Potto Home  Contact Us  
Next: Nozzle efficiency Up: gasDynamics Previous: Tables of Normal Shocks, Index

Since the key word ``large tank'' was used that
means that the stagnation temperature and
pressure are known and equal to the conditions in the tank.
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 calculated the Mach number after the shock, (point ``y''). From the Mach number after the shock, , the Mach number at the exit can be calculated utilizing the isentropic relationship. It has to be realized that for a large tank, the inside conditions are essentially the stagnation conditions (this statement is said without 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 error is less than %0.1). Thus, the stagnation temperature and pressure are known and . The star area (the throat area), , before the shock is known and given as well.
With this ratio utilizing the Table (5.1) or equation (4.48) or the GDCPotto, the Mach number, is about 2.197 as shown table below:
With this Mach number, the Mach number, can be obtained. From equation (5.22) or from Table (4.2) . With these values, the subsonic branch can be evaluated for the pressure and temperature ratios.
From Table (4.2) or from equation (4.11) the following Table for the isentropic relationship is obtained
Again utilizing the isentropic relationship the exit conditions can be evaluated. With known Mach number the new star area ratio, is known and the exit area can be calculated as with this area ratio, , one can obtain using the isentropic relationship as
Since the stagnation pressure is constant as well the stagnation temperature, the exit conditions can be calculated. P_exit = ( P_exit P_0 ) ( P_0 P_y) ( P_y P_x) ( P_x P_0) P_0
The exit temperature is
For the ``critical'' points "a" and "b" are the points that the shock doesn't occur and yet the flow achieve Mach equal 1 at the throat. In that case we don't have to go through that shock transition. Yet we have to pay attention that there two possible back pressures that can ``achieve'' it or target. The area ratio for both cases, is In the subsonic branch (either using equation or the isentropic Table or GDCPotto as
For the supersonic sonic branch It should be noted that the flow rate is constant and maximum for any point beyond the point "a" even if the shock is exist. The flow rate is expressed as following The temperature and pressure at the throat are:
The temperature at the throat reads
The speed of sound is
And the mass flow rate reads
It is interesting to note that in this case the choking condition is obtained ( ) when the back pressure just reduced to less than 5% than original pressure (the pressure in the tank). While the pressure to achieve full supersonic flow through the nozzle the pressure has to be below the 42% the original value. Thus, over 50% of the range of pressure a shock occores some where in the nozzle. In fact in many industrial applications, these kind situations exist. In these applications a small pressure difference can produce a shock wave and a chock flow. 
For more practical example^{6.1}
from industrial application point of view.
Solution
A solution procedure similar to what done in previous example
(6.1) can be used here.
The solution process starts at the nozzle's exit
and progress to the entrance.
The conditions at the tank are again the stagnation conditions. Thus, the exit pressure is between point ``a'' to point ``b''. It follows that there must exist a shock in the nozzle. Mathematically, there are two main possibles ways to obtain the solution. In the first method, the previous example information used and expanded. In fact, it requires some iterations by ``smart'' guessing the different shock locations. The area (location) that the previous example did not ``produce'' the ``right'' solution (the exit pressure was . Here, the needed pressure is only which means that the next guess for the shock location should be with a larger areaThe second (recommended) method is noticing that the flow is adiabatic and the mass flow rate is constant which means that the ratio of the (upstream conditions are known, see also equation (4.71)).
With the knowledge of the ratio which was calculated and determines the exit Mach number. Utilizing the Table (4.2) or the GDCPotto provides the following table is obtained
With these values the relationship between the stagnation pressures of the shock are obtainable e.g. the exit Mach number, , is known. The exit total pressure can be obtained (if needed). More importantly the pressure ratio exit is known. The ratio of the ratio of stagnation pressure obtained by
Looking up in the Table (4.2) or utilizing the GDCPotto provides
With the information of Mach number (either or ) the area where the shock (location) occurs can be found. First, utilizing the isentropic Table (4.2).
Approaching the shock location from the upstream (entrance) yields Note, as ``simple'' check this value is larger than the value in the previous example. 