## Chapter 11: Compressible Flow¶

### Example 11.1 Page no 420¶

In [1]:
# Example 11.1

from math import *

from __future__ import division

# Given

T1 = 273 + 15          # temperature in K

T2 = 273 + 90          # temperature in K

Cp = 0.24              # cp for air in kcal/kgK

# Solution

dh = Cp*(T2-T1)        # enthalpy per kg of air

H = 10*dh              # total enthallpy of 10 kg air

print "Total change in enthalpy for 10 kg air = ",round(H,0),"kcal"

Total change in enthalpy for 10 kg air =  180.0 kcal


### Example 11.2 Page no 420¶

In [3]:
# Example 11.2

from math import *

from __future__ import division

# Given

T1 = 273 + 15          # temperature in K

T2 = 273 + 90          # temperature in K

P1 = 40 + 101.3        # pressure in abs

P2 = 360 + 101.3       # presure in abs

Cv = 0.171             # Specific volume Coefficient of air

k = 1.4                # gas constant

# solution

dS = Cv*log((T2/T1)**k*(P2/P1)**(1-k))

S = 10*dS

print "Total change in enthalpy of 10 Kg of air =",round(S,3),"kcal/K"

Total change in enthalpy of 10 Kg of air = -0.255 kcal/K


### Example 11.3 Page no 421¶

In [4]:
# Example 11.3

from math import *

from __future__ import division

P1 = 10                  # pressure in psia

P2 = 30                  # pressure in psia

T1 = 460+110             # temperature in R

k =1.4                   # gas constant

T2 = T1*(P2/P1)**((k-1)/k)

t2 = T2-460              # final temperature of air

print "Final temperature if air = ",round(t2,1),"F"

Cv = 0.157               # coefficient of air

W = Cv*(T2-T1)           # work done per unit mass of oxygen

Tw = 10*W                # total work done on 10 slugs

print "Total work done on 10 slugs = ",round(Tw,0),"Btu"

Final temperature if air =  320.2 F
Total work done on 10 slugs =  330.0 Btu


### Example 11.4 Page no 426¶

In [5]:
# Example 11.4

from math import *

from __future__ import division

# Given

# for water

S =1           # specific gravity

rho = S*1000   # density in kg/m**3

bta = 2.2*10**9 # Bulk modulus of elasticity

# ethly alcohol

S1 =0.79           # specific gravity

rho2 = S1*1000   # density in kg/m**3

bta2 = 1.21*10**9 # Bulk modulus of elasticity

# for air

k = 1.4             # gas constant for air

R = 287             # universal gas constant

T = 273+20          # temperature in K

# Solution

C1 = sqrt(bta/rho)

C2 = sqrt(bta2/rho2)

print "Speed of sound in water =",round(C1,0),"m/s"

print "Speed of sound in ethly alcohol =",round(C2,0),"m/s"

C3 = sqrt(k*R*T)

print "Speed of sound in Air =",round(C3,0),"m/s"

Speed of sound in water = 1483.0 m/s
Speed of sound in ethly alcohol = 1238.0 m/s
Speed of sound in Air = 343.0 m/s


### Example 11.5 Page no 431¶

In [6]:
# Example 11.5

from math import *

from __future__ import division

# Given

P1 = 1.5        # pressure in psia

T1 = 40 + 460   # temperature in R

k = 1.4         # gas constant

R = 1716        # universal gas constant in ft.lb/slug R

V1 = 1500       # velocity in ft/s

# Solution

c1 = sqrt(k*R*T1)

M1 = V1/c1

M2 = sqrt((2+(k-1)*M1**2)/(2*k*M1**2-(k-1)))
print M2

P2 = P1*((1+k*M1**2)/(1+k*M2**2))

print "Pressure at downstream = ",round(P2,2),"psia"

T2 = T1*((1+0.5*(k-1)*M1**2)/(1+0.5*(k-1)*M2**2))

t2 = T2-460

print "Temperature at downstream = ",round(t2,1),"F"
V2 = M2*sqrt(k*R*t2)

print "Velocity downstream = ",round(V2,2),"ft/s"

0.753305702898
Pressure at downstream =  3.03 psia
Temperature at downstream =  157.3 F
Velocity downstream =  463.02 ft/s

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