# Chapter 16 : Mass Transfer¶

## Example 16.1 page : 276¶

In :
import math

# Variables
T = 25.+273;                        # Temperature in degK
p = 1.;                             # Pressure in atm
Va = 18.9;                         # Molecular volume of water vapour in cm**3/gm-mol
Vb = 29.9;                         # Molecular volume of air in cm**3/gm-mol
Ma = 18.;                           # Molecular weight of water vapour in gm/mol
Mb = 29.;                           # Molecular weight of air in gm/mol

# Calculations
Dab = 0.0043*(T**1.5)*math.sqrt((1/Ma)+(1/Mb))/(p*(Va**(1./3)+Vb**(1./3))**2);

# Results
print "The diffusion coefficient is %.3f cm**3/sec "%(Dab);

The diffusion coefficient is 0.200 cm**3/sec


## Example 16.2 page : 278¶

In :
import math

# Variables
T = 25.+273;                        # Temperature in degK
p = 1.;                             # Pressure in atm
Va = 96.;                           # Molecular volume of benzene in cm**3/gm-mol
Vb = 29.9;                         # Molecular volume of air in cm**3/gm-mol
Ma = 78.;                           # Molecular weight of benzene in gm/mol
Mb = 29.;                           # Molecular weight of air in gm/mol

# Calculations
Dab = 0.0043*(T**1.5)*math.sqrt((1/Ma)+(1/Mb))/(p*(Va**(1./3)+Vb**(1./3))**2);

# Results
print "The diffusion coefficient is %.3f cm**3/sec "%(Dab);

The diffusion coefficient is 0.082 cm**3/sec


## Example 16.3 page : 283¶

In :
import math

# Variables
x = 0.1/12;                            # thickness of still air layer in ft
T = 77.+460;                            # temperature in degR
p = 1.;                                 # Atmospheric pressure in atm
pa1 = 0.3;                             # Pressure of ammonia in still air in atm
pb1 = p-pa1;                           # pressure of air in atm
pa2 = 0;                               # pressure of ammonia in the absorption plane
pb2 = p-pa2;                           # pressure of air in absorption plane

# Calculations
pbm = (pb2-pb1)/(math.log(pb2/pb1));          # Logarithmic mean pressure
D = 0.914;                             # Diffusion coefficient for ammonia
R = 0.729;                             # Gas consmath.tant in ft**3-atm/lb-mole-degR
N = D*p*(pa1-pa2)/(R*T*x*pbm);

# Results
print "The amount of ammonia diffusing through the stagnant air is %.1f lb-mol/hr-ft**2"%(N);

The amount of ammonia diffusing through the stagnant air is 0.1 lb-mol/hr-ft**2


## Example 16.4 page : 287¶

In :
import math

# Variables
ri = 3./96;                            # Inner radius of pipe in ft
ro = 1./24;                            # Outer radius of pipe in ft
Ca1 = 0.0003;                         # Concentration at the inner hose of pipe in lb-mol/ft**2
Ca2 = 0;                              # Concentration at the outer surface

# Calculations
D = 0.7*10**-5;                        # Diffusion coefficient of hydrogen in rubber in ft**2/hr
N = 2*math.pi*D*(Ca1-Ca2)/math.log(ro/ri);     # Rate of diffusion in lb-mol/hr

# Results
print "The rate of diffusion iof hydrogen in rubber is %.2f*10**-8 lb-mole/hr"%(N*10**8);

# note : rounding off error.

The rate of diffusion iof hydrogen in rubber is 4.59*10**-8 lb-mole/hr


## Example 16.5 page : 296¶

In :
import math

# Variables
u = 0.0437;                             # Vismath.cosity in lb/hr-ft
rho = 0.077;                            # Density in lb-ft**2
D = 0.992;                              # Diameter of pipe in ft
v = 4.*3600;                             # Velocity in ft/sec
L = 6./12;                               # Length of pipe parallel to direction of air flow in ft
p = 14.7;                               # Atmospheric pressure in psi
T = 460.+65;                             # Temperature in degR

# Calculations
# Heat transfer equation for laminar flow of a flat surface
Nre = L*v*rho/u;                        # Reynolds number
Ns = u/(rho*D);                         # Schimdt mumber
Nnu = 0.662*(Ns)**(1./3)*math.sqrt(Nre);       # Nusselt number
hmc = Nnu*D/L;                          # Heat transfer coefficient
pv1 = 0.144;                            # Vapour pressure at 40% humidity
pv2 = 0.252;                            # Vapour pressure at saturation
pa1 = p-pv1;                            # Absolute pressure of air at 40% rel. humidity in psi
pa2 = p-pv2;                            # Absolute pressure of saturated air in psi
pbm = (pa1+pa2)/2;                      # Log mean pressure in psi
R = 1544.;                               # Universal gas consmath.tant in ft**3-psi/lbmol-degR
N = hmc*p*(pa1-pa2)*144/(R*T*pbm);

# Results
print "The amount of water evaporated per hour is %.4f lb mol/hr-ft**2"%(N);

The amount of water evaporated per hour is 0.0024 lb mol/hr-ft**2


## Example 16.6 page : 297¶

In :
# Variables
u = 0.047;                              # Vismath.cosity in lb/hr-ft
rho = 0.069;                            # Density in lb-ft**2
D = 0.992;                              # Diameter of pipe in ft
v = 7.5*3600;                           # Velocity in ft/sec
L = 2.;                                  # Length of pipe parallel to direction of air flow in ft
M = 0.992;                              # Molecular weight
p = 14.696;                             # Atmospheric pressure in psi
T = 460.+65;                             # Temperature in degR
M = 29.;                                 # molecular weight of air
M2 = 18.;                                # Molecular weight of water vapour
A = 4.;                                  # Area of water surface in ft**2

# Calculations
# Heat transfer equation for laminar flow of a flat surface
Nre = L*v*rho/u;                        # Reynolds number

# Assuming the case that of a fluid flowing parallel to a flat plate , jm = 0.0039
jm = 0.0039;
Ns = u/(rho*D);                         # Schimdt mumber
Gm = v*rho/M;                           # Mole flow rate
pv1 = 0.672;                            # Vapour pressure at 40% humidity
pv2 = 0.600;                            # Vapour pressure at saturation
pa1 = p-pv1;                            # Absolute pressure of air at 40% rel. humidity in psi
pa2 = p-pv2;                            # Absolute pressure of saturated air in psi
pbm = (pa1+pa2)/2;                      # Log mean pressure in psi
hmp = jm*Gm/(pbm*144*Ns**(2./3));             # Heat transfer coefficient in lbmol/ft**2-hr-psi
N = hmp*(pv1-pv2)*144;                # Mass transfer rate in lb mol/hr-ft**2
W = N*A*M2;

# Results
print "The amount of water evaporated per hour is %.3f lb mol/hr-ft**2"%(W);

The amount of water evaporated per hour is 0.119 lb mol/hr-ft**2


## Example 16.7 page : 300¶

In :
import math

# Variables
u = 3.82*10**-7;                          # Vismath.cosity in lb-sec/ft**2
rho = 2.3*10**-3;                         # Density in lbsec**2/ft**4
A = 1.;                                   # Area in ft**2
Cp = 0.24;                               # Specific heat capacity in abtu/lbm-degF
v = 4.*3600;                              # Velocity in ft/sec
k = 0.015;                               # Thermal conductivity in Btu/hr-ft-degF
p = 14.7;                                # Atmospheric pressure in psi
M = 29.;                                  # Avg. molecular weight of air
T1 = 70.+460;                             # Temperature of still air in degF
T2 = 90.+460;                             # temperature of surface of water in degF
L = 1.;                                   # For characteristic of 1 ft
D = 0.992;                               # Diffusivity in ft**2/sec

# Calculations
# Heat transfer equation for laminar flow of a flat surface
Ngr = 32.2*L**3*((T2/T1)-1)/(u/rho)**2;    # Grasshops number
Npr = u*3600*Cp*32.2/k;                  #  Prandtls number
Nnu = 0.75*(Ngr*Npr)**.25;                # Nusselt number
h = Nnu*k/L;                             # Heat transfer coefficient
Ns = u*3600/(rho*D);                     # Schimdt mumber
hmc = h*D*(Ns/Npr)**0.25/k;               # Heat transfer coe
pv1 = 0.18;                              # Vapour pressure at 40% humidity
pv2 = 0.69;                              # Vapour pressure at saturation
pa1 = p-pv1;                             # Absolute pressure of air at 40% rel. humidity in psi
pa2 = p-pv2;                             # Absolute pressure of saturated air in psi
pbm = (pa1+pa2)/2;                       # Log mean pressure in psi
R = 1544;                                # Universal gas consmath.tant in ft**3-psi/lbmol-degR
T = (T1+T2)/2;                           # Average temperature in degR
N = hmc*p*(pv2-pv1)*144/(R*T*pbm)*18;    # mass transfer rate in lbmol/hr-ft**2

# Results
print "The amount of water evaporated per hour is %.4f lb mol/hr-ft**2"%(N);

The amount of water evaporated per hour is 0.0873 lb mol/hr-ft**2


## Example 16.8 page : 303¶

In :
# Variables
Td = 70.+460;                               # Dry bulb temperature in degR
Tw = 60.+460;                               # Wet bulb temperature in degR
a = 0.26;                                  # Ratio of coefficients ie. h/hmw from table
L = 1059.9;                                # Latent heat Btu/lbmol
p = 14.7;                                  # Atmospheric pressure in psi
pa = 0.259;                                # Partial pressure of water in psi
Ma = 18.;                                   # Molecular weight of water vapour
Mb = 29.;                                   # Molecular weight of air

# Calculations
Wwb = pa*Ma/(Mb*(p-pa));                   # Absolte dry bulb humidity of air
Wdb = Wwb-(a*(Td-Tw)/L);                   # Absolte dry bulb humidity of air

# Results
print "The humidity of air at dry conditions is %.5f lbm/lbm of dry air"%(Wdb);

# rounding off error.

The humidity of air at dry conditions is 0.00868 lbm/lbm of dry air


## Example 16.9 page : 305¶

In :
# Variables
v = 20.;                                 # Velocity of air ammonia mixture in ft/sec
Npr = 0.72;                             # Prandtls number
Ns = 0.60;                              # Schimdt number
pbm = 14.7;                             # math.log mean pressure in psi
Mm = 29.;                                # Molecular weight of mixture
Mv = 17.;                                # Molecular weight of ammonia
Ma = 29.;                                # Molecular weight of air
Cp = 0.24;                              # specific heat capacity in Btu/lbm-degF
h = 8.;                                  # Heat transfer coefficient
p = 1.;                                  # Atospheric pressure in atm

# Calculations
hmp = h*Mv*(Npr/Ns)**(2./3)/(Cp*p*Ma);   # Mass transfer coefficient based on pressure

# Results
print "The mass transfer coefficient based on pressure is %.1f lbm/hr-ft**2-atm"%(hmp);

The mass transfer coefficient based on pressure is 22.1 lbm/hr-ft**2-atm