# Chapter 4 - Axial Flow Compressors & Fans¶

## Ex 4.1 Page 145¶

In [1]:
from __future__ import division
from math import tan, pi
#input data
b1=60#The angle made by the relative velocity vector at exit in degree
db=30#The turning angle in degree
dCx=100#The change in the tangential velocities in m/s
DR=0.5#Degree of reaction
N=36000#The speed of the compressor in rpm
P1=2#Inlet pressure in bar
T1=330#Inlet temperature in K
R=287#The universal gas constant in J/kg.K
Cp=1.005#The specific heat of air at constant pressure in kJ/kg.K
r=1.4#The ratio of specific heats of air

#calculations
b2=b1-db#The angle made by the relative velocity vector at entry in degree
a1=b2#Air flow angle at exit in degree as DR=0.5
U=(3.1415*D*N)/60#The blade mean speed in m/s
T2=((U*dCx)/(Cp*1000))+T1#The exit air temperature in K
P2=P1*(T2/T1)**(r/(r-1))#The exit air pressure in bar
dP=P2-P1#The pressure rise in bar
Ca=(2*U*DR)/(tan(b2*pi/180)+tan(b1*pi/180))#The axial velocity in m/s
A1=3.1415*D*b#The inlet flow area in m**2
d1=(P1*10**5)/(R*T1)#The inlet air density in kg/m**3
m=d1*A1*Ca#The amount of air handled in kg/s
W=m*Cp*(T2-T1)#The power developed in kW

#output
print '(a)Air flow angle at exit is %3i degree\n(b)The pressure rise is %3.2f bar\n(c)The amount of air handled is %3.2f kg/s\n(d)The power developed is %3.1f kW'%(a1,dP,m,W)
# The answer in the textbook is not correct.

(a)Air flow angle at exit is  30 degree
(b)The pressure rise is 0.61 bar
(c)The amount of air handled is 2.12 kg/s
(d)The power developed is 56.0 kW


## Ex 4.2 Page 147¶

In [2]:
from math import log10
#input data
P01=1#Atmospheric pressure at inlet in bar
T01=291#Atmospheric temperature at inlet in K
T02=438#Total head temperature in delivery pipe in K
P02=3.5#Total head pressure in delivery pipe in bar
P2=3#Staic pressure in delivery pipe in bar
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K
r=1.4#The ratio of specific heats of air

#calculations
T02s=T01*(P02/P01)**((r-1)/r)#Total isentropic head temperature in delivery pipe in K
np=((log10(P02/P01))/((r/(r-1))*(log10(T02/T01))))#Polytropic efficiency
T2=T02*(P2/P02)**((r-1)/r)#Static temperature in delivery pipe in K
C2=(2*Cp*(T02-T2))**(1/2)#The air velocity in delivery pipe in m/s

#output
print '(a)Total head isentropic efficiency is %0.1f %%\n(b)Polytropic efficiency %0.1f %%\n(c)The air velocity in delivery pipe is %3.2f m/s'%(nc*100,np*100,C2)

(a)Total head isentropic efficiency is 85.2 %
(b)Polytropic efficiency 87.5 %
(c)The air velocity in delivery pipe is 194.76 m/s


## Ex 4.3 Page 148¶

In [3]:
from math import atan, degrees
#input data
N=8#Number of stages
Po=6#Overall pressure ratio
T01=293#Temperature of air at inlet in K
nc=0.9#Overall isentropic efficiency
DR=0.5#Degree of reaction
Ca=100#Constant axial velocity in m/s
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K
r=1.4#The ratio of specific heats of air

#calculations
T0n1s=T01*(Po)**((r-1)/r)#The isentropic temperature of air leaving compressor stage in K
T0n1=((T0n1s-T01)/nc)+T01#The temperature of air leaving compressor stage in K
dta2ta1=(Cp*(T0n1-T01))/(N*U*Ca)#The difference between tan angles of air exit and inlet
sta1tb1=U/Ca#The sum of tan of angles of air inlet and the angle made by the relative velocity
b1=degrees(atan((dta2ta1+sta1tb1)/2))#The angle made by the relative velocity vector at exit in degree as the DR=1 then a2=b1
a1=degrees(atan(tan(b1*pi/180)-dta2ta1))#Air flow angle at exit in degree
W=Cp*(T0n1-T01)*10**-3#Power required per kg of air/s in kW

#output
print '(a)Power required is %3.2f kW\n(b)\n    (1)Air flow angle at exit is %.f degree \n    (2)The angle made by the relative velocity vector at exit is %.f degree'%(W,a1,b1)

(a)Power required is 218.73 kW
(b)
(1)Air flow angle at exit is 12 degree
(2)The angle made by the relative velocity vector at exit is 59 degree


## Ex 4.4 Page 149¶

In [4]:
#input data
W=4.5#Power absorbed by the compressor in MW
m=20#Amount of air delivered in kg/s
P01=1#Stagnation pressure of air at inlet in bar
T01=288#Stagnation temperature of air at inlet in K
np=0.9#Polytropic efficiency of compressor
dT0=20#Temperature rise in first stage in K
R=287#The universal gas constant in J/kg.K
Cp=1.005#The specific heat of air at constant pressure in kJ/kg.K
r=1.4#The ratio of specific heats of air

#calculations
T02=T01+dT0#Stagnation temperature of air at outlet in K
T0n1=((W*10**3)/(m*Cp))+T01#The temperature of air leaving compressor stage in K
P0n1=P01*(T0n1/T01)**((np*r)/(r-1))#Pressure at compressor outlet in bar
P1=(T02/T01)**((np*r)/(r-1))#The pressure ratio at the first stage
N=((log10(P0n1/P01)/log10(P1)))#Number of stages
T0n1T01=(P0n1/P01)**((r-1)/(np*r))#The temperature ratio at the first stage
T0n1sT01=(P0n1/P01)**((r-1)/r)#The isentropic temperature ratio at the first stage
nc=((T0n1sT01-1)/(T0n1T01-1))#The overall isentropic efficiency

#output
print '(a)Pressure at compressor outlet is %3.2f bar\n(b)Number of stages is %3.f\n(c)The overall isentropic efficiency is %0.1f %%'%(P0n1,N,nc*100)

(a)Pressure at compressor outlet is 6.12 bar
(b)Number of stages is   9
(c)The overall isentropic efficiency is 87.2 %


## Ex 4.5 Page 151¶

In [5]:
from math import log
#input data
DR=0.5#Degree of reaction
Po=5#The pressure ratio produced by the compressor
nc=0.87#The overall isentropic efficiency
T01=290#Inlet temperature in K
l=0.85#Work input factor
R=0.287#The universal gas constant in kJ/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K
r=1.4#The ratio of specific heats of air

#calculations
a2=b1#Air flow angle at entry in degree as DR=0.5
a1=b2#Air flow angle at exit in degree as DR=0.5
T0n1s=T01*(Po)**((r-1)/r)#The isentropic temperature of air leaving compressor stage in K
T0n1=((T0n1s-T01)/nc)+T01#The temperature of air leaving compressor stage in K
Ca=U/(tan(b2*pi/180)+tan(b1*pi/180))#The axial velocity in m/s
N=((Cp*(T0n1-T01))/(l*U*Ca*(tan(a2*pi/180)-tan(a1*pi/180))))#The number of stages
ds=(Cp*(10**-3)*log(T0n1/T01))-(R*log(Po))#Change in entropy in kJ/kg.K

#output
print '(a)The number of stages are %3.f\n(b)The change in entropy is %3.3f kJ/kg-K'%(N,ds)

(a)The number of stages are  12
(b)The change in entropy is 0.054 kJ/kg-K


## Ex 4.6 Page 152¶

In [6]:
#input data
D=0.6#Mean diameter of compressor in m
N=15000#Running speed of the compressor in rpm
dT=30#Actual overall temperature raise in K
PR=1.3#Pressure ratio of all stages
m=57#Mass flow rate of air in kg/s
nm=0.86#Mechanical efficiency
T1=308#Initial temperature in K
T2=328#Temperature at rotor exit in K
r=1.4#The ratio of specific heats of air
Cp=1.005#The specific heat of air at constant pressure in kJ/kg.K

#calculations
W=m*Cp*dT#Work done in kW
P=W/nm#Power required in kW
ns=((T1*((PR**((r-1)/r))-1))/(dT))#Stage efficiency
R=(T2-T1)/(dT)#Reaction ratio

#output
print '(a)Power required to drive the compressor is %3.3f kW\n(b)The stage efficiency is %0.2f %%\n(c)The degree of reaction is %3.2f'%(P,ns*100,R)

(a)Power required to drive the compressor is 1998.314 kW
(b)The stage efficiency is 79.92 %
(c)The degree of reaction is 0.67


## Ex 4.7 Page 153¶

In [7]:
#input data
Pr=2#The pressure ratio of first stage
P1=1.01#The inlet pressure in bar
T1=303#The inlet temperature in K
nc=0.83#Overall efficency of the compressor
pi=0.47#The flow coefficient
dCxCa=0.5#Ratio of change of whirl velocity to axial velocity
D=0.5#Mean diameter in m
r=1.4#The ratio of specific heats of air
Cp=1005#The specific heat of air at constant pressure in J/kg.K

#calculations
dT=T1*((Pr**((r-1)/r))-1)/nc#The Actual overall temperature raise in K
dCx=dCxCa*pi#The change of whirl velocity in m/s
U=(dT*Cp/dCx)**(1/2)#The mean blade speed in m/s
N=(U*60)/(3.1415*D)#Speed at which compressor runs in rpm
Cx2=(U+(dCx*U))/2#The whirl velocity at exit in m/s
Cx1=U-Cx2#The whirl velocity at entry in m/s
Ca=pi*U#The axial velocity in m/s
C1=((Ca**2)+(Cx1**2))**(1/2)#The inlet absolute velocity of air in m/s

#output
print '(a)The compressor speed is %3i rpm\n(b)The absolute velocity of air is %3.2f m/s'%(N,C1)

(a)The compressor speed is 22336 rpm
(b)The absolute velocity of air is 354.34 m/s


## Ex 4.8 Page 154¶

In [8]:
from math import acos, asin, sin,cos, sqrt, degrees, pi, atan, tan
from __future__ import division
#input data
N=9000#The rotational speed in rpm
dT0=20#The stagnation temperature rise in K
DhDt=0.6#The hub to tip ratio
l=0.94#The work donee factor
ns=0.9#The isentropic efficiency of the stage
C1=150#Inlet velocity in m/s
P01=1#The ambient pressure in bar
T01=300#The ambient temperature in K
Mr1=0.92#Mach number relative to tip
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in kJ/kg.K
r=1.4#The ratio of specific heats of air
g=9.81#Acceleration due to gravity in m/s**2

#calculations
T1=T01-((C1**2)/(2*Cp))#The inlet temperature in K
W1=Mr1*sqrt(r*R*T1)#The relative velocity at entry in m/s
b11=degrees(acos((C1)/(W1)))#The inlet rotor angle at tip in degree
Ut=W1*sin(b11*pi/180)#Tip speed in m/s
b12=degrees(atan((tan(b11*pi/180)))-((Cp*dT0)/(l*Ut*C1)))#The outlet rotor angle at tip in degree
P1=P01*(T1/T01)**(r/(r-1))#The inlet pressure in bar
d1=(P1*10**5)/(R*T1)#The density of air at the entry in kg/m**3
Dt=2*rt#The tip diameter in m
Dh=DhDt*(Dt)#The hub diameter in m
A1=(3.141/4)*((Dt**2)-(Dh**2))#The area of cross section at the entry in m**2
h=((Dt/2)-(Dh/2))#The height of the blade in m
A=2*3.1415*rm*h#The area of the cross section in m**2
m=d1*A*C1#The mass flow rate in kg/s
P03P01=(1+((ns*dT0)/T01))**(r/(r-1))#The stagnation pressure ratio
P=m*Cp*dT0*10**-3#The power required in kW
Uh=(3.1415*Dh*N)/60#The hub speed in m/s
b21=degrees(atan(Uh/C1))#The rotor air angle at entry in degree
b22=degrees(atan(tan(b21*pi/180)-((Cp*dT0)/(l*Uh*C1))))#The rotor air angle at exit in degree

#output
print '(a)\n    (1)The tip radius is %3.3f m\n    (2)The rotor entry angle at tip section is %3.1f degree\n    (3)The rotor exit angle at tip section is %3.2f degree\n(b)Mass flow entering the stage is %3.3f kg/s\n(c)\n    (1)The stagnation pressure ratio is %3.3f\n    (2)The power required is %3.2f kW\n(d)\n    (1)The rotor air angle at entry is %3.2f degree\n    (2)The rotor air angle at exit is %3.2f degree'%(rt,b11,b12,m,P03P01,P,b21,b22)
#the answer in the textbook is not correct.

(a)
(1)The tip radius is 0.292 m
(2)The rotor entry angle at tip section is 61.4 degree
(3)The rotor exit angle at tip section is 31.72 degree
(b)Mass flow entering the stage is 27.152 kg/s
(c)
(1)The stagnation pressure ratio is 1.226
(2)The power required is 545.75 kW
(d)
(1)The rotor air angle at entry is 47.74 degree
(2)The rotor air angle at exit is 13.35 degree


## Ex 4.9 Page 157¶

In [9]:
#input data
Ur=150#The blade root velocity in m/s
Um=200#The mean velocity in m/s
Ut=250#The tip velocity in m/s
dT0=20#The total change in temperature in K
Ca=150#The axial velocity in m/s
l=0.93#The work done factor
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K
r=1.4#The ratio of specific heats of air

#calculations
dtb1tb2=((Cp*dT0)/(l*Um*Ca))#The difference between the tangent angles of blade angles at mean
atb1tb2=((2*Rm*Um)/(Ca))#The sum of the tangent angles of blade angles at mean
b1m=degrees(atan((atb1tb2+dtb1tb2)/2))#The inlet blade angle in degree at mean
a2m=b1m#The exit air angle in degree as the Reaction at mean radius is 0.5
b2m=degrees(atan(tan(b1m*pi/180)-dtb1tb2))#The exit blade angle in degree at mean
a1m=b2m#The inlet air angle in degree as the reaction at mean radius is 0.5
rmrh=Um/Ur#The ratio of radii of mean and root velocities at hub
a1h=degrees(atan(tan(a1m*pi/180)*(rmrh)))#The inlet air angle in degree at hub
b1h=degrees(atan((Ur/Ca)-(tan(a1h*pi/180))))#The inlet blade angle in degree at hub
a2h=degrees(atan(tan(a2m*pi/180)*(rmrh)))#The outlet air angle in degree at hub
b2h=degrees(atan((Ur/Ca)-(tan(a2h*pi/180))))#The outlet blade angle in degree at hub
Rh=((Ca*(tan(b1h*pi/180)+tan(b2h*pi/180)))/(2*Ur))#The degree of reaction at the hub
rmrt=Um/Ut#The ratio of radii of mean and tip velocities at tip
a1t=degrees(atan(tan(a1m)*(rmrt)))#The inlet air angle in degree at tip
b1t=degrees(atan((Ut/Ca)-(tan(a1t*pi/180))))#The inlet blade angle in degree at tip
a2t=degrees(atan(tan(a2m)*(rmrt)))#The outlet air angle in degree at tip
b2t=degrees(atan((Ut/Ca)-(tan(a2t*pi/180))))#The outlet blade angle in degree at tip
Rt=((Ca*(tan(b1t*pi/180)+tan(b2t*pi/180)))/(2*Ut))#The degree of reaction at tip

#output
print '(a)At the mean\n    (1)The inlet blade angle is %3.2f degree\n    (2)The inlet air angle is %3.2f degree\n    (3)The outlet blade angle is %3.2f degree\n    (4)The outlet air angle is %3.2f degree\n    (5)Degree of reaction is %3.1f \n(b)At the root\n    (1)The inlet blade angle is %3.2f degree\n    (2)The inlet air angle is %3.2f degree\n    (3)The outlet blade angle is %3.2f degree\n    (4)The outlet air angle is %3.2f degree\n    (5)Degree of reaction is %3.3f\n(c)At the tip\n    (1)The inlet blade angle is %3.2f degree\n    (2)The inlet air angle is %3.2f degree\n    (3)The outlet blade angle is %3.2f degree\n    (4)The outlet air angle is %3.2f degree\n    (5)Degree of reaction is %3.3f\n'%(b1m,a1m,b2m,a2m,Rm,b1h,a1h,b2h,a2h,Rh,b1t,a1t,b2t,a2t,Rt)
#the answer in the textbook is not correct.

(a)At the mean
(1)The inlet blade angle is 45.76 degree
(2)The inlet air angle is 17.04 degree
(3)The outlet blade angle is 17.04 degree
(4)The outlet air angle is 45.76 degree
(5)Degree of reaction is 0.5
(b)At the root
(1)The inlet blade angle is 30.60 degree
(2)The inlet air angle is 22.23 degree
(3)The outlet blade angle is -20.26 degree
(4)The outlet air angle is 53.86 degree
(5)Degree of reaction is 0.111
(c)At the tip
(1)The inlet blade angle is -57.81 degree
(2)The inlet air angle is 72.92 degree
(3)The outlet blade angle is 79.66 degree
(4)The outlet air angle is -75.31 degree
(5)Degree of reaction is 1.168



## Ex 4.10 Page 160¶

In [10]:
#input data
Uh=150#The blade root velocity in m/s
Um=200#The mean velocity in m/s
Ut=250#The tip velocity in m/s
dT0=20#The total change in temperature in K
Ca1m=150#The axial velocity in m/s
l=0.93#The work done factor
N=9000#Rotational speed in rpm
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K
r=1.4#The ratio of specific heats of air

#calculations
dtb1tb2=((Cp*dT0)/(l*Um*Ca1m))#The difference between the tangent angles of blade angles at mean
atb1tb2=((2*Rm*Um)/(Ca1m))#The sum of the tangent angles of blade angles at mean
b1m=degrees(atan((atb1tb2+dtb1tb2)/2))#The inlet blade angle in degree at mean
a2m=b1m#The exit air angle in degree as the Reaction at mean radius is 0.5
b2m=degrees(atan(tan(pi/180*b1m)-dtb1tb2))#The exit blade angle in degree at mean
a1m=b2m#The inlet air angle in degree as the reaction at mean radius is 0.5
Dh=(Uh*60)/(3.141*N)#Hub diameter in m
Dm=(Um*60)/(3.141*N)#Mean diameter in m
Cx1m=Ca1m*tan(pi/180*a1m)#The whirl velocity at inlet at mean in m/s
Cx2m=Ca1m*tan(pi/180*a2m)#The whirl velocity at exit at mean in m/s
Cx1h=(Cx1m*(Dh/2)/(Dm/2))#The whirl velocity at inlet at hub in m/s
Cx2h=(Cx2m*(Dh/2)/(Dm/2))#The whirl velocity at exit at hub in m/s
K1=(Ca1m**2)+(2*(Cx1m**2))#Sectional velocity in m/s
Ca1h=((K1)-(2*(Cx1h**2)))**(1/2)#The axial velocity at hub inlet in (m/s)**2
K2=(Ca1m**2)+(2*(Cx2m**2))-(2*((Cx2h/(Dh/2))-(Cx1m/(Dm/2))))*(w*(Dm/2)**(2))#Sectional velocity in (m/s)**2
Ca2h=(K2-(2*Cx2h**2)+(2*((Cx2h/(Dh/2))-(Cx1h/(Dh/2))))*(w*(Dh/2)**(2)))**(1/2)#Axial velocity at hub outlet in m/s
a1h=degrees(atan(Cx1h/Ca1h))#Air angle at inlet in hub in degree
b1h=degrees(atan((Uh-Cx1h)/Ca1h))#Blade angle at inlet in hub in degree
a2h=degrees(atan(Cx2h/Ca2h))#Air angle at exit in hub in degree
b2h=degrees(atan((Uh-Cx2h)/Ca2h))#Blade angle at exit in hub in degree
W1=Ca1h/cos(pi/180*b1h)#Relative velocity at entry in hub in m/s
W2=Ca2h/cos(pi/180*b2h)#Relative velocity at exit in hub in m/s
Rh=((W1**2)-(W2**2))/(2*Uh*(Cx2h-Cx1h))#The degree of reaction at hub
Dt=(Ut*60)/(3.141*N)#Tip diameter in m
Cx1t=(Cx1m*(Dt/2)/(Dm/2))#The whirl velocity at inlet at tip in m/s
Cx2t=(Cx2m*(Dt/2)/(Dm/2))#The whirl velocity at exit at tip in m/s
Ca1t=(K1-(2*Cx1t**2))**(1/2)#Axial velocity at tip inlet in m/s
Ca2t=(K2-(2*Cx2t**2)+(2*((Cx2t/(Dt/2))-(Cx1t/(Dt/2))))*(w*(Dt/2)**(2)))**(1/2)#Axial velocity at tip outlet in m/s
a1t=degrees(atan(Cx1t/Ca1t))#Air angle at inlet in tip in degree
b1t=degrees(atan((Ut-Cx1t)/Ca1t))#Blade angle at inlet in tip in degree
a2t=degrees(atan(Cx2t/Ca2t))#Air angle at exit in tip in degree
b2t=degrees(atan((Ut-Cx2t)/Ca2t))#Blade angle at exit in tip in degree
W1=Ca1t/cos(pi/180*b1t)#Relative velocity at entry in tip in m/s
W2=Ca2t/cos(pi/180*b2t)#Relative velocity at exit in tip in m/s
Rt=((W1**2)-(W2**2))/(2*Ut*(Cx2t-Cx1t))#The degree of reaction at tip

#output
print '(a)At the mean\n    (1)The inlet blade angle is %3.2f degree\n    (2)The inlet air angle is %3.2f degree\n    (3)The outlet blade angle is %3.2f degree\n    (4)The outlet air angle is %3.2f degree\n    (5)Degree of reaction is %3.1f \n(b)At the root\n    (1)The inlet blade angle is %3.2f degree\n    (2)The inlet air angle is %3.1f degree\n    (3)The outlet blade angle is %3.1f degree\n    (4)The outlet air angle is %3.1f degree\n    (5)Degree of reaction is %3.1f\n(c)At the tip\n    (1)The inlet blade angle is %3.2f degree\n    (2)The inlet air angle is %3.2f degree\n    (3)The outlet blade angle is %3.2f degree\n    (4)The outlet air angle is %3.2f degree\n    (5)Degree of reaction is %3.1f\n'%(b1m,a1m,b2m,a2m,Rm,b1h,a1h,b2h,a2h,Rh,b1t,a1t,b2t,a2t,Rt)
# the answer in the textbook is not correct.

(a)At the mean
(1)The inlet blade angle is 45.76 degree
(2)The inlet air angle is 17.04 degree
(3)The outlet blade angle is 17.04 degree
(4)The outlet air angle is 45.76 degree
(5)Degree of reaction is 0.5
(b)At the root
(1)The inlet blade angle is 36.51 degree
(2)The inlet air angle is 12.5 degree
(3)The outlet blade angle is 12.5 degree
(4)The outlet air angle is 36.5 degree
(5)Degree of reaction is 0.5
(c)At the tip
(1)The inlet blade angle is 53.62 degree
(2)The inlet air angle is 22.05 degree
(3)The outlet blade angle is 22.05 degree
(4)The outlet air angle is 53.62 degree
(5)Degree of reaction is 0.5



## Ex 4.11 Page 163¶

In [11]:
#input data
N=3600#Running speed of blower in rpm
Dt=0.2#The rotor  tip diameter in m
Dh=0.125#The rotor hub diameter in m
P1=1.013#The atmospheric pressure in bar
T1=298#The atmospheric temperature in K
m=0.5#Mass flow rate of air in kg/s
db=20#The turning angle of the rotor in degree
b1=55#The inlet blade angle in degree
R=287#The universal gas constant in J/kg.K
nc=0.9#Total-to-total efficiency
P=0.25#Total pressure drop across the intake in cm of water
Cp=1005#The specific heat of air at constant pressure in J/kg.K
r=1.4#The ratio of specific heats of air
g=9.81#Acceleration due to gravity in m/s**2
ns=0.75#The stator efficiency
dw=1000#Density of water in kg/m**3

#calculations
d1=(P1*10**5)/(R*T1)#The density of air at inlet in kg/m**3
A=(3.141/4)*((Dt**2)-(Dh**2))#The area of flow in m**2
Ca=m/(d1*A)#The axial velocity of air in m/s
U=((3.141*(Dt+Dh)*N)/(2*60))#Mean rotor blade velocity in m/s
b2=b1-db#The outlet blade angle in degree
Cx2=U-(Ca*tan(pi/180*b2))#The whirl velocity at exit in m/s
Cx1=0#The whirl velocity at entry in m/s as flow at inlet is axial
dh0r=U*(Cx2-Cx1)#The actual total enthalpy rise across the rotor in J/kg
dh0sr=nc*dh0r#The isentropic total enthalpy rise across the rotor in J/kg
dP0r=(d1*dh0sr)*((10**-1)/(g))#The total pressure rise across the rotor in cm of water
P0=dP0r-P#Stagnation pressure at the rotor exit in cm of water
C2=((Ca**2)+(Cx2**2))**(1/2)#The absolute velocity at the exit in m/s
dPr=dP0r-((d1*((C2**2)-(Ca**2)))/2)*((10**-1)/g)#The static pressure across the rotor in cm of water
dhs=((C2**2)-(Ca**2))/2#The actual enthalpy change across the stator in J/kg
dhss=ns*dhs#The theoretical enthalpy change across the stator in J/kg
dPs=(d1*dhss)*((10**-1)/g)#The static pressure rise across the stator in cm of water
dP0s=-((dPs/((10**-1)/g))+((d1/2)*(Ca**2-C2**2)))*(10**-1/g)#The change in total pressure across the stator in cm of water
P03=P0-dP0s#Total pressure at stator inlet in cm of water
dh0ss=((dw*g*(P03/100))/d1)#Theoretical total enthalpy change across the stage in J/kg
ntt=dh0ss/dh0r#The overall total-to-total efiiciency
DR=dPr/(dPr+dPs)#The degree of reaction for the stage

#output
print '(a)Total pressure of air exit of rotor is %3.2f cm of water\n(b)The static pressure rise across the rotor is %3.2f cm of water\n(c)The static pressure rise across the stator os %3.2f cm of water\n(d)The change in total pressure across the stator is %3.2f cm of water\n(e)The overall total-to-total efficiency is %0.1f %%\n(f)The degree of reaction for the stage is %0.1f %%'%(P0,dPr,dPs,dP0s,ntt*100,DR*100)

(a)Total pressure of air exit of rotor is 4.80 cm of water
(b)The static pressure rise across the rotor is 3.66 cm of water
(c)The static pressure rise across the stator os 1.04 cm of water
(d)The change in total pressure across the stator is 0.35 cm of water
(e)The overall total-to-total efficiency is 79.3 %
(f)The degree of reaction for the stage is 77.8 %


## Ex 4.12 Page 166¶

In [12]:
#input data
Q=2.5#The amount of air which fan takes in m**3/s
P1=1.02#The inlet pressure of air in bar
T1=315#The inlet temperature of air in K
dH=0.75#The pressure head delivered by axial flow fan in m W.G
T2=325#The delivery temperature of air in K
R=287#The universal gas constant in J/kg.K
Cp=1.005#The specific heat of air at constant pressure in kJ/kg.K
r=1.4#The ratio of specific heats of air
g=9.81#Acceleration due to gravity in m/s**2

#calculations
d=(P1*10**5)/(R*T1)#The density of air in kg/m**3
m=d*Q#The mass flow rate of air in kg/s
W=m*Cp*(T2-T1)#Power required to drive the fan in kW
dP=((10**3)*g*dH)/(10**5)#The overall pressure difference in bar
P2=P1+(dP)#The exit pressure in bar
nf=((T1*(((P2/P1)**((r-1)/r))-1))/(T2-T1))#Static fan efficiency

#output
print '(a)Mass flow rate through the fan is %3.2f kg/s\n(b)Power required to drive the fan is %3.2f kW\n(c)Static fan efficiency is %0.2f %%'%(m,W,nf*100)

(a)Mass flow rate through the fan is 2.82 kg/s
(b)Power required to drive the fan is 28.35 kW
(c)Static fan efficiency is 63.31 %


## Ex 4.13 Page 167¶

In [13]:
#input data
b2=10#Rotor blade air angle at exit in degree
Dt=0.6#The tip diameter in m
Dh=0.3#The hub diameter in m
N=960#The speed of the fan in rpm
P=1#Power required by the fan in kW
pi=0.245#The flow coefficient
P1=1.02#The inlet pressure in bar
T1=316#The inlet temperature in K
R=287#The universal gas constant in J/kg.K
Cp=1.005#The specific heat of air at constant pressure in kJ/kg.K
r=1.4#The ratio of specific heats of air
g=9.81#Acceleration due to gravity in m/s**2

#calculations
A=(3.141/4)*((Dt**2)-(Dh**2))#Area of the fan at inlet in m**2
Dm=(Dt+Dh)/2#The mean rotor diameter in m
U=(3.141*Dm*N)/60#The mean blade speed in m/s
Ca=pi*U#The axial velocity in m/s
Q=A*Ca#The flow rate of air in m**3/s
d=(P1*10**5)/(R*T1)#Density of air in kg/m**3
dPst=((d*(U**2)*(1-((pi*tan(pi/180*b2))**2)))/2)*((10**5)/(g*(10**3)))*10**-5#Static pressure across the stage in m W.G
Wm=U*(U-(Ca*tan(pi/180*b2)))#Work done per unit mass in J/kg
m=d*Q#Mass flow rate in kg/s
W=m*Wm#Work done in W
no=W/(P*10**3)#Overall efficiency

#output
print '(a)THe flow rate is %3.3f m**3/s\n(b)Static pressure rise across the stage is %3.3f m W.G\n(c)The overall efficiency is %0.2f %%'%(Q,dPst,no*100)
# the answer for last part is not accurate.

(a)THe flow rate is 1.175 m**3/s
(b)Static pressure rise across the stage is 0.029 m W.G
(c)The overall efficiency is 67.35 %


## Ex 4.14 Page 169¶

In [14]:
#input data
b2=10#Rotor blade air angle at exit in degree
Dt=0.6#The tip diameter in m
Dh=0.3#The hub diameter in m
N=960#The speed of the fan in rpm
P=1#Power required by the fan in kW
pi=0.245#The flow coefficient
P1=1.02#The inlet pressure in bar
T1=316#The inlet temperature in K
R=287#The universal gas constant in J/kg.K
Cp=1.005#The specific heat of air at constant pressure in kJ/kg.K
r=1.4#The ratio of specific heats of air
g=9.81#Acceleration due to gravity in m/s**2

#calculations
A=(3.141/4)*((Dt**2)-(Dh**2))#Area of the fan at inlet in m**2
Dm=(Dt+Dh)/2#The mean rotor diameter in m
U=(3.141*Dm*N)/60#The mean blade speed in m/s
Ca=pi*U#The axial velocity in m/s
Q=A*Ca#The flow rate of air in m**3/s
d=(P1*10**5)/(R*T1)#Density of air in kg/m**3
b1=degrees(atan(U/Ca))#Rotor blade angle at entry in degree
dPst=((d*(U**2)*(1-((pi*tan(pi/180*b2))**2)))/2)#Static pressure rise across the stage in N/m**2
dPr=dPst#Static pressure rise across the rotor in N/m**2
Wm=U*(U-(Ca*tan(pi/180*b2)))#Work done per unit mass in J/kg
dP0st=d*Wm#Stagnation pressure of the stage in N/m**2
DR1=dPr/dP0st#Degree of reaction
DR2=(Ca/(2*U))*(tan(pi/180*b1)+tan(pi/180*b2))#Degree of reaction

#output
print '(a)Rotor blade angle at entry is %3.2f degree\n(b)Degree of reaction is %0.1f %%'%(b1,DR1*100)
# the answer for last part is not correct in the textbook.

(a)Rotor blade angle at entry is 76.23 degree
(b)Degree of reaction is 50.2 %


## Ex 4.15 Page 170¶

In [15]:
#input data
m=3#Mass flow rate of air in kg/s
P1=100*10**3#The atmospheric pressure in Pa
T1=310#The atmospheric temperature in K
nb=0.8#The efficiency of the blower
nm=0.85#The mechanical efficiency
P=30#The power input in kW
R=287#The universal gas constant in J/kg.K
g=9.81#Acceleration due to gravity in m/s**2
dw=1000#Density of water in kg/m**3

#calculations
no=nb*nm#Overall efficiency of the blower
d=(P1)/(R*T1)#The density of the air in kg/m**3
dP=((no*P*10**3)/m)*d#The pressure developed in N/m**2
dH=((dP)/(g*dw))*(10**3)#The pressure developed in mm W.G

#output
print '(a)Overall efficiency of the blower is %3.2f\n(b)The pressure developed is %3.2f mm W.G'%(no,dH)

(a)Overall efficiency of the blower is 0.68
(b)The pressure developed is 779.11 mm W.G


## Ex 4.16 Page 170¶

In [16]:
#input data
psi=0.4#Pressure coefficient
m=3.5#Mass flow rate of air in kg/s
N=750#The speed of fan in rpm
T1=308#The static temperature at the entry in K
Dh=0.26#The hub diameter in m
DhDt=1/3#The hub to tip ratio
P1=98.4*10**3#The static pressure at entry in Pa
nm=0.9#The mechanical efficiency
nf=0.79#Static fan efficiency
R=287#The universal gas constant in J/kg.K
Cp=1.005#The specific heat of air at constant pressure in kJ/kg.K
r=1.4#The ratio of specific heats of air
g=9.81#Acceleration due to gravity in m/s**2
dw=1000#Density of water in kg/m**3

#calculations
no=nm*nf#Overall efficiency
Dt=Dh/DhDt#The tip diameter in m
Dm=(Dt+Dh)/2#Mean rotor diameter in m
U=(3.141*Dm*N)/60#The mean blade speed in m/s
dPd=((U**2)/2)*psi#The ratio of change in pressure to density in J/kg
Wi=dPd*m#The ideal work in W
P=Wi/nm#The power required by the fan in W
d=P1/(R*T1)#The density of the air in kg/m**3
A=(3.141/4)*((Dt**2)-(Dh**2))#Area of cross section of the fan in m**2
Ca=m/(d*A)#The axial velocity of air in m/s
pi=Ca/U#The flow coefficient
tb1tb2=psi/(2*pi)#The difference between tangent angles of rotor inlet and exit angles
b2=degrees(atan((1-(dPd/U**2))/pi))#The exit rotor angle in degree
b1=degrees(atan((tan(b2*pi/180))+(tb1tb2)))#The inlet rotor angle in degree
dP=d*dPd#The pressure developed in N/m**2
dH=(dP/(dw*g))*10**3#Pressure developed in mm of W.G

#output
print '(a)The overall efficiency is %0.1f %%\n(b)The power required by the fan is %3.2f W\n(c)The flow coefficient is %3.2f\n(d)\n    (1)The rotor inlet angle is %3.2f degree\n    (2)The rotor exit angle is %3.2f degree\n(e)The pressure developed is %3.2f mm of W.G'%(no*100,P,pi,b1,b2,dH)
# the answer for part(d) is not correct in the textbook.

(a)The overall efficiency is 71.1 %
(b)The power required by the fan is 324.20 W
(c)The flow coefficient is 0.36
(d)
(1)The rotor inlet angle is 34.39 degree
(2)The rotor exit angle is 65.62 degree
(e)The pressure developed is 9.46 mm of W.G