Analyzing the direction of the streamline given the streamfunction $psi = x^2 - y^2$?
$begingroup$
The streamlines are represented by $psi = x^2 - y^2 = C$ where $C$ is a constant.We need to calculate the velocity and its direction at $(2,2)$, we need to sketch the streamlines and show the direction of flow.We found $q = uhat{i} + vhat{j}$ where $u = 4$ and $v = -4$ at point $(2,2)$, the slope is 1 so the velocity vector is oriented $45$ degree angle to the $x -$ axis.
We plot the streamlines that is the family of hyperbolas as below.But the main query is how can we decide the direction of the arrows?, as shown in the diagram one is moving upwards and the other downwards.
vectors fluid-dynamics
asked Mar 25 '18 at 13:23
BAYMAXBAYMAX
3,03521326
$endgroup$
add a comment |
$begingroup$
The streamlines are represented by $psi = x^2 - y^2 = C$ where $C$ is a constant.We need to calculate the velocity and its direction at $(2,2)$, we need to sketch the streamlines and show the direction of flow.We found $q = uhat{i} + vhat{j}$ where $u = 4$ and $v = -4$ at point $(2,2)$, the slope is 1 so the velocity vector is oriented $45$ degree angle to the $x -$ axis.
We plot the streamlines that is the family of hyperbolas as below.But the main query is how can we decide the direction of the arrows?, as shown in the diagram one is moving upwards and the other downwards.
vectors fluid-dynamics
asked Mar 25 '18 at 13:23
BAYMAXBAYMAX
3,03521326
$endgroup$
$begingroup$
For a 2D flow, $u = partial_ypsi$ and $v = -partial_xpsi$.
$endgroup$
– Chee Han
Mar 25 '18 at 20:55
$begingroup$
I'm not entirely sure, but I think you are on the right track here. By definition, streamlines are tangential to the flow field at a fixed time $t$. The flow field in this case is given by $(u,v) = -2(y,x)$ and I believe this yields the direction of the tangent vector for the level sets of streamlines. One last thing: you might also want to consider the case where $psi = textrm{negative constant}$.
$endgroup$
– Chee Han
Mar 25 '18 at 21:19
add a comment |
$begingroup$
The streamlines are represented by $psi = x^2 - y^2 = C$ where $C$ is a constant.We need to calculate the velocity and its direction at $(2,2)$, we need to sketch the streamlines and show the direction of flow.We found $q = uhat{i} + vhat{j}$ where $u = 4$ and $v = -4$ at point $(2,2)$, the slope is 1 so the velocity vector is oriented $45$ degree angle to the $x -$ axis.
We plot the streamlines that is the family of hyperbolas as below.But the main query is how can we decide the direction of the arrows?, as shown in the diagram one is moving upwards and the other downwards.
vectors fluid-dynamics
asked Mar 25 '18 at 13:23
BAYMAXBAYMAX
3,03521326
$endgroup$
The streamlines are represented by $psi = x^2 - y^2 = C$ where $C$ is a constant.We need to calculate the velocity and its direction at $(2,2)$, we need to sketch the streamlines and show the direction of flow.We found $q = uhat{i} + vhat{j}$ where $u = 4$ and $v = -4$ at point $(2,2)$, the slope is 1 so the velocity vector is oriented $45$ degree angle to the $x -$ axis.
We plot the streamlines that is the family of hyperbolas as below.But the main query is how can we decide the direction of the arrows?, as shown in the diagram one is moving upwards and the other downwards.
vectors fluid-dynamics
vectors fluid-dynamics
asked Mar 25 '18 at 13:23
BAYMAXBAYMAX
3,03521326
asked Mar 25 '18 at 13:23
BAYMAXBAYMAX
3,03521326
asked Mar 25 '18 at 13:23
BAYMAXBAYMAX
3,03521326
asked Mar 25 '18 at 13:23
BAYMAXBAYMAX
3,03521326
asked Mar 25 '18 at 13:23
BAYMAXBAYMAX
3,03521326
3,03521326
$begingroup$
For a 2D flow, $u = partial_ypsi$ and $v = -partial_xpsi$.
$endgroup$
– Chee Han
Mar 25 '18 at 20:55
$begingroup$
I'm not entirely sure, but I think you are on the right track here. By definition, streamlines are tangential to the flow field at a fixed time $t$. The flow field in this case is given by $(u,v) = -2(y,x)$ and I believe this yields the direction of the tangent vector for the level sets of streamlines. One last thing: you might also want to consider the case where $psi = textrm{negative constant}$.
$endgroup$
– Chee Han
Mar 25 '18 at 21:19
add a comment |
$begingroup$
For a 2D flow, $u = partial_ypsi$ and $v = -partial_xpsi$.
$endgroup$
– Chee Han
Mar 25 '18 at 20:55
$begingroup$
I'm not entirely sure, but I think you are on the right track here. By definition, streamlines are tangential to the flow field at a fixed time $t$. The flow field in this case is given by $(u,v) = -2(y,x)$ and I believe this yields the direction of the tangent vector for the level sets of streamlines. One last thing: you might also want to consider the case where $psi = textrm{negative constant}$.
$endgroup$
– Chee Han
Mar 25 '18 at 21:19
$begingroup$
For a 2D flow, $u = partial_ypsi$ and $v = -partial_xpsi$.
$endgroup$
– Chee Han
Mar 25 '18 at 20:55
$begingroup$
For a 2D flow, $u = partial_ypsi$ and $v = -partial_xpsi$.
$endgroup$
– Chee Han
Mar 25 '18 at 20:55
$begingroup$
I'm not entirely sure, but I think you are on the right track here. By definition, streamlines are tangential to the flow field at a fixed time $t$. The flow field in this case is given by $(u,v) = -2(y,x)$ and I believe this yields the direction of the tangent vector for the level sets of streamlines. One last thing: you might also want to consider the case where $psi = textrm{negative constant}$.
$endgroup$
– Chee Han
Mar 25 '18 at 21:19
$begingroup$
I'm not entirely sure, but I think you are on the right track here. By definition, streamlines are tangential to the flow field at a fixed time $t$. The flow field in this case is given by $(u,v) = -2(y,x)$ and I believe this yields the direction of the tangent vector for the level sets of streamlines. One last thing: you might also want to consider the case where $psi = textrm{negative constant}$.
$endgroup$
– Chee Han
Mar 25 '18 at 21:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As mentioned by Chee Han, $(u, v) = -2(y, x)$. To get a feeling for the plot of the streamlines it is often easiest to consider $psi = 0$, which corresponds to $pm y = pm x$, so the two lines $y=x$ and $y=-x$.
This hints at the fact that you are missing two families of lines, corresponding to negative $psi$-values. In fact, every value of $psi neq 0$ corresponds to a hyperbola.
A useful fact to remember is that, roughly speaking, streamlines close to each other point in similar directions. Abrupt changes of direction are impossible (due to the solution function depending continuously on the initial conditions).
If you probe the axes now to find the direction of the velocity there you find that it points up on the positive $x$-axis, left on the positive $y$-axis, down on the negative $x$-axis and right on the negative $y$-axis.
This information should be everything you need to plot all the streamlines.
answered May 3 '18 at 13:29
Lukas KoflerLukas Kofler
1,4212520
$endgroup$
add a comment |
$begingroup$
In continuation to the above comments; in the positive quadrant for instance, you may determine the direction of flux here by considering the velocities at these points.
This might be helpful. https://books.google.co.in/books?id=_Jo-T8nBghYC&pg=PA285&dq=hyperbola+streamline&hl=en&sa=X&ved=0ahUKEwj2speb2fbfAhVJ2LwKHVXtAAYQ6AEIDzAB#v=onepage&q=hyperbola%20streamline&f=false
answered Jan 18 at 6:34
Shatabdi SinhaShatabdi Sinha
19113
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
});
}
});
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%2fmath.stackexchange.com%2fquestions%2f2707374%2fanalyzing-the-direction-of-the-streamline-given-the-streamfunction-psi-x2%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
$begingroup$
As mentioned by Chee Han, $(u, v) = -2(y, x)$. To get a feeling for the plot of the streamlines it is often easiest to consider $psi = 0$, which corresponds to $pm y = pm x$, so the two lines $y=x$ and $y=-x$.
This hints at the fact that you are missing two families of lines, corresponding to negative $psi$-values. In fact, every value of $psi neq 0$ corresponds to a hyperbola.
A useful fact to remember is that, roughly speaking, streamlines close to each other point in similar directions. Abrupt changes of direction are impossible (due to the solution function depending continuously on the initial conditions).
If you probe the axes now to find the direction of the velocity there you find that it points up on the positive $x$-axis, left on the positive $y$-axis, down on the negative $x$-axis and right on the negative $y$-axis.
This information should be everything you need to plot all the streamlines.
answered May 3 '18 at 13:29
Lukas KoflerLukas Kofler
1,4212520
$endgroup$
add a comment |
$begingroup$
As mentioned by Chee Han, $(u, v) = -2(y, x)$. To get a feeling for the plot of the streamlines it is often easiest to consider $psi = 0$, which corresponds to $pm y = pm x$, so the two lines $y=x$ and $y=-x$.
This hints at the fact that you are missing two families of lines, corresponding to negative $psi$-values. In fact, every value of $psi neq 0$ corresponds to a hyperbola.
A useful fact to remember is that, roughly speaking, streamlines close to each other point in similar directions. Abrupt changes of direction are impossible (due to the solution function depending continuously on the initial conditions).
If you probe the axes now to find the direction of the velocity there you find that it points up on the positive $x$-axis, left on the positive $y$-axis, down on the negative $x$-axis and right on the negative $y$-axis.
This information should be everything you need to plot all the streamlines.
answered May 3 '18 at 13:29
Lukas KoflerLukas Kofler
1,4212520
$endgroup$
add a comment |
$begingroup$
As mentioned by Chee Han, $(u, v) = -2(y, x)$. To get a feeling for the plot of the streamlines it is often easiest to consider $psi = 0$, which corresponds to $pm y = pm x$, so the two lines $y=x$ and $y=-x$.
This hints at the fact that you are missing two families of lines, corresponding to negative $psi$-values. In fact, every value of $psi neq 0$ corresponds to a hyperbola.
A useful fact to remember is that, roughly speaking, streamlines close to each other point in similar directions. Abrupt changes of direction are impossible (due to the solution function depending continuously on the initial conditions).
If you probe the axes now to find the direction of the velocity there you find that it points up on the positive $x$-axis, left on the positive $y$-axis, down on the negative $x$-axis and right on the negative $y$-axis.
This information should be everything you need to plot all the streamlines.
answered May 3 '18 at 13:29
Lukas KoflerLukas Kofler
1,4212520
$endgroup$
As mentioned by Chee Han, $(u, v) = -2(y, x)$. To get a feeling for the plot of the streamlines it is often easiest to consider $psi = 0$, which corresponds to $pm y = pm x$, so the two lines $y=x$ and $y=-x$.
This hints at the fact that you are missing two families of lines, corresponding to negative $psi$-values. In fact, every value of $psi neq 0$ corresponds to a hyperbola.
A useful fact to remember is that, roughly speaking, streamlines close to each other point in similar directions. Abrupt changes of direction are impossible (due to the solution function depending continuously on the initial conditions).
If you probe the axes now to find the direction of the velocity there you find that it points up on the positive $x$-axis, left on the positive $y$-axis, down on the negative $x$-axis and right on the negative $y$-axis.
This information should be everything you need to plot all the streamlines.
answered May 3 '18 at 13:29
Lukas KoflerLukas Kofler
1,4212520
answered May 3 '18 at 13:29
Lukas KoflerLukas Kofler
1,4212520
answered May 3 '18 at 13:29
Lukas KoflerLukas Kofler
1,4212520
answered May 3 '18 at 13:29
Lukas KoflerLukas Kofler
1,4212520
1,4212520
add a comment |
add a comment |
$begingroup$
In continuation to the above comments; in the positive quadrant for instance, you may determine the direction of flux here by considering the velocities at these points.
This might be helpful. https://books.google.co.in/books?id=_Jo-T8nBghYC&pg=PA285&dq=hyperbola+streamline&hl=en&sa=X&ved=0ahUKEwj2speb2fbfAhVJ2LwKHVXtAAYQ6AEIDzAB#v=onepage&q=hyperbola%20streamline&f=false
answered Jan 18 at 6:34
Shatabdi SinhaShatabdi Sinha
19113
$endgroup$
add a comment |
$begingroup$
In continuation to the above comments; in the positive quadrant for instance, you may determine the direction of flux here by considering the velocities at these points.
This might be helpful. https://books.google.co.in/books?id=_Jo-T8nBghYC&pg=PA285&dq=hyperbola+streamline&hl=en&sa=X&ved=0ahUKEwj2speb2fbfAhVJ2LwKHVXtAAYQ6AEIDzAB#v=onepage&q=hyperbola%20streamline&f=false
answered Jan 18 at 6:34
Shatabdi SinhaShatabdi Sinha
19113
$endgroup$
add a comment |
$begingroup$
In continuation to the above comments; in the positive quadrant for instance, you may determine the direction of flux here by considering the velocities at these points.
This might be helpful. https://books.google.co.in/books?id=_Jo-T8nBghYC&pg=PA285&dq=hyperbola+streamline&hl=en&sa=X&ved=0ahUKEwj2speb2fbfAhVJ2LwKHVXtAAYQ6AEIDzAB#v=onepage&q=hyperbola%20streamline&f=false
answered Jan 18 at 6:34
Shatabdi SinhaShatabdi Sinha
19113
$endgroup$
In continuation to the above comments; in the positive quadrant for instance, you may determine the direction of flux here by considering the velocities at these points.
This might be helpful. https://books.google.co.in/books?id=_Jo-T8nBghYC&pg=PA285&dq=hyperbola+streamline&hl=en&sa=X&ved=0ahUKEwj2speb2fbfAhVJ2LwKHVXtAAYQ6AEIDzAB#v=onepage&q=hyperbola%20streamline&f=false
answered Jan 18 at 6:34
Shatabdi SinhaShatabdi Sinha
19113
answered Jan 18 at 6:34
Shatabdi SinhaShatabdi Sinha
19113
answered Jan 18 at 6:34
Shatabdi SinhaShatabdi Sinha
19113
answered Jan 18 at 6:34
Shatabdi SinhaShatabdi Sinha
19113
19113
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics 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%2fmath.stackexchange.com%2fquestions%2f2707374%2fanalyzing-the-direction-of-the-streamline-given-the-streamfunction-psi-x2%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
$begingroup$
For a 2D flow, $u = partial_ypsi$ and $v = -partial_xpsi$.
$endgroup$
– Chee Han
Mar 25 '18 at 20:55
$begingroup$
I'm not entirely sure, but I think you are on the right track here. By definition, streamlines are tangential to the flow field at a fixed time $t$. The flow field in this case is given by $(u,v) = -2(y,x)$ and I believe this yields the direction of the tangent vector for the level sets of streamlines. One last thing: you might also want to consider the case where $psi = textrm{negative constant}$.
$endgroup$
– Chee Han
Mar 25 '18 at 21:19