# CHAPTER 3: PROPERTIES OF STEAM¶

## Example 3.1 Page 30¶

In [2]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p=100#lb/in^2
x=0.8#lb
t1=164#degree C
t2=4.45 #ft^3
p1=0.016#ft^3
h1=493.4#C.H.U/lb
h2=165.9#C.H.U/lb
S=h2+h1#C.H.U/lb
w=(144*p)/1400*(t2-p1)#C.H.U/lb
H=h2+(x*h1)#C.H.U//lb
w1=(x*144*p)/1400*(t2-p1)#C.H.U

#CALCULATIONS
E=S-w#C.H.U/lb
IE=H-w1#C.H.U/lb

#RESULTS
print"The steam is total heat dry and satured is",round(E,3),"C.H.U/lb"
print"Total heat of wet steam",round(IE,3),"C.H.U/lb"

The steam is total heat dry and satured is 614.96 C.H.U/lb
Total heat of wet steam 524.135 C.H.U/lb


## Example 3.2 Page 31¶

In [3]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
t1=35#degree C
p=100#lb/in^2
L=435#C.H.U
L2=539.3#C.H.U
h1=165.9#H.C.U/lb
h2=493.4#C.H.U/lb
S=(h1-t1)#C.H.U
h3=304.1#C.H.U
h4=335#C.H.U/lb

#CALCULATIONS
X1=h3/h2#C.H.U/lb
X2=h4/L2#C.H.U/lb

#RESULTS
print"The heat giving to the water and steam is",round(X2,3),"C.H.U/lb"

The heat giving to the water and steam is 0.621 C.H.U/lb


## Example 3.3 Page 32¶

In [4]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p=35#lb/in^2
w=1425#lb
q=1474#lb
s1=126.7#C.H.U/lb
s2=28#C.H.U/lb
t1=5#degree C
t2=28#degree C
L1=521.4#C.H.U/lb
w1=245#lb
w2=0.2#lb

#CALCULATIONS
W=(s1-s2)+L1#C.H.U/lb
H=q*(t2-t1)#C.H.U/lb
T=H/W#lb

#RESULTS
print"The total equivalent is",round(T,3),"lb"

The total equivalent is 54.672 lb


## Example 3.4 Page 33¶

In [5]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p=100#lb/in^2
w=2400#lb
t1=15#degree C
s1=165.9#C.H.U/lb
x=0.9#lb
L2=493.4#C.H.U/lb
t2=65#degree
x4=0.8#lb
s3=64.8#C.H.U/lb
w1=2000#lb
w2=2400#lb
b1=12400#lb
b2=22000#lb
p1=4400#lb
n=421.65#lb
h1=w2*[s1+(x*L2)]#C.H.U/hr
h2=w1*[s1+(x4+L2)]#C.H.U/hr

#CALCULATIONS
T=w*[(s1-t1)+(x*L2)]#C.H.U/hr
T1=w1*[(s1-s3)+(x4*L2)]#C.H.U/hr
H=T+T1#C.H.U/hr
X=n/L2#C.H.U/lb

#RESULTS
print"The thermal capacity of the pipe is",round(X,3),"C.H.u/hr"

The thermal capacity of the pipe is 0.855 C.H.u/hr


## Example 3.5 Page 35¶

In [6]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
w1=4.5#lb
s1=45.5#lb
p1=165#lb/in^2
T=140#Degree C
h1=30#in
h2=4#in
p2=0.49#ln/in^2
T1=(w1+s1)#lb
T2=103.5#Degree C
T3=140#Degree
h3=0.48#in
x=0.988#berfore throttling
T=[103.12+537.1+h3*(T3-T2)]#C.H.U/lb
x1=0.012#lb of water
X=s1*x1#lb water
w=50#lb of steam

#CALCULATIONS
P=h2+h1#in of mercury
P1=s1*x1#lb/in^2
T4=w1+P1#lb
D=(w-T4)/w#lb

#RESULTS
print"The dryness of steam with a combined is",round(D,3),"lb"

The dryness of steam with a combined is 0.899 lb


## Example 3.6 Page 36¶

In [9]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
w=40.0#lb
w1=380.0#lb
t1=80.0#Degree
p=85.0#lb/in^2
p1=15.0#lb/in^2
W=w+w1#lb/hr
P=p+p1#lb/in^2
T=659.3#C.H.U/lb
d=10.0#h.p

#CALCULATIONS
H=W*T-w1*t1#C.H.U/hr
H1=(d*33000.0*60.0)/1400.0#C.H.U/hr
T1=H1/H*100.0#percent
D=w1/(w1+w)#C.H.U/hr
H2=(W*(99.6+D*539.3)-w1*t1)#C.H.U/hr
T2=H-H2#C.H.U/hr
H3=T2-H1#C.H.U/hr
E=(1400.0*H3)/(60.0*33000.0)#h.p

#RESULTS
print"The amount of radiations from the engine is",round(E,3),"h.p"

The amount of radiations from the engine is 11.311 h.p


## Example 3.10 Page 37¶

In [11]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
w=40.0#lb
w1=380.0#lb
t1=80.0#Degree
p=85.0#lb/in^2
p1=15.0#lb/in^2
W=w+w1#lb/hr
P=p+p1#lb/in^2
T=659.3#C.H.U/lb
d=10.0#h.p

#CALCULATIONS
H=W*T-w1*t1#C.H.U/hr
H1=(d*33000*60)/1400#C.H.U/hr
T1=H1/H*100#percent
D=w1/(w1+w)#C.H.U/hr
H2=(W*(99.6+D*539.3)-w1*t1)#C.H.U/hr
T2=H-H2#C.H.U/hr
H3=T2-H1#C.H.U/hr
E=(1400*H3)/(60*33000)#h.p

#RESULTS
print"The amount of radiations from the engine is",round(E,3),"h.p"

The amount of radiations from the engine is 11.311 h.p


## Example 3.12 Page 39¶

In [12]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p=120#lb/in^2
ts=264#degree C
T1=(273+130.6)#F
v=0.0171#ft^3/lb
L1=518.4#lb
T2=(273+171.9)#F
L2=487.4#lb
Cp=0.48#lb
L=0.0894/Cp#lb
Ts=T2*1.205#degree
ta=536-273#Degree C
T=649.9#C.H.U
S=131.2#C.H.U
w=(144*40)/1400*(10.49-v)#C.H.U
C=T-S#C.H.U
I=C-w#C.H.U
E=(704.7-57.8)#C.H.U
E1=E-606.5#C.H.U

#CALCULATIONS
E1=E-606.5#C.H.U
H=(704.7-T)#C.H.U

#RESULTS
print"Heat and internal energy after each operation is",round(H,3),"h.p"

Heat and internal energy after each operation is 54.8 h.p


## Example 3.17 Page 42¶

In [13]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
A=28.1#in Hg vacuum
a=0.93#lb/in^2
T=33#Degree
p=0.729#lb/in^2
P=-p+a#lb/in^2
p1=120000#lb
p2=28.1#in
a1=0.9#ln
p3=1000#lb
t=15#degree C
A1=(a1*(p1/(60*p3)))#lb/mim
v=(A1*96*306)/(144*P)#ft^3 of air per min
V=37.3+a1*610#C.H.U/lb

#CALCULATIONS
H=(V-T)#C.H.U
W=(H/t)*(p1/60)#gal/min

#RESULTS
print"The water per minute in cubic feet per minute passing to air extractor",round(W,3),"gal/min"

The water per minute in cubic feet per minute passing to air extractor 73773.333 gal/min