ES4400
THE UNIVERSITY OF WARWICK
FOURTH YEAR EXAMINATION: JUNE 2008
COMPUTATIONAL FLUID DYNAMICS
Time Allowed: 1 hour
Read carefully the instructions on the answer book and make sure that the particulars required
are entered on each answer book.
Calculators are not needed and are not permitted in this examination.
ANSWER 2 QUESTIONS.
If you have answered more than the required 2 questions in this examination, you will only be
given credit for your 2 best answers.
The numbers in the margin indicate approximately how many marks are available for each
part of a question.
This is a collection of exam questions with many of their solutions used in MA4G8/CY905,
ES440, and the maths 2nd year module on ODEs that might be appropriate for ES440,
CO906, CY905 exams/tests. Questions on exams/tests will be based upon these questions
and the supporting material in the lecture notes.
Questions (or parts of these questions) that might appear on the CY905 exam are noted by
boldface CY905. Whether a particular question is appropriate for CY905 in a given year is
also noted.
1. Types of Partial Differential Equations 2007-A1 CY905
a) Discuss the advantages and limitations of CFD with respect to experimental methods
in around 100 words.
(10 marks)
Answer:
(A) Advantages of CFD:
– CFD allows fluid properties and flow parameters, such as velocity, pressure,
and temperature to be evaluated at any point in the flow. This provides a
better understanding of the flow phenomenon.
– The use of properly calibrated CFD models can reduce or even eliminate the
need for expensive or large scale test facilities.
– CFD can be used to investigate congurations which may be too large to test
or which pose a signicant safety risk.
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ES4400
(B) Limitations of CFD: YMC answer:
– Generally, accuracy depends on algorithms and grid quality.
– Turbulence is rarely resolved except through use of some form of models.
– DNS is possible only for flows at relatively low Reynolds numbers and in
geo- metrically simple domains.
– LES remains a fairly computationally intensive technique.
– The complexity of turbulence makes it unlikely that any single turbulence
model will be able to represent all turbulent flows.
Limitations from my notes:
– Limited regimes Highest Reynolds, Froude and Rayleigh numbers are out
of reach.
– Complex geometries Unstructured grids can only take you so far. Moving
interfaces are even worse.
– Exotic regimes Regimes where kinetic theory must be mixed with continuum mechanics. Hypersonic. Solar corona.
– Combustion Reduced equations are very approximate.
– Radiation Coupling radiation to flows is still primitive.
– And all the issues of convergence and accuracy below.
b) Simple partial differential equations (PDEs) can be classified into three types: elliptic, parabolic and hyperbolic types. Describe each type of PDE and their traditional
solution methods in around 100 words each.
(i) Parabolic
Answer:
(5 marks)
• Parabolic equations describe time-dependent problems with involve signicant amount of dissipation.
• Examples are unsteady viscous flows and unsteady heat conduction.
• The prototype parabolic equation is the diusion equation.
• This type of problem is termed an initial-boundary-value problem.
• A disturbance at a point in the interior of the solution domain can only influence events at later times.
• Information travels only downstream in parabolic equations.
• Only one initial condition is normally required.
(ii) Elliptic
Answer:
(5 marks)
• Steady state or equilibrium problems are governed by elliptic equations.
• The prototype elliptic equation is Laplaces equation which describe the irrotational flow of an incompressible fluid and steady state conductive heat
transfer.
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• Since information may travel upstream as well as downstream, one cannot
apply conditions only at the upstream end of the flow.
• A unique solution to all elliptic equations can be obtained by specifying
conditions on all the boundaries of the solution domain.
• This type of problem is termed an boundary-value problem.
• An important feature of elliptic problems is that a disturbance in the interior
of the solution changes the solution everywhere else.
• Disturbance signals travel in all directions through interior solution.
• Consequently, the solutions to physical problems described by elliptic equations are always smooth even if the boundary conditions are discontinuous.
(iii) Hyperbolic
Answer:
(5 marks)
• Information propagates at nite speeds in two sets of directions.
• The prototype hyperbolic equation is the wave equation.
• Solution to all hyperbolic equations can be obtained by specifying two initial
condi- tions and one boundary condition.
• This type of problem is also termed an initial-boundary-value problem.
• Disturbance at a point can only influence a limited region in space.
• The speed of disturbance propagation through an hyperbolic problem is nite.
(25 marks total)
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2. Finite Difference Method 2007-A2, 2006-1 CY905
Discretisation of many equations requires the approximation of the first derivatives,
and the second derivative term,
∂ 2f
.
∂x2
∂f
,
∂x
a) Using diagrams where appropriate, explain briefly forward, backward, and central
∂f
difference approximations for the first derivative ( ), and comment on the accuracy
∂x
of each approximation.
(5 marks)
Answer:
• Forward difference: f (xi ) and f (xi + ∆x). First-order accuracy.
• Backward difference: f (xi − ∆x) and f (xi ). First-order accuracy.
• Central difference: f (xi − ∆x) and f (xi + ∆x). Second-order accuracy.
b) Consider a uniform grid with grid spacing ∆x. Using the central-difference schemes
∂f
(CDS), derive the second-order finite difference formulae for the first-order ( ) and
∂x
∂ 2f
(5 marks)
second-order ( 2 ) derivatives from Taylor series expansions.
∂x
Answer: Finite dierence formulas can be easily derived from Taylor series expansions. The widely used central dierence formula can be obtained from subtraction of
two Taylor series expansions; assuming uniformly spaced data we have
fi−1 = fi − ∆x
∂f
∆x2 ∂ 2 f
+
+ O(∆x3 ) ,
∂x
2 ∂x2
fi+1 = fi + ∆x
∆x2 ∂ 2 f
∂f
+
+ O(∆x3 ) ,
∂x
2 ∂x2
This leads to
∂f
fi+1 − fi−1
≈
,
∂x
2∆x
∂ 2f
fi+1 − 2fi + fi−1
≈
.
2
∂x
∆x2
∂f
c) Derive the fourth-order central-difference formula for the first-order ( ) derivative
∂x
from Taylor series expansions.
(5 marks)
d) Derive the modified wave-number for the second-order finite difference formula for
∂f
in part (b).
(10 marks)
∂x
Answer: Similarly
fi−2
∂f
4∆x2 ∂ 2 f
= fi − 2∆x
+
+ O(∆x3 ) ,
2
∂x
2 ∂x
∂f
∆x2 ∂ 2 f
+
+ O(∆x3 ) ,
∂x
2 ∂x2
∆x2 ∂ 2 f
∂f
= fi + ∆x
+
+ O(·§3 ) ,
2
∂x
2 ∂x
fi−1 = fi − ∆x
fi+1
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4∆x2 ∂ 2 f
∂f
+
+ O(·§3 ) ,
∂x
2 ∂x2
Now subtract the fi−2 from the fi+2 and the fi−1 from the fi+1 lines.
fi+2 = fi + 2∆x
fi+2 − fi−2 = 4∆x
∂f
8∆x3 ∂f
+
∂x
3 ∂x
fi+1 − fi−1 = 2∆x
∂f
∆x3 ∂f
+
∂x
3 ∂x
Subtract these two lines to get
8(fi+1 − fi−1 ) − fi+2 − fi−2 = (16 − 4)∆x
This leads to
To check:
∂f
∂x
−fi+2 + 8fi+1 − 8fi−1 + fi−2
∂f
≈
.
∂x
12∆x
8 fi+1 − fi−1 1 fi+2 − fi−2
4 1 ∂f
−
≈
−
6
2∆x
3
4∆
3 3 ∂x
(25 marks total)
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3. Ordinary Differential Equations 2006-2
A laminar (Prandtl) boundary layer can be solved by assuming that the local velocity vector
is u = (u, v) where u = ∂ψ/∂y and v = −∂ψ/∂x. The stream function ψ = U g(x)f (η)
where U is the free stream fluid velocity and η = y/g(x) is the similarity variable with
g(x) = (2νx/U )1/2 . After some manipulation and a similarity transformation, it can be
shown that a laminar boundary layer on a flat plate is self-similar and is governed by the
Blasius equation:
f 000 + f f 00 = 0
(1)
We wish to solve for f and its derivatives throughout the boundary layer.
Answer: Shooting to solve the Blasius boundary layer. A laminar boundary layer
on a flat plate is self-similar and is governed by above. f ∝ ψ the stream function and
f 0 = u/U where U is the free-stream velocity; and f 00 ∝ τ the shear stress. Boundary
conditions for the equations are derived from the physical boundary conditions.
We wish to solve for f and all its derivatives. Since one of the B.C. is prescribed at
η = ∞ we are required to solve a NL BV problem. Solution proceeds by breaking it up.
The solution will be advance from the prescribed condition at the wall η = 0 to η = ∞
Solutions have been found to converge very quickly for large η and marching from η = 0
to η = 10 has been shown to be sufficient for accurate solution. Two conditions at the
wall: f1 = 0 and f = 0. We must repeatedly solve the whole system and iterate to find the
value of f2 (0) that gives the required condition f1 = 1 at η = ∞. Two initial gueses were
(0)
(1)
made for f2 (0): f2 (0) = 1 and f2 (0) = 0.5 . From these two initial guesses two values
(0)
(0)
of f1 at ∞ were calculated: f1 (10) and f1 (10). Starting with these two calculations, the
secant method may be used to iterate toward an arbitrarily accurate value for f2 (0) based
on the following adaption
(α)
(α+1)
f2
(α)
(0) = f2 (0) +
(α−1)
f2 (0) − f2
(α)
f1 (10)
−
(0)
(α−1)
f1
(10)
(α)
1 − f1 (10)
An acceptable final solution of f2 (0) = f 00 (0) = 0.4690
a) Solution proceeds by breaking the third-order differential equation into a coupled set
of first-order differential equations. Convert (1) to a system of 3 first-order differential equations.
(9 marks)
0
00
Answer: Define: f1 = f , f2 = f . Then
f20 = −f f2
f10 = f2
f 0 = f1
(2)
b) Because this is a third-order problem, three boundary conditions at either η = 0 or
η = ∞ are needed. Boundary conditions for the equation are derived from the physical boundary conditions on the fluid velocity: two defined by the no-slip condition
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ES4400
at the wall and a third defined by the free stream conditions at large distances from
the wall.
u(η = ∞) = U
How does this translate into a boundary condition on f or one of its derivatives at
η = ∞?
(2 marks)
0
0
Answer: u(η = ∞) = U = U f (∞). Therefore f (∞) = 1.
c) There is an iteration that is taught in some fluids modules where it can be shown that
this leads to: f 00 (0) = 0.4690
What are the no-slip boundary conditions at η = 0 on u and v?
(2 marks)
d) How can these boundary conditions be rewritten in terms of boundary conditions on
ψ and thus f (η) and its derivatives?
(6 marks)
Answer: The three are summarised as
f (0) = f 0 (0) = 0,
f 0 (∞) = 1
.
e) Explain how to use these boundary conditions in a solution.
Answer: They would provide the values of f (0), f 0 (0) and f 00 (0) for the first iteration step.
f) Solve the ODE for two time steps (δ = ∆η = 0.1) using the Euler method in terms
of f (0), f1 (0) = f 0 (0), f2 (0) = f 00 (0) and δ.
(6 marks)
Answer:
f2 (δ) = f2 (0)
f1 (δ) = f2 (0)δ f0 (δ) = 0
f2 (2δ) = f2 (0)
f1 (2δ) = 2f2 (0)δ
f0 (δ) = f2 (0)δ 2
(25 marks total)
4. Partial Differential Equation 2006-3 CY905
Two-dimensional heat conduction can be described by the Poisson equation,
∂ 2T
∂ 2T
+
=f
∂x2
∂y 2
(3)
where T is the temperature and f (x, y) is the source term. We are interested in the steady
state solution in the unit square, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
a) Discretise the differential equation (3) using the central difference formula for the
second derivative and derive the finite difference equation.
(5 marks)
b) There are many different solution algorithms for two-dimensional problems. Explain
how to solve the system of linear algebraic equations including the ADI method and
comment on numerical errors.
(10 marks)
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c) Consider the one-dimensional case:
∂ 2T
∂ 2T
+
=f
∂x2
∂y 2
(4)
with the following boundary conditions: T (0) = 0, T (1) = 1.
Discretise the computational domain from x = 0 to x = 1 using N points (including
the boundary points). Discretise the differential equation (4) and write the system of
linear algebraic equations in matrix form AT = f . Explain how to solve this system
including the Tri-Diagonal Matrix Algorithm (TDMA) and comment on numerical
errors in comparison with the two-dimensional case in part (b).
(10 marks)
(25 marks total)
5. Advection-Diffusion 2006-4
The one-dimensional generic transport equation with constant velocity (u), constant fluid
properties (κ = ρ/Γ), and no source terms is given by:
∂f
∂ 2f
∂f
= −u
+κ 2
∂t
∂x
∂x
(5)
This equation is often used in the literature as a model equation for the Navier-Stokes equations. We assume that the spatial derivatives are approximated using central-difference
schemes (CDS) and that the grid is uniform in x-direction. Derive the algebraic equation
using the following methods and explain the accuracy and stability of each method.
a) Explicit Euler method.
(5 marks)
b) Implicit Euler method.
(5 marks)
c) Crank-Nicolson method.
(5 marks)
d) Solution of the Navier-Stokes equations is complicated by the lack of an independent
equation for the pressure, whose gradient contributes to each of the three momentum
equations. Furthermore, the continuity equation does not have a dominant variable
in incompressible flow. Mass conservation is a kinematic constraint on the velocity
field rather than a dynamic equation. Explain how to construct the pressure field both
in compressible and incompressible flows.
(10 marks)
(25 marks total)
6. Mathematical Classification of PDEs 2005-1
Simple partial differential equations (PDEs) can be classified into three types: elliptic,
parabolic and hyperbolic types
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a) Describe elliptic and parabolic types of PDE and their traditional solution methods
in around 100 words each. CY905
(10 marks)
b) The Navier-Stokes equations and its reduced forms can be classified similarly into
three types (elliptic, parabolic and hyperbolic). Determine the type of the following
PDEs:
(i) Steady Navier-Stokes equations for viscous flow;
(3 marks)
(ii) Unsteady Navier-Stokes equations for viscous flow;
(3 marks)
(iii) Steady boundary-layer equations;
(3 marks)
(iv) Steady Euler equations;
(3 marks)
(v) Unsteady Euler equations.
(3 marks)
(25 marks total)
7. Coordinate Transformation and Non-uniform Grid 2005-3 2007-B1 (Not CY905 in 2010)
The computation of flowfields in and around complex geometries, such as complete aircraft or automobiles, involves computational boundaries that do not coincide with coordinate lines in physical space. Generally, a mapping or transformation is used from physical
space (x, y) to a computational space (ξ, η).
∂u ∂v
+
= 0 into generalised coordinates.
(5 marks)
∂x ∂y
Using the chain rule, the derivatives in the continuity equation can be
(5 marks)
a) Transform the continuity equation
Answer:
written as
∂u ∂ξ ∂u ∂η
∂u
=
+
,
∂x
∂ξ ∂x ∂η ∂x
∂v
∂v ∂ξ ∂v ∂η
=
+
.
∂y
∂ξ ∂y ∂η ∂y
(6)
(7)
So, the continuity in the general coordinates is
(5 marks)
∂u ∂ξ ∂u ∂η ∂v ∂ξ ∂v ∂η
+
+
+
= 0.
∂ξ ∂x ∂η ∂x ∂ξ ∂y ∂η ∂y
(8)
∂U
∂V
∂x
∂y
∂y
∂x
b) Show that
+
= 0 on a conformal grid
= −
and
=
where
∂η ∂η ∂ξ
∂η
∂ξ
∂ξ
1 ∂ξ
∂η
1
∂ξ
∂η
U =
v+
u and V =
− u+
v . J is the Jacobian matrix of
J ∂y
∂y
J
∂y
∂y
the transformation. Use the following relations
∂ξ
1 ∂y
=
,
∂x
J ∂η
∂ξ
1 ∂x
=
,
∂y
J ∂η
∂η
1 ∂y
=−
,
∂x
J ∂ξ
and
∂η
1 ∂x
=
,
∂y
J ∂ξ
(10 marks)
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c) Consider a one-dimensional case, for example
anon-uniform grid, ξ = g(x).
Derive
∂f
∂ 2f
expressions for calculation of first-order
and second-order
deriva∂x
∂x2
∂ 2f
∂f
and 2 .
tives on a non-uniform grid using
(10 marks)
∂ξ
∂ξ
Answer: For the transformation ξ = g(x), we use the chain rule to transform the
derivatives to the new coordinate system
(5 marks)
dξ df
df
=
,
dx
dx dξ
df
df
= g0 .
dx
dξ
(9)
(10)
Similarly,
d2 f
d
0 df
g
,
=
dx2
dx
dξ
2
d2 f
00 df
0 2d f
=
g
+
(g
)
.
dx2
dξ
dξ 2
(11)
(12)
(25 marks total)
8. Coordinate Transformation and Non-uniform Grid 2005-5. Not my half of ES440, but
maybe last half of MA4G8.
Consider the resolution requirements for Direct Numerical Simulation (DNS).
a) First, for simplicity, we can look at the example of homogeneous isotropic turbulence. The Kolmogorov length scale is η = (ν 3 /)1 /4, the integral length scale is
l = q 3 /, and the Reynolds number is Re = q 4 /(ν). (Here, ν is the kinematic
viscosity, is the dissipation rate and q is a velocity scale.)
(i) Determine the ratio of the largest to the smallest scales that must be resolved in
each direction in a DNS of homogeneous isotropic turbulence as a function of
Re.
(5 marks)
(ii) Show that the number of grid points required for a three-dimensional DNS of homogeneous isotropic turbulence is proportional to Rem and determine the value
of m.
(7 marks)
b) The resolution requirements for turbulent channel flow are slightly different from
those for homogeneous isotropic turbulence. The ratio of the channel width to the
scale of the smallest eddies in the channel varies with Re0 .9.
(i) Determine the number of grid points required for a channel flow DNS as a function of Re.
(5 marks)
(ii) Kim, Moin & Moser (1987) used approximately 2 × 106 grid points in a DNS of
channel flow at Re = 3300. Determine the number of grid points required for a
DNS of channel flow at Re = 13800.
(8 marks)
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(25 marks total)
9. Resolution Requirements ES440-2010-4 Consider the resolution requirements for Direct
Numerical Simulation (DNS). Recall that the Kolmogorov length scale is η = (ν 3 /)1/4 .
Define q = urms as the scale of velocity fluctuations.
a) Homogeneous isotropic turbulence.
In this case the integral length scale is L = q 3 /, and the Reynolds number is Re =
qL/ν = q 4 /(ν).
(i) Assuming that the smallest dynamical scale in homogeneous, isotropic turbulence goes as η, determine the ratio of the largest to the smallest scales that in a
DNS of homogeneous isotropic turbulence as a function of Re.
(3 marks)
3
3/4 1/4
3
3/4
3/4
Answer: L/η = (q /)/(ν / ) = q /(ν) = Re
b) Channel flow
30.0
Leray
Yoshizawa
Kosovic
Measurements
A typical boundary layer mean velocity profile is shown. All scaling within the boundary layers is determined by the friction velocity
√
u∗ = −uv and the wall thickness viscosity
y + = ν/u∗ . The transition between the viscous sublayer and the log layer is at y = 10y + .
u+
20.0
10.0
0.0
0
1
10
100
1000
10000
y+
(i) What two length scale must be considered in resolving turbulent channel flow?
In which directions do they apply, which regions and where is refined mesh
spacing most needed?
(5 marks)
Answer: Both the boundary layers and the turbulence in the interior must be
resolved. The two relevant length scales are the local Kolmogorov scales and
y + in the boundary layer. Due to the shears in the boundary layers, refined mesh
spacing, as small as 2y + near the wall, is usually applied perpendicular to the
walls. The resolution of the interior flow comes from assuming homogeneous
isotropic turbulence, and is therefore the global Kolmogorov scale. Horizontal
resolution throughout is usually taken to be the global Kolmogorov scale.
(ii) In the boundary layer, what is the local Kolmogorov length scale in the boundary
layer in terms of u∗ and y + ? Assume that locally = u3∗ /y + .
(3 marks)
Answer: Because I just said that ALL scaling within the boundary layers is
determined by u∗ and y + , naturally η = y + . It can also be shown using the
formula for η above with = u3∗ /y +
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(iii) The other velocity scale in a channel flow is that of the velocity fluctuations, q.
Measurements find that u∗ /q ∼ Re−0.1 . How does the ratio of the channel width
to the scale of the smallest eddies in the boundary layer depend on Reynolds
number in DNS? Use a Reynolds number based on q and the channel width D.
For an LES, what would be good vertical resolution in terms of y + if we want to
properly represent the log-layer?
(9 marks)
Answer: Use η = y + . y + = ν/u∗ and D/y + = u∗ D/ν. Because u∗ ∼
q Re−0.1 , then D/y + = Re × Re−0.1 = Re0.9 .
For an LES, we want to properly represent the log-layer. Therefore 100y + .
(iv) Kim, Moin & Moser (1987) used approximately 2 × 106 grid points in a DNS
of channel flow at Re = 3300. Give a upper estimate the number of grid points
required for a DNS of channel flow at Re = 13200. Justify your estimate, then
suggest how fewer mesh points could be used.
(2 marks)
Answer: Reynolds number has increased by a factor of 4, therefore, the number
of mesh points must go up as (40.9 )3 = 42. I will take 43 = 64 as an answer.
Fewer points could be used because the number of vertical levels in the interior
can be reduced.
(25 marks total)
10. Taylor Series 2004-3. Not covered 2010
a) Using Taylors series show that the following can be written for a geographically
notated finite difference grid with a uniform spacing of ∆x:
−φW ∂φ
(i) U
+ HOT = U φP∆x
(5 marks)
∂x
−φW ∂φ
(ii) U
+ HOT = U φE2∆x
(5 marks)
∂x
∂ 2φ
P +φW
(iii) Γ 2 + HOT = Γ φE −2φ
(5 marks)
∆x2
∂x
b) Comment on the order of the above schemes as classically, and also more usually,
defined. Also, discuss the consistency of an algorithm using (ii) and (iii) above to
solve the equation
∂φ
∂ 2φ
U
=Γ 2
∂x
∂x
(10 marks)
(25 marks total)
11. Self-Advection or Inviscid Burgers 2004-4.
• The equation
∂u
∂u
∂u 1 ∂u2
+u
= 0 is equivalent to
+2
= 0.
∂t
∂x
∂t
∂x
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• Show that the following equation
∂u 1 ∂u2
+
=0
∂t
2 ∂x
(13)
can be discretized in the linearized Crank-Nicolson form below for a geographically
notated control volume
unew
− unew
1 old new P
P
old
+
u u
−
u
u
=0
new
E
W
∆t
2∆xp
In the above t is time and u the solution variable. The superscripts refer to old and
new time levels.
(25 marks total)
12. 1st and 2nd-order Schemes 2003-1.
a) Explain what is generally meant by the terms first- and second-order schemes.
(4 marks)
b) Using the middle control volume figure as a basis, produce sketches illustrating the
nature of the interpolation to control volume face ’W’ for the following schemes:
(i) 1st-order upwinding;
(ii) 2nd-order central differencing;
(iii) 2nd-order upwinding;
(iv) 3rd-order upwinding (QUICK);
(v) Skew upwinding.
(5 marks)
c) Sketch flow variable variations around a control volume for a ’kappa’ type scheme
(1 marks)
d) For the general flow direction identified, the ’W’ control volume fact interpolation
for the transported variable φ can be expressed as
φW = WW W φW W + WW φW + WP φP
where WW W , WW , WP are weighting functions associated with the grid nodes. For
schemes (i-iv) in part (b) produce and complete a table giving WW W , WW and WP .
Note: For scheme (iv) a full derivation is expected. For all schemes a uniform grid is
assume.
(15 marks)
(25 marks total)
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13. 1D Euler 2003-2.
The following simplified one-dimensional Euler-type equation is useful for studying the
performances of numerical schemes
∂φ
∂φ
+u
=S
∂t
∂x
Do the following
a) Show using the modified equation approach that when solving the above using a firstorder forwards Euler difference for the temporal term and a first-order backwards
difference for the spatial term, that the actual equation being solved takes broadly the
following general form
(25 marks total)
14. Transport 2003-3.
a) Show how the non-conservative transport equation below can be converted into the
following weak form.
2
∂φ
∂p
∂ φ ∂ 2φ
∂φ
+ ρv
=−
+µ
+
ρu
∂x
∂y
∂x
∂x2 ∂y 2
2
∂ φ ∂ 2φ
∂vφ
∂p
∂uφ
+ρ
=−
+µ
+
ρ
∂x
∂y
∂x
∂x2 ∂y 2
(6 marks)
b) Discretize the above about the two-dimensional, non-uniform, geographically notated control volume in the left frame of the figure below. In your discretization
apply second-order central differences to the spatial terms. The final form of the
discretized equation should appear as
AP φP = AW φW + AE φE + AN φN + AS φS + Sφ
where the A coefficients and S are defined as part of your solution.
14
(19 marks)
CONTINUED
ES4400
(25 marks total)
15. Time Advancement ES440-2009-2
a) Describe the following higher-order explicit methods. Each requires only two timelevels to be saved at any one time. Include sketches For 2nd-order RK, leap-frog and
2nd-order AB illustrating how the time-levels are related. Discuss their advantages
and disadvantages :
(i) Second-order Runge-Kutta
(ii) Leap-frog
(iii) Second-order Adams-Bashforth
(iv) Third-order Runge-Kutta (no sketch necessary)
(16 marks)
Answer:
(i) 2nd-order RK is self-starting, but it only marginally stable unless there is a dissipative term.
↓
↓
&
θ(n)
↓
θ0
θ(n+1)
→
f (t(n) , θ(n) )
.
&
0
→
f (t + τ, θ )
↓
.
←
← 0.5(f (θ0 ) + f (θ(n+1) )
→ f (t + τ, θ(n+1) )
(ii) Leap-frog is generally 2nd-order, but the separate steps can become out of phase
which can only be corrected with an occasional explicit Euler substep, which
means that overall the method is only 1st order accurate in time. It is also not
self-starting.
θ(0)
→
f (t(0) , θ(0) )
↓
,→
→
& Euler
(1) (1)
↓ f (t , θ )
←
θ(1)
↓
←↓
(2)
(2) (2)
θ
→
f (t , θ )
↓
↓
,→
&
(3) (3)
↓ f (t , θ
←
θ(3)
↓
←↓
(4)
(4) (4)
θ
→
f (t , θ )
↓
15
CONTINUED
ES4400
(iii) 2nd-order AB does not have phase errors, but it is not self-starting.
θ(0)
↓
θ(1)
↓
θ(2)
↓
θ(3)
→
f (t(0) , θ(0) )
.
Euler
↓
(1) (1)
→
f (t , θ ) ↓
(1)
. 1.5f (θ ) − 0.5f (θ(0) ) ↓
→
f (θ(2) ) ↓
. 1.5f (θ(1) ) − 0.5f (θ(1) ) ↓
→
f (θ(3) ) ↓
(iv) 3rd-order RK does not have phase errors and is self-starting.
b) Consider the advection equation
∂u
∂u
+ c(x)
= 0.
∂t
∂x
Write down an upwind scheme for this equation.
Answer:
un − uni−1
un+1
− uni
i
= −c i
τ
h
un − uni
un+1
− uni
i
= −c i+1
τ
h
(9 marks)
c>0
c<0
(25 marks total)
16. Advection ES904-2006-4.
a) Consider the advection equation CY905
∂u
∂
+
F (u) = 0
∂t
∂x
with
F (u) = cu
(i) Discretise this equation using forward-in-time (first-order explicit Euler) and
centered-in-space for a timestep τ and uniform mesh spacing h.
(4 marks)
Answer:
∂u
u(x, t + τ ) − u(x, t)
1
=
= − (u(x + h, t) − u(x − h, t))
∂t
τ
2h
(ii) Now discretise it using the Lax Method.
Answer:
u(x, t + τ ) =
(6 marks)
τ
cτ
(u(x + h, t) + u(x − h, t)) − (u(x + h, t)2u(x − h, t))
2
2h
16
CONTINUED
ES4400
b) Consider an advective-diffusion equation such as the Burgers equation CY905
∂u
∂u
∂ 2u
+u
=ν 2
∂t
∂x
∂x
∂u
∂ 1 2
∂u
or
+
u −ν
=0
∂t
∂x 2
∂x
∂
∂u
+
F (u) = 0
∂t
∂x
where ν =viscosity is the diffusive part. In this case ν = .001, the domain size is 1
and there are 60 meshpoints in a periodic domain.
when written in the form
(i) If h is the mesh spacing and the maximum of velocity is U = max(|u|), what
determines the time-step stability for the advection part of the Burgers equation?
(3 marks)
Answer: A CFL or Courant-Fredericks-Löwy condition τ = h/U
(ii) If both the advection and viscous terms are determined explicitly, what determines the timestep τ ?
(3 marks)
Answer: For a purely explicit scheme, the smallest time-scale determines the
time-step. In this case that is the viscous time-scale: τd = h2 /ν
(iii) It can be shown analytically that the Burgers equation forms shocks which form
a discontinuity in a finite time. The following two figures solve the Burgers
equation by the Lax method and the Lax-Wendroff method. Which figure is
which numerical method?
(2 marks)
Answer:Burgers
LeftGaussian
is Lax-Wendroff.
Right is Lax.
shock via Lax−W method
Burgers Gaussian shock via Lax method
0.1
0.1
Time= 0 to 1.2
0.08
0.06
0.06
0.04
0.04
amplitude
amplitude
0.08
0.02
0
−0.02
0.02
0
−0.02
−0.04
−0.04
−0.06
−0.06
−0.08
−0.08
−0.1
−0.5
Time= 0 to 0.6
−0.1
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
x
−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
x
(iv) Discuss the relative merits of Lax versus Lax-Wendroff, especially for cases that
might develop discontinuities.
(8 marks)
Answer: Lax is very dissipative but keeps things positive definite. Lax-Wendroff
advects structures better.
Student Version of MATLAB
Student Version of MATLAB
(v) Write a Lax method for solving the Burgers equation.
Answer:
τ
τ
u(x, t + τ ) = (u(x + h, t) + u(x − h, t)) − (u(x + h, t) − u(x − h, t))
2
4h
17
CONTINUED
(7 marks)
ES4400
ντ
(u(x + h, t) − 2u(x, t) + u(x − h, t))
h2
(vi) Consider the advection equation: (This is from CY905-2006-1e. The full LaxWendroff method has not been in my ES440 lecture notes since 2008 and the
only be examined at the level of knowing how it compares to other methods (it
is FTCS+a bit of leapfrog) as well as its advantages and disadvantages with
respect to the 1D compressible methods which were discussed in more detail.)
+
d
d
u + c u = 0.
dt
dx
In the case c = const the Lax-Wendroff scheme becomes:
Ujn+1 = Ujn −
µ2 n
µ n
n
n
) + (Uj−1
− 2Ujn + Uj+1
),
(Uj+1 − Uj−1
2
2
where µ = c/h. Use Fourier stability analysis to determine the stability limit for
this scheme.
Describe the form of the numerical instability that would be observed if h is
decreased such that the stability constraint is just slightly violated.
Answer:
• The Lax-Wendroff method is:
– Similar to second-order Runge-Kutta in that it makes a prediction at:
– An intermediate time t + τ /2 and
– Uses this to provide an improved advective or forcing term for Forward
Euler, or Forward in Time, Centered in Space FTCS.
– Or one can view the second step as a leap-frog step with the improved
advective term coming at t + τ /2.
• The steps are as follows:
– Lax method with half-steps in space and time: h/2, τ /2 for t = nτ :
(n+1/2)
(n)
(n)
u(xj + h/2, nτ + τ /2) = uj+1/2 = 21 [uj+1 + uj ] −
∆t (n)
(n)
[F − Fj ]
2∆x j+1
(n+1/2)
and analogously for uj−1/2
(n+1/2)
– Evaluation of new Fj+1/2
(n+1/2)
using uj+1/2 , which for (??) is
(n+1/2)
(n+1/2)
uj+1/2 −→ Fj+1/2
(n+1/2)
= cuj+1/2
– finally leapfrog (??) with a half-step in space and full step in time:
(n+1)
uj
(n)
= uj −
∆t (n+1/2)
(n+1/2)
[Fj+1/2 − Fj−1/2 ]
∆x
– The scheme still leaves behind dispersion.
– It does a better job of preserving shape than the Lax method.
18
CONTINUED
ES4400
– For this reason, this was one of the first methods used for shocks.
– Today, most shock capturing methods are based upon upwind methods.
– Stability: γ 2 (k) = |1 − λ2 (1 − cos kh) − iλ sin kh| where λ = cτ /h.
−
γ ≤ 1 again leads to λ = cτ /h ≤ 1.
(25 marks total)
17. Burgers ES904-2005-2.
a) Write the difference equation for the time advancement on a uniform mesh away
(n)
from any boundaries in these steps. Use ui = u(xi , tn ).
(i) Time derivative: First-order explicit forwards differencing
Answer:
(n+1)
(n)
(n)
ui
− ui
∂ui
≈
∂t
τ
(ii) First-derivative in the advection using central differencing
Answer:
(n)
(n)
(n)
ui+1 − ui−1
∂ui
≈
∂x
h2
(iii) Viscous diffusion: Central differencing.
Answer:
(n)
(n)
(n)
(n)
ui+1 − 2ui + ui−1
∂ 2 ui
≈
∂x2
h2
b) If h is the mesh spacing,maximum of velocity U = max(|u|) and U >> ν/h, what
determines the time-step stability for the method you just wrote down?
Answer: For a purely explicit scheme, the smallest time-scale determines the timestep. In this case that is the viscous time-scale: τd = ∆x2 /ν
(4 marks)
(4 marks)
(4 marks)
(5 marks)
c) Assume you wanted to use the Courant (CFL) condition to determine the timestep
for U >> ν/h:
(i) What type of numerical algorithm would you use?
Answer: Implicit or semi-implicit method
(4 marks)
(ii) Roughly, what is the CFL condition on the time-step for the advection algorithm
you just chose?
(4 marks)
Answer: ∆t < CFLh/U
(25 marks total)
18. Advection MA4G8-2007-1. CY905
a) What is the truncation error in each of the following approximations for a sufficiently
smooth function u?
19
CONTINUED
ES4400
−u(x − h) + u(x)
∂u
'
∂x
h
∂ 2u
−u(x − 2h) + 16u(x − h) − 30u(x) + 16u(x + h) − u(x + 2h)
(ii)
'
∂x2
16h2
b) Consider the advection equation
(i)
∂u
∂u
+ c(x)
= 0.
∂t
∂x
(i) Write down an upwind scheme for this equation.
(2 marks)
(2 marks)
(6 marks)
(ii) State the Courant, Friedrichs, and Lewy (CFL) condition and explain how it
holds as the mesh is refined.
(4 marks)
(25 marks total)
19. Advection MA4G8-2007-3. CY905
a) Which of the following temporal discretisations can exhibit numerical instability
when using a second-order central spatial discretisation of the heat equation? (State
all choices which are correct.)
(i) forward-in-time Euler
(ii) backward Euler
(iii) Crank-Nicolson
(2 marks)
∂u that uses grid points
b) Derive a fourth-order finite-difference approximation for
∂x j
j − 2, j − 1, j, j + 1 and j + 2, i.e.
2
X
d wl uj+l ,
u '
dx j l=−2
assuming that the grid points are equally spaced: xj − xj−1 = h.
(6 marks)
(i) What is the truncation error in this approximation?
(2 marks)
(25 marks total)
20. Effective wavenumbers. CY905 2010 exam. All here could be on a ES440 exam.
For functions on the real line discretised using a uniform mesh N with period L and mesh
spacing h = L/N , answer the following:
a) What is the maximum wavenumber? What will happen to the solution if energy
reaches this wavenumber?
(2 marks)
Answer: The maximum wavenumber is kmax = N π/L = π/h. If there are fluctuations on the scale of the mesh size, then this mode typically generates strong
oscillations in the solution.
20
CONTINUED
ES4400
b) The second-order central difference algorithm for first derivative on a periodic line
with a uniform mesh is
∂u
ui+1 − ui−1
≈
.
∂x
2h
sin(kh)
The effective wavenumber for this operator acting on u is kef f =
and
h
max(kef f ) is at kp = π/(2h).
Find: The slope as k → 0. Its maximum: max(kef f ). Its value at k = kmax .
Answer: dkef f /dz|k=0 = 1; max(kef f ) = 1/h; kef f (kmax ) = 0;
(3 marks)
c) A fourth-order central difference formula for the first derivative is
∂u −ui+2 + 8ui+1 − 8ui−1 + ui−2
≈
∂x i
12h
(i) Derive the effective wavenumber for this formula, which can be constructed
from two sine functions.
(5 marks)
Answer:
kef f =
8 sin(kh) − sin(2kh)
sin(kh)
=
[4 − cos(kh)]
6h
3h
(ii) Find the slope as k → 0.
(1 marks)
Answer: dkef f /dz = (8h cos(kh)−2h cos(2kh))/6h = (4 cos(kh)−cos(2kh))/3
so dkef f /dz|k=0 = (4 − 1)/3 = 1.
(iii) If the maximum of kef f is at kp , is kp greater or less than kp for the second-order
formula in (b-ii)?
(1 marks)
Useful numbers: cos(kp h) = −0.22 and sin(kp h) = 0.95.
Answer: kp is greater. That is kp > π/(2h).
(iv) What is max(kef f )?
Answer: max(kef f ) ≈ 1.336/h
(1 marks)
(v) What is kef f at k = kmax ?
Answer: kef f (kmax ) = 0;
(1 marks)
(vi) Sketch the effective wavenumbers for the 2nd and 4th-order central differences
for the first derivative.
(3 marks)
Answer: Compare CEN2 and CEN4.
21
CONTINUED
ES4400
21. Stability of Crank-Nicolson. CY905 2010 exam. Finding kef f is the only part that could
be examined for ES440.
Consider the heat equation in a periodic domain
∂ 2u
∂u
= κ 2.
∂t
∂x
Show that the Crank-Nicolson algorithm on a uniform mesh applied to the second-order
central difference operator for diffusion is stable using Neumann stability analysis. Written
in operator form the equations are
(I −
κτ
κτ
D)u(n+1) = (I + 2 D)u(n)
2
2h
2h
a) First show that the effective wavenumber of the diffusion operator: ∂ 2 u/∂x2 =
h−2 Du is kef f = 2 sin(kh/2)/h.
Answer: In Neumann stability analysis, Let u(x) = u(k)eikx , then
−ikh
1
− 2 + eikh )
−4
∂ 2u
ikx (e
2
=
Du
=
u(k)e
= u(k)eikx 2 sin2 (kh/2) = −u(x)kef
f
2
2
2
∂x
h
h
h
b) Now write the left and right-hand sides as finite-differences.
Answer:
(n+1)
ui
(n+1)
− λ(ui−1
(n+1)
− 2ui
(n+1)
(n)
+ ui−1 ) = ui
(n)
(4 marks)
(n)
+ λ(ui−1 − 2ui
(n)
+ ui−1 )
where λ = κτ /2h2 .
c) Then rewrite the left and right-hand sides using the effective wavenumbers for the
diffusion operator D .
(4 marks)
Answer: By substituting in the effective wavenumber for the diffusion operator, this
becomes
(n+1)
(n)
2
2
1 + λkef
= 1 − λkef
f ) ũk
f ũk
d) Conclude the proof.
Answer: Write
(5 marks)
(n+1)
ũk
=
2
1 − λkef
f
2
1 + λkef
f
(n)
ũ(n)
k = γ(k)ũk
Because sin2 (kh/2) > 0 for all k, γ(k) < 1 for all k. Therefore by iteration
(n)
(0)
(0)
ũk = γ n (k)ũk < ũk
(n)
Therefore ũk is bounded for all time and the solution is stable.
22
CONTINUED
ES4400
22. Industrial Elliptic Solvers ES440
Two-dimensional heat conduction can be described by the Poisson equation,
∂ 2T
∂ 2T
+
=f
∂x2
∂y 2
(14)
where T is the temperature and f (x, y) is the source term. We are interested in the steady
state solution in the unit square, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
a) Consider the one-dimensional differential equation
∂ 2T
=f
∂x2
(i) Derive the central difference formula for the second derivative by Taylor expanding T (x) on a uniform mesh ∆x and using T (x + ∆x) and T (x − ∆x).
Include the error.
(6 marks)
Answer: For h = ∆x
h2 ∂ 2 T
h3 ∂u
∂T
+
+
+ O(h4 )
T (x + h) = T (x) + h
∂x x 2! ∂x2 x 3! ∂x3 x
∂T
h2 ∂ 2 T
h3 ∂T
T (x − h) = T (x) − h
+
−
+ O(h4 )
2
3
∂x x 2! ∂x x 3! ∂x x
Add and then isolate ∂ 2 T /∂x2 result is:
Ti+1 − 2Ti + Ti−1
∂ 2T
=
+ O(∆x2 )
2
∂x
h2
(ii) Write the full difference equations to be solved.
Answer:
Ti+1 − 2Ti + Ti−1
= fi
h2
(iii) How many diagonals would appear?
Answer: 3 diagonals
(2 marks)
(2 marks)
(iv) Assume the following boundary conditions: T (0) = 0, T (1) = 1, N zones and
∆x = 1/N . How are the rows for the boundary conditions (0 and N ) modified?
(2 marks)
Answer:
T2 − 2T1
= f1
h2
1 − 2TN −1 + TN −2
= f1
h2
b) There are many different solution algorithms for two-dimensional problems. Diagram how the Nx Ny × Nx Ny matrix would look if you attempted direct inversion if
Nx = 4 and Ny = 4.
(5 marks)
23
CONTINUED
ES4400
Answer:

−2λ α
0
0
β
0
...

0
0
β
0
...
 α −2λ α

 0
α −2λ α
0
0
β
0 ...

 0
0
α −2λ α
0
0
β 0 ...

A=
0
0
α −2λ
0
0
0 β ...
 0

 β
0
0
0
0
−2λ α
0 0 β 0 ...

 0
β
0
0
0
α −2λ α 0 0 β 0 . . .

..
.















c) Explain how the difference formula just written is modified on different iteration
steps in ADI (Alternating Direction Implicit). Write the difference equation in terms
of the nodes ij and use α = 1/∆x2 and β = 1/∆x2 . Show one substep of the
iteration indicating which terms go on the right-hand side and which go on the left
for each substep.
(8 marks)
Answer: In ADI the full two-dimensional Crank-Nicolson Laplacian is replaced by
two one-dimensional solvers.
αu(i−1)j + αu(i+1)j + βui(j−1) + βui(j+1) −2(α + β)uij = fij .
{z
}
|
OLD
(25 marks total)
23. Industrial Elliptic Solvers ES440-2008-4
UNSTEADY PDE:
Consider the 1D generic transport equation with constant velocity (u), constant fluid properties (ρ, Γ ):
∂T
∂T
Γ ∂ 2T
= −u
+
∂t
∂x
ρ ∂x2
We assume that the spatial derivatives are approximated using central-difference schemes
(CDS) and that the grid is uniform. Derive the algebraic equation using the following
methods and explain the accuracy and stability of each method.
a) Derive the algebraic equation using the Explicit Euler method and explain the accuracy and stability. Also describe how the numerical solution can be obtained.
(10 marks)
Answer: We consider the 1D version:
∂f
∂f Γ ∂ 2 f
= −u
+
,
∂t
∂x ρ ∂x2
|{z}
|{z}
| {z }
U
N
24
L
CONTINUED
ES4400
where
U=
fin+1 − fin
,
∆t
N =u
∂f
,
∂x
L=ν
∂ 2f
.
∂x2
Explicit Euler Method
(1 marks)
un+1
− uni
i
= −N n + Ln
∆t
All of the fluxes and source terms are evaluated using known values at tn .
uni+1 − uni−1
uni+1 − 2uni + uni−1
un+1
− uni
i
= −u
+ν
.
∆t
2∆x
∆x2
The new variable value, un+1
, is:
i
uni+1 − uni−1
uni+1 − 2uni + uni−1
n+1
n
∆t.
ui = ui + −u
+ν
2∆x
∆x2
The above equation can be rewritten:
c n
c n
n+1
n
ui = d +
u + (1 − 2d) ui + d −
u ,
2 i−1
2 i+1
(1 marks)
(15)
(1 marks)
(16)
(1 marks)
(17)
• where we introduce the dimensionless parameters:
ν∆t
∆t
=
,
2
∆x
∆x2 /ν
u∆t
∆t
c=
=
.
∆x
∆x/u
d=
(18)
(19)
• c is called Courant number and is one of the key parameters in CFD.
• First-order accurate with truncation of O (∆t, ∆x2 )
Von Neumann Stability Analysis Conditionally stable:
d=
ν∆t
< 0.5.
∆x2
(1 marks)
(20)
The first condition leads to the limit on ∆t:
∆t =
∆x2
.
2ν
(21)
The requirement that d < 0.5 means that, each time ∆x is halved, ∆t has to be
reduced by a factor of four.
• This makes the scheme unsuitable for problems which do not require high temporal resolution.
25
CONTINUED
ES4400
b) Derive the algebraic equation using the implicit method weighted by θ.
Answer: Variable-weighted implicit Method
(5 marks)
(1 marks)
un+1
− uni
i
(22)
= θ [−N n + Ln ] + θ −N n+1 + Ln+1
∆t
All fluxes and sources are evaluated in terms of the unknown values at the new time
level tn+1 .
(1 marks)
uni+1 − uni−1
uni+1 − 2uni + uni−1
un+1
− uni
i
= θ −u
+ν
∆t
2∆x
∆x2
n+1
n+1
un+1
+ un+1
un+1
i+1 − 2ui
i−1
i+1 − ui−1
+ν
. (23)
+ θ −u
2∆x
∆x2
The above equation can be rewritten:
(1 marks)
uni+1 − uni−1
uni+1 − 2uni + uni−1
n+1
n
ui = ui + θ −u
+ν
∆t
2∆x
∆x2
n+1
n+1
un+1
un+1
+ un+1
i+1 − ui−1
i+1 − 2ui
i−1
+ θ −u
+ν
∆t. (24)
2∆x
∆x2
The above equation can be rewritten:
AW uni−1 + AP uni + AE uni+1 = Qi .
(1 marks)
(25)
• Second-order accurate with truncation of O (∆t2 , ∆x2 )
(1 marks)
• Unconditionally Stable.
c) When the weighting factor θ = 1/2, what is the method known as? Also describe
how the numerical solution can be obtained.
(5 marks)
Answer: Crank-Nicolson
1
1
un+1
− uni
i
= [−N n + Ln ] + −N n+1 + Ln+1
(26)
∆t
2
2
Resulting Finite Difference Equation (FDE) is
(1 marks)
ui+1 − 2ui + ui−1
= fi .
∆x2
N is the number of interior points, and u0 and uN +1 are boundary values. We have
N number of equations for u1 , u2 , . . . ui , . . . , uN −1 , uN .
u0 − 2u1 + u2 = f1 ,
u1 − 2u2 + u3 = f2 ,
..
.
ui−1 − 2ui + ui+1 = fi ,
..
.
uN −2 − 2uN −1 + uN = fN −1 ,
uN −1 − 2uN + uN +1 = fN .
26
CONTINUED
ES4400
The result of discretisation is a system of linear algebraic equations:

−2 1
 1 −2 1


1 −2 1



1 −2 1
1 −2

u1
u2
ui





  uN −1
uN


 
 
 
=
 
 
(1 marks)
f1 − u0
f2
fi
fN −1
fN − uN +1




.


An upper triangular form of the tridiagonal matrix may be obtained by computing
the new bi by
(1 marks)
bi = bi −
ai
ci−1 ,
bi−1
i = 2, 3, . . . , N,
and the new fi by
(1 marks)
fi = fi −
ai
fi−1 ,
bi−1
i = 2, 3, . . . , N,
then computing the unknowns from back substitution according to uN = fN /bN and
then
(1 marks)
uk =
fk − ck uk+1
,
bk
k = N − 1, N − 2, . . . , 2, 1.
d) Now, lets consider the two dimensional case. For Explicit Euler and Implicit Euler
methods, describe how the numerical solution to the equation below can be obtained.
∂T
∂T
Γ ∂ 2T
∂ 2T
∂T
= −u
−v
+
+
∂t
∂x
∂y
ρ ∂x2
∂y 2
(10 marks)
Answer: The answer should be based on the lecture handouts
(25 marks total)
24. Industrial ES440-2007-5
TURBULENT FLOW SIMULATION:
In turbulent flow calculations, there are three categories of numerical methods: Direct Numerical Simulation (DNS), Large Eddy Simulation (LES) and Reynolds-Averaged NavierStokes Simulation (RANS). Explain the following methods, making use of sketch(es) as
appropriate. Describe the relative strengths and weaknesses in around 100 words.
a) DNS
Answer:
(5 marks)
27
CONTINUED
ES4400
• The most accurate approach to turbulence is to solve the Navier-Stokes equations
without any models.
• It is also the simplest approach from the conceptual point of view.
• In DNS, all of the motions contained in the flow are resolved.
• The costs of a simulation scales as Re3L .
• So, DNS is possible only for flows at relatively low Reynolds numbers and in
geometrically simple domains.
• DNS contain very detailed information about the velocity, pressure, and any
other variable of interest at a large number of grid points.
• The wealth of information can then be used to develop a qualitative understanding of the physics of the flow or to concoct a quantitative model, such as RANS
type, which will allow other, similar, flows to be computed.
• Because it is more accurate, DNS is the preferred method whenever it is feasible.
• DNS is too expensive to be employed very often and cannot be used as a ’design’
tool.
b) LES
Answer:
(5 marks)
• LES is a technique intermediate between DNS and RANS.
• LES computes accurately the dynamics of the large-scale motions while approximating or modelling the small, subgrid scales of motions.
• By filtering the velocity field, the large-scale motions are separated from the
small-scale motions.
• Despite the fact that the small scales are modelled, LES remains a fairly computationally intensive technique but much less costly than DNS of the same flow.
• Subgrid-scale models are used to approximate the Reynolds stress.
c) RANS
Answer:
(5 marks)
• RANS is a practical approach to engineering problems.
• In Reynolds-averaged approaches to turbulence, all of the unsteadiness is averaged out.
• On averaging, the non-linearity of the Navier-Stokes equations gives rise to
terms that must be modelled.
• The complexity of turbulence makes it unlikely that any single turbulence model
will be able to represent all turbulent flows.
d) Applying LES filtering to the Navier-Stokes equations, derive the governing equations for the large-eddy motions of the flow, and explain the difference from the
Navier-Stokes equations.
(5 marks)
Answer:
28
CONTINUED
ES4400
• The large-scale field is defined with the aid of the filter function G:
Z
u(x) =
G(x − x0 )u(x0 )dx0 ,
(1 marks)
D
• Continuity equation:
(1 marks)
∂u ∂v ∂w
+
+
= 0,
∂x ∂y
∂z
• Navier-Stokes equation:
∂ui
∂
1 ∂ 2 ui
∂p
+
(ui uj ) = −
+
,
∂t
∂xj
∂xi
Re ∂xj ∂xj
∂ui
∂
∂
1 ∂ 2 ui
∂p
+
(ui uj ) = −
−
τij +
.
∂t
∂xj
∂xi
∂xj
Re ∂xj ∂xj
• Governing equations for large-scale motions:
(2 marks)
∂ui
∂p
∂
∂
1 ∂ 2 ui
+
(ui uj ) = −
−
τij +
.
∂t
∂xj
∂xi
∂xj
Re ∂xj ∂xj
• All the terms in the above equations are resolved in the LES except the residual
stress tensor
(1 marks)
τij = ui uj − ui uj .
e) Using dimensional analysis, explain how to model the subgrid-scale (SGS) stress,
and propose a simple SGS model.
(5 marks)
Answer:
• In the most commonly used model, developed by Smagorinsky, the eddy viscosity model is obtained by assuming that the small scales are in equilibrium, so
that energy production and dissipation are in balance. This yields an expression
of the form
(1 marks)
τij − 1/3δij τkk = −2νt Sij .
•
νt = (Cs ∆)2 |S|,
where Cs is the Smagorinsky constant, and ∆ is the filter width, |S| =
is the magnitude of large-scale strain-rate tensor
1 ∂ui ∂uj
Sij =
+
.
2 ∂xj
∂xi
p
2Sij Sij
• Smagorinsky model:
(1 marks)
(2 marks)
τij − 1/3δij τkk = −2νt Sij ,
νt = (Cs ∆)2 |S|,
29
CONTINUED
ES4400
• The filter width, ∆, represents the characteristic length scale of the largest subgridscale eddies, which are unresolved in the filtered momentum equations. ∆ was
defined as the geometric mean of the unidirectional filter width, i.e.,
(1 marks)
∆ = (∆1 ∆2 ∆3 )1/3 .
∂u
∂ 2u
∂u
+u
= µ 2.
∂t
∂x
∂x
By applying filtering to the Burgers equations, derive the filtered Burger equations
Answer:
f) A useful model is the one-dimensional Burgers equation.
• Filtered Burgers equation:
(9 marks)
(3 submarks)
∂u
∂ 2
1 ∂ 2u
+
u =−
∂t
∂x
Re ∂ 2 x
• Governing equations for large-scale motions:
(3 submarks)
∂u
∂
1 ∂ 2 ui
∂
+
(uu) = − τ +
.
∂t
∂x
∂x
Re ∂ 2 x
• All the terms in the above equations are resolved in the LES except the residual
stress tensor
(3 submarks)
τ = uu − u u.
(25 marks total)
25. Industrial ES440-2007-6
CAR AERODYNAMICS:
The size and dimensions of Formula One (F1) racing cars are tightly controlled by regulations. The strict regulations mean that the teams use very similar-sized cars. A typical
car is 4.67m long, 1.80m wide, and 0.95m high. The top speed of the F1 car is about 380
km/h. The kinematic viscosity of the air is 1.5 × 10−5 m2 /s.
a) Calculate the appropriate Reynolds number of the F1 car.
Answer:
(5 marks)
U = 40m/s,
380 × 103
= 105m/s,
U =
3600
Uh
105 × 1.0
=
= 7 × 106 .
Re =
ν
1.5 × 10−5
• Re in generic vehicle body study is 7.68 × 105 .
• Re of a family car is 2 × 106 .
30
CONTINUED
ES4400
b) Imagine that you are doing wind tunnel tests using a model car in the Warwick Engineering Wind Tunnel. The size of the test section of the wind tunnel is 1m by 1m,
and the maximum wind speed in the tunnel is 25m/s. Assuming that the maximum
blockage ratio permissible to avoid the wall effect is 1:10, then what should be the
size of the model car. Compare the Reynolds number with part a) of the question.
(5 marks)
Answer:
• The available area is 10% of the test section.
• Cross sectional area of the F1 car is 1.8 × 0.95 = 1.71m2 .
• Cross sectional area of the model car is 1.8α × 0.95α = 1.71α2 m2 = 0.1m2 .
• Scale is α = 4.135.
• The model car is 1.13m long, 0.44m wide, and 0.23 high.
Re =
UL
25 × 4.67/4.135
=
= 1.88 × 106 .
ν
1.5 × 10−5
c) In relation to the racing car aerodynamics, discuss the advantages and limitations of
CFD and wind tunnel tests, in around 100 words.
(5 marks)
Answer: YMC answer: The answer should be based on the lecture handouts and
include the following points:
• Full scale simulation,
• Reynolds number effect,
• Ground effect,
• Side wall effect,
• Easy visualisation of the results.
From my lecture notes. Introduction to numerical methods.
Limitations of CFD:
• Limited regimes Highest Reynolds, Froude and Rayleigh numbers are out of
reach.
• Complex geometries Unstructured grids can only take you so far. Moving interfaces are even worse.
• Exotic regimes Regimes where kinetic theory must be mixed with continuum
mechanics. Hypersonic. Solar corona.
• Combustion Reduced equations are very approximate.
• Radiation Coupling radiation to flows is still primitive.
• And all the issues of convergence and accuracy below.
Limitations of experimental measurments (wind tunnels):
• Incomplete measurements. Until recently at the highest Reynolds numbers
only 1D (time) data of velocity was available.
– With PIV: full 2D velocity snapshots can be obtained.
31
CONTINUED
ES4400
– But continuous in time at the highest speeds is still impossible.
– Full 3D is only possible at low Reynolds numbers.
• The measuring equipment might disturb the flow.
• The flow might be inaccessible - engine, turbine.
• Pressure, temperature probes still only at walls or bulky.
• Measurement might be too costly and/or time consuming.
• Need to corroborate data.
d) Now you are doing a numerical simulation of the F1 car. Considering the Reynolds
number in part a), which computational method for the turbulent flow would you
recommend for the F1 car simulation? Explain the reasons?
(5 marks)
Answer:
The answer should be based on the lecture handouts and include the
following points:
• Due to high Reynolds number, DNS is not possible, and full LES is not also
feasible.
• Steady RANS will not be very accurate due to the unsteadiness of the flow field,
• So, URANS for Full car simulation,
• LES for local simulation,
• DES can be a good alternative.
e) Determine the computational domain size and the appropriate boundary conditions
for the F1 car simulation.
(5 marks)
Answer:
The answer should be based on the lecture handouts and include the
following points:
• Computational domain should be large enough, and the effect of the domain size
has to be tested,
• Initial computational domain is at least 5 car lengths and 5 car heights and 5 can
widths,
• The location of the car inside the computational domain has to be decided carefully.
• Inflow, outflow boundary conditions,
• Moving boundary condition for the bottom wall,
• Slip boundary condition for side walls,
• Free stream boundary condition for the top wall,
• The effect of the boundary condition has to be tested.
(25 marks total)
32
CONTINUED
ES4400
26. Industrial ES440-2008-6
TURBULENT FLOW SIMULATION:
The Navier-Stokes equations for incompressible fluid are given below.
∇·u=0
∂~u
+ ρ(~u · ∇)~u = −∇p + µ∇2~u
∂t
where ~u is the velocity vector, ρ is the density, p is the pressure, and µ is the dynamic
viscosity of the fluid.
ρ
a) Give reasons why turbulent flows are more difficult to simulate than laminar flows. (2 marks)
Answer:
The answer should be based on the lecture handouts and include the
following points:
• Turbulent flows have a wide spectrum of length scales and time scales.
• Turbulent flows are unsteady flow.
b) The numerical methods used in turbulent flow simulations can be grouped into three
categories. Explain each category, making use of sketch(es) as appropriate, and also
describe the relative strengths and weaknesses of each category.
(6 marks)
Answer:
The answer should be based on the lecture handouts and include the
following points:
• DNS:
(2 marks)
– The most accurate approach to turbulence is to solve the Navier-Stokes
equations without any models.
– It is also the simplest approach from the conceptual point of view.
– In DNS, all of the motions contained in the flow are resolved.
– The costs of a simulation scales as Re3L .
– So, DNS is possible only for flows at relatively low Reynolds numbers and
in geometrically simple domains.
– DNS contain very detailed information about the velocity, pressure, and any
other variable of interest at a large number of grid points.
– The wealth of information can then be used to develop a qualitative understanding of the physics of the flow or to concoct a quantitative model, such
as RANS type, which will allow other, similar, flows to be computed.
– Because it is more accurate, DNS is the preferred method whenever it is
feasible.
– DNS is too expensive to be employed very often and cannot be used as a
’design’ tool.
• LES:
(2 marks)
33
CONTINUED
ES4400
– LES is a technique intermediate between DNS and RANS.
– LES computes accurately the dynamics of the large-scale motions while
approximating or modelling the small, subgrid scales of motions.
– By filtering the velocity field, the large-scale motions are separated from the
small-scale motions.
– Despite the fact that the small scales are modelled, LES remains a fairly
computationally intensive technique but much less costly than DNS of the
same flow.
– Subgrid-scale models are used to approximate the Reynolds stress.
• RANS:
(2 marks)
– RANS is a practical approach to engineering problems.
– In Reynolds-averaged approaches to turbulence, all of the unsteadiness is
averaged out.
– On averaging, the non-linearity of the Navier-Stokes equations gives rise to
terms that must be modelled.
– The complexity of turbulence makes it unlikely that any single turbulence
model will be able to represent all turbulent flows.
c) Explain the term Reynolds averaging.
Answer:
Z
ui (x) =
(2 marks)
T
ui (x, t)dt,
0
d) Applying Reynolds averaging to the Navier-Stokes equations, derive the ReynoldsAveraged Navier-Stokes equations, and explain the difference from the Navier-Stokes
equations.
(10 marks)
Answer:
The answer should be based on the lecture handouts and include the
following points:
• All the terms in the above equations are resolved in the RANS except the Reynolds
stress tensor
(4 marks)
The Reynolds averaging is defined as:
• Continuity equation:
(2 marks)
∂u ∂v ∂w
+
+
= 0,
∂x ∂y
∂z
• Navier-Stokes equation:
(4 marks)
∂ui
∂
∂p
∂
1 ∂ 2 ui
+
(ui uj ) = −
−
τij +
.
∂t
∂xj
∂xi
∂xj
Re ∂xj ∂xj
• With ∂ui /∂t = 0
τij = ui uj − ui uj = u0i u0j .
34
CONTINUED
ES4400
e) Using dimensional analysis, explain how to model the Reynolds stress, and write
down a simple model.
(5 marks)
Answer:
τij − 2/3kδij = −2νt Sij ,
1 0 0
k=
u u + v 0 v 0 + w0 w0 ,
2
νt = Cµ ql,
(27)
(28)
(29)
• Cµ is a dimensionless constant,
• q is the velocity scale,
• l is the length scale.
(25 marks total)
27. Industrial ES440-2008-6/2009-4
THE TURBULENT NAVIER-STOKES EQUATIONS:
Solution of the Navier-Stokes equations is complicated by the lack of an independent equation for the pressure, whose gradient contributes to each of the three momentum equations.
Furthermore, the continuity equation does not have a dominant variable in incompressible
flow.
a) Answer one of the following.
Answer: The answer should be based on the lecture handouts:
(5 marks)
Explain the solution algorithm for the fractional step method; OR
Explain the solution algorithm for the simple explicit time advance scheme; OR
Explain the solution algorithm for the pressure-correction methods.
b) Explain the multi-grid method.
(5 marks)
Not covered 2010
Answer: The answer should be based on the lecture handouts:
c) By filtering the Navier-Stokes equations, derive the LES filtered Navier-Stokes equations or URANS (Unsteady RANS) equations, and discuss the key differences from
the original Navier-Stokes equations.
(6 marks)
Specific filters not covered in 2010.
Answer:
• The large-scale field is defined with the aid of the filter function G:
Z
u(x) =
G(x − x0 )u(x0 )dx0 ,
D
35
CONTINUED
(1 marks)
ES4400
• Continuity equation:
(1 marks)
∂u ∂v ∂w
+
+
= 0,
∂x ∂y
∂z
• Filtered Navier-Stokes equation that are the governing equations for largescale motions:
(3 marks)
∂ui
∂
∂p
∂
1 ∂ 2 ui
+
(ui uj ) = −
−
τij +
.
∂t
∂xj
∂xi
∂xj
Re ∂xj ∂xj
• All the terms in the above equations are resolved in the LES except the residual
stress tensor
(1 marks)
τij = ui uj − ui uj .
d) Using dimensional analysis, explain how to model the subgrid-scale (SGS) stress in
LES, and write down a simple model.
(5 marks)
Answer:
• In the most commonly used model, developed by Smagorinsky, the eddy viscosity model is obtained by assuming that the small scales are in equilibrium, so
that energy production and dissipation are in balance. This yields an expression
of the form
(1 marks)
τij − 1/3δij τkk = −2νt Sij .
•
νt = (Cs ∆)2 |S|,
where Cs is the Smagorinsky constant, and ∆ is the filter width, |S| =
is the magnitude of large-scale strain-rate tensor
1 ∂ui ∂uj
Sij =
+
.
2 ∂xj
∂xi
p
2Sij Sij
• Smagorinsky model:
(1 marks)
(2 marks)
τij − 1/3δij τkk = −2νt Sij ,
νt = (Cs ∆)2 |S|,
• The filter width, ∆, represents the characteristic length scale of the largest subgridscale eddies, which are unresolved in the filtered momentum equations. ∆ was
defined as the geometric mean of the unidirectional filter width, i.e.,
(1 marks)
∆ = (∆1 ∆2 ∆3 )1/3 .
36
CONTINUED
ES4400
e) Explain how the van Driest damping function can be used for the coarse grid LES?
Answer:
(5 marks)
τij − 1/3δij τkk = −2νt Sij ,
νt = (Cs f ∆)2 |S|,
f is the van Driest damping function
(2 marks)
f = 1 − exp
y+
A
, A = 25.
where n is the wall normal distance.
(2 marks)
f) Discuss the underlying basis for k-formulations (including k − ).
(5 marks)
Answer: In k-formulations extra subgrid variables are carried with the flow and
evolved. The resolved scales are damped by a nonlinear viscosity formed by k and
a length scale. In the originally k-formulation the only additional variable was the
subgrid kinetic energy, whose equation includes production from the resolved scaled.
A length scale is formed from the mesh size and distance from the wall. In k − and
k − ω formulations (including Spalart-Almaras) a second quantity is evolved that
provides the damping for k. With k, the length scale forms the eddy damping of the
resolved scales.
28. Industrial ES911-2009-2
FRACTIONAL STEP METHOD
The Navier-Stokes equations for incompressible fluid can be written as follows:
∂ρui
∂p
= Hi −
∂t
∂xi
A A three-step fractional step method can be constructed based on the Crank-Nicolson
method.
a) Complete the following fractional step method by adding the third step, and the corresponding Poisson equation for pressure.
(15 marks)
δpn
(ρui )∗ − (ρui )n
= 12 H(uni ) + H(u∗i ) −
∆t
δxi
(ρui )∗∗ − (ρui )∗
δpn
= 12
∆t
δxi
b) Discuss the advantages of the fractional step method over the conventional pressurecorrection method.
(5 marks)
37
CONTINUED
ES4400
c) Discuss the role played by the continuity equation in relation to solution algorithms
for the incompressible Navier-Stokes equations.
(5 marks)
29. Industrial ES440-2009-6
CAR AERODYNAMICS
A simple, generic vehicle body is being used in the car aerodynamics community. All
the geometric quantities are normalised with the body height, H = 0.288m. The model is
2.928H long, and 1.35H wide. The Reynolds number, based on the incoming velocity and
the body height, H, is Re = 7.68 × 105 in the generic vehicle body study. The kinematic
viscosity of the air is 1.5−5 m2 /s.
a) Calculate the incoming velocity used in the generic vehicle body study. Also calculate the Reynolds number of a typical family car driving at the national speed limit
of 70 mph, and comment on how useful and relevant the generic vehicle body study
is to the real car aerodynamics.
(3 marks)
Answer:
U = 40m/s,
70 × 1.609 × 103
= 31.3m/s,
U =
3600
Uh
31.3 × 1.0
Re =
=
= 2 × 106 .
−5
ν
1.5 × 10
• Re in generic vehicle body study is 7.68 × 105 .
• Re of a family car is 2 × 106 .
b) In relation to the car aerodynamics, discuss the advantages of the non-dimensional
form of governing equations, in around 100 words.
(5 marks)
Answer: The answer should be based on the lecture handouts and include the following points:
• Computational domain should be large enough, and the effect of the domain size
has to be tested,
• Initial computational domain is at least 5 car lengths and 5 car heights and 5 can
widths,
• The location of the car inside the computational domain has to be decided carefully.
• Inflow, outflow boundary conditions,
• Moving boundary condition for the bottom wall,
• Slip boundary condition for side walls,
• Free stream boundary condition for the top wall,
38
CONTINUED
ES4400
• The effect of the boundary condition has to be tested.
c) In relation to the car aerodynamics, discuss the advantages of CFD and wind tunnel
tests , in around 100 words.
(5 marks)
Answer: YMC answer: The answer should be based on the lecture handouts and
include the following points:
• Full scale simulation,
• Reynolds number effect,
• Ground effect,
• Side wall effect,
• Easy visualisation of the results.
From my lecture notes. Introduction to numerical methods.
Limitations of CFD:
• Limited regimes Highest Reynolds, Froude and Rayleigh numbers are out of
reach.
• Complex geometries Unstructured grids can only take you so far. Moving interfaces are even worse.
• Exotic regimes Regimes where kinetic theory must be mixed with continuum
mechanics. Hypersonic. Solar corona.
• Combustion Reduced equations are very approximate.
• Radiation Coupling radiation to flows is still primitive.
• And all the issues of convergence and accuracy below.
Limitations of experimental measurments (wind tunnels):
• Incomplete measurements. Until recently at the highest Reynolds numbers
only 1D (time) data of velocity was available.
– With PIV: full 2D velocity snapshots can be obtained.
– But continuous in time at the highest speeds is still impossible.
– Full 3D is only possible at low Reynolds numbers.
• The measuring equipment might disturb the flow.
• The flow might be inaccessible - engine, turbine.
• Pressure, temperature probes still only at walls or bulky.
• Measurement might be too costly and/or time consuming.
• Need to corroborate data.
d) Now you are doing a numerical simulation of the generic vehicle body at Re =
7.68 × 105 . Which computational method for the turbulent flow would you recommend for this simulation? Explain the reasons for your choices.
(5 marks)
Answer: The answer should be based on the lecture handouts and include the following points:
39
CONTINUED
ES4400
• Due to high Reynolds number, DNS is not possible, and full LES is not also
feasible.
• Steady RANS will not be very accurate due to the unsteadiness of the flow field,
• So, URANS for Full car simulation,
• LES for local simulation,
• DES can be a good alternative.
e) Five boundary conditions (D-N-R-C-P) are used in CFD. Explain them briefly.
(5 marks)
Answer: The answer should be based on the lecture handouts and include the following points:
• Dirichlet (D): f = a
∂f
=a
• Neumann (N):
∂n
∂f
• Robin (R): αf + β
=a
∂n
∂f
• Cauchy (C): f = a and
=b
∂n
∂f
∂f
• Periodic (P): f (0) = f (l) and
(0) =
(l).
∂x
∂x
f) Determine the computational domain size and the appropriate boundary conditions
for the generic vehicle body simulation
(5 marks)
Answer: The answer should be based on the lecture handouts and include the following points:
• Computational domain should be large enough, and the effect of the domain size
has to be tested,
• Initial computational domain is at least 5 car lengths and 5 car heights and 5 can
widths,
• The location of the car inside the computational domain has to be decided carefully.
• Inflow, outflow boundary conditions,
• Moving boundary condition for the bottom wall,
• Slip boundary condition for side walls,
• Free stream boundary condition for the top wall,
• The effect of the boundary condition has to be tested.
30. Industrial ES440-2010-6
CAR AERODYNAMICS/ADAPTIVE MESH REFINEMENT
Flow around generic bodies: cars, aircraft, rail, must incorporate the true shape of the
bodies. This is accomplished by a variety of mesh refinement schemes. Two approaches
were discussed in lectures: Structured, curvilinear grids and unstructured grids.
40
CONTINUED
ES4400
a) Compare the two approaches in terms of how space is discretised, the respective data
management schemes, and how fluxes are calculated.
(6 marks)
Answer: A structured mesh can be transformed into a rectilinear mesh using a
conformal transformation. The mesh conforms to the shape of the underlying body.
Fluxes are determined between points on the mesh. Data is stored on a rectilinear
mesh. In an unstructured mesh, elements are tiled and clustered in space around
the object. Fluxes are calculated between faces of the elements. Data is nested
underneath the larger scale elements.
b) Currently, a self-consistent method of simultaneously simulating turbulent interior
flows coupled to turbulent boundary layers does not exist. Two methods frequently
used are DES and LES. Describe their separate approaches to addressing the boundary layer/interior flow coupling. Include what ad hoc assumptions are often made to
get meaningful results.
(6 marks)
Answer: In DES, two modelling approaches are used simultaneously. RANS or
Unsteady-RANS is applied to the inner boundary layer using all the parameterisations tuned for wind-tunnel flows past simple aerofoils. In the interior, LES is used.
The difficulty is in coupling the two methods in the two regions. Particularly difficult
is the effect of interior vortices impacting upon the boundary layer.
In LES, the coefficients and filtering in the interior and boundary layers can be different. Without filter functions, it is necessary to reduce the coefficient or shear
instabilities, which generate the turbulence, are suppressed.
c) What is the van Driest damping function and explain why it is used in some boundary
layer formulations.
(4 marks)
Answer: For example, in the Smagorinsky formulation if
τij − 1/3δij τkk = −2νt Sij ,
νt = (Cs f ∆)2 |S|,
then f is the van Driest damping function
+
y
f = 1 − exp
, A = 25.
A
(2 submarks)
where n is the wall normal distance.
It is used so as to force the eddy viscosity to go to zero in the k-formulations discussed above.
(2 subd) Discuss the advantages and limitations of CFD and why wind tunnel tests are still marks)
necessary, in around 50 words.
(5 marks)
Answer: CFD allows one to explore a wide variety of possible configurations and
parameter values without having to build experimental models. Even so, some experimental models still are necessary to validate the CFD and provide baseline values
for the CFD parameterisations.
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CONTINUED
ES4400
e) The following figure is from an LES flow around the nose of a car body.
This relatively course mesh is from Krajnovic
and Davidson (2005), discussed in class. Discuss what details might appear as the mesh
is further resolved and whether these substantially modify the large-scale flow. Would it
modify the drag prediction?
Answer: As the mesh is resolved, further vortices nested in the corner at the bottom of the nose will appear. These should not affect the overall flow, but additional
details on the nose where a substantial portion of the drag forms, could modify that
prediction.
(25 marks total)
31. Industrial ES440*
MILES FILTERING EXAMPLES. Methods developed post-Lax-Wendroff.
a) Consider the advection of a pulse, which for can now be simulated. For this 1D problem higher-order Riemann solvers can be used, as on the left. For higher dimensional
applications with more physics, alternative methods are needed, The other three examples are upwind, central differences and piecewise parabolic. Explain which picture goes with which method and discuss the advantages and disadvantages of the
two approaches as discussed in the lecture notes.
b) A common test of numerical methods is the one-dimensional shock tube problem
originally introduced by G.A Sod. A survey of several finite difference methods for
systems of nonlinear hyperbolic conservation laws. J. Comput. Phys., 27:131, 1997.
This is known as the Sod problem.
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CONTINUED
(4 marks)
ES4400
Consider the diagrams below. The one on the left is for the first-order MUSCL
scheme discussed in the lecture notes. The one on the right is for a piecewiseparabolic method.
The equations below the figures refer to that MUSCL scheme.
(i) Identify where all the terms in those equations are taken from in the MUSCL
diagram.
uLi+ 1 = ui + 0.5φ (ri ) (ui+1 − ui ) ,
2
uR
i+ 12
= ui+1 − 0.5φ (ri+1 ) (ui+2 − ui+1 ) ,
uLi− 1 = ui−1 + 0.5φ (ri−1 ) (ui − ui−1 ) ,
2
uR
= ui − 0.5φ (ri ) (ui+1 − ui ) ,
i− 1
2
ri (u) =
ui − ui−1
.
ui+1 − ui
Answer: See lecture notes
(ii) Now consider the final fluxes in the MUSCL formulation. What are the roles of
the factors ai+ 1 , φ(ri ) and ri (u) in the following equations.
2
∂F (ui (t))
∂F (ui+1 (t))
ai+ 1 (t) = max ρ
,ρ
∂u
∂u
2
∗
Fi−
1
2
i
h
io
1 nh L R
R
L
=
F ui+ 1 + F ui+ 1 − ai+ 1 ui+ 1 − ui+ 1
2
2
2
2
2
2
Answer: See lecture notes
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CONTINUED
ES4400
(iii) Now consider how the first-order MUSCL method compares with piecewiseparabolic methods. First in the determination of the fluxes on boundaries between zones. Second in the types of solutions achieved, as shown in the example
for how velocity is advected by the two methods in the Sod problem.
Answer: In the first-order methods the values and first-derivatives are continuous at the cell boundaries even though there are small velocity jumps. In the
piecewise-parabolic methods, second-derivatives are continuous as the velocities are also continuous.
32. Industrial CY905-2007-2a potential ES440 question
Resolving shocks in the compressible Navier-Stokes equations.
In one way of writing the compressible Navier-Stokes equations, the dependent variables
are velocity u, density ρ and temperature θ. This problem only considers the temperature
equation:
∂θ
p
κ1
1
+ (u · ∇)θ = −
∇·u+
4θ +
Φ
∂t
ρcv
cv ρ
ρcv
where Φ is a heat source and assume that the density ρ is constant.
Discontinuities (shocks) that would appear in the ideal equations are smoothed in the full
parabolic equations by the thermal diffusivity and fluid viscosity. However, solving the
full parabolic equations requires using a very small diffusive timestep.
This answer to this problem should discuss how the timestep can be increased to an
advective timestep. Including the following points:
a) What is the order of magnitude of the timestep required for stable molecular diffusion
assuming a uniform mesh spacing of h?
Answer: The diffusive timestep is determined by τ ≤ h2 /(κ/cv ρ).
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CONTINUED
ES4400
b) Describe two numerical approaches that would allow one to overcome the restriction
to a diffusive timestep and instead use the CFL condition for the determining the
timestep.
∗ If a particular numerical method is mentioned, be certain to explain what its accuracy
in time is and the conditions under which it can attain that accuracy.
Answer:
Some of the following could be used:
• The sharp interfaces need to be smoothed.
• In some methods such as monotone advection schemes, upwind methods, etc.
the numerics smooths the sharp interfaces with artificial numerical viscosities.
• Upwind is first-order in time locally, but globally is zeroth-order because its
direction can change with the local direction of velocity.
• Another approach is directly imposing artificial viscosities.
• MUSCL and piecewise-parabolic could also be discussed as in the lecture notes.
45
END