# Chapter 3 - Centrifugal Compressors & Fans¶

## Ex 3.1 Page 93¶

In [1]:
from __future__ import division
#input data
m=10#The mass flow rate of air into compressor in kg/s
P1=1#The ambient air pressure in compressor in bar
T1=293#The ambient air temperature in compressor in K
N=20000#The running speed of the compressor in rpm
nc=0.8#The isentropic efficiency of the compressor
P02=4.5#The total exit pressure from the compressor in bar
C1=150#The air entry velocity into the impeller eye in m/s
Cx1=0#The pre whirl speed in m/s
WS=0.95#The ratio of whirl speed to tip speed
Cp=1005#The specific heat of air at constant pressure in J/kg.K
R=287#The universal gas constant in J/kg.K
Dh=0.15#The eye internal diamater in m
r=1.4#Ratio of specific heats of air
d=1.189#The density of the air in kg/m**3

#calculations
T01=T1+((C1**2)/(2*Cp))#The stagnation temperature at inlet in K
P01=P1*(T01/T1)**(r/(r-1))#The stagnation pressure at inlet in bar
T02s=(T01)*(P02/P01)**((r-1)/r)#The temperature after isentropic compression from P01 to P02 in K
T=(T02s-T01)/nc#The actual rise in total temperature in K
W=Cp*(10**-3)*(T)#The work done per unit mass in kJ/kg
U2=((W*(10**3))/(WS))**(1/2)#The impeller tip speed in m/s
Dt=(U2*60)/(3.1415*N)#The impeller tip diameter in m
P=m*W#Power required to drive the compressor in kW
d1=((P1*10**5)/(R*T1))#The density of the air entry in kg/m**3
De=(((4*m)/(d*C1*3.14))+(Dh**2))**(1/2)#The eye external diameter in m

#output
print '(a)The actual rise in total temperature of the compressor is %3.1f K\n(b)\n      (1)The impeller tip speed is %3.2f m/s\n      (2)The impeller tip diameter is %3.2f m\n(c)The power required to drive the compressor is %3.1f kW\n(d)The eye external diameter is %0.1f cm'%(T,U2,Dt,P,De*100)

(a)The actual rise in total temperature of the compressor is 182.6 K
(b)
(1)The impeller tip speed is 439.55 m/s
(2)The impeller tip diameter is 0.42 m
(c)The power required to drive the compressor is 1835.4 kW
(d)The eye external diameter is 30.6 cm


## Ex 3.2 Page 95¶

In [2]:
from math import degrees, atan
#input data
Q1=20#Discharge of air to the centrifugal compressor in m**3/s
V1=Q1#Volume of rate is equal to the discharge in m**3/s
P1=1#Initial pressure of the air to the centrifugal compressor in bar
T1=288#Initial temperature of the air to the centrifugal compressor in K
P=1.5#The pressure ratio of compression in centrifugal compressor
C1=60#The velocity of flow of air at inlet in m/s
Cr2=C1#The radial velocity of flow of air at outlet in m/s
Dh=0.6#The inlet impeller diameter in m
Dt=1.2#The outlet impeller diameter in m
N=5000#The speed of rotation of centrifugal compressor in rpm
n=1.5#polytropic index constant in the given law PV**n
Cp=1005#The specific heat of air at constant pressure in J/kg.K

#calculations
U1=(3.14*Dh*N)/60#Peripheral velocity of impeller at inlet in m/s
b11=degrees(atan(C1/U1))#The blade angle at impeller inlet in degree
U2=(3.14*Dt*N)/60#Peripheral velocity of impeller top at outlet in m/s
T2=T1*(P)**((n-1)/n)#Final temperature of the air to the centrifugal compressor in K
Cx2=((Cp*(T2-T1))/U2)#The whirl component of absolute velocity in m/s
Wx2=U2-Cx2#The exit relative velocity in m/s
a2=degrees(atan(Cr2/Cx2))#The blade angle at inlet to casing in degree
b22=degrees(atan(Cr2/Wx2))#The blade angle at impeller outlet in degree
V2=(P1*V1*T2)/(T1*P*P1)#Volume flow rate of air at exit in m**3/s
Q2=V2#Volume flow rate is equal to discharge in m**3/s

#output
print '(a)The blade and flow angles\n   (1)The blade angle at impeller inlet is %3.1f degree\n   (2)The blade angle at inlet to casing is %3.1f degree\n   (3)The blade angle at impeller outlet is %3.2f degree\n(b)Breadth of the impeller blade at inlet and outlet\n   (1)The breadth of impeller blade at inlet is %3.3f m\n   (2)The Volume flow rate of air at exit is %3.2f m**3/s\n   (3)The breadth of impeller blade at outlet is %3.4f m'%(b11,a2,b22,b1,V2,b2)

#error in the first review is not printing the value of V2 which is corrected

(a)The blade and flow angles
(1)The blade angle at impeller inlet is 20.9 degree
(2)The blade angle at inlet to casing is 24.2 degree
(3)The blade angle at impeller outlet is 18.38 degree
(2)The Volume flow rate of air at exit is 15.26 m**3/s


## Ex 3.3 Page 97¶

In [3]:
#input data
m=14#The mass flow rate of air delivered to centrifugal compressor in kg/s
P01=1#The inlet stagnation pressure in bar
T01=288#The inlet stagnation temperature in K
P=4#The stagnation pressure ratio
N=200#The speed of centrifygal compressor in rps
ss=0.9#The slip factor
ps=1.04#The power input factor
ntt=0.8#The overall isentropic efficiency
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
pp=ss*ps*ntt#The pressure coefficient
U2=((Cp*T01*((P**((r-1)/r))-1))/pp)**(1/2)#Peripheral velocity of impeller top at outlet in m/s
D2=U2/(3.14*N)#The overall diameter of the impeller in m

#output
print 'The overall diameter of the impeller is %.f cm'%(D2*100)

The overall diameter of the impeller is 69 cm


## Ex 3.4 Page 98¶

In [4]:
from math import cos, pi, tan
#input data
D1=0.457#Impeller diameter at inlet in m
D2=0.762#Impeller diameter at exit in m
Cr2=53.4#Radial component of velocity at impeller exit in m/s
ss=0.9#Slip factor
N=11000#Impeller speed in rpm
P2=2.23#Static pressure at impeller exit in bar
T01=288#The inlet stagnation temperature in K
P01=1.013#The inlet stagnation pressure in bar
C1=91.5#Velocity of air leaving the guide vanes in m/s
a11=70#The angle at which air leaves the guide vanes in degrees
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K

#calculations
Cx1=C1*cos(a11*pi/180)#Inlet absolute velocity of air in tangential direction in m/s
Ca1=Cx1*tan(a11*pi/180)#Radial component of absolute velocity at inlet in m/s
U1=(3.14*D1*N)/(60)#Peripheral velocity of impeller at inlet in m/s
Wx1=U1-Cx1#Relative whirl component of velocity at inlet in m/s
W1=((Wx1**2)+(Ca1**2))**(1/2)#Relative velocity at inlet in m/s
T1=T01-((C1**2)/(2*Cp))#The inlet air temperature in K
a1=(r*R*T1)**(1/2)#The velocity of air in m/s
M1r=W1/a1#Initial relative mach number
U2=(3.14*D2*N)/60#Peripheral velocity of impeller top at exit in m/s
W=(ss*U2**2)-(U1*Cx1)#Work done by the compressor in kJ/kg
T02=(W/Cp)+T01#The outlet stagnation temperature in K
Cx21=ss*U2#Absolute whirl component of velocity with slip consideration in m/s
C2=((Cx21**2)+(Cr2**2))**(1/2)#The absolute velocity of air at exit in m/s
T2=T02-((C2**2)/(2*Cp))#The exit temperature of air in K
P02=P2*(T02/T2)**(r/(r-1))#The exit stagnation pressure of compressor in bar

#output
print '(1)The inlet relative mach number is %3.3f\n(2)The impeller total head efficiency is %0.1f %%'%(M1r,nc*100)

(1)The inlet relative mach number is 0.732
(2)The impeller total head efficiency is 90.9 %


## Ex 3.5 Page 100¶

In [5]:
#input data
P=4#The total pressure ratio
P01=1#The atmospheric pressure in bar
T01=298#THe atmospheric temperature in K
Dh=0.16#The hub diameter at impeller eye in m
Ca=120#The axial velocity at inlet in m/s
C1=Ca#The absolute velocity at inlet in m/s
sp=0.7#The pressure coefficient
C3=120#The absolute velocity at diffuser exit in m/s
m=8.3#The mass flow rate in kg/s
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K

#calculations
T1=T01-((C1**2)/(2*Cp))#The inlet temperature in K
P1=P01*(T1/T01)**(r/(r-1))#The inlet pressure in bar
d1=(P1*10**5)/(R*T1)#The inlet density of air in kg/m**3
Dt=(((4*m)/(3.14*d1*Ca))+(0.16**2))**(1/2)#The eye tip diameter in m
T=((T01)*((P**((r-1)/r))-1))/nc#The overall change in temperature in K
ssps=sp/nc#The product of slip factor and power factor
U2=(T*Cp/ssps)**(1/2)#Peripheral velocity of impeller top at exit in m/s
D2=(U2*60)/(3.14*N)#The impeller tip diameter in m
Uh=(3.14*Dh*N)/60#Peripheral velocity of eye hub in m/s
bh=degrees(atan(C1/Uh))#Blade angle at eye hub in degree
Ut=(3.14*Dt*N)/60#Peripheral velocity of eye tip in m/s
bt=degrees(atan(C1/Ut))#Blade angle at eye tip in degree
T03=T01+T#Temperature at the exit in K
T3=T03-((C3**2)/(2*Cp))#Exit static temperature in K
P3=(P*P01)*(T3/T03)**(r/(r-1))#Exit static pressure in bar
W=m*Cp*(T03-T01)*10**-6#Power required to drive the compressor in mW
#output
print '(a)The main dimensions of the impeller are\n    (1)Eye tip diameter is %3.3f m\n    (2)Impeller tip diameter is %3.3f m\n    (3)Blade angle at the eye hub is %3.2f degree\n       Blade angle at the eye tip is %3.2f degree\n(b)    (1)The static exit temperature is %3.1f K\n    (2)The static exit pressure is %3.3f bar\n(c)The power required is %3.3f mW'%(Dt,D2,bh,bt,T3,P3,W)

(a)The main dimensions of the impeller are
(1)Eye tip diameter is 0.325 m
(2)Impeller tip diameter is 0.528 m
(3)Blade angle at the eye hub is 40.98 degree
Blade angle at the eye tip is 23.15 degree
(b)    (1)The static exit temperature is 476.5 K
(2)The static exit pressure is 3.796 bar
(c)The power required is 1.549 mW


## Ex 3.6 Page 102¶

In [6]:
from math import sin
#input data
Dt=0.25#Tip diameter of the eye in m
Dh=0.1#Hub diameter of the eye in m
N=120#Speed of the compressor in rps
m=5#Mass of the air handled in kg/s
P01=102#Inlet stagnation pressure in kPa
T01=335#Inlet total temperature in K
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K

#calculations
d1=(P01*10**3)/(R*T01)#Density at the inlet of inducer in kg/m**3
Dm=(Dh+Dt)/2#Mean impeller diameter in m
C1=m/(d1*3.14*Dm*b)#Axial velocity component at the inlet in m/s
T11=T01-((C1**2)/(2*Cp))#Inlet temperature in K
P11=P01*(T11/T01)**(r/(r-1))#Inlet pressure in kPa
d11=(P11*10**3)/(R*T11)#Inlet density with mean impeller diameter an blade height in kg/m**3
C11=m/(d11*3.14*Dm*b)#Axial velocity component at inlet using mean blade values in m/s
T12=T01-((C1**2)/(2*Cp))#Initial temperature using modified axial velocity in K
P12=P01*(T12/T01)**(r/(r-1))#Initial pressure at inlet usin modified axial velocity in kPa
d12=(P12*10**3)/(R*T12)#Inlet density with modified axial velocity in kg/m**3
C12=m/(d12*3.14*Dm*b)#Axial velocity component at inlet using modified axial velocity in m/s
U1=3.14*Dm*N#Peripheral velocity of impeller at inlet in m/s
b1=degrees(atan(C12/U1))#The blade angle at impeller inlet in degree
W11=C12/sin(b1*pi/180)#Relative velocity at inlet in m/s
Mr11=W11/(r*R*T12)**(1/2)#Initial relative mach number
Ca=C12#Axial velocity at IGV in m/s
W12=Ca#Relative velocity at inlet usin IGV in m/s
a1=degrees(atan(Ca/U1))#Air angle at IGV exit in degree
C13=Ca/sin(a1*pi/180)#The velocity of flow of air at inlet in m/s
T13=T01-((C13**2)/(2*Cp))#Initial temperature using IGV in K
Mr12=W12/(r*R*T13)**(1/2)#Initial relative mach number using IGV

#output5
print '(1)Without using IGV\n    (a)The air angle at inlet of inducer blade is %3.2f degree\n    (b)The inlet relative mach number is %3.3f\n(2)With using IGV\n    (a))The air angle at inlet of inducer blade is %3.2f degree\n    (b)The inlet relative mach number is %3.3f'%(b1,Mr11,a1,Mr12)

(1)Without using IGV
(a)The air angle at inlet of inducer blade is 61.23 degree
(b)The inlet relative mach number is 0.377
(2)With using IGV
(a))The air angle at inlet of inducer blade is 61.23 degree
(b)The inlet relative mach number is 0.332


## Ex 3.7 Page 105¶

In [7]:
#input data
Cr2=28#Radial component of velocity at impeller exit in m/s
ss=0.9#The slip factor
U2=350#The impeller tip speed in m/s
A=0.08#The impeller area in m**2
T01=288#The ambient air temperature in K
P01=1#The ambient air pressure in bar
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K

#calculations
Cx2=ss*U2#outlet absolute velocity of air in tangential direction in m/s
C2=((Cx2**2)+(Cr2**2))**(1/2)#Axial velocity component at the outlet in m/s
T=(ss*(U2**2))/Cp#Total change in temperature in K
T02=T+T01#The final ambient air temperature in K
T2=T02-((C2**2)/(2*Cp))#The actual final air temperature in K
M2=(C2)/(r*R*T2)**(1/2)#Exit absolute mach number
P=((1+(ss*T/T01))**(r/(r-1)))#The overall pressure ratio
P02=P*P01#The final ambient pressure in bar
P2=P02*(T2/T02)**(r/(r-1))#The absolute final pressure in bar
d2=(P2*10**5)/(R*T2)#The final density of air at exit in kg/m**3
m=d2*A*Cr2#The mass flow rate in kg/s

#output
print '(a)The exit absolute mach number is %3.4f\n(b)The mass flow rate is %3.4f kg/s'%(M2,m)

(a)The exit absolute mach number is 0.8458
(b)The mass flow rate is 3.9423 kg/s


## Ex 3.8 Page 107¶

In [8]:
#input data
Dh=0.175#Hub diameter of the eye in m
Dt=0.3125#Tip diameter of the eye in m
m=20#Mass of the air handled in kg/s
N=16000#Speed of the compressor in rpm
T01=288#The ambient air temperature in K
P01=100#The ambient air pressure in kPa
Ca=152#The axial component of inlet velocity of eye in m/s
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K

#calculations
A=(3.14/4)*((Dt**2)-(Dh**2))#Annulus area of flow at the impeller eye in m**2
Ut=(3.1415*Dt*N)/60#Impeller eye tip speed in m/s
Uh=(3.1415*Dh*N)/60#Impeller eye hub speed in m/s
a1=90-20#Blade angle at inlet in degree
C1=Ca/sin(a1*pi/180)#The air entry velocity into the impeller eye in m/s
T1=T01-((C1**2)/(2*Cp))#The actual inlet air temperature in K
P1=P01*(T1/T01)**(r/(r-1))#The actual inlet air pressure in kPa
d1=P1/(R*T1)#The initial density of air at entry in kg/m**3
b1h=degrees(atan(Ca/(Uh-(Ca/tan(a1*pi/180)))))#Impeller angle at the hub in degree
b1t=degrees(atan(Ca/(Ut-(Ca/tan(a1*pi/180)))))#Impeller angle at the tip of eye in degree
Cx1=Ca/tan(a1*pi/180)#Inlet absolute velocity of air in tangential direction in m/s
Wx1=Ut-Cx1#Relative whirl component of velocity at inlet in m/s
W1=((Wx1**2)+(Ca**2))**(1/2)#Relative velocity at inlet in m/s
Mr1=W1/(r*R*T1)**(1/2)#Maximum mach number at the eye

#output
print '(a)\n    (1)The impeller eye tip speed is %3.2f m/s\n    (2)The impeller eye hub speed is %3.2f m/s\n    (3)The impeller angle at the hub is %i degree\n    (4)Impeller angle at the tip of eye is %3.2f degree\n(b)The maximum mach number at the eye is %3.2f'%(Ut,Uh,b1h,b1t,Mr1)

(a)
(1)The impeller eye tip speed is 261.79 m/s
(2)The impeller eye hub speed is 146.60 m/s
(3)The impeller angle at the hub is 59 degree
(4)Impeller angle at the tip of eye is 36.36 degree
(b)The maximum mach number at the eye is 0.77


## Ex 3.9 Page 109¶

In [9]:
#input data
P1=100#The air in take pressure in kPa
T1=309#The air in take temperature in K
H=0.750#Pressure head developed in mm W.G
P=33#Input power to blower in kW
nb=0.79#Blower efficiency
nm=0.83#Mechanical efficiency
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure 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
d=(P1*10**3)/(R*T1)#Density of air flow at inlet in kg/m**3
dP=dw*g*H#Total change in pressure in N/m**2
IW=dP/d#Ideal work done in J/kg
Wm=IW/nb#Actual work done per unit mass flow rate in J/kg
W=P*nm#Actual power input in kW
m=(W*10**3)/Wm#Mass flow rate in kg/s
Q=m/d#Volume flow rate in m**3/s
P2=P1+(dP/10**3)#The exit pressure of air in kPa
T2=T1+(Wm/(Cp))#The exit temperature of air in K

#output
print '(a)The mass flow rate of air is %3.3f kg/s\n(b)The volume flow rate of air is %3.2f m**3/s\n(c)\n    (1)The exit pressure of air is %3.2f kPa\n    (2)The exit temperature of air is %3.2f K'%(m,Q,P2,T2)

(a)The mass flow rate of air is 3.316 kg/s
(b)The volume flow rate of air is 2.94 m**3/s
(c)
(1)The exit pressure of air is 107.36 kPa
(2)The exit temperature of air is 317.22 K


## Ex 3.10 Page 110¶

In [10]:
#input data
H=0.075#Pressure developed by a fan in m W.G
D2=0.89#The impeller diameter in m
N=720#The running speed of the fan in rpm
b22=39#The blade air angle at the tip in degree
b2=0.1#The width of the impeller in m
Cr=9.15#The constant radial velocity in m/s
d=1.2#Density of air in kg/m**3
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure 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
IW=(dw*g*H)/d#Ideal work done in J/kg
U2=(3.1415*D2*N)/60#The impeller tip speed in m/s
Wx2=Cr/tan(b22*pi/180)#Relative whirl component of velocity at outlet in m/s
Cx2=U2-(Wx2)#Outlet absolute velocity of air in tangential direction in m/s
Wm=U2*Cx2#Actual work done per unit mass flow rate in J/kg
nf=IW/Wm#Fan efficiency
Q=3.1415*D2*b2*Cr#The discharge of the air by fan in m**3/s
m=d*Q#Mass flow rate of the air by the fan in kg/s
W=m*Wm*10**-3#Power required to drive the fan in kW
R=1-(Cx2/(2*U2))#Stage reaction of the fan
sp=2*Cx2/U2#The pressure coefficient

#output
print '(a)The fan efficiency is %0.1f %%\n(b)The Discharge of air by the fan is %3.3f m**3/s\n(c)The power required to drive the fan is %3.4f kW\n(d)The stage reaction of the fan is %0.2f %%\n(e)The pressure coefficient of the fan is %3.3f'%(nf*100,Q,W,R*100,sp)

(a)The fan efficiency is 82.1 %
(b)The Discharge of air by the fan is 2.558 m**3/s
(c)The power required to drive the fan is 2.2919 kW
(d)The stage reaction of the fan is 66.84 %
(e)The pressure coefficient of the fan is 1.326


## Ex 3.11 Page 111¶

In [11]:
#input data

b22=30#The blade air angle at the tip in degrees
D2=0.466#The impeller diameter in m
Q=3.82#The discharge of the air by fan in m**3/s
m=4.29#Mass flow rate of the air by the fan in kg/s
H=0.063#Pressure developed by a fan in m W.G
pi2=0.25#Flow coefficient at impeller exit
W=3#Power supplied to the impeller in kW
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1005#The specific heat of air at constant pressure in J/kg.K
g=9.81#Acceleration due to gravity in m/s**2
dw=10**3#Density of water in kg/m**3

#calculations
IW=Q*dw*g*H*(10**-3)#Ideal work done in kW
nf=IW/W#Fan efficiency
U2=(((W*10**3)/m)/(1-(pi2/tan(b22*pi/180))))**(1/2)#The impeller tip speed in m/s
Cr2=pi2*U2#The radial velocity at exit in m/s
Cx2=U2-(Cr2/tan(b22*pi/180))#Outlet absolute velocity of air in tangential direction in m/s
sp=2*Cx2/U2#Presuure coefficient of the fan
R=1-(Cx2/(2*U2))#Degree of reaction of the fan
N=(U2*60)/(3.141592*D2)#Rotational speed of the fan in rpm
b2=Q/(3.14*D2*Cr2)#Impeller width at the exit in m

#output
print '(a)The fan efficiency is %0.1f %%\n(b)The pressure coefficient is %3.3f\n(c)The degree of reaction of the fan is %0.1f %%\n(d)The rotational speed of the fan is %3.1f rpm\n(e)The impeller width at exit is %0.1f cm'%(nf*100,sp,R*100,N,b2*100)

(a)The fan efficiency is 78.7 %
(b)The pressure coefficient is 1.134
(c)The degree of reaction of the fan is 71.7 %
(d)The rotational speed of the fan is 1439.3 rpm
(e)The impeller width at exit is 29.7 cm


## Ex 3.12 Page 113¶

In [12]:
#input data
N=3000#The running speed of the blower in rpm
D2=0.75#The impeller diameter in m
Cr2=57#The radial velocity at exit in m/s
Cx1=0#Inlet absolute velocity of air in tangential direction in m/s
DR=0.58#Degree of reaction of the blower
nc=0.75#Total-to-total efficiency
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1.005#The specific heat of air at constant pressure in J/kg.K
T01=298#The inlet stagnation temperature in K
P01=1*101.325#The inlet stagnation pressure in kPa

#calculations
U2=(3.1415*D2*N)/60#The impeller tip speed in m/s
Cx2=2*(1-DR)*U2#Outlet absolute velocity of air in tangential direction in m/s
Wx2=U2-Cx2#Relative whirl component of velocity at outlet in m/s
b22=degrees(atan(Cr2/Wx2))#The blade air angle at the tip in degree
Wm=U2*Cx2*10**-3#Actual work done per unit mass flow rate when Cx1=0 in kW/(kg/s)
T=Wm/Cp#Total change in temperature in blower in K
P=(1+(nc*(T/T01)))**(r/(r-1))#Total pressure ratio in the blower
P02=P*P01#The outlet stagnation pressure from blower in kPa

#output
print '(a)The exit blade angle is %3.1f degree\n(b)The power input to the blower is %3.3f kW/(kg/s)\n(c)The exit stagnation pressure is %3.2f kPa'%(b22,Wm,P02)

(a)The exit blade angle is 71.7 degree
(b)The power input to the blower is 11.658 kW/(kg/s)
(c)The exit stagnation pressure is 112.06 kPa


## Ex 3.13 Page 114¶

In [13]:
#input data
D1=0.18#The impeller inner diameter in m
D2=0.2#The impeller outer diameter in m
C1=21#The absolute velocity at the entry in m/s
C2=25#The absolute velocity at the exit in m/s
W1=20#The relative velocity at the entry in m/s
W2=17#The relative velocity at the exit in m/s
N=1450#The running speed of the fan in rpm
m=0.5#The mass flow rate of the air in fan in kg/s
nm=0.78#The motor efficiency of the fan
d=1.25#The density of the air in kg/m**3
r=1.4#The ratio of specific heats of air
R=287#The universal gas constant in J/kg.K
Cp=1.005#The specific heat of air at constant pressure in J/kg.K

#calculations
U1=(3.14*D1*N)/60#Peripheral velocity of impeller at inlet in m/s
U2=(3.14*D2*N)/60#The impeller tip speed in m/s
dH=(((U2**2)-(U1**2))/2)+(((W1**2)-(W2**2))/2)#The actual total rise in enthalpy in kJ/kg
dH0=dH+(((C2**2)-(C1**2))/2)#The stage total isentropic rise in enthalpy in kJ/kg
dP0=d*dH0#The stage total pressure rise in N/m**2
dP=d*dH#The actual total rise in pressure in N/m**2
R=dP/dP0#The degree of reaction of the  fan
W=m*(dH0)#The work done by the fan per second in W
P=W/nm#The power input to the fan in W

#output
print '(a)The stage total pressure rise is %3.1f N/m**2\n(b)The degree of reaction of the fan is %3.3f\n(c)The power input to the fan is %3.1f W'%(dP0,R,P)

(a)The stage total pressure rise is 211.7 N/m**2
(b)The degree of reaction of the fan is 0.457
(c)The power input to the fan is 108.6 W


## Ex 3.14 Page 116¶

In [14]:
#input data
dH=0.14#Rise in static pressure of the air by fan in m of water
N=650#The running speed of the fan in rpm
P=85*0.735#Power consumed by the fan in kW
H1=0.75#The static pressure of the air at the fan in m of Hg
T1=298#The static pressure at the fan of air in K
m=260#Mass flow rate of air in kg/min
dHg=13590#Density of mercury in kg/m**3
dw=1000#Density of water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2
R=287#The universal gas constant in J/kg.K

#calculations
P1=dHg*g*H1*10**-3#The inlet static pressure in kPa
dP=dw*g*dH*10**-3#The total change in static pressures at inlet and outlet in kPa
P2=P1+dP#The exit static pressure in kPa
d1=(P1*10**3)/(R*T1)#The inlet density of the air in kg/m**3
Q=m/d1#The volume flow rate of air in fan in m**3/min

#output
print '(a)The exit static pressure of air in the fan is %3.2f kPa\n(b)The volume flow rate of the air is %3.1f m**3/min'%(P2,Q)

(a)The exit static pressure of air in the fan is 101.36 kPa
(b)The volume flow rate of the air is 222.4 m**3/min