# Chapter 9 Properties of the pure substance¶

## Ex:9.1 Pg:304¶

In [1]:
#Initialization of variables
T=32 #F
m=1 #lbm
J=778.16
#calculations
print "From steam tables,"
hf=0
p=0.08854 #psia
vf=0.01602 #ft**3/lbm
u=hf-p*144*vf/J
#results
print "Internal energy = %.7f Btu/lbm"%(u)
From steam tables,
Internal energy = -0.0002625 Btu/lbm

## Ex:9.2 Pg:305¶

In [2]:
#Initialization of variables
P=40 #psia
#calculations
print "from steam tables,"
hf=200.8 #Btu/lbm
hg=27 #Btu/lbm
T=495  #R
ds=(hf-hg)/T
#results
print "Change in entropy = %.3f Btu/lbm R"%(ds)
from steam tables,
Change in entropy = 0.351 Btu/lbm R

## Ex:9.3 Pg:305¶

In [3]:
#Initialization of variables
x=0.35
T=18 #F
#calculations
print "From table B-14,"
hf=12.12 #Btu/lbm
hg=80.27 #Btu.lbm
hfg=-hf+hg
h=hf+x*hfg
#results
print "specific enthalpy = %.1f Btu/lbm"%(h)
From table B-14,
specific enthalpy = 36.0 Btu/lbm

## Ex:9.4 Pg:306¶

In [4]:
#Initialization of variables
x=0.35
T=18 #F
T2=55.5 #F
#calculations
print "From table B-14,"
hf=12.12 #Btu/lbm
hg=80.27 #Btu.lbm
hfg=-hf+hg
h=hf+x*hfg
h2=85.68 #Btu/lbm
dh=h2-h
#results
print "Heat required = %.2f Btu/lbm"%(dh)
From table B-14,
Heat required = 49.71 Btu/lbm

## Ex:9.5 Pg:307¶

In [5]:
#Initialization of variables
P=1460 #psia
T=135 #F
P2=700 #psia
#calculations
print "From mollier chart,"
h=120 #Btu/lbm
x=0.83
#results
print "enthalpy = %d Btu/lbm"%(h)
print "\n Qulaity = %.2f"%(x)
From mollier chart,
enthalpy = 120 Btu/lbm

Qulaity = 0.83

## Ex:9.6 Pg:307¶

In [6]:
#Initialization of variables
m=1 #lbm
P1=144 #psia
P2=150 #psia
T1=360 #F
J=778.16
#calculations
print "From table 3,"
v1=3.160 #ft**3/lbm
h1=1196.5 #Btu/lbm
u1=h1-P1*144*v1/J
h2=1211.4 #Btu/lbm
u2=h2-P2*144*v1/J
dq=u2-u1
#results
print "Heat transferred = %.1f Btu/lbm"%(dq)
From table 3,
Heat transferred = 11.4 Btu/lbm

## Ex:9.7 Pg:307¶

In [7]:
#Initialization of variables
T1=100 #F
P2=1000 #psia
x=0.6
J=778.16
#calculations
print "From table 3,"
v=0.01613 #ft**3/lbm
P1=0.9 #psia
wrev=-v*(P2-P1)*144/J
dv=0.000051 #ft**3/lbm
wcomp=(P2+P1)/2 *dv*144/J
wact=wrev/x
#results
print "Work done = %.1f Btu/lbm"%(wrev)
print "\n In case 2, work required = %.1f Btu/lbm"%(wact)
From table 3,
Work done = -3.0 Btu/lbm

In case 2, work required = -5.0 Btu/lbm

## Ex:9.8 Pg:308¶

In [8]:
#Initialization of variables
pa=1000 #atm
ta=100 #F
#calculations
hf=67.97 #Btu/lbm
w=3 #Btu/lbm
ha=hf+w
print "from steam table 2,"
hc=1191.8 #Btu/lbm
qrev=hc-ha
#results
print "Heat transferred = %.1f Btu/lbm"%(qrev)
from steam table 2,
Heat transferred = 1120.8 Btu/lbm

## Ex:9.9 Pg:309¶

In [9]:
#Initialization of variables
P1=144 #psia
T1=400 #F
y=0.7
#calculations
print "From steam tables,"
h1=1220.4 #Btu/lbm
s1=1.6050 #Btu/lbm R
s2=1.6050 #Btu/lbm R
P2=3 #psia
sf=0.2008 #Btu/lbm R
sfg=1.6855 #Btu/lbm R
x=(s1-sf)/sfg
hf=109.37 #Btu/lbm
hfg=1013.2 #Btu/#bm
h2=hf+x*hfg
work=h1-h2
dw=y*work
h2d=h1-dw
#results
print "Work done = %d Btu/lbm"%(work)
print "\n work done in case 2 = %.1f Btu/lbm"%(dw)
print "\n Final state pressure = %d psia"%(P2)
print "The answer is a bit different due to rounding off error in textbook"
From steam tables,
Work done = 266 Btu/lbm

work done in case 2 = 186.8 Btu/lbm

Final state pressure = 3 psia
The answer is a bit different due to rounding off error in textbook

## Ex:9.10 Pg:309¶

In [10]:
#Initialization of variables
pb=14.696 #psia
pa=150 #psia
tb=300 #F
#calculations
print "From steam tables,"
hb=1192.8 #Btu/lbm
ha=hb
hf=330.51 #Btu/lbm
hfg=863.6 #Btu/lbm
x=(ha-hf)/hfg
#results
print "Quality of wet steam = %.1f percent"%(x*100)
From steam tables,
Quality of wet steam = 99.8 percent