Chapter 10:Turbulent Pipe Flow

Example 10.1 Page No216

In [12]:
import math

#variable initialisation

D=2                   #diameter in cm

v=0.0098              #viscosity of water

Re=2000               #Critical Reynolds number

rho=998               #density in kg/m^3

#solution

V=Re*v/10000/(D/100)   #Calculating V

L=(math.pi/4)*((D/100)**2)*V

print "The required Largest Discharge",round(L*1000*60,3),"L/min"

f=64/Re

u=V*(math.sqrt(f/8))   #Boundary shear stress

tou=rho*(u**2)

print "tou=",round(tou,4),"Pa"
The required Largest Discharge 1.847 L/min
tou= 0.0383 Pa

Example 10.2 Page No216

In [13]:
import math

#variable initialisation

f=0.02                                    #Friction factor

#Calculation

Ut=math.sqrt(f/8)                         #Shear velocity

print "U*=",Ut,"V"

Um=1.43*math.sqrt(f)+1                    #Maximum velocity

print "Um=",round(Um,5),"V"

r0=1

y=r0-0.3*r0

#Let us take (u-V/U*) as U

a=round(math.log(0.7,10),4)

U=5.75*a+3.75

#But,U*=0.05V

#Let's take u/V as V                     #Ratio of local velcotiy to mean velocity

V=U*(Ut)

print "u/V=",round(V+1,5)
U*= 0.05 V
Um= 1.20223 V
u/V= 1.14297

Example 10.5 Page No217

In [14]:
#variable initialization

f=0.035          #friction factor

Re=1E+5          #Reynolds number

d=10             #diameter in cm

Es=0.80          #roughness of pipe

#solution

ro=(d/2)/100     #Radius in m

delta=(65.6*ro)/(Re*math.sqrt(f))*1000 #Thickness of lamina sublayer

print "delta=",round(delta,3),"mm"  

Es_delta=Es/delta

print "Es_delta=",round(Es_delta,2)

print "This value being less than 6.0 and greater than 0.25,the pipe flow is in transition regime"
delta= 0.175 mm
Es_delta= 4.56
This value being less than 6.0 and greater than 0.25,the pipe flow is in transition regime

Example 10.6 Page No217

In [16]:
import math

#variable initialisation

r0=0.10                                 #Constant value

epsillon=0.0002                         #roughness magnitude in mm

v=1E-6                                  #velocity in m**2/s

D=0.2                                   #Diameter in mm

#Calculation

#Transition from the smooth pipe regime begins at,

Ref=17*(r0/epsillon)

Re=(2*(math.log(Ref,10))-0.8)*Ref      #friction factor

V=(Re*v)/D

print "Velocity at the upper limit  of smooth pipe regime is",round(V,1),"m/s"

#Fully rough turbulent flow begins at,

Ref=400*(r0/epsillon)

Re=(2*(math.log((r0/epsillon),10))+1.74)*Ref

V=(Re*v)/D

print "The flow will be at fully rough turbulent regime at V=",round(V,2),"m/s"
Velocity at the upper limit  of smooth pipe regime is 0.3 m/s
The flow will be at fully rough turbulent regime at V= 7.14 m/s

Example 10.7 Page No218

In [18]:
import math

#variable initialisation

r0=0.30                       #constant value

v1=4.50                       #velocity in m/s

v2=4.20                       #velocity in m/s

D=60                          #diameter in cm

#Calculation

D=D/100

u=(v1-v2)/(5.75*(math.log((r0/(0.20)),10)))#For smooth and rough pipe

V=(3.75*u)-v1                              #mean velocity V related to um as,

Q=math.pi/4*(D**2)*V                       #Discharge

print "Q=",round(-Q,3),"m**3/s"
Q= 0.958 m**3/s

Example 10.14 Page No220

In [19]:
import math

#variable initialisation

rho=804                      #Relative density in kg/m^3

mu=1.92E-3                   #viscosity in Pa.s.

D=20                         #Diameter in cm

v=3                          #Velocity in m/s

r0=0.10                      #Constant value

#Calculation

D=D/100#Converting into m

Re=rho*v*D/mu

#At the permissible height of roughness in the pipe the flow will be in smooth-turbulent regime.

f=0.0032+(0.221/(Re**0.237))

#Since the pipe will behave as smooth,at the limiting value

epsillon=r0/((Re*math.sqrt(f))/17)

print "Permissible height of surface roughness=",round(epsillon*1000,4),"mm"
Permissible height of surface roughness= 0.0556 mm

Example 10.15 Page No220

In [20]:
import math

#variable initialisation

r0=0.05                            #radius 

epsillon=0.0002                    #roughness magnitude in m

r1=0.075                           #radius

D1=10                              #Diameter in cm

D2=15                              #Diameter in cm

#Calculation

f1=round(1/((2*(math.log((r0/epsillon),10)))+1.74)**2,4)

f2=round(1/((2*(math.log((r1/epsillon),10)))+1.74)**2,4)

#P1/P2 is taken as P

P=(f1/f2)*((D2/D1)**5)             #Ratio of powers,

#Cost of pumping is proportional to power consumed.

print "cost of pumping in 10cm pipe/cost of pumping in 15 cm=",round(P,2)
cost of pumping in 10cm pipe/cost of pumping in 15 cm= 8.42

Example 10.18 Page No222

In [21]:
import math

#variable initialisation

v=1.0E-6                                #Velocity in m/s

epsillon_s=15E-3                        #roughness in m

epsillon2_s=0.2E-3                      #roughness in m

Q=4.0                                   #discharge in m

D1=1.5                                  #diameter in m

t=0.01                                  #thickness

g=9.81                                  #gravity 

L=1000                                  #length 

ga=9.79                                 #density

#Calculation

V1=round(Q/((math.pi/4)*(D1**2)),3)

Re1=round(V1*D1/v,3)

#From the emrpirical equivalent of colrebook equation,

f1=round((1/(1.14-(2*(math.log(((epsillon_s/D1)+(21.25/Re1**0.9)),10)))))**2,4)

#After the lining,

D2=D1-2*t                             #Diameter2 calculation

V2=round(Q/((math.pi/4)*(D2**2)),4)

Re2=round(V2*D2/v,3)

#From the emrpirical equivalent of colrebook equation,

f2=round((1/(1.14-(2*(math.log(((epsillon2_s/D2)+(21.25/Re2**0.9)),10)))))**2,4)

h_f1=round((f1*L*(V1**2))/(2*g*D1),3)

h_f2=round((f2*L*(V2**2))/(2*g*D2),6)

h_s=round(h_f1-h_f2,5)                #Saving in head

Ps=ga*Q*h_s                           #Saving in power

print"answer in the book is wrong.It should be as,"

print "Ps=",round(Ps,1),"kW"
answer in the book is wrong.It should be as,
Ps= 162.3 kW

Example 10.19 Page No222

In [22]:
import math

#variable initialisation

D=0.25                          #Diameter in m

v=1E-6                          #velocity

epsillon_s=0.15E-3              #roughness

h_f=0.025                       #hydraulic gradient

L=100                           #length in m

g=9.81                          #gravity

ga=9790                         #density

#calculation

#Re*math.sqrrt(f) is taken as Ref

#Re(math.sqrt(f))=(((V*D)/v)*((2*g*h_f*D)/(V**2*L)))

Ref=(((D)**(3/2))/v)*((2*g*h_f)**(1/2))

#By colerbrook formula,

f1=round((1/(1.14-(2*(math.log(((epsillon_s/D)+(9.35/Ref)),10)))))**2,5)

Re=round(Ref/math.sqrt(f1),3)

V=round((Re*v)/D,4)

Q=(math.pi/4)*(D**2)*V        #Discharge

print "Q=",round(Q*1000,1),"L/s"

h_f=h_f*100                   #head lost in 100 m length of pipe

P=ga*Q*h_f

print "P=",round(P/1000,2),"kW"
Q= 127.9 L/s
P= 3.13 kW

Example 10.20 Page No223

In [23]:
import math

#variable initialisation

L=1500                                #Length in m

Q=250                                 #discharge in L/s

g=9.81                                #gravity 

h_f=15                                #head loss

epsillon_s=0.12E-3                    #roughness

v=1E-6                                #viscosity in m^2/s

#Calculation

Q=Q/1000

#hf=fLV^2/2gD=(8*L*(Q**2))/((math.pi**2)*g)

D=((8*L*(Q**2))/((math.pi**2)*g))/h_f

#Re=1/D)(4*Q)/(math.pi*v)

#trial and error procedure is adpoted,

#1 st trail,assume

f=0.025

D1=(D*f)**(1/5)

Re=((4*Q)/(math.pi*v))/D1

f=round((1/(1.14-(2*(math.log(((epsillon_s/D1)+(21.25/Re**0.9)),10)))))**2,4)

#trial and error procedure is adpoted,

#2 nd trail,assume

f=0.0159

D1=(D*f)**(1/5)

Re=((4*Q)/(math.pi*v))/D1

f=round((1/(1.14-(2*(math.log(((epsillon_s/D1)+(21.25/Re**0.9)),10)))))**2,4)

#This value is practically same as assumbed hence no further trails are required.

#In practicce,the next larger size would be used.

print "The required diameter D=",round(D1*100,1),"cm"
The required diameter D= 38.3 cm

Example 10.21 Page No223

In [6]:
#variable initialisation

v=1E-6                        #Viscosity in m^2/s

V=1                           #Velcoity in m/s

L=100                         #Length in m

h_f=10/100                    #headloss

epsillon_s=0.45E-3            #roughness

g=9.81                        #gravity

#Calculation

#Re=V*D/v

Re=(V)/v                      #Reynolds number

f=0.0055*(1+(2000*epsillon_s+1)**(1/3))

#h_f=f*L*V^2/2gD

f1=(2*g*h_f)/(L*V**2)

#Using trial and error methods,F is found to be 0.0133,

f=0.0133

D=f/f1

print "D=",round(D,3),"m"

#This value is practically same as assumbed hence no further trails are required.

#In practicce,the next larger size would be used.
D= 0.678 m

Example 10.22 Page No224

In [8]:
import math

#variable initialisation

Rd=0.85                      #relative density of oil

L=150                        #length in m

h_f=67                       #head loss in m

v=0.1                        #viscosity in cm

Q=80                         #Discharge in L/s

g=9.81                       #gravity

#Calculation

Q=Q/1000#Converting into m

v=v/10000

#h_f=f*L*V**2/(2*g*D)

f=h_f*(math.pi**2)*g/(8*L*(Q**2))

#D=f/(844.66)**(1/5)

#Reynolds number,

Re=(4*Q)/(math.pi*v)

#Re=Re*(1/D)

#A trial and error procedure is used.

#1 st trial assume,

f1=0.02

D=(f1/f)**(1/5)

Re1=Re*(1/D)

#As Re<10^5,Blasius formula can be used.

#2nd trial,

f1=0.316/(Re1**(1/4))

D=round((f1/f)**(1/5)*100,1)

Re2=Re/D

#3 rs trial,

f1=0.316/(Re2**(1/4))

print "The required diameter is D=",round(D,2),"cm"

#This value is practically same as assumbed hence no further trails are required.

#In practicce,the next larger size would be used.
The required diameter is D= 11.7 cm

Example 10.23 Page No223

In [11]:
import math

#variable initialisation

rho=1.20                       #relative density

l=0.4                          #length in m

b=0.25                         #breadth in m

V=20                           #velcity in m/s

v=1.5E-5                       #viscosity

epsillon_s=0.05E-3             #roughness

ga=9.81*1.2                    #gravity

eta=0.60                       #efficiency

L=200                          #length in m

g=9.81                         #gravity

#Calculation

area=l*b                       #Calculating area

perimeter=2*(l+b)              #Calculating perimeter

Dh=4*(area/perimeter)

Reh=(V*Dh)/v

#By empirical equivalent of colrebook formula,

f=round((1/(1.14-(2*(math.log(((epsillon_s/Dh)+(21.25/Reh**0.9)),10)))))**2,4)

hf=(f*L*(V**2))/(2*g*Dh)        #Head loss

print "hf=",round(hf,1),"m(of air column)"

del_p=int(rho*g*hf)/1000        #pressure loss

print "del_p=",round(del_p,3),"kPa"

Q=V*l*b

P=ga*Q*hf/eta                   #Power required at 60% efficiency

print "P=",round(P/1000,1),"kW"
hf= 204.1 m(of air column)
del_p= 2.402 kPa
P= 8.0 kW
In [ ]: