Chapter 11:External Flow:Drag and Lift

Example 11.11-1, Page No:589

In [2]:
import math

#Variable Decleration
Fd=300 #Drag Force in N
A=2.07 #Frontal Aera in m^2
rho=1.204 #denisty of air in kg/m^3
V=95 #Velocity of the fluid around the body in km/h
C=3.6 #Conversion factor 

#Calculations
Cd=(2*Fd*C**2)/(rho*A*V**2) #Coefficient of Drag of the Car

#Result
print "The Coefficient of Drag of the Car is",round(Cd,2)
The Coefficient of Drag of the Car is 0.35

Example 11.11-2,Page No:599

In [4]:
import math

#Variable Decleration
W=1.85 #Width of the car in m
H=1.7 #Height of the car in m
Cd=0.3 #Drag Coefficient
rho=1.20 #Denisty of air in kg/m^3
V=95 #Velocity of the car in km/h
C=3.6 #Conversion Factor
L=18000 #Distance travelled by the car in one year in km
n_car=0.3 #Efficiency of the car in fraction
HV=44000 #Heating Value of the fuel in kJ/kg
rho_fuel=0.74 #Density of the fuel in kg/L
Unit_Cost=0.95 #Unit cost of fuel per litre in $
Hnew=1.55 #New design height in m

#Calculation
#Drag Force before Redesigning
Fd=Cd*W*H*rho*V**2*0.5*(1/C) #Drag Force in N
W_drag=Fd*L #Work Done to overcome the drag force in kJ/year
E_in=W_drag/n_car #Energy required in kJ/year
#Amount of fuel
Amount_of_fuel=E_in/(HV*rho_fuel) #Amount of fuel required in L/year
Cost=Amount_of_fuel*Unit_Cost #Total cost per year in $/year

#Reduction ratio
Reduction_Ratio=(H-Hnew)/H #Reduction ratio
#Fuel Reduction
Fuel_Reduction=Reduction_Ratio*Amount_of_fuel #Fuel reduced in L/year
Cost_Reduction=Reduction_Ratio*Cost #Cost Reduction in $/Year

#Result
print "The Reduction Ratio of the redesigned car is",round(Reduction_Ratio,3)
print "Therefore the cars height reduces the fuel consumption by",round(Reduction_Ratio*100),"%"
The Reduction Ratio of the redesigned car is 0.088
Therefore the cars height reduces the fuel consumption by 9.0 %

Example 11.11-3,Page No:604

In [5]:
import math

#Variable Decleration
L=5 #Length on the flat plate in m
V=2 #Full stream velocity in m/s
v=2.485*10**-4 #Kinematic Viscosity in m^2/s
rho=876 #Density of the fluid in kg/m^3

#Calculations
Rel=(V*L)/v #Reynolds Number
Cf=1.328*Rel**-0.5 #Average Friction Coefficient

#As pressure drag is zero Cd=Cf
Fd=Cf*L*rho*V**2*0.5 #Drag Force in N

#Result
print "The total Drag Force per Unit Width is",round(Fd),"N"
The total Drag Force per Unit Width is 58.0 N

Example 11.11-4,Page No:609

In [7]:
import math

#Variable Decleration
D=0.022 #Diameter of the pipe in m
rho=999.1 #Density of the fluid in kg/m^3
u=1.138*10**-3 #Dynamic Viscosity in kg/m.s
V=4 #Velocity in m/s
Cd=1 #Coefficent of Drag
L=30 #Width of the river in m

#Calculations
Re=(rho*V*D)/u #Reynolds Number
Fd=Cd*D*L*rho*V**2*0.5 #Drag Force in N

#Result
print "The drag force on the pipe is",round(Fd),"N"
#The answer in the textbook has been approximated to a large value
The drag force on the pipe is 5275.0 N

Example 11.11-5,Page No:616

In [15]:
import math

#Variable Decleration
m=70000 #Mass of the  airplane in kg
g=9.81 #Acceleration due to gravity in m/s^2
V_k=558 #Velocity of the airplane in km/h
rho=1.2 #Denisty of air in kg/m^3
Cl_max1=1.52 #Coefficient of lift case 1
Cl_max2=3.48 #Coefficient of lift case 2
A=150 #Area in m^2
rho_h=0.312 #Density at crusing altitude in kg/m^3
Cd=0.03 #Coefficient of drag at crusing altitude

#Calculations
W=m*g #Weight of the aircraft in N
V=V_k/3.6 #Velocity in m/s

#Part (A)
V_min1=((2*W)/(rho*Cl_max1*A))**0.5 #Minimum stall speed in m/s without flap
V_min2=((2*W)/(rho*Cl_max2*A))**0.5 #Minimum stall speed in m/s with flap
V_min1_safe=1.2*V_min1 #Safe minimum velocity to avoid stall in m/s without flap
V_min2_safe=1.2*V_min2 #Safe minimum velocity to avoid stall in m/s with flap

#Part(B)
Fl=W #Lift force required in N
Cl=(2*Fl)/(rho_h*A*V**2) #Coefficient of lift

#Part(C)
Fd=Cd*A*rho_h*V**2*0.5*10**-3 #Drag Force in kN
Thrust=Fd #thrust Force in kN
Power=Thrust*V #Power required in kW

#Result
print "The safe speed limits without and with flaps are",round(V_min1_safe,1),"m/s and",round(V_min2_safe,1),"m/s"
print "The lift coefficient is",round(Cl,2),"and the corresponding angle of attack is 10˚"
print "The power required to provide enough thrust is",round(Power),"kW"
#The final power answer has been rounded in the textbook
The safe speed limits without and with flaps are 85.0 m/s and 56.2 m/s
The lift coefficient is 1.22 and the corresponding angle of attack is 10˚
The power required to provide enough thrust is 2614.0 kW

Example 11.11-6, Page No:618

In [16]:
import math

#Variable Decleration
m=0.057 #mass of the tennis ball in kg
D=0.0637 #Diameter of the tennis ball in m
V_k=72 #Velocity with which th ball is hit in km/h
w_rpm=4800 #backspin given to the ball in rpm
Cl=0.21 #Coefficient of lift
rho=1.184 #Density of the fluid in kg/m^3
g=9.81 #Aceleration due to gravity in m/s^2

#Calculations
V=V_k/3.6 #Velocity of the ball in m/s
w=(w_rpm*2*pi)/60 #Angular velocity in rad/s

#non dimensional rate of rotation
#Changing the notation from the one used in the textbook to simplify
ror=(w*D)/(2*V) #Non dimnsional rate of rotation
A=4**-1*pi*D**2 #Frontal Area in m^2
Fl=Cl*A*rho*V**2*0.5 #Lift force in N
W=m*g #Weight of the ball in N
F=W-Fl #Combined force in N

#Result
print "The ball will drop due to a combined effect of lift and gravity with a force of",round(W,3),"N"
The ball will drop due to a combined effect of lift and gravity with a force of 0.559 N