# Chapter 2 - Blade Theory¶

## Ex 2.1 page 53¶

In :
#input data
c =2.25#Chord length of an aerofoil in m
l=13.5#Span of the aerofoil in m
C=125#Velocity of the aerofoil in m/s
Cl=0.465#Lift coefficient
Cd=0.022#Drag coefficient
d=1.25#Density of the air in kg/m**3

#calculations
A=c*l#Area of cross section of the aerofoil in m**2
W=Cl*d*((C**2)/2)*A*10**-3#Weight carried by the wings of aerofoil in kN
D=Cd*d*((C**2)/2)*A#Drag force on the wings of aerofoil in N
P=D*C*10**-3#Power required to the drive the aerofoil in kW

#output
print '''(a)Weight carrried by the wings is %3.2f kN
(b)Drag force on the wings of aerofoil is %3.2f N
(c)Power required to drive the aerofoil is %3.3f kW'''%(W,D,P)

(a)Weight carrried by the wings is 137.92 kN
(b)Drag force on the wings of aerofoil is 6525.46 N
(c)Power required to drive the aerofoil is 815.683 kW


## Ex 2.2 page 53¶

In :
from __future__ import division
#input data
W=980#The weight of the object being dropped by parachute in N
C=5#The maximum terminal velocity of dropping in m/s
d=1.22#The density of the air in kg/m**3
Cd=1.3#The drag coefficient of the parachute

#calculations
A=W/(Cd*d*((C**2)/2))#The area of cross section in m**2
D=((A*4)/(3.14))**(1/2)#Diameter of the parachute in m

#output
print 'The required diameter of the parachute is %3.2f m'%(D)

The required diameter of the parachute is 7.94 m


## Ex 2.3 page 54¶

In :
#input data
A=10*1.2#Area of the airplane wing in m**2
C=((240*10**3)/3600)#Velocity of the wing in m/s
F=20#Total aerodynamic force acting on the wing in kN
LD=10#Lift-drag ratio
d=1.2#Density of the air in kg/m**3

#calculations
L=(F)/(1.01)**(1/2)#The weight that the plane can carry in kN
Cl=(L*10**3)/(d*A*((C**2)/2))#Coefficient of the lift

#output
print '(1)The coefficient of lift is %3.3f\n(2)The total weight the palne can carry is %3.1f kN'%(Cl,L)

(1)The coefficient of lift is 0.622
(2)The total weight the palne can carry is 19.9 kN


## Ex 2.4 page 54¶

In :
#input data
m=25#Mass flow rate of the air in kg/s
d=1.1#Density of the air in kg/m**3
Ca=157#Axial flow velocity of the air in m/s
N=150#Rotational speed of the air in rev/s
sc=0.8#Pitch chord ratio

#calculations
A=(m)/(d*Ca)#The annulus area of flow in m**2
C=l/lc#The chord of the blades in m

#output

(b)The blade height is %3.2f m\n(c) (1)The pitch of the blades is %3.4f m
(2)The chord of the blades is %3.3f m\n(d)The number of the blades are %3.f'''%(rm,l,S,C,n)

(a)The mean radius of the blades is 0.212 m
(b)The blade height is 0.11 m
(c) (1)The pitch of the blades is 0.0294 m
(2)The chord of the blades is 0.037 m
(d)The number of the blades are  45


## Ex 2.5 page 56¶

In :
#input data
b1=45#Relative air angle at inlet in degree
b2=15#Relative air angle at oulet in degree
a1=b1#Cascade air angle at inlet in degree
a2=b2#Cascade air angle at outlet in degree

#calculations
en=a1-a2#Nominal deflection angle of the blade in degree
m=((0.23*(1)**2))+(0.1*a2/50)#An emperical constant for a circular arc camber where (2*a/c)=1
d=(m*(sc)**(1/2))*t#The deviation angle of the blade in terms of (degree*t)

#output
print '''(a)The nominal deflection angle is %i degree
(b)The blade camber angle is %3.2f degree
(c)The deviation angle is %3.2f degree
(d)The blade stagger is %3.2f degree'''%(en,t,d,ps)

(a)The nominal deflection angle is 30 degree
(b)The blade camber angle is 39.11 degree
(c)The deviation angle is 9.10 degree
(d)The blade stagger is 25.44 degree


## Ex 2.6 page 57¶

In :
from math import atan, degrees, tan
#input data
t=25#The camber angle of aero foil blades in degree
ps=30#The blade stagger angle in degree
sc=1#The pitch-chord ratio of the blades
In=5#The nominal value of incidence in degree

#calculations
a1n=In+a1#The nominal entry air angle in degree
a2n=degrees(atan((tan(a1n))-(1.55/(1.0+(1.5*sc)))))#The nominal exit air angle in degree

#output
(a)inlet is %3.1f degree
(b)exit is %3.1f degree
(2)The nominal air angles at -
(a)inlet is %3.1f degree
(b)exit is %3.2f degree'''%(a1,a2,a1n,a2n)
# the answer in the textbook is not correct.

(1)The cascade blade angles at -
(a)inlet is 42.5 degree
(b)exit is 17.5 degree
(2)The nominal air angles at -
(a)inlet is 47.5 degree
(b)exit is -12.69 degree


## Ex 2.7 page 58¶

In :
from math import atan, cos, tan, pi
#input data
C1=75#Velocity of air entry in m/s
a1=48#Air angle at entry in degree
a2=25#Air angle at exit in degree
cs=0.91#The chord-pitch ratio
P0m=(11*9.81*10**3)/10**3#The stagnation pressure loss in N/m**2
d=1.25#The density of the sair in kg/m**3

#calculations
Cp=(P0m/(0.5*d*C1**2))#The pressure loss coefficient
am=degrees(atan((tan(a1)+tan(a2))/2))#The mean air angle in degree
Cd=2*(1/cs)*(P0m/(d*C1**2))*((cos(pi/180*am))**3/(cos(pi/180*a1))**2)#The drag coefficient
Cl=(2*(1/cs)*cos(pi/180*am)*(tan(pi/180*a1)-tan(pi/180*a2)))-(Cd*tan(pi/180*am))#THe lift coefficient

#output
print '''(a)The pressure loss coefficient is %3.4f
(b)The drag coefficient is %3.4f\n(c)The lift coefficient is %3.3f'''%(Cp,Cd,Cl)
# the answer in the textbook is not correct.

(a)The pressure loss coefficient is 0.0307
(b)The drag coefficient is 0.0518
(c)The lift coefficient is 1.222


## Ex 2.8 page 59¶

In :
#input data
a1=40#The cascade air angle at entry in degree
a2=65#The cascade air angle at exit in degree
C1=100#Air entry velocity in m/s
d=1.25#The density of the air in kg/m**3
sc=0.91#The pitch-chord ratio of the cascade
P0m=(17.5*9.81*10**3)/10**3#The average loss in stagnation pressure across cascade in N/m**2

#calculations
Cp=(P0m/(0.5*d*C1**2))#The pressure loss coefficient in the cascade
am=atan((tan(a2)-tan(a1))/2)#The mean air angle in degree
Cd=2*(sc)*(P0m/(d*C1**2))*((cos(pi/180*am))**3/(cos(pi/180*a2))**2)#The drag coefficient
Cl=(2*(sc)*cos(pi/180*am)*(tan(pi/180*a1)+tan(pi/180*a2)))+(Cd*tan(pi/180*am))#THe lift coefficient

#output
print '(a)The pressure loss coefficient is %3.4f\n(b)The drag coefficient is %3.4f\n(c)The lift coefficient is %3.3f'%(Cp,Cd,Cl)
# the answer in the textbook is not correct.

(a)The pressure loss coefficient is 0.0275
(b)The drag coefficient is 0.1399
(c)The lift coefficient is 5.430


## Ex 2.9 page 60¶

In :
#input data
W=30000#The weight of the jet plane in N
A=20#The area of the wing in m**2
C=250*5/18#The speed of the jet plane in m/s
P=750#The power delivered by the engine in kW
d=1.21#Density of the air in kg/m**3

#calculations
L=W#The lift force on the plane is equal to the weight of the plane in N
Pd=0.65*P#The power required to overcome the drag resistance in kW
D=(Pd/C)*10**3#The drag force on the wing in N
Cd=D/(0.5*d*A*C**2)#The coefficient of drag for the wing
Cl=L/(0.5*d*A*C**2)#The coefficient of lift for the wing

#output
print '(a)The coefficient of lift on the wing is %3.3f\n(b)The coefficient of drag on the wing is %3.3f'%(Cl,Cd)

(a)The coefficient of lift on the wing is 0.514
(b)The coefficient of drag on the wing is 0.120