# Chapter 14 : Convective Mass Transfer¶

## Example 14.1 Page No : 574¶

In [1]:
import math

# Variables
D = 0.025;			    #Diameter of the cylinder in m
R = (2*10**-6);			#Rate of sublime in kg/s
C = (6*10**-6);			#Saturated vapour concentration in kmol/m**3
W = 128;			    #Molecular weight in kg/kmol

# Calculations
q = (R/W);			            #Molar transfer rate in k.mol/sm
h = (q/(math.pi*D*C));			#Convective mass transfer coefficient in m/s

# Results
print 'Convective mass transfer coefficient is %3.3f m/s'%(h)

Convective mass transfer coefficient is 0.033 m/s


## Example 14.2 Page No : 576¶

In [7]:
import math
# Variables
pA = -0.9;			#Partial pressure of water vapour in atm
t = 0.0025;			#Boundary layer thickness in m

# Calculations
y = 0;
pAs1 = math.exp(-33.35*y)-0.9;			#Partial pressure in atm
y = t;
pAs2 = math.exp(-33.35*y)-0.9;			#Partial pressure in atm
#partial derivative of pA wrt y is -33.35exp(y)-0.9
x = 0;
X = (-33.35*math.exp(x))-pA;			#Partial derivative value at x = 0
DAB = (0.26*10**-4)			#DAB value in m**2/s
h = (DAB*X)/(pAs2-pAs1);			#Local mass transfer coefficient in m/s

# Results
print 'Local mass transfer coefficient is %3.3f m/s'%(h)

Local mass transfer coefficient is 0.011 m/s


## Example 14.3 Page No : 583¶

In [3]:
# Variables
T = 27;			#Temperature of dry air in degree C
p = 1;			#Pressure of dry air in atm
L = 0.5;			#Length of the plate in m
v = 50;			#Velocity in m/s

# Calculations
DAB = (0.26*10**-4)			#DAB value in m**2/s
p = 1.16;			#Density in kg/m**3
u = (184.6*10**-7);			#Dynamic viscosity in N.s/m**2
Pr = 0.707;			#Prantl number
Sc = (u/(p*DAB));			#Schmidt number
Re = (p*v*L)/u;			#Reynolds number
jm = (0.0296*(Re**(-1./5)));			#jm value
h = (jm*v)/Sc**(2./3);			#Mass transfer coefficient of water vapour in m/s

# Results
print 'Mass transfer coefficient of water vapour is %3.3f m/s'%(h)

Mass transfer coefficient of water vapour is 0.118 m/s


## Example 14.4 Page No : 583¶

In [5]:
# Variables
T = 27;			#Temperature of swimming pool in degree C
h = 0.4;			#Relative humidity
v = 2;			#Speed of wind in m/s
v1 = (15.89*10**-6);			#Kinematic viscosity in m**2/s
p = 0.0436;			#Density in kg/m**3
DAB = (0.26*10**-4)			#DAB value in m**2/s
L = 15;			#Length in m

# Calculations
Sc = (v1/DAB);			#Schmidt number
Re = (v*L)/v1;			#Reynolds number
ShL = (((0.037*Re**(4./5))-870)*Sc**(1./3));			#Equivalent Schmidt number
h1 = (ShL*(DAB/L))/10**-3;			#Mass transfer coefficient for evaporation in mm/s

# Results
print 'Mass transfer coefficient for evaporation is %3.1f*10**-3 m/s'%(h1)

Mass transfer coefficient for evaporation is 4.4*10**-3 m/s


## Example 14.5 Page No : 585¶

In [6]:
# Variables
T = 25;			#Temperature of air in degree C
v = 3;			#Velocity im m/s
D = 0.01;			#Diameter of tube in m
L = 1;			#Length of tube in m

# Calculations
v1 = (15.7*10**-6);			#Kinematic viscosity in m**2/s
DAB = (0.62*10**-5)			#DAB value in m**2/s
Re = (v*D)/v1;			#Reynolds number
Sh = 3.66;			#Schmidt number
h = (Sh*DAB)/D;			#Average mass transfer coefficient in m/s

# Results
print 'Average mass transfer coefficient is %3.5f m/s'%(h)

Average mass transfer coefficient is 0.00227 m/s


## Example 14.6 Page No : 586¶

In [8]:
# Variables
T = 25;			#Temperature of air in degree C
v = 5;			#Velocity in m/s
D = 0.03;			#Diameter of tube in m
DAB = (0.82*10**-5)			#DAB value in m**2/s

# Calculations
v1 = (15.7*10**-6);			#Kinematic viscosity in m**2/s
Sc = (v1/DAB);			#Schnidt number
Re = (v*D)/v1;			#Reynolds number
h = (0.023*Re**(4./5)*Sc**(1./3)*DAB)/D;			#Mass transfer coefficient in m/s

# Results
print 'Mass transfer coefficient is %3.4f m/s'%(h)

Mass transfer coefficient is 0.0119 m/s


## Example 14.7 Page No : 589¶

In [9]:
import math
# Variables
Ta = 40.+273;			#Temperature of air in K
w = 100.;			    #Molecular weight in kg/k.mol
H = 120.;			    #Latent heat of vapourisation of volatile liquid in kJ/kg
p = 3530.;			    #Saturated vapour pressure in N/m**2
DAB = (0.2*10**-4);		#DAB value in m**2/s

# Calculations
p1 = 1.16;			#Density in kg/m**2
Cp = 1.007;			#Specific heat in J/kg.K
a = (22.5*10**-6);			#Diffusivity in m**2/s
X = ((H*100*p*10**-3)/(8.315*p1*Cp*(a/DAB)**(2./3)));			#X value for temperature
T = (Ta+math.sqrt((Ta**2-(4*X))))*0.5;			#Temperature in K

# Results
print 'Steady state temperature of cold water inside the pot is %3.1f K'%(T)

Steady state temperature of cold water inside the pot is 299.5 K


## Example 14.8 Page No : 590¶

In [10]:
# Variables
T = 22. + 273;			#Thermometer reading in K

# Calculations
p = 2617;			#Pressure in N/m**2
hfg = 2449;			#Enthalpy in kJ/kg
p1 = (p*18)/(8315*T);			#Density in kg/m**3
p2 = (1.0132*10**5)/(287*T);			#Density in kg/m**3
Cp = 1.008;			#Specific heat in kJ/kg.K
a = (26.2*10**-6);			#Diffusivity in m**2/s
DAB = (0.26*10**-4);			#DAB value in m**2/s
Ts = ((T-273)+((hfg*1000*p1)/(p2*Cp*1000)));			#True air temperature in degree C

# Results
print 'True air temperature is %3.2f degree C'%(Ts)

True air temperature is 60.99 degree C


## Example 14.9 Page No : 591¶

In [11]:
# Variables
T = 50.;			#Temperature of air stream in degree C
Tb = 22.;			#Bulb temperature in degree C

# Calculations
Tf = (T+Tb)/2;			#Film temperature in degree C
p = 1.14;			#Density in kg/m**3
Cp = 1.006;			#Specific heat in J/kg.K
Pr = 0.7;			#Prantl number
u = (2*10**-5);			#Dynamic viscosity in Ns/m**2
DAB = (0.26*10**-4);			#DAB value in m**2/s
Sc = (u/(p*DAB));			#Schmidt nuber
Le = (Sc/Pr);			#Lewis number
p1 = 0.01920;			#Density in kg/m**3
hfg = 2449;			#Enthalpy in kJ/kg
pA = 0.0064;			#Density in kg/m**3
psat = (1./12.23);			#Saturation density in kg/m**3
RH = (pA/0.0817)*100;			#Relative humidity

# Results
print 'Relative humidity of the airstream is %3.2f percent'%(RH)

Relative humidity of the airstream is 7.83 percent


## Example 14.10 Page No : 592¶

In [12]:
# Variables
Td = 27.;			#Dry bulb teperature in degree C
Tw = 17.;			#Wet bulb temperature in degree C
Pr = 0.74;			#Prantl number
Sc = 0.6;			#Schmidt number
Mv = 18.;			#Molecular weight of vapour
Ma = 29.;			#Molecular weight of air
Cp = 1004.;			#Specific heat in J/kg.K
p = (1.0132*10**5);			#Pressure in N/m**2

# Calculations
pv2 = 1917;			#Saturation presusre of air at 17 degree C in N/m**2
hfg = 2461;			#Enthalpy in kJ/kg
w2 = (Mv*pv2)/(Ma*(p-pv2));			#Weight in kg/kg of dry air
w1 = w2-((Cp*(Pr/Sc)**(2./3)*(Td-Tw))/(hfg*1000));			#Specific humidity of air in kg/kg of dry air

# Results
print 'Specific humidity of air is %3.5f kg/kg of dry air'%(w1)

Specific humidity of air is 0.00728 kg/kg of dry air


## Example 14.11 Page No : 592¶

In [4]:
# Variables
T = 27.;		    	#Temperature of swimming pool in degree C
Ts = 37.;			#Surface temperature in degree C
h = 0.4;			#Relative humidity
D1 = 5.;		    	#Dimension of swimming pool in m
D2 = 15.;			#Dimension of swimming pool in m
v = 2.;		    	#Speed of wind in m/s
v1 = (15.89*10**-6);			#Kinematic viscosity in m**2/s
p = 0.0436;			            #Density in kg/m**3
DAB = (0.26*10**-4)	    		#DAB value in m**2/s
Sc = (v1/DAB);		        	#Schmidt number
Re = (v*D2)/v1;		        	#Reynolds number
ShL = (((0.037*Re**(4./5))-870)*Sc**(1./3));			#Equivalent Schmidt number
h1 = (ShL*(DAB/D2));			#Mass transfer coefficient for evaporation in m/s

# Calculations
Psat = 3531.;			#Partial pressure of water vapour in N/m**2
pi = (0.4*6221);			#Saturation pressure of water vapour in N/m**2
pt = 101325.;			#Total pressure of air in N/m**2
pAs = (18*Psat)/(8361*(T+273));			#Density at the water surface in kg/m
pAi = (18*pi)/(8316*(T+273));			#Density at the water surface in kg/m
n = round((h1*(pAs-pAi)*3600*24),);			#Rate of evaporation of water in kg/m**2 day
L = (n*D1*D2);			#Total water loss from the swimming pool in kg/day

# Results
print 'Rate of evaporation of water is %3.1f kg/day'%(L)

# there is a rounding off error in textbook.

Rate of evaporation of water is 225.0 kg/day

In [ ]: