Exam January 2003.
Exercise 1
(In this kind of exercises, your answer does not necessarily need to be correct, but the arguments are
relevant as well!)
a)
Consider a velocity field, for which there exists a scalar stream function . Below there are
five statements. Discuss in a few words for each of them individually whether it is true or
false.
1
The flow must be two-dimensional.
In fact either 2D or 3D-axisymmetric. For general 3D incompressible flow there is a vector potential
form, i.e. u = curl Psi. For some: indeed, it can be 1D as well...
2
The flow must be irrotational.
Not relevant, think of solving the Blasius boundary layer flow using a stream function..
3
The flow must be incompressible.
Sure. Mass balance:
. Incompressible: is constant on the streamline, hence
. Expressing divergence in components of the stream function:
4
The flow must be frictionless.
Not relevant. see example for boundary layer flow, flow around a sphere, etc.etc.
5
The flow must be stationary.
Not relevant, as fe have seen in numerous examples.
2a
At any time, the velocity magnitude (hence length of the vectors) is A/2pi r, thus inverse proportional
with distance to the origin, & irrespective of the polar angle .
1) At t = 0, the velocity vectors are purely tangential, thus the stream lines are circles (flow field of a
point vortex).
3) At t =
, the velocity vectors are purely radial, thus the stream lines are straight lines from the
origin (flow field of a point source).
2) At t =
, the velocity vectors are rotated 45 degrees from radial, or halfway between
tangential and radial. The stream lines are spiralling (combination of source and vortex flow)
Exact: For a stream line we find in a Cartesian system: dx/u_x = dy/u_y. Thus in polar coordinates,
we get dr/u_r = rd /u_ . Integrating to find (r), we get
. More specific at
time t =
:
clearly shows the spiralling character.
Exercise 2
This problem was rather difficult, as I found out. The tangential
velocity relative to the location of the centre of a single vortex is
Gamma /(2 pi r), but I think you ought to know this.
... I could have added either:
a) The stream function of a point vortex at location (x0,y0) in
Cartesian coordinates is -Gamma/[2pi*((x-x0)^2+(y-y0)^2)^0.5] =
-Gamma/4pi((x-x0)^2+(y-y0)^2).
b) More helpful: for a point vortex at location (x0,y0) in Cartesian
coordinates you can derive (with the small graph on the right):
u = (u,v) = Gamma/(2pi r) (-sin(theta), cos(theta)),
thus (u,v) = Gamma/[2pi*((x-x0)^2+(y-y0)^2)] * [ -(y-y0), (x-x0)]
a)
Assume the velocity as (u,v)(x,y)
1) Boundary conditions:
- The velocity normal to the two walls has to be zero (no penetration): u(x=0) = 0; v(y=0) = 0.
- Since the flow is inviscid, the velocity parallel to the walls can be non-zero. So there is no
boundary condition there
- Some argued that the velocity on the wall of the
rotating cylinder has as well to fulfil these
conditions. Good job! (fortunately the line vortex
flow does this as well.
2) First consider the single vortex (#1) near the
horizontal wall only. We can fulfil the BC by
taking the mirror image of that vortex (try adding
up the two induced velocity vectors), producing
#2. (first picture). Now we can add the vertical wall, and again fulfill the BC by mirroring both #1
and #2, to find #3 and #4.
3) Using the above, one adds all four to find the velocity field: (u,v) = Gamma/2pi * (u1,v1) with
u1 = -(y-d)/((x-d)^2+(y-d)^2) + (y+d)/((x-d)^2+(y+d)^2) (y+d)/((x+d)^2+(y+d)^2) + (y-d)/((x+d)^2+(y-d)^2).
(#1, #2, #4, #3)
v1 = (x-d)/((x-d)^2+(y-d)^2) - (x+d)/((x+d)^2+(y-d)^2) +
(x+d)/((x+d)^2+(y+d)^2) - (x-d)/((x-d)^2+(y+d)^2).
(#1, #3, #4, #2)
which is quite a lot of formulae. However, many other answers were given credit as well...
Fortunately, knowing all the symmetries in the field, you didn't need much of this any more further!
b)
Purely straining formally implies:
1 Vorticity is zero.
2 Velocity is zero.
3 Velocity gradient is non-zero.
Of course, you didn't need to prove all of these, except for the last. Possible arguments are, for
instance:
1) The flow field of a point vortex is irrotational, just as that of four vortices, hence no vorticity
2) From symmetry, or from the BC above: at the origin, u=v=0.
3) The velocity gradient tensor consists of four components: du/dx, du/dy, dv/dx and dv/dy.
Since v = 0 for x = 0, we find dv/dx =0. The same holds for du/dy.
The flow is incompressible, i.e. div(u) = du/dx + dv/dy = 0, hence dv/dy = -du/dx.
What rests is to determine du/dx...
For this, symmetry shows that you need to consider only one vortex, and that all contribute in exactly
the same amount...
u = -(y-y0) Gamma/[2pi*((x-x0)2+(y-y0)2)] =>
du/dx|x=y=0 = d/dx( -(0-d) Gamma/[2pi*((x-d)2+(0-d)2)]) =
d/dx( d.Gamma/[2pi*((x-d)2+d2)]) =
2*(x-d) . -d.Gamma/[2pi*((x-d)2+d2)2]|x=0 =2d2Gamma/[2pi*(2d2)2] = Gamma/pi d2 QED
c
The dyed fluid element is getting deformed by the flow, in this case purely strained in the xdirection...The lower left corner will not move, the right side will move from a to d in time T
(Since it conserves its volume, at t =T , the top side has arrived at a^2/d...)
The transport is Lagrangian, i.e. the location of the side is X(t), with initial condition X(0) = a.
we get the original differential equation (ODE in English literature):
dX/dt = kX, which easily solves with the as X = a . e^(kt). Thus T is found: T = 1/k ln(d/a)
Many other answers based on the 'average velocity' between the two points got credit as well...
d
This is a problem which you should recognise from forces on
rotating cylinders. Even though the flow is not uniform anymore, it
is irrotational, thus we can apply the hint. Near the location of the
vortex, the velocity is that imposed by the three images: Either
evaluate the expression from a) at (x,y) = (d,d), or (if you didn't have
that one, and even more easy) add velocity components induced by
the three images componentwise:
U1x = Gamma/2pi 2d (#2) + 0 (#3) +
+ -Gamma(2pi 2sqrt(2)d)*1/Sqrt(2) (#4)
= Gamma/2pi 2d* (1-0.5) = Gamma/8pi d.
U1y = -U1x (from symmetry), or
U1x = 0 (#2) + -Gamma/2pi 2d (#3) +
+ Gamma(2pi 2sqrt(2)d)*1/Sqrt(2) (#4)
= Gamma/2pi 2d* (-1+0.5) = -Gamma/8pi d
Thus, for the force on the cylinder, FC, we find:
FCX = -rho Gamma^2/8pi d; FCY = -rho Gamma^2/8pi d. To keep the cylinder in place, we thus need
an opposite force FK = -FC .
e
After release, the vortex is not transported in direction of the force, but it is transported with the flow
present at location of the vortex, thus to larger x, and smaller y. While being advected, it comes
under stronger influence of image vortex #2, and weaker influence of #3,4. It thus increases in xvelocity, while the y velocity decreases (in magnitude). At larger time, images #3,4 have disappeared
to the left, and the only relevant image is #2. Thus the two form a vortex couple or vortex dipole, tht
moves to the right along the x-axis.
If you were at the appropriate lecture (#4 or #5), you have seen this as an experiment (a single vortex
near a wall moves along the wall).
f
We use the Lagrangian description, and similar to d:
dX/dt = Gamma / 2pi 2Y (#2) - Gamma / 2pi 2sqrt(X^2+Y^2) * Y/sqrt(X^2+Y^2) (#4)
= Gamma / 4pi Y - Gamma Y/4pi (X^2+Y^2) ; similar:
dY/dt = -Gamma / 2pi 2X (#3) + Gamma / 2pi 2sqrt(X^2+Y^2) * X/sqrt(X^2+Y^2) (#4)
= -Gamma / 4pi X + Gamma X/4pi (X^2+Y^2).
The quantity
is called a first integral; it is conserved during time.
Evaluating the time derivative of it:
d/dt (
)=
Since we need to show that this is zero, we can multiply by
, i.e.
.
To prove this, we first evaluate Y dX/dt + X dY/dt from above, throwing out Gamma / 4pi:
1)
Then we evaluate X dX/dt + Y dY/dt from above, again throwing out Gamma / 4pi:
2)
. Multiplying the latter
2
2
by XY, the former by (X +Y ), and subtracting, we prove the statement.
Evaluating the integral form at t = 0, we find the constant. Evaluating it for large X, the integral form
linearises to Y , i,e, Y = d/sqrt(2).
g
To make the flow zero at location of the vortex, we add a flow that is opposite to that as found in d:
(u,v) = (Gamma/8d, -Gamma/8d). Since the added flow is a pure strain, (u,v) = (kx, -ky), we thus find
k = -Gamma/8pi d^2
h
Besides the one at the origin, we get three extra stagnation points:
- The one at the vortex location
- One at the x- axis: there where the
outgoing flow by the vortex (getting
weaker with increasing x) exactly
balances the incomng flow as added
in g)
- A similar point on the y-axis (same
distance from the origin), where the
outgoing strain field balances the
incoming vortex flow.
It's relatively easy to find an equation
for these points, though it doesn't
have an analytical solution: you'll
find them around x = 2.3. They are
shown as red stars in the graph
below.
Problem 3
a
At rest all viscous stresses are zero!!!
Thus, the hydrostatic pressure of the ink, rho g h_fill, must be
compensated by the differential pressure over the curved interface at
the outflow of the nozzle, which thus is pointing outward (see picture).
This pressure is in general equal to sigma/R_eff, with 1/R_eff = 1/R_1
+ 1/R_2, with R_i the two principal radii of curvature. For a spherical
(radius R) surface, this comes down to 2sigma/R. The smallest
spherical surface fititng the nozzle is that with R = d_N /2; hence the
maximum pressure is 4sigma/d_N, hence h_fill,max = 4 sigma/(rho g
d_N). For the given parameters this is 1 m of height, which is much
more than the typical 5 cm of an ink cartridge.
However, when you drop it on a floor, the instantaneous acceleration is
considerably higher than g; for our parameters 20g suffices to
overcome the surface pressure...
b
If we suck in ink, the pressure gets lower than atmospheric. The ink interface at the nozzle will now
point inwards (see picture) In this case, we would suck in air, in case the under pressure gets more
than 4sigma/R. Hence the pressure drop over the flow through the ink channel must be less than that.
We assume Poiseuille flow passing through a straight channel. With the given pressure gradient,
multiplied with L we get the minimum filling time, realising that T1*Q= V_P
c
To get liquid ejected, the pressure must be larger than the under pressure from the suction phase.
Provided T2<T1, this is automatically realised. As soon as liquid flows out, the surface force gets
considerably smaller: There is not a spherical interface any more, but a cylindrical, which has only
half of the effective curvature. With the given liquid speed and length of the liquid cylinder coming
out, T2 simply follows from T2 = Length / velocity = pi d_N/U_N = 6.3 mu.s.
d
Re is U_N*d_N/nu = 40. Better estimates are possible, and are even higher.
This means that the accelerated flow (though the flow may be assumed stationary, the Lagrangian
term UdU/dx will be large) inside the nozzle will only show relatoivwely thin boundary layers which
aksdkjasd the flat velocity profile. Therefore, Bernoulli's law may be applied, to show that [p_0 +
U_N^2/(2 rho)]outflow [p_0+P_PO+U_P^2/2rho]plenum. At the outflow we know the pressure
(atmospheric and the high velocity. Because of the high contraction ratio, we can neglect the velocity
inside the plenum, hence we know the plenum pressure P_PO. Filling out numbers gives 50kPa.
e
Many answers were OK. The easiest goes as follows. The overpressure lasts only during time
interval T2 = fT1. At the same time the pressure driving flow through the ink channel; is
50kPa/10kPa =5 times that during suction. Therefore, V_B = 5f V_P.
f
With the hint V_B = V_P/2 and e), we find 5f = 0.5, thus f = 0.1.
The cycle time thus is T2*(1 + 1/f) = 11 T2 = 69 mu.s, or 1/69.10^-6 = 14.500 droplets /s
g
To print an A4 colour print, we need to print about 6*10 square inches, with a resolution of 720 dpi,
this are 60*720^2 = 31 million droplets. With a single nozzle at the droplet rate of f, this would take
more than half an hour. With 192 nozzles (64 per colour) doing the job, it only takes 33 seconds.