Chapter 14: INTERNAL AND EXTERNAL FLOW

Example 14.01, page: 319

In [2]:
from __future__ import division
import math

# Initialization  of  Variable
D = 4 #mm
V = 50 #m/s
l = 0.1 #m
p = 1.23 #kg/m3
mu = 1.79E-5 #Ns/m2

#calculations:
#reynolds number
Re = p*V*(D/1000)/mu
#from table 14.1
e = 0.0015 #mm
#e/D
e_D = e/D
#from chart 14.7
f = 0.028
#pressure drop
dP = f*(l/(D/1000))*(p/2)*V**2/1000

#Results
print "the pressure drop is",round(dP,3),"kPa"

the pressure drop is 1.076 kPa

Example 14.02, page: 322

In [3]:
from __future__ import division
import math

# Initialization  of  Variable
T = 60 #degF
p = 1.94 #slug/ft3
mu = 2.34E-5 #lbf.s/ft2
D = 0.0625 #ft
Q = 0.0267 #ft3/s
Df = 0.0417 #ft
z1 = 0 #ft
z2 = 20 #ft
P2 = 0 #gage
l13 = 15 #ft
l34 = 10 #ft
l45 = 5 #ft
l56 = 10 #ft
l67 = 10 #ft
l78 = 10 #ft

#calculations:
#Area at 1
A1 = math.pi*D**2/4
#fluid vel in pipe
V1 = Q/A1
#reynolds number
Re = p*V1*D/mu
#area at 2
A2 = math.pi*Df**2/4
#velocity of stream exciting from faucet
V2 = Q/A2
r = 62.4 #lbf/ft3
#total length
l = l13 + l34 + l45 + l56 + l67 + l78
#from Moody chart
f = 0.0215
#pressure drop
P1 = (r*z2 + 0.5*p*(V2**2 - V1**2) + p*f*(l/D)*(V1**2/2))/144
#loss coeff
Kl = 4*1.5 + 10 + 2
#entire Pressure drop
P1 = P1 + p*(V1**2/2)*Kl/144

#Results
print  "entire  pressure  drop  is", round(P1,1),"Psi"

entire  pressure  drop  is 30.4 Psi

Example 14.03, page: 324

In [4]:
from __future__ import division
import math

# Initialization  of  Variable
T = 140 #degF
r = 53.7 #lbf/ft3
p = 1.67 #slug/ft3
mu = 8E-5 #lbf.s/ft2
l = 799 #mile
D = 4 #ft
Q = 117 #ft3/s
V = 9.31 #ft/s
g = 32.2 #ft/s2

#calculations:
#reynolds number
Re = p*V*D/mu
#from fig 14.7
f = 0.0125
#pump head
hp = f*(l/D)*5280*(V**2/(2*g))
#power req
Wpdot = r*Q*hp/550

#Results
print  "the power added to the fluid by the pumps that drive this system is", round(Wpdot,0),"hp"
#answer wrong in book

the power added to the fluid by the pumps that drive this system is 202694.0 hp

Example 14.04, page: 325

In [5]:
from __future__ import division
import math

# Initialization  of  Variable
P1_rw = 0.2 #P1/rH2O in inch
V1 = 0
g = 32.2 #ft/s2
l = 20 #ft
D = 4 #in
P2 = 0
rair = 0.0709 #lbf/ft3
vair = 1.79E-4 #ft2/s
rw = 62.4 #lbf/ft2

#calculations:
#loss coeffs
KLent = 0.5 #entrance
KLelb = 1.5 #elbow
KL = KLent + 4*KLelb
#P1
P1 = P1_rw/12*rw
#assumption
f = 0.022
V = ((P1/rair)*2*g/(1 + (l*f/(D/12)) + KL))**0.5
#reynolds number
Re = V*D/(vair*12)
#Area
A = math.pi/4*(D/12)**2
#volumetric flow rate
Q = A*V

#Results
print  "Volumetric flow rate is", round(Q,3),"ft3/s"
#answer wrong in book

Volumetric flow rate is 0.903 ft3/s

Example 14.05, page: 325

In [6]:
from __future__ import division
import math

# Initialization  of  Variable
D = 60 #mm
d = 30 #mm
dP = 4 #kPa
mu = 1.19E-3 #N.s/m2
p = 789 #kg/m3

#calculations:
#area
Ao = math.pi*(d/1000)**2/4
b = d/D
#at this b,
Co = 0.613
#from fig 14.10 and Re at this value of b, we get
#volumetric flow rate
Q = Co*Ao*(2*dP*1000/(p*(1 - b**4)))**0.5

#Results
print  "Volumetric flow rate is", round(Q,5),"m3/s"
#answer wrong in book

Volumetric flow rate is 0.00142 m3/s

Example 14.06, page: 327

In [7]:
from __future__ import division
import math

# Initialization  of  Variable
U = 10 #ft/s
Tw = 60 #degF
Tg = 68 #degF
Rexc = 5E5
vW = 1.21E-5 #ft2/s
vA = 1.57E-4 #ft2/s
vG = 1.28E-2 #ft2/s

#calculations:
#water
Xc_W = vW*Rexc/U
dXc_W = 5*(vW*Xc_W/U)**0.5
#air
Xc_A = vA*Rexc/U
dXc_A = 5*(vA*Xc_A/U)**0.5
#glycerin
Xc_G = vG*Rexc/U
dXc_G = 5*(vG*Xc_G/U)**0.5

#Results
print "  Fluid\t\t v(ft2/s)\t\t Xc(ft)\t\t d(Xc)(ft)"
print "a)Water\t\t",vW,"\t\t ",round(Xc_W,3),"\t ", round(dXc_W,5)
print "b)Air\t\t",vA,"\t\t ",round(Xc_A,2),"\t\t ", round(dXc_A,4)
print "c)Glycerin\t",vG,"\t\t\t ",round(Xc_G,0),"\t ", round(dXc_G,2)

  Fluid		 v(ft2/s)		 Xc(ft)		 d(Xc)(ft)
a)Water		1.21e-05 		  0.605 	  0.00428
b)Air		0.000157 		  7.85 		  0.0555
c)Glycerin	0.0128 			  640.0 	  4.53

Example 14.07, page: 329

In [8]:
from __future__ import division
import math

# Initialization  of  Variable
l = 8 #ft
b = 4 #ft
U = 80.7 #ft/s
e = 0.003 #ft
p = 0.00238 #slug/ft3
mu = 3.74E-7 #lbf.s/ft2

#calculations:
A = l*b
#reynolds number
Re = p*U*l/mu
#e_D  = e/D
e_D = e/l
#from fig 14.14
Cd = 0.0066
#fraction drag
D = 0.5*p*U**2*l*b*Cd

#Results
print  "the drag on the top surface of the plywood is", round(D,2),"lbf"

the drag on the top surface of the plywood is 1.64 lbf

Example 14.08, page: 334

In [9]:
from __future__ import division
import math

# Initialization  of  Variable
U = 95.3 #ft/s
Cd1940 = 0.55
b1940 = 5.2 #ft
h1940 = 5.1 #ft
Cd2003 = 0.30
b2003 = 5.2 #ft
h2003 = 4.3 #ft
p = 0.00238 #slug/ft3

#calculations:
#Area
A1940 = b1940*h1940
A2003 = b2003*h2003
#Drag
D1940 = 0.5*p*U**2*A1940*Cd1940
D2003 = 0.5*p*U**2*A2003*Cd2003
#Power required
W1940dot = U*D1940/550
W2003dot = U*D2003/550

#Results
print  "drag  force on 1940 car is",round(D1940,0),"lbf and on 2003 is",round(D2003,1),"lbf"
print  "power  required  to  overcome  drag  force by 1940 car is",round(W1940dot,1),"hp and by 2003 car is",round(W2003dot,1),"hp"

drag  force on 1940 car is 158.0 lbf and on 2003 is 72.5 lbf
power  required  to  overcome  drag  force by 1940 car is 27.3 hp and by 2003 car is 12.6 hp

Example 14.09, page: 336

In [10]:
from __future__ import division
import math

# Initialization  of  Variable
b = 96 #ft
c = 7.5 #ft
W = 210 #lbf
U = 15 #ft/s
d = 0.5 #mi
p = 0.00238 #slug/ft3

#calculations:
#area
A = b*c
#lift coeff
CL = 2*W/(p*U**2*A)

#Results
print  "coeff.  of  lift  is",round(CL,2)

coeff.  of  lift  is 1.09