momentum equation to incorporate the gravity forces.
edited 5 hours ago
answered 6 hours ago
tpg2114tpg2114
13.6k23870
13.6k23870
$begingroup$
But without speaking about compressibility effects. Why $dfrac{Drho}{Dt} =0$ does not imply $rho = text{cst}$ everywhere at any time
$endgroup$
– IamNotaMathematician
6 hours ago
$begingroup$
Could you please give me an example when $dfrac{Drho}{Dt} = 0$ and the desnity is not constant everywhere?
$endgroup$
– IamNotaMathematician
6 hours ago
1
$begingroup$
@IamNotaMathematician Done. These examples come from inspecting the equation. I encourage you to understand what the terms mean in the equations you write out and to think about what happens under various situations (like steady, or no velocity, or spatial gradients exist, etc). Everything will fall out of the equations.
$endgroup$
– tpg2114
6 hours ago
$begingroup$
In the last example, I imagine that as follow: $vec u$ is a vector and $nabla rho$ (gradient) is a vector.The scalar product = 0 implies that they are perpendicular that the direction of density change is perpendicular to flow velocity.
$endgroup$
– IamNotaMathematician
5 hours ago
$begingroup$
@IamNotaMathematician I'd have to think about whether such a flow could exist and be stable. Remember, this is just the continuity equation, a real flow will have to satisfy the momentum equation also. A flow with velocity perpendicular to density gradient will end up generating baroclinic torque which will try to align the velocity and density gradients. So I suppose mathematically looking at this one equation, yes, flow perpendicular to density gradient is a possible solution. But I don't think it would hold for a real flow.
$endgroup$
– tpg2114
5 hours ago
add a comment |
$begingroup$
But without speaking about compressibility effects. Why $dfrac{Drho}{Dt} =0$ does not imply $rho = text{cst}$ everywhere at any time
$endgroup$
– IamNotaMathematician
6 hours ago
$begingroup$
Could you please give me an example when $dfrac{Drho}{Dt} = 0$ and the desnity is not constant everywhere?
$endgroup$
– IamNotaMathematician
6 hours ago
1
$begingroup$
@IamNotaMathematician Done. These examples come from inspecting the equation. I encourage you to understand what the terms mean in the equations you write out and to think about what happens under various situations (like steady, or no velocity, or spatial gradients exist, etc). Everything will fall out of the equations.
$endgroup$
– tpg2114
6 hours ago
$begingroup$
In the last example, I imagine that as follow: $vec u$ is a vector and $nabla rho$ (gradient) is a vector.The scalar product = 0 implies that they are perpendicular that the direction of density change is perpendicular to flow velocity.
$endgroup$
– IamNotaMathematician
5 hours ago
$begingroup$
@IamNotaMathematician I'd have to think about whether such a flow could exist and be stable. Remember, this is just the continuity equation, a real flow will have to satisfy the momentum equation also. A flow with velocity perpendicular to density gradient will end up generating baroclinic torque which will try to align the velocity and density gradients. So I suppose mathematically looking at this one equation, yes, flow perpendicular to density gradient is a possible solution. But I don't think it would hold for a real flow.
$endgroup$
– tpg2114
5 hours ago
$begingroup$
But without speaking about compressibility effects. Why $dfrac{Drho}{Dt} =0$ does not imply $rho = text{cst}$ everywhere at any time
$endgroup$
– IamNotaMathematician
6 hours ago
$begingroup$
But without speaking about compressibility effects. Why $dfrac{Drho}{Dt} =0$ does not imply $rho = text{cst}$ everywhere at any time
$endgroup$
– IamNotaMathematician
6 hours ago
$begingroup$
Could you please give me an example when $dfrac{Drho}{Dt} = 0$ and the desnity is not constant everywhere?
$endgroup$
– IamNotaMathematician
6 hours ago
$begingroup$
Could you please give me an example when $dfrac{Drho}{Dt} = 0$ and the desnity is not constant everywhere?
$endgroup$
– IamNotaMathematician
6 hours ago
1
1
$begingroup$
@IamNotaMathematician Done. These examples come from inspecting the equation. I encourage you to understand what the terms mean in the equations you write out and to think about what happens under various situations (like steady, or no velocity, or spatial gradients exist, etc). Everything will fall out of the equations.
$endgroup$
– tpg2114
6 hours ago
$begingroup$
@IamNotaMathematician Done. These examples come from inspecting the equation. I encourage you to understand what the terms mean in the equations you write out and to think about what happens under various situations (like steady, or no velocity, or spatial gradients exist, etc). Everything will fall out of the equations.
$endgroup$
– tpg2114
6 hours ago
$begingroup$
In the last example, I imagine that as follow: $vec u$ is a vector and $nabla rho$ (gradient) is a vector.The scalar product = 0 implies that they are perpendicular that the direction of density change is perpendicular to flow velocity.
$endgroup$
– IamNotaMathematician
5 hours ago
$begingroup$
In the last example, I imagine that as follow: $vec u$ is a vector and $nabla rho$ (gradient) is a vector.The scalar product = 0 implies that they are perpendicular that the direction of density change is perpendicular to flow velocity.
$endgroup$
– IamNotaMathematician
5 hours ago
$begingroup$
@IamNotaMathematician I'd have to think about whether such a flow could exist and be stable. Remember, this is just the continuity equation, a real flow will have to satisfy the momentum equation also. A flow with velocity perpendicular to density gradient will end up generating baroclinic torque which will try to align the velocity and density gradients. So I suppose mathematically looking at this one equation, yes, flow perpendicular to density gradient is a possible solution. But I don't think it would hold for a real flow.
$endgroup$
– tpg2114
5 hours ago
$begingroup$
@IamNotaMathematician I'd have to think about whether such a flow could exist and be stable. Remember, this is just the continuity equation, a real flow will have to satisfy the momentum equation also. A flow with velocity perpendicular to density gradient will end up generating baroclinic torque which will try to align the velocity and density gradients. So I suppose mathematically looking at this one equation, yes, flow perpendicular to density gradient is a possible solution. But I don't think it would hold for a real flow.
$endgroup$
– tpg2114
5 hours ago
add a comment |
$begingroup$
The general definition of an incompressible flow is $frac{Drho }{Dt}=0$ : the density of a fluid particle does not change along its path.
For example, if $overrightarrow{v}=v(x)overrightarrow{{{e}_{x}}}$ and $rho =rho (y)$ : the path lines are horizontal lines and on such a line, the density does not change.
The condition $rho =cst$ is a particular case ("incompressible fluid" rather than "incompressible flow").
But frequently, one mean $rho =cst$ when one speak of incompressible fluid !
Sorry for my poor english !
$endgroup$
add a comment |
$begingroup$
The general definition of an incompressible flow is $frac{Drho }{Dt}=0$ : the density of a fluid particle does not change along its path.
For example, if $overrightarrow{v}=v(x)overrightarrow{{{e}_{x}}}$ and $rho =rho (y)$ : the path lines are horizontal lines and on such a line, the density does not change.
The condition $rho =cst$ is a particular case ("incompressible fluid" rather than "incompressible flow").
But frequently, one mean $rho =cst$ when one speak of incompressible fluid !
Sorry for my poor english !
$endgroup$
add a comment |
$begingroup$
The general definition of an incompressible flow is $frac{Drho }{Dt}=0$ : the density of a fluid particle does not change along its path.
For example, if $overrightarrow{v}=v(x)overrightarrow{{{e}_{x}}}$ and $rho =rho (y)$ : the path lines are horizontal lines and on such a line, the density does not change.
The condition $rho =cst$ is a particular case ("incompressible fluid" rather than "incompressible flow").
But frequently, one mean $rho =cst$ when one speak of incompressible fluid !
Sorry for my poor english !
$endgroup$
The general definition of an incompressible flow is $frac{Drho }{Dt}=0$ : the density of a fluid particle does not change along its path.
For example, if $overrightarrow{v}=v(x)overrightarrow{{{e}_{x}}}$ and $rho =rho (y)$ : the path lines are horizontal lines and on such a line, the density does not change.
The condition $rho =cst$ is a particular case ("incompressible fluid" rather than "incompressible flow").
But frequently, one mean $rho =cst$ when one speak of incompressible fluid !
Sorry for my poor english !
answered 6 hours ago
Vincent FraticelliVincent Fraticelli
1,579125
1,579125
add a comment |
add a comment |
IamNotaMathematician is a new contributor. Be nice, and check out our Code of Conduct.
IamNotaMathematician is a new contributor. Be nice, and check out our Code of Conduct.
IamNotaMathematician is a new contributor. Be nice, and check out our Code of Conduct.
IamNotaMathematician is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f464208%2fincompressible-fluid-definition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
When just looking at the math: Say that $rho$ is constant. If the density is constant for all positions and all time then you have a blob in space (filling all the space where it it's position is defined) not moving. It is not changing shape, there are however certain types of ways the matter can flow. The derivative being $0$ means that you could have a $rho(vec{r})$ not constant, that just doesn't change wrt time. So it doesn't move but certain flow inside is still allowed. The density can be a time independent non constant function of position.
$endgroup$
– ty.
6 hours ago