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One Dimensional Piston Problem: Tel: 631-5736 Email:atassi.1@nd - Edu

This document describes the governing equations and solution methodology for a one-dimensional piston problem involving the motion of gas. The key details are: 1) The problem involves two first-order partial differential equations describing conservation of mass and momentum for an isentropic gas flow. 2) Characteristic equations are derived relating the Riemann invariants along characteristics to gas properties. 3) A simple wave solution is constructed for regions near the initial state, assuming no shocks occur. Expressions are derived relating gas velocity and speed of sound to piston velocity.

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134 views4 pages

One Dimensional Piston Problem: Tel: 631-5736 Email:atassi.1@nd - Edu

This document describes the governing equations and solution methodology for a one-dimensional piston problem involving the motion of gas. The key details are: 1) The problem involves two first-order partial differential equations describing conservation of mass and momentum for an isentropic gas flow. 2) Characteristic equations are derived relating the Riemann invariants along characteristics to gas properties. 3) A simple wave solution is constructed for regions near the initial state, assuming no shocks occur. Expressions are derived relating gas velocity and speed of sound to piston velocity.

Uploaded by

mohammadtari
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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UNIVERSITY OF NOTRE DAME

DEPARTMENT OF AEROSPACE AND MECHANICAL ENGINEERING


Professor H.M. Atassi AME-562
113 Hessert Center Mathematical Methods II
Tel: 631-5736
Email:atassi.1@nd.edu
One Dimensional Piston Problem
1 Governing equations
Consider a one-dimensional duct aligned with the x-axis. To the right of the piston, the
duct is full with a gas at rest. At time t=0 the piston starts moving with a velocity
u
p
(t). Let , u, and s describe the gas density, velocity and entropy. We assume an
isentropic process and therefore the gas properties are governed by the two rst order
partial dierential equations:

t
+ u

x
+
u
x
= 0 (1)
u
t
+ u
u
x
+
c
2

x
= 0 (2)
where c is the speed of sound. We assume the piston position is given by the function
X(t). The initial conditions are
t 0, u = 0 , c = c
0
(3)
and the boundary conditions are,
t > 0, u(X(t), t) = u
p
=

X(t). (4)
2 Characteristics Equations
I
+
:
dx
dt
= u + c C
+
(5)
I

:
dx
dt
= u c C

(6)
II
+
: du +
c

d = 0 on C
+
(7)
II

: du
c

d = 0 on C
+
(8)
1
Note that since the gas is isentropic, ( 1)
d

= 2
dc
c
. Therefore (7-8) can be rewritten
as
II
+
: du +
2
1
dc = 0 on C
+
(9)
II

: du
2
1
dc = 0 on C

(10)
Integrating 9 along C
+
and 10 along C

, gives
II
+
: u +
2
1
c = r

, on C
+
(11)
II

: u
2
1
c = s

, on C

(12)
where r

and s

constant along C
+
and C

, respectively. r

and s

are known as the


Riemann invariants.
3 Simple Waves
If one of the Riemann invariants is constant throughout the domain, the solution corre-
sponds to a wave motion in only one direction. In the present problem,
u =
r

+ s

2
(13)
c =
1
2
(r

) (14)
In general, the value of u and c will depend on the two parameters (r

, c

) indicating
two waves where the information is propagating along the two families of characteristics.
On the other hand, if s

=constant everywhere, then u and c receive information from


the characteristic C
+
only through the variation of r

from one characteristic to another.


Problems reduced to simple waves are much simpler to solve.
3.1 Proposition
The solution in a region adjacent to a constant state is always a simple wave solution.
4 Solution
We construct the solution assuming no breaking will occur. This will be examined later.
2
4.1 Steady State Region: t > 0, x > c
0
t
Along C
+
originating from the positive x-axis,
u +
2
1
c =
2
1
c
0
(15)
Similarly along C

originating from the positive x-axis,


u
2
1
c =
2
1
c
0
(16)
These equations imply that for the region x > c
0
t, we have
u = 0 c = c
0
(17)
The charcteristics in this region are straight lines whose equations are
x x
+
0
= c
0
t along C
+
(18)
x x

0
= c
0
t along C

, (19)
where x
+
0
and x

0
are the points of their intersection with the x-axis. Note in this region
there no waves and the gas remain quiescent.
4.2 Simple Wave Solution
All characteristics C

will originate from the positive x-axis. Thus they have the same
Riemann invariant as (16) shows. This means that we have the following relationship
u =
2
1
(c c
0
). (20)
Equation 20 is valid everywhere in the eld, assuming no shocks. Substituting 20 into
15 show that both u and c are constant along C
+
originating from the piston surface.
Hence, we have
u = u
p
=

X(t) (21)
c = c
p
(22)
Since both u and c are constant along C
+
we can integrate 5,
x = X() + (u
p
+ c
p
)(t ) (23)
or
x = X() + (c
0
+
+ 1
2

X())(t ) (24)
3
5 Piston Moving with a Constant Speed (-V)
Substituting

X by V , we get
x = V t + (c
0

+ 1
2
V )(t ) (25)
u = V (26)
c = c
0

1
2
V (27)
In the fan region dened by
(c
0

+ 1
2
V )t < x < tc
0
,
we have
dx
dt
= u + c = c
0
+
+ 1
2
u (28)
Moreover along C
+
, c +
1
2
u = constant. Therefore u and c are constant. As a result,
x = (c
0
+
+ 1
2
u)t (29)
and
u =
2
+ 1
(
x
t
c
0
) (30)
c =
2
+ 1
c
0
+
1
+ 1
x
t
(31)
6 Breaking
It is easy to show that breaking will occur if

X < 0.
4

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