Review: Grad, etc.
Pressure field
Figure 1 shows a Control Volume or circuit placed in a pressure field p(x, y). To evaluate
the pressure force on the CV, we need to examine the pressures on the CV faces.
Pressures on faces of
Control Volume
Pressure Field
y
p
x
Figure 1: Pressure field and Control Volume.
Pressure force – infinitesimal rectangular CV
Let us now assume the CV is an infinitesimal rectangle, with dimensions dx and dy. We
wish to compute the net pressure force (per unit z depth) on this CV,
−dF~ ′ =
�
(p n)
ˆ ds
where ds is the CV side arc length, either dx or dy depending on the side in question. As
shown in Figure 2, the pressures across the opposing faces 1,2 and 3,4 are related by using
the local pressure gradients ∂p/∂x and ∂p/∂y.
dy
9
p + 9p dy
y 2
9
p − 9p dx
x 2
4
j
n
9
p + 9p dx
x 2
p
Axis unit vectors
Normal vectors
Face pressures
y
i
2
1
9
3
dy
p
9
p−
y 2
x
dx
Figure 2: Infinitesimal CV surface pressures and normal vectors.
The pressure force is then computed by evaluating the integral as a sum over the four faces.
−dF~ ′ =
=
�
(p n̂) ds
�
�
1
�
+ (p n̂) ds
�
∂p dx
ı̂ dy +
−p +
∂x 2
�
2
�
�
+ (p n̂) ds
�
+ (p n̂) ds
3
�
∂p dx
p+
ı̂ dy +
∂x 2
1
�
�
�
4
�
∂p dy
−p +
̂ dx +
∂y 2
�
�
∂p dy
p+
̂ dx
∂y 2
=
�
�
∂p
∂p
ı̂ +
̂ dx dy
∂x
∂y
−dF~ ′ = (∇p) dx dy
(1)
where ∇p is a convenient shorthand for the gradient expression in the parentheses. This
final result for any 2-D infinitesimal CV can be stated as follows:
2-D :
× d(area)
-d(force/depth) = (pressure gradient)
(2)
For a 3-D infinitesimal “box” CV, a slightly more involved analysis gives
−dF~ =
3-D :
�
∂p
∂p
∂p ˆ
ı̂ +
̂ +
k
∂x
∂y
∂z
�
dx dy dz= (∇p) dx dy dz
-d(force) = (pressure gradient)
× d(volume)
(3)
(4)
Pressure force – finite CV
We now wish to integrate the pressure gradient over a finite CV.
��
(∇p) dx dy =
�� �
�
∂p
∂v
ı̂ +
̂
∂x
∂y
dx dy
As shown in Figure 3, this is equivalent to summing the pressure forces on all the infinitesimal
CVs in the finite CV’s interior. We then note that the contributions of all the interior faces
( p ) dx dy = Sum ( p n ) ds
Δ
interior face integrations cancel
( p n) ds
Figure 3: Summation of pressure gradient over interior of a finite CV
cancel, since these have directly opposing n̂ normal vectors, leaving only the boundary faces
in the overall summation which the give the net pressure force on the CV.
��
(∇p) dx dy =
2
�
(pn̂) ds
This general result is known as the Gradient Theorem, and it applies to any scalar field f ,
not just fluid pressure fields. The general Gradient Theorem in 3-D is
dA
n
���
∇f dx dy dz=
��
f n̂ dA
f
dx dy dz
f
Δ
where the volume integral is over the interior of the CV, and the area integral is over the
surface of the CV.
Constant pressure field contribution
It is useful to note that adding a constant pressure pC to the pressure field p(x, y) has no net
effect on the calculation of F~ ′ . Because the grad operation involves derivatives, any constant
component in p will disappear, e.g.
∇ (p + pC ) = ∇p + ∇pC = ∇p
(5)
Similarly, the normal vector integral of a constant around a closed CV is zero.
�
pC n
ˆ ds = 0
These properties can be used to see the important feature of a pressure field. We can take
the four face pressures shown in Figure 1 and define their average
pavg =
1
(p1 + p2 + p3 + p4 )
4
and subtract this constant pavg from each pressure to give the local deviations Δp, as shown
in Figure 4.
pn
Δ p n =( p− pavg ) n
0
pavg
Figure 4: Average pressure subtracted out to give local pressure force deviations
From (5) we see that
∇ (Δp) = ∇p
so that these pressure deviations have exactly the same F~ ′ as the full pressure. Removal of
the large constant pavg makes it easier to visualize the net magnitude and direction of the
pressure force acting on the CV.
3
Review: Div, Curl, etc.
Velocity field
~ (x, y). To evaluate
Figure 1 shows a Control Volume or circuit placed in a velocity field V
the volume flow through the CV, we need to examine the velocities on the CV faces. It is
convenient to work directly with the x,y velocity components u, v.
Velocities on faces of
Control Volume
or Circuit
Velocity Field
y
Velocity components
V
v
u
x
Figure 1: Velocity field and Control Volume.
Volume outflow – infinitesimal rectangular CV
Let us now assume the CV is an infinitesimal rectangle, with dimensions dx and dy. We
wish to compute the net volume outflow (per unit z depth) out of this CV,
dV̇ ′ =
� �
~ ·n
V
ˆ ds =
�
�
(u nx + v ny ) ds
where ds is the CV side arc length, either dx or dy depending on the side in question.
Figure 2 shows the normal velocity components which are required. The velocities across
the opposing faces 1,2 and 3,4 are related by using the local velocity gradients ∂u/∂x and
∂v/∂y.
y
dy
Normal vectors
Normal velocity components
9
v + 9v dy
y
4
9
u + 9u dx
x
u
n
2
1
v
3
x
dx
Figure 2: Infinitesimal CV with normal velocity components.
The volume outflow is then computed by evaluating the integral as a sum over the four faces.
dV̇ ′ =
~ · n̂ ds
V
��
�
�
1
+
V~ · n̂ ds
��
�
1
�
2
+
V~ · n̂ ds
��
�
�
3
+
V~ · n̂ ds
��
�
�
4
=
�
=
�
�
dV̇ ′ =
or
�
�
−u dy +
�
∂u
u+
dx dy +
∂x
�
�
−v dx +
�
�
∂v
v+
dy dx
∂y
�
∂u ∂v
dx dy
+
∂x ∂y
~
∇·V
�
dx dy
(1)
~ is a convenient shorthand for the velocity divergence in the parentheses. This
where ∇ · V
final result for any 2-D infinitesimal CV can be stated as follows:
d(volume outflow/depth) = (velocity divergence)
2-D :
×
d(area)
(2)
For a 3-D infinitesimal “box” CV, a slightly more involved analysis gives
dV̇ =
3-D :
�
�
�
�
∂u ∂v ∂w
~ dx dy dz
dx dy dz = ∇ · V
+
+
∂x ∂y
∂z
d(volume outflow) = (velocity divergence)
×
(3)
d(volume)
(4)
Volume outflow – infinitesimal triangular CV
Although statements (2) and (4) were derived for a rectangular infinitesimal CV, they are in
fact correct for any infinitesimal shape. For illustration, consider the volume outflow from a
right-triangular CV, pictured in Figure 3. One complication is that the unit normal vector
n̂ on the hypotenuse is not parallel to one of the axes. However, its nx ,ny components are
easily obtained from the geometry as shown in the figure.
y
Velocity components
Normal vectors
n y = dx
ds
n
9
9
v + v dy
y 2
dy
9
u + 9u dx
x 2
u
dy
n x = ds
1
ds
dy
dx
3
v
2
x
dx
Figure 3: Infinitesimal triangular Control Volume.
The volume outflow per unit depth is now
dV̇ ′ =
=
~ ·n
V
ˆ ds
��
�
�
�
�
−u dy +
1
+
�
ˆ ds
V~ · n
��
�
�
−v dx +
�
��2
+
ˆ ds
V~ · n
��
∂u dx
u+
∂x 2
2
�
�
�
3
dy
+
ds
�
∂v dy
v+
∂y 2
�
�
dx
ds
ds
=
dV̇ ′ =
or
�
�
∂u ∂v
+
∂x ∂y
~
∇·V
�
�
dx dy
2
dx dy
2
The area of the triangular CV is dx dy/2, so statement (2) is still correct for this case.
Circulation – infinitesimal rectangular circuit
We now wish to compute the clockwise circulation dΓ on the perimeter of the infinitesimal
rectangular CV, called a circuit by convention.
�
−dΓ =
~ · d~s =
V
� �
V~ · ŝ ds
�
This now requires knowing the tangential velocity components, shown in Figure 4.
y
9
u + 9u dy
y
4
9
v + 9v dx
x
v
dy
Tangential vectors
Tangential velocity components
s
1
2
u
3
x
dx
Figure 4: Infinitesimal CV with tangential velocity components.
The circulation is then computed by evaluating the integral as a sum over the four faces.
−dΓ =
or
��
=
�
=
�
− dΓ =
~ · ŝ ds
V
�
�
�
1
−v dy +
�
+
�
��
v+
~ · ŝ ds
V
�
�
�
2
+
∂v
dx dy +
∂x
V~ · ŝ ds
��
� �
�
u dx +
�
�3
+
V~ · ŝ ds
��
−u −
�
�
�
4
∂u
dy dx
∂y
�
∂v ∂u
dx dy
−
∂x ∂y
∇ × V~ · k̂ dx dy
�
(5)
~ · kˆ is a convenient (?) shorthand for the z-component of the velocity curl in
where ∇ × V
the parentheses. For a triangular circuit, equation (5) would have dx dy/2 as expected. The
general result for any infinitesimal circuit can be stated as follows:
�
�
-d(circulation) = (normal component of velocity curl)
3
×
d(area)
(6)
Volume outflow – finite CV
We now wish to integrate the velocity divergence over a finite CV.
�� �
~
∇·V
�
dx dy =
�� �
∂u ∂v
+
∂x ∂y
�
dx dy
As shown in Figure 5, this is equivalent to summing the volume outflows from all the infinitesimal CVs in the finite CV’s interior, either rectangular or triangular in shape, as needed
to conform to the boundary. We then note that the contributions of all the interior faces
(
V ) dx dy = Sum
( V n ) ds
interior face integrations cancel
( V n ) ds
Δ
Figure 5: Summation of divergence over interior of a finite CV
cancel, since these have directly opposing n̂ normal vectors, leaving only the boundary faces
in the overall summation which the give the net outflow out of the CV.
�� �
~
∇·V
�
dx dy =
� �
V~ · n
ˆ ds
�
This general result is known as Gauss’s Theorem, and it applies to any vector field ~υ , not
just fluid velocity fields. The general Gauss’s Theorem in 3-D is
dA
n
���
∇· ~υ dx dy dz =
��
~υ · n
ˆ dA
υ
dx dy dz
υ
where the volume integral is over the interior of the CV, and the area integral is over the
surface of the CV.
Circulation – finite circuit
We now wish to integrate the normal component of the curl over the interior of a finite
circuit.
�
�� �
�� �
�
∂v
∂u
~ · kˆ dx dy =
−
dx dy
∇×V
∂x ∂y
As shown in Figure 6, this is equivalent to summing the circulations from all the infinitesimal circuits in the circuit’s interior, either rectangular or triangular in shape, as needed to
4
(
V ) k dx dy = Sum V ds
V ds
interior face integrations cancel
Δ
Figure 6: Summation of curl over interior of a finite circuit
conform to the boundary. We then note that the contributions of all the interior faces cancel,
since these have directly opposing ŝ tangential vectors, leaving only the boundary faces in
the overall summation.
�� �
~ · kˆ dx dy =
∇×V
�
� �
~ · ŝ ds
V
�
This general result is known as Stokes’s Theorem, and it applies to any vector field ~υ , not
just fluid velocity fields. The general Stokes’s Theorem in 3-D is
ds
υ
��
(∇× ~υ) · n
ˆ dA =
�
n
dA
~υ · d~s
ds
υ
υ
where the area integral is over any “potato chip” surface spanned by the circuit, and the line
integral is over the circuit itself. The circuit and spanned surface do not need to be flat.
Constant velocity field contribution
It is useful at this point to observe that adding a constant vector V~C to the velocity field
~ (x, y) has no net effect on the calculation of V̇ ′ or Γ. Because the div and curl operations
V
~ will disappear, e.g.
involve derivatives, any constant component in V
~ +V
~C
∇· V
�
~ +V
~C
∇× V
�
�
�
= ∇ · V~
~C = ∇ · V
~
+ ∇·V
~C = ∇ × V
~
= ∇ × V~ + ∇ × V
(7)
(8)
Similarly, the normal or tangential integral of a constant around a closed CV or circuit is
zero.
�
~C · n
V
ˆ ds = 0
�
V~C · sˆ ds = 0
These properties can be used to help “visualize” the divergence or curl of the velocity field.
We can take the four face velocities shown in Figure 1 and define their average
�
�
~2 + V
~3 + V
~4
~avg = 1 V
~1 + V
V
4
5
~avg from each velocity to give the local deviations ΔV~ , as shown
and subtract this constant V
in Figure 7.
Vavg
V
ΔV = V− Vavg
Figure 7: Average velocity subtracted out to give local velocity deviations
From (7) and (8) we see that
∇ · ΔV~
�
∇× ΔV~
�
~
= ∇·V
�
~
= ∇× V
�
so that these velocity deviations have exactly the same V̇ ′ and Γ as the full velocities.
~avg makes it easier to visualize a net outflow or circulation
Removal of the large constant V
~ distributions, along with their corresponding
around the perimeter. Figure 8 shows four V
ΔV~ distributions. The latter readily indicate zero or nonzero V̇ ′ or Γ, or equivalently, zero
or nonzero divergence or curl.
Δ
V =0
Δ
V =0
Δ
V =0
Δ
V =0
Δ
V =0
Δ
V =0
Δ
V =0
Δ
V =0
Figure 8: Example velocity distributions around a CV or circuit.
6
Fluids – Lecture 4 Notes
1. Dimensional Analysis – Buckingham Pi Theorem
2. Dynamic Similarity – Mach and Reynolds Numbers
Reading: Anderson 1.7
Dimensional Analysis
Physical parameters
A large number of physical parameters determine aerodynamic forces and moments. Specifically, the following parameters are involved in the production of lift.
Parameter
Lift per span
Angle of attack
Freestream velocity
Freestream density
Freestream viscosity
Freestream speed of sound
Size of body (e.g. chord)
Symbol
L′
α
V∞
ρ∞
µ∞
a∞
c
Units
mt−2
—
lt−1
ml−3
ml−1 t−1
lt−1
l
For an airfoil of a given shape, the lift per span in general will be a function of the remaining
parameters in the above list.
L′ = f (α, ρ∞ , V∞ , c, µ∞ , a∞ )
(1)
In this particular example, the functional statement has N = 7 parameters, expressed in a
total of K = 3 units (mass m, length l, and time t).
Dimensionless Forms
The Buckingham Pi Theorem states that this functional statement can be rescaled into an
equivalent dimensionless statement
Π1 = f¯( Π2 , Π3 . . . ΠN −K )
having only N −K = 4 dimensionless parameters. These are called Pi products, since they
are suitable products of the dimensional parameters. In the particular case of statement (1),
suitable Pi products are:
L′
= cℓ
1
ρ V2c
2 ∞ ∞
= α
= α
Π1 =
lift coefficient
Π2
angle of attack
ρ∞ V∞ c
µ∞
V∞
=
a∞
Π3 =
= Re
Π4
= M∞ Mach number
Reynolds number
The dimensionless form of statement (1) then becomes
¯ Re, M∞ )
cℓ = f(α,
1
(2)
We see that the original 6 dimensional parameters which influence L′ has been reduced to
only 3 dimensionless parameters which influence cℓ .
Benefits of non-dimensionalization
The reduction of parameter count is potentially a huge simplification. Consider an exaustive
lift-measurement experiment where the effect of all parameters is to be determined. Let’s
assume that in this experiment we need to give each parameter 5 distinct values in order
to adequately ascertain its effect on the lift. If we work with the 6 dimensional parameters
in statement (1), then the number of possible parameter combinations and experimental
runs required is 56 = 15625 (!). But if we work with the 3 dimensionless parameters in
statement (2), the number of parameter combinations and experimental runs is only 53 = 125,
which is more than a hundredfold reduction in effort. Nondimensionalization is clearly a
powerful technique for minimizing experimental effort.
The benefits of non-dimensionalization also extend to theoretical work. Deducing a statement
such as (2) at the outset can be useful to guide subsequent detailed analysis. Theoretical
results are also usually more concise and clear when presented in dimensionless form.
Derivation of dimensionless forms
Anderson 1.7 has details on how the Pi product combinations can be derived for any complex
situation using linear algebra. In many cases, however, the products can be obtained by
physical insight, or perhaps by inspection. Several rules can be applied here:
• Any parameter which is already dimensionless, such as α, is automatically one of the
Pi products.
• If two parameters have the same units, such as V∞ and a∞ , then their ratio (M∞ in
this case) will be one of the Pi products.
• A power or simple multiple of a Pi product is an acceptable alternative Pi product. For
example, (V∞ /a∞ )2 is an acceptable alternative to V∞ /a∞ , and ρ∞ V∞2 is an acceptable
alternative to 12 ρ∞ V∞2 . Which particular forms are used is a matter of convention.
• Combinations of Pi products can replace the originals. For example, the 3rd and 4th
products in the example could have been defined as
ρ∞ a∞ c
= Re/M∞
µ∞
V∞
=
= M∞
a∞
Π3 =
Π4
which is workable alternative, but perhaps less practical, and certainly less traditional.
Dynamic Similarity
It is quite possible for two differently-sized physical situations, with different dimensional
parameters, to nevertheless reduce to the same dimensionless description. The only requirement is that the corresponding Pi products have the same numerical values.
2
Airfoil flow example
Consider two airfoils which have the same shape and angle of attack, but have different sizes
and are operating in two different fluids. Let’s omit the ()∞ subscript for clarity.
Airfoil 1 (sea level)
α1 = 5◦
V1 = 210m/s
ρ1 = 1.2kg/m3
µ1 = 1.8 × 10−5 kg/m-s
a1 = 300m/s
c1 = 1.0m
Airfoil 2 (cryogenic tunnel)
α2 = 5◦
V2 = 140m/s
ρ2 = 3.0kg/m3
µ2 = 1.5 × 10−5 kg/m-s
a2 = 200m/s
c2 = 0.5m
Airfoil 2 − Cryogenic tunnel
Airfoil 1 − Sea level air
The Pi products evaluate to the following values.
Airfoil 1
α1 = 5◦
Re1 = 1.4 × 107
M1 = 0.7
Airfoil 2
α2 = 5◦
Re2 = 1.4 × 107
M2 = 0.7
Since these are also the arguments to the f¯ function, we conclude that the cℓ values will be
the same as well.
f¯(α1 , Re1 , M1 ) = f¯(α2 , Re2 , M2 )
cℓ 1 = cℓ 2
When the nondimensionalized parameters are equal like this, the two situations are said to
have dynamic similarity. One can then conclude that any other dimensionless quantity must
also match between the two situations. This is the basis of wind tunnel testing, where the
flow about a model object duplicates and can be used to predict the flow about the fullsize object. The prediction is correct only if the model and full-size objects have dynamic
similarity.
3
Lecture F4 Mud: Dimensional Analysis
(37 respondents)
1. I don’t understand what � products are. How do you get them. (3 students)
A � product is a dimensionless combination of parameters which influence a physical
situation (e.g. flow over an airfoil). They can be obtained by inspection, by trial &
error, by insight, or by linear algebra as a last resort (see Anderson 1.7). The choices
are not unique. Instead of V� /a� , we could have chosen a� /V� . But V /a is more
traditional, so that’s what we use.
2. Can I get additional � products just by combining the ones I already have?
(1 student)
No. You can get alternative � products that way, but not additional ones. Here are
two possible sets you can use:
Set 1
�1
�2
�3
Set 2
= �
= �� V� c/µ�
= V� /a�
�1
�2
�3
= �2
= �� a� c/µ�
= �a� /V�
Set 2 is usable, but Set 1 is traditional and easier to use (that’s why it’s traditional!).
3. What is the function ḡ(c� , �, M� , Re) = 0? ( student)
This is a conceptual way to say that the four parameters interrelate in some way
for a physical situation. The amazing thing is that with Dimensional Analysis (and
Buckingham Pi Theorem) we are able to identify these parameters without knowing
what the dependencies actually are quantitatively.
4. It seems we’re extrapolating a lot of info from small observations? (3 stu­
dents)
Yep. Dimensional analysis is amazingly powerful.
5. How does Buckinghams’ Theorem relate to Dimensional Analysis? (1 stu­
dent)
B’s Theorem is one step of Dimensional Analysis. Other steps include identifying
parameters, deriving other dimensionless quantities (Cp , etc.)
6. When two airfoils have the same � products, what do you mean that they
are the “same”? (1 student)
They may look different, but they are the same only when expressed, examined, or
plotted in the dimensionless variables. For example, L1 =
� L2 , but c�1 = c�2 .
7. What is this stuff used for? (3 students)
Lots of uses. One use is applying wind tunnel data to full size. To get full-size lift L1
from the wind-tunnel lift L2 , we would compute
CL2 = L2 / 0.5�2 V22 S2
C L1 = C L2
L1 = 0.5�1 V12 S1
8. How does the air viscosity vary with temperature? (1 student)
It’s a complicated relationship. Look up “Sutherland’s Law” for gasses. We can fre­
quently approximate this as µ � T , so gas viscosity always increases with increasing
temperature. Liquid viscosity usually decreases with increasing temperature.
9. What does Reynolds Number mean? (1 student)
It’s a measure of how “viscous” or “sticky” the flow is, or how big inertial forces are
relative to viscous or friction forces. This comparison depends not only on the viscosity
of the fluid, but also on density, velocity, and body size. The Reynolds Number lumps
all these effects into one significant parameter.
10. Can you bring more demos to class? (1 student)
I will if it’s feasible. Unfortunately, in-class wind tunnel demos are not very practical.
The small tunnel we had in class is too small for most demos of interest.
11. No mud (18 students)
Fall 2003
Unified Engineering
Fluids Problem F4
F4.
A rectangular wing of chord c and span b is operating in low-speed flow. Its drag
depends on the following parameters:
D = f (�, �� , V� , µ� , b, c)
Determine the dimensionless parameters (Pi products) which determine the drag coefficient
CD .
Fluids – Lecture 6 Notes
1. Control Volumes
2. Mass Conservation
3. Control Volume Applications
Reading: Anderson 2.3, 2.4
Control Volumes
In developing the equations of aerodynamics we will invoke the firmly established and timetested physical laws:
Conservation of Mass
Conservation of Momentum
Conservation of Energy
Because we are not dealing with isolated point masses, but rather a continuous deformable
medium, we will require new conceptual and mathematical techniques to apply these laws
correctly.
One concept is the control volume, which can be either finite or infinitesimal. Two types of
control volumes can be employed:
1) Volume is fixed in space (Eulerian type). Fluid can freely pass through the volume’s
boundary.
2) Volume is attached to the fluid (Lagrangian type). Volume is freely carried along with the
fluid, and no fluid passes through its boundary. This is essentially the same as the free-body
concept employed in solid mechanics.
V
V
Eulerian Control Volume
fixed in space
Lagrangian Control Volume
moving with fluid
Both approaches are valid. Here we will focus on the fixed control volume.
Mass Conservation
Mass flow
Consider a small patch of the surface of the fixed, permeable control volume. The patch has
1
area A, and normal unit vector n̂. The plane of fluid particles which are on the surface at
time t will move off the surface at time t + Δt, sweeping out a volume given by
ΔV = Vn A Δt
~ ·n
where Vn = V
ˆ is the component of the veleocity vector normal to the area.
V
n^
V
Vn
swept volume
n^
Vn Δt
V Δt
A
V Δt
A
The mass of fluid in this swept volume, which evidently passed through the area during the
Δt interval, is
Δm = ρ ΔV = ρ Vn A Δt
The mass flow is defined as the time rate of this mass passing though the area.
mass flow = m
˙ = lim
Δt→0
Δm
= ρ Vn A
Δt
The mass flux is defined simply as mass flow per area.
mass flux =
ṁ
= ρVn
A
Mass Conservation Application
The conservation of mass principle can now be applied to the finite fixed control volume,
but now it must allow for the possibility of mass flow across the volume boundary.
d
(Mass in volume) = Mass flow into volume
dt
Using the previous relations we have
��
d ���
~ ·n
ρ dV = − ρ V
ˆ dA
dt
(1)
where the negative sign is necessary because n̂ is defined to point outwards, so an inflow is
~ ·n
ˆ is positive.
where −V
Using Gauss’s Theorem and bringing the time derivative inside the integral we have the
result
�
��� �
�
�
∂ρ
~ dV = 0
+ ∇ · ρV
∂t
2
This relation must hold for any control volume whatsoever. If we place an infinitesimal
control volume at every point in the flow and apply the above equation, we can see that the
whole quantity in the brackets must be zero at every point. This results in the Continuity
Equation
�
�
∂ρ
+ ∇ · ρ V~ = 0
(2)
∂t
which is the embodiment of the Mass Conservation principle for fluid flow. The steady flow
version is:
�
�
~ = 0
∇ · ρV
(steady flow)
For low-speed flow, steady or unsteady, the density ρ is essentially constant, which gives the
very great simplification that the velocity vector field has zero divergence.
∇ · V~ = 0
(low-speed flow)
Control Volume Applications
Steady Surface-Integral Form
All of the above forms of the continuity equation are used in practice. The surface-integral
form (1) with the steady assumption,
��
~ ·n
ρV
ˆ dA = 0
(3)
is particularly useful in many engineering applications.
Use of equation (3) first requires construction of the fixed control volume. The boundaries are
typically placed where the normal velocity and the density are known, so the surface integral
can be evaluated piecewise. Placing the boundary along a solid surface is particularly useful,
since here
~ ·n
V
ˆ = 0
(at solid surface)
and so that part of the boundary doesn’t contribute to the integral. In the figure, we note
that
~ ·n
V
ˆ dA = V cos θ dA = Vn dA = V dAs
~.
where dAs is the area of the surface element projected onto a plane normal to V
V
n^
Vn
V
θ
or
A
A
As
The form of the integrand using dAs is most useful on a portion of the boundary which has
~ (in magnitude and direction). The contribution of this boundary
a constant value of ρV
portion to the overall integral (3) is then
��
~ ·n
ρV
ˆ dA = ρV
��
dAs = ρV As
3
~)
(portion with constant ρV
V
V
boundary portion
with constant flow
As
~
Note that the boundary portion does not need to be flat for As to be well defined. Only ρV
must be a constant.
Channel Flow Application
Placing the control volume inside a pipe or channel of slowly-varying area, we now evaluate
equation (3).
��
~ ·n
ρV
ˆ dA = −ρ1 V1 A1 + ρ2 V2 A2 = 0
ρ1 V1 A1 = ρ2 V2 A2
~ ·n
The negative sign for station 1 is due to V
ˆ = −V1 at that location. The control volume
faces adjacent to walls do not contribute to the integral, since their normal vectors are
ˆ = 0.
perpendicular to the local flow and therefore have V~ · n
n^
V. n^ = 0
A1
n^
A2
V2
n^
V1
Placing plane 2, say, at any other location, gives the general result that
ρV A = constant
inside a channel. The product ρV A is also recognized as the constant channel mass flow.
4
Lecture F6 Mud: Control Volumes, Mass Conservation
(33 respondents)
1. The 18.02 math stuff in the lecture was rather complex. (3 students)
Yes, but it’s tough to avoid when doing Fluids. I’ll try to review the math concepts as
they come up.
2. How do you calculate the divergence? (1 student)
∂ =
Using its definition. In the general continuity equation the vector in question is �V
�u ı̂ + �v �ˆ + �w k̂, so we calculate:
∂ ) = �(�u) + �(�v) + �(�w)
� · (�V
�x
�y
�z
If � is a constant (low-speed flow), it can be taken outside the derivatives, but not in
general.
3. How can one increase volume and decrease density in the steady flow equa­
tion? (1 student)
2
In a steady low speed flow (M�
� 1), the only way to decrease the gas density is to add
2
heat to it, as in the heater-in-pipe PRS question. In high-speed flow (M�
not small),
density will increase locally wherever the pressure increases, such as just in front of
an airfoil leading edge. In liquids the density is pretty much constant no matter what
happens, short of boiling.
4. Didn’t understand how you got
I’ll go over this again on Monday.
−�1 V1 A1 + �2 V2 A2 = 0 (1 student)
5. There’s a typo in the notes (“mass flux” should be “mass flow” in one
equation). (1 student)
Thanks. It’s fixed. The Fluids notes this term will be hot out of the oven, so typo
reports are welcome.
6. No mud (26 students)
Fall 2003
Unified Engineering
Fluids Problem F6
F6.
Determine the streamline shapes of the follwing 2-D velocity field.
u=
−y
x2 + y 2
v=
x
x2 + y 2
Assuming the density is constant, does this flow obey the mass conservation law?
Fluids – Lecture 7 Notes
1. Momentum Flow
2. Momentum Conservation
Reading: Anderson 2.5
Momentum Flow
Before we can apply the principle of momentum conservation to a fixed permeable control
volume, we must first examine the effect of flow through its surface. When material flows
through the surface, it carries not only mass, but momentum as well. The momentum flow
can be described as
−→
−→
momentum flow = (mass flow) × (momentum /mass)
where the mass flow was defined earlier, and the momentum/mass is simply the velocity
~ . Therefore
vector V
−→
~ = ρ V~ · n
momentum flow = m
˙ V
ˆ A V~ = ρ Vn A V~
�
�
~ ·n
where Vn = V
ˆ as before. Note that while mass flow is a scalar, the momentum flow is a
~ . The momentum flux vector is defined simply
vector, and points in the same direction as V
as the momentum flow per area.
−→
~
momentum flux = ρ Vn V
V
n^
.
mV
.
m
A
ρ
Momentum Conservation
Newton’s second law states that during a short time interval dt, the impulse of a force F~
applied to some affected mass, will produce a momentum change dP~a in that affected mass.
When applied to a fixed control volume, this principle becomes
dP~a
= F~
dt
(1)
dP~
˙
˙
+ P~ out − P~ in = F~
dt
(2)
F
V
.
P in
P(t)
.
P out
1
In the second equation (2), P~ is defined as the instantaneous momentum inside the control
volume.
���
~ dV
P~ (t) ≡
ρV
˙
The P~ out is added because mass leaving the control volume carries away momentum provided
˙
by F~ , which P~ alone doesn’t account for. The P~ in is subtracted because mass flowing into
the control volume is incorrectly accounted in P~ , and hence must be discounted. Both terms
are evaluated by a surface integral of the momentum flux over the entire boundary.
˙
˙
P~ out − P~ in =
��
�
�
~ ·n
~ dA
ρ V
ˆ V
~ ·n
The sign of V
ˆ automatically accounts for both inlow and outflow.
Applied forces
The force F~ consists of two types.
Body forces. These act on fluid inside the volume. The most common example is the gravity
force, along the gravitational acceleration vector ~g .
���
F~gravity =
ρ ~g dV
Surface forces. These act on the surface of the volume, and can be separated into pressure
and viscous forces.
��
F~pressure =
−p n
ˆ dA
The viscous force is complicated to write out, and for now will simply be called F~viscous .
ρg
−pn
Fviscous
−pn
Fviscous
Integral Momentum Equation
Substituting all the momentum, momentum flow, and force definitions into Newton’s second
law (2) gives the Integral Momentum Equation.
d
dt
���
~ dV +
ρV
��
~ ·n
~ dA =
ρ V
ˆ V
�
�
2
��
−p n
ˆ dA +
���
ρ ~g dV + F~viscous
(3)
Along with the Integral Mass Equation, this equation can be applied to solve many problems
involving finite control volumes.
Differential Momentum Equation
The pressure surface integral in equation (3) can be converted to a volume integral using the
Gradient Theorem.
��
���
p n̂ dA =
∇p dV
The momentum-flow surface integral is also similarly converted using Gauss’s Theorem. This
integral is a vector quantity, and for clarity the conversion is best done on each component
ˆ we have
separately. After substituting V~ = u ı̂ + v ̂ + w k,
��
~ ·n
ρ V
ˆ
�
u ı̂ + v ̂ + w kˆ dA = ı̂
� �
�
���
~ u dV
∇ · ρV
���
∇ · ρ V~ w dV
+ ̂
+ kˆ
���
�
�
�
�
~ v dV
∇ · ρV
�
�
The x-component of the integral momentum equation (3) can now be written strictly in
terms of volume integrals.
��� �
�
�
�
∂(ρu)
~ + ∂p − ρgx − (Fx )viscous dV = 0
+ ∇ · ρuV
∂t
∂x
(4)
This relation must hold for any control volume whatsoever. If we place an infinitesimal
control volume at every point in the flow and apply equation (4), we can see that the whole
quantity in the brackets must be zero at every point. This results in the x-Momentum
Equation
�
�
∂(ρu)
~ = − ∂p + ρgx + (Fx )viscous
+ ∇ · ρuV
(5)
∂t
∂x
and the y- and z-Momentum Equations follow by the same process.
�
�
∂(ρv)
~ = − ∂p + ρgy + (Fy )viscous
+ ∇ · ρv V
∂t
∂y
(6)
�
�
∂(ρw)
~ = − ∂p + ρgz + (Fz )viscous
+ ∇ · ρw V
(7)
∂t
∂z
These three equations are the embodiment of the Newton’s second law of motion, applied
at every point in the flowfield. The steady flow version has the ∂/∂t terms omitted.
3
Lecture F7 Mud: Momentum Flow, Momentum Conservation
(34 respondents)
1. Can you give some examples? (3 students)
Besides the PRS question, there wasn’t time for other examples, since I wanted to
cover the stuff in one organized swoop. The next lecture F9 will revisit the material,
and deal mostly with applications.
� ’s should have? (1 students)
2. How do you know what signs the V
� = V ı̂. If it points opposite
If the velocity points in the direction of the x-axis, then V
�
to the x-axis, then V = −V ı̂.
3. Unclear on the physical principles/concepts behind the equations (2 students)
� )/dt. Because the affected
The main underlying concept is Newton’s Law F� = d(ma V
mass that we are applying this equation to is moving through our control volume, we
need to bring in the related concept of momentum flow to do the accounting properly.
The rest is mainly math manipulation.
�i get subtracted? Seems it also gets kicked by F� dt. (1 student)
4. Why did dmi V
�
Actually, dmi Vi does not get kicked by F� dt, since it’s initially outside the volume,
and F� dt by its definition acts only on what was inside the volume. And since the
� ) integral incorrectly includes this dmi V
�i , we have to subtract it from the total
d(mV
momentum tally.
5. Why did you set the whole thing in the brackets to zero? What do the
individual terms mean? (2 student)
You are referring to:
�
�
∂(ρu)
� + ∂p − ρgx − (Fx )viscous = 0
+ ∇ · ρuV
∂t
∂x
It’s equal to zero because I moved all the terms to the lefthand side. This in effect says
d(ma u)
− Fx = 0
dt
� )/dt for the mass inside control
The first two terms represent the x-component of d(ma V
volume before the impulse. The last three terms represent x-component of F� .
6. When can we set some of the terms in the final equation to zero?
(1
student)
Depends on the situation. If flow is steady, then d()/dt = 0. If gravity is not important
(most aero applications), then drop the g term. If viscous forces are not significant,
then assume F�visc = 0. The two most important terms which are almost always retained
� · n)
� and the pressure force −pˆ
are the momentum flow ρ(V
n.
ˆ V
7. What are the important formulas we need to take away from this? (1 student)
One key thing is the integral equation (2) in the notes. This will be exersized in F9
and later.
Another key thing is the differential equations (4)-(6). These are the point-equivalents
of (2), and will be the starting point of the next block of lectures.
8. Math is very difficult to follow. Various questions on ∇, ∇ · (), etc. Can
we review some of 18.02? Do you have links to relevant 18.02 material? (6
students)
I’ve been stating relevant vector theorems and identities before they are used in class.
I’m trying to decide what else I can do in a time-effective manner.
18.02 links are at http://student.mit.edu/catalog/m18a.html
Also try googling “div grad curl
”
9. No mud (12 students)
Fluids – Lecture 8 Notes
1. Streamlines
2. Pathlines
3. Streaklines
Reading: Anderson 2.11
Three types of fluid element trajectories are defined: Streamlines, Pathlines, and Streaklines.
They are all equivalent for steady flows, but differ conceptually for unsteady flows.
Streamlines
Streamline equations
A streamline is defined as a line which is everywhere parallel to the local velocity vector
ˆ Define
~ (x, y, z, t) = u ı̂ + v ̂ + w k.
V
d~s = dx ı̂ + dy ̂ + dz k̂
~ , we
as an infinitesimal arc-length vector along the streamline. Since this is parallel to V
must have
~ = 0
d~s × V
(w dy − v dz) ı̂ + (u dz − w dx) ˆ + (v dx − u dy) ˆk = 0
Separately setting each component to zero gives three differential equations which define the
streamline. The three velocity components u, v, w, must be given as functions of x, y, z
before these equations can be integrated. To set the constants of integration, it is sufficient
to specify some point xo , yo, zo through which the streamline passes,
y
y
ds
V
ds
V
v
dy
dx
xo yo z o
xo yo
x
z
u
x
2−D streamline
3−D streamline
In 2-D we have dz = 0 and w = 0, and only the k̂ component of the equation above is
non-trivial. It can be written as an Ordinary Differential Equation for the streamline shape
y(x).
dy
v
=
dx
u
Again, u(x, y) and v(x, y) must be given to allow integration, and xo , yo must be given to
set the integration constants. In a numerical integration, xo , yo would serve as the initial
values.
1
Streamtubes
Consider a set of xo , yo , zo points arranged in a closed loop. The streamlines passing
through all these points form the surface of a streamtube. Because there is no flow across the
surface, each cross-section of the streamtube carries the same mass flow. So the streamtube
is equivalent to a channel flow embedded in the rest of the flowfield.
y
y
xo yo z o
z
xo yo
x
x
2−D streamtube
3−D streamtube
In 2-D, a streamtube is defined by two streamlines passing through two specified xo , yo
points. The flow between these two streamlines carries the same mass flow/span at each
cross-section, and can be considered as a 2-D channel flow embedded in the rest of the
flowfield.
Pathlines
The pathline of a fluid element A is simply the path it takes through space as a function
of time. An example of a pathline is the trajectory taken by one puff of smoke which is
carried by the steady or unsteady wind. This path is fully described by the three position
functions xA (t), yA (t), zA (t), which can be computed by integrating the three velocity-field
components u(x, y, z, t), v(x, y, z, t), w(x, y, z, t) along the path. The integration is started
at time to , from the element’s initial position xo , yo , zo (e.g. the smoke release point), and
proceeds to some later time t.
xA (t) = xo +
yA (t) = yo +
zA (t) = zo +
� t
�
�
dτ
�
dτ
to
�
� t
�
�
dτ
to
� t
to
u xA (τ ), yA (τ ), zA (τ ), τ
v xA (τ ), yA (τ ), zA (τ ), τ
w xA (τ ), yA (τ ), zA (τ ), τ
The dummy variable of integration τ runs from to to t.
Streaklines
A streakline is associated with a particular point P in space which has the fluid moving past
it. All points which pass through this point are said to form the streakline of point P . An
example of a streakline is the continuous line of smoke emitted by a chimney at point P ,
which will have some curved shape if the wind has a time-varying direction.
2
Unlike a pathline, which involves the motion of only one fluid element A in time, a streakline
involves the motion of all the fluid elements along its length. Hence, the trajectory equations
for a pathline are applied to all the fluid elements defining the streakline.
The figure below illustrates streamlines, pathlines, and streaklines for the case of a smoke
being continuously emitted by a chimney at point P , in the presence of a shifting wind. One
particular smoke puff A is also identified. The figure corresponds to a snapshot when the
wind everywhere is along one particular direction.
s
eous
n
a
t
e
tan lin
ins tream
s
streakline from point P
(smoke line)
instantaneous
wind velocity
V
A
P
streakline at
successive times
pathline of fluid element A
(smoke puff)
In a steady flow, streamlines, pathlines, and streaklines all cooincide. In this example they
would all be marked by the smoke line.
3
Lecture F8 Mud: Momentum Flow, Momentum Conservation
(34 respondents)
1. Can you give some examples? (3 students)
Besides the PRS question, there wasn’t time for other examples, since I wanted to
cover the stuff in one organized swoop. The next lecture F9 will revisit the material,
and deal mostly with applications.
� ’s should have? (1 students)
2. How do you know what signs the V
� = V ı̂. If it points opposite
If the velocity points in the direction of the x-axis, then V
�
to the x-axis, then V = −V ı̂.
3. Unclear on the physical principles/concepts behind the equations (2 students)
� )/dt. Because the affected
The main underlying concept is Newton’s Law F� = d(ma V
mass that we are applying this equation to is moving through our control volume, we
need to bring in the related concept of momentum flow to do the accounting properly.
The rest is mainly math manipulation.
�i get subtracted? Seems it also gets kicked by F� dt. (1 student)
4. Why did dmi V
�
Actually, dmi Vi does not get kicked by F� dt, since it’s initially outside the volume,
and F� dt by its definition acts only on what was inside the volume. And since the
� ) integral incorrectly includes this dmi V
�i , we have to subtract it from the total
d(mV
momentum tally.
5. Why did you set the whole thing in the brackets to zero? What do the
individual terms mean? (2 student)
You are referring to:
�
�(�u)
� + �p − �gx − (Fx )viscous = 0
+ � · �uV
�t
�x
It’s equal to zero because I moved all the terms to the lefthand side. This in effect says
d(ma u)
− Fx = 0
dt
� )/dt for the mass inside control
The first two terms represent the x-component of d(ma V
volume before the impulse. The last three terms represent x-component of F� .
6. When can we set some of the terms in the final equation to zero?
(1
student)
Depends on the situation. If flow is steady, then d()/dt = 0. If gravity is not important
(most aero applications), then drop the g term. If viscous forces are not significant,
then assume F�visc = 0. The two most important terms which are almost always retained
� · n)
� and the pressure force −pˆ
are the momentum flow �(V
n.
ˆ V
7. What are the important formulas we need to take away from this? (1 student)
One key thing is the integral equation (2) in the notes. This will be exersized in F9
and later.
Another key thing is the differential equations (4)-(6). These are the point-equivalents
of (2), and will be the starting point of the next block of lectures.
8. Math is very difficult to follow. Various questions on �, � · (), etc. Can
we review some of 18.02? Do you have links to relevant 18.02 material? (6
students)
I’ve been stating relevant vector theorems and identities before they are used in class.
I’m trying to decide what else I can do in a time-effective manner.
Try googling “div grad curl”
9. No mud (12 students)
Fall 2003
Unified Engineering
Fluids Problem F8
F8. A cascade of vanes turns the airflow around each corner of the Wright Brothers Wind
Tunnel. Assume the flow parameters are:
constant air density:
upstream air velocity:
vane spacing:
vane span:
upstream flow angle:
downstream flow angle:
� = 1.2 kg/m3
V = 30 m/s
h = 0.4 m
b = 2.5 m
45�
−45�
Determine the force on each vane. Again, clearly draw a suitable control volume for your
analysis.
h
V
Fluids – Lecture 9 Notes
1. Momentum-Integral Simplifications
2. Applications
Reading: Anderson 2.6
Simplifications
For steady flow, the momentum integral equation reduces to the following.
��
�
�
~ ·n
~ dA =
ρ V
ˆ V
��
−p n
ˆ dA +
���
ρ ~g dV + F~viscous
(1)
Defining h as the height above ground, we note that ∇h is a unit vector which points up, so
that the gravity acceleration vector can be written as a gradient.
~g = −g ∇h
Using the Gradient Theorem, this then allows the gravity-force volume integral to be converted to a surface integral, provided we make the additional assumption that ρ is nearly
constant throughout the flow.
���
ρ ~g dV =
���
−ρ g ∇h dV =
��
−ρgh n̂ dA
We can now combine the pressure and gravity contributions into one surface integral.
��
−p n
ˆ dA +
���
��
→
ρ ~g dV
−(p + ρgh) n
ˆ dA
Defining a corrected pressure pc = p+ρgh, the Integral Momentum Equation finally becomes
��
�
�
~ ·n
~ dA =
ρ V
ˆ V
��
−pc n
ˆ dA + F~viscous
(2)
Aerodynamic analyses using (2) do not have to concern themselves with the effects of gravity,
since it does not appear explicitly in this equation. In particular, the velocity field V~ will
not be affected by gravity. Gravity enters the problem only in a secondary step, when the
true pressure field p is constructed from pc by adding the “tilting” bias −ρgh.
h
h
h
h
+
=
g
pc
aerodynamic
p
net
(gravity neglected)
1
ρg h
aerostatic
(gravity alone)
For clarity, we will from now on refer to pc simply as p. The understanding is that the
forces and moments computed using this p will not include the contributions of buoyancy.
However, buoyancy is relatively easy to compute, and can be added as a secondary step
if appropriate. It should be noted that buoyancy is rarely significant in aerodynamic flow
situations, one exception being lighter-than-air vehicles like blimps.
One situation where gravity is crucial is a flow with a free surface, like water waves past a
ship. Here, gravity enters not in equation (2), but in the free-surface boundary condition.
Free-surface flows are not relevant to most aeronautical problems.
Applications
Basic Procedures
Application of the integral momentum equation (2) uses the same basic techniques as for
the integral continuity equation. Both can use the same control volume, and both demand
that the integrals are evaluated for the entire surface of the control volume. There are three
significant differences, however:
1) Momentum is a vector. Each of the three x, y, and z components of equation (2) is
independent, and must be treated separately.
2) A volume boundary lying along a streamline now has a nonzero −pn̂ contribution.
3) A volume boundary lying along a solid surface now has a nonzero −pn̂ contribution, and
possibly also a viscous shear ~τ contribution if viscous effects are deemed significant.
Effect of body in flow
Equation (2) assumes that there is no internal force acting on the fluid other than gravity.
If there is a solid body in the flow, it is therefore necessary to construct the control volume
such that the body is on the volume’s exterior. This is easily done by placing the body in
an indentation cdef g of the volume surface, as shown in the figure.
p1 = p
ay
streamline far aw
.
dm
a
p2 = p
p=p
b
u2
u1
e
d
c
f
g
i
h
streamline far away
The surface integrals are now broken up into the individual pieces: the outermost part abhi,
the body surface def , and the “cut” surfaces cd and f g.
��
( ) dA =
��
abhi
( ) dA +
��
( ) dA +
def
��
cd
2
( ) dA +
��
fg
( ) dA
Surfaces cd and f g have the same flow variables, but opposite n̂ vectors, so that all their
contributions are equal and opposite, and hence cancel.
��
( ) dA +
��
cd
( ) dA = 0
fg
~ ′ on the body is the integrated pressure and shear force distriThe resultant force/span R
bution on the body surface. Because the body surface and volume surface have opposite
~ ′ is precisely equal and opposite to all the def surface integrals for the
normals n,
ˆ this R
control volume.
��
~′
( ) dA = −R
def
R’
p
Forces on
airfoil
τ
d
e
d
e
f
f
Forces on
control volume
τ
p
−R’
An intuitive explanation is that the aerodynamic force exerted by the fluid on the body is
exactly equal and opposite to the force which the body exerts on the fluid.
Collecting the remaining integrals produces the very general result
��
�
�
~ dA +
ρ V~ · n̂ V
abhi
��
~′
p n̂ dA = −R
(3)
abhi
which gives the resultant force on the body only in terms of the surface integrals on the
outer surface of the volume.
Drag relations
To obtain the drag/span, we define the x direction to be aligned with V~∞ , and take the
x-component of equation (3).
��
�
��
�
~ ·n
ρ V
ˆ u dA +
abhi
pn
ˆ · ı̂ dA = −D ′
(4)
abhi
The pressure on the outer surface is uniform and equal to p∞ , and hence the pressure integral is zero. The momentum-flux integral is zero on the top and bottom boundaries, since
3
these are defined to be along streamlines, and hence have zero momentum flux. Only the
momentum flux on the inflow and outflow planes remain.
��
��
ρ2 u22 dA2 −
bh
ρ1 u21 dA1 = −D ′
(5)
ia
Using the concept of a streamtube, we see that planes 1 and 2 have the same mass flows,
with
ρ1 u1 dA1 = ρ2 u2 dA2 = dṁ
as shown in the figure. Therefore both the plane 1 and plane 2 integrals can be lumped into
one single plane 2 integral.
� b
D′ =
ρ2 u2 (u1 − u2 ) dA2
(6)
h
Equation (6) gives the drag only in terms of the freestream speed u1 , and the downstream
wake “profiles” ρ2 and u2 . These downstream profiles can be measured in a wind tunnel,
and numerically integrated in equation (6) to obtain the drag.
A physical interpretation of (6) can be obtained by writing it as
′
D =
�
b
(u1 − u2 ) dṁ
h
This is simply the summation of the momentum flow “lost” by all the infinitesimal streamtubes in the flow. The lost momentum appears as drag on the body.
4
Lecture F9 Mud: Momentum Theorem Applications
(30 respondents)
�
1. What does −pn̂ dA physically represent? (1 student)
The net force that the pressure distribution applies to the control volume.
2. Why is momentum flow added to the pressure force to get the total force?
(3 students)
I wouldn’t phrase it that way. One way to write the governing relation is:
�
~′ =
−pˆ
n dA − R
�
~ ·n
ρV
ˆ V~ dA
�
~ ′ ) that are
Or in words: The pressure force ( −pˆ
n dA) plus internal body force (−R
applied
to the volume are balanced by a net momentum flow out of the volume
�
~
~ dA).
( ρV · n
ˆV
3. What is −pn̂ · ı̂ ? (1 student)
−pˆ
n is the local pressure force on the control volume boundary.
n · ı̂ is just the x-component of this force.
−pˆ
4. Explain the wing-above-ground example better? (3 students)
If you put the left, right, and top control volume boundaries far from the wing, only
the pressure along the bottom boundary contributes to the momentum integral (in
~ ′ = L′ ̂ as given. Along the bottom boundary, we also have
mud point 2). We set R
n̂ = −̂. The integral boils down to
�
−(p − p∞ ) ̂ dA = −L̂
bottom
The lefthand side is precisely the force that the overpressure imparts to the ground,
and it’s equal to −L̂. Intuitively, the wing’s downward push inside the volume must
be balanced by the ground’s upward push on the bottom boundary, since there is no
net momentum flow (see mud point 2 above). Since the ground is pushing up on the
air with a force of L′ , the air must be pushing down on the ground with a force of L′ .
5. Math too dense. Too many variables. Too much too fast. (5 students)
I realize all the integrals are intimidating. You should put most of your effort at
grasping the underlying physical concepts (e.g. mud point 2 above). The integrals
are just expressions which have to be evaluated to get numerical answers out of the
physical principle. Doing the homework problems is good practice for this. See me or
the grad TA’s if you need help.
6. No mud (13 students)
Unified Engineering
Fluids Problem F10
Fall 2003
F10.
a) Determine the streamline shapes of the following 2-D velocity field (closely related to HW
problem F6).
u = −y
v=x
b) Evaluate Du/Dt and Dv/Dt, and determine the pressure gradient �p. Assume the
density � is constant (low speed flow).
c) Using your result from b), determine the pressure field p(x, y) to within an additive
constant.
Fluids – Lecture 10 Notes
1. Substantial Derivative
2. Recast Governing Equations
Reading: Anderson 2.9, 2.10
Substantial Derivative
Sensed rates of change
The rate of change reported by a flow sensor clearly depends on the motion of the sensor.
For example, the pressure reported by a static-pressure sensor mounted on an airplane in
level flight shows zero rate of change. But a ground pressure sensor reports a nonzero rate
as the airplane rapidly flies by a few meters overhead. The figure illustrates the situation.
p
p1 (t)
wing location at
t = to
p2 (t)
p2 (t)
p1 (t)
t
to
Note that although the two sensors measure the same instantaneous static pressure at the
same point (at time t = to ), the measured time rates are different.
p1 (to ) = p2 (to )
dp2
dp1
6
(to )
(to ) =
dt
dt
but
Drifting sensor
We will now imagine a sensor drifting with a fluid element. In effect, the sensor follows the
element’s pathline coordinates xs (t), ys (t), zs (t), whose time rates of change are just the
local flow velocity components
dxs
= u(xs , ys , zs , t) ,
dt
dys
= v(xs , ys , zs , t) ,
dt
pressure
field
dzs
= w(xs , ys , zs , t)
dt
ps
pathline
V
ps (t)
Dp
Dt
dps
dt
t
sensor drifting with local velocity
Consider a flow field quantity to be observed by the drifting sensor, such as the static pressure
p(x, y, z, t). As the sensor moves through this field, the instantaneous pressure value reported
by the sensor is then simply
ps (t) = p (xs (t), ys (t), zs (t), t)
1
(1)
This ps (t) signal is similar to p2 (t) in the example above, but not quite the same, since the
p2 sensor moves in a straight line relative to the wing rather than following a pathline like
the ps sensor.
Substantial derivative definition
The time rate of change of ps (t) can be computed from (1) using the chain rule.
dps
∂p dxs
∂p dys
∂p dzs
∂p
=
+
+
+
dt
∂x dt
∂y dt
∂z dt
∂t
But since dxs /dt etc. are simply the local fluid velocity components, this rate can be expressed using the flowfield properties alone.
dps
∂p
∂p
∂p
∂p
Dp
=
+ u
+ v
+ w
≡
dt
∂t
∂x
∂y
∂z
Dt
The middle expression, conveniently denoted as Dp/Dt in shorthand, is called the substantial
derivative of p. Note that in order to compute Dp/Dt, we must know not only the p(x, y, z, t)
field, but also the velocity component fields u, v, w(x, y, z, t).
Although we used the pressure in this example, the substantial derivative can be computed
for any flowfield quantity (density, temperature, even velocity) which is a function of x, y, z, t.
D( )
∂( )
∂( )
∂( )
∂( )
∂( )
~ · ∇( )
≡
+ u
+ v
+ w
=
+ V
Dt
∂t
∂x
∂y
∂z
∂t
The rightmost compact D/Dt definition contains two terms. The first ∂/∂t term is called
the local derivative. The second V~ · ∇ term is called the convective derivative. In steady
flows, ∂/∂t = 0, and only the convective derivative contributes.
Recast Governing Equations
All the governing equations of fluid motion which were derived using control volume concepts
can be recast in terms of the substantial derivative. We will employ the following general
vector identity
∇ · (a~v ) = ~v · ∇a + a ∇·~v
which is valid for any scalar a and any vector ~v .
Continuity equation
Applying the above vector identity to the divergence form continuity equation gives
�
�
∂ρ
~
+ ∇ · ρV
= 0
∂t
∂ρ
~ · ∇ρ + ρ ∇· V
~ = 0
+ V
∂t
Dρ
~ = 0
+ ρ ∇· V
Dt
(2)
The final result above is called the convective form of the continuity equation. A physical
interpretation can be made if it’s written as follows.
2
1 Dρ
~
= ∇·V
ρ Dt
−fractional density rate = velocity divergence
fractional volume rate = velocity divergence
−
or . . .
For a fluid element of given mass, the volume must vary as 1/density, which gives the second
interpretation above. Both interpretations are illustrated in the left figure below, where the
~ > 0. In low
fluid element expands when it flows through a flowfield region where ∇ · V
speed flows and in liquid flows the density is essentially constant, so that Dρ/Dt = 0 and
~ = 0.
by implication ∇· V
Momentum equation
The divergence form of the x-momentum equation is
�
�
∂(ρu)
~ = − ∂p + ρgx + (Fx )viscous
+ ∇ · ρuV
∂t
∂x
Applying the vector identity again, and also cancelling some terms by use of the continuity
equation (2), produces the convective form of the momentum equation. The y- and zmomentum equations are also derived the same way.
Du
∂p
= −
+ ρgx + (Fx )viscous
Dt
∂x
Dv
∂p
= −
+ ρgy + (Fy )viscous
ρ
Dt
∂y
Dw
∂p
ρ
= −
+ ρgz + (Fz )viscous
Dt
∂z
ρ
(3)
(4)
(5)
The Du/Dt etc. substantial derivatives are recognized as the acceleration components experienced by a fluid element. This leads to a simple physical interpretation or these equations
as Newton’s law applied to a fluid element of unit volume.
mass/volume × acceleration = total force/volume
The element’s mass/volume is simply the density ρ, and the total force/volume consists of
the buoyancy-like pressure gradient force, the gravity force, and the viscous force.
Δρ = D ρ Δ t
Dt
ΔV = DV Δ t
Dt
t + Δt
V
t + Δt
t
t
p + ρ g +Fviscous
.V
Δ−
Δ
expanding fluid element
accelerating fluid element
3
Lecture F10 Mud: Momentum Theorem Applications
(37 respondents)
1. What’s the difference between D()/Dt and d()/dt ? (1 student)
What we’re after is the time rate of change of some flowfield quantity like p(x, y, z, t),
as seen by a sensor moving as a fluid element. Just writing dp/dt is very sloppy math,
since p depends on four variables x, y, z, t, not just on t. So we must work with the
four partial derivatives of p. The correct expression for the time rate of change seen
by the sensor is
∂p
∂p
∂p
∂p
+ u
+v
+w
∂t
∂x
∂y
∂z
but this is long and tedious to write. So we use the shorthand Dp/Dt for this whole
expression.
2. Not clear how Dρ/Dt differs from ∂ρ/∂t. (2 students)
The partial derivative ∂ρ/∂t merely gives the rate of change of density seen by fixed
sensor sitting at one particular x, y, z location. But we’re after the rate seen by a sensor
which moves with the flow velocity, so the sensor’s x, y, z are continually changing.
Using the chain rule then brings in the three spatial partial derivatives ∂ρ/∂x, y, z
whose contributions are added to the ∂ρ/∂t term. See point 1 above.
3. Why does time dependence depend on where you’re at? (1 student)
Look at the first figure in the F10 notes and try to see why the p1 (t) signal has a different
shape than the p2 (t) signal. In particular, we’re after the t derivative of a signal, which
are different for the two sensors, even when the sensors are instantaneously at the same
point.
4. What is the substantial derivative used for? (3 students)
A physicist might use it for formulating equations which describe some physical process
happening to a fluid element (combustion, precipitate formation, heat conduction, . . . ).
~ (x, y, z, t)
An experimentalist might use it to determine fluid accelerations from the V
field measured in the lab frame. A computational aerodynamicist might use it to
formulate the equations of motion to allow numerical solution, etc, etc, etc.
5. What are some example applications? (2 students)
In the current problem set you will use D/Dt to deduce the pressure field from a given
velocity field. We will come back to it occasionally later in the course.
6. What is F~viscous ? How do you calculate it? (2 students)
F~viscous is the force applied to the control volume boundary by viscous stresses, the
main one being the viscous shear stress τ . The full expression for F~viscous using the
velocities and viscosity is long and tedious, and not really important at this point. We
will look at it in detail later in the course when dealing with viscous flows.
7. How do you know that −∂p/∂x + ρgx + (Fx )viscous is force/volume? (1 student)
First of all, the units of each term is N/m3 , which is one clue. Second, the pressure
gradient also appears in the primary hydrostatic buoyancy equation:
(Fx )buoyancy = −(∂p/∂x) × volume
which clearly shows that −∂p/∂x is equal to the applied force/volume.
8. Clarify the interpretations of Control-Volume vs. Substantial Derivative
forms of continuity. (3 students)
Look at the last figure in the F10 notes to help visualization. See also mud point 2
above. The control-volume (CV) form is
�
�
∂ρ
~
= −∇ · ρV
∂t
or . . . “The rate of change of mass inside a fixed CV is equal to the mass flow out of
the CV, all per unit volume”
The substantial derivative form is
1 Dρ
~
= −∇ · V
ρ Dt
or . . . “Fractional rate of change of density of a moving fluid element is equal and
opposite to the expansion rate of the fluid element”
9. When asked to find the derivative (.e.g acceleration), will it be specified
what the sensor motion should be? (1 student)
The substantial derivative (sensor moves with local V~ ) is the only physically-meaningful
way to get the acceleration of the fluid, if you want to apply F = ma for instance. In
practice, physical considerations dictate how the derivative should be defined.
10. Can the substantial derivative be used to get 2nd derivatives? (1 student)
Yes, but it’s not necessary in practice, since second time derivatives do not appear
in physically-derived equations of fluid motion. Second derivatives like ∂ 2 ()/∂t2 might
appear, but only after some elaborate mathematical manipulations of the equations. In
8.01, for example, F = ma could have been written as dF/dt = m da/dt = m d2 u/dt2 ,
but the d2 u/dt2 here is completely artificial.
11. Many sections in Chapter 3 of Anderson apply to Pset questions. Can you
add these to the reading list? (1 student)
Chapter 2 is for “tool building”, while Chapters 3 onward are more concerned with
applications of these tools. If you find it helpful to read ahead, then by all means do
so. It’s not practical for me to assign reading not only for the basic material, but also
all later applications which appear in the book. The reading list for each lecture would
be huge, with lots of material extraneous to what we’re trying to focus on.
12. No mud (15 students)
Fall 2003
Unified Engineering
Fluids Problems F10–F11
F11.
a) The flow inside a boundary layer near a wall has a simple linear velocity distribution
u = Cy, v = 0. Determine the vorticity and strain rate of the fluid.
b) A fluid element initially has a square shape. Determine and sketch its shape a short time
later. Again using a sketch, describe how its shape relates to rotation and shearing element
motions.
y
u = Cy
x
Fluids – Lecture 11 Notes
1. Vorticity and Strain Rate
2. Circulation
Reading: Anderson 2.12, 2.13
Vorticity and Strain Rate
Fluid element behavior
When previously examining fluid motion, we considered only the changing position and
velocity of a fluid element. Now we will take a closer look, and examine the element’s
changing shape and orientation
.
Consider a moving fluid element which is initially rectangular, as shown in the figure. If
the velocity varies significantly across the extent of the element, its corners will not move in
unison, and the element will rotate and become distorted.
y
V(y+dy)
V(y)
x
element at time
z
t
element at time
t + Δt
In general, the edges of the element can undergo some combination of tilting and stretching.
For now we will consider only the tilting motions, because this has by far the greatest
implications for aerodynamics.
The figure below on the right shows two particular types of element-side tilting motions. If
adjacent sides tilt equally and in the same direction, we have pure rotation. If the adjacent
sides tilt equally and in opposite directions, we have pure shearing motion.
Both of these motions have strong implications. The absense of rotation will lead to a great
simplification in the equations of fluid motion. Shearing together with fluid viscosity produce
shear stresses, which are responsible for phenomena like drag and flow separation.
tilting
stretching
Rotation
(vorticity)
General edge movement
Shearing motion
(strain rate)
Tilting edge movements
1
Side tilting analysis
Consider the 2-D element in the xy plane, at time t, and again at time t + Δt.
element at time
y
6u dy Δ t
6y
t
u Δt
Δθ2
6v dx Δ t
6x
v Δt
dy
v
A
t + Δt
−Δθ1
u + 6u dy
6y
B
element at time
v + 6v dx
6x
u
dx
C
x
Points A and B have an x-velocity which differs by ∂u/∂y dy. Over the time interval Δt
they will then have a difference in x-displacements equal to
∂u
dy Δt
∂y
ΔxB − ΔxA =
and the associated angle change of side AB is
−Δθ1 =
ΔxB − ΔxA
∂u
=
Δt
dy
∂y
assuming small angles. A positive angle is defined counterclockwise. We now define a time
rate of change of this angle as follows.
dθ1
∂u
Δθ1
= lim
= −
Δt→0 Δt
dt
∂y
Similar analysis of the angle rate of side AC gives
∂v
dθ2
=
dt
∂x
Vorticity
The angular velocity of the element, about the z axis in this case, is defined as the average
angular velocity of sides AB and AC.
�
1 dθ1 dθ2
ωz =
+
dt
2 dt
�
�
1 ∂v ∂u
=
−
2 ∂x ∂y
�
The same analysis in the xz and yz planes will give a 3-D element’s angular velocities ωy
and ωx .
�
�
�
�
1 ∂w ∂v
1 ∂u ∂w
,
ωx =
−
−
ωy =
2 ∂z
∂x
2 ∂y
∂z
2
These three angular velocities are the components of the angular velocity vector
.
~ω = ωx ı̂ + ωy ̂ + ωz kˆ
However, since 2~ω appears most frequently, it is convenient to define the vorticity vector ξ~
as simply twice ~ω .
ξ~ = 2~ω =
�
�
∂w ∂v
ı̂ +
−
∂y
∂z
�
�
∂u ∂w
̂ +
−
∂z
∂x
�
�
∂v ∂u
k̂
−
∂x ∂y
The components of the vorticity vector are recognized as the definitions of the curl of V~ ,
hence we have
~
ξ~ = ∇ × V
Two types of flow can now be defined:
1) Rotational flow. Here ∇ × V~ 6= 0 at every point in the flow. The fluid elements move and
deform, and also rotate.
~ = 0 at every point in the flow. The fluid elements move
2) Irrotational flow. Here ∇ × V
and deform, but do not rotate.
The figure contrasts the two types of flow.
Irrotational flows
Rotational flows
Strain rate
Using the same element-side angles Δθ1 , Δθ2 , we can define the strain of the fluid element.
strain = Δθ2 − Δθ1
This is the same as the strain used in solid mechanics. Here, we are more interested in the
strain rate, which is then simply
d(strain)
dΔθ1
∂v
∂u
dΔθ2
≡ εxy =
−
=
+
dt
dt
dt
∂x
∂y
Similarly, the strain rates in the yz and zx planes are
εyz =
∂w
∂v
+
∂y
∂z
εzx =
,
Circulation
3
∂u
∂w
+
∂z
∂x
Consider a closed curve C in a velocity field as shown in the figure on the left. The instantaneous circulation around curve C is defined by
Γ ≡ −
�
C
~ · d~s
V
In 2-D, a line integral is counterclockwise by convention. But aerodynamicists like to define
circulation as positive clockwise, hence the minus sign.
dA
ds
ξ
V
C
A
Circulation is closely linked to the vorticity in the flowfield. By Stokes’s Theorem,
Γ ≡ −
�
C
~ · d~s = −
V
�� �
A
~ ·n
ˆ dA = −
∇×V
�
��
A
ˆ dA
ξ~ · n
where the integral is over the area A in the interior of C, shown in the above figure on the
right, and n
ˆ is the unit vector normal to this area. In the 2-D xy plane, we have ξ~ = ξ kˆ and
ˆ
n
ˆ = k, in which case we have a simpler scalar form of the area integral.
Γ = −
��
A
ξ dA
(in 2-D)
From this integral one can interpret the vorticity as –circulation per area, or
ξ = −
dΓ
dA
Irrotational flows, for which ξ = 0 by definition, therefore have Γ = 0 about any contour
inside the flowfield. Aerodynamic flows are typically of this type. The only restriction on
this general principle is that the contour must be reducible to a point while staying inside
the flowfield. A contours which contains a lifting airfoil, for example, is not reducible, and
will in general have a nonzero circulation.
Γ=0
Γ>0
4
Fluids – Lecture 11 Notes
1. Introduction to Compressible Flows
2. Thermodynamics Concepts
Reading: Anderson 7.1 – 7.2
Introduction to Compressible Flows
Definition and implications
A compressible flow is a flow in which the fluid density ρ varies significantly within the
flowfield. Therefore, ρ(x, y, z) must now be treated as a field variable rather than simply
a constant. Typically, significant density variations start to appear when the flow Mach
number exceeds 0.3 or so. The effects become especially large when the Mach number
approaches and exceeds unity.
The figure shows the behavior of a moving Lagrangian Control Volume (CV) which by
definition surrounds a fixed mass of fluid m. In incompressible flow the density ρ does not
change, so the CV’s volume V = m/ρ must remain constant. In the compressible flow
case, the CV is squeezed or expanded significantly in response to pressure changes, with ρ
changing in inverse proportion to V. Since the CV follows the streamlines, changes in the
CV’s volume must be accompanied by changes in the streamlines as well. Above Mach 1,
these volumetric changes dominate the streamline pattern.
+p
+p
increasing
pressure
−p
increasing
pressure
volume
constant
V
−p
volume
decreases
V
Lagrangian
control volume
Lagrangian
control volume
(Incompressible)
Incompressible
Compressible
Many of the relations developed for incompressible (i.e. low speed) flows must be revisited
and modified. For example, the Bernoulli equation is no longer valid,
p+
1 2
ρV 6= constant
2
since ρ = constant was assumed in its derivation. However, concepts such as stagnation
pressure po are still usable, but their definitions and relevant equations will be different from
the low speed versions.
Some flow solution techniques used in incompressible flow problems will no longer be applicable to compressible flows. In particular, the technique of superposition will no longer be
generally applicable, although it will still apply in some restricted situations.
Perfect gas
A perfect gas is one whose individual molecules interact only via direct collisions, with no
1
other intermolecular forces present. For such a perfect gas, p, ρ, and the temperature T are
related by the following equation of state
p = ρRT
where R is the specific gas constant. For air, R = 287J/kg-K◦ . It is convenient at this point
to define the specific volume as the limiting volume per unit mass,
ΔV
1
=
ΔV→0 Δm
ρ
υ ≡ lim
which is merely the reciprocal of the density. In general, the nomenclature “specific X” is
synonymous with “X per unit mass”. The equation of state can now be written as
pυ = RT
which is the more familiar thermodynamic form.
The appearance of the temperature T in the equation of state means that it must vary
within the flowfield. Therefore, T (x, y, z) must be treated as a new field variable in addition
to ρ(x, y, z). In the moving CV scenario above, the change in the CV’s volume is not only
accompanied by a change in density, but by a change in temperature as well.
The appearance of the temperature also means that thermodynamics will need to be addressed. So in addition to the conservation of mass and momentum which were employed in
low speed flows, we will now also need to consider the conservation of energy. The following
table compares the variables and equations which come into play in the two cases.
Incompressible flow
Compressible flow
variables:
V~ , p
~ ,p,ρ,T
V
equations:
mass , momentum
−→
−→
mass , momentum , energy , state
Thermodynamics Concepts
Internal Energy and Enthalpy
The law of conservation of energy involves the concept of internal energy, which is the sum
of the energies of all the molecules of a system. In fluid mechanics we employ the specific
internal energy, denoted by e, which is defined for each point in the flowfield. A related
quantity is the specific enthalpy, denoted by h, and related to the other variables by
h = e + pυ
The units of e and h are (velocity)2 , or m2 /s2 in SI units.
For a calorically perfect gas, which is an excellent model for air at moderate temperatures,
both e and h are directly proportional to the temperature. Therefore we have
e = cv T
h = cp T
where cv and cp are specific heats at constant volume and constant pressure, respectively.
h − e = pυ = (cp − cv )T
2
and comparing to the equation of state, we see that
cp − cv = R
Defining the ratio of specific heats, γ ≡ cp /cv , we can with a bit of algebra write
1
R
γ−1
γ
=
R
γ−1
cv =
cp
so that cv and cp can be replaced with the equivalent variables γ and R. For air, it is handy
to remember that
1
= 2.5
γ−1
γ = 1.4
γ
= 3.5
γ−1
(air)
First Law of Thermodynamics
Consider a thermodynamic system consisting of a small Lagrangian control volume (CV)
moving with the flow.
t + dt
e + de
h + dh
time
t
...
time
e
h
...
δw
δq
Over the short time interval dt, the CV undergoes a process where it receives work δw and
heat δq from its surroundings, both per unit mass. This process results in changes in the
state of the CV, described by the increments de, dh, dp . . . The first law of thermodynamics
for the process is
δq + δw = de
(1)
This states that whatever energy is added to the system, whether by heat or by work, it
must appear as an increase in the internal energy of the system.
The first law only makes a statement about de. We can deduce the changes in the other
state variables, such as dh, dp, . . . if we assume special types of processes.
1. Adiabatic process, where no heat is transferred, or δq = 0. This rules out heating of
the CV via conduction though its boundary, or by combustion inside the CV.
2. Reversible process, no dissipation occurs, implying that work must be only via volumetric compression, or dw = −p dυ. This rules out work done by friction forces.
3. Isentropic process, which is both adiabatic and reversible, implying −p dυ = de.
3
e2
T2
ρ2
e + de
e
p dυ
...
e1
T1
ρ1
...
Isentropic flow process
from state 1 to state 2
Isentropic relations
Aerodynamic flows are effectively inviscid outside of boundary layers. This implies they have
negligible heat conduction and friction forces, and hence are isentropic. Therefore, along the
pathline followed by the CV in the figure above, the isentropic version of the first law applies.
−p dυ = de
(2)
This relation can be integrated after a few substitutions. First we note that
dυ = d
� �
1
ρ
= −
dρ
ρ2
and with the perfect gas relation
de = cv dT =
1
R dT
γ−1
the isentropic first law (2) becomes
p
dρ
1
=
R dT
2
ρ
γ−1
1 ρR
dρ
=
dT
ρ
γ−1 p
dρ
1 dT
=
ρ
γ−1 T
The final form can now be integrated from any state 1 to any state 2 along the pathline.
1
ln T + const.
γ−1
ρ = const. × T 1/(γ−1)
ln ρ =
ρ2
=
ρ1
�
T2
T1
�1/(γ−1)
(3)
From the equation of state we also have
p2 T1
ρ2
=
ρ1
p1 T2
which when combined with (3) gives the alternative isentropic relation
p2
=
p1
�
T2
T1
4
�γ/(γ−1)
(4)
Lecture F11 Mud: Vorticity, Strain, Circulation
(41 respondents)
1. Can you explain the 2nd figure in the notes more? (1 student)
This shows the fluid element at two successive times, t and t + �t. Differences in
velocity at two nodes will cause the nodes to not move in unison, resulting in tilting
of the edges. Try re-reading the notes and Anderson.
2. How are �∂1 and �∂2 defined? (1 student)
These are angles that the edges rotate, or tilt, during the time interval �t. For conve­
nience, we draw the initial element to be aligned with the x and y axes, so the angles
can be conveniently measured from those axes.
3. What does �∂˙ mean? (1 student)
This is the same as d(�∂)/dt, or the rotation rate of the edge.
4. In the definition of �z , Where did the �u/�y and �v/�x come from? (1
student)
These velocity derivatives are equal to −�∂˙1 and �∂˙2 as obtained from the element
edge tilting analysis.
5. Where did the 1/2 factor in the strain just go away? (1 student)
It’s just a matter of convention in how you define ωxy . Anderson doesn’t use the 1/2.
Other fluids books keep the 1/2. Either way is OK, as long as you’re consistent with
the definition when using it later on.
6. What’s the squiggly vorticity symbol? (3 students)
This is the Greek letter � (or “xi”). Uppercase version is �, and even harder to write.
7. What’s the difference between vorticity and rotation? (3 students)
None, really. Vorticity is just twice the rotation rate of the fluid element.
�ξ = 2 ξ�
This just gets rid of the annoying factor of 1/2 in the �
ξ definition. It also gives a
simpler raltionship between the velocity and vorticity:
ξ
�ξ = � × V
again without a factor of 2 or 1/2.
8. What does vorticity of a fluid element describe? (1 student)
ξ (x, y, z) using the culr oper­
Vorticity is uniquely determined from the velocity field V
ation (see above). Nonzero curl at a point means that a fluid element at a point is
rotating.
9. What’s the (math) interpretation of curl? (1 student)
The physical interpretation of twice fluid element rotation rate is easier to see I think.
10. When is a flow rotational or irrotational? (1 student)
Aerodynamic flows which are uniform upstream, as most often occurs, are irrotational
everywhere downstream, outside the boundary layers. The fluid elements outside the
boundary layers do not rotate, even though they might take highly-curved paths when
going around the body.
11. What does circulation do to a fluid element? (1 student)
Circulation is not defined for a point or infinitesimal fluid element, but rather it’s
defined for a finite-size loop or circuit. Vorticity, on the other hand, is defined for a
point. Circulation is in effect the integrated or cumulative contribution of all vorticity
inside the loop:
��
� = −
� dA
12. What happens to � if the loop is right on the airfoil? (1 student)
� is nonzero and has the same value for any loop enclosing the lifting airfoil, provided
the loop stays outside the boundary layer. � is zero for any loop not enclosing the
airfoil, provided again that the loop stays outside the boundary layer.
13. Can you review 18.02 material which is used in UE Fluids? (3 students)
Yes, but this would have to be scheduled off hours. I can’t steal time from lectures or
recitations. Please give the grad TA’s possible times on your schedule.
14. No mud (16 students)
Spring 2004
Unified Engineering
Fluids Problems F11–F14
F11+12. Air is drawn at high speed out of a large reservoir through a duct of constant area
A, which contains a radiator delivering a known Q̇ to the flow (in Watts). The heating and
friction of the duct walls are negligible.
.
Q
h o 1 � 1 p 1 V1
h o �o po
h o 2 � 2 p 2 V2
The known flow quantities are:
ho = ho1 reservoir total enthalpy
�o reservoir total density
po reservoir total pressure
p2 outlet pressure (drives the flow)
The remaining six unknown quantities inside the duct are:
ho2 outlet total enthalpy
V1 inlet velocity
V2 outlet velocity
�1 inlet density
�2 outlet density
p1 inlet pressure
A total of six equations are needed to solve for the six unknowns. One of these equations is
the isentropic relation between the reservoir and station 1,
p1
=
po
�
ho1 − 12 V12
ho
��/(�−1)
and two additional ones are the state equations at stations 1 and 2.
�−1
1
�1 ho1 − V12
2
�
�
�−1
1
=
�2 ho2 − V22
2
�
p1 =
p2
�
Write down the remaining three equations by constructing a suitable control volume and
applying the integral mass, momentum, energy equations. (Do not try to solve the six
equations — it gets very messy!)
Fluids – Lecture 12 Notes
1. Energy Conservation
Reading: Anderson 2.7, 7.4, 7.5
Energy Conservation
Application to fixed finite control volume
The first law of thermodynamics for a fixed control volume undergoing a process is
V
.
E out
E(t)
.
E in
.
Q
dE = δQ + δW
dE
˙
+ Ėout − Ėin = Q̇ + W
dt
(1)
.
W
where the second rate form is obtained by dividing by the process time interval dt, and
including the contribution of flow in and out of the volume.
Total energy
˙ will go towards the kinetic energy as well as the internal energy
In general, the work rate W
of the fluid inside the CV, in some unknown proportion. This is ambiguity is resolved by
defining the total specific energy, which is simply the sum of internal and kinetic specific
energies.
1
1
eo = e + V 2 = cv T + V 2
2
2
This is the overall energy/mass of the fluid seen by a fixed observer. The e part corresponds
to the molecular motion, while the V 2 /2 part corresponds to the bulk motion.
=
total energy
+
internal energy
kinetic energy
We now define the overall system energy E to include the kinetic energy
E =
���
ρ eo dV
˙ can now include all work.
so that W
Energy flow
The Ėout and Ėin terms in equation (1) account for mass flow through the CV boundary,
which carries not only momentum, but also thermal and kinetic energies. The internal energy
flow and kinetic energy flow can be described as
internal energy flow = (mass flow) × (internal energy/mass)
kinetic energy flow = (mass flow) × (kinetic energy/mass)
1
where the mass flow was defined earlier. The internal energy/mass is by definition the specific
internal energy e, and the kinetic energy/mass is simply V 2 /2. Therefore
~ ·n
m
˙ e = ρ V
ˆ Ae
= ρ Vn A e
�
� 1
1
1
~ ·n
kinetic energy flow = m
˙ V2 = ρ V
ˆ A V 2 = ρ Vn A V 2
2
2
2
�
internal energy flow =
�
where Vn = V~ · n.
ˆ Note that both types of energy flows are scalars.
V
n^
.
me
.
m 12 V 2
A
,
ρ , e , 12 V 2
internal energy flow
kinetic energy flow
The net total energy flow rate in and out of the volume is obtained by integrating the internal
and kinetic energy flows over its entire surface.
Ėout
− E˙ in =
��
��
�
~ ·n
ρ V
ˆ
1
e+ V2
2
�
dA =
��
�
�
~ ·n
ρ V
ˆ eo dA
Heat addition
Two types of heat addition can occur.
Body heating. This acts on fluid inside the volume. Examples are combustion, and internal
absorption of radiation (e.g. microwaves). Whatever the source, the body heating is given
by a specific heating rate q̇, with units of power/mass.
Q̇body =
���
ρ q˙ dV
Surface heating. The surface can transmit heat flux into or out of the volume. This is usually
associated with viscous action, and like the viscous work rate it is complicated to write out.
It will be simply called Q̇viscous .
ρg
.
Wviscous
.
Qviscous
V
V
−pn
−pn
V
.
ρq
Work by applied forces
~ is
The work rate (or power) performed by a force F~ applied to fluid moving at velocity V
˙ = F~ · V
~
W
2
We will consider two types of applied forces.
Body forces. These act on fluid inside the volume. The most common example is the gravity
force, along the gravitational acceleration vector ~g . The work rate done on the fluid inside
the entire CV is
���
˙
~ dV
Wgravity =
ρ ~g · V
Surface forces. These act on the surface of the volume, and can be separated into pressure
and viscous forces. The work rate done by the pressure is
˙ pressure =
W
��
~ dA
−p n
ˆ·V
The work rate done by the viscous force is complicated to write out, and for now will simply
be called Ẇviscous .
Integral Energy Equation
Substituting all the energy flow, heating, and work term definitions into equation (1) gives
the Integral Energy Equation.
d
dt
���
ρ eo dV +
��
~ ·n
ρ V
ˆ eo dA =
�
�
���
ρ q˙ dV +
��
~ dA +
−p n
ˆ·V
˙ viscous
+ Q̇viscous + W
���
~ dV
ρ ~g · V
(2)
Total enthalpy
We now define the total enthalpy as
1
p 1
p
ho = h + V 2 = e + + V 2 = eo +
2
ρ 2
ρ
We can now combine the two surface integrals in equation (2) into one enthalpy-flow integral.
The result is the simpler and more convenient Integral Enthalpy Equation which does no
longer explicitly containes the pressure work term.
�� �
���
���
�
d ���
~ ·n
~ dV + Q̇viscous + W
˙ viscous (3)
ρ eo dV +
ρ V
ˆ ho dA =
ρ ~g · V
ρ q˙ dV +
dt
In most applications we can assume steady flow, and also neglect gravity and viscous work.
Equation (3) then simplifies to
��
���
~ ·n
ρ V
ˆ ho dA =
�
�
ρ q˙ dV + Q̇viscous
(4)
Along with the Integral Mass Equation and the Integral Momentum Equation, equation (4)
can be applied to solve many thermal flow problems involving finite control volumes. The
Q̇viscous term typically represents heating or cooling of the fluid by a solid wall.
Differential Enthalpy Equation
The enthalpy-flow surface integral in equation (3) can be converted to a volume integral
using Gauss’s Theorem.
��
ρ V~ · n
ˆ ho dA =
�
�
3
���
~ ho dV
∇ · ρV
�
�
Omitting the viscous terms for simplicity, the integral enthalpy equation (3) can now be
written strictly in terms of volume integrals.
��� �
�
�
∂(ρeo )
~ ho − ρq̇ − ρ~g · V
~
+ ∇ · ρV
∂t
�
dV = 0
This relation must hold for any control volume whatsoever, and hence the whole quantity
in the brackets must be zero at every point.
�
�
∂(ρeo )
~ ho = ρq̇ + ρ~g · V~
+ ∇ · ρV
∂t
which applies at every point in the flowfield. Adding ∂p/∂t to both sides and combining it
on the left side gives the Differential Enthalpy Equation in “divergence form”.
�
�
∂ (ρho )
~ ho = ∂p + ρq̇ + ρ~g · V
~
+ ∇ · ρV
∂t
∂t
(5)
We can recast this equation further by expanding the lefthand side terms.
ho
�
�
�
�
�
∂ρ
∂ho
~
~ · ∇ho
+ ∇ · ρV
+ ρ
+ V
∂t
∂t
�
=
∂p
~
+ ρq̇ + ρ~g · V
∂t
The quantity in the brackets is exactly zero by the mass continuity equation, and the quantity
in the braces is simply Dho /Dt, so that the result can be written compactly as the Differential
Enthalpy Equation in “convective form”.
1 ∂p
Dho
~
=
+ q̇ + ~g · V
Dt
ρ ∂t
(6)
More specifically, equation (6) gives the rate of change of total enthalpy along a pathline, in
terms of three simple terms.
Steady, adiabatic flows
Assume the following three conditions are met:
Steady Flow. This implies that ∂p/∂t = 0
Adiabatic Flow. This implies that q̇ = 0
~
Negligible gravity. This implies that we can ignore ~g · V
With these assumptions, equation (6) becomes simply
Dho
= 0
Dt
(7)
so that the total enthalpy is constant along a streamline. If all streamlines originate at a common upstream reservoir or uniform flow, which is the usual situation in external aerodynamic
flows, then
1
ho = h + V 2 = constant
(8)
2
throughout the flowfield. Equation (8) in essence replaces the full differential enthalpy equation (5), which is an enormous simplification. Note also its similarity to the incompressible
Bernoulli equation. However, equation (8) is more general, since it is valid for compressible
flows, and is also valid in the presence of friction.
4
Fluids – Lecture 12 Notes
1. Stream Function
2. Velocity Potential
Reading: Anderson 2.14, 2.15
Stream Function
Definition
~ as the partial derivatives
Consider defining the components of the 2-D mass flux vector ρV
¯ y):
of a scalar stream function, denoted by ψ(x,
ρu =
∂ψ¯
∂y
,
ρv = −
∂ψ¯
∂x
For low speed flows, ρ is just a known constant, and it is more convenient to work with a
scaled stream function
ψ̄
ψ(x, y) =
ρ
~.
which then gives the components of the velocity vector V
u =
∂ψ
∂y
,
v = −
∂ψ
∂x
Example
Suppose we specify the constant-density streamfunction to be
ψ(x, y) = ln
�
x2 + y 2 =
1
ln(x2 + y 2)
2
which has a circular “funnel” shape as shown in the figure. The implied velocity components
are then
∂ψ
y
∂ψ
−x
u =
= 2
,
v = −
= 2
2
∂y
x +y
∂x
x + y2
which corresponds to a vortex flow around the origin.
ψ
y
x
vortex flow example
1
Streamline interpretation
The stream function can be interpreted in a number of ways. First we determine the differential of ψ¯ as follows.
∂ψ¯
∂ψ¯
dψ¯ =
dx +
dy
∂x
∂y
dψ¯ = ρu dy − ρv dx
Now consider a line along which ψ¯ is some constant ψ¯1 .
¯ y) = ψ¯1
ψ(x,
Along this line, we can state that dψ¯ = dψ¯1 = d(constant) = 0, or
ρu dy − ρv dx = 0
dy
v
=
dx
u
→
¯ y) are
which is recognized as the equation for a streamline. Hence, lines of constant ψ(x,
streamlines of the flow. Similarly, for the constant-density case, lines of constant ψ(x, y) are
streamlines of the flow. In the example above, the streamline defined by
ln
�
x2 + y 2 = ψ1
can be seen to be a circle of radius exp(ψ1 ).
y
−
−
dψ = 0
−
ψ = ψ1
V
v
dy
u
dx
x
Mass flow interpretation
Consider two streamlines along which ψ¯ has constant values of ψ¯1 and ψ¯2 . The constant
mass flow between these streamlines can be computed by integrating the mass flux along
any curve AB spanning them. First we note the geometric relation along the curve,
n
ˆ dA = ı̂ dy − ˆdx
and the mass flow integration then proceeds as follows.
m
˙ =
�
B
A
~ ·n
ˆ dA =
ρV
�
B
A
(ρu dy − ρv dx) =
�
B
A
dψ¯ = ψ¯2 − ψ¯1
Hence, the mass flow between any two streamlines is given simply by the difference of their
stream function values.
2
y
−
−
ψ + dψ
B
V
−
−
−
−
ψ = ψ2
V
dy
dA
n^
−
n^
ψ
A
x
dx
ψ = ψ1
Continuity identity
¯ y). Computing the divergence
~ (x, y) specified by some ψ(x,
Consider the mass flux field ρV
of this field we have
�
�
�
�
�
�
¯
¯
∂(ρv)
∂
∂ψ
∂
∂ψ
∂ 2 ψ̄
∂ 2 ψ̄
∂(ρu)
~
−
=
−
= 0
∇ · ρV
=
+
=
∂x
∂y
∂x ∂y
∂y ∂x
∂y ∂x
∂x ∂y
¯ y) will automatically satisfy the steady mass
so that any mass flux field specified via ψ(x,
continuity equation. In low speed flow, a similar computation shows that any velocity field
specified via ψ(x, y) will automatically satisfy
~ = 0
∇·V
which is the constant-density mass continuity equation. Because of these properties, using
the stream function to define the velocity field can give mathematical simplification in many
fluid flow problems, since the continuity equation then no longer needs to be addressed.
Velocity Potential
Definition
~ as the gradient of a scalar velocity
Consider defining the components of the velocity vector V
potential function, denoted by φ(x, y, z).
~ = ∇φ = ı̂ ∂φ + ˆ ∂φ + kˆ ∂φ
V
∂x
∂y
∂z
If we set the corresponding x, y, z components equal, we have the equivalent definitions
u =
∂φ
∂x
,
v =
∂φ
∂y
,
w =
∂φ
∂z
Example
For example, suppose we specify the potential function to be
� �
φ(x, y) = arctan
x
y
which has a corkscrew shape as shown in the figure. The implied velocity components are
then
∂φ
−x
∂φ
y
= 2
,
v =
= 2
u =
2
∂x
x +y
∂y
x + y2
3
φ
y
x
vortex flow example
which corresponds to a vortex flow around the origin. Note that this is exactly the same
velocity field as in the previous example using the stream function.
Irrotationality
If we attempt to compute the vorticity of the potential-derived velocity field by taking its
curl, we find that the vorticity vector is identically zero. For example, for the vorticity
x-component we find
ξx ≡
∂w
∂v
∂ ∂φ
∂ ∂φ
∂2φ
∂2φ
−
=
−
=
−
= 0
∂y
∂z
∂y ∂z
∂z ∂y
∂y∂z
∂z∂y
and similarly we can also show that ξy = 0 and ξz = 0. This is of course just a manifestation
of the general vector identity curl(grad) = 0 . Hence, any velocity field defined in terms
of a velocity potential is automatically an irrotational flow . Often the synonymous term
potential flow is also used.
Directional Derivative
In many situations, only one particular component of the velocity is required. For example,
for computing the mass flow across a surface, we only require the normal velocity component.
~ · n.
This is typically computed via the dot product V
ˆ In terms of the velocity potential, we
have
∂φ
~ ·n
V
ˆ = ∇φ · n
ˆ =
∂n
where the final partial derivative ∂φ/∂n is called the directional derivative of the potential
along the normal coordinate n. The figure illustrates the relations.
φ = φ4
φ = φ3
φ = φ2
V= φ
Δ
n
φ = φ1
V . n^ = 6φ
6n
In general, the component of the velocity along any direction can be obtained simply by
taking the directional derivative of the potential along that same direction.
4
Lecture F12 Mud: Energy Conservation
1. What’s the difference between dE/dt and Ėflow ? (1 student)
dE/dt is the rate of change of energy contained by the mass already in the volume.
Ėflow is haw fast energy is being brought into or out of the volume by inflowing and
outflowing mass.
˙ viscous ? (1 student)
2. What’s the difference between Q̇viscous and W
Imagine a viscous friction force F�f rubbing on the surface of the CV, much like a
boundary layer “rubs” on the surface of an airfoil. This force does work if there is
˙ viscous = F�f · V
� . But if there’s a friction
fluid velocity at the point of application, W
force, there must also be frictional heating present (a boundary layer is always slightly
warmer than the effectively-inviscid fluid next to it). This excess heat will flow into
the CV, and constitute Q̇viscous .
3. How do the physical definitions of h and ho and ho� differ? (1 student)
The static enthalpy h measures only the energy of the molecules as seen be an observer
moving with the mean velocity. The total enthalpy ho is the energy seen by a fixed
observer (it includes the mean kinetic energy). ho� is just the ho of the freestream
flow.
4. Shouldn’t �p/�t come into play in the last PRS? How about the velocity?
(1 student)
No. This was a steady problem. There is no variation of the flowfield with time, so
�p/�t = 0.
� . This would come into play if the channel was
The other term you refer to is �g · V
� are perpendicular, so the dot
tilted up or down. With a horizontal channel, �g and V
product is zero. But even with a tilted channel, gravity effects in thermal problems
are usually miniscule.
5. How can the energy equation be violated? (1 student)
I’m not sure what you mean. In practice it can’t be violated. The First Law always
works.
6. In the PRS, how is the momentum flow �= 0 while the energy flows = 0? (1
student)
That’s how it works out for the flow case shown. Momentum is a vector, so the opposite
flow jets have a different effect on it than on the scalar energy.
7. How can we have a q̇ without a �T at the boundary? (1 student)
The heat can “sneak” into the interior of the CV in the form of chemical bonds in the
fuel molecules brought inside, and then released as the q̇ term. This chemical energy
isn’t counted in the usual measure of e or h, which depend only the molecular kinetic
and vibration energies. Another way energy can sneak into the CV is in the form of
photons, which aren’t part of the energy accounting until they are absorbed by matter.
Heat flow by �T at the boundary constitutes Q̇viscous in our notation.
8. What does the porous plug do? Is it a strainer? (1 student)
The plug exerts a retarding friction force on the fluid passing through it. The screen
in your window does the same thing to air passing through it. In each case there must
be a pressure drop across the plug or screen to force the fluid against the friction.
9. In PRS 2, how do you know that no heat is removed from the CV as the
flow is forced through the plug? (1 student)
Once steady state is reached, the isolated plug cannot deliver or remove heat, otherwise
it would continually get colder or warmer. There is no heat flow out of or into the plug
to sustain this.
10. Wouldn’t the temperature be slightly lower downstream of the plug? (1
student)
You are correct. The air downstream has been expanded to a larger specific volume
v because of the pressure drop. So its � is smaller, hence V downstream is larger by
continuity. For a constant ho = h + V 2 /2, a larger V implies a lower h and hence a
lower T = h/cp .
11. No mud (6 students)
Lecture F12 Mud: Stream Function, Potential Function
(25 respondents)
1. The book claims “�¯2 = �¯1 represents mass flow perpendicular to the page”.
Isn’t 2-D flow in the plane of the page? (1 student)
It actually says “mass flow per unit depth perpendicular to the page”. The “perpen­
dicular” refers to the unit depth, not to the flow.
2. What’s the difference between �¯ and �? (2 students)
�¯ is the most general type of stream function, and gives the mass flux components:
¯
��/�y
= �u
¯
− ��/�x
= �v
For low speed flows where � is a constant, it is convenient to absorb the constant �
¯
factor into the stream function by defining � = �/�.
We now have:
��/�y = u
− ��/�x = v
It’s important to remember that � can be used only for low speed flows, while �¯ has
no such restriction.
The difference �¯2 − �¯1 gives the mass flow between two streamlines, while �2 − �1
gives the volume flow between two streamlines (but only in low-speed flow).
�
3. Where does �(x, y) = ln x2 + y 2 come from? (1 student)
I just made it up. The neat thing about a stream function is that no matter how
complicated �(x, y) might get, the resulting u(x, y) and v(x, y) that you get from it
satisfy the mass continuity equation
ρ � �u/�x + �v/�y = 0
�·V
and hence represent a physically possible flow.
4. Is there are graphical relation between streamlines and stream function
lines? (1 student)
Lines of constant stream function are the same as streamlines.
5. I don’t understand how the funnel-shaped � and the spiral-shaped ∂ repre­
sent the same vortex flow? (1 student)
Both functions produce the same u(x, y) and v(x, y), and hence they both represent
the same physical flowfield.
6. Will be dealing mostly with 2-D flows? (1 student)
The approach in Unified Fluids is to present concepts in the simplest way possible, so
that the understanding isn’t lost in unnecessary complexity. So we use the smallest
number of spatial dimensions in the examples and applications. Sometimes 1-D, usually
2-D, and occasionally 3-D if it’s unavoidable.
7. What’s the difference between � and ∂? (1 student)
They are alternative ways to define the velocity field u(x, y) and v(x, y). There are
important differences also. For example, � is usable only in 2-D, while ∂ easily extends
to 3-D with minimal complication.
¯ �, ∂? (1 student)
8. Is there a physical interpretation of �,
¯ has several interpretations: streamlines, mass flow, as presented in the notes. ∂ is
psi
a bit harder to interpret.
9. Why would we use � or ∂? (1 student)
When applicable, they almost always produce a tremendous mathematical simplifica­
tion of a fluid flow problem. This makes solving the equations much easier and/or
faster.
10. Are there conditions under which these methods work? (1 student)
Yes. The main restrictions are: � and �¯ are usable only in 2-D
∂ is usable only for irrotational flows.
11. Seems like we’re learning only endless equations and symbols. Where’s the
justification? (1 student)
So far we’ve focused almost entirely on “tools and concepts”. Applications will come
next. Anderson lays out this strategy.
12. Can you give a numerical example of calculating a directional derivative?
(1 student)
A directional derivative is normally computed using the dot product.
�∂/�n = �∂ · n̂
For example, say
∂(x, y) = − arctan(y/x)
and we want to know �∂/�n at the point (x, y) = (0, 1), along a line tilted 45� up from
horizontal. Along this direction
1
1
n̂ = � ı̂ + � ψ̂
2
2
and the gradient at the chosen point is
�∂ = 1 ı̂ + 0 ψ̂
Hence,
13. No mud (10 students)
1
�∂/�n = �∂ · n
ˆ = �
2
Fall 2003
Unified Engineering
Fluids Problem F12
F12. For the two flows given by . . .
�(x, y) = arctan (y/x)
�(x, y) = x2 + y 2
a) Determine the velocity fields, and sketch the streamlines.
b) Determine the volume flow rate through a circle of radius r.
c) Which of these flows is not feasible to set up in a lab? Explain.
Fluids – Lecture 13 Notes
1. Bernoulli Equation
2. Uses of Bernoulli Equation
Reading: Anderson 3.2, 3.3
Bernoulli Equation
Derivation – 1-D case
The 1-D momentum equation, which is Newton’s Second Law applied to fluid flow, is written
as follows.
ρ
∂u
∂u
∂p
+ ρu
= −
+ ρgx + (Fx )viscous
∂t
∂x
∂x
We now make the following assumptions about the flow.
• Steady flow: ∂/∂t = 0
• Negligible gravity: ρgx ≃ 0
• Negligible viscous forces: (Fx )viscous ≃ 0
• Low-speed flow: ρ is constant
These reduce the momentum equation to the following simpler form, which can be immediately integrated.
du
dp
+
= 0
dx
dx
dp
1 d(u2 )
ρ
+
= 0
2
dx
dx
1 2
ρ u + p = constant ≡ po
2
ρu
The final result is the one-dimensional Bernoulli Equation, which uniquely relates velocity
and pressure if the simplifying assumptions listed above are valid. The constant of integration
po is called the stagnation pressure, or equivalently the total pressure, and is typically set by
known upstream conditions.
Derivation – 2-D case
The 2-D momentum equations are
∂u
∂u
∂u
∂p
+ ρu
+ ρv
= −
+ ρgx + (Fx )viscous
∂t
∂x
∂y
∂x
∂v
∂v
∂v
∂p
ρ
+ ρu
+ ρv
= −
+ ρgy + (Fy )viscous
∂t
∂x
∂y
∂y
ρ
Making the same assumptions as before, these simplify to the following.
∂u
∂u
∂p
+ ρv
+
= 0
∂x
∂y
∂x
∂v
∂v
∂p
ρu
+ ρv
+
= 0
∂x
∂y
∂y
ρu
1
(1)
(2)
Before these can be integrated, we must first restrict ourselves only to flowfield variations
along a streamline. Consider an incremental distance ds along the streamline, with projections dx and dy in the two axis directions. The speed V likewise has projections u and
v.
y
p + dp
u + du
v + dv V
p
u
v
line
m
strea
v
dy
dx
u
x
Along the streamline, we have
or
dy
v
=
dx
u
u dy = v dx
(3)
We multiply the x-momentum equation (1) by dx, use relation (3) to replace v dx by u dy,
and combine the u-derivative terms into a du differential.
∂u
∂u
∂p
dx +
dx = 0
ρu
dx + ρv
∂x
∂y
∂x
�
�
∂u
∂u
∂p
ρu
dx +
dy +
dx = 0
∂x
∂y
∂x
∂p
dx = 0
ρu du +
∂x
1 � 2�
∂p
dx = 0
(4)
ρd u +
2
∂x
We multiply the y-momentum equation (2) by dy, and performing a similar manipulation,
we get
∂v
∂v
∂p
dy + ρv
dy +
dy
∂x
∂y
∂y
�
�
∂v
∂v
∂p
ρv
dx +
dy +
dy
∂x
∂y
∂y
∂p
ρv dv +
dy
∂y
� �
∂p
1
dy
ρ d v2 +
2
∂y
ρu
= 0
= 0
= 0
= 0
Finally, we add equations (4) and (5), giving
�
∂p
∂p
1 � 2
ρ d u + v2 +
dx +
dy = 0
2
∂x
∂y
�
1 � 2
ρ d u + v 2 + dp = 0
2
2
(5)
which integrates into the general Bernoulli equation
1
ρ V 2 + p = constant ≡ po
2
(along a streamline)
(6)
where V 2 = u2 + v 2 is the square of the speed. For the 3-D case the final result is exactly
the same as equation (6), but now the w velocity component is nonzero, and hence V 2 =
u2 + v 2 + w 2 .
Irrotational Flow
Because of the assumptions used in the derivations above, in particular the streamline relation (3), the Bernoulli Equation (6) relates p and V only along any given streamline. Different
streamlines will in general have different po constants, so p and V cannot be directly related
between streamlines. For example, the simple shear flow on the left of the figure has parallel
flow with a linear u(y), and a uniform pressure p. Its po distribution is therefore parabolic
as shown. Hence, there is no unique correspondence between velocity and pressure in such
a flow.
y
y
V
V
po
po
Rotational flow
Irrotational flow
� = ∇φ and V 2 = |∇φ|2 , then po takes on the same
However, if the flow is irrotational, i.e. if V
value for all streamlines, and the Bernoulli Equation (6) becomes usable to relate p and V in
the entire irrotational flowfield. Fortunately, a flowfield is irrotational if the upstream flow
is irrotational (e.g. uniform), which is a very common occurance in aerodynamics. From the
uniform far upstream flow we can evaluate
1
po = p∞ + ρV∞2 ≡ po∞
2
and the Bernoulli equation (6) then takes the more general form.
1
ρ V 2 + p = po∞
2
(everywhere in an irrotational flow)
(7)
Uses of Bernoulli Equation
Solving potential flows
Having the Bernoulli Equantion (7) in hand allows us to devise a relatively simple two-step
solution strategy for potential flows.
� = ∇φ using the
1. Determine the potential field φ(x, y, z) and resulting velocity field V
3
governing equations.
2. Once the velocity field is known, insert it into the Bernoulli Equation to compute the
pressure field p(x, y, z).
This two-step process is simple enough to permit very economical aerodynamic solution
methods which give a great deal of physical insight into aerodynamic behavior. The alter� (x, y, z) and
native approaches which do not rely on Bernoulli Equation must solve for V
p(x, y, z) simultaneously, which is a tremendously more difficult problem which can be approached only through brute force numerical computation.
Venturi flow
Another common application of the Bernoulli Equation is in a venturi , which is a flow tube
with a minimum cross-sectional area somewhere in the middle.
A1
A2
V2
V1
p
po
p1
p2
x
Assuming incompressible flow, with ρ constant, the mass conservation equation gives
A1 V1 = A2 V2
(8)
This relates V1 and V2 in terms of the geometric cross-sectional areas.
V2 = V1
A1
A2
Knowing the velocity relationship, the Bernoulli Equation then gives the pressure relationship.
1
1
p1 + ρV12 = po = p2 + ρV22
(9)
2
2
Equations (8) and (9) together can be used to determine the inlet velocity V1 , knowing only
the pressure difference p1 − p2 and the geometric areas. By direct substution we have
V1 =
�
�
�
�
2(p1 − p2 )
ρ [(A1 /A2 )2 − 1]
A venturi can therefore by used as an airspeed indicator, if some means of measuring the
pressure difference p1 − p2 is provided.
4
Fall 2003
Unified Engineering
Fluids Problem F13
F13. A venturi has a minimum throat area of 0.7 times the inlet/outlet areas. The water
tank is open to ambient atmospheric pressure. Determine the sea-level wind speed V which
is needed at the inlet to raise the water column 10 cm.
V
10 cm
Lecture F13 Mud: Bernoulli Equation
(25 respondents)
1. How is ρu du/dx = 21 ρ d(u2 )/dx? (2 students)
You can see this easily just by differentiating d(u2 )/dx via the product rule.
2. How is ρu du = 21 ρ d(u2 )? (2 students)
Take the differential d(u2 ) using the product rule. This is essentially the same operation
as in mud 1 above.
3. How did you go from 21 ρ d(u2)/dx + dp/dx = 0 to 21 ρ u2 + p = C ? (1 student)
Via an indefinite integration in x:
Z 1
ρ d(u2)/dx + dp/dx = 0 dx
2
4. How did you know you can multiply x-momentum by dx? (1 student)
You can always multiply equations by anything you want, whether it’s useful or not
to do so. Bernoulli in the 1700’s figured out that multiplying by dx actually gets you
somewhere.
~ should you use in Bernoulli? ∇φ? (1 student)
5. What form of V
In aerodynamic flows, the velocity used in Bernoulli is usually defined via φ(x, y, z).
There are exceptions, however, such as when a pitot probe is used inside a boundary
~ is not obtained from any
layer to actually measure the varying po (y). In this case, V
φ(x, y, z).
6. Is po the same as p∞ ? (1 student)
No. In aerodynamic flows, po is usually po∞ , which is the total pressure far away, while
p∞ is the static pressure far away. They are related to the far-away speed V∞ , via
Bernoulli:
1
po∞ = p∞ + ρ V∞2
2
7. Is po constant for all streamlines outside the boundary layer? (1 student)
Yep.
8. What is Anderson’s term “quasi-1-dimensional” on page 185? (1 student)
There is a slight variation in the flow velocity direction across the width of a pipe
which varies in cross-section, so this flow is not strictly 1-D. But we neglect this slight
transverse variation, and call the flow “quasi-1-D”. Note that the only “truly-1-D” flow
is a uniform flow, which isn’t particularly interesting.
9. Why does the air speed up over a wing? (1 student)
That’s the velocity field which is needed to satisfy the equations of mass and momentum
conservation for the moving fluid. A somewhat similar question is “Why does a mass
accelerate when you apply a force to it?” It just does. That’s physical reality as
described by Newton’s Laws.
10. If we can’t say that high velocity causes low pressure, what’s a good layman’s interpretation of Bernoulli’s Equation? (1 student)
I really don’t know a good way to do this. Anderson on page 183 makes an energy
interpretation, but I don’t think that would convince most laymen.
11. A venturi must narrow down and expand again fairly gently, correct? (1
student)
Yes. If the area changes are rapid, then the boundary layers on the inside surface will
separate (like on a stalled airfoil), and the effective areas will change drastically. The
pressures inside the venturi will then differ a great deal from what’s predicted from the
geometric areas.
12. No mud (15 students)
Fluids – Lecture 14 Notes
1. Helmholtz Equation
2. Incompressible Irrotational Flows
Reading: Anderson 3.7
Helmholtz Equation
Derivation (2-D)
If we neglect viscous forces, the x- and y-components of the 2-D momentum equation can
be written as follows.
∂u
∂u
∂u
−1 ∂p
+ u
+ v
=
+ gx
∂t
∂x
∂y
ρ ∂x
∂v
∂v
∂v
−1 ∂p
+ u
+ v
=
+ gy
∂t
∂x
∂y
ρ ∂y
(1)
(2)
We now take the curl of this momentum equation by performing the following operation.
∂
y-momentum (2)
∂x
∂
−
x-momentum (1)
∂y
If we assume that ρ is constant (low speed flow), the two pressure derivative terms cancel.
Since the gravity components gx and gy are generally constant, these also disappear when
the curl’s derivatives are applied. Using the product rule on the lefthand side, the resulting
equation is
∂ ∂v ∂u
−
∂t ∂x ∂y
!
∂ ∂v ∂u
+ u
−
∂x ∂x ∂y
+
!
!
∂ ∂v ∂u
+ v
−
∂y ∂x ∂y
!"
#
∂v ∂u ∂u ∂v
= 0
−
+
∂x ∂y ∂x ∂y
We note that the quantity inside the parentheses is merely the z-component of the vorticity
ξ ≡ ∂v/∂x − ∂u/∂y, so the above equation can be more compactly written as
"
∂ξ
∂ξ
∂ξ
∂u ∂v
+ u
+ v
+ ξ
+
∂t
∂x
∂y
∂x ∂y
#
= 0
We further note that the quantity in the brackets is the divergence of the velocity, which in
low speed flow must be zero because of mass conservation.
∂u ∂v
~ = 0
+
≡ ∇·V
∂x ∂y
(mass conservation equation)
This gives the following final result.
∂ξ
∂ξ
∂ξ
+ u
+ v
= 0
∂t
∂x
∂y
Dξ
= 0
or . . .
Dt
1
(3)
Equation (3) is the 2-D form of the Helmholtz Equation, which governs the vorticity field
ξ(x, y, t) in inviscid flow.
Interpretation and Implications
Consider a microscopic sensor drifting with the flow (along a pathline) near an airfoil. The
sensor’s time-trace signal ξs (t) is the vorticity at the sensor’s instantaneous location. Equation (3) implies that this vorticity signal ξs (t) will have zero time rate of change, since we
know that
dξs
Dξ
=
= 0
dt
Dt
Hence, the vorticity along a pathline must be constant.
ξs = constant
ξs
dξ s = 0
dt
ξs(t)
t
Vorticity sensor moving along streamline
Constant sensor output
Furthermore, this constant value must be determined far upstream of the airfoil where the
~∞ is uniform (either zero or some
streamline originates. If the freestream flow velocity V
constant), then
~∞ = 0
ξs = ∇ × V
This is true for all streamlines which originate in the uniform upstream flow, so that the
entire flowfield must be irrotational, as shown in the figure.
~ = 0
ξ(x, y, t) ≡ ∇ × V
(if upstream flow is uniform)
The one exception is that ξ 6= 0 for any streamline which is affected by viscous forces. For
these streamlines the Helmholtz equation (3) does not hold, since here the viscous forces are
not negligible, as was assumed at the outset.
ξ=0
here
implies . . .
ξ=0
everywhere downstream, except . . .
.
.
.
ξ=0
inside
boundary
layers
For 3-D flows, it is possible to derive a more general 3-D Helmholtz equation. From this
we can also conclude that 3-D flows which are initially uniform are irrotational downstream.
These 3-D derivations are considerably more cumbersome, and will not be attempted here.
2
Incompressible, Irrotational Flows
Governing Equations
The mass conservation equation for an incompressible flow states that the velocity field has
zero divergence.
~ = 0
∇·V
(4)
The Helmholtz equation implies that an inviscid flow which is uniform upstream must be
irrotational, and can therefore be expressed in terms of a potential function.
~ = ∇φ
V
Substituting this into the divergence equation (4) gives
∇ · (∇φ) = ∇2 φ = 0
(5)
This is Laplace’s Equation. In Cartesian coordinates, with φ = φ(x, y, z), Laplace’s equation
is explicitly given by
∇2 φ =
∂2φ
∂2φ
∂2φ
+
+
= 0
∂x2
∂y 2
∂z 2
In cylindrical coordinates, with φ = φ(r, θ, z), it has the form
∂φ
1 ∂
r
∇ φ =
r ∂r
∂r
2
!
+
1 ∂2φ
∂2φ
+
= 0
r 2 ∂θ2
∂z 2
For 2-D problems, the stream function can be employed in lieu of the potential function.
The requirement that the flow be irrotational leads to
∂u
∂ ∂ψ
∂ ∂ψ
∂v
−
= −
−
= 0
∂x
∂y
∂x ∂x
∂y ∂y
∂2ψ
∂2ψ
+
= 0
∂x2
∂y 2
(6)
Hence, if the stream function is employed, it must also satisfy Laplace’s equation.
Superposition
Laplace’s equation is linear . If φ1 (x, y, z), φ2 (x, y, z) are valid solutions, then their sum
φ3 (x, y, z) = φ1 + φ2
is another valid solution. The corresponding velocities can therefore be obtained via vector
summation.
~3 (x, y, z) = ∇φ3 = ∇ (φ1 + φ2 ) = V
~1 + V
~2
V
This is the principle of superposition, which allows constructing complex flowfields from any
number of relatively simple components. The figure shows an example of two uniform flows
being superimposed into a third uniform flow. Stream functions can be superimposed in the
same manner. The pressure field in each case is obtained using Bernoulli’s equation.
p1 (x, y, z) = po −
1
ρ |∇φ1 |2
2
,
p2 (x, y, z) = po −
3
1
ρ |∇φ2 |2
2
. . . etc
φ1 = x
φ 2 = 0.5 y
φ 3 = x + 0.5 y
+
=
Boundary Conditions
In order to solve Laplace’s equation, it is necessary to apply boundary conditions at all
boundaries of the flowfield. For most aerodynamic problems these fall into two categories.
Infinity Boundary Conditions
The flow far away from the body must approach the freestream velocity. Choosing the x
axis to be aligned with the freestream direction, we require
∂φ
= V∞
(at infinity)
u =
∂x
If a stream function is used, the corresponding boundary condition would be
∂ψ
u =
= V∞
(at infinity)
∂y
Wall Boundary Conditions
The flow adjacent to the wall is physically constrained to flow parallel, or tangent to the
wall. If the velocity vector is tangent, then its normal component must clearly be zero.
~ · n̂ = (∇φ) · n̂ = ∂φ = 0
V
∂n
(on wall)
The boundary condition on the alternative stream function is
∂ψ
= 0
V~ · n̂ =
∂s
(on wall)
where s is the arc length along the surface. This can also be specified as
ψ(s) = constant
(on wall)
6φ = V
6x
n^
V
6φ = 0
6n
6φ = V
6x
6φ = V
6x
4
6φ = V
6x
Unified Engineering
Fluids Problem F14
Fall 2003
F14. �1 (x, y) and �2 (x, y) are known to be physically-possible flows (i.e. satisfy mass con­
servation), and their corresponding pressure fields p1 (x, y) and p2 (x, y) are known via the
Bernoulli equation.
a) A third flow is now defined by �3 (x, y) = �1 + �2 . Explain how you would obtain its
corresponding pressure field p3 .
b) Yet another flow �4 = ��1 /�x is defined. Is this a physically-possible flow?
Lecture F14 Mud: Helmholtz, Laplace’s Equation
(33 respondents)
1. What is the physical meaning of ∇2 φ, or ∇2 in general? (2 students)
~ , or the velocity divergence. Also, for
In the case of fluid flow, ∇2 φ is the same as ∇ · V
~ , or the vorticity. Laplace’s equation also appears
2-D flow, ∇2 ψ is the same as ∇ × V
in a great number of other situations in math and physics, such as in Electricity and
Magnetism.
2. Can we use Laplace Transforms to solve for streamline shapes? (1 student)
No. Laplace’s equation and Laplace transforms are totally unrelated. Laplace himself
dabbled in lots of different stuff.
3. How do you actually solve ∇2 φ = 0? (1 student)
There are lots of different techniques, many employing superposition. In aerodynamics,
one class of techniques are called panel methods, which we will discuss later.
4. What’s the Helmoltz equation used for? (1 student)
A major use is to prove that aerodynamic flows are irrotational, like we did in class.
5. Can we plot streamlines by plotting ψ(x, y) = constant? (1 student)
Yep.
6. What’s the difference between φ and ψ? (2 students)
Each is used to generate a velocity field, but in two different ways. I suggest reviewing
the past few lectures.
7. How do you define where the boundary layer starts? (1 student)
One way to define the edge of a boundary layer is: “Normal distance from the wall
beyond which the vorticity is below some very small threshold.” At the edge the
2
vorticity decreases extremely fast, roughly as e−n , so the precise threshold level is not
too important.
8. How can the wall be another streamline when it has boundary layers on it?
(1 student)
Good question. When solving ∇2 φ = 0 or ∇2 ψ = 0 for the flow around a body we
assume that the boundary layers are negligibly thin. This is by necessity, since we’ve
assumed that viscous effects are neglible. In flow situations where the viscous effects
are actually not neglible, such as in the flow around a cylinder in the Fluids Lab, the
idealized inviscid flow that we predict from the solution of ∇2 φ = 0 will not be very
realistic. Using your data you will compare the ideal inviscid flow to the actual flow.
9. Why does ∂φ/∂n = 0 mean that the flow is perpendicular to n̂? (1 student)
~ ·n
∂φ/∂n = 0 is the same as V
ˆ = 0, which means the normal velocity component Vn
is zero. This is the definition of “perpendicularity”.
10. What’s the difference in boundary conditions between infinity and at a wall?
(1 student)
They specify different things. ∂φ/∂x = V∞ states that the horizontal velocity is equal
to V∞ . ∂φ/∂n = 0 states than the normal velocity (Vn in an earlier lecture) is zero.
11. The superposition example was like the flow around a torpedo nose. How
is that different than the flow around a missile nose? (1 student)
I picked a torpedo since the nose shape in the example is sort of rounded. To get the
flow to take on a pointy nose shape like on a missle requires superimposing something
else besides a single source.
12. No mud (19 students)
Fluids – Lecture 15 Notes
1. Uniform flow, Sources, Sinks, Doublets
Reading: Anderson 3.9 – 3.12
Uniform Flow
Definition
A uniform flow consists of a velocity field where V~ = uı̂ + v̂ is a constant. In 2-D, this
velocity field is specified either by the freestream velocity components u∞ , v∞ , or by the
freestream speed V∞ and flow angle α.
u = u∞ = V∞ cos α
v = v∞ = V∞ sin α
2
Note also that V∞2 = u2∞ + v∞
. The corresponding potential and stream functions are
φ(x, y) = u∞ x + v∞ y = V∞ (x cos α + y sin α)
ψ(x, y) = u∞ y − v∞ x = V∞ (y cos α − x sin α)
V
v
α
u
Zero Divergence
A uniform flow is easily shown to have zero divergence
~ = ∂u∞ + ∂v∞ = 0
∇·V
∂x
∂y
since both u∞ and v∞ are constants. The equivalent statement is that φ(x, y) satisfies
Laplace’s equation.
∇2 φ =
∂ 2 (u∞ x + v∞ y)
∂ 2 (u∞ x + v∞ y)
+
= 0
∂x2
∂y 2
Therefore, the uniform flow satisfies mass conservation.
Zero Curl
A uniform flow is also easily shown to be irrotational, or to have zero vorticity.
~ ≡ ξ~ =
∇×V
�
�
∂u∞ ˆ
∂v∞
−
k = 0
∂x
∂y
1
The equivalent irrotationality condition is that ψ(x, y) satisfies Laplace’s equation.
∇2 ψ =
∂ 2 (u∞ y − v∞ x)
∂ 2 (u∞ y − v∞ x)
+
= 0
∂x2
∂y 2
Source and Sink
Definition
A 2-D source is most clearly specified in polar coordinates. The radial and tangential velocity
components are defined to be
Vr =
Λ
2π r
Vθ = 0
,
where Λ is a scaling constant called the source strength. The volume flow rate per unit span
V̇ ′ across a circle of radius r is computed as follows.
′
V̇ =
�
2π
0
~ ·n
V
ˆ dA =
�
2π
0
Vr r dθ =
�
2π
0
Λ
r dθ = Λ
2π r
Hence we see that the source strength Λ specifies the rate of volume flow issuing outward
from the source. If Λ is negative, the flow is inward, and the flow is called a sink .
y
y
Vθ
r
x
Cartesian representation
The cartesian velocity components of the source or sink are
Λ
x
2
2π x + y 2
y
Λ
v(x, y) =
2
2π x + y 2
u(x, y) =
and the corresponding potential and stream functions are as follows.
�
Λ
Λ
ln r
φ(x, y) =
ln x2 + y 2 =
2π
2π
Λ
Λ
ψ(x, y) =
arctan(y/x) =
θ
2π
2π
2
θ
Vr
x
It is easily verified that apart from the origin location (x, y) = (0, 0), these functions satisfy
∇2 φ = 0 and ∇2 ψ = 0, and hence represent physically-possible incompressible, irrotational
flows.
Singularities
The origin location (0, 0) is called a singular point of the source flow. As we approach this
point, the magnitude of the radial velocity tends to infinity as
Vr ∼
1
r
Hence the flow at the singular point is not physical, although this does not prevent us from
using the source to represent actual flows. We will simply need to ensure that the singular
point is located outside the flow region of interest.
Uniform Flow with Source
Two or more incompressible, irrotational flows can be combined by superposition, simply by
adding their velocity fields or their potential or stream function fields. Superposition of a
uniform flow in the x-direction and a source at the origin therefore has
x
Λ
+ V∞
2
2π x + y 2
y
Λ
v(x, y) =
2
2π x + y 2
u(x, y) =
or
or
�
Λ
Λ
ln x2 + y 2 + V∞ x =
ln r + V∞ r cos θ
2π
2π
Λ
Λ
arctan(y/x) + V∞ y =
θ + V∞ r sin θ
ψ(x, y) =
2π
2π
φ(x, y) =
The figure shows the streamlines of the two basic flows, and also the combined flow.
The bullet-shaped heavy line on the combined flow corresponds to the dividing streamline,
which separates the fluid coming from the freestream and the fluid coming from the source.
If we replace the dividing streamline by a solid semi-infinite body of the same shape, the
flow about this body will be the same as the flow outside the dividing streamline in the
superimposed flow.
3
Uniform Flow with Source and Sink
We now superimpose a uniform flow in the x-direction, with a source located at (−ℓ/2, 0),
and a sink of equal and opposite strength located at (+ℓ/2, 0), plus a freestream.
ψ =
Λ
(θ1 − θ2 ) + V∞ r sin θ
2π
y
ψ
r1
r2
θ2
θ1
x
l
The figure on the right shows the streamlines of the combined flow. The heavy line again
indicates the dividing streamline, which traces out a Rankine oval . All the streamlines inside
the oval originate at the source on the left, and flow into the sink on the right. The net
volume outflow from the oval is zero. Again, the dividing streamline could be replaced by a
solid oval body of the same shape. The flow outside the oval then corresponds to the flow
about this body.
Doublet
Consider a source-sink pair with strengths ±Λ, located at (∓ℓ/2, 0). Now let the separation
distance ℓ approach zero, while simultaneously increasing the source and sink strengths such
that the product κ ≡ ℓΛ remains constant. The resulting flow is a doublet with strength κ.
ψ =
lim −
ℓ→0
κ=const.
κ
κ sin θ
Δθ = −
2π ℓ
2π r
y
y
ψ
r
Δθ
x
x
l
4
A similar limiting process can be used to produce the doublet’s potential function.
φ =
κ cos θ
2π r
The streamline shapes of the doublet are obtained by setting
ψ = −
where
κ sin θ
= c = constant
2π r
r = d sin θ
κ
d = −
2πc
In polar coordinates this is the equation for circles of diameter d, centered on x, y = (0, ±d/2).
Nonlifting Flow over Circular Cylinder
Flowfield definition
We now superimpose a uniform flow with a doublet.
κ sin θ
κ
ψ = V∞ r sin θ −
= V∞ r sin θ 1 −
2π r
2π V∞ r 2
�
or
where
�
R2
ψ = V∞ r sin θ 1 − 2
r
2
R ≡ κ/(2πV∞ )
�
�
This corresponds to the flow about a circular cylinder of radius R.
The radial and tangential velocities can be obtained by differentiating the stream function
as follows.
�
1 ∂ψ
R2
Vr =
= V∞ cos θ 1 − 2
r ∂θ
r
�
�
R2
∂ψ
= −V∞ sin θ 1 + 2
Vθ = −
∂r
r
5
�
Surface velocities and pressures
On the surface of the cylinder where r = R, we have
Vr = 0
Vθ = −2V∞ sin θ
The maximum surface speed of 2V∞ occurs at θ = ±90◦ .
The surface pressure is then obtained using the Bernoulli equation
p(θ) = po −
�
1 � 2
ρ Vr + Vθ2
2
Substituting Vr = 0 and Vθ (θ), and using the freestream value for the total pressure,
po = p∞ +
1 2
ρV
2 ∞
gives the following surface pressure distribution.
p(θ) = p∞ −
�
1 2�
ρV∞ 1 − 4 sin2 θ
2
The corresponding pressure coefficient is also readily obtained.
Cp (θ) ≡
p(θ) − p∞
= 1 − 4 sin2 θ
1
2
ρV
∞
2
6
Lecture F15 Mud: Uniform flow, sources, sinks, doublets
(27 respondents)
1. In the freestream+source case, if V∞ increases, will the picture change? (1
student)
The streamline pattern depends only on the ratio Λ/V∞ . So let’s say you double both
Λ and V∞ , all the velocities will double, but the streamline pattern won’t change. If
you change only V∞ , then the streamline pattern will change.
2. How do we model more complicated flows like airfoils or wings? (2 students)
With more complicated arrangements of sources, sinks, vortices, and doublets. The
most complicated 3-D models might have thousands of sources or doublets. The necessary bookkeeping must be done by computer programs.
3. How do we select the source, sink, doublet, etc. required to model a specific
flow? (2 students)
In our examples we superimposed a source or doublet of prescribed strength with a
freestream of prescribed strength. We then obtained the shape of the implied body
via the dividing streamline. In aerodynamic applications, the problem is somewhat in
reverse – the shape of the body (airfoil or wing) is a given, and we have to determine
out the source and doublet strengths which are necessary to produce a flow about the
body. One technique for doing this is thin airfoil theory
, which we will address later in
considerable detail. Other techniques are panel methods, which are rather complicated,
so we will only mention these in passing.
4. In the doublet construction, when is Λ = +κ/ℓ, and when is Λ = −κ/ℓ? (1
student)
The left source has positive Λ = κ/ℓ, and the right source (sink, actually), has a
negative Λ = −κ/ℓ. We picked the signs this way to get a doublet which has the
flow along the x-axis going to the left. Picking opposite signs would simply reverse
the doublet’s flow direction. A right-going doublet superimposed with a right-going
uniform flow will not give the flow about a cylinder.
5. How do we know the value of κ so we can apply the results for the cylinder?
(1 student)
You can set κ from the radius and freestream speed:
κ = 2πR2 V∞
6. How does ψ go to −(κ/2π) sin θ/r in the limit? (1 student)
You gave to plug in the flowfield geometry, and take the limit. Andreson does this on
pp. 221–222.
7. Does the Cp (θ) = 1 − 4 sin2 θ correspond to what we will plot for our lab? (1
student)
Yes, it’s one of the curves.
8. What makes this flow non-lifting? (1 student)
It’s symmetrical between top and bottom, so there’s no net y-force.
9. How do you take friction into account? (1 student)
Predicting viscous separated flow on a bluff body like the cylinder is extremely difficult.
In contrast, prediction of viscous flow about a streamline shape is much easier and
faster. Either one is beyond the scope of Unified Fluids, however. This is addressed to
some extent in 16.110 (Senior/Grad aerodynamics course), and in a few other courses.
10. Confused about PRS question at the end. How did you get h = Λ/V∞ ? (6
students)
The source’s contribution to the velocity decreases with distance as 1/r. Hence, the
velocity very far downstream is just V∞ ı̂ (i.e. only the freestream part), since far away
the contribution of the source decreases to nothing.
The
volume flow rate between the two streamlines is therefore computed to be v̇ =
�
~ ·n
V
ˆ dA = V∞ h. But the volume flow rate between the two streamlines is equal
to Λ, which is what the source cranks out. Hence v̇ = V∞ h = Λ, or h = Λ/V∞ . I
apologize again for the typos on the PRS answer list.
11. No mud (10 students)
Fluids – Lecture 16 Notes
1. Vortex
2. Lifting flow about circular cylinder
Reading: Anderson 3.14 – 3.16
Vortex
Flowfield Definition
A vortex flow has the following radial and tangential velocity components
Vr = 0
,
Vθ =
C
r
where C is a scaling constant. The circulation around any closed circuit is computed as
Γ ≡ −
�
~ · d~s = −
V
�
Vθ r dθ = −
�
θ2
θ1
C
r dθ = −C (θ2 − θ1 )
r
y
y
V
dθ
ds
r dθ
x
x
The integration range θ2 −θ1 = 2π if the circuit encircles the origin, but is zero otherwise.
Γ =
�
−2πC , (circuit encircles origin)
0
, (circuit doesn’t encircle origin)
y
y
θ1 θ
2
θ1 θ
2
x
x
In lieu of C, it is convenient to redefine the vortex velocity field directly in terms of the
circulation of any circuit which encloses the vortex origin.
Vθ = −
1
Γ
2π r
A positive Γ corresponds to clockwise flow, while a negative Γ corresponds to counterclockwise flow.
Cartesian representation
The cartesian velocity components of the vortex are
Γ
y
2
2π x + y 2
x
Γ
v(x, y) = −
2
2π x + y 2
u(x, y) =
and the corresponding potential and stream functions are as follows.
Γ
Γ
arctan(y/x) = −
θ
2π
2π
�
Γ
Γ
ln x2 + y 2 =
ln r
ψ(x, y) =
2π
2π
φ(x, y) = −
Singularity
As with the source and doublet, the origin location (0, 0) is called a singular point of the
vortex flow. The magnitude of the tangential velocity tends to infinity as
Vθ ∼
1
r
Hence, the singular point must be located outside the flow region of interest.
Lifting Flow over Circular Cylinder
Flowfield definition
We now superimpose a uniform flow with a doublet and a vortex.
�
R2
ψ = V∞ r sin θ 1 − 2
r
�
+
Γ
ln r
2π
This corresponds to the flow about a circular cylinder of radius R as before, but now a
top/bottom assymetry is introduced by the vortex.
2
The radial and tangential velocities can be obtained by differentiating the stream function
as follows.
�
1 ∂ψ
R2
Vr =
= V∞ cos θ 1 − 2
r ∂θ
r
�
�
∂ψ
R2
Vθ = −
= −V∞ sin θ 1 + 2
∂r
r
�
−
Γ
2π r
−
�
Surface velocities and pressures
On the surface of the cylinder where r = R, we have
Vr = 0
Vθ = −2V∞ sin θ −
Γ
2π R
The corresponding surface pressure coefficient follows.
Cp (θ) = 1 −
V2
= 1 − 4 sin2 θ −
V∞2
�
Γ
2π V∞ R
�2
2Γ
sin θ
π V∞ R
�
(1)
Forces
The resultant force/span is obtained by integrating the pressure forces over the surface of
the cylinder.
�
�
′
′
′
~
R ≡ D ı̂ + L ̂ =
−p n
ˆ dA =
−(p − p∞ ) n
ˆ dA
(2)
The constant p∞ which has been subtracted from p in the integrand does not change the
integrated result. This follows from the general identity
�
(constant) n̂ dA = 0
which holds for any closed body.
y
ny
dθ
R
θ
n^
nx
x
Breaking up the resultant force/span (2) into separate x- and y-components, and dividing
by 21 ρV∞2 2R, we obtain expressions for the drag and lift coefficients.
cd
cℓ
1
−Cp nx dA
=
2R
�
1
=
−Cp ny dA
2R
�
3
Using the cylinder geometry relations
nx = cos θ
,
ny = sin θ
,
dA = R dθ
and substituting the Cp (θ) result (1) gives
�
cd
1
=
2
−1 + 4 sin θ +
�
cℓ
1 � 2π
1 � 2π
−Cp sin θ dθ =
−1 + 4 sin2 θ +
=
2 0
2 0
�
�
0
1
−Cp cos θ dθ =
2
2π
�
0
2π
2
�
Γ
2πV∞ R
Γ
2πV∞ R
�2
�2
+
�
+
�
�
2Γ
sin θ cos θ dθ
πV∞ R
�
�
2Γ
sin θ sin θ dθ
πV∞ R
�
After evaluating the integrals we obtain the final results.
cd = 0
Γ
V∞ R
cℓ =
The equivalent dimensional forms are
D′ = 0
′
L
(3)
= ρ V∞ Γ
(4)
The result of zero drag (3) is known as d’Alembert’s Paradox , since it’s in direct conflict
with the observation that D ′ > 0 for all real bodies in a uniform flow. The explanation is of
course that viscosity has been neglected. The lift result (4) is known as the Kutta-Joukowski
Theorem, which will turn out to be valid for a 2-D body of any shape, not just for a circular
cylinder.
Real Cylinder Flows
Real viscous flow about a circular cylinder at large Reynolds numbers exhibits large amounts
of flow separation and drag. Normally the flow is symmetric between top and bottom, and
hence the lift is zero. However, if the cylinder has a rotational velocity, the separation is
pushed aft on the aft-going side and pushed forward on the forward-going side, resulting in
a flow assymetry. This assymetry has an associated nonzero circulation and a corresponding
lift. This phenomenon is known as the Magnus effect. Although the lift generated by
a rotating cylinder can match or exceed the lift achievable by a wing of similar size, the
cylinder is not a satisfactory lifting device because of its unavoidably large drag.
L’
Ω
boundary layer
separation
A rotating sphere also exhibits the Magnus effect, and here it has a strong influence on
many ball sports. The curveball pitch in baseball, the diving topspin volley in tennis, and
the sideways curving flight of a sliced golf ball are all due to the Magnus effect.
4
Lecture F16 Mud: Vorticity, Strain, Circulation
(33 respondents)
1. What is the physical meaning of �2 �, or �2 in general? (2 students)
∂ , or the velocity divergence. Also, for
In the case of fluid flow, �2 � is the same as � · V
∂ , or the vorticity. Laplace’s equation also appears
2-D flow, �2 � is the same as � × V
in a great number of other situations in math and physics, such as in Electricity and
Magnetism.
2. Can we use Laplace Transforms to solve for streamline shapes? (1 student)
No. Laplace’s equation and Laplace transforms are totally unrelated. Laplace himself
dabbled in lots of different stuff.
3. How do you actually solve �2 � = 0? (1 student)
There are lots of different techniques, many employing superposition. In aerodynamics,
one class of techniques are called panel methods, which we will discuss later.
4. What’s the Helmoltz equation used for? (1 student)
A major use is to prove that aerodynamic flows are irrotational, like we did in class.
5. Can we plot streamlines by plotting �(x, y) = constant? (1 student)
Yep.
6. What’s the difference between � and �? (2 students)
Each is used to generate a velocity field, but in two different ways. I suggest reviewing
the past few lectures.
7. How do you define where the boundary layer starts? (1 student)
One way to define the edge of a boundary layer is: “Normal distance from the wall
beyond which the vorticity is below some very small threshold.” At the edge the
2
vorticity decreases extremely fast, roughly as e−n , so the precise threshold level is not
too important.
8. How can the wall be another streamline when it has boundary layers on it?
(1 student)
Good question. When solving �2 � = 0 or �2 � = 0 for the flow around a body we
assume that the boundary layers are negligibly thin. This is by necessity, since we’ve
assumed that viscous effects are neglible. In flow situations where the viscous effects
are actually not neglible, such as in the flow around a cylinder in the Fluids Lab, the
idealized inviscid flow that we predict from the solution of �2 � = 0 will not be very
realistic. Using your data you will compare the ideal inviscid flow to the actual flow.
9. Why does ��/�n = 0 mean that the flow is perpendicular to n̂? (1 student)
∂ ·n
��/�n = 0 is the same as V
ˆ = 0, which means the normal velocity component Vn
is zero. This is the definition of “perpendicularity”.
10. What’s the difference in boundary conditions between infinity and at a wall?
(1 student)
They specify different things. ��/�x = V� states that the horizontal velocity is equal
to V� . ��/�n = 0 states than the normal velocity (Vn in an earlier lecture) is zero.
11. The superposition example was like the flow around a torpedo nose. How
is that different than the flow around a missile nose? (1 student)
I picked a torpedo since the nose shape in the example is sort of rounded. To get the
flow to take on a pointy nose shape like on a missle requires superimposing something
else besides a single source.
12. No mud (19 students)
Unified Engineering
Fluids Problem F16
Fall 2003
F16. �1 (x, y) and �2 (x, y) are known to be physically-possible flows (i.e. satisfy mass con­
servation), and their corresponding pressure fields p1 (x, y) and p2 (x, y) are known via the
Bernoulli equation.
a) A third flow is now defined by �3 (x, y) = �1 + �2 . Explain how you would obtain its
corresponding pressure field p3 .
b) Yet another flow �4 = ��1 /�x is defined. Is this a physically-possible flow?
Fluids – Lecture 17 Notes
1. Flowfield prediction
2. Source Sheets
Reading: Anderson 3.17
Flowfield Prediction
Problem definition
The flowfield examples used so far were used to demonstrate the basic ideas behind the
method of superposition. We chose some combination of elementary flows (uniform flow,
sources, vortices, etc.), and then determined the resulting flowfield. The corresponding body
shape was determined from the shape of the dividing streamline. However, such an approach
is not practical for engineering applications, where we want to specify the body shape, rather
than have it as an outcome. The problem can therefore be stated as follows.
Given: Body shape Y (x), Freestream velocity V~∞
Determine: Superposition of suitable elementary flows which produce the velocity field
~ (x, y) about the body.
V
It turns out that sources, vortices, and doublets are not ideally suited to this task because
of their strong singularities. The constraint that these singularities must be inside the body
is difficult to meet, especially if the body is very slender. For this reason we now define
slightly more elaborate elementary flows which are smoother, and therefore better suited to
representing smooth bodies.
Source Sheets
Definition
Consider a sequence of flows where a single source of strength Λ is repeatedly subdivided
into smaller sources which are evenly distributed along a line segment of length ℓ. The limit
of this subdivision process is a source sheet of strength λ = Λ/ℓ.
Λ
→
2 ×
Λ
2
→
4 ×
Λ
4
→
8 ×
Λ
8
...
λ
The units of Λ are length2 /time, while the units of λ are length/time (or velocity). Note
that the total source strength is not changed in this process.
The limiting process shown above has assumed that the sheet is straight, and that the
sources are uniformly subdivided and uniformly distributed along the sheet. Neither of
these assumptions are required. The subdivided sources can be distributed along any chosen
curve, in any chosen density. Hence, the source sheet can be curved, and its strength λ can
vary along the sheet.
1
Properties
Consider an infinitesimal length ds of the sheet. The infinitesimal source strength of that
piece is dΛ = λ ds, and the corresponding potential at some field point P at (x, y) is
dφ =
dΛ
λ
ln r =
ln r ds
2π
2π
where r is the distance between point (x, y) and the point on the sheet.
dφ
dΛ = λ ds
r
λ(s)
P
(x,y)
ds
0
l
The potential of the entire sheet at point P is then obtained by integrating the infinitesimal
contributions all along the sheet.
φ(x, y) =
�
0
ℓ
λ
ln r ds
2π
The shape of the sheet and the λ(s) distribution must be specified before this integral can
be evaluated. The velocity of the sheet is then obtained by taking the gradient of the result.
Note that to build up the entire flowfield, the integral must be evaluated for each point P in
the xy plane. In practice this is not necessary, since for engineering purposes the velocity is
required only at a small set of points, such as on the surface of a body to allow computation
of the pressure and the resultant force.
Consider now a simpler straight source sheet extending from (−ℓ/2, 0) to (ℓ/2, 0), with a
constant strength λ.
y
P
y
λ
ds
s
−l/2
x
x
l/2
The potential and the velocity components at point P are given by
λ ℓ/2 �
φ(x, y) =
(1)
ln (x − s)2 + y 2 ds
2π −ℓ/2
�
� �
�
�
λ ℓ/2
x−s
∂φ
λ ℓ/2 ∂
2
2
u(x, y) =
=
ds
ln (x − s) + y ds =
∂x
2π −ℓ/2 ∂x
2π −ℓ/2 (x − s)2 + y 2
�
� �
�
�
λ ℓ/2
y
∂φ
λ ℓ/2 ∂
2
2
ln (x − s) + y ds =
ds
v(x, y) =
=
∂y
2π −ℓ/2 ∂y
2π −ℓ/2 (x − s)2 + y 2
�
These integrals can be evaluated, although the resulting expressions are cumbersome, and
not too important for our purposes here. The really interesting result is for the normal
2
velocity v(x, y) very close to the sheet, either just above at y = 0+ , or just below at y = 0− .
After the necessary integration, we find that
v(x, 0+ ) =
λ
2
,
v(x, 0− ) = −
λ
2
λ
(flat, isolated source sheet)
V
V . n^ = λ/2
V . n^ = −λ/2
The normal velocity is then simply a constant λ/2 directed outward. But if any other
singularity or freestream is present, this additional velocity will be superimposed on each
side of the sheet. For example, if the sheet is immersed in a freestream, we will have
v(x, 0+ ) =
λ
+ v∞
2
,
v(x, 0− ) = −
λ
+ v∞
2
By taking the difference between the top and bottom points, any such additional velocity is
removed, giving the very general normal-velocity jump condition for any source sheet in any
situation.
~ · n̂ = λ
(2)
v(x, 0+ ) − v(x, 0− ) ≡ ΔV
λ
ΔV . n^ = λ
V
u
v
The advantage of using source sheets rather than sources to represent a flowfield is illustrated
in the figure below, which shows source sheets superimposed on a uniform flow to the right.
In each case the sheet’s strength λ is set so as to cancel the freestream’s component normal to
the sheet, giving a net zero normal flow. Hence, the sheet is ideally suited for representing a
solid surface of a body, since it can impose the physically necessary flow-tangency condition
~ · n̂ = 0 by suitably adjusting the sheet’s strength λ.
V
Modeling approach
The fact that the velocity field of a source sheet is smooth, without the troublesome 1/r
3
singularity of a point source, allows us to place some number of such sheets (or panels) end
to end on the surface of the body. We then determine the strengths λj of all the panels
j = 1, 2, . . . n such that the flow is tangent everywhere on the surface of the body. The
superposition also incidentally produces some flow inside the body, but this is not physical
and is simply ignored.
λ2
λj
V
λ1
λn
uniform flow + n source panels
resulting flowfield
This use of source sheets in this manner to represent a flow is the basis of the panel method ,
which is widely used to compute the flow about aerodynamic bodies of arbitrary shape. The
approach presented here is actually suitable only for non-lifting bodies such as fuselages.
For airfoils, wings, and other lifting bodies, vortices must be added in some form to enable
circulation to be represented. This modification will be treated later.
Solution technique
It is important to realize that each panel strength λj cannot be set independently of the
~ and hence the flow tangency
others. With more than one panel present, the velocity V
~ ·n
ˆ = 0 at any point i on the surface is influenced not only by that panel’s λi ,
condition V
but also by the strengths λj of all the other panels. In tensor notation this can be written
as
�
�
~ ·n
~∞ · n
V
ˆ = Aij λj + V
ˆi
i
where Aij is called the aerodynamic influence matrix , which can be computed once the
geometry of all the panels is decided.
P
Vi
V
λj
V
P
~ ·n
Requiring that V
ˆ = 0 for each of the n panel midpoints gives the following.
~∞ · n
Aij λj = −V
ˆi
This is a n × n linear system for the λj unknowns, which can be solved numerically using
matrix solution methods such as Gaussian elimination. With the λj determined, the velocity
and pressure (via Bernoulli) can then be computed at any point in the flowfield and on the
surface of the body. Forces are then computed by integrating the surface pressures. This
completes the aerodynamic analysis problem.
4
Lecture F17 Mud: Vorticity, Strain, Circulation
(27 respondents)
1. In the freestream+source case, if V� increases, will the picture change? (1
student)
The streamline pattern depends only on the ratio �/V� . So let’s say you double both
� and V� , all the velocities will double, but the streamline pattern won’t change. If
you change only V� , then the streamline pattern will change.
2. How do we model more complicated flows like airfoils or wings? (2 students)
With more complicated arrangements of sources, sinks, vortices, and doublets. The
most complicated 3-D models might have thousands of sources or doublets. The nec­
essary bookkeeping must be done by computer programs.
3. How do we select the source, sink, doublet, etc. required to model a specific
flow? (2 students)
In our examples we superimposed a source or doublet of prescribed strength with a
freestream of prescribed strength. We then obtained the shape of the implied body
via the dividing streamline. In aerodynamic applications, the problem is somewhat in
reverse – the shape of the body (airfoil or wing) is a given, and we have to determine
out the source and doublet strengths which are necessary to produce a flow about the
body. One technique for doing this is thin airfoil theory , which we will address later in
considerable detail. Other techniques are panel methods, which are rather complicated,
so we will only mention these in passing.
4. In the doublet construction, when is � = +�/ψ, and when is � = −�/ψ? (1
student)
The left source has positive � = �/ψ, and the right source (sink, actually), has a
negative � = −�/ψ. We picked the signs this way to get a doublet which has the
flow along the x-axis going to the left. Picking opposite signs would simply reverse
the doublet’s flow direction. A right-going doublet superimposed with a right-going
uniform flow will not give the flow about a cylinder.
5. How do we know the value of � so we can apply the results for the cylinder?
(1 student)
You can set � from the radius and freestream speed:
� = 2�R2 V�
6. How does � go to −(�/2�) sin κ/r in the limit? (1 student)
You gave to plug in the flowfield geometry, and take the limit. Andreson does this on
pp. 221–222.
7. Does the Cp (κ) = 1 − 4 sin2 κ correspond to what we will plot for our lab? (1
student)
Yes, it’s one of the curves.
8. What makes this flow non-lifting? (1 student)
It’s symmetrical between top and bottom, so there’s no net y-force.
9. How do you take friction into account? (1 student)
Predicting viscous separated flow on a bluff body like the cylinder is extremely difficult.
In contrast, prediction of viscous flow about a streamline shape is much easier and
faster. Either one is beyond the scope of Unified Fluids, however. This is addressed to
some extent in 16.110 (Senior/Grad aerodynamics course), and in a few other courses.
10. Confused about PRS question at the end. How did you get h = �/V� ? (6
students)
The source’s contribution to the velocity decreases with distance as 1/r. Hence, the
velocity very far downstream is just V�ı̂ (i.e. only the freestream part), since far away
the contribution of the source decreases to nothing.
The
volume flow rate between the two streamlines is therefore computed to be v̇ =
�
π ·n
V
ˆ dA = V� h. But the volume flow rate between the two streamlines is equal
to �, which is what the source cranks out. Hence v̇ = V� h = �, or h = �/V� . I
apologize again for the typos on the PRS answer list.
11. No mud (10 students)
Fall 2003
Unified Engineering
Fluids Problem F17
F17. A vortex flow is given by
u1 (x, y) =
x2
y
+ y2
v1 (x, y) =
−x
+ y2
x2
A uniform flow in the x-direction is given by
u2 (x, y) = V�
v2 (x, y) = 0
Superimpose these two flows, determine the pressure field, and find the x, y location of the
point of maximum pressure.
Fluids – Lecture 18 Notes
1. Prediction of Lift
2. Vortex Sheets
Reading: Anderson 3.17
Prediction of Lift
Limitations of Source Sheets
A point source has zero circulation about any circuit. Evaluating Γ using its definition we
have
Γ ≡ −
�
~ · d~s = −
V
�
Vr dr = −
�
r2
r1
Λ
Λ
dr = −
(ln r2 − ln r1 ) = 0
2πr
2π
which gives zero simply because r1 = r2 for any closed circuit, whether the origin is enclosed
or not. A source sheet, which effectively consists of infinitesimal sources, must have zero
circulation as well.
y
r1
r2
Γ=0
Γ=0
V
ds
dr
λ
x
Λ
This zero-circulation property of source sheets has severe consequences for flow representation. Any aerodynamic model consisting only of a freestream and superimposed source
sheets will have Γ = 0, and hence L′ = 0 as well. Hence, lifting flows cannot be represented
by source sheets alone.
This limitation is illustrated if we use source panels to model a flow expected to produce
lift, such as that on an airfoil at an angle of attack. Examination of the streamlines reveals
that the rear dividing streamline leaves the airfoil off one surface as shown in the figure. The
model also predicts an infinite velocity going around the sharp trailing edge.
source sheet model
Γ=0
L’ = 0
V
reality
Γ>0
L’ > 0
1
smooth flow−off
(Kutta condition)
On real airfoils the flow always flows smoothly off the sharp trailing edge, with no large
local velocities. This smooth flow-off is known as the Kutta condition, and it must be
faithfully duplicated in any flow model which seeks to predict the lift correctly. Changing
the streamline pattern to force the flow smoothly off the trailing edge requires the addition
of circulation, which implies that vortices must be included in the flow representation in
some manner.
Vortex Sheets
Definition
Consider a sequence of flows where a single vortex of strength Γ is repeatedly subdivided
into smaller vortices which are evenly distributed along a line segment of length ℓ. The limit
of this subdivision process is a vortex sheet of strength γ = Γ/ℓ.
Γ
2 ×
→
Γ
2
→
4 ×
Γ
4
8 ×
→
Γ
8
...
γ
Like with the source sheet strength λ, the units of γ are length/time (or velocity).
Properties
The analysis of the vortex sheet closely follows that of the source sheets. The potential of
the vortex sheet at point P is
φ(x, y) = −
�
ℓ
0
γ
θ ds
2π
dφ
P
d Γ= γ ds
(x,y)
θ
γ (s)
ds
0
l
For a straight vortex sheet extending from (−ℓ/2, 0) to (ℓ/2, 0), with a constant strength γ,
the potential and the velocity components at point P are given by
γ ℓ/2
φ(x, y) =
2π −ℓ/2
�
γ ℓ/2
∂φ
=
u(x, y) =
∂x
2π −ℓ/2
�
γ ℓ/2
∂φ
=
v(x, y) =
∂y
2π −ℓ/2
�
y
ds
x−s
�
�
�
y
γ ℓ/2
∂
y
− arctan
ds =
ds
∂x
x−s
2π −ℓ/2 (x − s)2 + y 2
�
�
�
∂
−x
y
γ ℓ/2
− arctan
ds =
ds
∂y
x−s
2π −ℓ/2 (x − s)2 + y 2
− arctan
2
As with the earlier source sheets, these integrals are cumbersome to evaluate in general. But
if we evaluate very close to the sheet, either just above at y = 0+ , or just below at y = 0− ,
the tangential velocity becomes very simple.
u(x, 0+ ) =
γ
2
u(x, 0− ) = −
,
γ
γ
2
(flat, isolated vortex sheet)
V . s^ = γ / 2
V
V . s^ = −γ / 2
The tangential velocity is then simply a constant γ/2 directed clockwise around the sheet.
By taking the difference between the upper and lower points at some x location,we obtain a
very general tangential-velocity jump condition for any vortex sheet in any situation.
~ · ŝ = γ
u(x, 0+ ) − u(x, 0− ) ≡ ΔV
(1)
The figure shows the vortex sheet with a freestream superimposed. The surface velocity
vector pattern is very complicated, but the tangential velocity jump across the sheet is a
constant equal the γ at all points.
ΔV . s^ = γ
V
u
v
γ
The advantage of using vortex sheets rather than vortices to represent a flowfield is illustrated
in the figure below. The left figure shows a vortex sheet superimposed on a uniform flow. The
vortex sheets smoothly deforms the flowfield in the manner required to impose circulation
and lift. The right figure shows the same airfoil as before, but now vortex panels have been
used instead of source panels to represent the flow. The nature of the vortex panels permits
the Kutta condition to be imposed, giving smooth flow off the trailing edge. The airfoil now
has the expected amount of lift.
3
Modeling approach and solution technique
As illustrated with the airfoil example, vortex panels provide an alternative way to model
the flow about a body, both for lifting and non-lifting bodies. The solution approach is
ˆ = 0 flow tangency condition is imposed for
nearly the same as with source panels. The V~ · n
each panel, but now the additional Kutta condition at the trailing edge is also imposed. The
resulting linear system is then solved for all the panel strengths γj . The surface velocities,
surface pressures, and overall forces can then be computed.
Types of panels used in practice
Vortex panels are by far the most widely used for 2-D problems, such as the flow about an
airfoil. Vortex panels can represent lifting or nonlifting flows equally well, so there is little
reason to use the more restrictive source panels.
For 3-D problems, however, vortex panels run into serious difficulties. The main problem
is that the sheet strength ~γ is now a vector lying in the sheet. The associated tangential
velocity of magnitude γ/2 is also in the sheet, and perpendicular to ~γ . In contrast, the source
panel strength λ is still a scalar in 3-D.
λ
3−D source sheet
γ
3−D vortex sheet
µ
3−D doublet sheet
Because ~γ is now a vector, it is not well suited for solution in a panel method. Instead, 3-D
panel methods employ doublet sheets, whose doublet strength µ (unrelated to viscosity) is a
scalar quantity, and can also represent lifting flows. The figure above conceptually shows a
doublet sheet. The axis of each infinitesimal doublet is oriented normal to the sheet, rather
than along the x-axis as in our previous examples.
Doublet sheets alone are sufficient to represent the flow about any lifting or nonlifting 3-D
body. However, most modern 3-D panel methods actually employ a combination of source
sheets and doublet sheets. Compared to using only doublet sheets, the source+doublet sheet
combination turns out to give the best accuracy for a given computational time. The details
of such combined methods are far beyond scope here.
4
Fall 2003
Unified Engineering
Fluids Problem F18
�
F18. Wind with velocity V� is flowing over a mountain ridge have the shape Y (x) = Cx.
The flow is to be modeled by superimposing a uniform flow with a source located at some
location x, y = (d, 0).
� �
�(x, y) = V� y +
ln (x − d)2 + y 2
4κ
y
r
V
�
�
x
d
a) Determine both the source’s location d, and the strength �, with the conditions:
u=0
at x, y = (0, 0)
�
v/u = dY /dx at x, y = (d, Cd)
The second condition simply requires that the flow direction on the ridge surface directly
above the source is parallel to the ridge surface.
b) A sailplane flying in the slope lift upwind of the ridge requires a vertical velocity of at
least v � 1m/s to stay aloft. For a wind speed of V� = 15m/s (33 mph) and ridge size
scale C = 500m, determine the maximum flyable radius r(λ) inside which the sailplane can
sustain flight. Plot the r(λ) boundary superimposed on a plot of the ridge.
Lecture F18 Mud: Vortex, Lift
(38 respondents)
1. What’s the difference between r and R? (1 student)
r is a general radius, as in r-κ polar coordinates. R is the specific cylinder’s radius.
2. How can we define C as −�/2�? (1 student)
For a vortex defined by V� = C/r, we calculate its circulation to be � = −2�C. So C
and � are effectively equivalent, since they differ only by the constant factor of −2�.
We can use them interchangeably. Since � is more aerodynamically significant (directly
related to vorticity, lift, etc.), we choose to use � instead of C to define the vortex:
V� = −�/(2�r).
3. I don’t understand �. (1 student)
I don’t know how to explain it better without just repeating the lecture material.
Maybe try reading Anderson 2.13 again? One suggestion is to clearly distinguish the
definition from the implications. Two equivalent definitions are
��−
�
θ · dθs
V
or
��−
��
�θ · n
ˆ dA
Some implications are:
� = 0 for any circuit in irrotational flow not enclosing a body.
� = L� /ξV� for any circuit enclosing an airfoil with lift L� .
4. For the vortex flowfield, is � nonzero only if origin is enclosed in the xy
plane, as opposed to the yz plane? (1 student)
First of all, the vortex flow is a 2-D flow in the xy or rκ plane. So in the real 3-D world
it’s actually a “roller” threaded on the z axis. We can take any circuit which encloses
the z axis and get the same � value. The circuit does not have to lie flat in the xy
plane.
5. In the notes, the total stream function for the uniform flow + doublet +
vortex has −(�/2�) ln r for the vortex part. Shouldn’t that be +(�/2�) ln r?
(1 student)
You are correct. It’s a typo which has been corrected in the web notes.
6. How can D � = 0? (1 student)
It can be seen to be zero by symmetry. The speed |V� (κ)| and hence the pressure p(κ)
are always the same on corresponding left and right points, for any combination of
V� , doublet strength �, and vortex strength �. So all the x-components forces which
produce D � add up to exactly zero.
7. When did we neglect viscosity? (1 student)
We neglected viscosity way back when we assumed irrotational flow, and started using
θ = �π and the Bernoulli equation. Once you go this route you are destined to have
V
D � = 0 for any closed 2-D body (d’Alembert’s Paradox).
We acknowledge that the inviscid assumption is quite bad for a bluff body with flow
separation. But it is very good for streamlined bodies like airfoils, at least if mainly
L� and M � are of interest.
8. On the board you wrote
Cp = 1 −
Vr2 + V�2
V2
=
1
−
V�2
V�2
Shouldn’t that be (Vr + V� )2 ? (1 student)
No. I had it correct. V is the magnitude of the total velocity vector, and so V 2 =
Vr2 + V�2 by Pythagoras.
9. Why is pˆ
n dA = (p − p� )ˆ
n dA? (1 student)
We can add any constant (p� in this case) to this integrand and not change the result.
This is because of the following math identity for any closed integration contour:
�
(constant) n̂ dA = 0
The physical interpretation is that a constant pressure (p� in this case) cannot exert
a net force on a closed body.
10. What’s the difference between lifting and non-lifting flows in the way you
model them? (1 student)
Good question! A non-lifting flow can be modeled by sources alone, although vortices
can also be used. A lifting-flow model must include vortices in one way or another,
since to get lift you must have � =
� 0, and have � �= 0 you must have vortices present.
Sources always have a zero circulation and hence by themselves cannot represent a
lifting flow.
11. Lost in the math. Do we need to understand all that algebra? (4 students)
I admit the math is relatively complicated. But there’s no way to effectively describe
fluid flows without it. I try to use physical pictures when possible, but these are a
qualitative aid, not a substitute for the math.
Understanding the starting assumptions and interpreting the final result are vastly
more important than being able to recite the algebra in between.
12. No mud (13 students)
Fluids – Lecture 19 Notes
1. Compressible Channel Flow
Reading: Anderson 10.1, 10.2
Compressible Channel Flow
Quasi-1-D Flow
A quasi-one-dimensional flow is one in which all variables vary primarily along one direction,
say x. A flow in a duct with slowly-varying area A(x) is the case of interest here. In practice
this means that the slope of the duct walls is small. Also, the x-velocity component u
dominates the y and z-components v and w.
dr
dx
1
u
v,w
y
r
x
x
z
A
A(x)
Quasi−1−D Flow
1−D Flow
Governing equations
Application of the integral mass continuity equation to a segment of the duct bounded by
any two x locations gives
−ρ1 u1
��
��
1
~ ·n
ρV
ˆ dA = 0
dA + ρ2 u2
��
x
2
dA = 0
−ρ1 u1 A1 + ρ2 u2 A2 = 0
1
2
The quasi-1-D approximation is invoked in the second line, with u and ρ assumed constant
on each cross-sectional area, so they can be taken out of the area integral.
Since stations 1 or 2 can be placed at any arbitrary location x, we can define the duct mass
flow which is constant all along the duct, and relates the density, velocity, and area.
ρ(x) u(x) A(x) ≡ m
˙ = constant
(1)
If we assume that the flow in the duct is isentropic, at least piecewise-isentropic between
shocks, the stagnation density ρo and stagnation speed of sound ao are both constant. This
allows the normalized ρ and u to be given in terms of the Mach number alone.
ρ
=
ρo
γ −1 2
1+
M
2
�
�−
1
γ−1
�
�− 1
γ −1 2
ρu
M
= M 1+
ρo ao
2
�−
γ −1 2
u
Ma
=
= M 1+
M
ao
ao
2
�
1
2
γ+1
2(γ−1)
(2)
(3)
(4)
1.5
u
ao
ρ
ρo
1.0
ρu
ρo ao
0.5
0.0
0.0
0.5
1.0
1.5
M
2.0
The figure shows these variables, along with the normalized mass flux , or ρu product, all
plotted versus Mach number.
The significance of ρu is that it represents the inverse of the duct area, or
1
ρu
A ∼
It is evident that the maximum possible mass flux occurs at a location where locally M = 1.
This can be proven by computing
�
d
ρu
dM ρo ao
�
=
�
1−M
2
γ −1 2
1+
M
2
��
�−
γ−3
2(γ−1)
which is clearly zero at M = 1. Therefore, the duct must have a local minimum, or throat,
wherever M = 1.
Sonic conditions
In the development above, the stagnation conditions ρo and ao were used to normalize the
various quantities. For compressible duct flows, it is very convenient to also define sonic
conditions which can serve as alternative normalizing quantities. These are defined by a
hypothetical process where the flow is sent through a duct of progressively reduced area
until M = 1 is reached, shown in the figure along with the familiar stagnation process.
Duct Flow
Variables
A
ρ
a
M
Isentropic
Stagnation
Process
A
ρ
a
M
Isentropic
Sonic−flow
Process
A=
ρ = ρo
a = ao
M =0
A
ρ
a
M
A =A*
ρ = ρ*
a = a*
M =1
The resulting quantities at the hypothetical sonic throat are denoted by a ()∗ superscript.
The advantage of the sonic-flow process is that it produces a well-defined sonic throat area
2
A∗ , while for the stagnation process A tends to infinity, and cannot be used for normalization.
The ratios between the stagnation and sonic conditions are readily obtained from the usual
isentropic relations, with M = 1 plugged in. Numerical values are also given for γ = 1.4 .
ρ∗
=
ρo
�
a∗
=
ao
�
p∗
=
po
�
γ −1
1+
2
�−
1
γ−1
γ −1
1+
2
�− 1
γ −1
1+
2
�−
2
= 0.6339
= 0.9129
γ
γ−1
= 0.5283
The sonic flow area A∗ can be obtained from the constant mass flow equation (1). For the
sonic-flow process we have
m
˙ = ρuA = ρ∗ u∗ A∗
and we also note that u∗ = a∗ since M = 1 at the sonic throat. Therefore,
ρ∗ a∗
ρ∗ ρo a∗ ao
A
=
=
A∗
ρ u
ρo ρ ao u
Using the previously-defined expressions produces
�
A
1
2
γ −1 2
=
1+
M
∗
A
M γ +1
2
�
γ+1
�� 2(γ−1)
(5)
This is the area-Mach relation, which is plotted in the figure below for γ = 1.4, and is also
available in tabulated form. It uniquely relates the local Mach number to the area ratio
A/A∗ , and can be used to “solve” compressible duct flow problems. If the duct geometry
A(x) is given, and A∗ is defined from the known duct mass flow and stagnation quantities,
then M(x) can be determined using the graphical technique shown in the figure, or using
the equivalent numerical table.
5.0
4.5
4.0
3.5
A 3.0
A* 2.5
2.0
1.5
1.0
0.5
0.0
0.0
0.2
0.4
0.6
0.8
1.0
3
1.2
M 1.4
1.6
1.8
2.0
Once M(x) is determined, any remaining quantity of interest, such as ρ(x), u(x), p(x), etc.,
can be computed from the isentropic or adiabatic relations such as (2) and (3).
Note that for any given area A(x), two solutions are possible for the given mass flow: a
subsonic solution with M < 1, and a supersonic solution with M > 1. Which solution
corresponds to the actual flow depends on whether the flow upstream of that x location is
subsonic or supersonic.
There is also the possibility of shock waves appearing in the duct. This introduces additional
complications which will be considered later.
4
Lecture F19 Mud: Quasi-1D-Flow
1. Why was ho constant? (1 student)
The flow is assumed to be adiabatic (no heat addition via combustion, etc). The flow
is also assumed to be frictionless, so �o and po are constant as well.
2. Why did M increase linearly in the PRS question? (1 student)
I just declared that it was. The objective was then to deduce the duct area A(x) which
would generate this given linear M (x) distribution.
3. Which way will the flow go after the sonic throat? Will it go back to
subsonic (M < 1), or go to supersonic (M > 1)? (1 student)
It depends on what pressure is imposed at the very exit. If the exit pressure is low
enough, the flow will accelerate to supersonic after the throat. Otherwise, it will stay
subsonic. We’ll cover this in the next (and last) lecture.
4. What would happen if you try making A < A� ? (1 student)
˙ which would in turn decrease
This would force a reduction in the channel mass flow m,
A� to be equal to or less than the new minimum A. Ya can’t trick it.
5. What properties determine the boundary layer thickness? (1 student)
Primarily the Reynolds number Re, and the velocity gradient du/dx. Increasing the
Re reduces the boundary layer (BL), and vice versa. A positive du/dx, or accelerating
flow, tends to reduce the BL thickness. A negative du/dx, or decelerating flow, tends
to increase the BL thickness.
6. You drew the effective area slightly within the BL? Is this intentional? (1
student)
Yes. The area reduction is not as big as the full BL thickness. It’s typically only about
1/5 to 1/3 of the BL thickness. This will be looked at in subsequent fluids courses,
which treat BL’s in more detail.
7. No mud (3 students)
Unified Engineering
Fluids Problems F19
Spring 2004
F19.
The pressure-tank reservoir for a paint sprayer contains air at 50 psi (3.45 × 10 5 Pa), and
T = 300 K◦. Air is to be supplied to the spray gun at a rate of 10 grams/s. What is the
absolute minimum diameter of the air hose that’s required to supply this air?
What must be true about the difference
�u − �l at the two surface points at the
trailing edge of a lifting airfoil?
1. �u − �l < 0
2. �u − �l = 0
3. �u − �l > 0
4. No way to know for sure from
given information
L’
�u
�l
What must be true about the difference
�u − �l at the two surface points at the
trailing edge of a lifting airfoil?
1. �u − �l < 0
2. �u − �l = 0
3. �u − �l > 0
4. No way to know for sure from
given information
L’
�u
�l
If D�/Dt = 0 in a steady inviscid flow, what
must be strictly true about the �(x, y) field?
1. � = 0 everywhere
2. � = 0 along any streamline
3. � = const. everywhere
4. � = const. along any streamline
A source of strength � is in a uniform flow
V�. What is the spacing height h of the di­
viding streamlines infinitely far downstream?
1. h = 0
2. h = �/V�
3. h = 2�/V�
4. h = �
5. Cannot be determined from
given information
V
�
h
Which of the following renders the isen­
tropic relations invalid.
1. Flow is unsteady
2. Velocity is very large
3. * Gas is non-perfect
4. Changes between states 1 and 2
are finite (rather than infinitesimal)
5. Not sure
Which of the following is NOT an ex­
ample of a �w or a �q process?
1. Bit of fuel inside CV ignites
2. * Gravity acts on descending CV
3. Friction acts along motion of CV
4. Pressure acts on contracting CV
5. Not sure
Note:
Only static internal energy changes
were being considered, not kinetic
energy changes
Two fluid jets of the same density � and
internal energy e flow as shown. What
is the internal energy flow integral for
the control volume?
⎛⎛
δ ˆ ⎝ e dA
� �V·n
�
�
1. 2�VA e
2. −2�VA e
3. * 0
4. not sure
V
control volume
A
n
A
V
n
Two fluid jets of the same density �
flow as shown. What is the kinetic en­
ergy flow integral for the control vol­
ume?
�
�
⎛⎛
1 2
�δ
⎝
� V·n
ˆ V dA
2
1. �V3A
2. −�V3A
3. * 0
4. not sure
V
control volume
A
n
A
V
n
Air is forced through a porous plug.
What’s the expected ho(x) distribution?
1.
2.
3.
ho
x
ho
x
ho
4. Not sure
2. *
x
Air flows out of a nozzle from a pres­
surized tank at room temperature. The
air comes out cold. What is the jet’s
density relative to ambient?
1. * Higher
2. Same
3. Lower
4. not sure
The spacing of the cars is 40m and 4m
before and after the “shock”. The shock
appears stationary to a stopped pedes­
trian. How fast is the obstructing ve­
hicle moving?
1. 3.3 m/s
2. * 3.0 m/s
3. 2.7 m/s
4. not sure
30 m/s
V=?
What is true about the temperatures
T1, T2, T3, (or enthalpies h1, h2, h3)
at the points shown?
1. T1 > T2 > T3
2. * T1 < T2 < T3
3. T1 < T2 = T3
4. Not enough information given
5. Not sure
1 2
3
Which flow pattern is physically cor­
rect?
1.
2.
3. Not sure
1. *
The two flows have the same wall shape.
Which flow has a larger M1 ?
1.
2.
M1
M1
3. Not sure
1. *
The two flows have the same M1. Which
flow has a larger turning angle?
1.
2.
M1
M1
3. Not sure
2. *
Which is a physically-possible shock flow?
1.
2.
3.
4.
5. All are valid
6. Not sure
2. *
A fan expands from M1 � 1 to near vac­
uum. Approximately, what is the ex­
pected turning angle?
1. � = 45�
2. � = 90�
3. * � = 130�
4. not sure
M1 ~ 1
p1
�=?
p2 ~ 0
A flow in a duct accelerates smoothly
from M = 0.5 to M = 1.5. What must
the duct look like?
M
x
1.
2.
3.
4.
5. Not sure
2. *
Low-speed flow is drawn from a reser­
voir by the exit pressure pe.
pr = 100000 Pa
�r = 1 kg/m3
A = 1 m2
Athroat = 0.5 m2
If pe = 99000 Pa, what is m?
˙
1. 100 kg/s
2. * 45 kg/s
3. 22.5 kg/s
4. not sure
large
reservoir
pr
�r
A throat
A
.
m
pe