Chapter 9 : Steady Flow of Compressible Fluids

Example 9.1 Page No : 268

In [1]:
	
#Initialization of variables
T2 = 30. 	#C
T1 = 20. 	#C
cv = 716.
m = 15. 	#kg
cp = 1003.
	
#calculations
di = cv*(T2-T1)
dU = di*m
dh = cp*(T2-T1)
dH = dh*m
	
#Results
print "Change in Internal energy  =  %d N m"%(dU)
print " Change in Enthalpy  =  %d Nm"%(round(dH,-3))

Change in Internal energy  =  107400 N m
 Change in Enthalpy  =  150000 Nm

Example 9.2 Page No : 268

In [9]:
	
#Initialization of variables
cv = 716.
m = 15. 	#kg
cp = 1003.
T1 = 20.+273 	#K
k = 1.4
ratio = 0.4
	
#calculations
T2 = int((T1)*(1/ratio)**(k-1))
P1 = 95. 	#kN/m**2
P2 = int(P1*T2/(T1)/ratio)
di = round(cv*(T2-T1),-2)
dU = di*m
dh = round(cp*(T2-T1),-2)
dH = dh*m


#Results
print "Final temperature  =  %d K"%(T2)
print " Final pressure  =  %d kN/m**2"%(P2)
print " Change in Internal energy  =  %d N m"%(dU)
print " Change in Enthalpy  =  %d Nm"%(dH)

#The answers are a bit different due to rounding off error.please check.

Final temperature  =  422 K
 Final pressure  =  342 kN/m**2
 Change in Internal energy  =  1386000 N m
 Change in Enthalpy  =  1941000 Nm

Example 9.3 Page No : 271

In [11]:
import math 
	
#Initialization of variables
k = 1.4
R = 1773.
v = 600. 	#fps
T = 660. 	#K
P = 100. 	#psia
cp = 6210.
g = 32.2
	
#calculations
c = math.sqrt(k*R*T)
M = v/c
rho = k*P*144/c**2
Ps = P*144 + 0.5*(rho)*v**2 *(1+ 0.25*M**2)
Ts =  (cp/g *T + v**2 /(2*g))*g/cp
	
#Results
print "Stagnation pressure  =  %d lb/ft**2"%(Ps)
print " Stagnation temperature  =  %.f R"%(Ts)


# note :Please check the units of the answer.

Stagnation pressure  =  16736 lb/ft**2
 Stagnation temperature  =  689 R

Example 9.4 Page No : 275

In [4]:
import math 
	
#Initialization of variables
g = 32.2
A = 0.0218 	#ft**2
P1 = 25.6 	#psia
T1 = 540. 	#K
k = 1.4
R = 1715.
	
#calculations
G = g*A*P1*144/math.sqrt(T1) *math.sqrt(k/R *(2/(k+1))**((k+1)/(k-1)))
	
#Results
print "Flow rate   =  %.2f lb/s"%(G)

Flow rate   =  1.84 lb/s

Example 9.5 Page No : 278

In [15]:
import math 
	
#Initialization of variables
P1 = 50.    	#psia
T1 = 540.   	#K
g = 32.2
R = 1715.
k = 1.4
P3 = 13.5 	    #psia
A2 = 0.0218 	#ft**2
cp = 6000.
	
#calculations
Pc = 0.528*P1
V32 = round(R*T1/g *k/(k-1) *(1- (P3/P1)**((k-1)/k)),-2)
V3 = round(math.sqrt(V32*2*g),-1)
G3 = g*A2*P1*144/math.sqrt(T1) *math.sqrt(k/R *(2/(k+1))**((k+1)/(k-1)))
T3 =  T1 - V3**2 /(2*cp) 
gam3 =  g*P3*144/(R*T3)
gam2 =  (Pc/P3 *gam3**k )**(1/k)
V2 = G3/gam2/A2
T2 =  (V3**2 -V2**2)/(2*cp) + T3
A3 = G3/gam3/V3
D3 =  math.sqrt(4/math.pi *A3)
G2 = G3

#Results
print " velocity at section 3  =  %d fps"%(V3)
print " Flow rate at section 3  =  %.3f lb/s"%(G3)
print " temperature at section 3   =  %d R"%(T3)
print " velocity at section 2  =  %d fps"%(V2)
print " Flow rate at section 2  =  %.3f lb/s"%(G2)
print " temperature at section 2   =  %d R"%(T2)
print " Required Diameter  =  %.2f in"%(D3*12)

# book anwers are wrong. please check.

 velocity at section 3  =  1420 fps
 Flow rate at section 3  =  3.596 lb/s
 temperature at section 3   =  371 R
 velocity at section 2  =  1041 fps
 Flow rate at section 2  =  3.596 lb/s
 temperature at section 2   =  449 R
 Required Diameter  =  2.18 in

Example 9.6 Page No : 281

In [16]:
	
#Initialization of variables
P1 = 10.     	#psia
T1 = 460+40. 	#R
R = 1715.
k = 1.4
V1 = 1400.   	#fps
	
#calculations
rho1 = P1/(R*T1)
c1 = math.sqrt(k*R*T1)
M1 = V1/c1
P2 =  P1 * (2*k*M1**2  - (k-1))/(k+1)
V2  = V1*((k-1)*M1**2 +2)/((k+1)*M1**2) 
rho2 = rho1*V1/V2
T2 = P2/rho2/R
P22 = 122.5
V22 = 286
T22 = 328
	
#Results
print "Pressure at point 2  =  %.1f psia and %.1f N/m**2"%(P2,P22)
print " Velocity at point 2  =  %d fps and %d m/s"%(V2,V22)
print " Temperature at point 2  =  %d R and %d K"%(T2,T22)
print ("Similarly it can be done for SI units")

# note : ronding off error. please check.

Pressure at point 2  =  17.4 psia and 122.5 N/m**2
 Velocity at point 2  =  947 fps and 286 m/s
 Temperature at point 2  =  588 R and 328 K
Similarly it can be done for SI units

Example 9.7 Page No : 286

In [17]:
import math 
	
#Initialization of variables
A = 140. 	#in**2
P = 48. 	#lb/in**2
mu = 3.78*10**-7
g = 32.2
G = 100. 	#lb/s
p = 80. 	#lb/in**2
T = 65.+460 	#R
k = 1.4
R = 1715.
	
#calculations
Rh = A/P /12
R1 = G*4*Rh/ (mu*g*A/144)
R2 = R1
f = 0.0083
gam1 = p*g*144/(R*T)
V1 = G*144/gam1/A
c = math.sqrt(k*R*T)
M1 = V1/c
M2 = 1/math.sqrt(k)
D = 4*Rh
L =  ((1-M1**2 /M2**2)/(k*M1**2) - 2*math.log(M2/M1) )*D/f
Ln = 500 	#ft
P2 = math.sqrt((p*144)**2 - G**2 *R*T/(g**2 *(A/144)**2 *f*Ln/D))
Pa = 12.2
	
#Results
print "Max. length  =  %d ft"%(round(L,-1))
print " Pressure required  =  %.1f psia"%(P2/144 -Pa)

Max. length  =  1260 ft
 Pressure required  =  67.1 psia

Example 9.8 Page No : 287

In [2]:
import math 
	
#Initialization of variables
G = 100. 	#lb/s
g = 32.2
V2 = 300. 	#fps
V1 = 250. 	#fps
	
#calculations
Qh =  (V2**2 -V1**2)/(2*g)
Q = Qh*G
	
#Results
print "Thermal energy added  =  %.2f ft lb/s"%(Q)

Thermal energy added  =  42701.86 ft lb/s

Example 9.9 Page No : 290

In [19]:
	
#Initialization of variables
gam1 = 0.41
g = 32.2
V1 = 250. 	#fps
R1 = 8.2*10**6
f = 0.0083
A = 0.97 	#ft**2
G = 100. 	#lb/s
k = 1.4
P = 80. 	#pressure - psia
ratio = 0.8
R = 1715
	
#calculations
rho1 = gam1/g
X  =  G**2 /(gam1*A)**2 + 2*k/(k-1) *(P*144/rho1)
P2 = (k-1)/2/k *(X*ratio*rho1 - G**2 /(g**2 *A**2 *ratio*rho1))
L = 563 	#ft
rho2 = ratio*rho1
V2 = G/(rho2*g*A)
T2 = P2/(rho2*R)
	
#Results
print "Length  =  %d ft"%(L)
print " velocity  =  %.f fps"%(V2)
print " Temperature  =  %d R"%(T2)

# note : rounding off error

Length  =  563 ft
 velocity  =  314 fps
 Temperature  =  524 R