Fluid Analysis
A simulated response in ANSYS
By Terence James Haydock
1
Overview
• Introduction
• Background
o Density and pressure
o Example and Problem
• ANSYS Tutorial
• Summary
• References
2
Introduction
Aim: To calculate a fluid analysis problem via analytical methods
and compare such results with a numerical analysis with ANSYS
3
Background
A fluid is a material that offers no permanent resistance to change of
shape i.e. its flow matches the shape of the vessel to which its found.
The general laws of mechanics for solids apply equally to fluids.
Acting force, equilibrium, momentum and energy all have the same
meaning, hence, Newton’s law are just as valid for solids and fluids.
4
Density and Pressure
Density (ƿ) is found by dividing mass by volume and its SI units are kg/m^3. For
example, water has a density of 1000kg/m^3. The density of other liquids are
defined as a ratio of water (relative density or specific gravity) e.g. oil has a
relative density of 0.9.
When a fluids is not in motion is exerts pressure in all directions. The unit of
pressure is N/m^2, though it is usually quoted in bar e.g. 100kN/m^2 which equals
10^5N/m^2 or 10^5Pa. Consider a vertical tube of water, at a height h, if the
density of a liquid is ƿ and the cross sectional area is A, then the volume of liquid
is Ah and its mass is ƿAh. The weight of the liquid is ƿgAh. The pressure at any
cross section due to liquid is:
P = force / area
= ƿgAh / A
= ƿgh
5
Cont.
The height h is also known as the pressure head, thus, pressure maybe given in
terms of head. All liquids have the same pressure at the same level (2 & 3) and as
such, the required pressure is given by the column of liquid above 2. If the column
is sealed (vacuum above liquid) the value is known as absolute pressure, if open,
then the liquid gives a gauge value, both are related by:
Absolute pressure = gauge pressure + atmospheric pressure
6
The stresses in the walls of a pressure vessel rely on the difference between
external and internal pressures, with external pressure been atmospheric,
thus, gauge pressure must also be used. In fluid mechanics the change
in pressure is required whether the separate values are gauge
or absolute.
Example
The pressure at the top of a mountain is found to be 0.7bar. Express this in
metres of water. What would be the reading of a mercury barometer at this
point? Relative density of mercury, 13.6
Solution: The density of water maybe taken as 1000kg/m^3. Working in SI
units, the equivalent head h is given by:
pgh = 0.7 x 10^5 or
h=
0.7 𝑥 10^5
𝑝𝑔
=
0.7 𝑥 10^5
=
1000 𝑥 9.81
7.14m
The corresponding head for mercury is thus:
7.14 / 13.6 = 0.525m or525mm
7
Problem
The cross section of a circular pipe is occupied by a constant flow of water. The
water gradually expands due to varied diameters at either end, with one end
been 0.2m diameter and the other with a 0.3m diameter. If the velocity at the
first section is 5m/s, find:
a)
The volume flow rate,
b) The velocity at the second section,
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Area A1
Area A2
v2
v1
Use subscripts for the two sections
a)
The volume flow rate Q is:
Q  v1 A1  5 *

4
* 0.2 2  0.1571m3 / s
b) The equation of continuity is: Q  v1 A1  v2 A2
So the second velocity, after transposition is:
A1
( / 4) * 0.2 2
v2  v1 *
 5*
 2.222m / s
2
A2
( / 4) * 0.3
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Area A2
v2
Area A1
z2
v1
If the question was further expanded
and we were to describe the second area
of 0.3m as been 2m above the first area
of 0.2m, then what would the pressure
head be?
z1
2
Tip: Bernoulli's equation would be used,
as follows:
10
2
v1
p1
v2
p2
z1 

 z2 

2 g pg
2 g pg
Transpose the equation and note the level difference ( z2 – z1 ), is 2m
pressure difference and expressed as the head is as follows:
p1 p2
v2
v1
2.222 2  52

 z2  z1 

 2
 0.977m
pg pg
2g 2g
2 * 9.81
2
2
The pressure difference is 0.977m of water greater at the first section.
In N/m^2 this becomes:
p1  p2  pg * 0.977  1000 * 9.81* 0.977
 9.5 *10^3N / m 2
or
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9.5kN / m 2
ANSYS Tutorial
Geometry
We start of with our first sketch, apply the correct
dimensions of 0.2m and 0.3m, to represent the inlet
and outlet. The distance in-between is set at 0.5m
12
Mesh
As shown, the mesh has been applied using the advanced size function, on: proximity and
curvature, with the rest of the options left at default. The result can be seen in a well
structured mesh.
13
Right-click on
Default Domain in
the Outline tree
Select Rename
The domain name
can now be edited.
Change the domain
name to Pipe
Set the Material to Water.
The available materials can
be found in the drop-down menu
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Our next aim is to define the inlet an outlet of the pipe. By
right clicking on pipe, going to inset and then boundary, we
are given the options to do so. There are other variables to
choose from however, this options are for more
complicated examples.
Click the Boundary Details
tab. Enter a value of 5 for
Normal Speed. The default
units are [m s^-1]
15
Set Relative Pressure to 0
[Pa]. This is relative to the
domain Reference Pressure,
which is 1 [atm]
Click Start Run to begin the solution process. 45 iterations are required to reduce
the RMS residuals to below the target of 1.0x10-4. The pressure monitor points
approach steady values
The Solver Control
options set various
parameters that are
used by the solver and
can
affect
the
accuracy
of
the
results. The default
settings
are
reasonable, but will
not be correct for all
simulations. In this
case
the
default
settings will be used,
but you will still look
at what those defaults
are.
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An expression will
be used to define
the monitor point. 7.
Set
Option
to
Expression
Enter
the
expression:
areaAve(Pressure)
@inlety in the
Expression
Value
field The expression
calculates the area
weighted average of
pressure at the
boundary inlet.
When the results are loaded, CFD-Post displays the outline (wireframe) of
the model. The icons on the viewer toolbar control how the mouse
manipulates the view
You can create
many different
objects in CFDPost. The Insert
menu shows a
full list, but
there are toolbar
shortcuts for all
items.
Such as: Location: Points, Lines, Planes, Surfaces, Volumes
17
Summary
The results shown in ANSYS are very accurate when compared to the analytical
method we applied earlier. As such, we have proved the correct results for both
analytical and numerical application. Learning objectives covered in this
workshop relate to fluid analysis, with knowledge gained in density and pressure
when applied to real world application, furthermore the use of ANSYS for
verification has also proven to be a small example its potential.
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References
Norman. E. et al. Advanced Design and Technology. 1995
Longman Group UK Limited, ISBN 0 582 014638
ANSYS 13.0 Help files
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