ME6601: Introduction to Fluid Mechanics

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Module 5 - Kinematics IV

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Slide Index

Slide 1 - Kinematics IV - Fluid-material Lines

Slide 2 - Fluid-material lines (continued)]

Slide 3 - Streaklines

Slide 4 - Streaklines - solving a streakline problem

Slide 5 - Streaklines - solving a streakline problem (continued)

Slide 6 - Streakline example

Slide 7 - Streakline example (continued)

Slide 8 - Streakline example (continued)

Slide 9 - Streakline example (continued)

Slide 10 - Streakline example (continued)

Slide 11 - Streakline example (continued)

Slide 12 - Kinematics IV summary

Slide 1 of 12

Module Five - Kinematics IV - Fluid-material lines

A fluid-material line, as its name suggests, is a line identifying a particular piece of the continuum, i.e., it always marks the same fluid material for all time.
If the material-coordinate displacement field bold capital X as a function of capital X subscript 0 and lowercase t is known, then a particular fluid-material line may be identified by specifying two of its initial coordinates as functions of the third, i.e.,
Capital Y subscript 0 equals lowercase f subscript capital Y, as a function of capital X subscript 0.
Capital Z subscript 0 equals lowercase f subscript capital Z, as a function of capital X subscript 0.
This provides us with the initial position of the line of interest.

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Slide 2 of 12

Fluid-material lines (continued)

The position of the fluid-material line, at any subsequent time, is then obtained from
Capital bold X subscript capital F equals capital bold X as a function of capital X subscript zero, lowercase f subscript capital Y as a function of capital X subscript 0, lowercase f subscript capital Z as a function of capital X subscript 0, and time, lowercase t.
We can eliminate X₀ from these three equations to get
Capital Y subscript capital F equals capital Y subscript capital F as a function of capital X subscript capital F and lowercase t. Capital Z subscript capital F equals capital Z subscript capital F as a function of capital X subscript capital F and lowercase t.

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Slide 3 of 12

Streaklines

A particular type of fluid-material line which is of much utility is the streakline, which is the locus of all particles which have passed a specified ("tagging") location in some interval of time.
The use of dye, smoke or hydrogen bubbles to generate streaklines is common in practice.
As we have already note in our previous module, in steady flow, streaklines (generated by stationary taggers), streamlines and pathlines through the same points are all identical.

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Slide 4 of 12

Streaklines - solving a streakline problem

Suppose the tagging location is at spatial position lowercase x superscript star.
A fluid particle is then tagged when, at time lowercase t subscript one 1,
Capital bold faced X as a function of capital X subscript 0 and lowercase t subscript 1 equals lowercase x superscript star, regardless of lowercase t subscript 1.
We invert this relation to obtain
Lowercase bold x superscript star and subscript 0 equals lowercase bold x superscript star and subscript zero as a function of both lowercase x superscript star and lowercase t,
which is an equation for the initial positions of all particles that will be tagged up to time t.

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Slide 5 of 12

Streaklines - solving a streakline problem (continued)

Now that we know the lowercase t equals 0 position of the streakline, its position at any subsequent time lowercase t is given by
Capital bold-faced X subscript capital F equals capital bold X as a function of both capital X superscript star and subscript 0 superscript star, and lowercase t.

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Slide 6 of 12

Streakline example

The previous discussion is best reinforced with a worked example, which we present now. So that the problem does not reduce to one which is trivial, we'll consider a steady flow, but with a moving tagger.
[D1]
Example: plane stagnation-point flow with a tagger that is moving to the right with constant speed U.

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Slide 7 of 12

Streakline example (continued)

We assume the tagger to be initially located at the spatial location (lowercase x superscript star, lowercase y superscript star) equals (negative lowercase x subscript 1, lowercase y subscript 1).
Since it moves horizontally with constant speed U, at any time t its position is
Lowercase x superscript star equals negative lowercase x subscript one plus capital U times lowercase t. Lowercase y superscript star equals lowercase y subscript 1.
For plane stagnation-point flow, we have already determined
Capital X equals capital X subscript 0 times lowercase e to the power of capital K times lowercase t. Capital Y equals capital Y subscript 0 times lowercase e to the power of negative capital K times lowercase t.

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Slide 8 of 12

Streakline example (continued)

Equating the right-hand sides of these two sets of equations at time lowercase t subscript 1
Negative lowercase x subscript one plus capital U times lowercase t subscript one equals capital X subscript 0 times lowercase e to the power of capital K times lowercase t subscript one.
Lowercase y subscript one equals capital Y subscript 0 times lowercase e to the power of negative K times lowercase t subscript one.
Eliminating t subscript 1 using the second equation, i.e., t subscript 1 equals (1 divided by capital K) times (The natural log of capital Y subscript 0 divided by lowercase y subscript 1), the first equation becomes (solving for X subscript 0)
Capital X subscript 0 equals (lowercase y subscript one over capital Y subscript 0) times, open parenthesis, negative lowercase x subscript one plus (capital U divided by capital K) times the natural log of capital Y subscript 0 divided by lowercase y subscript 1, end parentheses.

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Slide 9 of 12

Streakline example (continued)

Using this in the equations for capital X and capital Y (and using an capital S subscript to denote streakline),
Capital X subscript capital S equals (lowercase y subscript 1 divided by capital Y subscript 0) times (negative lowercase x subscript 1, plus (capital U divided by capital K), times (the natural log of capital Y subscript 0 divided by lowercase y subscript 1) ), times lowercase e superscript capital K lowercase t
Capital Y subscript capital S = capital Y subscript 0 times lowercase e superscript negative capital K lowercase t.
Eliminating capital Y subscript 0 using the second equation, we obtain, finally
Capital X subscript capital S equals (Lowercase y subscript 1 divided by capital Y subscript capital S), times negative lowercase x subscript 1, plus (capital U divided by capital K), times (the natural log of capital Y subscript capital S divided by lowercase y subscript 1) ), times lowercase e superscript capital K lowercase t
the streakline equation.

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Slide 10 of 12

Streakline example (continued)

This is a parametric form of the curve, yielding a family of curves for various times t.
Let's evaluate this for the time when the tagger has reached the x-location lowercase x subscript 1: to determine the time corresponding to this, we may use either the above expression or
Lowercase x subscript 1 equals negative lowercase x subscript 1 plus capital U times lowercase t, arrow pointing to lowercase t equals 2 times x subscript 1 divided by capital U.

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Slide 11 of 12

Streakline example (continued)

Using this value of lowercase t in the equation for capital X subscript capital S:

Capital X subscript S equals (Lowercase y subscript 1 divided by capital Y subscript capital S), times (negative lowercase x subscript 1, plus (capital U divided by capital K) times (The natural log of capital Y subscript capital S divided by lowercase y subscript 1) times lowercase e superscript (2 times capital K times x subscript i divided by capital U)).
This is the equation of the streakline at this time.
[D2]
[D3]

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Slide 12 of 12

Kinematics IV summary

We finish our study of kinematics by studying fluid material lines, which are employed often in experimental work.
In particular, we consider the streakline, working an example problem employing the plane stagnation-point flow considered earlier.

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