Examples of problems that can be solved with the Gas Dynamics  daemon includes isentropic flows through variable area channels and flows with normal and oblique shocks. In addition, tables such as (a) Isentropic flow tables, (b) Normal shock tables, (c) Oblique shock charts (delta-theta tables), and (d) Prandtl-Meyer  functions are also provided for the advanced users.

c. Governing Equations: The steady state mass, energy and entropy balance equations (see Fig 2) assume simplified expressions as the newly defined properties are introduced. 

For an isentropic flow, for instance, it can be shown that the energy balance equation reduces to Tt_i=Tt_e, the entropy balance reduces to pt_i=pt_e and mass balance reduces to A*_i=A*_e . Because these properties are part of the State, almost all isentropic flow problems can be solved with the help of the state panel of the gas dynamic daemon.
 

Choose He from the state selector, enter the given variables and Calculate State-1. The fully evaluated state is shown in Fig. 2. Note that A*, Tt, pt etc. are properties of State-1, even though to evaluate those properties we had to imagine an isentropic stagnation chamber somewhere else. If the gas expanded from a real stagnation chamber (see Fig. 2) in an isentropic manner, the stagnation properties at State-1 would be identical to the corresponding quantities within the chamber. Moreover, the velocity being zero inside the reservoir, the stagnation properties and static properties are identical there. We use this fact in the following example.

After launching the Gas Dyanmics daemon, choose State-0 as the reservoir. Air is the default gas. Enter p0 (500 kPa), T0 (500 K) and Vel0 (0 m/s) and  Calculate. 

As the back pressure is gradually reduced from 500 kPa, the flow rate is expected to increase until Mach=1 is reached at the exit at which point communications between the reservoir and the back region of the nozzle is cut off. Let State-1 represent the exit state when this critical velocity is achieved. Enter Mach1(=1), p_t1 ('=p_t0'), T_t1 ('=T_t0'), and A1 (40 cm^2). A Calculate produces the critical exit pressure as p1=264 kPa and mdot1=3.61 kg/s .

Now depending on how the back pressure, p_b, compares with p1, different situations can be studied. 

(i) p_t>=p_b>p1 : An exit pressure of p1 =264 kPa produces choking at the throat. Hence for an exit pressure above this value, the flow will be subsonic everywhere. The back pressure and the exit pressure must be equal for such subsonic flows. Given a back pressure, the exit state can be evaluated as the exit pressure (=p_b), exit area, T_t (=temperature of the stagnation chamber) and p_t (=pressure of the stagnation chamber) are all known. Once the exit state has been evaluated, state at any other location can be obtained from the known local area, p_t (=p_t0), T_t (=T_t0) and A* (=A_exit). The solution has no ambiguity as the flow is known to be subsonic everywhere. A trivial limit of this case is no flow when p_b=p_t. 

As an example, change the exit pressure to p1=375 kPa, above the critical value of 264 kPa. A Calculate produces Mach2=0.653 and mdot1=3.195 kg/s. Suppose we would like to know the state at a location where A=80 cm2. Choose State-2, enter A2(=80 cm2), p_t2 ('=p_t0'), T_t2 ('=T_t0'), and either mdot2 ('=mdot1') or A*2('=A*1'). A Calculate produces the complete state with  p2=475 kPa , M2=0.267 and T2=493 K . 

(ii) p_b=p1: This is the limiting situation of the previous case when the flow just reaches sonic velocity at the throat. The exit state for this condition has been already evaluated. The procedure to determine the State at any given location  (known area) remains identical to the one discussed in the previous case. 

(iii) p_b<p1: With the back pressure lower than the lowest possible exit pressure, the nozzle is under-expanded. Further expansion occurs outside the nozzle through isentropic expansion waves (outside the domain of our discussion). Unaffected by these structures outside the nozzle, the solution procedure inside the nozzle remains the same as in case (ii). 

What happens if you allow exit pressure to drop below the critical pressure? Change p1 to, say, 125 kPa, much  below the critical pressure of 264 kPa.  A Calculate produces a supersonic exit velocity with Mach1=1.56 , with A* smaller than A indicating a converging/diverging nozzle.  For this converging nozzle, as the back pressure is reduced below 264 kPa,  the exit state, and hence the mass flow rate and exit velocity,  remain unchanged because the sonic velocity makes it impossible for the pressure information, which travels at the speed of sound, to propagate upstream. The nozzle, under such conditions, is considered to be choked with the exit state invariant with any further drop of back pressure.  

After launching the Gas Dynamics daemon, choose State-0 as the reservoir. Air is the default gas. Enter p0 (500 kPa), T0 (500 K) and Vel0 (0 m/s), and  Calculate . 

Let State-1 represent the throat condition. To determine the exit condition under which choking occurs at the throat, enter Mach1(=1), A1 (40 cm2), p_t1 ('=p_t0') and T_t1 ('=T_t0'). A Calculate produces p1=264 kPa and mdot1=3.61 kg/s . (identical to the choked converging nozzle discussed earlier). 

Choose State-2   to represent the exit condition. Enter  A2 (80 cm2), A*2 ('=A1'),  p_t2 ('=p_t0') and T_t2 ('=T_t0'). A Calculate to produces MN2=0.306 , p2=468 kPa and mdot1=3.6 kg/s . However, the message panel also displays (see Fig. 3) that a supersonic solution with MN2=2.197 exists.

Let State-3   represent the supersonic exit solution. Enter  Mach3 (2.197), A3 ('=A2'),  p_t3 ('=p_t0') and T_t3 ('=T_t0'). A Calculate produces p3=46.98 kPa and mdot1=3.61 kg/s. 

Now depending on how the back pressure compares with p2 and p3, different situations may arise. 

(i) p_b<=p_t and p_b>p2 : An exit pressure of p2 =468 kPa produces choking at the throat. Hence for an exit pressure above this value, the flow will be subsonic everywhere. This situation has been already discussed for the converging nozzle example in the earlier section.

(ii) p_b=p2: This is the limiting situation of the previous case when the flow just reaches sonic velocity at the throat. The exit state for this condition has been already evaluated. The procedure to determine the State at any given location  (known area) remains identical to the one discussed in the case (i).

(iii) p_b=p3: This is the supersonic branch of the isentropic solution where the flow keeps accelerating at the expense of pressure until this low value of pressure is reached a the exit. The exit state for this condition has been already evaluated. The procedure to determine the state at any given location (known area) remains the same as in the case (i), except one should watch out for multiple solutions, one for the converging side (subsonic) and one for the diverging side (supersonic). The subsonic solution is always displayed with the supersonic Mach number appearing on the Message Panel.

(iv) p2>p_b>p3: If the back pressure is slightly higher than p3, isentropic expansion (as in case iii)  may still occur in a nozzle. The exit pressure being less than the back pressure, the nozzle is said to be over-expanded. The pressure recovery immediately outside the nozzle takes place through what is known as oblique shock waves (outside the domain of our discussion). If the back pressure is increased to a particular value, p4 (to be determined), a normal shock wave (to be discussed later) stands at the exit. 

(v) p2>p_b>p4: The normal shock moves inside the diverging section  of the nozzle. As pressure is increased to p2, the shock moves to the throat where it disappears (Mach number=1 at the throat). The solution procedure for flow with a normal shock is discussed in the next section.

(v) p3>p_b: With the back pressure lower than the exit pressure, the nozzle is under-expanded. Further expansion occurs outside the nozzle through isentropic expansion waves (outside the domain of our discussion). Unaffected by these structures outside the nozzle, the solution procedure inside the nozzle remains the same as in case (iii). 

In this case air expands isentropically until it encounters the normal shock, a discontinuity through which the flow turns subsonic.  State-0 and State-1  remain identical to those in section-e. Following the procedure outlined in section e-(iii), we can evaluate the state, say State-2, just before the shock. Enter A2 (60 cm2), A* 2('=A1'), p_t2 ('=p_t0'),  T_t2('=T_t0'), and Calculate. The subsonic solution for the converging part with Mach2=0.43 is displayed, while the Message Panel shows the supersonic Mach number 1.854 (right after you click Calculate). To obtain the supersonic solution, un-check A*2 and enter MN2 as 1.854. Note that the supersonic solution produces the same value of A*2 (40 cm2) and mdot2=3.61 kg/s . 

Switch panel to the Tables and select the Shock Table. Enter M_i as 1.854 and Calculate M_e=0.6048, p_te/p_ti=0.788. Now get back to the States panel and choose State-3   to represent the State immediately after the shock. Enter  Mach3 (0.6048), A3 ('=A2'), p_t3 ('=p_t0*0.788') and T_t3 ('=T_t0'). A Calculate produces the same mass flow rate but a larger critical area as expected,  A*3=50.73cm2 , p3=307 kPa and mdot3=3.61 kg/s. 

To obtain the exit condition,  State-4, enter  A4 (80 cm2), A*4('=A*3),  p_t4 ('=p_t3') and T_t4 ('=T_t0'). A Calculate produces p4=352 kPa and mdot4=3.61 kg/s. 

One must set State-2 and 3 to reflect the new location of the shock before using the Super-Calculate button.

Go back to State-2, Change A2 (70 cm2), make MN2 an unknown, and enter A*2 ('=A1'). The alternative Mach number is reported on the message panel as 2.043. Now make A*2 an unknown and enter MN2 (2.043) and Calculate the State fully.

Go back to the Shock Table. Enter M_i as 2.043 and Calculate M_e=0.5701, p_te/p_ti=0.700. Now get back to State-3    and correct  MN3 (0.5701) and p_t3 ('=p_t0*0.7') . A Calculate produces the same mass flow rate but a larger critical area as expected,  A*3=57.08 cm2 , p3=280 kPa and mdot3=3.61 kg/s . A Super-Calculate finishes the rest of the calculations yielding p4=300.8 kPa .