Chapter 9:Drag and Lift on Immersed Bodies

Example 9.2 Page No198

In [9]:
import math

#variable initialisation

D=20                         #diameter in cm

rho_s=2.8*998                #relative density

rho_w=998                    #relative density of water in kg/m^3

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

g=9.81                       #gravity constant in m/s^2

rho_f=998

#solution

#assume Re>3*E+5.then

C_D=0.20

#At terminal velocity V-0t,submerged weight=Drag

D=D/100                     #converting D in m

V_ot=round(math.sqrt((4/3)*(D/C_D)*((g)*((rho_s/rho_f)-1))),3)

print "V_ot=",V_ot,"m/s"

Re=(V_ot*(D/100))/v         #calculating Re

#from Re,its clear,Re>3*10^5

print "Hence the assumption is correct and V_ot=",V_ot,"m/s"
                    
V_ot= 4.852 m/s
Hence the assumption is correct and V_ot= 4.852 m/s

Example 9.3 Page No199

In [10]:
import math

#variable initialisation

rho_f=998                  #density of water in kg/m^3

mu=1E-3                    #coefficient of viscosity in Pa.s

g=9.81                     #gravity in m/s^2

ga_s=2.65*998              #particles RD

ga_f=998                   #relative density in kg/m^3

#solution

#Stoke's law is valid upto Re=1.0

#For maximum size particles that obey stoke's law,

#Using fall velocity and stoke's law,

V_ot=(1/18)*g*(mu/rho_f)*((ga_s/ga_f)-1)        #terminal velocity

V_o=round(math.pow(V_ot,1/3)*1000,2)            #taking cube root

print "V_ot=",V_o,"mm/s"

D=(mu)/(rho_f*V_o/1000)                          #size

print "D=",round(D*1000,4),"mm"
V_ot= 9.66 mm/s
D= 0.1037 mm

Example 9.4 Page No199

In [11]:
#variable initialisation

ga_s=2.60*998           #relative density of sphere

D=2.0                   #Diameter in mm

v=1.25                  #velocity in cm/s

ga_f=917                #density of oil in kg/m^3 

g=9.81                  #gravity in m/s^2

#solution

#Assuming validity of Stoke's law,

#using Fall Velocity equation,

mu=round((((D/1000)**2)*((ga_s*g)-(ga_f*g)))/(18*(v/100)),3)

print "coefficient of dynamic viscosity mu=",mu,"Pa.s"

#Reynold's number,

Re=round((ga_f*(v/100)*(D/1000))/(mu),3)

print "Reynolds number is",Re

print Re," < 1.0"

print "Hence the assumed Stoke's law is valid"
coefficient of dynamic viscosity mu= 0.293 Pa.s
Reynolds number is 0.078
0.078  < 1.0
Hence the assumed Stoke's law is valid

Example 9.5 Page No199

In [12]:
import math

#variable initialisation

D=0.2                        #diameter in mm

rho_f=1.20                   #density of air in kg/m^3

rho_s=998                    #density of water in kg/m^3

g=9.81                       #gravity in m/s^2

C_D=4.20                     #Drag coefficient

D1=2.0                       #Diameter in mm

C_D1=0.517                   #Drag coefficient

#solution

D=D/1000                     #converting D to m

print "For 0.2 mm rain drop:"

V_ot=round(math.sqrt((((4*g*D)/(3*C_D))*((rho_s/rho_f)-1))),2)

print "V_ot=",V_ot,"m/s"

print "For 2.0 mm rain drop:"

D1=D1/1000                   #converting D to m

V_ot=round(math.sqrt((((4*g*D1)/(3*C_D1))*((rho_s/rho_f)-1))),2)

print "V_ot=",V_ot,"m/s"
For 0.2 mm rain drop:
V_ot= 0.72 m/s
For 2.0 mm rain drop:
V_ot= 6.48 m/s

Example 9.6 Page No199

In [13]:
import math

#variable initialisation

V_o=60                  #velocity in Km/h

C_d=0.35                #Drag coefficient

C_d1=0.30               #reduced Drag coefficient 

A=1.6                   #area in m^2

rho=1.2 

#solution

V_o=(V_o*1000)/(60*60)   #converting V_o into m/s

#power required to overcome wind resistance by the car

F_D=round(C_d * A * rho * ((V_o**2)/2),2)

p=F_D*V_o                #power 

print "a)power=",round(p/1000,3),"kW"

V_o=(p*2)/(C_d1*A*rho)   #Speed

V=math.pow(V_o,1/3)*18/5 #taking cuberoot

print "b)Vo=",round(V,2),"Km/h"
a)power= 1.556 kW
b)Vo= 63.16 Km/h

Example 9.9 Page No200

In [14]:
import math

#variable initialisation

g=9.81               #gravity in m/s^2

h=2                  #height in m

F_D=1000             #total load in N

rho=1.2              #density in kg/m^3

#solution

C_d=1.33                       #for the hemisphere with concave frontal surface

V_ot=round(math.sqrt(2*g*h),2) #Terminal Velocity

D=F_D*(4/3.14)*(2/(C_d*rho*(V_ot**2)))#minimum size
       
print "D=",round(math.sqrt(D),2),"m,say 6.5 m"
D= 6.38 m,say 6.5 m

Example 9.10 Page No201

In [15]:
#variable initialisation

rho_air=1.2                      #relative ddensity in kg/m^3

v=1.5E-5                         #velocityin m^2/s

C_D=1.20                         #drag coefficient

V0=80                            #velocity in km/h

D=0.05                           #Diameter in m

L=1                              #length in m

S=0.21                           #Strouhal number

#Calculation

V0=80*1000/3600                  #velcoity in m/s

Re=V0*D/v                        #Reynolds number

A=L*D#area

F_D=C_D*A*(rho_air*(V0**2))/2    #Drag force for unit length of cable

print "F_D=",round(F_D,2),"N/metre length of cable"

n=S*(V0/D)                       #frequency of vortex shedding"

print "n=",round(n,1),"Hz"
F_D= 17.78 N/metre length of cable
n= 93.3 Hz

Example 9.11 Page No201

In [16]:
#variable initialisation

rho_air=1.2                      #relative ddensity in kg/m^3

v=1.5E-5                         #velocityin m^2/s

C_D=0.33                         #drag coefficient

V0=80                            #velocity in km/h

D=2.5                            #Diameter in m

L=50                             #length in m

S=0.21                           #Strouhal number

#Calculation

V0=80*1000/3600                  #velcoity in m/s

Re=V0*D/v                        #Reynolds number

A=L*D                            #area

#Calculation

V0=80*1000/3600                  #velcoity in m/s

Re=V0*D/v                        #Reynolds number

A=L*D                            #area

F_D=C_D*A*(rho_air*(V0**2))/2    #Force on the chimney

M0=F_D*(L/2)

print "M0=",round(M0/1000,1),"kN.m"
M0= 305.6 kN.m

Example 9.14 Page No202

In [18]:
import math

#variable initialisation

l=2.0            #length in m

w=1.5            #width in m

C_d=0.20         #Drag coefficient

C_l=0.60         #lift coefficient

V_o=30           #Velocity in km/hr

rho=998          #density in kg/m^3

#solution

V_o=(V_o*1000/3600)

A=l*w                                         #Calculating area

F_D=round(((C_d*A*(rho*(V_o**2))/2)/1000)-.01,2)  #Drag force

F_L=round(((C_l*A*(rho*(V_o**2))/2)/1000)-.01,2)  #Lift force

F=round(math.sqrt((F_D**2)+(F_L**2)),2)       #Resultant Force

print "F=",F,"kN"

P=F_D*V_o                                     #Power required to tow the plate

print "P=",round(P,1),"kW"                    #when we roundoff,points vary
F= 65.74 kN
P= 173.2 kW

Example 9.17 Page No204

In [19]:
import math

#variable initialisation

l=2 #length in m

w=1.2 #width in m

C_d=0.15 #Drag coefficient

C_l=0.75 #lift coefficient

V_o=50 #Velocity in km/hr

rho=1.2  

#solution

V_o=round((V_o*1000)/3600,5)

A=l*w                                            #Calculating area

F_L=round((C_l*A)*((rho*(V_o**2))/2),1)          #Lift force

print "F_L=",F_L,"N"

F_D=round((C_d)*(A)*((rho*(V_o**2))/2),2)        #Drag Force

print "F_D=",F_D,"N"

F=math.sqrt((F_D**2)+(F_L**2))

print "F=",round(F,2),"N"

theta=math.atan(round((F_L/F_D),0))/0.0175       #Installation of F with free stream

print "theta=",int(theta),"degrees,42 minutes"

P=F_D*V_o                                        #Power Expended

print "P=",round(P,1),"W"
F_L= 208.3 N
F_D= 41.67 N
F= 212.43 N
theta= 78 degrees,42 minutes
P= 578.8 W

Example 9.18 Page No204

In [20]:
#from sympy import asin

import math

#variable initialisation

D=1.2                                                 #Diameter in m

N=210                                                 #rotations in rpm

V0=10                                                 #velocity of air stream

L=9                                                   #length in m

rho=1.2                                               #relative density

#calculation

Vc=round((math.pi*D*N)/60,2)                          #Tangential velocity due to rotation

T=round(2*math.pi*(D/2)*Vc,2)                         #Circulation

print "T=",T,"m^2/s"

F_L=L*rho*V0*T                                        #Lift force

print "F_L=",round(F_L/1000,3),"kN"

C_L=2*math.pi*(Vc/V0)                                 #Lift coefficient

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

theta=round(math.degrees(asin(-(Vc/V0)*(1/2))),2)     #Stagnation point location

theta1=360+theta

print "theta=",theta1,"degrees,(Stagnation point S2)"

theta2=180+(-theta)

print "theta=",theta2,"degrees,(Stagnation point S1)"
T= 49.73 m^2/s
F_L= 5.371 kN
C_L= 8.29
theta= 318.74 degrees,(Stagnation point S2)
theta= 221.26 degrees,(Stagnation point S1)
In [ ]: