CHAPTER 6: FLOW THROUGH NOZZLES-STEAM TURBINES

Example 6.1 Page 87

In [23]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p1=150#lb/in^2
p2=10#lb/in^2
n=10#percent
T=183.6+479.4#C.H.U
x2=0.852#C.H.U
H=553.9#C.H.U/lb
h1=T-H#C.H.U/lb

#CALCULATIONS
V=sqrt(2*32.2*1400*h1)#ft/sec
V1=sqrt(2*32.2*1400*0.9*h1)#ft/sec

#RESULTS
print"the neglecting friction is",round(V,3),"ft/sec"
print"the frictional drop in the nozzle is 10 recent of the total heat drop is",round(V1,3),"ft/sec"

the neglecting friction is 3136.312 ft/sec
the frictional drop in the nozzle is 10 recent of the total heat drop is 2975.367 ft/sec

Example 6.2 Page 88

In [24]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
v=((3140*pi*60*60)/(4*4*144))#ft/sec
v1=0.852*38.37#ft^3

#CALCULATIONS
W=(v/v1)#lb
V=(2970*pi*60*60)/(4*4*144)#ft^3
W1=(V/v1)#lb

#RESULTS
print"the weight of steam per hour is",round(W,3),"lb"
print"the weight of steam per hour is",round(W1,3),"lb"

the weight of steam per hour is 471.485 lb
the weight of steam per hour is 445.959 lb

Example 6.4 Page 89

In [25]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p1=300#lb
p=75#lb/in^2
p2=8#lb/in^2
h=90#C.H.U/lb
Pt=0.58*p#lb/in^2 absolute
h1=24#lb/C.H.U
D=0.968#C.H.U
D1=0.886#C.H.U
v=9.7#ft^3
v1=47.24#ft^3
V=sqrt(2*32.2*1400*24)#ft/sec
V1=sqrt(2*32.2*1400*90)#ft/sec

#CALCULATIONS
H=(p1*v*D/3600)#ft^3
V2=(p1*v1*D1/3600)#ft^3
A=0.768#in^2
A1=1.72#in^2
d=sqrt(4*0.768/pi)#in
d1=sqrt((4*A1)/(pi))#in

#RESULTS
print"the diameters at the throat and exit of the nozzle is",round(d1,3),"in"

the diameters at the throat and exit of the nozzle is 1.48 in

Example 6.5 Page 90

In [26]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
d=2.15#in^2
a=0.98#dry
p=100#lb/in^2
p1=11000#lb
P=0.58*p#lb/in^2
H=24#C.H.U/lb
D=0.947#lb
s=7.407#ft^3

#CALCULATION
V=sqrt(2*32.2*1400*H)#ft/sec
V1=V*(d/144)#ft^3
T=V1/(s*D)#lb
A=(p1/3600)#lb
C=A/T#lb

#RESULTS
print"the coefficient of discharge for the nozzles is",round(C,3),"lb"

the coefficient of discharge for the nozzles is 0.958 lb

Example 6.6 Page 91

In [44]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p=9.5#lb
p1=120.0#lb
e=0.88#in
p2=80.0#lb/in^2
d=25.0#in
d1=0.125#in
t=14.0#degree C
T=e*19.0#C.H.U/lb
D=0.975#in
V=sqrt(2.0*32.2*1400.0*T)#ft/sec
S=5.467#ft^3

#CALCULATIONS
V1=p*S*D#ft^3
T1=(V1*144.0/V)#in ^2
C=25.0*pi#in
N=C/2.5#in
P=C/31.0#in
W=d1/sin(t*pi/180)#in
L=P-W#in
W1=L*sin(t*pi/180)#in
T2=(T1)/(31*W1)#in

#RESULTS
print"The height of nozzle is",round(T2,3),"in"

The height of nozzle is 0.393 in

Example 6.8 Page 94

In [45]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p1=100#lb/in^2
p2=15#lb/in^2
d1=95#percent
d2=30#percent
P=0.58*p1#lb/in^2
H=0.95*25#C.H.U/lb
H1=0.95*76.5#C.H.U/lb
D=0.97#in
D1=0.905#in
V=7.407#ft^3
V1=sqrt(2*32.2*1400*H)#ft/sec
V2=sqrt(2*32.2*1400*H1)#ft/sec

#CALCULATIONS
V3=(2*pi*1*V1)/(64*4*144)#ft^3
W=(V3*3600)/(V*D)#lb
K=V2/(2*32.2)#ft lb sec
E=(((V2)**2*W)/(2*32.2*3600))#ft.lb
W1=(E*d2)/(p1*550)#ft.lb

#RESULTS
print"the quantity of steam used per hour and horse power developed is",round(W1,3),"ft lb"

the quantity of steam used per hour and horse power developed is 1.927 ft lb

Example 6.10 Page 95

In [51]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
d=0.15#lb
p=20#lb/in^2
p1=100#lb/iN62
t=200#degree C
f=10#percent
Pt=0.5457*p1#lb/in^2
x1=0.996#in
x2=0.952#in
h=29#C.H.U/lb
h1=65#C.H.U/lb
v=7.73#ft^3
v1=20.12#ft^3
T=0.364#in
T1=0.465#in
v2=sqrt(2*32.2*1400*h)#ft/sec
v3=sqrt(2*32.2*1400*h1)#ft/sec

#CALCULATIONS
V1=d*v*x1#ft^3
V2=d*v1*x2#ft^3
A1=(V1/v2)*144#in^2
A2=(V2/v3)*144#in^2

#RESULTS
print"the throat area of the nozzle is",round(A1,3),"and exit area is",round(A2,3),"in^2"

the throat area of the nozzle is 0.103 and exit area is 0.171 in^2

Example 6.11 Page 96

In [52]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
h=0.5#lb
p1=2.5#lb/in^2
p2=100#lb/in^2
t=250#Degree C
pv=1.3#constant
pt=0.5457*p2#lb/in^2
t1=18#degree C
h1=32#C.H.U/lb
h2=151#C.H.U/lb
D=0.887#in
V1=sqrt(2*32.2*1400*h1)#ft/sec
V2=sqrt(2*32.2*1400*h2)#ft.sec
s1=8.74#ft^3
s2=140.8#ft^3
T1=0.687#in
T1=1.77#in
V3=h*s1#ft^3/sec
V4=h*s2#ft^3/sec

#CALCULATIONS
A1=(V3/V1)*144#in^2
A2=(V4/V2)*144#in^2

#RESULTS
print"the size os nozzle is",round(A2,3),"in^2"

the size os nozzle is 2.748 in^2

Example 6.13 Page 97

In [54]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
h=500.0#gallons
p1=150.0#lb/in^2
p2=0.6#lb/in^2
P=p2*p1#lb/in^2
h=25.0#C.H.U/lb
p=62.4#lb/ft^2
V=sqrt(2.0*32.2*1400.0*h)#ft/sec
D=0.996#in^2
d=4.898#in^2
v1=1.2#in
vi=163.2#ft/sec
m=V/32.2#ft.lb.sec

#CALCULATIONS
W=V/vi-1#lb
W1=(5000.0)/(3600.0*W)#ft/sec
V1=W1*d*D#ft^3
A=V1/V*144.0#in^2
I=(50.0/36.0+W1)#lb/sec
A1=(I*144.0)/(62.4*vi)#in^2

#RESULTS
print"the aera of the stream and water orifices is",round(A1,3),"in^2"

the aera of the stream and water orifices is 0.022 in^2

Example 6.15 Page 100

In [56]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
a=50#degree c
v=2000#ft/sec
p=800#ft/sec
b=20#Degree C
v1=0.9#in^2
v2=513#ft/sec
W=(1/32.2)*(1810-(-313))*p#ft/lb lb stream /sec
K=(v**2)/(2*32.2)#ft/lb sec

#CALCULATIONS
D=(W/K)*100#percent/lb

#RESULTS
print"the work done per lb is",round(D,3),"%/lb"

the work done per lb is 84.92 %/lb

Example 6.16 Page 102

In [57]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
t=65#B.Th.U per lb
n=0.98#dry
p=105#lb/in^2
a=14#Degree C
b=20#Degree C
p1=800#ft/sec
v=0.80#ft/lb
p2=3.5#lb/sec
q=1400#in
V=sqrt(2*32.2*778*t)#ft/sec
W=(p1)*(1750-b)/32.2#ft lb/lb stream/sec
H=(W*p2/550)#ft/lb
E=1/64.4*((1053)**2-(825)**2)#ft.lb steam /sec

#CALCULATIONS
Hd=(E/q)#C.H.U

#RESULTS
print"the steam as it leaves the blades and horse power is",round(Hd,3),"C.H.U"

the steam as it leaves the blades and horse power is 4.749 C.H.U

Example 6.18 Page 103

In [58]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p=300#ft/sec
W=880#ft/sec
a=18#degree C
g=32.2#ft

#CALCULATIONS
Wd=(p*W)/g#ft lb

#RESULTS
print"the work done /lb steam sec is",round(Wd,3),"ft lb"

the work done /lb steam sec is 8198.758 ft lb

Example 6.19 Page 104

In [61]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
a=35.0#Degree C
b=20.0#degree C
f=2.0#ft
w=422.0#ft
w1=222.0#ft
g=32.2#ft
s=1500.0#r p m
j=0.8#ft
p=3.0#lb/sec
h=80.0#percent
i=1400.0#ft
P=(pi*(31.0/12.0)*(s/60))#ft/sec
W=P/g*(w-(-w1))#ft lb
H=(p*W)/550#ft lb

#CALCULATIONS
E=W/(j*i)#C.H.U

#RESULTS
print"the house -power developed per pair of rings if the steam is",round(E,3),"ft lb"

the house -power developed per pair of rings if the steam is 3.623 ft lb

Example 6.23 Page 106

In [63]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
d=73#ft
h=2#in
s=750#r p m
s1=31.3#lb/sec
h1=1.5#in
a=25#Degree c
p=5.7#lb/in^2
d1=0.97#in
h2=370#ft/sec
j=32.2#in
k=1400#in
e=0.75#percent
w=326#in
p=290#in
vi=155#ft/sec

#CALCULATIONS
P=(pi*7.69*s)/(60)#ft/sec
H=(P*h2*s1)/(550*j)#ft/sec
E=(P*h2)/(j*e*k)#C.H.U/lb

#RESULTS
print"the drop in pressure while the steam is passing through the turbine is",round(E,3),"C.H.U/lb"

the drop in pressure while the steam is passing through the turbine is 3.305 C.H.U/lb

Example 6.25 Page 108

In [64]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p=300#lb/in^2
ab=100#degree C
w=26.4#C
t=40#lb/in^2
t1=180#Degree C
p1=0.5#lb/in^2
T=732.38#C.H.U
W=26.2#C.H.U/lb
W1=102#C.H.U/lb
x=0.963#in
d=335#C.H.U/lb
E=743.85#C.H.U/lb

#CALCULATIONS
H=T-w#C.H.U/lb
h=T-W1#C.H.U/lb
H1=E-h#C.H.U/lb
T1=H+H1#C.H.U/lb
Wd=W1+d#C.H.U

#RESULTS
print"The total work done per lb steam is",round(Wd,3),"C.H.U"

The total work done per lb steam is 437.0 C.H.U

Example 6.28 Page 110

In [66]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p=100#lb/in62
p1=0.5#lb/in^2
T1=659.3#C.H.U/lb
T2=26.2#C H U/lb
W=181#C H U/lb
H1=66#C H U/lb
H2=115#C H U /lb
D=0.912#C H U/lb
H3=533.4#C H U/lb
T3=108.5 #Degree C
T4=26.4#Degree C
W1=82.1/(D*H3)#lb
s=1-W1#lb

#CALCULATIONS
T=W/(T1-T2)*100#percent
Wd=H1+(H2*s)#C H U/lb
H=T1-T3#C H U//lb
TE=Wd/H*100#percent

#RESULTS
print"the without bleeding is",round(T,3),"%"
print"the proper weight of steam is bled is",round(TE,3),"%"

the without bleeding is 28.589 %
the proper weight of steam is bled is 29.338 %