Gas Dynamics Software - Web-based Analysis of High Speed Flows

Tutorial (TOC) > Daemons > Gas Dynamics > Examples

Ex-1 Steam at 250^oC and quality 95% is flowing through a duct with a velocity of 250 m/s. Determine (a) the stagnation (total) properties (temperature, pressure, quality and density). (b) What-if scenario: How would the answers change if steam was treated as a perfect gas?

Solution The gas dynamics daemon only handles perfect gases as working fluid. To use the phase-change model, treat this problem as a single-flow open steady problem. The appropriate daemon can be found at TEST. Daemons. Systems. Open. SteadyState. Generic. SingleFlow. PCModel.

Let State-0 and State-1 represent the stagnation and the given states.

State-1: Enter Vel1 (250 m/s), x1 (95%), T1 (250^oC), and Calculate. The stagnation enthalpy, j, is calculated as 2747 kJ/kg. and p1 is 3973 kPa.

State-0: Enter Vel0 (0 m/s), j0 ('=j1'), s0 ('=s1') and Calculate. The stagnation properties are calculated as: T0=260^oC , p0=4678 kPa , x0=97% and rho0=24.348 kg/m3 .

For the what-if study, select the ..Specific.GasDynamics daemon. Select H2O. Calculate. The stagnation properties are calculated as part of the state: T_t1=266^oC , p_t1=4512 kPa. The answers can also be obtained by using Table Panel. Because Mach1 is calculated as 0.4414, enter M_i=0.4414 in the Table Panel to obtain p/p_t=0.88, T/T_t=0.969, A/Astar=1.477.

# TEST-codes for Ex-1;
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>
# Generic>SingleFlow>PC Model;

States   {
   State-0: H2O;
   Given:       { s0= "s1" kJ/kg.K;   Vel0= 0.0 m/s;   z0= 0.0 m;   j0= "j1" kJ/kg;   }

   State-1: H2O;
   Given:       { T1= 250.00003 deg-C;   x1= 95.0 %;   Vel1= 250.0 m/s;   z1= 0.0 m;   }
   }

# TEST>Daemons>Systems>Open>SteadyState>
# Specific>GasDynamics;
States    {
    State-1: H2O;
    Given:       { p1= 3973.0 kPa;   T1= 250.0 deg-C;   Vel1= 250.0 m/s;   z1= 0.0 m;   }
    }

Fig. 1.1 Image of the Daemons.Systems...SingleFlow.H2O page solving Ex. 1.

Fig. 1.2 Image of the Isentropic Table in the Systems...GasDynamics daemon solving Ex. 1.

Ex-2 Steam at 250^oC and quality 95% is flowing through a duct with a velocity of 250 m/s. Determine the change in area required to isentropically accelerate the flow to 500 m/s.

Solution The change in area is independent of the mass flow rate. Let us assume a mass flow rate of of 1 kg/s through the duct, obtain the answer and then see if the answer depends on the choice of the mass flow rate. We use the same daemon as in the previous example.

State-1: Enter Vel1 (250 m/s), x1 (95%), T1 (250^oC), mdot1 (1 kg/s) and Calculate. A1, is calculated as 1.91 cm^2 .

State-2: Enter mdot2 ('=mdot1'), j2 ('=j1'), s2 ('=s1'), Vel2 (500 m/s) and Calculate to obtain A2=1.509 cm^2 . This is a reduction of 21%.

Now change the mass flow rate mdot1 to 2 kg/s and Calculate and Super-Calculate. The new areas are: A1=3.828 cm^2. and A2=3.018 cm^2., also a 21% reduction.

# TEST-codes for Ex-2;
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>
# Generic>SingleFlow>PC Model;

States    {
    State-1: H2O;
    Given:       { T1= 250.00003 deg-C;   x1= 95.0 %;   Vel1= 250.0 m/s;   z1= 0.0 m;   mdot1= 1.0 kg/s;   }

    State-2: H2O;
    Given:       { s2= "s1" kJ/kg.K;   Vel2= 500.0 m/s;   z2= 0.0 m;   j2= "j1" kJ/kg;   mdot2= "mdot1" kg/s;   }
    }

Ex-3 Steam at 250^oC and quality 95% is flowing at a rate of 1 kg/s through a duct with a velocity of 250 m/s. Determine (a) the throat area where the flow becomes critical (i.e. Mach number=1). What-if scenario: (b) How would the answer change if the gas dynamics tables was used for H2O?

Solution This problem is very similar to the previous one. We use the same daemon as in the previous example.

We will use different guesses for Vel2 until the throat area, indicated by convergence and divergence on two sides, is found.

# TEST-codes for Ex-3;
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>
# Generic>SingleFlow>PC Model;

States    {
    State-1: H2O;
    Given:       { T1= 250.00003 deg-C;   x1= 95.0 %;   Vel1= 250.0 m/s;   z1= 0.0 m;   mdot1= 1.0 kg/s;   }

    State-2: H2O;
    Given:       { s2= "s1" kJ/kg.K;   Vel2= 462.0 m/s;   z2= 0.0 m;   j2= "j1" kJ/kg;   mdot2= "mdot1" kg/s;   }
    }

# TEST>Daemons>Systems>Open>SteadyState>
# Specific>GasDynamics;
States    {
    State-1: H2O;
    Given:       { p1= 3973.0 kPa;   T1= 250.0 deg-C;   Vel1= 250.0 m/s;   z1= 0.0 m;   mdot1= 1.0 kg/s;   }

    State-2: H2O;
    Given:       { z2= 0.0 m;   T_t2= "T_t1" deg-C;   p_t2= "p_t1" kPa;   Mach2= 1.0 UnitLess;   }
    }

State-1: Enter Vel1 (250 m/s), x1 (95%), T1 (250^oC), mdot1 (1 kg/s) and Calculate. A1, is calculated as 1.91 cm^2

State-2: Enter mdot2 ('=mdot1'), j2 ('=j1'), s2 ('=s1') and the first guess for Vel2 as 400 m/s. Calculate to obtain A2=1.511 cm^2. Now change Vel2 to 500 m/s and Calculate A2=1.509 cm^2 . Likewise for Vel2= 600 m/s A2=1.663 cm^2 . A plot of A2 vs. Vel2 shows that the area decreases and then increases again as Vel2 increases.

Now that we have approximately located the neck of the nozzle, we can refine the solution as follows. Vel2= 550 m/s A2=1.57 cm^2 . Vel2= 450 m/s A2=1.49 cm^2 . Vel2= 475 m/s A2=1.491 cm^2 . Vel2= 425 m/s A2=1.495 cm^2 . The best answer can be shown to be 462 m/s.

Note that although the throat area depends on the mass flow rate, the sonic velocity can be shown, in a similar manner, to be independent of the flow rate.

For the what-if study, launch the Gas Dynamics daemon and evaluate State-1 and State-2 as described in the TEST-codes above. For State-2, Mach2=1, T_t2=T_t1 (energy balance), p_t1=p_t2 (entropy balance), and Astar2=Astar1 (mass balance). The throat area is calculated as A2=Astar1= 1.646 cm^2. Yet another way is to go to the Table Panel after State-1 is evaluated and find A/Astar for M_i=M1=0.4414 producing A/Astar=1.4812. Therefore, A2=Astar=A1/1.4812=2.433/1.4812=1.646 cm^2 .

Ex-4 An aircraft is flying at a cruising speed of 275 m/s at an altitude of 32,000 ft where the atmospheric pressure is 54 kPa and the temperature is 256 K. The ambient air is first decelerated in a diffuser before it enters the compressor. Assuming isentropic flow, determine (a) the stagnation pressure at the compressor inlet and (b) the compressor work per unit mass if the compressor delivers air at a stagnation pressure of 700 kPa to the combustion chamber. (c) What-if scenario: How would the answer change if the cruising speed was increased to 350 m/s?

# TEST-codes for Ex-4;
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>
# Specific>GasDynamics;

States {
                 State-1: Air;
                Given:       { p1= 54.0 kPa;   T1= 256.0 K;
                           Vel1= 275.0 m/s;    z1= 0.0 m;   mdot1= 1.0 kg/s;   }

                State-2: Air;
                Given:       { s2= "s1" kJ/kg.K;   Vel2= 0.0 m/s;
                            z2= 0.0 m;   j2= "j1" kJ/kg;   mdot2= 1.0 kg/s;   }

                State-3: Air;
                Given:       { s3= "s2" kJ/kg.K;   Vel3= 0.0 m/s;
                            z3= 0.0 m;   p_t3= 700.0 kPa;
                            mdot3= "mdot1" kg/s;                 }
}

Analysis {
Device-A: i-State = State-2; e-State = State-3;
Given: { Qdot= 0.0 kW; T_B= -17.15 deg-C; }
}

Solution Launch the gas dynamics daemon by navigating to the Daemons. Systems. Open. SteadyState. Specific. GasDynamics page.

Let State-1 represent the i-State and State-2 the e-State for the diffuser, and State-2 represent the i-State and State-3 the e-State for the compressor. Instead of basing the analysis on unit mass, base it on a unit mass flow rate of 1 kg/s.

State-1: Enter mdot1 (1 kg/s), p1 (54 kPa, it is not always quite accurate to assume the inlet pressure to be equal to the ambient pressure), T1(256 K), Vel1 (275 m/s) and Calculate. Leave z1 at its default value of zero (you could change it to 32,000 fl, but it does not change between the inlet and exit). The stagnation pressure is calculated as 87 kPa .

State-2: Enter Vel2 (=0), T_t2 ('=T_t1') (or j2=j1), p_t2 ('=p_t1') (or s2=s1), and mdot2 ('=mdot1'). Calculate.

State-3: Enter Vel3 (=0), p_t3 (700 kPa), mdot3 ('=mdot1'), s3('=s2'), and Calculate. The stagnation temperature is calculated as 532 K.

On the Analysis panel, load State-2 as the i-State and State-3 as the e-State of the compressor. Enter the known device variable Qdot (=0), and T_B as 256 K (its value does not affect the calculations as the heat transfer is assumed to be zero). A Calculate and Super-Calculate produce Wdot_ext=-240 kW. Notice that Sdot_gen (is calculated as part of the solution to be zero, as expected for an isentropic (adiabatic and reversible) flow.

For the what-if study, go to States panel, choose State-1, change Vel1 to its new value (350 m/s) and Calculate . Get back to the Analysis panel and do a Super-Calculate . All the answers are updated. The new value for Wdot_ext=-216 kW. (the power requirement dropped because of the higher stagnation pressure of the incoming air.)

In this problem, even if you assume the velocity to be non-zero at the inlet and exit of the compressor, the answers do not change. Also note that, State-2 is not necessary, as it becomes internal to the system, if one constructs a control volume around the diffuser and the compressor bundled together.

Fig. 4.1 Image of Device Panel of the Gas Dynamics daemon. Note that this panel is identical to the one found on any
Single Flow Device daemons.

Ex-5 Air at 250^oC and 300 kPa enters a converging/diverging nozzle with a velocity of 50 m/s. If the exit Mach number is 2 and the exit area is 10 cm2, (a) determine the mass flow rate. (b) What-if scenario: How would the answers change if the gas was argon instead?

# TEST-codes for Ex-5;
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>
# Specific>GasDynamics;

States {
               State-1: Air;
               Given:       { p1= 300.0 kPa;   T1= 250.0 deg-C;
                          Vel1= 50.0 m/s;   z1= 0.0 m;   }

               State-2: Air;
               Given:       { z2= 0.0 m;   T_t2= "T_t1" deg-C;
                          p_t2= "p_t1" kPa;   Mach2= 2.0 UnitLess;
                          A2= 10.0 cm^2;   }
}

Solution Launch the gas dynamics daemon by navigating to the Daemons. Systems. Open. SteadyState. Specific. GasDynamics page.

State-1: Enter Vel1 (250 m/s), p1 (95%) and T1 (250^oC), and Calculate.

State-2: Enter Mach2 ('=mdot1'), A2, T_t2 ('=T_t1'), p_t2 ('=p_t1') and Calculate to obtain mdot2=0.32 kg/s .

Now change the working fluid to Ar and Super-Calculate. The new mass flow rate is calculated as 0.44 kg/s.

Ex-6 A rocket nozzle operates supersonically with a constant chamber pressure and temperature of 3 MPa and 1500 K respectively. It is designed for an atmospheric pressure of 100 kPa with a throat area of 5 cm². Determine (a) the mass flow rate, (b) exit velocity, (c) the thrust at sea level, and (d) the thrust in space (0 kPa). Assume the exhaust gas to behave as a perfect gas with k=1.3 and R=0.4 kJ/kg.K. (e) What-if scenario: How would the answers change if the reservoir temperature was 2000 K?

# TEST-codes for Ex-6;
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>
# Specific>GasDynamics;

States    {
    State-0: Custom;
    Given:       { p0= 3.0 MPa;   T0= 1500.0 K;   Vel0= 0.0 m/s;   z0= 0.0 m;   R0= 0.4 kJ/kg.K;   k0= 1.3 UnitLess;   }

    State-1: Custom;
    Given:       { z1= 0.0 m;   T_t1= "T_t0" deg-C;   p_t1= "p_t0" kPa;   Mach1= 1.0 UnitLess;   A1= 5.0 cm^2;   R1= "R0" kJ/kg.K;   k1= "k0" UnitLess;
        }

    State-2: Custom;
    Given:       { p2= 100.0 kPa;   z2= 0.0 m;   T_t2= "T_t1" deg-C;   p_t2= "p_t1" kPa;   Astar2= "Astar1" m^2;   R2= "R1" kJ/kg.K;   k2= "k1" UnitLess;
        }
    }

Solution Launch the gas dynamics daemon by navigating to the Daemons. Systems. Open. SteadyState. Specific. GasDynamics page.

Select Custom Gas from the fluid selector. Since the nozzle produces supersonic flow, the throat must have a Mach number of 1. Also, isentropic flow means stagnation temperature and pressure, and critical area must remain constant as a consequence of satisfying the energy, entropy and mass balance equations.

State-0: (Stagnation State) Enter k0, R0, p0 (3 MPa), T0 (1500 K) and Vel0 (0), and Calculate.

State-1: (Nozzle Throat) Enter k1=k0, R1=R0, T_t1=T_t0, p_t1=p_t0, M1=1, and Calculate. The mass flow rate is found as mdot1=1.29 kg/s .

State-2: (Nozzle Exit) Enter k2=k0, R2=R0, T_t2=T_t0, p_t2=p_t0, p2=100 kPa, and Calculate. The exit velocity is calculated as Vel2=1682 m/s.

The thrust can be calculated at the sea level by evaluating the expression '=(p2-100)*A2+mdot2*Vel2/1000' on I/O Panel as 2.172 kPa. Note that since the back pressure is same as the exit pressure there is no contribution from the pressure term.

In space the flow inside the nozzle remains unchanged since the lower back pressure cannot not be propagated upstream against a supersonic flow. The thrust can be calculated from the expression '=(p2-0)*A2+mdot2*Vel2/1000' on I/O Panel as 2.38 kPa.

Ex-7 A converging/diverging nozzle is designed to operate with an exit Mach number of 2. The nozzle is supplied from an air reservoir at 4 MPa and 1500 K. Determine (a) ideal back pressures for isentropic expansion, (b) maximum back pressure to choke the nozzle, (c) range of back pressures over which a normal shock appears in the nozzle, and (d) range of back pressures for supersonic flow at the nozzle exit.

# TEST-codes for Ex-7;
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>
# Specific>GasDynamics;

   States    {
    State-0: Air;
    Given:       { p0= 4.0 MPa;   T0= 1500.0 K;   Vel0= 0.0 m/s;   z0= 0.0 m;   }

    State-1: Air;
    Given:       { z1= 0.0 m;   T_t1= "T_t0" deg-C;   p_t1= "p_t0" kPa;   Mach1= 2.0 UnitLess;   }

    State-2: Air;
    Given:       { z2= 0.0 m;   T_t2= "T_t1" deg-C;   p_t2= "p_t1" kPa;   Mach2= 0.372 UnitLess;   }

    State-3: Air;
    Given:       { p3= "4.5*p1" kPa;   z3= 0.0 m;   T_t3= "T_t0" deg-C;   Mach3= 0.5773 UnitLess;   }
    }

Solution Launch the gas dynamics daemon by navigating to the Daemons. Systems. Open. SteadyState. Specific. GasDynamics page.

Air is the default gas. In this problem we take advantage of the Normal Shock Table, which is combined with the Isentropic Table in Table Panel.

State-0: (Stagnation State) Enter p0 (4 MPa), T0 (1500 K) and Vel0 (0), and Calculate.

State-1: (Supersonic Isentropic Exit) Enter T_t1=T_t0, p_t1=p_t0, M1=2, and Calculate. The exit pressure p1=511 kPa is an ideal back pressure for supersonic isentropic flow.

However, this is not the only possible isentropic solution. To obtain the subsonic isentropic solution, go to Table Panel and enter M_i=2. This produces A/Astar=1.6877. Now make M_i an unknown and enter A/Astar as known (simply un-check M_i and check the checkbox of A/Astar) and Calculate. The subsonic solution is displayed as M_i=0.372 (and the supersonic solution, M_i=2, is displayed on Message Panel).

State-2: (Subsonic Isentropic Exit) Enter T_t2=T_t0, p_t2=p_t0, M2=0.372 kPa, and Calculate. The exit pressure p2=3635 kPa is an ideal back pressure for subsonic isentropic flow in the nozzle.

If the back pressure is increased above p2, the flow will no longer be chocked at the throat. Therefore, the maximum back pressure for chocking is 3635 kPa.

As the back pressure is lowered below 3635 kPa, a normal shock appear downstream of the throat. As the pressure is reduced further the shock moves towards the exit. In the extreme case, the shock stands at the exit where the Mach number before the shock is 2.0 (due to isentropic flow upstream). The state after the normal shock, State-3, can be calculated if the pressure ratio can be obtained from the Normal Shock Table. On Table Panel enter M_i=2 and Calculate M_e=0.577 and p_e/p_i =4.5.

State-3: (Shock exit flow, shock inlet being State-1) Enter T_t2=T_t0, p_2=4.5*p1, M2=0.577 kPa, and Calculate. The exit pressure p2=2301 kPa is an back pressure for which a normal shock stands at the exit. Therefore the range is 3635 kPa-2301 kPa. Below 2301 kPa at the exit oblique shocks are formed until the pressure reaches down to the isentropic value 511 kPa. If the exit pressure is lowered further, Prandtl Meyer expansion waves are formed at the exit.

The reservoir temperature has no effect in any of the answers. This can be verified by changing T0 and pressing the Super-Calculate button.

Ex-8 A parallel flow at M=2.0 passes over a wedge of 8^o half-angle. Using the oblique shock chart, (a) find M2, p2/p1, T2/T1, pt_2/pt_1. (b) Also find the half-angle above which the shock will become detached. Assume k=1.3.

# TEST-codes for Ex-8;
# Daemon Path: TEST>Daemons>Systems>Open>SteadyState>
# Specific>GasDynamics;

Solution Launch the gas dynamics daemon by navigating to the Daemons. Systems. Open. SteadyState. Specific. GasDynamics page.

Select Custom from the gas selector. Enter k=1.3 and Calculate. Now go to Table Panel and select the Delta-Theta Table. Enter M_i=2 and delta=8 and Calculate M_e=1.759 and theta=36.37 degrees. The normal component of the incident Mach number can be obtained by evaluating M1_n =M1 Sin(theta) on I/O Panel as '=2*Sin(36.37)'=1.186. Now switch to the Normal Shock Table and enter M_i=1.186. The desired ratios now can be obtained from the shock wave results as p_e/p_i=1.46, T_e/T_T_i=1.0927, and p_t,e/p_t,i=0.99.

Go back to Delta-Theta Table. Increase delta without changing M_i. The weak shock angle is displayed on the theta widget and the strong shock angle is displayed on Message Panel. Keep increasing delta until the strong and weak solutions become almost equal at delta=22.6^o.