RS 2 Streampathlines

8/20/2019 RS 2 Streampathlines

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MEC3451Review Sheet - Week 2

Monash UniversityMech. & Aerosp. Eng.

Problem 1

The velocity field of a flow is given by v = 2x2t ex + [4y(t− 1 ) + 2x2t] ey m/s, wherex and y are in metres and t is in seconds.

For fluid particles on the x axis, determine the speed and direction of flow.

Solution: The given velocity field is:

u = 2x2t ; v = 4y(t − 1) + 2x2t .

The speed at any point is:

V =√

u2 + v2 ,

=

8x4t2 + 16x2yt (t − 1) + 16y2(t − 1)2

1/2

.

On the x-axis where y = 0 at any t ≥ 0 therefore,

V =√

8x2t .

The direction of flow at any point is the unit vector along the velocity direction:

v = 1

V v =

2x2tex + 4y(t − 1) + 2x2tey[8x4t2 + 16x2yt (t − 1) + 16y2(t − 1)2]1/2

.

On the x-axis at any t≥

0,

v = 1

V v =

2x2tex + 2x2tey√

8x2t,

= 1√

2(ex + ey) .

Problem 2

Compute and sketch the streamline and pathline that pass through (x, y) = (1, 1) at

t = 1 for the velocity field v = x ex − yt ey.

Solution:

The equation for a streamline in 2D at a fixed time t is:

dx

u =

dy

v .

At time t = 1, the given velocity field is u = x, v = −y. Substituting these in the


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above equation,

dx

x =

dy

−y ;ˆ

dxx

=ˆ

dy−y ;ln x = − ln y + c ;

ln xy = c ;

xy = exp c ;

xy = c1 .

We are interested in the streamline passing through (1, 1). Substituting the co-ordinates in the above equation results in c1 = 1. Therefore, the streamline is

given by,xy = 1 .

The equation for a pathline in 2D is,

dx

dt = u;

dy

dt = v .

For a pathline t is not fixed, but varies. Integrating in time,

dx

dt = x ;

dxx

= dt ;ˆ xx0

dx

x =

ˆ tt0

dt ;

ln x/x0 = t − t0 ;

t = ln x/x0 + t0 . (1)

(Definite integrals can be used instead of working with integration constants.)

Similarly,dy

dt = −yt ;

dy

y = −tdt ;

ˆ yy0

dy

y =

ˆ tt0

−tdt ;

ln y/y0 = −(t2 − t02)

2 ;

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t2 = −2 ln y/y0 + t02 . (2)Substitute from equation 1 into 2 to get the pathline equation:

(ln x/x0 + t0)2 = t0

2

−2 ln y/y0 .

Problem 3

The x and y components of a velocity field are given by u = x2y and v = −xy2.Determine the equation for the streamlines of this flow and compare it with those foranother velocity field where the components are u = −x and v = y. Is the flow inthe first velocity field different from the second? Explain

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Problem 4

A flying airplane produces a swirling flow near the end of its wings. In certaincircumstances this flow can be approximated by the velocity field u = −Ky/(x2 + y2)and v = K x/(x2 + y2), where K is a constant depending on the shape of the airplane,and x and y are measured from the centre of the swirl.

(a) Show for this flow that the velocity magnitude is inversely proportional to thedistance from the origin.

(b) Show that the streamlines are circles.

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Problem 5

Consider the flow field where u = τ/t2 and v = xy/(τ ), where and τ are constantparameters with the units of length and time, respectively.

(a) Find the equation of the streamline for this flow passing through a point (x0, y0)

at t0.

Solution: Equation for a streamline in 2D at fixed time is:

dx

u =

dy

v .

At time t = t0: u = τ/t20, v = xy/τ . Substituting these in the above

equation,

dxτ

t2

0

= dyxyτ

;

xdx

(τ/t0)2 =

dy

y .

For clarity, let K = (τ/t0)2. Then, we have

xdx

K =

dy

y ;

ˆ xx0

xdx

K =

ˆ yy0

dy

y ;

x2 − x022K

= ln y/y0 . (3)

(b) Rearrange the equation to express it in dimensionless form, by defining dimen-sionless position (x∗, y∗) = (x/,y,) and the dimensionless time t∗ = t/τ .

Solution: Define:

x∗ = x

; y∗ =

y

; t∗

0 =

t0

τ .

Hence, we can substitute for x = x∗; y = y∗ in the equation obtained inthe previous part:

(x∗2 − x0∗2)2 = 2K ln y∗/y0∗ ;

= 2(τ

t0)2 ln y∗/y0

∗ ;

= 2(

t∗0)2 ln y∗/y0

∗ ;

(x

∗2

− x0∗2

)t

∗

0

2

= 2 ln y

∗

/y0∗

;

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or,

y∗ = y0∗ e(x

∗2−x0∗

2)t∗0

2/2 .

(c) Sketch the streamlines to obtain an image of the flow field at t∗0. Show that thex-axis is a streamline.

Solution: To show that the x-axis is a streamline, choose a point on thex-axis as (x0, y0) and then find the equation of the streamline through thatpoint, and show that it is the equation for the x-axis i.e. y = 0. For example,if we choose (x∗0, y0) = (1, 0) and substitute in the dimensionless equationabove, we see that the streamline equation through this point is y∗ = 0.

(d) Find the equation of the pathline for a fluid particle that passes through the

point (x0, y0) at t0. Rearrange to express the equation in dimensionless form.

Solution: u = τ t2

; v = xyl τ

dx

dt = u

= τ

t2 ;

dx = τ

t2 dt ;

ˆ x

x0

dx =

ˆ t

t0

τ

t2 dt ;

x − x0 = τ

1

t − 1

t0

. (4)

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dy

dt = v

= xy

τ ;

dyy

= xdtτ

;

=

x0τ

+

1

t − 1

t0

dt ;

ˆ yy0

dy

y =

ˆ tt0

x0τ

+

1

t − 1

t0

dt ;

ln y/y0 =

x0τ

+ 1

t0

(t − t0)− ln t/t0 . (5)

Equations 4 and 5 together gives the path line.

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RS 2 Streampathlines