The Calculation of Hydrodynamic Shocks by Numerical Methods - A Tutorial

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• Publié par: Anfo
• Nombre de lectures: 16
• Langue: English

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THE CALCULATION OF HYDRODYNAMIC
SHOCKS BY NUMERICAL METHODS:
A TUTORIAL







Dr. Paul H. Rosenthal, Professor
California State University, Los Angeles







This paper is an extract of my thesis submitted in partial fulfillment of the requirements for the
Master of Arts degree at Temple University, June 1, 1953.

It is being published online to honor the memory of Drs. Robert Richtmyer and John Von
Neumann whose generous support and advice made it possible.










June 1, 1953
INDEX

Page
I. SHOCK HYDRODYNAMICS 1
1. Introduction
2. Lagrangian and Eularian system
3. Equations of motion
4. Energy equation
5. Properties of shook

II. AN ILLUSTRATION; AN ANALYTIC SOLUTION OF A SIMPLE PROBLEM 4
1. A plane piston problem
2. Definition of terms
3. Hugoniot’s equation for shocks
4. Analytic solution of advancing shock front
5. Status of advanced methods of solution

III. A COMPLEX HYDRODYNAMICAL PROBEM 9
1. Description
2. Definition of terms
3. Discussion of properties
4. Equations of motion
5. Conservation equations
6. Viscosity method of integration
7. Stability condition

IV. SOLUTION BY NUMERICAL METHODS 18
1. Partial difference operator as an approximation to the partial derivative
2. Lagrangian mesh
3. Centering of the difference equations
4. Derivation of the difference equations
5. Form for the .equations of viscosity and stability

V. COMPUTATION METHOD 29
1. System and order of computation
2. Storage requirements

VI. BIBLIOGRAPHY 40 I
SHOCK HYDRODYNAMICS

1. Introduction
The field of hydrodynamics of compressible fluids has many problems of great interest to the
physicist and engineer. Until the advent of the modern digital computer, no practical method
existed for the solution of a complex problem involving many interactions.

In the investigation of the flow of compressible fluids on a digital computer, it is necessary to
solve the equations of fluid motion by stepwise numerical procedures. A method for solution
of the equations that automatically handles all types of shock fronts is necessary for a
successful solution on a computer. Such a method has recently been developed by Von
Neumann and Richtmyer (1950) of the Institute of Advanced Study. This paper is a result of
the extensive use of this method for a solution of complex problems in shock hydrodynamics
on the Univac I computer system. The method is ideally suited for programming on digital
computers. It has given reasonable answers for many complex problems with a reasonable
expenditure of computer time.

Lagrangian and Eularian System 2.
There are two methods of numerically solving a hydrodynamic system. With the Eularian
system, we us a fixed mesh in space and study the fluid as it passes. In the Lagrangian
system, we study the motion of individual particles so that our mesh points move with the
particles they are associated with. The pressure (P), specific volume (V),
internal energy (E), and position (R) of each point is studied as it varies in time. The
[1]
Lagrangian system has proved very useful when a linear mesh is possible and numerical
computation is used. These concepts will be defined more carefully later on.
This paper uses the Lagrangian system.

3. Equations of Motion
Each mesh point (j) can be thought of a particle between R and R+dR. For each point,
therefore, we have that M = 1/v dT. Using
2 2 th F = M x d r/dt , where F is the force on the j element. (1.1) j j
and F δP/δR x δR (1.2) j =
the basic equation of motion is therefore that
2 2 d r/dt = -V x δP/δR (1.3)
This simply states that the acceleration of any mesh point is equal to minus the specific
volume times the pressure gradient at that point.

4. Energy Equation
If E is defined as the specific internal energy, then we can derive, using the principle of
conservation of energy, that dE/dT = -P x dV/dt (1.4)

More elaborate formulas will be given later for computation of these quantities when used to
forming an estimate of the accuracy of the computation. These checks use the laws of
conservation of energy

[2]
5. Properties of Shock Waves
The formation of shock waves can be demonstrated using equation (1.3). When a
pressure wave is introduced into medium the particles in the high pressure region travel faster
than the particles in the low or medium pressure regions.

t = 1 t = 2 t = 3
P P P
R R R


The front of the wave gradually gets steeper as time progresses. After a sufficient
time, the wave front gets infinitely steep and a true mathematical discontinuity is present. If
however, viscosity is taken into account, it is seen that physically no true discontinuity is
present. In the integration scheme used a non-linear viscosity term is introduced to produce a
shock wave of sufficient width for the differential equations to hold.

[3]
II
AN ILLUSTRATION:
AN ANALYTIC SOLUTION OF A PROBLEM

1. A plane piston problem
As an example of the analytic solution of the shock equations, we will consider a
steady -- state shock. Given a long pipe containing a fluid; the piston is pushing from one
side; the velocity of the piston is constant; after a sufficient time. We have the following
situation:
Moving Material Undisturbed Material
Shock Front

When the velocity of the piston and properties of the undisturbed material are known, all
other information can be derived.
The problem, we will do is a simplification of the general plane piston problem
described above. The boundary conditions at the shock front are given by the Rankine-
2Huguenot equations.
2. Definition of Terms
In the definitions that follow, U is defined with respect to a set of axes fixed to the
pipe. On the other hand, the D and D are measured with respect to the shock front. 1 2
M Mass of material per unit cross-sectional area flowing through the shock wave
in unit time.

U velocity of the shock wave.
D + U velocity of material before passing through the shock way 1
D + U velocity of material after passing through the shock way 2
[4]

ρ and ρ density of fluid before and after passing through the shock way 1 2
V and V specific volume of fluid before and after passing through the shock wave 1 2
P and P pressure of fluid before and after passing through the shock wave 1 2
ξ and ξ specific internal energy before and after passing through the shock wave 1 2
E and E specific total energy before and after passing through the shock wave 1 2
3. Huguenots equations for shocks
+ U and D + In this section, we will solve the problem for the special case where D1 2
U are constant.
Conservation of matter
M = D ρ = D ρ (2.1) 1 1 2 2
Conservation of momentum
M (D – D ) = P – P 2 1 1 2
(2.2)
If we consider the mass of fluid M, passing through the shock wave as a particle, the force
pushing is P minus P and the rate of change of momentum is M (D – D ) 1 2 2 1
Conservation of energy
M (E – E ) = D P – D P (2.3) 2 1 1 1 2 2
The work done per unit area and time (D P – D P is equal to the rate of change of energy. 1 1 2 2)
2If M D /2 is the kinetic energy and M ξ is the total internal energy of fluid passing through
the shock wave, then equation (2.3) follows immediately from
22DD21Mε +M− ε +=Dρ Dρ11 - 22 ( ) ( )

[5]
From equation (2.2) and (2.3) we derive that
E – E = P V – P V 2 1 1 1 2 2
(2.5)
By using equation (2.1) and (2.2), then applying (2.5) with suitable rearranging we have
(P + P )/ 2 = (ξ – ξ )/ (V – V ) (2.6) 1 2 2 1 1 2
Equation of State
The material we will consider is a perfect gas. Its equation of state is
ξ = PV/(γ – 1)
We will for simplicity of calculation take γ = 2. Then the relation between energy, pressure,
and volume becomes
P = ξ/V

Analytic Solution of Advancing Shock Front 4.
We will solve the problem of section 1 using the following numbers.
The velocity of the piston is 4 cm/sec
For the undisturbed material we will take
3 V = 3 cm /gm 1
D + U = 0 1
E , ξ = 0 1 1
P = 0 1

[6]
Substituting these values in the equations of the previous section we have
D – D = 4 (2.8) 2 1
P = ξ /V (2.9) 2 2 2
P /2 = ξ /(3-V ) (2.10) 2 2 2
D /3 = D /V (2.11) 1 2 2
And
221DD1 12 −+=ε 22D 42 42
These equations can be solved simultaneously. The values obtained are
V = 1 2
U = 6
ξ = 8 2
P = 8 2
M = 2
We have therefore determined that the shock wave moves at 1 ½ times speed of the
material behind it. The gas is compressed 3 fold