Homework Set #1 of Course AerE 541 - Iowa State University · Homework Set #3 of Course AerE 541 1....
Transcript of Homework Set #1 of Course AerE 541 - Iowa State University · Homework Set #3 of Course AerE 541 1....
Homework Set #3 of Course AerE 541
1. Consider the incompressible, irrotational flow where the potential function is:
22ln YXK ,
where K is an arbitrary constant.
(a) What is the velocity field for this flow?
What is the magnitude and direction of
the velocity at ( 0,2 ), at ( 2,2 ) and
at( 2,0 )
(b) Is the flow physically possible? Is the
flow field irrotational?
(c) What is the stream function for this flow?
Sketch the stream line pattern.
(d) Sketch the lines of constant potential.
How do the iso-potential lines related to
the streamlines?
(e) For the region shown in the figure, evaluate ?)( AdV
and ?sdV
to
demonstrate that Stokes’s theorem is valid.
(1,1)
(1,0) (0,0)
X
Y
2. In an ideal, 2-D incompressible irrotational flow field, the fluid is flowing past a wall
with a sink of strength K per unit length at the origin as shown in the Figure. The potential
function of a 2-D sink is 22ln2
YXK
. At infinity the flow is parallel to wall and of
uniform velocity U .
(a) Determine the location of the stagnation point 0X at the wall in terms of U and K.
(b) Find the pressure distribution along the wall as a function of X. Taking the free stream
statitc pressure at infinity to be p , express the pressure coefficient as a function of 0/ XX .
(c) Sketch the resulting pressure distribution.
wall Sink of strength K at X=0
X
U
p Y
3. Superposition of a uniform stream ( V ) and a point source (strength q ) located at the origin
produces the flow over a smooth blunt-nosed body. The blunt-nosed body is usually called
Rankine nose. The radius of the
Rankine nose at Z is
V
qr
(f) To express r
Z as a function
of .
(g) To express the radius of the
Rankine nose (r
r) as a
function of , and plot the
curve of r
rchanging with
r
Z .
(h) Derived the flow velocity (
2
2
V
V) on the surface of the Rankine nose as a function of
, and plot the curve of V
Vchanging with
r
Z .
(i) Derived the pressure coefficient (2
2
1
V
ppC p
) on the surface of the Rankine nose as
a function of , and plot the curve of 2
2
1
V
ppC p
changing with r
Z.
Solution:
The stream function for the combined flow is:
cos4
sin2
22
qRV
The corresponding velocity field is:
sin
4cos
2
VV
R
qVVR
At stagnation point 0;0 VVR
Z
Y
rr
X
i.e., 04
cos2
R
qVVR
and 0sin VV
0sin RVV or0
0
04
cos2
R
qVVR
V
qRstag
4
2 It is impossible
0
4cos
2R
qVVR
V
qRstag
4 Possible solution
Stream line passing the stagnation point:
4cos
4sin
2
22 qqRV
stagnation
Therefore, the equation of the streamline passing the stagnation point is given by:
2cos
2
)1(cos
2
)1(cossin
2
)1(cos
sinsin2
)1(cos
sin
)1(cos
2
sin
)1(cos
2sin
2
)1(cos4
0)1(cos4
sin2
22
22
2
22
r
r
r
R
r
V
q
V
qR
V
q
V
q
R
qRV
Since )2/sin(
cos
2
1)2/cos(
)2/cos()2/sin(2
cos
2
)1(cos
sin
costan/
r
ZrZ
Therefore:
(a). )2/sin(
cos
2
1
r
Z
(b). 2
cos2
)1(cos
r
r
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
0 30 60 90 120 150 180
r/r
Z/r
, deg.
r/r
, Z
/r
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
-2 0 2 4 6 8 10 12
Z/r
r/r
The stream line passing the stagnation point can be treated as the surface of a solid body since no
flow will pass the stream line according to the definition of streamline.
On the surface of Rankin nose:
2sin
)1(cos
2
V
qR
Thus,
sin
)]2/(sin[cos)1(cos2
sincos
4cos 2
2
2
VV
VV
VR
qVVR
Therefore:
)2/(sin)2/(sincos21)2/(sin)2/(sincos2cossin)( 4242222 V
V
)2/(sin)2/(sincos21)( 422 V
V
)2/(sin)2/(sincos2
)]2/(sin)2/(sincos21[1
)(1
2
1
42
42
2
2
V
V
V
ppC p
4. Consider air flowing past a hemisphere resting on a flat surface, as shown in the Figure.
Neglecting the effects of viscosity, if the internal pressure is ip , find an expression for the
pressure force on the hemisphere. At
what angle location should a hole be
cut in the surface of the hemisphere
so that the net pressure force acting
on the hemisphere will be zero.
Solution:
On the surface of the sphere, the
local pressure is:
)sin4
91()sin
4
91(
2
1 222 qPVPP ;
2
2
1 Vq
The area of a small element on the sphere can be expressed as:
RddRdA sin
The x-component of the force (i.e., along x-direct) acting on the
small element at the outer surface:
ddRqP
ddRPRddRPdAPdFx
cossin)]sin4
91([
cossinsincossincossin
222
22
Therefore, the lift force the due to the external pressure acting on the
surface of the hemisphere is
)16
11()
16
271((]
8
3
4
9
2)[(2
])32
4sin
4
2sin
8
3(
4
9)
4
2sin
2)([(2
]sin4
9sin)[(2
sin)]sin4
91([2
cossin)]sin4
91([
cossin)]sin4
91([
222
00
2
0
4
0
22
0
222
0
2
2
222
0
2
2
222
qPRqPRqqPR
qqPR
dqdqPR
dqPR
ddqPR
ddRqPLoutside
If the pressure inside the semi-sphere shell is insideP , then, the total lift force (i.e., along x-
component) acting on the semi-sphere due to the inner pressure will be insideinside PRL 2
If the net lift acting on the semi-sphere shell is to be zero:
qPPqPRPRLL insideinsideoutsideinside16
11)
16
11(22
On the surface of the sphere, the local pressure is: )sin4
91( 2 qPP
Therefore, at the location where the hole is made will have:
O
O
qPqPP
120
60
2
3sin
4
3sin
16
27sin
4
9
16
11)sin
4
91(
16
11)sin
4
91(
2
22
2