Chapter 11 - The ideal gas and mixture relationships

Example 1 - Pg 184

In [1]:
#calculate the work done
#Initialization of variables
n=1.3
T1=460+60. #R
P1=14.7 #psia
P2=125. #psia
R=1545.
M=29.
#calculations
T2=T1*(P2/P1)**((n-1)/n)
wrev=R/M *(T2-T1)/(1-n)
#results
print '%s %d %s' %("Work done =",wrev,"ft-lbf/lbm")
print '%s' %("The answer is a bit different due to rounding off error in textbook")

Work done = -58988 ft-lbf/lbm
The answer is a bit different due to rounding off error in textbook

Example 2 - Pg 184

In [2]:
#calculate the Change in kinetic energy
#Initialization of variables
P2=10 #psia
P1=100 #psia
T1=900 #R
w=50 #Btu/lbm
k=1.39
cp=0.2418
#calculations
T2=T1*(P2/P1)**((k-1)/k)
T2=477
KE=-w-cp*(T2-T1)
#results
print '%s %.1f %s' %("Change in kinetic energy =",KE,"Btu/lbm")

Change in kinetic energy = 52.3 Btu/lbm

Example 3 - Pg 190

In [3]:
#calculate the Final temperature
#Initialization of variables
T1=900 #R
P1=100 #psia
P2=10 #psia
#calculations
print '%s' %("From table B-9")
pr1=8.411
pr2=pr1*P2/P1
T2=468 #R
#results
print '%s %.1f %s' %("Final temperature =",T2,"R ")

From table B-9
Final temperature = 468.0 R

Example 4 - Pg 190

In [4]:
#calculate the final temperature and pressure, work done and enthalpy
#Initialization of variables
cr=6
p1=14.7 #psia
t1=60.3 #F
M=29
R=1.986
#calculations
print '%s' %("from table b-9")
vr1=158.58 
u1=88.62 #Btu/lbm
pr1=1.2147
vr2=vr1/cr
T2=1050 #R
u2=181.47 #Btu/lbm
pr2=14.686
p2=p1*(pr2/pr1)
dw=u1-u2
h2=u2+T2*R/M
#results
print '%s %d %s' %("final temperature =",T2,"R")
print '%s %.1f %s' %("\n final pressure =",p2," psia")
print '%s %.2f %s' %("\n work done =",dw," Btu/lbm")
print '%s %.1f %s' %("\n final enthalpy =",h2," Btu/lbm")

from table b-9
final temperature = 1050 R

 final pressure = 177.7  psia

 work done = -92.85  Btu/lbm

 final enthalpy = 253.4  Btu/lbm

Example 5 - Pg 193

In [5]:
#calculate the mole fractions of oxygen and nitrogen, Average molecular weight and partial pressures, densities, volumes
#Initialization of variables
m1=10. #lbm
m2=15. #lnm
p=50. #psia
t=60.+460 #R
M1=32.
M2=28.02
R0=10.73 
#calculations
n1=m1/M1
n2=m2/M2
x1=n1/(n1+n2)
x2=n2/(n1+n2)
M=x1*M1+x2*M2
R=1545/M
V=(n1+n2)*R0*t/p
rho=p/(R0*t)
rho2=M*rho
p1=x1*p
p2=x2*p
v1=x1*V
v2=x2*V
#results
print '%s' %("part a")
print '%s %.3f %s %.3f %s' %("Mole fractions of oxygen and nitrogen are",x1," and",x2," respectively")
print '%s' %("part b")
print '%s %.1f' %("Average molecular weight = ",M)
print '%s' %("part c")
print '%s %.2f %s' %("specific gas constant =",R,"psia ft^3/lbm R")
print '%s' %("part d")
print '%s %.1f %s' %("volume of mixture =",V,"ft^3")
print '%s %.5f %s %.3f %s' %("density of mixture is",rho, "mole/ft^3 and",rho2, "lbm/ft^3")
print '%s' %("part e")
print '%s %.2f %s %.2f %s' %("partial pressures of oxygen and nitrogen are",p1,"psia and",p2,"psia respectively")
print '%s %.2f %s %.2f %s' %("partial volumes of oxygen and nitrogen are",v1,"ft^3 and",v2,"ft^3 respectively")

part a
Mole fractions of oxygen and nitrogen are 0.369  and 0.631  respectively
part b
Average molecular weight =  29.5
part c
specific gas constant = 52.40 psia ft^3/lbm R
part d
volume of mixture = 94.6 ft^3
density of mixture is 0.00896 mole/ft^3 and 0.264 lbm/ft^3
part e
partial pressures of oxygen and nitrogen are 18.43 psia and 31.57 psia respectively
partial volumes of oxygen and nitrogen are 34.87 ft^3 and 59.74 ft^3 respectively

Example 6 - Pg 195

In [6]:
#calculate the gravimetric and ultimate analysis
#Initialization of variables
m1=5.28
m2=1.28
m3=23.52
#calculations
m=m1+m2+m3
x1=m1/m*100
x2=m2/m*100
x3=m3/m*100
C=12./44 *m1/ m*100
O=(32./44 *m1 + m2)/m*100
N=m3/m*100
#results
print '%s %.1f %s %.1f %s %.1f %s' %("From gravimetric analysis, co2 =",x1,"percent , o2 =",x2,"percent and n2 =",x3,"percent")
print '%s %.2f %s %.2f %s %.2f %s' %("\n From ultimate analysis, co2 =",C,"percent , o2 =",O," percent and n2 =",N," percent")

From gravimetric analysis, co2 = 17.6 percent , o2 = 4.3 percent and n2 = 78.2 percent

 From ultimate analysis, co2 = 4.79 percent , o2 = 17.02  percent and n2 = 78.19  percent

Example 7 - Pg 197

In [7]:
#calculate the Entropy of the mixture
#Initialization of variables
import math
x1=1/3.
n1=1.
n2=2.
x2=2/3.
p=12.7 #psia
cp1=7.01 #Btu/mole R
cp2=6.94 #Btu/mole R
R0=1.986
T2=460+86.6 #R
T1=460. #R
p0=14.7 #psia
#calculations
p1=x1*p
p2=x2*p
ds1= cp1*math.log(T2/T1) - R0*math.log(p1/p0)
ds2= cp2*math.log(T2/T1) - R0*math.log(p2/p0)
S=n1*ds1+n2*ds2
#results
print '%s %.2f %s' %("Entropy of mixture =",S,"Btu/R")
print '%s' %("\n the answer given in textbook is wrong. please check using a calculator")

Entropy of mixture = 8.27 Btu/R

 the answer given in textbook is wrong. please check using a calculator

Example 8 - Pg 198

In [8]:
#calculate the change in internal energy and entropy
#Initialization of variables
import math
c1=4.97 #Btu/mol R
c2=5.02 #Btu/mol R
n1=2
n2=1
T1=86.6+460 #R
T2=50.+460 #R
#calculations
du=(n1*c1+n2*c2)*(T2-T1)
ds=(n1*c1+n2*c2)*math.log(T2/T1)
#results
print '%s %d %s' %("Change in internal energy =",du," Btu")
print '%s %.3f %s' %("\n Change in entropy =",ds,"Btu/R")

Change in internal energy = -547  Btu

 Change in entropy = -1.037 Btu/R

Example 9 - Pg 198

In [9]:
#calculate the Pressure of the mixture
#Initialization of variables
n1=1
n2=2
c1=5.02
c2=4.97
t1=60. #F
t2=100. #F
R0=10.73
p1=30. #psia
p2=10. #psia
#calcualtions
t=(n1*c1*t1+n2*c2*t2)/(n1*c1+n2*c2)
V1= n1*R0*(t1+460)/p1
V2=n2*R0*(t2+460)/p2
V=V1+V2
pm=(n1+n2)*R0*(t+460)/V
#results
print '%s %.1f %s' %("Pressure of mixture =",pm," psia")

Pressure of mixture = 12.7  psia

Example 10 - Pg 199

In [10]:
#calculate the change in entropy of gas 
#Initialization of variables
import math
T2=546.6 #R
T1=520 #R
T3=560 #R
v2=1389.2
v1=186.2
R0=1.986
c1=5.02
c2=4.97
n1=1
n2=2
v3=1203
#calculations
ds1=n1*c1*math.log(T2/T1) + n1*R0*math.log(v2/v1)
ds2=n2*c2*math.log(T2/T3)+n2*R0*math.log(v2/v3)
ds=ds1+ds2
#results
print '%s %.3f %s' %("Change in  entropy for gas 1 =",ds1," Btu/R")
print '%s %.3f %s' %("\n Change in  entropy for gas 1 =",ds2,"Btu/R")
print '%s %.3f %s' %("\n Net change in entropy =",ds,"Btu/R")
print '%s' %("The answer is a bit different due to rounding off error in the textbook")

Change in  entropy for gas 1 = 4.242  Btu/R

 Change in  entropy for gas 1 = 0.331 Btu/R

 Net change in entropy = 4.572 Btu/R
The answer is a bit different due to rounding off error in the textbook

Example 11 - Pg 200

In [11]:
#calculate the Final temperature and change in entropy of air and water.
#Initialization of variables
import math
m1=1. #lbm
m2=0.94 #lbm
M1=29.
M2=18.
p1=50. #psia
p2=100. #psia
t1=250 +460. #R
R0=1.986
cpa=6.96
cpb=8.01
#calculations
xa = (m1/M1)/((m1/M1)+ m2/M2)
xb=1-xa
t2=t1*(p2/p1)**(R0/(xa*cpa+xb*cpb))
d=R0/(xa*cpa+xb*cpb)
k=1/(1.-d)
dsa=cpa*math.log(t2/t1) -R0*math.log(p2/p1)
dSa=(m1/M1)*dsa
dSw=-dSa
dsw=dSw*M2/m2
#results
print '%s %d %s' %("Final temperature =",t2," R")
print '%s %.3f %s %.5f %s' %("\n Change in entropy of air =",dsa," btu/mole R and",dSa, "Btu/R")
print '%s %.4f %s %.5f %s' %("\n Change in entropy of water =",dsw,"btu/mole R and",dSw," Btu/R")
print '%s' %("The answers are a bit different due to rounding off error in textbook")

Final temperature = 851  R

 Change in entropy of air = -0.115  btu/mole R and -0.00395 Btu/R

 Change in entropy of water = 0.0757 btu/mole R and 0.00395  Btu/R
The answers are a bit different due to rounding off error in textbook

Example 12 - Pg 202

In [12]:
#calculate the volume occupied and mass of steam
#Initialization of variables
T=250. + 460 #R
p=29.825 #psia
pt=50 #psia
vg=13.821 #ft^3/lbm
M=29.
R=10.73
#calculations
pa=pt-p
V=1/M *R*T/pa
ma=V/vg
xa=p/pt
mb=xa/M *18./(1.-xa)
#results
print '%s %.2f %s' %("In case 1, volume occupied =",V," ft^3")
print '%s %.2f %s' %("\n In case 1, mass of steam =",ma," lbm steam")
print '%s %.3f %s' %("\n In case 2, mass of steam =",mb," lbm steam")

In case 1, volume occupied = 13.02  ft^3

 In case 1, mass of steam = 0.94  lbm steam

 In case 2, mass of steam = 0.918  lbm steam

Example 13 - Pg 203

In [13]:
#calcualte the percentage
#Initialization of variables
ps=0.64 #psia
p=14.7 #psia
M=29.
M2=46.
#calculations
xa=ps/p
mb=xa*9./M *M2/(1-xa)*100
#results
print '%s %.1f %s' %("percentage =",mb,"percent")

percentage = 65.0 percent

Example 14 - Pg 203

In [14]:
#calculate the partial pressure of water vapor
#Initialization of variables
ps=0.5069 #psia
p=20 #psia
m1=0.01
m2=1
M1=18.
M2=29.
#calculations
xw= (m1/M1)/(m1/M1+m2/M2)
pw=xw*p
#results
print '%s %.3f %s' %("partial pressure of water vapor =",pw,"psia")

partial pressure of water vapor = 0.317 psia