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)
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
b1=Q1/(2*3.14*(Dh/2)*C1)#The breadth of impeller blade at inlet in m
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
b2=Q2/(2*3.14*(Dt/2)*Cr2)#The breadth of impeller blade at outlet in m
#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)
#comments
#error in the first review is not printing the value of V2 which is corrected
#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)
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
nc=(T01/(T02-T01))*(((P02/P01)**((r-1)/r))-1)#Total head isentropic efficiency
#output
print '(1)The inlet relative mach number is %3.3f\n(2)The impeller total head efficiency is %0.1f %%'%(M1r,nc*100)
#input data
N=16500#The running speed ofradial blade of a centrifugal compressor in rpm
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
nc=0.78#The adiabatic 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=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)
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
b=(Dt-Dh)/2#Impeller blade height 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)
#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
nc=0.9#Total head isentropic efficiency
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)
#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)
#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)
#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)
#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)
#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)
#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)
#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)