3.13. An inviscid, perfect gas is contained in a rigid sphere of radius L. Show that the natural frequencies of spherically symmetric oscillations of the gas are given by tan(wL/a.) = coLloo. 3.21....

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 i
otational 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


 ...