expansion of cylindrical shock waves with counterpressure … 2016. 4. 1. · section 2.2 when the...

Bachelor Thesis

Expansion of cylindrical shock waves withcounterpressure produced by instantaneous release

of energy

ausgeführt am Institut für Strömungsmechanik und Wärmeübertragung (E322)

der Technischen Universität Wien

im Rahmen der Lehrveranstaltung

166.600 Bachelorarbeit

betreut von

Ao.Univ.Prof. Dr. Herbert Steinrück und Dr. Georg Meyer

von

Clemens Gößnitzer

Matrikelnummer: 1126267

Große Neugasse 22 – 24/1/15

1040 Wien

Wien, im September 2014

Contents

Abstract iii

Nomenclature v

1 Introduction 1

2 Physical model 32.1 Solution for strong shock waves . . . . . . . . . . . . . . . . . . . . . 5

2.2 Extension to shock waves with counter-pressure . . . . . . . . . . . . 8

2.2.1 Expansion with respect to small τ . . . . . . . . . . . . . . . . 10

2.2.2 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . 13

2.2.3 Energy balance . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Numerical implementation 19

4 Numerical solutions 214.1 Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5 Conclusions 29

Bibliography 31

i

Abstract

In this thesis, the expansion of cylindrical shock waves produced by instantaneous

release of energy per unit length along a line is investigated, as it can occur due to

a line explosion or a spark discharge. First, the conditions of a strong shock wave

are presumed, i.e. the pressure of the surrounding is – compared to the pressure

behind the shock – very low and thus can be neglected. Afterwards the equations

are expanded to consider and calculate the impact of the counter-pressure. Since

those equations are complex, they are simplified by asymptotic expansion for small

times and only the leading and first order terms are considered. The numerical

implementation of this problem is described, and all obtained results are ploted and

discussed, and conclusions are given.

iii

Nomenclature

Superscripts� dimensional

�partial derivate with respect to η

· partial derivate with respect to τ

h homogeneous part of the differential equation system

p particular part of the differential equation system

(m) discretisation step with respect to τ

(n) discretisation step with respect to η

Subscripts

0 order O(τ 0)

1 order O(τ 1)

s immediately behind the shock wave

u undisturbed surrounding

Symbols

∆η step size

η radius scaling function

γ ratio of specific heats

φ velocity scaling function

v

Nomenclature

Π non-dimensional product to determine characteristic values

ψ density scaling function

ρ density

τ time scaling function

A parameter dependent on τ ; constant

a speed of sound

cv specifit heat capacity at constant volume

e internal energy

Ec characteristic energy per unit length

Ecyl energy per unit length inside the cylinder

Etot total energy per unit length in the system

Eu energy per unit length as pressure-volume work

f pressure scaling function

lc characteristic length

M molecular weight of air

m mass per unit length

Ma Mach number

P perimeter of the shock wave cylinder

p pressure

R radius of shock wave

r radius

Ra common gas constant for air

Ru universal gas constant

vi

Nomenclature

T temperature

t time

tc characteristic time

Tref reference temperature to define the internal energy

U speed of shock wave

u velocity

vii

1 Introduction

The instantaneous release of energy along a straight line is investiaged as it happens

in the first stage of an explosion or in spark discharge. Afterwards the expansion

of the resulting shock wave is described and calculated. Figure 1.1 shows the shock

wave cylinder for two different times.

Ei

Ei

t0 = 0

U

p, u, ρ

U

shock wave

ps, ρs pu, ρu

p, u, ρ

pu, ρu

R

r

t1 > 0

Figure 1.1: Model of the cylindrical shock wave.

The results of this thesis will be embedded and used in a research project [3] [4],

where the ignition in gas engines will be modeled. The aim of this thesis is to provide

a profile of pressure, density and velocity at a certain time.

Lin [10] extended G.I. Talylor’s analysis [2] of strong spherical explosions to strong

cylindrical shock waves, presented a set of ordinary differential equations for a sim-

ilarity solution and solved it nummerically. Here, we follow the scaling of [10] and

consider the case of a non-vanishing counter-pressure.

Plooster [7] investigated this problem, too. The main difference between his ap-

1

1 Introduction

proach and the one used in this thesis is Plooster’s definition of the fluid: he does

not consider it as an ideal gas and uses an equation of state called Saha ionization

equation which considers the dissociation and ionization of the fluid. Furthermore, he

does not scale time and radius, respectively. Therefore, no similarity solution exists

and the results are purely numerical.

Ö. Ekici et al. [8] present a model for fast spark discharge with focus on the spark

discharge mechanism. However, we will focus on the fluid mechanical aspects.

In chapter 2,the governing equations will be introduced. In section 2.1, the problem

will be simplified in order to get a preliminary solution. Section 2.2 will provide us

with a more general approach to solving this problem. First, a system of partial

differential equations will be solved, and with the expansion made in section 2.2.1,

said system will become a system of ordinary differential equations. The boundary

conditions described in section 2.2.2 are set by jump conditions for pressure, density

and velocity. Some parameters we introduce can be calculated with an energy balance,

as described in section 2.2.3.

In chapter 3, the numerical implementation of the equations previously obtained

will be presented. The impact of the number of steps used for calculations and its

effect on the precision of the results is investigated.

In chapter 4, a numerical solution with chosen parameters will be showed. In

section 4.1, the limitations of the solutions will be discussed. Any numerical solution

needs to correspond with the general laws of physics, these restrictions then limit

the possible range of certain parameters. In section 4.2, the numerical solutions are

presented and plotted.

In chapter 5, the solution obtained is discussed and some conclusions are given.

2

2 Physical model

In order to solve the problem presented in chapter 1, the following will be as-

sumed [10]:

• constant specific heat capacities,

• friction and heat conduction and radiation are neglected.

• The problem given is considered to be cylindrically symmetrical.

The fundamental equations for solving the given problem are the Euler equation of

momentum and the equation of continuity. With time t, radius r, pressure p, velocity

u and density ρ, it reads:

∂u

∂ t+ u

∂u

∂r= −

1

ρ

∂p

∂r, (2.1)

∂ρ

∂ t+ u

∂ρ

∂r+ ρ

�∂u

∂r+

u

r

�= 0. (2.2)

The total derivate (derivate in frame of reference of fluid particle) of the entropy s

is constant as long as the particle is not hit by a shock [9], where γ is the ratio of

specific heats:

�∂

∂ t+ u

∂

∂r

��p

ργ

�= 0. (2.3)

Furthermore, the shock front is located at r = R. Therefore, the solutions to equa-

tions (2.1) to (2.3) must fulfill the jump conditions over the shock wave which follow

from [5] and [10]. With the speed of the shock U , and the entities before and after

the shock – subscript u and s, respectively – the following equations are obtained:

3

2 Physical model

pspu

=2 γ

γ + 1Ma2 −

γ − 1

γ + 1, (2.4)

ρsρu

=γ + 1

γ − 1 + 2Ma−2 , (2.5)

us

U=

2

γ + 1

�1−Ma−2

�, (2.6)

U =dR

dt, (2.7)

with the speed of sound a and the Mach number Ma

a =

�γ puρu

, (2.8)

Ma =U

a. (2.9)

For the formulation of the energy balance, a system whose boundaries are just outside

of the shock wave is chosen. At this boundary, the gas is still at rest and thus, the

increase of energy per unity time of the control volume is equal to the internal energy

of the mass which is passed over with the shock wave moving boundary, see figure 2.1.

tt+ dt

dEcyl = ρu U P eu dt

R

R + dR

Ecyl

T , ρ, e, u Tu, ρu, eu

shock wave

system boundary

U = dR/dt

Figure 2.1: System boundary and energy per unit length transfer into the system.

dEtot = ρu U P eu dt, (2.10)

4

2.1 Solution for strong shock waves

with the perimeter of the chosen system P , specific heat at constant volume cv and

the definition of internal energy per unit length e

e(T ) = cv (T − Tref ), (2.11)

and thus

dEtot

dt= ρu U P cv

�Tu − Tref

�. (2.12)

The energy per unit length in the system Ecyl consists of the internal energy per unit

length and the kinetic energy per unit length in the system. By integrating over the

area of the cross section of the shock wave cylinder, Ecyl conforms with:

Ecyl =

�

A

�ρ e+

1

2ρ u2

�dA =

�

A

�ρ cv (T − Tref ) +

1

2ρ u2

�dA. (2.13)

By setting the reference temperature of internal energy Tref to Tu, the energy per

unit length in the system remains constant, see equation (2.12). Therfore, the total

energy per unit length Etot can be defined:

Etot = 2 π

� R

0

ρ

�cv�T − Tu

�+

1

2u2

�r dr, (2.14)

or, using the ideal gas equation (see also [1]):

Etot = 2 π

� R

0

�p− puγ − 1

+1

2ρ u2

�r dr. (2.15)


First, it is assumed that the pressure in front of the shock wave has no influence on

the shock or the fluid behind it, i.e. the pressure after the shock wave is significantly

higher than the pressure before, ps � pu, or, in other words, the Mach number is

5

2 Physical model

high, Ma � 1. Because of lack of a reference length and time, it is possible to

introduce a similarity varible η and the partial differential equations (2.1) to (2.3)

become ordinary differential equations. By scaling pressure, density and velocity the

solutions obtained are non-dimensional. First, partially dimensional scaling functions

are introduced. Later, another set of completely non-dimensional scaling functions is

used. Introducing the dimensional scaling functions for pressure f and velocity φ, the

non-dimensional scaling function for density ψ, and the similarity variables η which

scales radius r and τ which scales time t [10]:

p(r, t) =pu

R2(τ)f(η), (2.16)

u(r, t) =1

R(τ)φ(η), (2.17)

ρ(r, t) = ρu ψ(η), (2.18)

r = R(τ) η(r, t), (2.19)

t = tc τ(t). (2.20)

Reference time tc and reference length lc are introduced here and will be defined in

section 2.2 when the counter-pressure is reintroduced into the problem. Inserting the

scaling functions into equations (2.1) to (2.3), following equations are obtained after

rearranging terms and putting the equations in standard variable separated form:

R ˙R

tc=

�φ φ� +

puρu

f �

ψ

��φ+ η φ�

�−1, (2.21)

R ˙R

tc=

φ

η+

�φ� +

φ

η

�ψ

η ψ� , (2.22)

R ˙R

tc=

�γ f φ

ψ�

ψ− φ f �

��γ η f

ψ�

ψ− η f �

− 2 f

�−1

. (2.23)

Given that the left part of equations (2.21) to (2.23) is only dependent on time,

whereas the right part is only a funtion of η, both parts must be constant. The

resulting constant is being called A:

6


A : =R ˙R

tc. (2.24)

The time independent constant A has the dimension of a velocity multiplied by a

length. With this definition, it is possible to introduce a set of completely non-

dimensional scaling functions [10]:

f(η) =A2

a2f(η), (2.25)

φ(η) = Aφ(η). (2.26)

With these new functions, equations (2.21) to (2.23) become, after sorting the indi-

vidual terms with respect to the derivate:

γ−1 f �− (η − φ)ψ φ� = φψ, (2.27)

−ψ φ� + (η − φ)ψ� = φψ η−1, (2.28)

−(η − φ)ψ f � + γ (η − φ) f ψ� = 2 f ψ. (2.29)

In order to describe the problem more clearly, the system of ordinary differential

equations is written in matrix-vector form:

Lx� = n (2.30)

with

L =

γ−1 −(η − φ)ψ 0

0 −ψ η − φ

−(η − φ)ψ 0 γ f (η − φ)

,

x =�f(η) φ(η) ψ(η)

��,

n =�φψ φψ η−1 2 f ψ

��.

7

2 Physical model

This system of ordinary differential equations can be solved subjected to the boundary

conditions given by equations (2.4) to (2.6). Inserting the scaling functions into the

boundary conditions, it reads as follows:

f(1) =2 γ

γ + 1−

γ − 1

γ + 1Ma−2, (2.31)

φ(1) =2

γ + 1−

2

γ + 1Ma−2, (2.32)

ψ(1) =γ + 1

γ − 1−

2 (γ + 1)

(γ − 1)2Ma−2. (2.33)

For the similarity solution a very high Mach number is assumed, Ma � 1. Therefore,

the boundary conditions simplify to:

f(1) ∼2 γ

γ + 1, (2.34)

φ(1) ∼2

γ + 1, (2.35)

ψ(1) ∼γ + 1

γ − 1(2.36)

The imparted energy per unit length must remain constant in the system, therefore

A can be calculated by inserting the scaling functions into equation (2.15), with the

assumption that ps � pu:

Etot = 2 π ρu A2

� 1

0

�f

γ (γ − 1)+

φ2ψ

2

�η dη. (2.37)

2.2 Extension to shock waves with counter-pressure

For this extension, the counter-pressure in front of the shock wave is not neglected

any longer. Therefore, the scaling functions (2.16) to (2.18) change and are now

explicitly time, i.e. τ , dependent as follows:

p(r, t) =pu

R2(τ)f(η, τ), (2.38)

8


u(r, t) =1

R(τ)φ(η, τ), (2.39)

ρ(r, t) = ρu ψ(η, τ). (2.40)

Because of the time dependence in the scaling functions f , φ and ψ, there are addi-

tional terms in equations (2.21) to (2.23):

R ˙R

tc=

�φ φ� +

puρu

f �

ψ+

R2

tc

˙φ

��φ+ η φ�

�−1, (2.41)

R ˙R

tc=

φ

η+

�φ� +

φ

η

�ψ

η ψ� +R2

η tc

ψ

ψ� , (2.42)

R ˙R

tc=

�γ f φ

ψ�

ψ− φ f � +

R2

tc

�γ f ψ

ψ−

˙f

��γ η f

ψ�

ψ− η f �

− 2 f

�−1

. (2.43)

The left part of equations (2.41) to (2.43) is only dependent on time, whereas the

right part is merely a function of η and τ . Thus, parameter A can only depend on τ :

A(τ) : =R(τ) ˙R(τ)

tc. (2.44)

With this definition, the completely non-dimensional scaling functions conform with:

f(η, τ) =A2(τ)

a2f(η, τ), (2.45)

φ(η, τ) = A(τ)φ(η, τ), (2.46)

A(τ) =l2ctcA(τ), (2.47)

R(τ) = lc R(τ). (2.48)

Inserting the scaling functions into equations (2.41) to (2.43) and ordering terms with

respect to the derivates, it follows:

γ−1 f �− (η − φ)ψ φ� = φψ −

R2

Aψ φ−

R2

A2ψ φ A, (2.49)

9

2 Physical model

−ψ φ� + (η − φ)ψ� = φψ η−1 +R2

Aψ, (2.50)

−(η − φ)ψ f � + γ f(η − φ)ψ� = 2 f ψ +R2

A

�γ f ψ − ψ f

�− 2

R2

A2f ψ A. (2.51)

Putting this system of differential equations in matrix form, it reads as follows:

Lx� = n+R2

AK x+

R2

A2m A, (2.52)

with

L =

γ−1 −(η − φ)ψ 0

0 −ψ η − φ

−(η − φ)ψ 0 γ f (η − φ)

,

x =�f(η, τ) ψ(η, τ) ψ(η, τ)

��,

n =�φψ φψ η−1 2 f ψ

��,

K =

0 −ψ 0

0 0 1

−ψ 0 γ f

,

m =�−φψ 0 −2 f ψ

��.

2.2.1 Expansion with respect to small τ

Now all functions and variables are expanded with respect to τ . Entities of oder

O(τ 0) have subscript 0, entities of order O(τ 1) have subsrcipt 1:

f(η, τ) ∼ f0(η) + τ f1(η), (2.53)

φ(η, τ) ∼ φ0(η) + τ φ1(η), (2.54)

ψ(η, τ) ∼ ψ0(η) + τ ψ1(η), (2.55)

R(τ) ∼√τ (R0 + τ R1) , (2.56)

A(τ) ∼ A0 + τ A1. (2.57)

10


Now two different systems of equations have to be solved: one with order O(τ 0), and

another one with order O(τ 1). Expanding equation (2.52) with respect to the order

of τ will become for the order O(τ 0):

L0 x�0 = n0, (2.58)

with

L0 =

γ−1 −(η − φ0)ψ0 0

0 −ψ0 η − φ0

−(η − φ0)ψ0 0 γ f0 (η − φ0)

,

n0 =�φ0 ψ0 φ0 ψ0 η−1 2 f0 ψ0

��,

which equals the system described in section 2.1, and for the order O(τ 1):

L0 x�1 = −L1 x

�0 + n1 +

R20

A0K0 x1 +

R20

A20

m0 A1, (2.59)

with

R20 = 2A0, (2.60)

and, thus

L0 x�1 = −L1 x

�0 + n1 +K0 x1 +m0 A1, (2.61)

with

L1 =

0 φ1 ψ0 − (η − φ0)ψ1 0

0 −ψ1 −φ1

φ1 ψ0 − (η − φ0)ψ1 0 γ (f1 (η − φ0)− f0 φ1)

,

11

2 Physical model

n1 =�φ1 ψ0 + φ0 ψ1 (φ1 ψ0 + φ0 ψ1) η−1 2 (f1 ψ0 + f0 ψ1)

��,

K0 =

0 −2ψ0 0

0 0 2

−2ψ0 0 2 γ f0

,

m0 =�−2φ0 ψ0 A

−10 0 −4 f0 ψ0 A

−10

��.

In order to simplify the structure of this equation system, x1 is singled out:

L0 x�1 = −L1 x

�0 + n1 +K0 x1 +m0 A1, (2.62)

L0 x1 : = −L1 x�0, (2.63)

N0 x1 : = n1, (2.64)

L0 x�1 =

�L0 +N0 +K0

�x1 +m0 A1, (2.65)

P0 : = L0 +N0 +K0, (2.66)

with

L0 =

0 −φ�

0 ψ0 (η − φ0)φ�0

0 ψ�0 φ�

0

−γ (η − φ0)ψ�0 γ f0 ψ�

0 − f �0 ψ0 (η − φ0) f �

0

,

N0 =

0 ψ0 φ0

0 ψ0 η−1 φ0 η−1

2ψ0 0 2 f0

.

Now the equation system in matrix form is more structured:

L0 x�1 = P0 x1 +m0 A1, (2.67)

with

12


P0 =

0 −ψ0 − φ�

0 ψ0 (η − φ0)φ�0 + φ0

0 ψ0 η−1 + ψ�0 φ0 η−1 + φ�

0 + 2

−γ (η − φ0)ψ�0 γ f0 ψ�

0 − f �0 ψ0 (η − φ0) f �

0 + 2 f0 (1 + γ)

.

2.2.2 Boundary conditions

The boundary conditions dependent on the Mach number Ma were given by equa-

tions (2.31) to (2.33). In order to be able to expand those equations, a correlcation

of Ma−2is necessary. The lowest order of Ma−2

is O(τ 1):

Ma−2 =

�a

dR/dt

�2

, (2.68)

dR

dt=

dR

dτ

lctc, (2.69)

Ma−2∼

4 a2 t2cR2

0 l2c

τ. (2.70)

Therefore, the boundary conditions conform with:

f(1, τ) ∼ f0(1) + τ f1(1), (2.71)

φ(1, τ) ∼ φ0(1) + τ φ1(1), (2.72)

ψ(1, τ) ∼ ψ0(1) + τ ψ1(1), (2.73)

for the order O(τ 0):

f0(1) =2 γ

γ + 1, (2.74)

φ0(1) =2

γ + 1, (2.75)

ψ0(1) =γ + 1

γ − 1, (2.76)

and for the order O(τ 1):

13

2 Physical model

f1(1) = −γ − 1

γ + 1

4 a2 t2cR2

0 l2c

, (2.77)

φ1(1) = −2

γ + 1

4 a2 t2cR2

0 l2c

, (2.78)

ψ1(1) = −2(γ + 1)

(γ − 1)24 a2 t2cR2

0 l2c

. (2.79)

Setting a tc/lc = 1 (see also section 2.2.3), the boundary conditions in new scaling for

η = 1 read as follows:

f1(1) = −2 (γ − 1)

γ + 1A−1

0 , (2.80)

φ1(1) = −4

γ + 1A−1

0 , (2.81)

ψ1(1) = −4 (γ + 1)

(γ − 1)2A−1

0 . (2.82)

2.2.3 Energy balance

The energy per unit length inside the shock wave cylinder consists of internal energy

per unit length and kinetic energy per unit length. By integrating over the area, Ecyl

equals:

Ecyl = 2 π

� R

0

�p

γ − 1+

1

2ρ u2

�r dr. (2.83)

Since the counter-pressure in front of the shock wave is not neglected anymore, a

definition of an energy per unit length Eu that reflects the counter-pressure can be

defined as:

Eu = −2 π

� R

0

puγ − 1

r dr. (2.84)

The total energy per unit length Etot in the system is composed of the energy per

unit length in the cylinder Ecyl, and the energy per unit length Eu. As it was

14


already showed in equations (2.11) to (2.15), the total energy per unit length remains

constant:

Etot = Ecyl + Eu = const. (2.85)

Using the scaling functions, Ecyl, Eu and Etot can be written as:

Ecyl = 2 πρu l4ct2c

A2

� 1

0

I η dη, (2.86)

Eu = −π pu l2c R

2, (2.87)

Etot = πρu l4ct2c

�2A2

� 1

0

I η dη −a2 t2cl2c

R2

�, (2.88)

with

I =f

γ (γ − 1)+

φ2 ψ

2.

Scaling Etot, Ecyl and Eu with a characteristic energy per unit length Ec, it reads:

Etot = Ec Etot, (2.89)

Ecyl = Ec Ecyl, (2.90)

Eu = Ec Eu. (2.91)

Looking at equations (2.88) and (2.90), three non-dimensional products can be found

to determine the characteristic quantities:

Π1 =Etot

Ec

(2.92)

Π2 =a tc

lc(2.93)

Π3 =ρu l4cEtot t2c

. (2.94)

15

2 Physical model

Setting Π1 = Π2 = Π3 = 1 without loss of generality, it reads:

Ec = Etot, (2.95)

tc =

�Etot

ρua−2, (2.96)

lc =

�Etot

ρua−1. (2.97)

By expanding the total, i.e. initial, energy per unit length Etot and the energy per

unit length inside the shock wave cylinder Ecyl, the energies per unit length conform

with:

Etot ∼ Etot,0 + τ Etot,1, (2.98)

Ecyl ∼ Ecyl,0 + τ Ecyl,1. (2.99)

The energy per unit length Eu scales with R2, therefore it is of the order O(τ 1) and

higher, and only the first order will be considered:

Eu ∼ τ Eu,1. (2.100)

Therefore, and with the scaling for the energy per unit length, the leading oder terms

are:

Etot,0 = Ecyl,0 = 1, (2.101)

Etot,0 = 2 πA20

� 1

0

I0 η dη, (2.102)

with

I0 =f0

γ (γ − 1)+

φ20 ψ0

2.

16


The total energy per unit length remains constant. Thus, the first order terms of Ecyl

and Eu, i.e. Etot,1, must equal zero:

Etot,1 = Ecyl,1 + Eu,1 = 0, (2.103)

Etot,1 = 2 πA20

� 1

0

I1 η dη + 4 πA0 A1

� 1

0

I0 η dη −π

γ(γ − 1)R2

0, (2.104)

with

I1 =f0

γ (γ − 1)+ φ0 φ1 ψ0 +

φ20 ψ1

2.

Now A0 can be expressed using equation (2.102):

A0 =1�

2 π� 1

0 I0 η dη. (2.105)

The value of A1 has to be know in oder to be able to calculate the correct value of

I1:

A1 =π

2 γ (γ − 1)A0 R

20 − πA3

0

� 1

0

I1 η dη. (2.106)

As the system of equations is linear in the order O(τ 1) terms, two values for A1

can be set at will and solve those two problems individually. If A1 = 0, the problem

is linear and homogeneous, therefore this solution will have the index h, whereas the

particular solution will have the index p. Ah1 = 0 and Ap

1 = 1 will be chosen. When

the solution to both problems is known, the real value of A1 can be calculated by

superposing the two solutions. The superposed solution must also fulfill the boundary

conditions. Therefore, the boundary condition for the homogeneous solution are

chosen according to equations (2.80) to (2.82), and for the particular solution the

boundary conditions are set zero, xp1(1) = 0. Now A1 can be calculated using the

physical condition, that the velocity must be zero u = 0 at η = 0. That means that

there is no source of mass at the center of the shock wave cylinder, i.e. the energy

per unit length balance can be satisfied. Equations (2.39) and (2.46) show that this

17

2 Physical model

means that φ1(0) = 0. A1 will be chosen accordingly:

x1 = xh1 + A1 x

p1, (2.107)

φ1 = φh1 + A1 φ

p1, (2.108)

A1 = −φh1(0)

φp1(0)

. (2.109)

18

3 Numerical implementation

The numerical method used to solve the system of ordinary differential equations

presented before is the explicit Euler method. The BLAS functions were provided

by the GNU Scientific Library. The first system, i.e. equation (2.58), is non-linear,

the second one, i.e. equation (2.67), is linear. First, the system of the order O(τ 0)

problem is solved by starting at η = 1 and integrating to η = 0, then A0 is calculated

using equation (2.105). The integral in this equation is solved by the GNU Scientific

Library provided function using the QAG adaptive integrating algorithm with the 61

point Gauß-Kronrod rules [6]. Then the solution to the second system is obtained

with two chosen values for A1 (Ah1 = 0 and Ap

1 = 1) as described in section 2.2

starting again at η = 1 integrating to η = 0. After getting those solutions, the exact

value of A1 is calculated according to equation (2.109). The two systems of ordinary

differential equations are as follows:

L0 x�0 = n0, (3.1)

L0 x�1 = P0 x1 +m0 A1. (3.2)

These equations are discretised:

L(n)0

x(n)0 − x(n−1)

0

∆η= n(n)

0 , (3.3)

L(n)0

x(n)1 − x(n−1)

1

∆η= P (n)

0 x(n)1 +m(n)

0 A1. (3.4)

The only unknowns in these equations are x(n−1)0 and x(n−1)

1 , respectively:

x(n−1)0 = x(n)

0 −L(n)0

−1n(n)

0 ∆η, (3.5)

19

3 Numerical implementation

x(n−1)1 = x(n)

1 −L(n)0

−1�P (n)

0 x(n)1 +m(n)

0 A1

�∆η. (3.6)

Figure 3.1 shows how the systems of differential equations were solved. Noteworthy

is that the integrating was done starting at η = 1 to η = 0.

x

η0 η(n)η(n−1)

x(n−1) ≈ x(n) − x(n)� ∆η

∆η

x(n)

x(n−1)

∆x

Figure 3.1: Explicit Euler method for integrating.

The number of steps, i.e. the size of ∆η, may not have an influence on the results.

Therefore, comparing two numerical solutions with different numbers of steps will

show if there is a difference between those solutions. If this difference is very small

or inexistent, the number of calucation steps is high enough. Figure 3.2 shows that

the numerical calculation does not change noticeably choosing ∆η1 = 1× 10−4, i.e.

n1 = 1× 104, or ∆η2 = 5× 10−6, i.e. n2 = 2× 105.

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

η

f 0

(a) ∆η1 = 1× 10−4, n1 = 1× 104

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

η

f 0

(b) ∆η2 = 5× 10−6, n2 = 2× 105

Figure 3.2: Comparison of f0 for two different step sizes.

20

4 Numerical solutions

4.1 Limitations

The approach to solve this problem is only valid when τ � 1. The three scaling

functions have to fulfill the following physical conditions:

• The pressure p at η = 1 must be higher than the pressure of the undisturbed

surrounding pu.

• The density ρ at η = 1 must be higher than the density of the undisturbed

surrounding ρu

• Looking at figures 4.2 and 4.8, it is possible to find a τmax,φ so that the velocity

inside the cylinder is zero. This would mean that the shock wave has become

too weak to be described correctly by the equations obtained.

Therefore, three conditions are set for τmax. The first condition assumes that

the pressure directly behind the shock wave may not be below the pressure of the

undisturbed surrounding:

p(r, t)

pu=

A2

R2f(η(r, t), τ(t)), (4.1)

p(R, τmax,f ) = pu, (4.2)

f(1, τmax,f ) =R2(τmax,f )

A2(τmax,f ). (4.3)

The second condition is that the density immediately behind the shock wave may

not fall below the density in the undisturbed surrounding:

21


ρ(r, t)

ρu= ψ(η(r, t), τ(t)), (4.4)

ρ(R, τmax,ψ) = ρu, (4.5)

ψ(1, τmax,ψ) = 1. (4.6)

The third condition for τmax says that the velocity inside the shock wave cylinder

may not be less than zero. If the velocity is zero, that means that the shock wave

has decayed completely, and we cannot fulfil the shock wave equations anymore.

Therefore, the assumptions made in chapter 2 are not valid any longer.

4.2 Results

The following parameters were used for the results presented in this section:

• Tu = 300K,

• pu = 1× 105 Pa,

• γ = 1.4,

• Ei = 5× 10−2Jm

−1,

• Ru = 8.314 JK−1

mol−1

,

• M = 2.897 kg mol−1

,

• n = 2× 106.

Using these parameters for the numerical calculations, the calculated parameters

conform with:

• A0 = 0.504,

• A1 = 1.983,

• lc = 5.976× 10−4m,

• R0 = 1.004,

• R1 = 0.988,

• tc = 1.721× 10−6s,

• τmax,f = 0.853,

• τmax,φ = 0.182,

• τmax,ψ = 0.042.

22

4.2 Results

Table

4.1

:R

esults

ofnum

eric

alcalc

ula

tio

n.

ηf 0

φ0

ψ0

f 1φ1

ψ1

fh 1

φh 1

ψh 1

fp 1

φp 1

ψp 1

1.00

+1.16

7+0.83

3+6.00

0−0.66

1−3.30

7−11

9.0

−0.66

1−3.30

7−11

9.0

+0.00

0+0.00

0+0.00

00.95

+0.84

2+0.76

1+3.18

2+2.39

6−3.10

9−3.74

8−0.68

4−4.42

7−16.28

+1.55

3+0.66

5+6.32

30.90

+0.67

3+0.69

8+1.88

4+1.75

6−3.03

3+12.49

−3.75

7−6.44

8+2.20

3+2.78

0+1.72

2+5.18

90.85

+0.57

8+0.64

3+1.19

3+0.85

8−2.98

7+12.15

−6.98

6−9.31

4+6.14

0+3.95

5+3.19

0+3.03

50.80

+0.52

2+0.59

3+0.78

6+0.18

5−2.92

3+9.38

0−9.87

2−12.90

+7.65

9+5.07

1+5.03

3+0.86

80.75

+0.48

8+0.54

9+0.52

8−0.26

5−2.82

5+6.77

9−12.35

−17.11

+8.81

3+6.09

5+7.20

3−1.02

50.70

+0.46

6+0.50

8+0.35

6−0.55

3−2.69

5+4.74

8−14.45

−21.87

+9.71

0+7.00

7+9.67

1−2.50

20.65

+0.45

3+0.46

8+0.23

8−0.73

5−2.54

0+3.24

3−16.19

−27.20

+10.17

+7.79

4+12.43

−3.49

40.60

+0.44

5+0.43

1+0.15

6−0.84

6−2.36

7+2.15

7−17.60

−33.17

+10.09

+8.45

2+15.53

−4.00

00.55

+0.44

1+0.39

4+0.10

0−0.91

2−2.18

2+1.38

9−18.72

−39.90

+9.46

6+8.98

1+19.02

−4.07

30.50

+0.43

8+0.35

8+0.06

2−0.95

0−1.99

1+0.86

0−19.57

−47.62

+8.39

5+9.39

0+23.00

−3.80

00.45

+0.43

6+0.32

2+0.03

6−0.97

1−1.79

5+0.50

7−20.19

−56.63

+7.03

0+9.69

1+27.65

−3.28

90.40

+0.43

6+0.28

6+0.02

0−0.98

2−1.59

7+0.28

1−20.61

−67.44

+5.53

9+9.90

0+33.20

−2.65

10.35

+0.43

5+0.25

0+0.01

0−0.98

7−1.39

8+0.14

4−20.89

−80.83

+4.07

8+10.03

+40.05

−1.98

40.30

+0.43

5+0.21

4+0.00

5−0.98

9−1.19

9+0.06

7−21.05

−98.08

+2.77

3+10.11

+48.85

−1.36

40.25

+0.43

5+0.17

9+0.00

2−0.99

0−0.99

9+0.02

7−21.14

−12

1.5

+1.70

6+10.16

+60.77

−0.84

70.20

+0.43

5+0.14

3+0.00

1−0.99

0−0.79

9+0.00

9−21.18

−15

5.8

+0.91

6+10.18

+78.17

−0.45

80.15

+0.43

5+0.10

7+0.00

0−0.99

0−0.59

9+0.00

2−21.20

−21

1.8

+0.40

0+10.19

+10

6.5

−0.20

10.10

+0.43

5+0.07

1+0.00

0−0.99

0−0.40

0+0.00

0−21.20

−32

2.1

+0.12

2+10.19

+16

2.2

−0.06

10.05

+0.43

5+0.03

6+0.00

0−0.99

0−0.20

0+0.00

0−21.20

−64

9.3

+0.01

5+10.19

+32

7.3

−0.00

80.00

+0.43

5+0.00

0+0.00

0−0.99

0+0.00

0+0.00

0−21.20

−∞

+0.00

0+10.19

+∞

+0.00

0

23


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

η

f 0

Figure 4.1: Solution of f0.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.15

0.3

0.45

0.6

0.75

0.9

η

φ0

Figure 4.2: Solution of φ0.

24

4.2 Results

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

3

4

5

6

η

ψ0

Figure 4.3: Solution of ψ0.

0 0.2 0.4 0.6 0.8 1−22

−16.5

−11

−5.5

0

5.5

η

fh 1

(a)

0 0.2 0.4 0.6 0.8 10

2.1

4.2

6.3

8.4

10.5

η

fp 1

(b)

Figure 4.4: Homogeneous (a) and particular (b) solution of f1.

25


0 0.2 0.4 0.6 0.8 1−100

−80

−60

−40

−20

0

η

φh 1

(a)

0 0.2 0.4 0.6 0.8 10

20

40

60

80

100

ηφp 1

(b)

Figure 4.5: Homogeneous (a) and particular (b) solution of φ1.

0 0.2 0.4 0.6 0.8 1−120

−90

−60

−30

0

30

η

ψh 1

(a)

0 0.2 0.4 0.6 0.8 1−5

−2.5

0

2.5

5

7.5

η

ψp 1

(b)

Figure 4.6: Homogeneous (a) and particular (b) solution of ψ1.

26

4.2 Results

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1

−0.5

0

0.5

1

1.5

2

2.5

η

f 1

Figure 4.7: Solution of f1.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

η

φ1

Figure 4.8: Solution of φ1.

27


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−120

−100

−80

−60

−40

−20

0

20

η

ψ1

Figure 4.9: Solution of ψ1.

28

5 Conclusions

The simplifications made in chapter 2.2.1 were necessary because it was not possible

to solve the system of partial differential equations numerically directly. The problems

occuring were primarily the following: when η approaches zero, the results for φ were

physically impossible.

Although there are two different equations to calculate the value of A1, imple-

menting those two equations numerically will give the same result. This shows that

the energy per unit length balance and the physical assumption that the velocity at

η = 0, respectively, are valid and interchangeable

The rather small values of τmax indicate, that terms of higher order than O(τ 1)

become significant quite soon. Therefore,they have to be considered in calcualtions

after a rather short period of time.

However, the solution procedure outlined before may be extended and used in a

simple numerical scheme integrating with respect to time: By setting a solution for

a very small value of ∆τ � τmax as the new order O(τ 0) solution, τ can be increased

step by step:

x(0) = x0 +∆τ x1 (5.1)

x(m)0 : = x(m−1)

(5.2)

x(m) = x(m)0 +∆τ x(m)

1 . (5.3)

This simple extension allows to calculate x for τ > τmax without having to calculate

higher order terms. The vadility of this extension has to be checked seperately and

will not be discussed in this thesis.

29

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31

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