# Chapter 14 : Equilibrium With Gravity Or Centrifugal Force Osmotic Equilibrium Equilibrium With Surface Tension¶

### Example 14.1 Page: 379¶

In :
import math

T = 300.        #[K] Temperature of the natural gas well
R = 8.314       #[J/(mol*K)] universal gas consmath.tant
z_1 = 0         #[m]
y_methane_surf = 85./100            #[mol%]
y_ethane_surf = 10/100.             #[mol%]
y_propane_surf = 5/100.             #[mol%]
P = 2.                              #[MPa] Total equilibrium pressure
z_2 = 1000.                         #[m] Depth of the well

M_methane = 16./1000                #[kg/mol]
M_ethane = 30./1000                 #[kg/mol]
M_propane = 44./1000                #[kg/mol]

g = 9.81                            #[m/s**(2)]

f_methane_1 = y_methane_surf*P      #[MPa]
f_ethane_1 = y_ethane_surf*P        #[MPa]
f_propane_1 = y_propane_surf*P      #[MPa]

f_methane_2 = f_methane_1*math.exp((-M_methane*g*(z_1-z_2))/(R*T))          #[MPa]
f_ethane_2 = f_ethane_1*math.exp((-M_ethane*g*(z_1-z_2))/(R*T))             #[MPa]
f_propane_2 = f_propane_1*math.exp((-M_propane*g*(z_1-z_2))/(R*T))          #[MPa]

P_2 = (f_methane_2 + f_ethane_2 + f_propane_2 )                             #[MPa]

y_methane_2 = f_methane_2/P_2
y_ethane_2 = f_ethane_2/P_2
y_propane_2 = f_propane_2/P_2

print "The mol fraction of the methane at the depth 1000m is %f"%(y_methane_2)
print "The mol fraction of the ethane at the depth 1000m is %f"%(y_ethane_2)
print "The mol fraction of the propane at the depth 1000m is %f"%(y_propane_2)

The mol fraction of the methane at the depth 1000m is 0.840351
The mol fraction of the ethane at the depth 1000m is 0.104461
The mol fraction of the propane at the depth 1000m is 0.055187


### Example 14.2 Page: 380¶

In :
import math

T = 288.                #[K] Atmospheric temperature
R = 8.314               #[J/(mol*K)] universal gas consmath.tant
z_2 = 15000.            #[m] Thickness of the atmosphere
z_1 = 0.                #[m] Surface
y_N2_1 = 0.79
y_O2_1 = 0.21
M_N2 = 28./1000         #[kg/mol]
M_O2 = 32./1000          #[kg/mol]

g = 9.81                #[m/s**(2)]

a = math.exp(-(M_N2-M_O2)*g*(z_2-z_1)/(R*T))
yi2_by_yi1 = 1/(y_N2_1 + y_O2_1/a)

print " Concentration of the nitrogen at the top of atmosphere with respect to the concentration of nitrogen \
at the surface of the earth is         yi2_by_yi1 = %0.2f"%(yi2_by_yi1)

 Concentration of the nitrogen at the top of atmosphere with respect to the concentration of nitrogen  at the surface of the earth is         yi2_by_yi1 = 1.05


### Example 14.3 Page: 381¶

In :
import math

T = 288.                #[K] Atmospheric temperature
R = 8.314               #[J/(mol*K)] Universal gas consmath.tant
z_2 = 10.               #[m] Height of the reactor
z_1 = 0.                #[m] Surface
g = 9.81                #[m/s**(2)] Accelaration due to gravity

y_N2_1 = 0.79
y_O2_1 = 0.21
M_N2 = 28./1000         #[kg/mol]
M_O2 = 32./1000         #[kg/mol]

a = math.exp(-(M_N2-M_O2)*g*(z_2-z_1)/(R*T))
yi2_by_yi1 = 1/(y_N2_1 + y_O2_1/a)

print " Concentration of the nitrogen at the top of reactor with respect to the concentration of nitrogen \
at the bottom of reactor is         yi2_by_yi1 = %f"%(yi2_by_yi1)

 Concentration of the nitrogen at the top of reactor with respect to the concentration of nitrogen  at the bottom of reactor is         yi2_by_yi1 = 1.000034


### Example 14.4 Page: 382¶

In :
import math

T = 300.                #[K] Temperature of the centrifuge
R = 8.314               #[J/(mol*K)] Universal gas consmath.tant
y_UF6_238_1 = 0.993     # Mole fraction of UF6 with 238 isotope of uranium in feed
y_UF6_235_1 = 0.007     # Mole fraction of UF6 with 235 isotope of uranium in feed
M_UF6_238 = 352./1000       #[kg/mol] Molecular weight of UF6 with 238 isotope of uranium
M_UF6_235 = 349./1000       #[kg/mol] Molecular weight of UF6 with 235 isotope of uranium
r_in = 2./100               #[m] Interanal raddi of the centrifuge
r_out = 10./100             #[m] outer raddi of the centrifuge
f = 800.                    #[revolution/second] Rotational frequency of centrifuge

a = math.exp((M_UF6_235-M_UF6_238)*(2*3.141592*f)**(2)*(r_out**(2)-r_in**(2))/(2*R*T))

A = 1./(y_UF6_235_1 + y_UF6_238_1/a)

B = 1./A

print "The ratio of the mole fraction of UF6 with uranium 235 isotope) at the 2 cm radius to\
that at the 10 cm radius is %0.3f"%(B)

The ratio of the mole fraction of UF6 with uranium 235 isotope) at the 2 cm radius to that at the 10 cm radius is 1.156


### Example 14.5 Page: 384¶

In :
import math

x_water_1 = 0.98            # mole fraction of water in phase 1 i.e. in seawater
x_water_2 = 1.              # mole fraction of water in the phase 2 i.e. in water
R = 10.73                   #[(psi*ft**(3))/(lbmol*R)] Universal gas consmath.tant
T = 500.                    #[R] temperature
v_water_1 = 18/62.4         # [ft**(3)/(lbmol)]

delta_P = (-(R*T)*math.log(x_water_1))/v_water_1#[psi]

print "The pressure difference between the two phases is %0.1f psi"%(delta_P)

The pressure difference between the two phases is 375.7 psi


### Example 14.6 Page: 386¶

In :
import math

T = 100.            #[C] Temperature of the outside
P_outside = 1.      #[atm]
T = 0.05892         #[N/m] From metric steam table (7, page 267)

d_1 = 0.001         #[m]

delta_P_1 = (4*T)/d_1           #[Pa]

d_2 = 10**(-6)                  #[m]

delta_P_2 = (4*T)/d_2           #[Pa]

d_3 = 0.01*10**(-6)             #[m]
delta_P_3 = (4*T)/d_3           #[Pa]

print "Pressure difference with the change in radius of the drop of the water is given as in the following table"
print "            Diameter of the droplet d_iin meter                   Pressure difference  P_inside - P_outside in atm"
print "                     %0.2e                                                     %0.2e"%(d_1,delta_P_1)
print "                     %0.2e                                                     %0.2e"%(d_2,delta_P_2)
print "                     %0.2e                                                     %0.2e"%(d_3,delta_P_3)

Pressure difference with the change in radius of the drop of the water is given as in the following table
Diameter of the droplet d_iin meter                   Pressure difference  P_inside - P_outside in atm
1.00e-03                                                     2.36e+02
1.00e-06                                                     2.36e+05
1.00e-08                                                     2.36e+07


### Example 14.7 Page: 387¶

In :
import math

P_NBP = 1.                          #[atm]
Temp =273.15+100                    #[C] Temperature
D = 0.01*10**(-6)                   #[m] Diameter of the condensation nuclei( due to impurity )
T = 0.05892                         #[N/m] Surface tension between water drops and gas
R = 8.314                           #[J/(mol*K)]

v_water_liquid = 1/958.39*0.018     #[m**(3)/mol]

I = math.exp(v_water_liquid*(4*T/D)/(R*Temp))

P_gas_minus_P_NBP = (I-1)*P_NBP         #[atm]
delta_P = P_gas_minus_P_NBP*1.01325     #[bar]

delta_P_1 = delta_P*100*0.1450377       #[psi]

print "The equilibrium pressure at which the steam begin to condence at this temperature on \
the nuclei is %f psi above the normal boiling point."%(delta_P_1)

The equilibrium pressure at which the steam begin to condence at this temperature on  the nuclei is 2.253760 psi above the normal boiling point.


### Example 14.8 Page: 388¶

In :
import math

Temp = 273.15+100                   #[K] Temperature of the water drop
R = 8.314                           #[J/(mol*K)] Universal gas consmath.tant
D = 0.01*10**(-6)                   #[m] Diameter of the water drop
P_g = 0.15                          #[bar] guage pressure
T = 0.05892                         #[N/m] Surface tension between water drop and gas

v_water_liquid = 0.018/958.39       #[m**(3)/mol]
P_NBP = 1.013                       #[bar]
P_gas = 1.013+0.15                  #[bar]

P_1 = P_gas + 4*T/D                 #[bar]

def f2(P):
return v_water_liquid*P**(0)

def f3(P):
return (R*Temp)/P

delta_g = (delta_g_1 - delta_g_2)

print "The liquid can lower its free energy %0.2f J/mol by Changing to gas,"%(delta_g)
print "So that even at 0.15 bar above the normal boiling point, a drop of this small size is unstable and will quickly evaporate."

The liquid can lower its free energy 14.25 J/mol by Changing to gas,
So that even at 0.15 bar above the normal boiling point, a drop of this small size is unstable and will quickly evaporate.


### Example 14.9 Page: 390¶

In :
from scipy.optimize import fsolve
import math

Temp = 904.7                #[R] Temperature of the pure liquid water
P_NBP = 400.                #[psia] Saturation pressure of the pure liquid water at the given temperature
T = 1.76*10**(-4)           #[lbf/inch] Surface tension of water
R = 10.73                   #[(psi*ft**(3))/(lbmol*R)]

v_water_liquid = 18*0.01934             #[ft**(3)/lbmol]
D = 10**(-5.)                           #[inch]

def f(p):
return  v_water_liquid*(p - P_NBP)-(R*Temp)*math.log((p+4*T/D)/P_NBP)
P_liquid = fsolve(f,300)

P_inside = P_liquid + 4*T/D             #[psia]

print "The liquid pressure at which these boiling nuclei will begin to grow and intiate boiling is %0.1f psia"%(P_liquid)
print "At this external pressure the pressure inside the bubble is %0.1f psia"%(P_inside)

The liquid pressure at which these boiling nuclei will begin to grow and intiate boiling is 328.6 psia
At this external pressure the pressure inside the bubble is 399.0 psia

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