# Chapter 10 - Vapors¶

## Example 1 - Pg 154¶

In [1]:
#calculate the specific volume, enthalpy and entropy of the gas
#Initialization of variables
import math
p=3000. #psia
T=250. #F
#calculations
print '%s' %("From table 1, keenan and keynes,")
vf=0.01700
print '%s' %("From table 4,")
dvf=-18.3*math.pow(10,-5)
v=vf+dvf
print '%s' %("From table 1,")
hf=218.48
print '%s' %("From table 4,")
dhf=6.13
h=hf+dhf
sf=0.3675
dsf=-4.34*math.pow(10,-3)
s=sf+dsf
#results
print '%s %.5f %s' %("Specific volume =",v," cu ft/lb")
print '%s %.2f %s' %("\n Enthalpy =",h," Btu/lb")
print '%s %.4f %s' %("\n Entropy  =",s," Btu/lb per deg R")

From table 1, keenan and keynes,
From table 4,
From table 1,
From table 4,
Specific volume = 0.01682  cu ft/lb

Enthalpy = 224.61  Btu/lb

Entropy  = 0.3632  Btu/lb per deg R


## Example 2 - Pg 157¶

In [2]:
#calculate the state of the gas and moisture content
#Initialization of variables
h=1100 #Btu/lb
P=100 #psia
#calculations
print '%s' %("From table 2 of keenan and keynes,")
hg=1187.2 #Btu/lb
hfg=888.8 #Btu/lb
y=-(h-hg)/hfg*100
#results
print '%s %.2f %s %.2f %s' %("The state is",P,"psia with a moisture content of",y,"percent")

From table 2 of keenan and keynes,
The state is 100.00 psia with a moisture content of 9.81 percent


## Example 3 - Pg 157¶

In [4]:
#calculate the state of steam
#Initialization of variables
print '%s' %("From table 1 of keenan and keynes,")
v1=0.2688
#calculations
v2=3.060
p2=200 #psia
t2=600 #F
#results
print '%s %d %s %d %s' %("State of steam is",p2, "psia and",t2,"F")

From table 1 of keenan and keynes,
State of steam is 200 psia and 600 F


## Example 4 - Pg 157¶

In [5]:
#calculate the state of steam
#Initialization of variables
print '%s' %("From table 2 of keenan and keynes,")
t1=439.60 #F
u1=1118.4 #Btu/lb
#calculations
p2=380 #psia
#results
print '%s %d %s %.2f %s' %("The state of steam is saturated at",p2,"psia and",t1,"F")

From table 2 of keenan and keynes,
The state of steam is saturated at 380 psia and 439.60 F


## Example 5 - Pg 157¶

In [6]:
#calcualte the state of steam and moisture content
#Initialization of variables
print '%s' %("From table 2 of keenan and keynes,")
p1=1 #in of Hg
s=1.9812
#calculations
sf=2.0387
sfg=1.9473
y=-(s-sf)/sfg*100
#results
print '%s %d %s %.2f %s' %("The state is",p1, "in of Hg with a moisture content of",y, "percent")

From table 2 of keenan and keynes,
The state is 1 in of Hg with a moisture content of 2.95 percent


## Example 6 - Pg 161¶

In [8]:
#calculate the final state of steam and temperature. Also calculate the final enthalpy
#Initialization of variables
print '%s' %("From table 1 of keenan and keynes,")
h1=1204.8 #Btu/lb
q=174. #Btu/lb
#calculations
h2=h1+q
p2=30. #psia
t2=720. #F
#results
print '%s %d %s %d %s' %("Final state of steam is",p2,"psia and",t2," F")
print '%s %.1f %s' %("\n Final enthalpy is",h2,"Btu/lb")

From table 1 of keenan and keynes,
Final state of steam is 30 psia and 720  F

Final enthalpy is 1378.8 Btu/lb


## Example 7 - Pg 161¶

In [9]:
#calculate the final specific volume and final state of steam
#Initialization of variables
print '%s' %("From table 1 of keenan and keynes,")
p=70 #psia
x=0.1
p2=198 #psia
#calculations
v1=6.206
v2=0.017
vx=v1-x*(v1-v2)
t2=1400 #F
#results
print '%s %.3f %s' %("Final specific volume =",vx,"cu ft")
print '%s %d %s %d %s' %("\n Final state is",p2,"psia and",t2,"F")

From table 1 of keenan and keynes,
Final specific volume = 5.587 cu ft

Final state is 198 psia and 1400 F


## Example 8 - Pg 162¶

In [10]:
#calculate the final state of steam
#Initialization of variables
print '%s' %("From table 1 of keenan and keynes,")
p=400 #psia
t1=700 #F
p2=85 #psia
#calculations
s2=1.6398 #units/lb
t2=350 #F
#results
print '%s %d %s %d %s' %("Final state of steam is",p2,"psia and",t2,"F")

From table 1 of keenan and keynes,
Final state of steam is 85 psia and 350 F


## Example 9 - Pg 162¶

In [11]:
#calculate the work of compression and heat removed
#Initialization of variables
import math
p1=20. #psia
p2=140. #psia
J=778.
t2=150. #F
t1=30. #F
#calculations
print '%s' %("From Table A-3,")
v1=2.0884 #cu ft/lb
v2=0.33350 #cu ft/lb
h2=95.709
h1=81.842
n=math.log(p2/p1) /math.log(v1/v2)
W=(p2*v2-p1*v1)*144/(1-n)
du=h2-h1 + (p1*v1-p2*v2)*144/J
Q=du+W/J
s2=0.17718
s1=0.18126
Q2=((t2+t1)/2 +460) *(s2-s1)
#results
print '%s %d %s' %("Work of compression =",W,"ft-lb")
print '%s %.3f %s' %("\n Heat removed per pound of refrigerant =",Q," Btu/lb")
print '%s %.4f %s' %("\n Heat removed in case 2 =",Q2," Btu")

From Table A-3,
Work of compression = -11671 ft-lb

Heat removed per pound of refrigerant = -2.046  Btu/lb

Heat removed in case 2 = -2.2440  Btu


## Example 10 - Pg 163¶

In [12]:
#calculate the specific enthalpy of exhaust steam
#Initialization of variables
print '%s' %("From table 1 of keenan and keynes,")
intt=440000 #lb/hr
out=255000 #lb/hr
p1=400 #psia
t1=700 #F
p2=35 #psia
t2=290 #F
vel=500 #ft/s
hp=44000 #hp
ent=1362.7 #Btu/lb
#calculations
ein=ent*intt
eout=hp*2544 + out*1183 + 925000.
h2= (ein-eout)/185000.
#results
print '%s %d %s' %("Specific enthalpy of exhaust steam =",h2,"Btu/lb")

From table 1 of keenan and keynes,
Specific enthalpy of exhaust steam = 1000 Btu/lb


## Example 11 - Pg 165¶

In [13]:
#calcualte the final state of steam and loss of available energy
#Initialization of variables
print '%s' %("From table 1 of keenan and keynes,")
h1=1351.1 #Btu/lb
p1=600. #psia
t1=700. #F
p2=234. #psia
h2=1.6865
h1=1.5875
t3=101.74
#calculations
t2=660. #F
loss= (h2-h1)*(t3+459.69)
#results
print '%s %d %s %d %s' %("Final state of steam is",p2," psia and",t2,"F")
print '%s %.1f %s' %("\n Loss of available energy =",loss,"Btu/lb")

From table 1 of keenan and keynes,
Final state of steam is 234  psia and 660 F

Loss of available energy = 55.6 Btu/lb


## Example 12 - Pg 165¶

In [14]:
#calculate the state of vapor and also the quality
#Initialization of variables
print '%s' %("From table 2 of keenan and keynes,")
p1=98.87 #psia
p2=31.78 #psia
t1=80 #F
h2=26.365 #btu/lb
h1=11.554 #btu/lb
hfg=67.203 #btu/lb
#calculations
x=(h2-h1)/hfg*100
#results
print '%s %.2f %s %.2f %s' %("The state of vapor leaving is",p2, "psia with a quality of",x, " percent")

From table 2 of keenan and keynes,
The state of vapor leaving is 31.78 psia with a quality of 22.04  percent


## Example 13 - Pg 167¶

In [15]:
#calculate the mean state in the line and also the moisture content
#Initialization of variables
ps=216 #psig
pb=29.12 #in of Hg
p2=0.4 #in
t2=244 #F
#calculations
pa=0.491*pb
pabs=pa + p2*0.491
plb=pa+ ps
hcal=1166.5 #Btu/lb
h2=1200.1 #Btu/lb
h3=831.9 #Btu/lb
y=-(hcal-h2)/h3*100
#results
print '%s %.1f %s %.2f %s' %("Mean state in the line is",plb," psia with a moisture content of",y,"percent")

Mean state in the line is 230.3  psia with a moisture content of 4.04 percent