# CHAPTER08:NORMAL SHOCK WAVES AND RELATED TOPICS¶

## Example E01 : Pg 256¶

In :
# All the quantities are expressed in SI units
from math import sqrt,pi
R = 287.;
gam = 1.4;
V_inf = 250.;

# (a)
# At sea level
T_inf = 288.;

# the velocity of sound is given by
a_inf = sqrt(gam*R*T_inf);

# thus the mach number can be calculated as
M_inf = V_inf/a_inf;

print"(a)The Mach number at sea level is:M_inf =",M_inf

# similarly for (b) and (c)
# (b)
# at 5km
T_inf = 255.7;

a_inf = sqrt(gam*R*T_inf);

M_inf = V_inf/a_inf;

print"(b)The Mach number at 5 km is: M_inf = ",M_inf

# (c)
# at 10km
T_inf = 223.3;

a_inf = sqrt(gam*R*T_inf);

M_inf = V_inf/a_inf;

print"(c)The Mach number at 10 km is: M_inf =",M_inf

(a)The Mach number at sea level is:M_inf = 0.73491785465
(b)The Mach number at 5 km is: M_inf =  0.779955236945
(c)The Mach number at 10 km is: M_inf = 0.834623638772


## Example E02 : Pg 256¶

In :
# All the quantities are expressed in SI units
import math
T = 320;            # static temperature
V = 1000;           # velocity
gam = 1.4;          # ratio of specific heats
R = 287;            # universal gas constant

# the speed of sound is given by
a = math.sqrt(gam*R*T);

# the mach number can be calculated as
M = V/a;

print"The Mach number is:\nM =",M

The Mach number is:
M = 2.78881717658


## Example E03 : Pg 257¶

In :
# All the quantities are expressed in SI units

gam = 1.4;                    # ratio of specific heats

# (a)
M = 2;                        # Mach number

# the ratio of kinetic energy to internal energy is given by
ratio = gam*(gam-1)*M*M/2;

print"(a)The ratio of kinetic energy to internal energy is:",ratio

# similarly for (b)
# (b)
M = 20;

ratio = gam*(gam-1)*M*M/2;

print"\n(b)The ratio of kinetic energy to internal energy is:",ratio

(a)The ratio of kinetic energy to internal energy is: 1.12

(b)The ratio of kinetic energy to internal energy is: 112.0


## Example E04 : Pg 259¶

In :
# All the quantities are expressed in SI units

M = 2.79;        # Mach number
T = 320;         # static temperature from ex. 7.3
p = 1;           # static pressure in atm
gam = 1.4;

# from eq. (8.40)
T0 = T*(1+((gam-1)/2*M*M));

# from eq. (8.42)
p0 = p*((1+((gam-1)/2*M*M))**(gam/(gam-1)));

print"The total temperature and pressure are:\nT0 =",T0,"K","\nP0 =",p0,"atm"

The total temperature and pressure are:
T0 = 818.1824 K
P0 = 26.7270201929 atm


## Example E05 : Pg 260¶

In :
# All the quantities are expressed in SI units
import math
M = 3.5;        # Mach number
T = 180;         # static temperature from ex. 7.3
p = 0.3;           # static pressure in atm
gam = 1.4;
R = 287;

# from eq. (8.40)
T0 = T*(1+((gam-1)/2*M*M));

# from eq. (8.42)
p0 = p*((1+((gam-1)/2*M*M))**(gam/(gam-1)));

a = math.sqrt(gam*R*T);
V = a*M;

# the values at local sonic point are given by
T_star = T0*2/(gam+1);
a_star = math.sqrt(gam*R*T_star);
M_star = V/a_star;

print"T0 =",T0,"K","\nP0 =",p0,"atm","\nT* =",T_star,"k","\na* =",a_star,"m/s","\nM* =",M_star

T0 = 621.0 K
P0 = 22.8816894716 atm
T* = 517.5 k
a* = 455.995065763 m/s
M* = 2.06418738617


## Example E06 : Pg 263¶

In :
# All the quantities are expressed in SI units

p_inf = 1;
p1 = 0.7545;
M_inf = 0.6;
gam = 1.4;

# from eq. (8.42)
p0_inf = p_inf*((1+((gam-1)/2*M_inf*M_inf))**(gam/(gam-1)));

p0_1 = p0_inf;

# from eq. (8.42)
ratio = p0_1/p1;

# from appendix A, for this ratio, the Mach number is
M1 = 0.9;

print"The mach number at the given point is:\nM1 =",M1

The mach number at the given point is:
M1 = 0.9


## Example E07 : Pg 268¶

In :
# All the quantities are expressed in SI units
import math
T_inf = 288;                                        # freestream temperature
p_inf = 1;                                          # freestream pressure
p1 = 0.7545;                                        # pressure at point 1
M = 0.9;                                            # mach number at point 1
gam = 1.4;                                          # ratio of specific heats
R=1.;#
# for isentropic flow, from eq. (7.32)
T1 = T_inf*((p1/p_inf)**((gam-1)/gam));

# the speed of sound at that point is thus
a1 = math.sqrt(gam*R*T1);

# thus, the velocity can be given as
V1 = M*a1;

print"The velocity at the given point is:\nV1 =",V1,"m/s"

The velocity at the given point is:
V1 = 17.3590326624 m/s


## Example E08 : Pg 268¶

In :
# All the quantities are expressed in SI units
import math
u1 = 680;                            # velocity upstream of shock
T1 = 288;                            # temperature upstream of shock
p1 = 1;                              # pressure upstream of shock
gam = 1.4;                           # ratio of specific heats
R = 287;                             # universal gas constant

# the speed of sound is given by
a1 = math.sqrt(gam*R*T1)

# thus the mach number is
M1 = 2;

# from Appendix B, for M = 2, the relations between pressure and temperature are given by
pressure_ratio = 4.5;                # ratio of pressure accross shock
temperature_ratio = 1.687;           # ratio of temperature accross shock
M2 = 0.5774;                         # mach number downstream of shock

# thus the values downstream of the shock can be calculated as
p2 = pressure_ratio*p1;
T2 = temperature_ratio*T1;
a2 = math.sqrt(gam*R*T2);
u2 = M2*a2;

print"p2 =",p2,"atm","\nT2 =",T2,"K","\nu2 =",u2,"m/s"

p2 = 4.5 atm
T2 = 485.856 K
u2 = 255.114727639 m/s


## Example E09 : Pg 271¶

In :
# All the quantities are expressed in SI units

p1 = 1;                                            # ambient pressure upstream of shock

# (a)
# for M = 2;
p0_1 = 7.824*p1;                                   # total pressure upstream of shock
pressure_ratio = 0.7209;                           # ratio of total pressure accross the shock
p0_2 = pressure_ratio*p0_1;                        # total pressure downstream of shock

# thus the total loss of pressure is given by
pressure_loss = p0_1 - p0_2;

print"The total pressure loss is:\n(a)P0_loss=",pressure_loss,"atm"

# similarly
# (b)
# for M = 4;
p0_1 = 151.8*p1;
pressure_ratio = 0.1388;
p0_2 = pressure_ratio*p0_1;

# thus the total loss of pressure is given by
pressure_loss = p0_1 - p0_2;

print"\n(b)P0_loss =",pressure_loss,"atm"

The total pressure loss is:
(a)P0_loss= 2.1836784 atm

(b)P0_loss = 130.73016 atm


## Example E10 : Pg 272¶

In :
# All the quantities are expressed in SI units

M_inf = 2.;                     # freestream mach number
p_inf = 2.65e4;                # freestream pressure
T_inf = 223.3;                 # freestream temperature

# from Appendix A, for M = 2
p0_inf = 7.824*p_inf;          # freestream total pressure
T0_inf = 1.8*T_inf;            # freestream total temperature

# from Appendix B, for M = 2
p0_1 = 0.7209*p0_inf;          # total pressure downstream of the shock
T0_1 = T0_inf;                 # total temperature accross the shock is conserved

# since the flow downstream of the shock is isentropic
p0_2 = p0_1;
T0_2 = T0_1;

# from Appendix A, for M = 0.2 at point 2
p2 = p0_2/1.028;
T2 = T0_2/1.008;

p2_atm = p2/102000;

print"The pressure at point 2 is:p2 =",p2_atm,"atm",

The pressure at point 2 is:p2 = 1.42546466011 atm


## Example E11 : Pg 273¶

In :
# All the quantities are expressed in SI units

M_inf = 10;                    # freestream mach number
p_inf = 2.65e4;                # freestream pressure
T_inf = 223.3;                 # freestream temperature

# from Appendix A, for M = 2
p0_inf = 0.4244e5*p_inf;       # freestream total pressure
T0_inf = 21*T_inf;             # freestream total temperature

# from Appendix B, for M = 2
p0_1 = 0.003045*p0_inf;        # total pressure downstream of shock
T0_1 = T0_inf;                 # total temperature downstream of shock is conserved

# since the flow downstream of the shock is isentropic
p0_2 = p0_1;
T0_2 = T0_1;

# from Appendix A, for M = 0.2 at point 2
p2 = p0_2/1.028;
T2 = T0_2/1.008;

p2_atm = p2/102000;

print"The pressure at point 2 is: p2 =",p2_atm,"atm"

The pressure at point 2 is: p2 = 32.6599307622 atm


## Example E13 : Pg 274¶

In :
# All the quantities are expressed in SI units

p1 = 4.66e4;                                # ambient pressure
M = 8;                                      # mach number

# from Appendix B, for M = 8
p0_2 = 82.87*p1;                            # total pressure downstream of the shock

# since the flow is isentropic downstream of the shock, total pressure is conserved
ps_atm = p0_2/101300;                       # pressure at the stagnation point

print"The pressure at the nose is:\np_s =",ps_atm,"atm"

The pressure at the nose is:
p_s = 38.1218361303 atm


## Example E14 : Pg 274¶

In :
# All the quantities are expressedin SI units
import math
p1 = 2527.3;                    # ambient pressure at the altitude of 25 km
T1 = 216.66;                    # ambient temperature at the altitude of 25 km
p0_1 = 38800;                   # total pressure
gam = 1.4;                      # ratio of specific heats
R = 287;                        # universal gas constant
pressure_ratio = p0_1/p1;       # ratio of total to static pressure

# for this value of pressure ratio, mach number is
M1 = 3.4;

# the speed of sound is given by
a1 = math.sqrt(gam*R*T1)

# thus the velocity can be calculated as
V1 = M1*a1;

print"The Velocity of the airplane is:\nV1 =",V1,"m/s"

The Velocity of the airplane is:
V1 = 1003.16703558 m/s