Chapter 11:Thermodynamic relations Equilibrium and stability

Ex11.3:pg-436

In [1]:
import math
Tb = 353.0 # boiling point of benzene in K
T = 303.0 # Operational temperature in K
R = 8.3143 #Gas constant
P = 101.325*math.exp((88/R)*(1.0-(Tb/T)))

print "\n Example 11.3"
print "\n Vapour pressure of benzene is ",P ," kPa"
#The answers vary due to round off error

 Example 11.3

 Vapour pressure of benzene is  17.6682592008  kPa

Ex11.4:pg-436

In [2]:
import math
T = (3754-3063)/(23.03-19.49) # Temperature at triple point in K
P = math.exp(23.03-(3754/195.2)) # Pressure at triple point
R = 8.3143 # Gas constant
Lsub = R*3754 # Latent heat of sublimation
Lvap = 3063*R # Latent heat of vaporisation
Lfu = Lsub-Lvap # Latent heat of fusion

print "\n Example 11.4"
print "\n Temperature at triple point is ",T ," K"
print "\n Pressure at triple point is ",P ," mm Hg"
print "\n\n Latent heat of sublimation is ",Lsub ," kJ/kg mol"
print "\n Latent heat of vapourization is is ",Lvap ," kJ/kg mol"
print "\n Latent heat of fusion is ",Lfu ," kJ/kg mol"
#The answers vary due to round off error

 Example 11.4

 Temperature at triple point is  195.197740113  K

 Pressure at triple point is  44.631622076  mm Hg


 Latent heat of sublimation is  31211.8822  kJ/kg mol

 Latent heat of vapourization is is  25466.7009  kJ/kg mol

 Latent heat of fusion is  5745.1813  kJ/kg mol

Ex11.6:pg-438

In [4]:
R = 8.3143 # Gas constant in kJ/kg-mol-K
N1 = 0.5 # Mole no. of first system
N2 = 0.75 # Mole no. of second system
T1 = 200 # Initial temperature of first system in K
T2 = 300 # Initial temperature of second system in K
v = 0.02 # Total volume in m**3
print "\n Example 11.6\n"
Tf = (T2*N2+T1*N1)/(N1+N2)
Uf_1 = (3.0/2.0)*(R*N1*Tf)*(10**-3)
Uf_2 = (3.0/2.0)*(R*N2*Tf)*(10**-3)
pf = (R*Tf*(N1+N2)*(10**-3))/v
Vf_1 = R*N1*(10**-3)*Tf/pf
Vf_2 = v-Vf_1
print "\n Energy of first system is ",Uf_1 ," kJ,\n Energy of second system is ",Uf_2 ," kJ,\n Volume of first system is ",Vf_1 ," m**3,\n Volume of second system is ",Vf_2 ," m**3,\n Pressure is ",pf ," kN/m**2,\n Temperature is ",Tf ," K."
#The answers vary due to round off error

 Example 11.6


 Energy of first system is  1.6212885  kJ,
 Energy of second system is  2.43193275  kJ,
 Volume of first system is  0.008  m**3,
 Volume of second system is  0.012  m**3,
 Pressure is  135.107375  kN/m**2,
 Temperature is  260.0  K.

Ex11.10:pg-446

In [3]:
import math
R = 0.082 # Gas constant in litre-atm/gmol-K
m = 1.5 # Mass flow rate in kg/s
p1 = 1.0 # Pressure in atm
t2 = 300.0 # Temperature after compression in K
p2 = 400.0 # Pressure after compression in atm
Tc = 151.0 # For Argon in K
pc = 48.0 # For Argon in atm
print "\n Example 11.10 "
a = 0.42748*((R*1000)**2)*((Tc)**2)/pc
b = 0.08664*(R*1000)*(Tc)/pc
# By solving equation v2**2 - 49.24*v2**2 + 335.6*v2 - 43440 = 0
v2 = 56.8 # In cm**3/g mol
v1 = (R*1000)*(t2)/p1
delta_h = -1790 # In J/g mol
delta_s = -57 # In J/g mol
Q = (t2*delta_s*(10**5)/39.8)/(3600*1000)
W = Q - (delta_h*(10**5)/39.8)/(3600*1000)
print "\n Power required to run the compressor = ",W ," kW, \n The rate at which heat must be removed from the compressor = ",Q ," kW"
# Answers vary due to round off error.

 Example 11.10 

 Power required to run the compressor =  -10.6853713009  kW, 
 The rate at which heat must be removed from the compressor =  -11.9346733668  kW