Chapter 9 - Properties of the pure substance¶

Example 1 - Pg 149¶

In [1]:
#calculate the Internal energy
#Initialization of variables
T=32 #F
m=1 #lbm
J=778.16
#calculations
print '%s' %("From steam tables,")
hf=0
p=0.08854 #psia
vf=0.01602 #ft^3/lbm
u=hf-p*144*vf/J
#results
print '%s %.7f %s' %("Internal energy =",u," Btu/lbm")

From steam tables,
Internal energy = -0.0002625  Btu/lbm


Example 2 - Pg 149¶

In [2]:
#calculate the Change in entropy
#Initialization of variables
P=40 #psia
#calculations
print '%s' %("from steam tables,")
hf=200.8 #Btu/lbm
hg=27 #Btu/lbm
T=495.  #R
ds=(hf-hg)/T
#results
print '%s %.3f %s' %("Change in entropy =",ds,"Btu/lbm R")

from steam tables,
Change in entropy = 0.351 Btu/lbm R


Example 3 - Pg 149¶

In [3]:
#calculate the specific enthalpy
#Initialization of variables
x=0.35
T=18. #F
#calculations
print '%s' %("From table B-14,")
hf=12.12 #Btu/lbm
hg=80.27 #Btu.lbm
hfg=-hf+hg
h=hf+x*hfg
#results
print '%s %.1f %s' %("specific enthalpy =",h,"Btu/lbm")

From table B-14,
specific enthalpy = 36.0 Btu/lbm


Example 4 - Pg 149¶

In [4]:
#calculate the heat required
#Initialization of variables
x=0.35
T=18 #F
T2=55.5 #F
#calculations
print '%s' %("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 '%s %.2f %s' %("Heat required =",dh,"Btu/lbm")

From table B-14,
Heat required = 49.71 Btu/lbm


Example 5 - Pg 149¶

In [5]:
#calculate the enthalpy and Quality
#Initialization of variables
P=1460. #psia
T=135. #F
P2=700. #psia
#calculations
print '%s' %("From mollier chart,")
h=120 #Btu/lbm
x=0.83
#results
print '%s %d %s' %("enthalpy =",h,"Btu/lbm")
print '%s %.2f' %("\n Quality =",x)

From mollier chart,
enthalpy = 120 Btu/lbm

Quality = 0.83


Example 6 - Pg 149¶

In [6]:
#calculate the Heat transferred
#Initialization of variables
m=1 #lbm
P1=144. #psia
P2=150. #psia
T1=360. #F
J=778.16
#calculations
print '%s' %("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 '%s %.1f %s' %("Heat transferred =",dq,"Btu/lbm")

From table 3,
Heat transferred = 11.4 Btu/lbm


Example 7 - Pg 150¶

In [7]:
#calculate the work done and work required
#Initialization of variables
T1=100. #F
P2=1000. #psia
x=0.6
J=778.16
#calculations
print '%s' %("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 '%s %.1f %s' %("Work done =",wrev,"Btu/lbm")
print '%s %.1f %s' %("\n In case 2, work required =",wact," Btu/lbm")

From table 3,
Work done = -3.0 Btu/lbm

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


Example 8 - Pg 150¶

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

from steam table 2,
Heat transferred = 1120.8  Btu/lbm


Example 9 - Pg 151¶

In [9]:
#calculate the work done and Final state pressure
#Initialization of variables
P1=144 #psia
T1=400 #F
y=0.7
#calculations
print '%s' %("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 '%s %d %s' %("Work done =",work,"Btu/lbm")
print '%s %.1f %s' %("\n work done in case 2 =",dw,"Btu/lbm")
print '%s %d %s' %("\n Final state pressure =",P2,"psia")

From steam tables,
Work done = 266 Btu/lbm

work done in case 2 = 186.8 Btu/lbm

Final state pressure = 3 psia


Example 10 - Pg 154¶

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

From steam tables,
Quality of wet steam = 99.8 percent