Answer To: 3.13. An inviscid, perfect gas is contained in a rigid sphere of radius L. Show that the natural...
David answered on Dec 27 2021
SOLUTION 3.13:
For inviscid perfect gas,
(t) x x(t)u
Differentiating the above equation, we get,
(t) x x(t)u
Applying time – dependent Bernoulli,
2 2 0
0
0
0
( )1 1
.0
2 2
( )
1
( ') '
t
P tP
ux x u u x u
P t
u
L
u P t dt
L
So, 0
P
u t
L
Then,
2
1
( ) ( )
( )
2 2
a u t au t
x t
a
So,
, .
2
u t x
x t a u t
This implies,
021 0 1 0
2 2
P t
uL u au t
Hence,
0
02
1 1
2 2
1
2
P t
L a u u
P t
Lu u
Now, solving this,
0
0
2
0
P
u
, for 0t
0
2
u t
T
L
and
0
u t
u t
u
So, 21
dU
u
dT
, with 0 0U
Now,
2
0
0
0
1 1
log
1 2 1
tanh
2
du u
dT c
u u
u t
u
L
L
a
SOLUTION 3.21:
Given that, the gas is in steady, homentropic and irrotational flow.
Since, 0u , it is incompressible.
For incompressible fluid, the Euler Equation can be written as:
du P
dt
Consider that the pressure is only varying with x .
dP
P
dx
Thus,
1
( )
du dP
dt dx
dx dP
du
dt
dP
u du i
Now, we know that: P k
, where, k is constant.
Therefore,
1
( )
k
P
P
K
P
ii
K
From equation (i) and (ii), we get,
1
dP
u du
P
k
Integrating on both sides,
1
11
1
2
11
1
2
11 1
2
1 1
1
2
1 1
1
2
constant
2 1
1
constant
2 1
constant
2 1
constant
2 1
constant
2 1
dP
udu
P
k
u k P
u k P
P Pu
u P
u P
SOLUTION 3.22:
We are given that p c .
This means that: p c .
Since the individual fluid elements in this case, still conserve their mass, thus we have the
equation for conservation of mass as:
0u
t
Now, consider the momentum equation:
u
u u p
t
Here, now we substitute the value of p , in above equation, we get:
u
u u c
t
Simplifying, we get:
1
2
P
P x x
Now, we substitute
P
a
. We get the following equation:
2 1a
a x x
The above equation holds good even if x is replaced by t .
Now, consider the equations:
022
1 1
aa
u
P
a
Adding and subtracting the above equations and eliminating the constant 0a , yields a set of
two equations, respectively:
2
0
1
2
0
1
a
u a u
t x
a
u a u
t x
The above equations are known as evolution equations.
SOLUTION 3.23:
Consider the equation below:
0
1
1
2
u F x u a t
When, u F , such that, 0
1
1
2
x u a t
.
Now,
'
1
1 1 '
2
Fu
x
tF
Given,
1
,0 1 tanh
2
x
u x U
L
.
Thus, putting this initial condition, we get,
1
1 tanh
2
F U
L
...