Chapter 3 : Flow through constant area duct adiabatic flow

Example 3.1 page : 11

In [1]:
import math 

				
#Input data
M1 = 0.25 				#Mach number at entrance
M2 = 1 				#Mach number at exit
D = 0.04 				#inner tude diameter in m
f = 0.002 				#frictional factor

				
#Calculation
X1 = 8.537 				#frictional consmath.tant fanno parameter at entry from gas tables @M1 = 0.25
X2 = 0 				#frictional consmath.tant fanno parameter at exit from gas tables @M2 = 1
X = X1-X2 				#overall frictional consmath.tant fanno parameter i.e. (4*f*L)/D
L = (X*D)/(4*f) 				#Length of the pipe in m

				
#Output
print 'A)Length of the pipe is %3.3f m'%(L)

A)Length of the pipe is 42.685 m

Example 3.2 page : 11

In [2]:
import math 

				
#Input data
M1 = 0.1 				#Mach number at entrance
M2 = 0.5 				#Mach number at a section
M3 = 1 				#Mach number at critical condition
D = 0.02 				#Diameter of duct in m
f = 0.004 				#Frictional factor

				
#Calculation
X1 = 66.922 				#frictional consmath.tant fanno parameter from gas tables @M1 = 0.1
X2 = 1.069 				#frictional consmath.tant fanno parameter from gas tables @M2 = 0.5
X3 = 0 				#frictional consmath.tant fanno parameter from gas tables @M3 = 1
X4 = X1-X3 				#				#frictional consmath.tant fanno parameter from M2 = 0.1 to M3 = 1
L1 = (X4*D)/(4*f) 				#Length of the pipe in m
X5 = X2-X3 				#frictional consmath.tant fanno parameter from M2 = 0.5 to M3 = 1
L2 = (X5*D)/(4*f) 				#Addition length of the pipe required to accelerate into critical condition in m
L = L1-L2 				#Length of the pipe required to accelerate the flow from M1 = 0.1 to M2 = 0.5 in m

				
#Output 
print 'A)Length of the pipe required to accelerate the flow from M1 = %3.1f to M2 = %3.1f is %3.3f m \
\nB)Additional length required to accelerate into critical condition is %3.5f m'%(M1,M2,L,L2)

A)Length of the pipe required to accelerate the flow from M1 = 0.1 to M2 = 0.5 is 82.316 m 
B)Additional length required to accelerate into critical condition is 1.33625 m

Example 3.3 page : 12

In [3]:
import math 

				
#Input data
D = 0.05 				#inner pipe diameter in m
Po = 10 				#Stagnation Pressure at reservoir in bar
To = 400 				#Stagnation temperature at reservoir in K
f = 0.002 				#frictional factor 
M1 = 3 				#Mach number at entrance
M2 = 1 				#Mach number at end of pipe
R = 287 				#Gas consmath.tant in J/kg-K
k = 1.4 				#Adiabatic consmath.tant

				
#Calculation
X1 = 0.522 				#frictional consmath.tant fanno parameter from gas tables @M1 = 3
X2 = 0 				#frictional consmath.tant fanno parameter from gas tables @M2 = 1
X = X1-X2 				#overall frictional consmath.tant fanno parameter
L = (X*D)/(4*f) 				#Length of the pipe in m
p1 = 0.0272 				#Pressure ratio from gas tables (M = 3,k = 1.4,isentropiC)
P1 = p1*Po 				#Static pressure at entrance in bar
t1 = 0.3571 				#Temperature ratio from gas tables (M = 3,k = 1.4,isentropic)
T1 = t1*To 				#Static temperature at entrance in K
d1 = (P1*10**5)/(R*T1) 				#Density of air in kg/m**3, P1 in Pa
a1 = math.sqrt(k*R*T1) 				#Sound velocity in m/s
C1 = a1*M1 				#air velocity in m/s
A1 = (math.pi*D**2)/4 				#Cross sectional area of pipe in m**2
m = d1*A1*C1 				#Mass flow rate in kg/s

				
#Output
print 'A)Length of the pipe is %3.2f m \
\nB)Mass flow rate is %3.4f kg/s'%(L,m)

A)Length of the pipe is 3.26 m 
B)Mass flow rate is 0.9363 kg/s

Example 3.4 page : 13

In [6]:
import math 

				
#Input data
C1 = 235. 				#Velocity at entrance in m/s
P1 = 13. 				#Static Pressure at entry in bar
P2 = 10. 				#Static Pressure at a point in duct in bar
T1 = 543. 				#Static temperature at entry in Kelvin
D = 0.15 				#inner duct diameter in m
f = 0.005 				#frictional factor
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Gas consmath.tant in J/kg-K

				
#Calculation
a1 = math.sqrt(k*R*T1) 				#Sound velocity in m/s
M1 = C1/a1 				#Mach number at entry
p1 = 2.138 				#Static Pressure ratio from gas tables (fanno flow tables,k = 1.4,M = 0.5)
Pt = P1/p1 				#Static critical pressure in bar
t1 = 1.143 				#Static temperature ratio from gas tables (fanno flow tables,k = 1.4,M = 0.5) 
Tt = T1/t1 				#Static critical temperature in K
c1 = 0.534 				#Velocity ratio from gas tables (fanno flow tables,k = 1.4,M = 0.5)
Ct = C1/c1 				#Critical velocity in m/s
p2 = 1.644 				#Pressure ratio from gas tables (fanno flow tables,k = 1.4)
M2 = 0.64 				#Mach number from gas tables (fanno flow tables,k = 1.4,p2)
c2 = 0.674 				#Velocity ratio from gas tables (fanno flow tables,k = 1.4,p2)
C2 = Ct*c2 				#Air velocity at P2 in m/s
t2 = 1.109 				#Temperature ratio from gas tables (fanno flow tables,k = 1.4,p2)
T2 = t2*Tt 				#Satic temperature at P2 is K
X1 = 1.06922 				#frictional consmath.tant fanno parameter from gas tables @M1
X2 = 0.353 				#frictional consmath.tant fanno parameter from gas tables @M2
X = X1-X2 				#overall frictional consmath.tant fanno parameter
L = (X*D)/(4*f) 				#Length of the pipe in m

				
#Output
print 'A)Temperature and velocity at section of the duct where the pressure has dropped to %3i bar \
\ndue to friction are %3.1f K and %3.2f m/s \
\nB)The Distance between two section is %3.3f m'%(P2,T2,C2,L)

A)Temperature and velocity at section of the duct where the pressure has dropped to  10 bar 
due to friction are 526.8 K and 296.61 m/s 
B)The Distance between two section is 5.372 m

Example 3.5 page : 14

In [9]:
import math 

				
#Input data
P1 = 120. 				#Static pressure at entrance in bar
T1 = 313. 				#Static temperature at entry in Kelvin
M1 = 2.5 				#Mach number at entrance
M2 = 1.8 				#Mach number at exit
D = 0.2 				#inner pipe diameter in m
f = 0.01/4 				#frictional factor
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Gas consmath.tant in J/kg-K

				
#Calculation
a1 = math.sqrt(k*R*T1) 				#Sound velocity in m/s
C1 = a1*M1 				#air velocity in m/s
p1 = 0.292 				#Static Pressure ratio from gas tables (fanno flow tables,k = 1.4,M = 2.5)
Pt = P1/p1 				#Static critical pressure in kPa
t1 = 0.533 				#Static temperature ratio from gas tables (fanno flow tables,k = 1.4,M = 2.5)
Tt = T1/t1 				#Static critical temperature in K
c1 = 1.826 				#Velocity ratio from gas tables (fanno flow tables,k = 1.4,M = 2.5)
Ct = C1/c1 				#Critical velocity in m/s
X1 = 0.432 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1 = 3
X2 = 0 				#frictional consmath.tant fanno parameter from gas tables @M2 = 1
X3 = X1-X2 				#overall frictional consmath.tant fanno parameter
L1 = (X3*D)/(4*f) 				#Maximum length of the pipe in m
p2 = 0.474 				#Static Pressure ratio from gas tables (fanno flow tables,k = 1.4,M = 1.8)
P2 = Pt*p2 				#Static pressure in kPa
t2 = 0.728 				#static temperature ratio from gas tables (fanno flow tables,k = 1.4,M = 1.8)
T2 = Tt*t2 				#Static temperature in K
c2 = 1.536 				#Velocity ratio from gas tables (fanno flow tables,k = 1.4,M = 1.8)
C2 = c2*Ct 				#Critical velocity in m/s
X4 = 0.242 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M = 1.8
X5 = X4-X2 				#overall frictional consmath.tant fanno parameter
L2 = (X5*D)/(4*f) 				#Length between sonic and oulet section
L = L1-L2 				#Length of the pipe in m

				
#Output
print 'A)Maximum length of the pipe is %3.2f m \
\nB)Properties of air at sonic condition:    \
\nPressure is %3i kPa    \
\nTemperature is %3.2f K    \
\nVelocity is %3.1f m/s \
\nC)Length of the pipe is %3.1f m \
\nD)Properties of air at M2 = %3.1f:    \
\nPressure is %3i kPa    \
\nTemperature is %3.2f K    \
\nVelocity is %3.2f m/s'%(L1,Pt,Tt,Ct,L,M2,P2,T2,C2)

# note : rounding off error.

A)Maximum length of the pipe is 8.64 m 
B)Properties of air at sonic condition:    
Pressure is 410 kPa    
Temperature is 587.24 K    
Velocity is 485.5 m/s 
C)Length of the pipe is 3.8 m 
D)Properties of air at M2 = 1.8:    
Pressure is 194 kPa    
Temperature is 427.51 K    
Velocity is 745.77 m/s

Example 3.6 page : 15

In [11]:
import math 

				
#Input data
M1 = 0.25 				#Mach number at entrance
ds = 0.124 				#Change in entropy in kJ/kg-K
P1 = 700. 				#Static pressure at entrance in bar
T1 = 333. 				#Static temperature at entry in Kelvin
D = 0.05 				#inner pipe diameter in m
f = 0.006 				#frictional factor
k = 1.4 				#Adiabatic consmath.tant
R = 0.287 				#Gas consmath.tant in kJ/kg-K

				
#Calculation
p1 = math.exp(ds/R) 				#Ratio of Stagnation pressure at inlet to outlet 
t1 = 0.987 				#Ratio of Static Temperature to Stagnation temperature at entry from gas tables @M1
To1 = T1/t1 				#Stagnation temperature at entry in K
p2 = 0.957 				#Ratio of Static pressure to Stagnation pressure at entry from gas tables @M1
Po1 = P1/p2 				#Stagnation pressure at entry in kPa
Po2 = Po1/p1 				#Stagnation pressure at exit in kPa
a1 = math.sqrt(k*R*10**3*T1) 				#Sound velocity in m/s, R in J/kg
C1 = a1*M1 				#air velocity in m/s
p3 = 4.3615 				#Static Pressure ratio from gas tables (fanno flow tables,k = 1.4,M = 0.25)
Pt = P1/p3 				#Static critical pressure in kPa
t1 = 1.185 				#Static temperature ratio from gas tables (fanno flow tables,k = 1.4,M = 0.25)
Tt = T1/t1 				#Static critical temperature in K
c1 = 0.272 				#Velocity ratio from gas tables (fanno flow tables,k = 1.4,M = 0.25)
Ct = C1/c1 				#Critical velocity in m/s
p4 = 2.4065 				#Pressure ratio at entry from gas tables @M1,k
Pot = Po1/p4 				#Stagnation pressure at critical state in kPa
X1 = 8.537 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k
p5 = Po2/Pot 				#Pressure ratio 
M2 = 0.41 				#Mach number at exit from gas tables @p5
p6 = 2.629 				#Pressure ratio at exit from gas tables @p5
P2 = Pt*p6 				#Exit pressure in kPa
t2 = 1.161 				#Temperature ratio at exit from gas tables @p5
T2 = Tt*t2 				#Exit temperature in K
c2 = 0.4415 				#Velocity ratio at exit from gas tables @p5
C2 = Ct*c2 				#Exit velocity in m/s
X2 = 2.141 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2,k
X3 = X1-X2 				#overall frictional consmath.tant fanno parameter
L = (X3*D)/(4*f) 				#Length of the pipe in m

				
#Output
print 'A)Mach number at exitsection 2) is %3.2f  \
\nB)Properties at exitsection 2):    \
\nPressure is %3.2f kPa    \
\nTemperature is %3i K    \
\nVelocity is %3.3f m/s \
\nC)Length of the duct is %3.3f m'%(M2,P2,T2,C2,L)

A)Mach number at exitsection 2) is 0.41  
B)Properties at exitsection 2):    
Pressure is 421.94 kPa    
Temperature is 326 K    
Velocity is 148.432 m/s 
C)Length of the duct is 13.325 m

Example 3.7 page : 17

In [12]:
import math 

				
#Input data
M1 = 0.25 				#Initial Mach number 
M2 = 0.75 				#Final mach number 
P1 = 1.5 				#Inlet pressure in bar
T1 = 300. 				#Inlet temperature in K
k = 1.4 				#Adiabatic consmath.tant
R = 0.287 				#Gas consmath.tant in kJ/kg-K

				
#Calculation
a1 = math.sqrt(k*R*10**3*T1) 				#Sound velocity in m/s, R in J/kg
C1 = a1*M1 				#air velocity in m/s
p1 = 4.3615 				#Pressure ratio at entry from gas tables @M1,k
Pt = P1/p1 				#Static critical pressure in kPa
c1 = 0.272 				#Velocity ratio from gas tables (fanno flow tables,k = 1.4,M1)
Ct = C1/c1 				#Critical velocity in m/s
p2 = 1.385 				#Pressure ratio at exit from gas tables @M2,k
P2 = Pt*p2 				#Exit pressure in bar
c2 = 0.779 				#Velocity ratio at exit from gas tables @M2,k
C2 = Ct*c2 				#Exit velocity in m/s

				
#Output
print 'Final pressure and velocity are %3.4f bar and %3.2f m/s'%(P2,C2)

Final pressure and velocity are 0.4763 bar and 248.58 m/s

Example 3.8 page : 17

In [15]:
import math 

				
#Input data
T1 = 333. 				#Inlet temperature in K
D = 0.05 				#inner duct diameter in m
f = 0.005/4 				#frictional factor
L = 5. 				#Length of the pipe in m
Pt = 101. 				#Exit pressure in kPa, Pt = P2 Since flow is choked 
M2 = 1. 				#Mach number at exit math.since pipe is choked 
k = 1.4 				#Adiabatic consmath.tant
R = 0.287 				#Gas consmath.tant in kJ/kg-K

				
#Calculation
X = (4*f*L)/D 				#frictional consmath.tant fanno parameter 
M1 = 0.6 				#Inlet mach number 
t1 = 1.119 				#Temperature ratio at entry from fanno flow gas tables @M1,k
Tt = T1/t1 				#Static critical temperature in K
at = math.sqrt(k*R*10**3*Tt) 				#Sound velocity in m/s, R in J/kg
Ct = at 				#air velocity in m/s
d_t = Pt/(R*Tt) 				#Density at exit in kg/m**3
At = math.pi*D**2/4 				#Critical area in m**2
m = d_t*At*Ct 				#Mass flow rate in kg/s

				
#Output
print 'A)Mach number at inlet is %3.1f  \
\nB)Mass flow rate is %3.5f kg/s \
\nC)Exit temperature is %3.3f K'%(M1,m,Tt)

# note : rouding off error.

A)Mach number at inlet is 0.6  
B)Mass flow rate is 0.80291 kg/s 
C)Exit temperature is 297.587 K

Example 3.9 page : 18

In [18]:
import math 

				
#Input data
m = 8.25 				#Mass flow rate in kg/s
M1 = 0.15 				#Mach number at entrance
M2 = 0.5 				#Mach number at exit
P1 = 345. 				#Static pressure at entrance in kPa
T1 = 38.+273 				#Static temperature at entry in Kelvin
f = 0.005 				#frictional factor
k = 1.4 				#Adiabatic consmath.tant
R = 0.287 				#Gas consmath.tant in kJ/kg-K

#calculation
d1 = (P1*10**3)/(R*10**3*T1) 				#Density of air in kg/m**3, P1 in Pa
a1 = math.sqrt(k*R*10**3*T1) 				#Sound velocity in m/s, R in J/kg
C1 = a1*M1 				#air velocity in m/s
A1 = m/(d1*C1) 				#Inlet area in m**2
D = (math.sqrt((4*A1)/(math.pi)))*10**3 				#inner duct diameter in mm
p1 = 7.3195 				#Static Pressure ratio from gas tables (fanno flow tables,k = 1.4,M = 0.15)
Pt = P1/p1 				#Static critical pressure in kPa
t1 = 1.1945 				#Static temperature ratio from gas tables (fanno flow tables,k = 1.4,M = 0.15)
Tt = T1/t1 				#Static critical temperature in K
c1 = 0.164 				#Velocity ratio from gas tables (fanno flow tables,k = 1.4,M = 0.15)
Ct = C1/c1 				#Critical velocity in m/s
p2 = 0.984 				#Pressure ratio at entry from gas tables (fanno flow tables,k = 1.4,M = 0.15)
Po1 = P1/p2 				#Stagnation pressure at entry in kPa
p3 = 3.928 				#Stagnation pressure ratio at entry from gas tables (fanno flow tables,k = 1.4,M = 0.15)
Pot = Po1/p3 				#Stagnation pressure at critical state in kPa
X1 = 28.354 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k
p5 = 2.138 				#Pressure ratio at exit from gas tables (fanno flow tables,k = 1.4,M2)
P2 = Pt*p5 				#Exit pressure in kPa
t2 = 1.143 				#Temperature ratio at exit from gas tables (fanno flow tables,k = 1.4,M2)
T2 = Tt*t2 				#Exit temperature in K
c2 = 0.534 				#Velocity ratio at exit from gas tables (fanno flow tables,k = 1.4,M2) 
C2 = Ct*c2 				#Exit velocity in m/s
p6 = 1.34 				#Stagnation pressure ratio at exit from gas tables (fanno flow tables,k = 1.4,M2)
Po2 = Pot*p6 				#Stagnation pressure at exit in kPa
SPL = Po1-Po2 				#Stagnation Pressure lose in kPa
X2 = 1.069 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2,k
X3 = X1-X2 				#overall frictional consmath.tant fanno parameter
L = (X3*D*10**-3)/(4*f) 				#Length of the duct in m

				#verification
a2 = math.sqrt(k*R*10**3*T2) 				#Sound velocity in m/s, R in J/kg
M2_v = C2/a2 				#air velocity in m/s

				
#Output
print 'A)Length of the duct is %3.2f m \
\nB)Diameter of the duct is %3i mm \
\nC)Pressure and diameter at exit are %3.2f kPa, and %3i mm respectively \
\nD)Stagnation Pressure lose is %3i kPa \
\nE)using exit velocity %3.2f m/s, \
\ntemperature %3.2f K \
\nMach number is found to be %3.2f'%(L,D,P2,D,SPL,C2,T2,M2_v)

# note : rouding off error.

A)Length of the duct is 308.85 m 
B)Diameter of the duct is 226 mm 
C)Pressure and diameter at exit are 100.77 kPa, and 226 mm respectively 
D)Stagnation Pressure lose is 231 kPa 
E)using exit velocity 172.65 m/s, 
temperature 297.59 K 
Mach number is found to be 0.50

Example 3.10 page : 19

In [20]:
import math 

				
#Input data
M1 = 0.25 				#Mach number at entrance
f = 0.01/4 				#frictional factor
D = 0.15 				#inner pipe diameter in m
p1 = 0.8 				#Stagnation pressure ratio at exit to entry when loss in stagnation pressure is 20%
M3 = 0.8 				#Mach number at a section

				
#Calculation
p2 = 2.4065 				#Ratio of Stagnation pressure at entry from gas tables @M1,k = 1.4
X1 = 8.537 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1
p3 = p1*p2 				#Ratio of Stagnation pressure at exit
M2 = 0.32 				#Exit mach number at p1 = 0.8
X2 = 4.447 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2
L1 = (X1*D)/(4*f) 				#Length of the pipe in m
L2 = (X2*D)/(4*f) 				#Length of the pipe in m
L = L1-L2 				#Overall length of the duct in m
p4 = 1.038 				#Stagnation pressure ratio from M = 1 to M3
PL = (1-(p4/p2))*100 				#Percentage of stagnation pressure from inlet to section at which M3 in percent

				
#Output
print 'A)Length of the pipe is %3.2f m \
\nB)Mach number at this exit is %3.2f \
\nC)Percentage of stagnation pressure from inlet to section at which M = %3.1f is %3.2f percent \
\nD)Maximum length to reach choking condition is %3.3f m'%(L,M2,M3,PL,L1)

A)Length of the pipe is 61.35 m 
B)Mach number at this exit is 0.32 
C)Percentage of stagnation pressure from inlet to section at which M = 0.8 is 56.87 percent 
D)Maximum length to reach choking condition is 128.055 m

Example 3.11 page : 20

In [21]:
import math 

				
#Input data
D = 0.3 				#inner duct diameter in m
P1 = 10. 				#Static pressure at entrance in bar
T1 = 400. 				#Static temperature at entry in Kelvin
M1 = 3. 				#Mach number at entrance
M2 = 1. 				#Mach number at exit
k = 1.3 				#Adiabatic consmath.tant
R = 287. 				#Specific Gas consmath.tant in J/kg-K, wrong printing in question
f = 0.002 				#frictional factor

				
#Calculation
p1 = 0.233 				#Pressure ratio from gas tables (M = 3,k = 1.4,isentropic)
Pt = P1/p1 				#Static pressure at entrance in bar
t1 = 0.489 				#Temperature ratio from gas tables (M = 3,k = 1.4,isentropic)
Tt = T1/t1 				#Static temperature at entrance in K
X1 = 0.628 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k = 1.3
L1 = (X1*D)/(4*f) 				#Length of the pipe in m
d_t = (Pt*10**5)/(R*Tt) 				#Density at critical state in kg/m**3, Pt in Pa
at = math.sqrt(k*R*Tt) 				#Sound velocity in m/s, R in J/kg 
Ct = at 				#air velocity in m/s
At = (math.pi*D**2)/4 				#Critical area in m**2
m = d_t*At*Ct 				#Mass flow rate in kg/s

				
#Output
print 'A)Length of the pipe is %3.2f m \
\nB)Mass flow rate is %3.3f kg/s'%(L1,m)

A)Length of the pipe is 23.55 m 
B)Mass flow rate is 713.891 kg/s

Example 3.12 page : 21

In [23]:
import math 

				
#Input data
M1 = 0.25 				#Mach number at entrance
f = 0.04/4 				#frictional factor
D = 0.15 				#inner duct diameter in m
p1 = 0.9 				#Stagnation pressure ratio at exit to entry when loss in stagnation pressure is 10%
ds = 190 				#/Change in entropy in J/kg-K
k = 1.3 				#Adiabatic consmath.tant
R = 287 				#Specific Gas consmath.tant in J/kg-K, wrong printing in question

				
#Calculation
p2 = 2.4064 				#Ratio of stagnation pressures at inlet to critical state from gas tables fanno flow tables @M1,k = 1.3
X1 = 8.537 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k = 1.3
p3 = p1*p2 				#Ratio of stagnation pressures at exit to critical state from gas tables fanno flow tables @M1,k = 1.3
M2 = 0.28 				#Mach number at p1 = 0.9 from gas tables @p3
X2 = 6.357 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2,k = 1.3
X3 = X1-X2 				#overall frictional consmath.tant fanno parameter
L1 = (X3*D)/(4*f) 				#Length of the pipe in m
p4 = math.exp(ds/R) 				#Ratio of Stagnation pressure at entry to Stagnation pressure where ds = 190 
p5 = p1/p4 				#Ratio of Stagnation pressures where ds = 190 to critical state
M3 = 0.56 				#Mach number where ds = 190
X4 = 0.674 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M3,k = 1.3
X5 = X1-X4 				#overall frictional consmath.tant fanno parameter
L2 = (X5*D)/(4*f) 				#Length of the pipe in m

				
#Output
print 'A)Length of the pipe is %3.3f m \
\nB)Length of the pipe would require to rise entropy by %3i J/kg-K is %3.5f m \
\nC)Mach number is %3.2f'%(L1,ds,L2,M3)

A)Length of the pipe is 8.175 m 
B)Length of the pipe would require to rise entropy by 190 J/kg-K is 29.48625 m 
C)Mach number is 0.56

Example 3.13 page : 22

In [25]:
import math 

				
#Input data
Po1 = 200. 				#Stagantion pressure at inlet in kPa
To1 = 303. 				#Stagnation temperature at inlet in K
M1 = 0.2 				#Inlet Mach number from diagram
D = 0.025 				#inner tude diameter in m(mismath.sing data)
M2 = 0.8 				#Outlet Mach number 
f = 0.005/4 				#frictional factor
R = 287. 				#Gas consmath.tant in J/kg-K

				
#Calculation
t1 = 0.992 				#Static to Stagnation temperature ratio at entry from gas tables (M1,k = 1.4,isentropic)
T1 = To1*t1 				#Static temperature in K
p1 = 0.973 				#Static to Stagnation pressure ratio at entry from gas tables (M1,k = 1.4,isentropic)
P1 = Po1*p1 				#Static pressure in kPa
p2 = 2.964 				#Stagnation pressure ratio at inlet to critical state from gas tables (M1,k = 1.4,fanno flow)
Pot = Po1/p2 				#Stagnation pressure at critical state in kPa
X1 = 14.533 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k = 1.4
p3 = 1.038 				#Stagnation pressure ratio at outlet to critical state from gas tables (M1,k = 1.4,fanno flow)
Po2 = Pot*p3 				#Stagnation pressure at exit in kPa
X2 = 0.073 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2,k = 1.4
X3 = X1-X2 				#overall frictional consmath.tant fanno parameter
L1 = (X3*D)/(4*f) 				#Length of the pipe in m
SPL = (1-(p3/p2))*100 				#Percentage decrease in stagnation pressure in percent
ds = R*math.log(Po1/Po2) 				#Change of entropy in kJ/kg-K

				
#Output
print 'A)Length of the pipe is %3.1f m \
\nB)Percentage decrease in stagnation pressure is %3.2f percent \
\nC)Change of entropy is %3.3f kJ/kg-K'%(L1,SPL,ds)

A)Length of the pipe is 72.3 m 
B)Percentage decrease in stagnation pressure is 64.98 percent 
C)Change of entropy is 301.133 kJ/kg-K

Example 3.14 page : 23

In [27]:
import math 

				
#Input data
D1 = 0.03 				#Inlet duct diameter in m
D2 = 0.015 				#Throat diameter of  duct in m 
Po1 = 750. 				#Stagantion pressure at inlet in kPa
To1 = 450. 				#Stagnation temperature at inlet in K
f = 0.02/4 				#frictional factor
L = 0.25 				#Length of the duct in m
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Gas consmath.tant in J/kg-K

				
#Calculation
ar = (D1/D2)**2 				#Ratio of areas
M1 = 2.94 				#Mach number at inlet from gas tables (ar,k = 1.4,isentropic)
p1 = 0.0298 				#Static to Stagnation pressure ratio at entry from gas tables (M1,k = 1.4,isentropic)
P1 = Po1*p1 				#Static pressure at inlet in kPa
t1 = 0.367 				#Static to Stagnation temperature ratio at entry from gas tables (M1,k = 1.4,isentropic)
T1 = To1*t1 				#Static temperature at inlet in K
a1 = math.sqrt(k*R*T1) 				#Sound velocity in m/s
C1 = a1*M1 				#Air velocity at inlet in m/s
X1 = 0.513 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k = 1.4
p2 = 0.226 				#Static to Critical pressure ratio at inlet from gas tables,fanno flow tables @M1,k = 1.4
Pt = P1/p2 				#Critical pressure in kPa
c1 = 1.949 				#Static to Critical velocity ratio at inlet from gas tables,fanno flow tables @M1,k = 1.4
Ct = C1/c1 				#Critical velocity in m/s
t2 = 0.439 				#Static to Critical temperature ratio at inlet from gas tables,fanno flow tables @M1,k = 1.4
Tt = T1/t2 				#Critical temperature in K
L1 = (X1*D1)/(4*f) 				#Length of the pipe from inlet to critical state in m
L2 = L1-L 				#Length of the pipe from required point to critical state in m
X2 = (4*f*L2)/D2 				#frictional consmath.tant fanno parameter
M2 = 2.14 				#Mach number at inlet from gas tables (X2,k = 1.4,fanno flow)
p3 = 0.369 				#Static to Critical pressure ratio at outlet from gas tables,fanno flow tables @M2,k = 1.4
P2 = Pt*p3 				#Exit pressure in kPa
c2 = 1.694 				#Static to Critical velocity ratio at outlet from gas tables,fanno flow tables @M2,k = 1.4
C2 = Ct*c2 				#Exit velocity in m/s
t3 = 0.623 				#Static to Critical temperature ratio at outlet from gas tables,fanno flow tables @M2,k = 1.4
T2 = t3*Tt 				#Exit temperature in K

				
#Output
print 'A)Maximum length of the pipe is %3.4f m \
\nB)Condition of air at exit:    \
\nPressure is %3.2f kPa    \
\nVelocity is %3.2f m/s    \
\nTemperature is %3.2f K'%(L1,P2,C2,T2)

A)Maximum length of the pipe is 0.7695 m 
B)Condition of air at exit:    
Pressure is 36.49 kPa    
Velocity is 658.25 m/s    
Temperature is 234.37 K

Example 3.15 page : 25

In [30]:
import math 

				
#Input data
f = 0.002 				#frictional factor
C1 = 130. 				#Air velocity at inlet in m/s
T1 = 400. 				#Inlet temperature at inlet in K
P1 = 250. 				#Inlet pressure at inlet in kPa
D = 0.16 				#Inlet duct diameter in m
p1 = 0.8 				#Stagnation pressure ratio at exit to entry when loss in stagnation pressure is 20%
L1 = 35. 				#Length of duct from inlet to required section
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Gas consmath.tant in J/kg-K

				
#Calculation
a1 = math.sqrt(k*R*T1) 				#Sound velocity in m/s
M1 = C1/a1 				#Mach number at inlet
p2 = 0.9295 				#Static to Stagnation pressure ratio at entry from gas tables (M1,k = 1.4,isentropic)
Po1 = P1/p2 				#Stagantion pressure at inlet in kPa
Po2 = 0.8*Po1 				#Stagantion pressure at outlet in kPa
p3 = 1.89725 				#Stagnation pressure ratio at inlet to critical state from gas tables (M1,k = 1.4,fanno flow)
Pot = Po1/p3 				#Stagnation pressure at critical state in kPa
X1 = 4.273 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k = 1.4
p4 = 3.33725 				#Static Pressure ratio from gas tables (fanno flow tables,k = 1.4,M = 0.5)
Pt = P1/p4 				#Static critical pressure in kPa
t1 = 1.175 				#Static temperature ratio from gas tables (fanno flow tables,k = 1.4,M = 0.5) 
Tt = T1/t1 				#Static critical temperature in K
c1 = 0.347 				#Velocity ratio from gas tables (fanno flow tables,k = 1.4,M = 0.5)
Ct = C1/c1 				#Critical velocity in m/s
p5 = Po2/Pot 				#Pressure ratio
M2 = 0.43 				#Mach number at p1 = 0.8
X2 = 1.833 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2,k = 1.4
X3 = X1-X2 				#overall frictional consmath.tant fanno parameter
L2 = (X3*D)/(4*f) 				#Length of the pipe in m, (from required section to critical state) 
L3 = (X1*D)/(4*f) 				#Length of the pipe in m, (from required inlet to critical state) 
L4 = L3-L1 				#Length of the pipe in m
X4 = (4*f*L3)/D 				#frictional consmath.tant fanno parameter
M3 = 0.39 				#Mach number at L1 = 35m
p6 = 2.767 				#Static to Critical pressure ratio at outlet from gas tables,fanno flow tables @M3,k = 1.4
P2 = Pt*p6 				#Exit pressure in kPa
t2 = 1.1645 				#Static to Critical temperature ratio at outlet from gas tables,fanno flow tables @M3,k = 1.4
T2 = Tt*t2 				#Exit temperature in K
c2 = 0.42087 				#Static to Critical velocity ratio at outlet from gas tables,fanno flow tables @M3,k = 1.4
C2 = Ct*c2 				#Exit velocity in m/s

				
#Output
print 'A)Length of pipe required for p = %3.1f m is %3.3f m \
\nB)Properties of air at section %3i from inlet:    \
\nTemperature is %3.3f K    \
\nPressure is %3.2f kPa    \
\nVelocity is %3.1f m/s \
\nC)Maximum length of the pipe is %3.2f m'%(p1,L2,L1,T2,P2,C2,L3)

# rounding off error.

A)Length of pipe required for p = 0.8 m is 48.800 m 
B)Properties of air at section  35 from inlet:    
Temperature is 396.426 K    
Pressure is 207.28 kPa    
Velocity is 157.7 m/s 
C)Maximum length of the pipe is 85.46 m

Example 3.16 page : 26

In [32]:
import math 

				
#Input data
D = 0.3 				#inner pipe diameter in m
Q = 1000. 				#Discharge in m**3/min
P2 = 150. 				#Exit pressure in kPa
T2 = 293. 				#Exit temperature in K
L1 = 50. 				#Length of the pipe in m
f = 0.005 				#frictional factor
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Gas consmath.tant in J/kg-K

				
#Calculation
A = math.pi*D**2/4 				#Area of duct in m**2 
C2 = Q/(A*60) 				#Exit air velocity in m/s
a2 = math.sqrt(k*R*T2) 				#Sound velocity in m/s
M2 = C2/a2 				#Exit mach number 
p1 = 1.54 				#				#Static to Critical pressure ratio at outlet from gas tables,fanno flow tables @M2,k = 1.4
Pt = P2/p1 				#Critical pressure in kPa
t1 = 1.10 				#Static to Critical temperature ratio at outlet from gas tables,fanno flow tables @M2,k = 1.4
Tt = T2/t1 				#Critical temperature in K
X1 = 0.228 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2,k = 1.4
L2 = (X1*D)/(4*f) 				#Length of the pipe in m
L2 = L1+L2 				#Overall length of pipe from inlet to critical state in m
X2 = (4*f*L2)/D 				#frictional consmath.tant fanno parameter for M1
M1 = 0.345 				#Inlet Mach number from gas tables fanno flow tables @X2,k = 1.4
p2 = 3.14 				#Static to Critical pressure ratio at inlet from gas tables,fanno flow tables @M1,k = 1.4
P1 = Pt*p2 				#Static pressure at inlet in kPa
t2 = 1.17 				#Static to Critical temperature ratio at inlet from gas tables,fanno flow tables @M1,k = 1.4
T1 = Tt*t2 				#Static temperature at inlet in K

				
#Output
print 'A)Mach number at the exit is %3.3f \
\nB)Inlet pressure and temperature are %3.3f kPa and %3.2f K'%(M2,P1,T1)

A)Mach number at the exit is 0.687 
B)Inlet pressure and temperature are 305.844 kPa and 311.65 K

Example 3.17 page : 27

In [34]:
import math 

				
#Input data
D = 0.0254 				#inner pipe diameter in m
f = 0.003 				#frictional factor
M1 = 2.5 				#Inlet Mach number 
To1 = 310. 				#Stagnation temperature at inlet in K
P1 = 0.507 				#Static pressure at inlet in kPa
M2 = 1.2 				#Exit mach number 
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Gas consmath.tant in J/kg-K

				
#Calculation
t1 = 0.4444 				#Static to Stagnation temperature ratio at entry from gas tables (M1,k = 1.4,isentropic)
T1 = To1*t1 				#Static temperature at inlet in K
p1 = 0.05853 				#Static to Stagnation pressure ratio at entry from gas tables (M1,k = 1.4,isentropic)
Po1 = P1/p1 				#Stagantion pressure at inlet in kPa
a1 = math.sqrt(k*R*T1) 				#Sound velocity at inlet in m/s, R in J/kg
C1 = a1*M1 				#air velocity at inlet in m/s
c1 = 2.95804 				#Static to Critical velocity ratio at inlet from gas tables,isothermal tables @M1,k = 1.4
Ctt = C1/c1 				#Critical velocity at isothermal state in m/s
p2 = 0.33806 				#Static to Critical pressure ratio at inlet from gas tables,isothermal @M1,k = 1.4
Ptt = P1/p2 				#Critical pressure at isothermal state in bar
p3 = 3.61691 				#Stagnation pressure ratio at inlet to isothermal state from gas tables,isothermal tables @M1,k = 1.4
Pott = Po1/p3 				#Critical pressure at isothermal state in K
t2 = 1.968748 				#Stagnation temperature ratio at inlet to isothermal state from gas tables,isothermal tables @M1,k = 1.4
Tott = To1/t2 				#Critical temperature at isothermal state in K
X1 = 1.28334 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k = 1.4
c2 = 1.4186 				#Static to Critical velocity ratio at exit from gas tables,isothermal tables @M2,k = 1.4
C2 = Ctt*c2 				#Exit velocity in m/s
p4 = 0.7043 				#Static to Critical pressure ratio at inlet from gas tables,isothermal @M2,k = 1.4
P2 = Ptt*p4 				#Exit pressure in bar
p5 = 1.07026 				#Stagnation pressure ratio at inlet to isothermal state from gas tables,isothermal tables @M2,k = 1.4
Po2 = Pott*p5 				#Stagnation pressure at exit in bar 
t3 = 1.127 				#Stagnation temperature ratio at inlet to isothermal state from gas tables,isothermal tables @M2,k = 1.4
To2 = Tott*t3 				#Stagnation temperature at exit in bar
T2 = T1 				#Exit temperature in K, Since isothermal flow
X2 = 0.19715 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2,k = 1.4
X3 = X1-X2 				#Overall frictional consmath.tant fanno parameter
L1 = (X3*D)/(4*f) 				#Length of the pipe in m
d1 = (P1*10**5)/(R*T1) 				#Density of air in kg/m**3, P1 in Pa
A1 = (math.pi*D**2)/4 				#Cross sectional area of pipe in m**2
m = d1*A1*C1 				#Mass flow rate in kg/s

				
#Output
print 'At M = %3.1f : A)Static pressure and static temperature are %3.5f bar and %3.3f K respectively \
\nB)Stagnation pressure and temperature are %3.4f bar and %3.3f K respectively \
\nC)Velocity of air is %3.3f m/s \
\nD)Distance of the section from innlet is %3.3f m \
\nE)Mass flow rate is %3.5f kg/s'%(M2,P2,T2,Po2,To2,C2,L1,m)

At M = 1.2 : A)Static pressure and static temperature are 1.05626 bar and 137.764 K respectively 
B)Stagnation pressure and temperature are 2.5632 bar and 177.458 K respectively 
C)Velocity of air is 282.078 m/s 
D)Distance of the section from innlet is 2.299 m 
E)Mass flow rate is 0.38217 kg/s

Example 3.18 page : 29

In [38]:
import math 

				
#Input data
D = 0.12 				#inner duct diameter in m
f = 0.004 				#frictional factor
M1 = 0.4 				#Inlet Mach number 
P1 = 300. 				#Static pressure at inlet in kPa
T1 = 310. 				#Static temperature at inlet in K
M2 = 0.6 				#Exit mach number
k = 1.4 				#Adiabatic consmath.tant

				
#Calculation
p1 = 2.118 				#Static to Critical pressure ratio at inlet from gas tables,fanno flow tables @M1,k = 1.4
Pt = P1/p1 				#Critical pressure in kPa
X1 = 1.968 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k = 1.4
p2 = 1.408 				#Static to Critical pressure ratio at outlet from gas tables,fanno flow tables @M2,k = 1.4
P2 = Pt*p2 				#Exit pressure in kPa
X2 = 0.299 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2,k = 1.4
X3 = X1-X2 				#Overall frictional consmath.tant fanno parameter
L1 = (X3*D)/(4*f) 				#Length of the pipe in m
T2 = T1 				#Exit temperature in K, Since isothermal flow
Ttt = T1 				#Critical temperature at critical state, Since isothermal flow 
Mtt = 1/math.sqrt(k) 				#Limiting Mach number
L2 = (X1*D)/(4*f) 				#Length of the duct required to attain limiting mach number in m

				
#Output
print 'A)Length of the duct required to chnage the mach number to %3.1f is %3.4f m \
\nB)Pressure and temperature at M = %3.1f is %.f kPa and %3i K respectively \
\nC)Length of the duct required to attain limiting mach number is %3.3f m \
\nD)State of air at limiting mach number %3.3f is subsonic'%(M2,L1,M2,P2,T2,L2,Mtt)

# rounding off error.

A)Length of the duct required to chnage the mach number to 0.6 is 12.5175 m 
B)Pressure and temperature at M = 0.6 is 199 kPa and 310 K respectively 
C)Length of the duct required to attain limiting mach number is 14.760 m 
D)State of air at limiting mach number 0.845 is subsonic

Example 3.19 page : 29

In [1]:
import math 
from numpy import roots

				
#Input data
m = 0.32 				#Mass flow rate in kg/s
L = 140. 				#Length of the pipe in m
P1 = 800. 				#Inlet pressure in N/m**2, wrong units in textbook
T1 = 288. 				#Inlet temperature in K
P2 = 600. 				#Outlet pressure in N/m**2, wrong units in textbook
f = 0.006 				#frictional factor
R = 287. 				#Gas consmath.tant in J/kg-K

				
#Calculation
				#Umath.sing Adiabatic Equation d = 1/((((((math.pi*(d/2)**2)**2)/(2*m**2*R*T))*(P1**2-P2**2))-(math.log(P1/P2)))/(2*f*L)) and converting into 5th degree polynomial of d
a = (math.pi**2*(P1**2-P2**2))/(32*m**2*R*T1) 				#Coefficient of power 5
b = math.log(P1/P2) 				#Coefficient of power 1
c = 2*f*L 				#Coefficient of consmath.tant
#p5 = poly([-c -b 0 0 0 a],'d','coeff') 				#Solving polynomial of degree 5
d = roots([a,0,0,0,-b, -c]) 				#Command to find roots

				
#Output
print ("Possible values for diameter of pipe are:") 				#Displays whatever within paranthesis 
print (d) 															#To print lay roots
print 'Therefore Diameter of the pipe is : %.1f m'%d[0].real

Possible values for diameter of pipe are:
[ 0.71338855+0.j          0.20228465+0.67301992j  0.20228465-0.67301992j
 -0.55897893+0.39355091j -0.55897893-0.39355091j]
Therefore Diameter of the pipe is : 0.7 m

Example 3.20 page: 30

In [43]:
import math 

				
#Input data
Q = 225./60 				#Discharge in m**3/s
T2 = 293. 				#Exit temperature in K
P2 = 1.25 				#Exit pressure in bar
L1 = 30. 				#Length of the pipe in m
D = 0.15 				#Duct diameter in m
f = 0.02/4 				#frictional factor
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Gas consmath.tant in J/kg-K

				
#Calculation
A = math.pi*D**2/4 				#area in m**2
C2 = Q/A 				#Exit air velocity in m/s
a2 = math.sqrt(k*R*T2) 				#Exit sound velocity in m/s 
M2 = C2/a2 				#Exit mach number 
p1 = 1.703 				#Static to Critical pressure ratio at outlet from gas tables,fanno flow tables @M2,k = 1.4
Pt = P2/p1 				#Critical pressure in bar
c1 = 0.654 				#Static to Critical velocity ratio at outlet from gas tables,fanno flow tables @M2,k = 1.4
Ct = C2/c1 				#Critical velocity in m/s
t1 = 1.114 				#Static to Critical temperature ratio at outlet from gas tables,fanno flow tables @M2,k = 1.4
Tt = T2/t1 				#Critical temperature in K
X1 = 0.417 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k = 1.4
X2 = (4*f*L1)/D 				#frictional consmath.tant fanno parameter
X3 = X1+X2 				#overall frictional consmath.tant fanno parameter
M1 = 0.32 				#Mach number at entrance
p2 = 3.385 				#Static to Critical pressure ratio at inlet from gas tables,fanno flow tables @M1,k = 1.4
P1 = Pt*p2 				#Static pressure at inlet in bar
c2 = 0.347 				#Static to Critical velocity ratio at inlet from gas tables,fanno flow tables @M1,k = 1.4
C1 = Ct*c2 				#Air velocity at inlet in m/s
t2 = 1.176 				#Static to Critical temperature ratio at inlet from gas tables,fanno flow tables @M1,k = 1.4
T1 = Tt*t2 				#Static temperature at inlet in K

				
#Output
print 'Required Inlet Condition:    \
\nPressure is %3.4f bar    \
\nVelocity is %3.3f m/s    \
\nTemperature is %3.1f K'%(P1,C1,T1)

Required Inlet Condition:    
Pressure is 2.4846 bar    
Velocity is 112.593 m/s    
Temperature is 309.3 K

Example 3.21 page : 31

In [44]:
import math 

				
#Input data
D1 = 0.134 				#Inlet duct diameter in m
Po1 = 7 				#Stagnation pressure at inlet in bar
P1 = 0.245 				#Static pressure at 5*D1 i.e. L1 in bar
P2 = 0.5 				#Static pressure at 33*D1 i.e. L2 in bar
D2 = 0.0646 				#throat diameter in m 
L1 = 5*D1 				#Length of nozzle till section-1 in m
L2 = 33*D1 				#Length of nozzle till section-2 in m 

				
#Calculation
ar = (D1/D2)**2 				#Ratio of areas
p1 = P1/Po1 				#Pressure ratio
APR1 = p1*ar 				#Area Pressure ratio i.e. (A1*P1)/(At*Po1)
M1 = 2.54 				#Mach number at inlet from isentropic gas tables @APR1 
X1 = 0.44 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M1,k = 1.4
APR2 = 0.3073 				#Area Pressure ratio i.e. (A2*P2)/(At*Po1)
M2 = 1.54 				#Exit mach number
X2 = 0.151 				#frictional consmath.tant fanno parameter from gas tables,fanno flow tables @M2,k = 1.4
X3 = X1-X2 				#overall frictional consmath.tant fanno parameter
L3 = L2-L1 				#Length of the nozzle (Section-1 to Section-2) in m 
f = (X3*D1)/(4*L3) 				#frictional factor

				
#Output
print 'A)Mach number at %3.3f m and %3.3f m are %3.2f and %3.2f respectively \
\nB)Mean value of friction between two sections is %3.5f'%(L1,L2,M1,M2,f)

A)Mach number at 0.670 m and 4.422 m are 2.54 and 1.54 respectively 
B)Mean value of friction between two sections is 0.00258