# CHAPTER10:COMPRESSIBLE FLOW THROUGH NOZZLES DIFFUSERS AND WIND TUNNELS¶

## Example E01 : Pg 345¶

In :
# All the quantities are expressed in Si units
area_ratio = 10.25;                        # exit to throat area ratio
p0 = 5;                                    # reservoir pressure in atm
T0 = 333.3;                                # reservoir temperature

# from appendix A, for an area ratio of 10.25
Me = 3.95;                                 # exit mach number
pe = 0.007*p0;                             # exit pressure
Te = 0.2427*T0;                            # exit temperature

print"Me =",Me
print"pe =",pe,"atm"
print"Te =",Te,"K"

Me = 3.95
pe = 0.035 atm
Te = 80.89191 K


## Example E02 : Pg 346¶

In :
# All the quantities are expressed in Si units

area_ratio = 2.;                            # exit to throat area ratio
p0 = 1.;                                    # reservoir pressure in atm
T0 = 288.;                                  # reservoir temperature

# (a)
# since M = 1 at the throat
Mt = 1.;
pt = 0.528*p0;                             # pressure at throat
Tt = 0.833*T0;                             # temperature at throat

# from appendix A for supersonic flow, for an area ratio of 2
Me = 2.2;                                  # exit mach number
pe = 1./10.69*p0;                           # exit pressure
Te = 1./1.968*T0;                           # exit temperature

print"At throat: Mt =",Mt
print"\nAt throat: pt =",pt,"atm"
print"\nAt throat: Tt = ",Tt,"K"
print"\nAt throat: For supersonic exit:",Me
print"\nAt throat: pe =",pe,"atm"
print"\nAt throat: Te = ",Te,"K"

# (b)
# from appendix A for subonic flow, for an area ratio of 2
Me = 0.3;                                  # exit mach number
pe = 1/1.064*p0;                           # exit pressure
Te = 1/1.018*T0;                           # exit temperature

print"\nFor subrsonic exit:",Me
print"\npe=",pe,"atm"
print"\nTe=",Te,"K"

At throat: Mt = 1.0

At throat: pt = 0.528 atm

At throat: Tt =  239.904 K

At throat: For supersonic exit: 2.2

At throat: pe = 0.0935453695042 atm

At throat: Te =  146.341463415 K

For subrsonic exit: 0.3

pe= 0.93984962406 atm

Te= 282.907662083 K


## Example E03 : Pg 346¶

In :
# All the quantities are expressed in Si units

area_ratio = 2.;                            # exit to throat area ratio
p0 = 1.;                                    # reservoir pressure in atm
T0 = 288.;                                  # reservoir temperature
pe = 0.973;                                # exit pressure in atm

p_ratio = p0/pe;                           # ratio of reservoir to exit pressure

# from appendix A for subsonic flow, for an pressure ratio of 1.028
Me = 0.2;                                  # exit mach number
area_ratio_exit_to_star = 2.964;           # A_exit/A_star

# thus
area_ratio_throat_to_star = area_ratio_exit_to_star/area_ratio;           # A_exit/A_star

# from appendix A for subsonic flow, for an area ratio of 1.482
Mt = 0.44;                                 # throat mach number

print"Me =",Me
print"Mt =",Mt

Me = 0.2
Mt = 0.44


## Example E04 : Pg 352¶

In :
# All the quantities are expressed in SI units
import math
p0 = 30.*101000.;                            # reservoir pressure
T0 = 3500.;                                 # reservoir temperature
R = 520.;                                   # specific gas constant
gam = 1.22;                                # ratio of specific heats
A_star = 0.4;                              # rocket nozzle throat area
pe = 5529.;                                 # rocket nozzle exit pressure equal to ambient pressure at 20 km altitude

# (a)
# the density of air in the reservoir can be calculated as
rho0 = p0/R/T0;

# from eq.(8.46)
rho_star = rho0*(2/(gam+1))**(1/(gam-1));

# from eq.(8.44)
T_star = T0*2/(gam+1);
a_star = math.sqrt(gam*R*T_star);
u_star = a_star;
m_dot = rho_star*u_star*A_star;

# rearranging eq.(8.42)
Me = math.sqrt(2/(gam-1)*(((p0/pe)**((gam-1)/gam)) - 1));
Te = T0/(1+(gam-1)/2*Me*Me);
ae = math.sqrt(gam*R*Te);
ue = Me*ae;

# thus the thrust can be calculated as
T = m_dot*ue;
T_lb = T*0.2247;

# (b)
# rearranging eq.(10.32)
Ae = A_star/Me*((2/(gam+1)*(1+(gam-1)/2*Me*Me))**((gam+1)/(gam-1)/2));

print"(a)The thrust of the rocket is:T =" ,T/1e6, "N"
print"\n(b)The nozzle exit area is",T_lb

print"\nAe =",Ae, "m2"

(a)The thrust of the rocket is:T = 2.1702872295 N

(b)The nozzle exit area is 487663.540469

Ae = 16.7097500627 m2


## Example E05 : Pg 353¶

In :
# All the quantities are expressed in SI units
import math
p0 = 30.*101000.;                            # reservoir pressure
T0 = 3500.;                                 # reservoir temperature
R = 520.;                                   # specific gas constant
gam = 1.22;                                # ratio of specific heats
A_star = 0.4;                              # rocket nozzle throat area

# the mass flow rate using the closed form analytical expression
# from problem 10.5 can be given as
m_dot = p0*A_star*math.sqrt(gam/R/T0*((2/(gam+1))**((gam+1)/(gam-1))));

print"The mass flow rate is: m_dot =",m_dot, "kg/s"

The mass flow rate is: m_dot = 586.100122081 kg/s


## Example E06 : Pg 356¶

In :
# All the quantities are expressed in SI units
M = 2.;                        # Mach number
# for this value M, for a normal shock, from Appendix B
p0_ratio = 0.7209;
# thus
area_ratio = 1./p0_ratio;
print"The diffuser throat to nozzle throat area ratio is: =",area_ratio

The diffuser throat to nozzle throat area ratio is: = 1.38715494521