# Chapter 7 : Steam generators¶

## Ex: 7.1 Pg: 510¶

In :
from math import sqrt
#Input data
p1=10#Initial pressure in bar
T1=300+273#Initial temperature in K
p2=2#Final pressure in bar
m=1#Mass flow rate of steam in kg/s

#Calculations
px=(0.546*p1)#Critical pressure in bar
ho=3052.2#Enthalpy in kJ/kg
so=7.1229#Entropy in kJ/kg.K
sx=so#Entropy in kJ/kg.K
hx=2905.9#Enthalpy in kJ/kg
vx=0.4125#Specific volume in m**3/kg
Vx=(44.72*sqrt(ho-hx))#Critical velocity in m/s
Ax=(vx/Vx)*10**4#Minimum area of the nozzle in sq.cm

#Output
print "Minimum area of the nozzles is %3.3f sq.cm"%(Ax)

Minimum area of the nozzles is 7.626 sq.cm


## Ex: 7.2 Pg: 511¶

In :
from math import sqrt
#Input data
p1=10#Initial pressure in bar
T1=300+273#Initial temperature in K
p2=1#Final pressure in bar
x=0.15#Friction loss of the isentropic enthalpy drop
ms=1#Steam flow rate in kg/s
d=25#Exit diameter of the nozzles in mm

#Calculations
px=(0.546*p1)#Critical pressure in bar
h1=3052.2#Enthalpy in kJ/kg
s1=7.1276#Entropy in kJ/kg
s2s=s1#Entropy in kJ/kg
h2s=2916.2#Enthalpy in kJ/kg
Vx=(44.72*sqrt(h1-h2s))#Critical velocity in m/s
h3s=2605#Enthalpy in kJ/kg
V1=(44.72*sqrt((h1-h2s)+(0.85*(h2s-h3s))))#Velocity in m/s
s3s=s1#Entropy in kJ/kg
x3s=(s3s-1.3025)/6.0579#Dryness fraction
h3s=(417.46+(x3s*2258.01))#Enthalpy in kJ/kg
h2s3=((1-x)*(h2s-h3s))#Enthalpy in kJ/kg
h3=h2s-h2s3#Enthalpy in kJ/kg
x3=(h3-417.46)/2258.01#Dryness fraction
v3=(0.001043+(x3*1.694))#Specific volume in m**3/kg
v2s=0.416#Specific volume in m**3/kg
vx=v2s#Specific volume in m**3/kg
Ax=(ms/Vx)*vx*10**4#Minimum area in cm**2
A1=(ms*v3)/V1*10**4#Area in cm**2
n=(A1*4)/(3.14*(d/10)**2)#Number of nozzles

#Output
print " Minimum area of the nozzles is %3.2f cm**2 \n the number of nozzles are %3.0f"%(Ax,n)

 Minimum area of the nozzles is 7.98 cm**2
the number of nozzles are   4


## Ex: 7.3 Pg: 512¶

In :
from math import sqrt
#Input data
p1=7.8#Pressure in bar
t1=180+273#Temperature in K
p2=1.03#pressure in bar
m=3.6#flow rate of air in kg/s
g=1.4#Ratio of specific heats
R=287#Characteristic gas constant in J/kg.K
cp=1.005#Specific heat in kJ/kg.K

#Calculations
pxpo=(2/(g+1))**(g/(g-1))#Ratio of pressure
px=pxpo*p1#Critical pressure in bar
txto=(2/(g+1))#Ratio of temperatures
tx=t1*txto#Critical temperature in K
vx=(R*tx)/(px*10**5)#Critical specific volume in m**3/kg
Vx=sqrt(g*R*tx)#Critical velocity in m/s
Ax=((m*vx)/Vx)*10**6#Critical area in mm**2
tot1=(p1/p2)**((g-1)/g)#Ratio of temperatures
t1i=t1/tot1#Temperature in K
v1=(R*t1i)/(p2*10**5)#Specific volume in m**3/kg
V1=44.72*sqrt(cp*(t1-t1i))#Velocity in m/s
A1=((m*v1)/V1)*10**6#Area in mm**2

#Output
print " Area of throat is %3.0f mm**2 \n Exit area is %d mm**2"%(Ax,A1)

 Area of throat is 2430 mm**2
Exit area is 4029 mm**2


## Ex: 7.4 Pg: 514¶

In :
from math import sqrt
#Input data
p1=3.8#pressure in bar
T1=450+273#Tempereture in K
p2=1#pressure in bar
m=16#Flow rate in kg/s
Cd=0.98#coefficient of discharge
nv=0.93#nozzile effeciency
cp=1.11#Specific heat in kJ/kg.K
g=1.333#Ratio of specific heats

#Calculations
pxpo=(2/(g+1))**(g/(g-1))#Pressure ratio
px=pxpo*p1#Critical pressure in bar
TxTo=2/(g+1)#Temperature ratio
Tx=T1*TxTo#Critical temperature in K
Vx=44.72*sqrt(cp*(T1-Tx))#critical velocity in m/s
R=(cp*(g-1)*1000)/g#Characteristic gas constant in J/kg.K
vx=(R*Tx)/(px*10**5)#Critical specific volume in m**3/kg
ws=(m/Cd)#Mass flow rate in kg/s
Ax=(ws*vx)/Vx#Critical area in m**2
T1sTo=(p2/p1)**((g-1)/g)#Temperature ratio
T1s=T1*T1sTo#Temperature in K
T1i=(T1-(nv*(T1-T1s)))#Temperature in K
v1=(R*T1i)/(p2*10**5)#Specific volume in m**3/kg
V1=44.72*sqrt(cp*(T1-T1i))#Velocity in m/s
A1=(ws*v1)/V1#Area in m**2

#Output
print " Throat raea is %3.4f m**2 \n Exit arae is %3.4f m**2"%(Ax,A1)

 Throat raea is 0.0286 m**2
Exit arae is 0.0370 m**2


## Ex: 7.5 Pg: 515¶

In :
from math import sqrt
#Input data
p1=20#pressure in bar
T1=300+273#Tempereture in K
p2=3#pressure in bar
m=0.3#Flow rate in kg/s
Cd=0.98#Coefficient of discharge
Cv=0.92#Coefficient of velocity

#Calculations
vo=0.1255#Specific volume in m**3/kg
px=(0.546*p1)#Critical pressure in bar
vx=(p1/px)**(1/n)*vo#Critical specific volume in m**3/kg
Vx=sqrt(n*px*10**5*vx)#Critical velocity in m/s
Ax=((m*vx)/Vx)*10**6#Critical area in m**2
v1vo=(p1/p2)**(1/n)#Ratio of specific volumes
v1=(vo*v1vo)#Specific volume in m**3/kg
V1=sqrt(2*((n/(n-1))*10**5*((p1*vo)-(p2*v1))))#Velocity in m/s
A1=((m*v1)/V1)*10**6#Area in mm**2
ho=3050#Enthalpy in kJ/kg
hx=2920#Enthalpy in kJ/kg
h1s=2650#Enthalpy in kJ/kg
ws=(m/Cd)#Flow rate in kg/s
Vsx=44.72*sqrt(ho-hx)#Velocity in m/s
V1s=44.72*sqrt(ho-h1s)#Velocity in m/s
Vo1=(V1s*Cv)#Velocity in m/s
hoh1=(V1/44.72)**2#Change in enthalpy in kJ/kg
h1=ho-hoh1#Enthalpy in kJ/kg
x1=(h1-561.47)/2163.8#Dryness fraction
vo1=(0.001073+(x1*0.6047))#Specific volume in m**3/kg
Ao1=((ws*vo1)/Vo1)*10**6#Exit nozzle area in mm**2
Vox=(Vsx*Cv)#Velocity in m/s
hohx=(Vox/44.72)**2#Change in enthalpy in kJ/kg
hox=(ho-hohx)#Enthalpy in kJ/kg
vox=0.22#Specific volume in m**3/kg
Aox=((ws*vox)/Vox)*10**6#Critical area in m**2

#Output
print "(a) Area of throat is %3.1f mm**2 \n Exit area is %3.1f mm**2 \n\n (b) Area of throat is %3.1f mm**2 \n Exit area is %3.1f mm**2"%(Ax,A1,Aox,Ao1)
#In textbook, Ao1 is given wrong.

(a) Area of throat is 112.6 mm**2
Exit area is 184.5 mm**2

(b) Area of throat is 143.6 mm**2
Exit area is 219.0 mm**2


## Ex: 7.6 Pg: 517¶

In :
from math import sqrt
#Input data
p1=5#Pressure of steam in bar
V=100#Velocity in m/s
p2=1.5#Exit pressure in bar
At=1280#Throat area in mm**2
Ae=1600#Exit area in mm**2
rp=0.58#Critical pressure ratio

#Calculations
ho=2749#Enthalpy in kJ/kg
so=6.822#Entropy in kJ/kg.K
px=(rp*p1)#Critical pressure in bar
sx=so#Entropy in kJ/kg.K
xx=(sx-1.660)/5.344#Dryness fraction
hx=(556+(xx*2168))#Enthalpy in kJ/kg
Vx=sqrt(((ho+((V**2*10**-3))/2)-hx)*(2/10**-3))#Velocity in m/s
vx=(xx*0.6253)#Specific volume in m**3/kg
w=(At*10**-6*Vx)/vx#Mass flow rate in kg/s
s1s=sx#Entropy in kJ/kg.K
x1s=(so-1.434)/5.789#Dryness fraction
h1s=(467+x1s*2226)#ENthalpy in kJ/kg
z=((Vx**2*10**-3)/2)-hx#z value
#By iteratio scheme
x1=0.932#Dryness fraction
v1=1.080#Specific volume in m**3/kg
h1=2542#Enthalpy in kJ/kg
V1=652.2#Velocity in m/s
nn=((hx-h1)/(hx-h1s))#Nozzle efficiency

#Output
print " Mass flow rate is %3.3f kg/s \n Nozzle efficiency is %3.3f"%(w,nn)

 Mass flow rate is 0.966 kg/s
Nozzle efficiency is 0.971


## Ex: 7.7 Pg: 819¶

In :
from math import sqrt
#Input data
p1=5#Pressure in bar
T1=200+273#Temperature in K
p2=2#Pressure in bar
m=0.3#Mass flow rate in kg/s

#Calculations
vo=0.4249#Specific volume in m**3/kg
ho=2855.4#Enthalpy in kJ/kg
so=7.0592#Entropy in kJ/kg.K
x1=0.972#Dryness fraction
h1=(504.7+x1*2201.9)#Enthalpy in kJ/kg
v1=x1*0.8857#Specific volume in m**3/kg
V1=44.72*sqrt(ho-h1)#Velocity in m/s
A1=((m*v1)/V1)*10**6#Area in mm**2
rp=(p1/p2)**(1/n)#Specific volume ratio
vR=(vo*rp)#Specific volume in m**3/kg
VR=sqrt(2*((n/(n-1))*(p1*vo-p2*vR)*10**5))#Velocity in m/s
AR=((m*vR)/VR)*10**6#Area in mm**2
TR=T1/(p1/p2)**((n-1)/n)#Temperature in K
tR=(TR-273)#Temperature in degree C
ts=120.23#Saturation temperature at pressure p1 in degree C
ds=ts-tR#Degree of subcooling in degree C
ps=1.4327#Saturation pressure at tR in bar
dsu=(p2/ps)#Degree of supersaturation

#Output
print " (a) Exit area when the flow is in equilibrium throughout is %3.0f mm**2 \n (b) Exit area when the flow is supersaturated is %3.1f mm**2 \n (i) The degree of supercooling is %3.2f degree C \n (ii) The degree of supersaturation is %3.3f"%(A1,AR,ds,dsu)

 (a) Exit area when the flow is in equilibrium throughout is 398 mm**2
(b) Exit area when the flow is supersaturated is 435.4 mm**2
(i) The degree of supercooling is 10.38 degree C
(ii) The degree of supersaturation is 1.396


## Ex: 7.8 Pg: 520¶

In :
from math import sqrt
#Input data
p1=5#Pressure in bar
T1=200#Temperature in degree C
p2=2#Pressure in bar
m=0.3#Mass flow rate in kg/s
nn=0.92#Nozzle efficiency
cp=1.925#mean specific heat in kJ/kg.K
x=[0,2.308,1943]#pv*10**3 = 2.308(h-1943)

#Calculations
vo=0.4249#Specific volume in m**3/kg
ho=2855.4#Enthalpy in kJ/kg
so=7.0592#Entropy in kJ/kg.K
x1=0.972#Dryness fraction
h1=(504.7+x1*2201.9)#Enthalpy in kJ/kg
v1=x1*0.8857#Specific volume in m**3/kg
V1=44.72*sqrt(ho-h1)#Velocity in m/s
h=ho-h1#Change in enthalpy in kJ/kg
hoq=nn*h#Change in enthalpy in kJ/kg
VQ=44.72*sqrt(hoq)#Velocity in m/s
toq=(hoq/cp)#Temperature difference in degree C
tQ=(T1-toq)#Temperature in degree C
TQ=tQ+273#Temperature in K
vQ=((p1*100*vo)/(T1+273))*(TQ/T1)#Specific volume in m**3/kg
A1=((m*vQ)/VQ)*10**6#Area in mm**2
vQ=(x[(1)]*(ho-hoq-x[(2)]))/(10**3*p2)#Specific volume in m**3/kg
A11=((m*vQ)/VQ)*10**6#Area in mm**2

#Output
print "Exit area is %3.1f mm**2 which upon checking is %3.0f mm**2"%(A1,A11)

Exit area is 403.2 mm**2 which upon checking is 400 mm**2


## Ex: 7.9 Pg: 422¶

In :
from math import sqrt, atan,sin,cos,pi,degrees
#Input data
V1=1000#Speed in m/s
Vb=400#Peripheral velocity in m/s
a=20#Nozzle angle in degree
m=0.75#Mass flow in kg/s
f=80#Percentage reduction of relative velocity

#Calculations
V=342#Velocity from E7.9 in m/s
Vr1=V/sin(b1*pi/180)#Velocity in m/s
dVw=(2*Vr1*cos(b1*pi/180))#Velocity in m/s
Pt=(m*dVw)#Tangential thrust in N
WD=(Pt*Vb)/1000#Diagram power in kW
nD=(WD/(0.5*m*V1**2*10**-3))*100#Diagram efficiency in percent
Pa=0#Axial thrust in N
Vr2=(f/100)*Vr1#Velocity in m/s
Pa2=m*sin(b1*pi/180)*(Vr1-Vr2)#Axial thrust in N
WD2=(m*(Vr1+Vr2)*cos(b1*pi/180)*Vb)/1000#Diagram power in kW
nD2=(WD2/(0.5*m*V1**2*10**-3))*100#Diagram efficiency in percent

#Output
print " Blade Angle is %3.2f degrees \n\n Neglecting the friction effects \n Tangential force is %3.2f N \n Axial thrust is %d N \n Diagram efficiency is %3.1f percent \n\n Considering the friction effects \n Axial thrust is %3.1f N \n Diagram Power is %3.2f kW \n Diagram efficiency is %3.2f percent"%(b1,Pt,Pa,nD,Pa2,WD2,nD2)

 Blade Angle is 32.36 degrees

Neglecting the friction effects
Tangential force is 809.49 N
Axial thrust is 0 N
Diagram efficiency is 86.3 percent

Considering the friction effects
Axial thrust is 51.3 N
Diagram Power is 291.42 kW
Diagram efficiency is 77.71 percent


## Ex:7.10 Pg: 523¶

In :
from math import atan, sin, cos, degrees, ceil,pi
#Input data
a=20#Nozzle angle in degrees
V1=330#Velocity of steam in m/s
f=0.8#Friction factor
nn=0.85#Nozzle efficiency
p1=20#Pressure in bar
T1=250+273#Temperature in K
p2=0.07#Pressure in bar
rf=1.06#Reheat factor

#Calculations
Vr1=((V1*sin(a*pi/180))/sin(b1*pi/180))#Velocity in m/s
Vr2=(f*Vr1)#Velocity in m/s
dVw=(Vr1*cos(b1*pi/180))+(Vr2*cos(b2*pi/180))#Vecoity in m/s
WD=(dVw*Vb)/1000#Workdone in kJ/kg
nb1=((2*dVw*Vb)/V1**2)*100#Efficiency in percent
nst=(nn*nb1)#Efficiency in percent
nin=(nst*rf)*100#Efficiency in percent
h1=2902.3#Enthalpy in kJ/kg
s1=6.5466#Entropy in kJ/kg.K
x2s=(s1-0.5582)/7.7198#Dryness fraction
h2s=(163.16+x2s*2409.54)#Enthalpy in kJ/kg
h12=(0.7041*(h1-h2s))#Change in enthalpy in kJ/kg
n=ceil(h12/WD)#Number of stages

#Output
print " (a) Work done in the stage per kg of steam is %3.2f kJ/kg \n Stage efficiency is %3.1f percent \n\n (b) Number of stages are %3.0f"%(WD,nst,n)

 (a) Work done in the stage per kg of steam is 42.56 kJ/kg
Stage efficiency is 66.4 percent

(b) Number of stages are  15


## Ex: 7.10 Pg: 525¶

In :
from __future__ import division
from math import atan, sin, cos, degrees, ceil,pi
#Input data
d=800#Diameter in mm
N=3000#Speed in rpm
V1=300#Velocity in m/s
a=20#Nozzle angle in degrees
f=0.86#Frictional factor
T=140#Axial thrust in N

#Calculations
Vb=((3.14*(d/1000)*N)/60)#Velocity in m/s
Vr1=(V1*sin(a*pi/180))/sin(b1*pi/180)#Velocity in m/s
Vr2=f*Vr1#Velocity in m/s
w=(T/((Vr1*sin(b1*pi/180))-(Vr2*sin(b2*pi/180))))#Mass flow rate in kg/s
dVw=(Vr2*cos(b2*pi/180))+(Vr1*cos(b1*pi/180))#Velocity in m/s
P=(w*dVw*Vb*10**-3)#Power developed in kW

#Output
print "Power deveoped in the blading is %3.2f kW"%(P)

Power deveoped in the blading is 355.89 kW


## Ex: 7.12 Pg: 525¶

In :
from __future__ import division
from math import sqrt, cos, pi, atan,sin, degrees
#Input data
p1=15#Pressure in bar
T1=300+273#Temperature in K
p2=10#Pressure in bar
nn=95#Nozzle efficiency in percent
a=20#Nozzle angle in degrees
x=5#The blade exit angle is 5 degrees less than the inlet angle
f=0.9#Friction factor
m=1350#Steam flow rate in kg/h

#Calculations
h1=3038.9#Enthalpy in kJ/kg
s1=6.9224#Entropy in kJ/kg.K
s2=s1#Entropy in kJ/kg.K
t2s=250#Temperature in degree C
h2s=2943.1#Enthalpy in kJ/kg
V1=44.72*sqrt((nn/100)*(h1-h2s))#Velocity in m/s
Vb=V1*(cos(a*pi/180)/2)#Velocity in m/s
Vr1=((V1*sin(a*pi/180))/sin(b1*pi/180))#Velocity in m/s
Vr2=(f*Vr1)#Velocity in m/s
dVw=(Vr1*cos(b1*pi/180))+(Vr2*cos(b2*pi/180))#Velocity in m/s
dVa=(Vr1*sin(b1*pi/180))-(Vr2*sin(b2*pi/180))#Velocity in m/s
Pa=(m/3600)*dVa#Axial thrust in N
Pt=(m/3600)*dVw#Tangential thrust in N
WD=(Pt*Vb*10**-3)#Diagram Power in kW
dn=((WD*1000)/((1/2)*(m/3600)*V1**2))*100#Diagram efficiency in percent

#Output
print " (a) Axial thrust is %3.2f N \n  Tangential thrust is %3.2f N \n\n (b) Diagram Power is %3.3f kW \n\n (c) Diagram Efficiency is %3.1f percent"%(Pa,Pt,WD,dn)

 (a) Axial thrust is 11.56 N
Tangential thrust is 146.85 N

(b) Diagram Power is 29.437 kW

(c) Diagram Efficiency is 86.3 percent


## Ex: 7.13 Pg: 527¶

In :
from __future__ import division
from math import sqrt, cos, pi, atan,sin, degrees

#Input data
V1=600#Velocity in m/s
a=16#Nozzle angle in degrees
b2=18#Exit angle in degrees
aa1=22#Exit angle in degrees
b4=36#Exit angle in degrees
m=5#Steam flow rate in kg/s
f=0.85#Friction coefficient

#Calculations
b1=degrees(atan((V1*sin(a*pi/180))/((V1*cos(a*pi/180))-Vb)))#Exit angle in degrees
Vr1=((V1*sin(a*pi/180))/sin(b1*pi/180))#Velocity in m/s
Vr2=(f*Vr1)#Velocity in m/s
a1=degrees(atan((Vr2*sin(pi/180*b2))/((Vr2*cos(pi/180*b2))-Vb)))#Angle in degrees
V2=((Vr2*sin(pi/180*b2))/sin(pi/180*a1))#Velocity in m/s
V3=(f*V2)#Velocity in m/s
dVw1=(Vr1*cos(pi/180*b1))+(Vr2*cos(pi/180*b2))#Velocity in m/s
dVa1=(V1*sin(pi/180*a))-(V2*sin(pi/180*a1))#Velocity in m/s
b3=degrees(atan((V3*sin(pi/180*aa1))/((V3*cos(pi/180*aa1))-Vb)))#Angle in degrees
Vr3=((V3*sin(pi/180*aa1))/sin(pi/180*b3))#Velocity in m/s
Vr4=(f*Vr3)#velocity in m/s
dVw2=(Vr3*cos(pi/180*b3))+(Vr4*cos(pi/180*b4))#Velocity in m/s
dVa2=(V3*sin(pi/180*aa1))-(Vr4*sin(pi/180*b4))#Velocity in m/s
udVw=(dVw1+dVw2)#Total velocity in m/s
udVa=(dVa1+dVa2)#Total velocity in m/s
Pt=(m*udVw*10**-3)#tangential thrust in kN
Pa=(m*udVa*10**-3)#Axial thrust in kN
WD=(Pt*Vb)#Power developed in kW
nd=((2*udVw*Vb)/V1**2)*100#Diagram efficiency in percent

#Output
print " (a) the tangential thrust is %3.3f kW \n (b) Axial thrust is %3.2f kN \n (c) Power developed is %3.2f kW \n (d) Diagram efficiency is %3.2f percent"%(Pt,Pa,WD,nd)

 (a) the tangential thrust is 5.355 kW
(b) Axial thrust is 0.29 kN
(c) Power developed is 642.54 kW
(d) Diagram efficiency is 71.39 percent


## Ex: 7.14 Pg: 529¶

In :
from __future__ import division
#Input data
a=17#Nozzle angle in degrees
f=0.9#Friction factor
a2=90#Axial angle in degrees

#Calculations
dVw=1040#Velocity in m/s from Velocity triangles Fig. E.7.14
V1=575#Velocity in m/s from Velocity triangles Fig. E.7.14
V4=75#Velocity of steam exiting stage in m/s from Velocity triangles Fig. E.7.14
WD=(dVw*Vb)/1000#Diagram work in kJ/kg
nd=((WD*1000)/((1/2)*V1**2))*100#Diagram efficiency in percent

#Output
print " (a) Absolute velocity of steam leaving the stage is %3.0f m/s \n (b) the diagram work is %3.0f kJ/kg \n (c) the diagram efficiency is %3.2f percent"%(V4,WD,nd)

 (a) Absolute velocity of steam leaving the stage is  75 m/s
(b) the diagram work is 130 kJ/kg
(c) the diagram efficiency is 78.64 percent


## Ex: 7.15 Pg: 530¶

In :
from __future__ import division
from math import sqrt, atan,sin,cos,pi,degrees
#Input data
p1=35#Pressure in bar
T1=350+273#Temperature in K
p2=0.07#Pressure in bar
x=1/4#Fraction of drop in isentropic enthalpy
a=20#Nozzle angle in degrees
nn=88#Nozzle efficiency in percent
y=0.2#Velocity ratio
f=0.9#Friction coefficienct
In=75#Internal efficiency of the turbine in percent

#Calculations
h1=3106.4#Enthalpy in kJ/kg
s1=6.6643#Entropy in kJ/kg.K
x2s=(s1-0.5582)/7.7198#dryness fraction
h2s=(163.16+x2s*2409.54)#Enthalpy in kJ/kg
dh=(h1-h2s)#Change in enthalpy in kJ/kg
h13s=x*dh#Change in enthalpy in kJ/kg
h13=(nn/100)*h13s#Change in enthalpy in kJ/kg
V1=(44.72*sqrt(h13))#Velocity in m/s
Vb=(y*V1)#Velocity in m/s
b1=degrees(atan((V1*sin(pi/180*a))/((V1*cos(pi/180*a))-Vb)))#Angle in degrees
Vr1=((V1*sin(pi/180*a))/sin(pi/180*b1))#Velocity in m/s
Vr2=(f*Vr1)#Velocity in m/s
dVw1=(Vr1*cos(pi/180*b1))+(Vr2*cos(pi/180*b2))#Velocity in m/s
V2=sqrt((Vr2*sin(pi/180*b2))**2+((Vr2*cos(pi/180*b2))-Vb)**2)#Velocity in m/s
V3=f*V2#Velocity in m/s
b3=degrees(atan((V3*sin(pi/180*b2))/((V3*cos(pi/180*b2))-Vb)))#Angle in degrees
Vr3=((V3*sin(pi/180*b2))/sin(pi/180*b3))#Velocity in m/s
Vr4=f*Vr3#Velocity in m/s
dVw2=(Vr3*cos(pi/180*b3))+(Vr4*cos(pi/180*b4))#Velocity in m/s
dVw=(dVw1+dVw2)#Velocity in m/s
nb1=((2*dVw*Vb)/V1**2)*100#Efficiency in percent
ns=(nn*nb1)/100#Efficiency in percent
ht=(In/100)*dh#Total change in enthalpy in kJ/kg
hc=(ns/100)*h13s#Total change in enthalpy in kJ/kg
pp=(hc/ht)*100#Percentage of enthalpy

#Output
print " Efficiency of first stage is %3.2f percent \n Percentage of the total power developed by the turbine is %3.2f percent"%(ns,pp)

 Efficiency of first stage is 66.67 percent
Percentage of the total power developed by the turbine is 22.22 percent


## Ex: 7.16 Pg: 532¶

In :
from math import sqrt, atan,sin,cos,pi,degrees
#Input data
R=50#Percentage of reaction
b1=35#Angle in degrees
q=b1#Angle in degrees
b2=20#Angle in degrees
a=b2#Angle in degrees
N=1500#Speed in rpm
d=0.67#Mean diameter in m
p=1.5#Pressure in bar
x=0.96#Dryness fraction
w=3.6#Flow rate in kg/s

#Calculations
Vb=(3.14*d*N)/60#Velocity in m/s
V1=(Vb*(sin(pi/180*180-b1)/sin(pi/180*b1-b2)))#Veocity in m/s
Vr1=(Vb*(sin(pi/180*b2)/sin(pi/180*b1-b2)))#Velocity in m/s
dVw=(V1*cos(pi/180*a))+(Vr1*cos(pi/180*q))#Velocity in m/s
v1=(0.001052+x*1.15937)#Specific volume in m**3/kg
hb=((w*v1)/(3.14*d*V1*sin(pi/180*a)))*1000#Required height in mm
P=(w*dVw*Vb)/1000#Power developed in kW

#Output
print " (a) the required height of blading is %3.1f mm \n (b) the power developed by the ring is %3.3f kW"%(hb,P)

 (a) the required height of blading is 127.2 mm
(b) the power developed by the ring is 2.368 kW


## Ex: 7.17 Pg: 533¶

In :
from math import sqrt, atan,sin,cos,pi,degrees
#Input data
N=400#Speed in rpm
P=5#Power in MW
m=6#Flow rate in kg/kWh
a=b2#Angle in degrees
x=1.35#Velocity ratio
p=1.2#Pressure in bar
x1=0.95#Steam quality
Dh=12#Ratio of Dm and hb

#Calculations
w=(m*P*1000)/3600#Mass flow rate in kg/s
v1=(0.0010468+x1*1.454)#Specific volume in m**3/kg
Vb=((3.14*Dh*(hb/1000)*N)/60)#velocity in m/s
V1=(x*Vb)#Velocity in m/s
dVw=((2*V1*cos(pi/180*a))-Vb)#velocity in m/s
WD=(w*dVw*Vb*10**-3)#Diagram power in kW

#Output
print " (a) Blade height is %3.0f mm \n (b) the diagram power is %3.2f kW"%(hb,WD)

 (a) Blade height is 138 mm
(b) the diagram power is 15.42 kW


## Ex: 7.18 Pg: 534¶

In :
from math import sqrt, atan,sin,cos,pi,degrees
#Input data
N=3000#Speed in rpm
x=0.56#Velocity ratio
v=0.65#Specific volume in m**3/kg
h=25#Mean height in mm

#Calculations
V1=(Vb/x)#Velocity in m/s
Vr2=V1#Velocity in m/s
Dm=(Vb*60)/(3.14*N)#Diameter in m
w=((3.14*Dm*h*V1*sin(pi/180*a))/v)/1000#Mass flow rate in kg/s
ws=(w*3600)#Mass flow rate in kg/hr
Vr1=((V1*sin(pi/180*a))/sin(pi/180*b1))#Velocity in m/s
dhmb=(1/2)*(Vr2**2-Vr1**2)/1000#Change in enthalpy in kJ/kg
dsta=(2*dhmb)#Change in enthalpy of stage in kJ/kg
dsta5=(n*dsta)#Total Change in enthalpy of stage in kJ/kg
Dp=(w*dsta5)#Diagram power in kW

#Output
print " Mass flow rate of steam is %3.3f kg/s \n Useful enthalpy drop is %3.2f kJ/kg \n The diagram power is %3.1f kW"%(w,dsta5,Dp)

 Mass flow rate of steam is 4.698 kg/s
Useful enthalpy drop is 117.80 kJ/kg
The diagram power is 553.4 kW


## Ex: 7.19 Pg: 535¶

In :
from math import sqrt, atan,sin,cos,pi,degrees
#Input data
P=8#Power in MW
N=5000#Speed in rpm
p=40#pressure in bar
T=500#Temperature in degree C
p2=0.1#Pressure in bar
In=0.85#Internal efficiency of turbine
nm=0.96#Mechanical efficiency
nn=0.92#Nozzle efficiency
a=15#Nozzle angle in degrees

#Calculations
V1=(2*Vb)/cos(pi/180*a)#Velocity in m/s
dh=((V1/44.72)**2/nn)#Change in enthalpy in kJ/kg
h1=3445.3#Enthalpy in kJ/kg
s1=7.0901#Entropy in kJ/kg.K
s2=s1#Entropy in kJ/kg.K
x2=(s2-0.6493)/7.5009#Dryness fraction
h2s=(191.83+x2*2392.8)#Enthalpy in kJ/kg
h12s=(h1-h2s)#Change in enthalpy in kJ/kg
n=(h12s/dh)#Number of stages
w=((P*1000)/(In*nm))/h12s#Mass flow rate in kg/s
h13=(nn*dh)#Change in enthalpy in kJ/kg
h3=h1-h13#Enthalpy in kJ/kg
v3=0.17#Specific volume in m**3/kg
A=(w*v3)/V1#Area in m**2
hm=(A/(((Vb*60)/N)*sin(pi/180*a)))*1000#Height in mm

#Output
print " (a) the number of impulse stages are%3.0f \n (b) the nozzle height is %3.1f mm"%(n,hm)

 (a) the number of impulse stages are  6
(b) the nozzle height is 2.4 mm


## Ex: 7.20 Pg: 537¶

In :
from math import sqrt, atan,sin,cos,pi,degrees
from __future__ import division
#Input data
p=1.5#Pressure in bar
x1=0.9#Dryness fraction
m=7#Steam flow rate in kg/s
N=3000#Turbine speed in rpm
x=0.7#Velocity ratio
y=0.75#Velocity ratio
a=20#Exit angle in degrees
b2=a#Angle in degrees
hx=1/10#Fraction of height

#Calculations
v=0.001052+x1*1.15937#Specific volume in m**3/kg
Dm=((m*v*60)/(3.14*hx*y*3.14*N))**(1/3)#Diameter in m
hb=Dm*1000*hx#Height in mm
Vb=(3.14*Dm*N)/60#Velocity in m/s
dVw=((2*x*Vb*cos(pi/180*a)/sin(pi/180*a))-Vb)#Velocity in m/s
P=(m*dVw*Vb)/1000#Power developed in kW

#Output
print " Height of the moving blades at exit is %3.1f mm \n Power developed in the blade row is %3.2f kW"%(hb,P)

 Height of the moving blades at exit is 58.3 mm
Power developed in the blade row is 166.70 kW


## Ex: 7.21 Pg: 538¶

In :
from math import sqrt, atan,sin,cos,pi,degrees
#Input data
p=40#Pressure in bar
T=500#Temperature in degree C
p1=0.1#Pressure in bar
a=16#Nozzle angle in degrees
N=3000#Speed in rpm

#Calculations
h1=3445.3#Enthalpy in kJ/kg
s1=7.0901#Entropy in kJ/kg.K
s2=s1#Entropy in kJ/kg.K
x2s=(s2-0.6493)/7.5009#Dryness fraction
h2s=(191.83+x2s*2392.8)#Enthalpy in kJ/kg
V1=44.72*sqrt(h1-h2s)#Velocity in m/s
Vb=V1*(cos(pi/180*a)/2)#Velocity in m/s
Dm=(Vb*60)/(3.14*N)#Diameter in m
V2=44.72*sqrt((h1-h2s)/2)#Velocity in m/s
Vb2=V2*cos(pi/180*a)#Velocity in m/s
Dm2=(Vb2*60)/(3.14*N)#Diameter in m
V3=44.72*sqrt((h1-h2s)/4)#Velocity in m/s
Vb3=V3*(cos(pi/180*a)/2)#Velocity in m/s
Dm3=(Vb3*60)/(3.14*N)#Diameter in m
V4=44.72*sqrt(h1-h2s)#Velocity in m/s
Vb4=V4*(cos(pi/180*a)/4)#Velocity in m/s
Dm4=(Vb4*60)/(3.14*N)#Diameter in m
V5=44.72*sqrt((h1-h2s)/(2*4))#Velocity in m/s
Vb5=V5*cos(pi/180*a)#Velocity in m/s
Dm5=(Vb5*60)/(3.14*N)#Diameter in m

#Output
print "The mean diameter of the wheel if the turbine were of \n (a) single impulse stage is %3.2f m \n (b) single 50 percent reaction stage is %3.1f m \n (c) four pressure (or Rateau) stages is %3.2f m \n (d) one two-row Curtis stage is %3.3f m \n (e) four stage 50 percent reaction stages is %3.2f m"%(Dm,Dm2,Dm3,Dm4,Dm5)

The mean diameter of the wheel if the turbine were of
(a) single impulse stage is 4.74 m
(b) single 50 percent reaction stage is 6.7 m
(c) four pressure (or Rateau) stages is 2.37 m
(d) one two-row Curtis stage is 2.370 m
(e) four stage 50 percent reaction stages is 3.35 m


## Ex: 7.22 Pg: 539¶

In :
from math import sqrt, atan,sin,cos,pi,degrees,ceil
from __future__ import division
#Input data
p=150#Pressure in bar
T=600#Temperature in degree C
Vb=300#Velocity in m/s
nn=95#Nozzle efficiency in percent
a=15#Nozzle angle in degrees
a1=25#Angle in degrees

#Calculations
h1=3582.3#Enthalpy in kJ/Kg
s1=6.6776#Entropy in kJ/kg.K
s2=s1#Entropy in kJ/kg.K
x2s=(s2-0.6493)/7.5009#Dryness fraction
h2s=(191.83+x2s*2392.8)#Enthalpy in kJ/kg
h12s=(h1-h2s)#Difference in enthalpy in kJ/kg
V1=(Vb*2)/cos(pi/180*a)#Velocity in m/s
dhs=(V1/44.72)**2/(nn/100)#Change in enthalpy in kJ/kg
n1=ceil(h12s/dhs)#Number of stages
V2=(Vb/cos(pi/180*a1))#Velocity in m/s
dhs2=(V2/44.72)**2/(nn/(2*100))#Change in enthalpy in kJ/kg
n2=h12s/dhs2#Number of stages
V3=(Vb*4)/cos(pi/180*a)#Velocity in m/s
dhs3=(V3/44.72)**2/(nn/100)#Change in enthalpy in kJ/kg
dhhs3=(h12s-dhs3)#Change in enthalpy in kJ/kg
n3=dhhs3/dhs#Number of stages
n4=dhhs3/dhs2#Number of stages

#Output
print "Number of stages required in \n (a) all simple impulse stages are %3.0f \n (b) all 50 percent reaction stages are %3.0f \n (c) a 2-row Cutris stage follwed by simple impulse stages are %3.0f \n (d) a 2-row Cutris stage followed by 50 percent reaction stages are %3.0f"%(n1,n2,n3,n4)

Number of stages required in
(a) all simple impulse stages are   8
(b) all 50 percent reaction stages are  13
(c) a 2-row Cutris stage follwed by simple impulse stages are   3
(d) a 2-row Cutris stage followed by 50 percent reaction stages are   6


## Ex: 7.23 Pg: 541¶

In :
from __future__ import division
#Input data
p1=20#Pressure in bar
T=400#Temperature in degree C
p2=0.1#Pressure in bar
n=4#Number of stages
ns=75#Stage efficiency in percent

#Calculations
h16s=(3250-2282)#Change in enthalpy in kJ/kg
h12s=(h16s/n)#Change in enthalpy in kJ/kg
p=[0,8,2.6,0.6]#pressures in bar from Mollier chart
h12=(ns/100)*h12s#Change in enthalpy in kJ/kg
h23s=(3060-2800)#Change in enthalpy in kJ/kg
h23=(ns/100)*h23s#Change in enthalpy in kJ/kg
h34s=(2870-2605)#Change in enthalpy in kJ/kg
h34=(ns/100)*h34s#Change in enthalpy in kJ/kg
h45s=(2680-2410)#Change in enthalpy in kJ/kg
h45=(ns/100)*h45s#Change in enthalpy in kJ/kg
h5=2470#Enthalpy in kJ/kg
rf=(h12s+h23s+h34s+h45s)/h16s#Reheat factor
nth=((h12+h23+h34+h45)/h16s)*100#Internal efficiency in percent
nin=(ns*rf)#Internal efficiency in percent

#Output
print " The interstage pressures are %d bar, %3.1f bar, %3.1f bar \n The reheat factor is %3.3f \n The turbine internal efficiency is %3.1f percent"%(p[(1)],p[(2)],p[(3)],rf,nin)

 The interstage pressures are 8 bar, 2.6 bar, 0.6 bar
The reheat factor is 1.071
The turbine internal efficiency is 80.3 percent


## Ex: 7.25 Pg: 544¶

In :
from math import sqrt, atan,sin,cos,pi,degrees
#Input data
n=20#Number of stages
P=12#Power in MW
p=15#Pressure in bar
T=350#Temperature in degree C
p1=0.14#Pressure in bar
ns=75#Stage efficiency in percent
rf=1.04#Reheat factor
p2=1#Pressure in bar
a=20#Angle in degrees
v=0.7#Velocity ratio

#Calculations
nint=(ns/100)*rf#Internal efficiency
dhs=855#Enthalpy in kJ/kg
dha=ceil(nint*dhs)#Actual enthalpy change in kJ/kg
w=(P*1000)/dha#Mass flow rate in kg/s
Vb=(sqrt((dha/n)/((((2/v)*cos(pi/180*a))-1)*10**-3)))#Velocity in m/s#
vg=1.694#Specific volume in m**3/kg
Dm=sqrt((w*vg)/(3.14*h*(Vb/v)*sin(pi/180*a)))#Diameter in m
N=((Vb*60)/(3.14*Dm))#Speed in rpm

#Output
print " (a) the flow rate of steam required is %3.2f kg/s \n (b) the mean blade diameter is %3.3f m \n (c) the speed of the rotor is %3.0f rpm"%(w,Dm,N)
#In textbook, Vb is given wrong as 141.4 m/s instead of 140.6 m/s. Hence the answers are different.

 (a) the flow rate of steam required is 17.99 kg/s
(b) the mean blade diameter is 1.302 m
(c) the speed of the rotor is 2065 rpm


## Ex: 7.26 Pg: 546¶

In :
from __future__ import division
from math import sqrt, atan,sin,cos,pi,degrees
#Input data
V1=600#Velocity in m/s
a=16#Nozzle angle in degrees
b=[0,18,21,35]#Exit angles in degrees
m=5#Steam flow rate in kg/s
h=25#Nozzle height in mm
v=0.375#Specific volume in m**3/kg
p=25#Pitch in mm
t=0.5#Thickness in mm
kb=0.9#Constant

#Calculations
l=((m*v)/(sin(pi/180*a)*V1*(h/1000)*kb))#Length of the nozzle arc in m #Length of the nozzle arc is calculated wrong as 0.454m instead of 0.5 m
b1= degrees(atan((V1*sin(pi/180*a))/((V1*cos(pi/180*a))-Vb)))#Angle in degrees
Vr1=((V1*sin(pi/180*a))/sin(pi/180*b1))#Velocity in m/s
Vr2=kb*Vr1#Velocity in m/s
V2=sqrt(Vr2**2+Vb**2-2*Vr2*Vb*cos(pi/180*b[(1)]))#Velocity in m/s
V3=291#Velocity in m/s
b3=degrees(atan((V3*sin(pi/180*b[(2)]))/((V3*cos(pi/180*b[(2)]))-Vb)))#Angle in degrees
Vr3=((V3*sin(pi/180*b[(2)]))/sin(pi/180*b3))#Velocity in m/s
Vr4=(Vr3*kb)#Velocity in m/s
hb1=((m*v*(h/1000))/(l*((p/1000)*sin(pi/180*b[(1)])-(t/1000))*Vr2))*1000#Height in mm
hn=((m*v*(h/1000))/(l*((p/1000)*sin(pi/180*b[(2)])-(t/1000))*V3))*1000#Height in mm
hb2=((m*v*(h/1000))/(l*((h/1000)*sin(pi/180*b[(3)])-(t/1000))*Vr4))*1000#Height in mm

#Output
print " Blade heights at the exit of each row: \n First row of moving blades is %3.1f mm \n Fixed row of guide blades is %3.1f mm \n Second row of moving blades is %3.1f mm"%(hb1,hn,hb2)

 Blade heights at the exit of each row:
First row of moving blades is 29.4 mm
Fixed row of guide blades is 37.8 mm
Second row of moving blades is 40.6 mm


## Ex: 7.27 Pg: 548¶

In :
from math import sqrt, atan,sin,cos,pi,degrees, ceil
#Input data
P=200#Power in MW
p=180#Pressure in bar
T=550#Temperature in degree C
P1=600#Power in MW
p1=300#Pressure in bar
T1=580#Temperature in degree C
nt=90#Turbine efficiency in percent

#Calculations
h1=3430#Enthalpy in kJ/kg
h2s=3040#Enthalpy in kJ/kg
h12s=(h1-h2s)#Enthalpy in kJ/kg
h12=(nt/100)*h12s#Enthalpy in kJ/kg
h2=3070#Enthalpy in kJ/kg
v2=0.06#Specific volume in m**3/kg
h4=3560#Enthalpy in kJ/kg
h3s=2000#Enthalpy in kJ/kg
h13s=(h1-h3s)#Enthalpy in kJ/kg
h13=(nt/100)*h13s#Enthalpy in kJ/kg
w=(P*10**3)/h13#Mass flow rate in kg/s
Vbm=350#Velocity in m/s
N=3000#Speed in rpm
a=25#Angle in degrees
Dm=(Vbm*60)/(3.14*N)#Diameter in m
hD=0.3#Assuming (hb/Dm)max
hb=(hD*Dm)#Height in m
Ab=(3.14*Dm*hb*0.9*sin(pi/180*a))#Flow area in m**2
V1=(Vbm/cos(pi/180*a))#Velocity in m/s
Vo=(Ab*V1)#Volume flow rate in m**3/s
v=(Vo/w)#Specific volume in m**3/kg
h5s=2456#Enthalpy in kJ/kg
p5=0.36#Pressure in bar
h45s=(h4-h5s)#Enthalpy in kJ/kg
h45=(nt/100)*h45s#Enthalpy in kJ/kg
h5=h4-h45#Enthalpy in kJ/kg
x5=0.952#Dryness fraction
h56s=(h5-2340)#Enthalpy in kJ/kg
h56=(nt/100)*h56s#Enthalpy in kJ/kg
h6=h5-h56#Enthalpy in kJ/kg
v6=18#Specific volume in m**3/kg
mm=(Vo/v6)#Maximum mass flow that one stage can accommodate in kg/s
np=(w/mm)#Number of parallel exhausts
rp=(p1/4)#Reheat pressure in bar
xh1=3410#Enthalpy in kJ/kg
xh2s=3015#Enthalpy in kJ/kg
xh12s=xh1-xh2s#Enthalpy in kJ/kg
xh12=(nt/100)*xh12s#Enthalpy in kJ/kg
xv2=0.035#Specific volume in m**3/kg
xh4=3060#Enthalpy in kJ/kg
xh3s=1960#Enthalpy in kJ/kg
xh13s=xh1-xh3s#Enthalpy in kJ/kg
xh3=(xh1-xh13s)#Enthalpy in kJ/kg
xw=(P1*10**3)/xh13s#Mass flow rate in kg/s
xvm=(Vo/xw)#Maximum specific volume in m**3/kg
Vf=(xw*xv2)#Volume flow rate in m**3/s
xh5s=2300#Enthalpy in kJ/kg
xh45s=xh4-xh5s#Enthalpy in kJ/kg
xh45=(nt/100)*xh45s#Enthalpy in kJ/kg
xh5=xh4-xh45s#Enthalpy in kJ/kg
xv5=1.25#Specific volume in m**3/kg
xx5=0.86#Dryness fraction
xh6s=2050#Enthalpy in kJ/kg
xh56s=xh5-xh6s#Enthalpy in kJ/kg
xh56=(nt/100)*xh56s#Enthalpy in kJ/kg
xh6=(xh5-xh56)#Enthalpy in kJ/kg
xv6=12#Specific volume in m**3/kg
xx6=0.792#Dryness fraction
xmm=(Vo/xv6)#Maximum mass flow in kJ/kg
xnp=ceil(xw/xmm)#Number of parallel exhausts

#Output
print "Number of parallel exhausts in : \n (a)condition a are %d \n (b)condition b are %d"%(np,xnp)

Number of parallel exhausts in :
(a)condition a are 4
(b)condition b are 8


## Ex: 7.28 Pg: 552¶

In :
#Input data
P=100#Power in MW
T=550#temperature in degree C
p=0.1#Pressure in bar
m=500000#Mass flow rate in kg/h at rated load
mo=25000#Mass flow rate in kg/h at zero load

#Calculations
b=(m-mo)/(P*10**3)#Steam rate in kg/kWh
s1=(mo/y1)+b#Steam rate in kg/kWh
s2=(mo/y2)+b#Steam rate in kg/kWh
s3=(mo/y3)+b#Steam rate in kg/kWh
s4=(mo/y4)+b#Steam rate in kg/kWh
h1=3511#Enthalpy in kJ/kg
xs1=6.8142#Entropy in kJ/kg.K
xs2=xs1#Entropy in kJ/kg.K
x2s=(xs2-0.6493)/7.5009#Dryness fraction
h2s=191.83+x2s*2392.8#Enthalpy in kJ/kg
nR=((h1-h2s)/(h1-191.83))*100#Rankine efficiency in percent
nac=((P*10**3*3600)/(m*(h1-191.83)))*100#Actual efficiency in percent
nTG=((P*10**3*3600)/(m*(h1-h2s)))*100#Turbogenerator efficiency in percent

#Output
print "(a) Steam rate at: \n One-fourth load is %3.2f kg/kWh \n Half load is %3.2f kg/kWh \n Three-fourth load is %3.2f kg/kWh \n Full load is %3.1f kg/kWh \n\n (b) Rankine cycle efficiency is %3.1f percent \n (c) Actual efficiency at full load is %3.1f percent \n (d) The turbogenerator efficiency at full load is %3.1f percent"%(s1,s2,s3,s4,nR,nac,nTG)

(a) Steam rate at: