Chapter 7 : Steady, One-Dimensional, Irreversible Flow

Example 7.1 Page No : 213

In [1]:
	
#initialisation of variables
r= 1.5
f= 0.025   # friction factor
	
#CALCULATIONS
r1= (2/f)*(r**2-1)
	
#RESULTS
print  'ratio L/D2 = %.f'%(r1)
ratio L/D2 = 100

Example 7.2 Page No : 214

In [2]:
import math	
#initialisation of variables
a= 6.    	#degrees angle
r= 1.5
l= 100.  	#ft
f= 0.025
K= 0.15
	
#CALCULATIONS
R= r**4-1
R1= 1/math.tan(math.radians(a/2))*(1-(1./r))
p1= f*l
p2= 2.5*(l-p1)/l
p3= (1-r**2)**2
p4= K*p3
pt= p4+p2
	
#RESULTS
print  ' lowest ratio = %.2f'%(R)
print  ' contribtuion of friction in pipe = %.3f lbf/ft**2'%(p1)
print  ' contribtuion of diffuser in pipe = %.3f lbf/ft**2'%(p2)
print  ' stagnant pressure drop = %.3f lbf/ft**2'%(p3)
print  ' contribtuion of friction in pipe after reduction = %.3f lbf/ft**2'%(pt)

# note : rounding off error
 lowest ratio = 4.06
 contribtuion of friction in pipe = 2.500 lbf/ft**2
 contribtuion of diffuser in pipe = 2.438 lbf/ft**2
 stagnant pressure drop = 1.562 lbf/ft**2
 contribtuion of friction in pipe after reduction = 2.672 lbf/ft**2

Example 7.3 Page No : 219

In [3]:
import math 
	
#initialisation of variables
d= 4. 	    #in galvanised iron pipe diameter
q= 0.5 	    #ft**3/sec  flow rate
w= 62.4 	#lb/ft**3 density
u= 2.7*10**-5 	#lbf sec/ft**2 viscosity
e= 0.0005 	#ft
g= 32.1 	#ft/sec**2 acceleration
f= 0.0235
lt= 400. 	#ft long
	
#CALCULATIONS
V= 4*q/(math.pi*(d/12)**2)
Re= w*V*(d/12)/(u*g)
r= e/(d/12)
dz= (V**2/(2*g))*(1.7+f*lt/(d/12))
	
#RESULTS
print  ' mean flow velocity = %.2f ft/sec'%(V)
print  ' Reynolds number = %.2e'%(Re)
print  ' Relative roughness = %.4f'%(r)
print  ' difference in the levels of water = %.1f ft'%(dz)
 mean flow velocity = 5.73 ft/sec
 Reynolds number = 1.38e+05
 Relative roughness = 0.0015
 difference in the levels of water = 15.3 ft

Example 7.4 Page No : 220

In [5]:
import math 
	
#initialisation of variables
d= 4.    	#in
v= 6.64 	#ft/sec
	
#CALCULATIONS
Q= math.pi*0.25*(d/12)**2*v
	
#RESULTS
print  'Flow rate= %.3f ft**3/sec'%(Q)
Flow rate= 0.579 ft**3/sec

Example 7.5 Page No : 221

In [6]:
	
#initialisation of variables
d= 0.366 	#ft
i= 12
	
#CALCULATIONS
pd= d*i
	
#RESULTS
print  'Required pipe diameter = %.2f in'%(pd)
Required pipe diameter = 4.39 in

Example 7.6 Page No : 222

In [5]:
	
#initialisation of variables
Ps1= 1050. 	#lbf/ft**2
fr= 10.7
p= 36.6 	#lbf/ft**2
p1= 195. 	#lbf/ft**2
fr1= 16.
fr2= 1.8
	
#CALCULATIONS
deltap = (p+957+p1+Ps1)
p2= round(fr*p)
dp= Ps1-p2
lc= round(dp/p)
sp= Ps1+p1-p*(fr1+fr2)
lc1= sp/p

#RESULTS
print  ' Pressure = %.f lbf/ft**2'%(round(deltap,-1))
print  ' pressure difference = %.f lbf/ft**2'%(dp)
print  ' Loss coefficient = %.f '%(lc)
print  ' Loss coefficient = %.1f '%(lc1)
 Pressure = 2240 lbf/ft**2
 pressure difference = 658 lbf/ft**2
 Loss coefficient = 18 
 Loss coefficient = 16.2 

Example 7.7 Page No : 232

In [4]:
import math 
	
#initialisation of variables
p1= 50. 	#lbf/in**2 pressure
R= 96.3 	#ft lbf/lbm R
T= 80. 	    #F temperature
p2= 20. 	#lbf/in**2 pressure
r= 1.31
u= 2.34*10**-7 	#lbf sec/ft**2
e= 0.00005 	    #ft
m= 5.*10**4 	#lbm/sec
d= 1.5 	        #ft
g= 32.2 	    #ft/sec**2
f= 0.113
	
#CALCULATIONS
w1= p1*144/(R*(460+T))
V1= 4*(m/3600)/(math.pi*w1*d**2)
Ma1= V1/(r*R*g*(460+T))**0.5
Re= w1*V1*d/(u*g)
dx= (((1/(r*Ma1**2))*10*(1-(p2/p1)**2))+math.log(p2/p1))*d/f
	
#RESULTS
print  ' density = %.3f lbm/ft**3'%(w1)
print  ' mean flow velocity = %.1f ft/sec'%(V1)
print  ' Match number = %.4f '%(Ma1)
print  ' Reynolds number = %.2e '%(Re)
print  ' Length of pipe = %.2e ft'%(dx)
 density = 0.138 lbm/ft**3
 mean flow velocity = 56.8 ft/sec
 Match number = 0.0383 
 Reynolds number = 1.56e+06 
 Length of pipe = 5.79e+04 ft

Example 7.9 Page No : 238

In [5]:
import math 
	
#initialisation of variables
r= 1.4
R= 53.3 	#ft lbf/lbm R
g= 32.2 	#ft/sec**2
T1= 410. 	#R temperature
v= 2500. 	#ft/sec steadility
P1= 628. 	#lbf/in**2 pressure
	
#CALCULATIONS
v1= int(math.sqrt(r*g*R*T1))
Ma1= round(v/v1,2)
Ts1= int(T1*(1+0.5*(r-1)*Ma1**2))
Ps1= P1*(1+0.5*(r-1)*Ma1**2)**(r/(r-1))
Ps2= Ps1*((r+1)/(2*r*Ma1**2-r+1))**(1/(r-1))*(0.5*(r+1)*Ma1**2/(1+0.5*(r-1)*Ma1**2))**(r/(r-1))

#RESULTS
print  ' acoustic velocity = %.f ft/sec'%(v1)
print  ' Match number = %.2f '%(Ma1)
print  ' Stagnition temperature = %.f R'%(Ts1)
print  ' Stagnition pressure = %.f lbf/ft**2'%(Ps1)
print  ' Stagnition pressure = %.f lbf/ft**2'%(Ps2)

# note : answer in book is wrong. Please check manually.
 acoustic velocity = 992 ft/sec
 Match number = 2.52 
 Stagnition temperature = 930 R
 Stagnition pressure = 11069 lbf/ft**2
 Stagnition pressure = 5435 lbf/ft**2

Example 7.10 Page No : 245

In [6]:
#initialisation of variables
p2= 67.2 	#lbf/in**2 pressure
p1= 63. 	#lbf/in62 pressure
r= 1.4
n= 0.6      # efficiency
T1= 870. 	#R temperature
ma1= 0.8 	#ft/sec mach number

#CALCULATIONS
dt= (p2/p1)**((r-1)/r)-1
dt1= dt/n
T2= T1*(1+dt1)
Ts1= T1*(1+0.5*(r-1)*ma1**2)
ps1= p1*(1+0.5*(r-1)*ma1**2)**(r/(r-1))
ps2= p2*(Ts1/T2)**(r/(r-1))
dp= ps1-ps2

#RESULTS
print  ' dT = %.5f '%(dt)
print  ' dT1 = %.5f '%(dt1)
print  ' Temperature = %.f R'%(T2)
print  ' Temperature = %.1f R'%(Ts1)
print  ' Pressure = %.1f lbf/in**2'%(ps1)
print  ' Pressure = %.1f lbf/in**2'%(ps2)
print  ' pressure difference = %.1f lbf/in**2'%(dp)
 dT = 0.01861 
 dT1 = 0.03102 
 Temperature = 897 R
 Temperature = 981.4 R
 Pressure = 96.0 lbf/in**2
 Pressure = 92.0 lbf/in**2
 pressure difference = 4.0 lbf/in**2

Example 7.11 Page No : 246

In [7]:
import math 

#initialisation of variables
r= 1.4
ma3= 3. 	#ft/sec mach number
ps= 80. 	#lbf/ft**2 pressure
Ts= 840. 	#R temperature
r1= 53.3 	#ft lbm/ft**3
A3= 2.   	#in**2 flow area
g= 32.2 	#ft/sec**2
ma1= 1.6

#CALCULATIONS
R= (1+(r-1)*0.5*ma3**2)**(r/(r-1))
p3= ps/R
R1= 1+0.5*(r-1)*ma3**2
T3= Ts/R1
w3= p3*144/(r1*T3)
V3= ma3*math.sqrt(r*r1*g*T3)
m= w3*V3*A3/144
ra= ((r+1)/(2*r*ma1**2-(r-1)))**(1/(r-1))*(0.5*(r+1)*ma1**2/(1+0.5*(r-1)*ma1**2))**(r/(r-1))
ps2= ps*ra
dp= ps-ps2

#RESULTS
print  ' outlet pressure = %.2f lbf/in**2'%(p3)
print  ' outlet temperature = %.f R'%(T3)
print  ' mass flow rate = %.3f lbm/sec'%(m)
print  ' ps2 = %.1f lbf/in**2'%(ps2)
print  ' Stagnation pressure loss = %.1f lbf/in**2'%(dp)

# rounding off error
 outlet pressure = 2.18 lbf/in**2
 outlet temperature = 300 R
 mass flow rate = 0.694 lbm/sec
 ps2 = 71.6 lbf/in**2
 Stagnation pressure loss = 8.4 lbf/in**2