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ME 306 Fluid Mechanics II
Part 1
Potential Flow
These presentations are prepared by
Dr. Cüneyt Sert
Department of Mechanical Engineering
Middle East Technical University
Ankara, Turkey
csert@metu.edu.tr
You can get the most recent version of this document from Dr. Sert’s web site.
Please ask for permission before using them to teach. You are not allowed to modify them.
1 - 2
Inviscid (Frictionless) Flow
- Continuity and Navier-Stokes equations govern the flow of fluids.
- For incompressible flows of Newtonian fluids they are
2 𝑉
- These equations can be solved analytically only for a few problems.
- They can be simplified in various ways.
- Common fluids such as water and air have small viscosities.
- Neglecting the viscous term (zero shear force) gives the Euler’s equation.
- Still difficult get a general analytical solution for the unknowns 𝑝 and 𝑉.
𝜇 𝑤𝑎𝑡𝑒𝑟 = 10
− 3 Pa s
𝜇 𝑎𝑖𝑟 = 2 × 10
− 5 Pa s
Inviscid Flow (cont’d)
- Bernoulli Equation (BE) is Euler’s equation written along a streamline.
2
- 𝑧 = constant along a streamline
Exercise: Starting from the Euler’s equation derive the BE.
Airfoil
Viscous flow close to the object and in the
wake of it (non negligible shear forces)
Adapted from
http://web.cecs.pdx.edu/~gerry/class/ME
Inviscid flow away from the
object (negligible shear forces)
Inviscid and Irrotational Flow
- To simplify further we can assume the flow to be irrotational.
𝜉 = 2 𝜔 = 𝛻 × 𝑉 = 0 (irrotational flow)
- Question: What’s the logic behind irrotationality assumption?
- Irrotationality is about velocity gradients.
or
𝑧
𝜃
𝑟
𝑟
𝑧
𝜃
𝜃
𝑟
𝑧
Vorticity (ksi) Angular velocity
- One special irrotational flow is when all velocity gradients are zero.
- An example is uniform flow such as 𝑢 = 𝑈, 𝑣 = 0 , 𝑤 = 0.
- In many flow fields there will be uniform-like flow regions.
Exercise: Sketch the developing flow inside a pipe with uniform entrance and
show the uniform and non-uniform flow regions.
1 - 5
Inviscid and Irrotational Flow (cont’d)
𝑥
𝑦
𝑧
𝑢 = 𝑈
𝑤 = 0
Airfoil
Close to the body velocity gradients
are high, shear forces are high and
flow becomes rotational.
Away from the body, flow has small velocity
gradients (uniform-like flow), small shear
forces and can remain irrotational.
Uniform
approach
velocity
(irrotational)
- In an inviscid flow net shear force acting on a fluid element is zero.
- Only pressure and body forces act on the fluid element. But they cannot cause
rotation because
- pressure forces act perpendicular to the element’s surface.
- body forces act through element’s center of gravity.
Exercise: Show how a fluid element will rotate inside the developing flow region
of a pipe with uniform entrance.
1 - 6
Inviscid and Irrotational Flow (cont’d)
Airfoil
Uniform
upstream
flow
(irrotational)
In an inviscid flow, a fluid element
that originates from an irrotational
flow region will remain irrotational.
- In general, flow fields are composed of both
- irrotational regions with negligible shear forces
- and rotational regions with considerable shear forces
- Sometimes rotational regions will be very thin such as high speed external flow
over an airfoil.
- But still neglecting them and assuming the flow to be totally irrotational would
yield unrealistic results.
Exercise: What will happen if we assume pipe flow with uniform entrance to be
inviscid and irrotational?
Inviscid and Irrotational Flow (cont’d)
Assuming external flow over a body to
be inviscid and irrotational everywhere
will result in zero air drag, which is not
correct. This is known as d’Alambert’s
paradox.
𝐹 drag =?
Note: There are other
factors that can cause
rotation, but they are
not as common as
viscous effects.
Exercise: Repeat the exercise of Slide 1-3 (derive BE) for irrotational flow.
- In an irrotational flow BE is valid between any two points of the flow field, not
necessarily two points on the same streamline.
BE for Irrotational Flow
1
3
𝑝
𝑉
2
1
=
𝑝
𝑉
2
2
=
𝑝
𝑉
2
3
Inviscid, irrotational
flow over an object
1 - 13
Exercise : The two-dimensional flow of a nonviscous, incompressible fluid in the
vicinity of a corner is described by the stream function
2 sin(2𝜃)
where 𝜓 has units of m
2 /s when 𝑟 is in meters. Assume the fluid density is
1000 kg/m
3 and the 𝑥𝑦 plane is horizontal.
a) Determine, if possible, the corresponding velocity potential.
b) If the pressure at point 1 on the wall is 30 kPa, what is the pressure at point 2?
Reference: Munson’s book.
Potential Flow Exercises (cont’d)
𝑥
𝑦
𝜃
1
2
1 m
0.5 m
1 - 14
Exercise : A horizontal slice through a tornado is
modeled by two distinct regions. The inner or core
region ( 0 < 𝑟 < 𝑅) is modeled by solid body rotation.
The outer region (𝑟 > 𝑅) is modeled as an irrotational
region of flow. The flow is 2D in the 𝑟𝜃-plane, and the
components of the velocity field are given by
𝑉 𝑟 = 0 , 𝑉 𝜃 =
𝜔𝑟 0 < 𝑟 < 𝑅
𝜔𝑅
2
𝑟
𝑟 > 𝑅
where 𝜔 is the magnitude of the angular
velocity in the inner region. The ambient
pressure (far away from the tornado) is
equal to 𝑝 ∞
. Obtain the shown nondimen-
sional pressure distribution.
Reference: Çengel’s book.
𝑟
𝑥
𝑦
𝜃
Inner
region
Outer region
0 1 2 3 4 5
𝑟/𝑅
𝑝 − 𝑝 ∞
𝜌𝜔
2 𝑅
2
0
Inner
region Outer region
Potential Flow Exercises (cont’d)
1
2
3
1
2
3
1
2
3
- To obtain complicated flow fields we can combine elementary ones such as
- Uniform flow
- Line source/sink
- Vortex
Superposition of Elementary Potential Flows
- Laplace’s equation is a linear PDE.
- Superposition can be applied to both velocity potential and streamfunction.
Potential flow 1 Potential flow 2 Potential flow 3
1. Uniform Flow
- Consider uniform flow in the 𝑥𝑦 plane in +𝑥 direction.
- Let’s find the equation for velocity potential.
= 0 → 𝑓 = constant
- Taking 𝑓 = 0 for simplicity
- Constant 𝜙 lines correspond to constant 𝑥 lines, i.e. lines parallel to the 𝑦 axis.
Exercise : Show that streamfunction equation is 𝜓 = 𝑈𝑦
1 - 17
1. Uniform Flow (cont’d)
- Constant 𝜙 and constant 𝜓 lines are shown below.
𝑥
𝑦
𝜙
=
𝜙
0
𝜙
=
𝜙
1
𝜓 = 𝜓 0
𝜓 = 𝜓 1
Exercise : Determine the equations of 𝜙 and 𝜓
for uniform flow in a direction making an angle
of 𝛽 with the 𝑥 axis.
𝑥
𝑦
𝛽
𝑈
- Consider the 2D flow emerging at the origin of the 𝑥𝑦 plane and going radially
outward in all directions with a total flow rate per depth of 𝑞.
- Conservation of mass: 𝑞 = 2 𝜋𝑟 𝑉 𝑟
𝑟 decreases with 𝑟, i.e. effect of source diminishes with 𝑟.
- Origin is a singular point with 𝑉 𝑟 → ∞, which is not physical, so don’t get too close.
1 - 18
2. Line Source at the Origin
View from the top
𝑥
𝑦
Constant
𝜙 lines
Streamlines
𝑞 : Flow rate per depth
(Strength of source [m
2 /s])
𝑟
𝜃
𝑥
𝑦
- Let’s find the equation for velocity potential.
𝑟
ln(𝑟) + 𝑓(𝜃)
𝜃
= 0 → 𝑓 = constant
- Taking 𝑓 = 0 for simplicity
ln(𝑟)
- Constant 𝜙 lines correspond to constant 𝑟 lines as shown in the previous slide.
Exercise : Show that the streamfunction equation is 𝜓 =
𝑞
2 𝜋
- To study a line sink for which the flow is radially inward towards a point, simply
use a negative 𝑞 value.
2. Line Source (cont’d)
- Consider a line source that is located NOT at the origin.
- Equations for 𝜙 and 𝜓 change as follows
ln 𝑟 1
1
or equivalently using 𝑥 and 𝑦 coordinates
ln 𝑥 − 𝑎
2
2
arctan
2. Line Source (cont’d)
Some useful relations
𝑥 = 𝑟 cos(𝜃)
𝑦 = 𝑟 sin(𝜃)
𝑟 = 𝑥
2
2
𝜃 = arctan
𝑦
𝑥
𝑥
𝑦
𝑟 1
1
𝑎
𝑟
(𝑥, 𝑦)
Exercise : Elementary components of a potential flow of water is shown below.
Find the velocity and pressure at point A if the pressure at infinity is 100 kPa.
1 - 25
Exercises for Elementary Potential Flows
Exercise : For the previous problem determine the equations of velocity potential
and streamfunction by superimposing elementary flows. Find the velocity at point
A by differentiating both 𝜙 and 𝜓 equation.
𝑈 = 3 m/s
𝑞 = 10 m
2 /s
𝑎 = 0. 8 m
𝑥 𝑏 = 0. 6 m
𝑦
Sink (−𝑞) Source (𝑞)
A
Exercise: Study the flow obtained by the combination of uniform flow in 𝑥
direction and a source at the origin. Obtain the location of the stagnation point(s)
and draw the stagnation streamline.
1 - 26
Source in a Uniform Flow
(Flow Past a Half Body)
Uniform flow: 𝑢 = 𝑈 , 𝑣 = 0 , 𝜙 = 𝑈𝑥 , 𝜓 = 𝑈𝑦
Source: 𝑉 𝑟
𝑞
2𝜋𝑟
𝜃
𝑞
2 𝜋
ln 𝑟 , 𝜓 =
𝑞
2 𝜋
𝑥
𝑦
𝑈
𝑞
Flow Past a Half Body (cont’d)
- Flow outside the stagnation streamline resembles a flow over
a body with a blunt nose.
- Equation of the half body is given by the equation of the
stagnation streamline.
𝑥
𝑦
𝑠
Movie
Flow Over Half Body
𝑈
𝑞/2𝜋𝑟
Stagnation point
𝑠
Stagnation
streamline
Flow Past a Half Body (cont’d)
Exercise : Consider the top part of a half body. Draw speed vs. 𝜃 and pressure vs. 𝜃
using the following values: 𝜌 = 1000 kg/m
3 , 𝑈 = 5 m/s , 𝑞 = 10 m
2 /s and
∞
= 100 kPa.
Exercise: A 64 km/h wind blows toward a hill that can be approximated with the
top part of a half body. The maximum height of the hill approaches 60 m.
a) What is the air speed at a point directly above the origin (at point 2)?
b) What is the elevation of point 2?
c) What is the pressure difference between point 2 and point 1 far from the hill?
Reference: Munson’s book
64 km/h
2
𝑥
60 m
1
- Superposition of
- a source of strength 𝑞 at 𝑥 = −𝑐,
- a sink of strength −𝑞 at 𝑥 = 𝑐 and
- uniform flow of magnitude 𝑈.
𝑢𝑛𝑖
𝑠𝑜𝑢
𝑠𝑖𝑛𝑘
𝑞
2 𝜋
ln 𝑟 1
𝑞
2 𝜋
ln(𝑟 2
𝑢𝑛𝑖
𝑠𝑜𝑢
𝑠𝑖𝑛𝑘
𝑞
2 𝜋
1
𝑞
2 𝜋
2
1 - 29
A Source and a Sink in Uniform Flow
(Flow Past a Rankine oval)
𝑥
𝑥
𝑦
𝑐 𝑐
A
𝑟 1 𝑟 2 𝜃 1
𝜃 2
1 - 30
Flow Past a Rankine oval (cont’d)
Exercise : Determine the location of the stagnation points of the shown Rankine
oval. Determine its length and thickness of the oval. Plot the variation of speed
and pressure (with respect to 𝑝 ∞ ) on it as a function of 𝜃.
𝑞
𝑥
𝑦
−𝑞
𝑐 𝑐
Doublet
- Superposition of
- a source of strength 𝑞 at the orgin (moved from – 𝑥 axis to the origin),
- a sink of strength −𝑞 at the origin (moved from +𝑥 axis to the origin) ,
- Consider the limiting case of the source and sink of Slide 1 - 29 approaching to the
origin. Skipping the details we can get
𝑑𝑜𝑢𝑏𝑙𝑒𝑡
cos 𝜃 , 𝜓 𝑑𝑜𝑢𝑏𝑙𝑒𝑡
sin 𝜃
where 𝑑 is the strength of the doublet.
- Velocity field is given by
𝑟
2
cos 𝜃
𝜃
2
sin 𝜃
𝑥
𝑦
Constant 𝜙 lines
Streamlines
A Doublet in Uniform Flow (Flow Past a Cylinder)
- Superposition of
- a doublet of strength 𝑑 at the origin.
- uniform flow of magnitude 𝑈 in +𝑥 direction.
𝑢𝑛𝑖
𝑑𝑜𝑢𝑏𝑙𝑒𝑡
𝑑
2 𝜋𝑟
cos(𝜃)
𝑢𝑛𝑖
𝑑𝑜𝑢𝑏𝑙𝑒𝑡
𝑑
2 𝜋𝑟
sin(𝜃)
𝑥
𝑦
𝑑
1 - 37
Flow Past a Cylinder (cont’d)
Exercise : When a small circular cylinder is placed in a uniform stream, a
stagnation point is created on the cylinder. If a small hole is located at this point,
the stagnation pressure, can be measured and used to determine the approach
velocity, 𝑈 (similar to a Pitot tube).
a) Show how 𝑝 𝑠𝑡𝑎𝑔
and 𝑈 are related. Pressure far away is 𝑝 ∞
b) If the cylinder is misaligned by an angle 𝛼, but the measured pressure is still
interpreted as the stagnation pressure, use potential flow theory to determine an
expression for the ratio of the true velocity, 𝑈, to the predicted velocity, 𝑈′. Plot
this ratio as a function of 𝛼 for the range 0 ° < 𝑎 < 20 °.
Reference: Munson’s book.
𝑈
𝑅
Stagnation point
𝑈
𝛼
1 - 38
Flow Past a Rotating Cylinder
- Superposition of
- a doublet of strength 𝑑 located at the origin,
- CCW rotating irrotational vortex of strength Γ
located at the origin,
- uniform flow of magnitude 𝑈.
- This will result in
2
2
cos 𝜃 +
2
2
sin 𝜃 −
ln 𝑟
- This is the potential flow that resembles the flow over a rotating cylinder of
radius 𝑅.
with 𝑅 = 𝑑/𝑈
𝑥
𝑦
𝑑
Γ
Flow Past a Rotating Cylinder (cont’d)
Exercise : For the flow shown above, obtain the following results
𝑉 𝜃 𝑐𝑦𝑙
= −2𝑈 sin 𝜃 +
Γ
2𝜋𝑅
𝑐𝑦𝑙
∞
𝜌𝑈
2
1 − 4 sin
2 𝜃 +
2Γ
sin 𝜃 −
Γ
2
Integrate the above pressure distribution to get the following results
𝑑𝑟𝑎𝑔
𝑙𝑖𝑓𝑡 = −𝜌Γ𝑈 (per unit depth)
𝑥
𝑦
𝑈
𝑑
Γ
𝑅
Flow Past a Rotating Cylinder (cont’d)
Γ
4 𝜋𝑈𝑅
< 1
Γ
4 𝜋𝑈𝑅
Γ
4 𝜋𝑈𝑅
= 1
- Streamlines and stagnation points for different circulation values.
White’s book
𝑅
- Magnus Effect: A rotating body in a uniform flow will have a net lift force on it (1853).
- Direction of the lift force depends on the direction of 𝑈 and Γ.
Exercise: Determine the direction of
the lift force.
1 - 41
Magnus Effect
Γ
𝑈
𝑈
Γ
Top
𝑉 𝜃 𝑡𝑜𝑝
= − 2 𝑈 +
Γ
2𝜋𝑅
Opposite signs
Velocity is low.
Pressure is high.
Bottom
𝑉 𝜃 𝑏𝑜𝑡𝑡𝑜𝑚
= 2 𝑈 +
Γ
2𝜋𝑅
Same signs
Velocity is high.
Pressure is low.
1 - 42
Magnus Effect (cont’d)
Exercise : Magnus effect acts not only on cylinders but also on other rotating bodies
such as spheres. It can be used to explain how a spinning ball moves in a curved
trajectory. A football player wants to make a penalty kick as sketched below. Will a
CW or a CCW spin do the trick?
Exercise : Watch the following movies
https://www.youtube.com/watch?v=2OSrvzNW9FE (Suprising applications of Magnus effect)
http://www.youtube.com/watch?v=23f1jvGUWJs (Magnus force on Veritasium channel)
http://www.youtube.com/watch?v=2pQga7jxAyc (Enercon's rotor ship. Audio in German)
http://www.youtube.com/watch?v=wb5tc_nnMUw (Roberto Carlos knows the Magnus force)
Magnus Effect (cont’d)
- Warning: Although potential flow theory can predict the direction of lift force
due to Magnus effect correctly, it may give quite inaccurate values for its
magnitude. We’ll come back to this in the next chapter.
Exercise: In 1920s Anton Flettner built a series of rotor ships that are propelled
by rotating cylinders driven by electric motors. Read about Flettner’s ship at
rexresearch.com/flettner/flettner.htm and understand how it works.
One of Flettner’s original rotating cylinder ships https://en.wikipedia.org/wiki/E-Ship_
Kutta Condition (Lift on an Airfoil)
- Magnus effect applies not only to cylinders but any closed shape.
- Consider the flow over a slender body with a sharp trailing edge, such as an airfoil.
- An airfoil is designed to generate small drag and high lift force.
- There are two stagnation points, 𝑠 1 and 𝑠 2
- Experiments show that the streamlines leave the trailing edge smoothly as shown
above, known as the Kutta condition.
𝑠 1
𝑠 2
1 - 49
Numerical Solution of Potential Flow
- Obtaining complicated flow fields by superposing elementary ones is limited.
- To study potential flows on arbitrary geometries one can perform a numerical
solution.
- Consider a flow inside an expanding duct (coordinates are in meters).
- Potential flow inside the duct can be obtained by solving Laplace’s equation
2 𝜓
2
2 𝜓
2
= 0 , with proper boundary conditions
10 m/s
5 m/s
(0,1)
(0,2)
(1,1)
(2,0)
(3,0)
(3,2)
𝑥
𝑦
1 - 50
Numerical Solution of Potential Flow (cont’d)
- Boundary conditions are (study in the given order)
Bottom wall is a streamline. 𝜓
should be constant there. Let’s set
it to zero.
bottom
At the inlet 𝑢 = 10.
Therefore
𝜕𝜓
𝜕𝑦
𝜓 varies linearly from 0 to 10.
left
Top wall is a streamline. 𝜓 should be constant
there. In order to have 10 m
2 /s flow rate per
depth between the top and bottom walls
top
At the exit 𝑢 = 5.
Therefore
𝜕𝜓
𝜕𝑦
𝜓 varies linearly from 0 to 10.
𝜓 right = 5𝑦
1
2
4
Numerical Solution of Potential Flow (cont’d)
- Finite Difference Method can be used to get the numerical solution.
- First we discretize the problem domain into a set of nodes.
- Following mesh has 55 nodes with Δ𝑥 = Δ𝑦 = 1 / 3.
- 27 of the nodes are at the boundaries and 𝜓 is known at these nodes.
- 28 of the nodes are inside the domain and 𝜓 needs to be calculated at them.
Δ𝑥
Δ𝑦
𝑖, 𝑗
𝑖 + 1 , 𝑗 𝑖 − 1 , 𝑗
𝑖, 𝑗 + 1
𝑖, 𝑗 − 1
5 point computational stencil
Numerical Solution of Potential Flow (cont’d)
- Discretized form of the Laplace’s equation at node (𝑖, 𝑗) is
𝑖+ 1 ,𝑗
𝑖,𝑗
𝑖− 1 ,𝑗
2
𝑖,𝑗+ 1
𝑖,𝑗
𝑖,𝑗− 1
2
- For Δ𝑥 = Δ𝑦, discretized equation for node (𝑖, 𝑗) becomes
𝑖+ 1 ,𝑗
𝑖− 1 ,𝑗
𝑖,𝑗+ 1
𝑖,𝑗− 1
𝑖,𝑗
- This equation needs to be written for all non-boundary nodes.
- For nodes with boundary neighbors, some 𝜓 values are known and they need to be
transferred to the right-hand-side of the equation.
- At the end we’ll get a system of 28 equations for 28 unknowns and solve it.
2 𝜓
2
𝑖,𝑗
2 𝜓
2
𝑖,𝑗
- Following solution is obtained using a mesh with Δ𝑥 = Δ𝑦 = 0. 2.
- After obtaining the 𝜓 values at the nodes, velocity components can be obtained
𝑖,𝑗
𝑖,𝑗+ 1
𝑖,𝑗− 1
𝑖,𝑗
𝑖+ 1 ,𝑗
𝑖− 1 ,𝑗
- Different formulas need to be used at the boundary nodes.
Numerical Solution of Potential Flow (cont’d)
White’s book
𝝍 = 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 1 0.0 0 10.00 10.00 10.
8.00 8.02 8.04 8.07 8.12 8.20 8.30 8.41 8.52 8.62 8.71 8.79 8.85 8.91 8.95 9.
6.00 6.03 6.06 6.12 6.22 6.37 6.58 6.82 7.05 7.26 7.44 7.59 7.71 7.82 7.91 8.
4.00 4.03 4.07 4.13 4.26 4.48 4.84 5.24 5.61 5.93 6.19 6.41 6.59 6.74 6.88 7.
2.00 2.02 2.05 2.09 2.20 2.44 3.08 3.69 4.22 4.65 5.00 5.28 5.50 5.69 5.85 6.
𝝍 = 0 .00 0.00 0.00 0.00 0.00 0.00 1.33 2.22 2.92 3.45 3.87 4.19 4.45 4.66 4.84 5.
0.00 1.00 1.77 2.37 2.83 3.18 3.45 3.66 3.84 4.
0.00 0.80 1.42 1.90 2.24 2.50 2.70 2.86 3.
0.00 0.63 1.09 1.40 1.61 1.77 1.89 2.
0.00 0.44 0.66 0.79 0.87 0.94 1.
0.00 0.00 0.00 0.00 0.00 0.
Red ones are
boundary values Black ones are
calculated
Numerical Solution of Potential Flow (cont’d)
- After obtaining 𝑢 and 𝑣, pressures
can be calculated using the Bernoulli
equation.
modifications need to be done.
- Red node is NOT at ∆𝑥 distance from
the central node. What can be done?
White’s book
𝜓 = 10
4
6
8
Lower wall
Upper
wall
𝐶 𝑝 =
𝑝 − 𝑝 𝑖𝑛
𝜌𝑉 𝑖𝑛
2 / 2
𝑉 𝑖𝑛
𝑖𝑛
