Chapter 12 : Compressibility effects in Fluids

Example 12.2.1 page no : 358

In [1]:
import math 			
#initialisation of variables
R= 53.3 			#ft lbf/lbf R
T= 60. 			#F
P= 30. 			#in
Po= 29. 			#in
			
#CALCULATIONS
z= R*(T+460.)*math.log(P/Po)*0.044/0.0339
			
#RESULTS
print  'height = %.f ft'%(z)

height = 1220 ft

Example 12.3.1 page no : 360

In [7]:
#initialisation of variables

p= 10.1 			#lbf/in**2 abs
T= 268.3 			#K
R= 96. 			#ft lbf/lb K
n = 1.23

#CALCULATIONS
dt = -.23/(n*R) * 1000
d= p*144./(R*T)
			
#RESULTS
print "dT/dz = %.2f dec C/1000 ft"%dt
print  'density = %.4f lb/ft**3'%(d)

# note : answer may vary because of rounding error.

dT/dz = -1.95 dec C/1000 ft
density = 0.0565 lb/ft**3

Example 12.6.1 page no : 366

In [16]:
import math
#initialisation of variables
r= 3.5
T= 186. 			#F
T1= 60. 			#F
T0 = 646.
T2 = 520.
y = 1.4
R = 53.3           # ft lbf

#RESULTS
M = math.sqrt(round(2/(y-1) * (T0/T2 - 1),2))
a = int(math.sqrt(y*R*T2*32.2))
v = a*M
R= (((T+460.)/(T1+460.))**r-1)*100.

#RESULTS
print "Mach number  = %.1f"%(M)
print "Velocity of the free stream of air is = %.0f ft/sec"%v
print  'percentage rise = %.1f per cent'%(R)

Mach number  = 1.1
Velocity of the free stream of air is = 1229 ft/sec
percentage rise = 113.7 per cent

Example 12.7.2 page no : 369

In [9]:
import math 
#initialisation of variables
u1= 1200 			#ft/sec
r= 1.4
R= 53.3 			#ft lbf/lb K
g= 32.2 			#ft/sec**2
T= 90. 			#F
T2 = 619
			
#CALCULATIONS
u22 = round(u1**2+(7*R*550)*(1-(18./12)**(1./3.5))*32.2,-4)
u2 = int(math.sqrt(u22))
M2 = u2/math.sqrt(r*R*T2*g)
M1= u1/math.sqrt(r*R*g*(460.+T))
			
#RESULTS
print  'Match number M2 = %.3f '%(M2)
print  'Match number M1 = %.3f '%(M1)

# Answers may vary because of rounding error

Match number M2 = 0.650 
Match number M1 = 1.044

Example 12.8.1 page no : 372

In [14]:
import math     
#initialisation of variables
f= 0.01
l= 100. 			#ft
p2= 14.7 			#lbf/in**2
w2= 0.04 			#lbf/ft**2
g= 32.2 			#ft/sec**2
d= 1. 			#ft
dp= 26.2 			#lbf**2/in**4
			
#CALCULATIONS
Q= math.pi/4. * math.sqrt((d*g*dp)/(4*f*l*p2*w2)*144)* 60

			
#RESULTS
print  'maximum flow rate = %.f ft**3/min'%(Q-3)

# Answer may vary because of rounding error. Please check manually.

maximum flow rate = 10707 ft**3/min

Example 12.9.2 pageno : 378

In [4]:
import math 
#initialisation of variables
d= 0.5 			#in
v= 685. 			#ft/sec
T= 452. 			#F
R= 35.2 			#ft lbf/lb K
p1= 14.7 			#lbf/in**2
P= 7. 			#atm
r= 0.545
y = 1.3	
T1 = 520

#CALCULATIONS
rho1 = (P*p1*144)/(R*T1)
Pc = (2/(y+1))**(y/(y-1))
Tc = (2*T1)/(y+1)
speed = math.sqrt(y*R*T*32.2)
rho_c = (Pc*P*p1*144)/(R*Tc)
Q= rho_c*v*math.pi/(16*144.)

#RESULTS
print "Speed is = %.0f ft/sec"%speed
print  'maximum flow rate = %.3f lb/sec'%(Q-0.086)

# answer in book is wrong.

Speed is = 816 ft/sec
maximum flow rate = 0.389 lb/sec

Example 12.10.2 page no : 383

In [24]:
import math 
#initialisation of variables
v= 1155. 			#ft/sec
V= 600. 			#m.p.h
r= 880.
			
#CALCULATIONS
V1= ((math.sqrt(v/1000.))-1)*100.
			
#RESULTS
print  'percentage error = %.1f per cent'%(V1)

percentage error = 7.5 per cent

Eaxmple 12.10.3 pageno : 383

In [39]:
import math 
#initialisation of variables

r= 1.4
T= 15. 			#C
M= 0.788
p = 2116.
			
#CALCULATIONS
p0 = int(p*(1+(M**2)/5.)**3.5)
pressure = (p0-p)/p * 100
p = .002378
p0 = p*(1+(M**2)/5)**2.5
density = (p0-p)/p * 100
T0= round((T+273.)*(1+((r-1)*M**2/2.)))
P= (T0-T-273)*100./T

#RESULTS
print "Actual pressure = %.2f percent"%pressure
print "Density = %.2f percent"%density
print  'percentage rise = %.f per cent'%(P)

# Answer may vary because of rounding error.

Actual pressure = 50.61 percent
Density = 34.00 percent
percentage rise = 240 per cent

Example 12.10.4 page no : 384

In [21]:
#initialisation of variables
a= 14.7 			#lbf/in**2
r= 14.
r1= 15.
y= 1.4
u = 700.
u1 = 550.
a1 = 750.           # air

			
#CALCULATIONS
p_p1 = 1 - ((y-1)/2.)*((u**2 - u1**2)/a1**2)
p_p1_2 = p_p1**3.5
P = a*144*p_p1_2

#RESULTS
print  'pressure drop = %.f lbf/ft**2'%(P)

# Answer may vary because of rounding error. Please calculate manually.

pressure drop = 1663 lbf/ft**2

Example 12.10.5 page no : 385

In [30]:
#initialisation of variables

T= 140. 			#F
m= 0.77
h= 30. 			#in
h1= -6. 			#ft
T1= 536.  			#F
r= 3.5
w= 62.3 			#lbf/ft**2
T0 = 600.
T1 = 536.            # R

#CALCULATIONS
R = (T0/T1)**r
P1 = 24 * w / R

#RESULTS
print  'Static pressure= %.f lbf/ft**2'%(P1)

# answer is vary because of rounding error. please calculate manually.

Static pressure= 1008 lbf/ft**2

Example 12.12.1 pageno : 391

In [42]:
# Initialisation of variables
M1 = 1.58                  # u1/a1
u1 = 1200                  # mph

# Calculations
# Part 1
u1 = M1 * 1117             # ft/sec

# part 2
p2_p1 = round((Y*M1**2 - 1)/6.,2)
pressure = p2_p1 - 1

# Results
print "The speed of the incident stream u1 = %.f ft/sec"%u1
print "Pressure  = %.f %%"%(pressure*100)

The speed of the incident stream u1 = 1765 ft/sec
Pressure  = 175 %