Page 145 - 48Fundamentals of Compressible Fluid Mechanics

107

First, the exit Mach number has to be determined. This Mach number can
be calculated by utilizing the isentropic relationship from the large tank to shock
(point “x”). Then the relationship developed for the shock can be utilized to cal-
culated the Mach number after the shock, (point “y”). From the Mach number
, the Mach number at the exit can be calculated utilizing the
isentropic relationship.
after the shock,
It has to be realized that for a large tank the inside conditions are essen-
tially the stagnation conditions (This statement said without a a proof, but can be
shown that the correction is negligible for a typical dimension ratio that is over 100.
For example, in the case of ratio of 100 the Mach number is 0.00587 and the er-
ror is less than %0.1). Thus, the stagnation temperature and pressure are known
. The star area (the throat area), , before the shock
is known and given as well.

%

and
utilizing the Table (5.1) or equation (4.49) or the GDC–
3

is about 2.197 as shown table below:

With this ratio

Potto, the Mach number,

can be ob-

tained. From equation (5.22) or from Table (5.1)

0 !( 3 . With these values,
the Mach number,
With this Mach number,
the subsonic branch can be evaluated for the pressure and temperature ratios.

(

%(*%(

'%((

% 3'

(

%

3

From Table (??) or from equation (4.11) the following table for the isentropic

relationship is obtained
(
( 3*3

!(

Again utilizing the isentropic relationship the exit conditions can be evalu-

ated. With known Mach number the new star area ratio, is known and the
exit area can be calculated as

3%

%3 !( '%'%

0 (

% *3%

!

, one can obtain using the isentropic rela-
tionship as
#
#
%%'%
*%%
3
with this area ratio,
%'%'%