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A Short Course on

Introduction to Mass Transfer
One two-hour lecture each week
from October 1 to October 26, 2012
Ampere A. Tseng* and Miroslav Raudensky
Heat Transfer and Fluid Flow Laboratory
Brno University of Technology
Brno, Czech Republic
E-mail: [email protected]
References:
• A. F. Mills, “Mass Transfer,” Prentice Hall, Upper Saddle River, NJ,
USA 2001.
• K. Asano, “Mass Transfer, from Fundamentals to Modern Industrial
Applications," Wiley, Weinheim, Germany, 2006.
• D. Basmadjian, “Mass Transfer, Principles and Applications," CRC
Press, Boca Raton FL, USA 2003.
________________

* On leave from Arizona State University, Tempe, Arizona, USA

What is Mass Transfer?
• Mass transfer is mass in transit as the result of a species
concentration difference in a mixture. Must have a mixture
of two or more species for mass transfer to occur. It occurs
in many processes, such as absorption, evaporation,
adsorption, drying, precipitation, membrane filtration, and
distillation
• Mass transfer is the net movement of mass from one
location, usually meaning a stream, phase, fraction or
component, to another. It involves two mechanism: diffusion
and convection.

• Diffusion or mass diffusion is due to random molecular
motion (can be simplified as Brownian motion).
• Mass transfer by diffusion is analogous to heat transfer by
conduction.

Mass Diffusion
Mass diffusion refers
to the mass transfer that
occurs across a
stationary fluid or solid
in which a concentration
gradient exists. The
Consider two species A and B at
the same T and p, but initially
mass transport is driven
separated by a partition.
by the gradient without
In contrast, convention refers requiring bulk motion.
Eventually, uniform
to mass transfer that occurs
across a moving fluid in which concentration for all
species (A & B) are
concentration gradient also
achieved.
exist.

Diffusion Versus Convection
Dissolution of a sugar cube in a
cup of black coffee. Initially, the
concentration of sugar
molecules in significant only in
the vicinity of the cube, which is
a slow process and dominated
by diffusion.
By stirring the coffee with a
spoon, sugar molecules are
transferred to or mixed with the
coffee much faster, which is
dominated by convection.

Diffusion still contributes to mass
transfer, regardless of whether
the coffee is stirred or not.

Definitions and Units
c

Total molar concentrat ion of the mixture [mol/m 3 ]

c A M olar concentrat ion of species A (  ρ A /M A ) [mol/m 3 ]
D A Diffusion coefficien t of species A[m2 /s]
h

Convective heat transfer coefficien t [W/m 2 - K]

h m Convective mass transfer coefficien t [m/s]
jAx M ass diffusion flux of species A in x - direction [kg/m 2 - s]
(the amount of a species diffused per unit area per unit time)
j*Ax M olar diffusion flux of species A in x - direction [mol/m 2 - s]
J Ax M ass flow rate of species A in x - direction [kg/s]
J *Ax M olar flow rate of species A in x - direction [mol/s]
m A M ass fraction of species A (   A / , xA M A /M , m A  1)
M

M ean molecular weight of the mixture (  /c  x A M A )
[atomic mass unit (symbol u, or Da)  1.66  10-27 kg or  1 g/mol]

M A M olecular weight of species A [u or Da]

Definitions and Units
p

Total pressure of the mixture [N/m 2 of Pa]

p A Partial pressure of species A [N/m 2 of Pa]
R u Universal or ideal gas constant (  the product of Boltzmann' s
constant and Avogadro' s constant, 8.314426 J/mol - K), pV  nR u T,
where n, p, V, and T are the number of the amount, pressure [Pa], volume
T

[m 3 ], and absolute temperature [K] of the gas.
Temperature [C or K]

V

Volume [m 3 ]

x A M olar fraction of species A (  c A /c, mA M /M A & x A  1)



Total density of the mixture [kg/m 3 ]

A

Partial density of species A (    m A ) [kg/m 3 ]

Re  ρUL/μ

Subscripts

Nu = hL/k
Pr = c pμ/k

A, B, C.... Species

Sh = h m L/D A-B

x, y, z........ Coordinates

Sc = μ/(ρD A-B )

Composition of Dry Air
Elements
Volume [%]
Weight [%]
Nitrogen (N2)
78,09
75,51
Oxygen (O2)
20,95
23,16
Argon (Ar)
0,93
1,28
Carbon dioxide (CO2)
0,033
0,05
Neon (Ne)
0,0018
0,0012
Helium (He)
0,000524
0,000072
Methane (CH4)
0,0002
0,0001
Krypton (Kr)
0,000114
0,0003
Hydrogen (H2)
0,00005
0,000001
Xenon (Xe)
0,0000087
0,00004
Not included in above dry atmosphere:
Water vapor (H2O)
~0.40% over full atmosphere, typically 1%-4% at surface
References:
[1] - Vzduch. In: Wikipedia: the free encyclopedia [online]. Praha: Wikimedia Foundation, 2012, 30. 6.
2012 [cit. 2012-09-10]. Dostupné z:
http://cs.wikipedia.org/wiki/Slo%C5%BEen%C3%AD_vzduchu#Slo.C5.BEen.C3.AD_vzduchu
[2] - Amosphere of Earth. In: Wikipedia: the free encyclopedia [online]. San Francisco (CA):
Wikimedia Foundation, 2012, 5 October 2012 [cit. 2012-10-10]. Dostupné z:
http://en.wikipedia.org/wiki/Atmosphere_of_Earth#Composition

Composition of Dry Air
On a volume basis, the composition of dry air is 78.1% N 2 ,
20.9% O2 , and 0.9% Ar.
Since equal volumes of gases contain the same number of
molecules, the mass fraction of O2 can be found, for example,
as mO2 

xO2 M O2
xN 2 M N 2  xO2 M O2  x Ar M Ar

(0.209)(32)

 0.231
(0.781)( 28)  (0.209)(32)  (0.009)( 46)
The corresponding mass fraction is m N 2  0.755, mO2  0.231,
m Ar  0.014.

mol

There is 1,009 kg of nitogen, 0,272 kg of oxygen and 0,013 kg of carbon dioxide
and rare gases in 1 m3 of air.

Mass Flux of Species A by Mass Diffusion
A diffusion flux is defined as the amount of a species diffused per unit
area per unit time and denoted as jAx , indicating that species A diffuses
in the x direction. Let' s first considerin g a one - dimensiona l diffusion
case. Based on Fick' s law of diffusion, one can obtain that
dm
d
jAx   DA A  jAx   DA A , if ρ is constant
dx
dx
where  is the mass density of the solution, D A , the diffusion coefficien t
of species A in the solution, and m A , the mass fraction of species A. In the
case of three - dimensiona l diffusion, one equation is for each direction :
jAx   DA
or

m A
m
m
, jAy   DA A , jAz   DA A
x
y
z

jA   DAm A

 jA   DA A , if ρ is a constant.

The last of the above equations is in vector form. It has been assumed that
the material is isotropic, i.e.,  and D A are direction - independen t.

Molar Flux of Species A due to Diffusion
Fick' s law of diffusion can also be written as
dx A
j  cD A
dx
where j*Ax is the molar diffusion flux , c is the molar density of the
*
Ax

solution (moles per unit volume), x A , the mole fraction of species A.
In the case of three - dimensiona l diffusion, one can find that :
*

jA  cD Ax A
*

where jA is the molar diffusion flux vecto r. If the molar density of
the solution, c, is constant, the above equations can be written as
*

jA   DAc A
where c A  cx A is the moles of species A per unit volume of the solution,
i.e., the molar concentrat ion of species A. In a dilute solution, the molar
density of the solution, c, is essentiall y constant.

Standard Temperature & Pressure, STP
It is commonly define solubility and other fluid
properties at a standard condition. In chemistry,
IUPAC (International Union of Pure and Applied
Chemistry) established standard temperature and
pressure, abbreviated as STP, as a temperature of
0 °C (273 K or 32 °F) and an absolute pressure of
100 kPa (14.504 psi, 0.986 atm, 1 bar). One kmol
of gas at STP occupies 22.414 m3.

Mass Diffusion
Through a Plan
Wall

Let' s consider a one - dimensiona l mass
diffusion problem first. The attached
figure shows a plane wall of surface area,
A, with two surfaces located at z  0 and
z  L, respectively. An elemental control
volume, V  Az is located between z
and z  z. If the concentrat ion is
unchanging in time and there are no
chemical reactions producing or
comsuming species A, the principle of
conservation of mass species requires
that the flow of species A, denoted J A
[kg/s], across of the surfaces z
and z  z should be equal.

Mass Diffusion Through a Plan Wall
One has : J A z  J A z  z  jAz  A  constant, where J A is mass flow
of species A. Substituting jAz by Fick's law, i.e., jAz   DAdm A / dz, to
the above equation, one can obtain that
JA
dm
J
 jAx   DA A  A dz   DAdm A  constant
A
dz
A
In egrating across the wall and assuming that ρ and D Aare constant,
m A, L
JA L
JA
dz   DA  dm A  L  DA ( m A,0  m A,L )

0
mA, 0
A
A
( m A, 0  m A, L )
Then we obtain the result : J A 
. This solution is analog
L /( DA A)

to Ohm' s law, I  E/R, where I  J A , E  m A,0  m A,L , and R  L /( DA A).
The mass flow, J [kg/s] is also analog to heat flow, Q [J/s].
If the preceding analysis is repeated on a molar basis, the results becomes
( x A, 0  x A , L )
J 
, where J *A is the molar flow of species A [kmol/s].
L /( cD A A)
*
A

Homework #2
a) Find the three components of Fick’s law based on
cylindrical coordinates, r, , z for both mass and
molar flux of species A due to diffusion?

Components of the mass diffusion flux, jA in
cylinderic al coordinates, jAy   DAm A :
m A
m A
m A
jAr   DA
, jA   DA
, jAz   DA
r
r
z
*

Components of the molar diffusion flux, jA  cD Ax A :
x A *
x A *
x A
j  cD A
, jA  cD A
, jAz  cD A
r
r
z
b) Concentrations at Interface??
*
Ar

Concentrations at Interface
Although temperature is continuous
across a phase, concetration is usually
dicontinuous. In order to clearly define
concentrat ion, the subscripts u and s
denote the regions on the two sides of
the interface.
As shown in the water - air interface figure, u and s represent the water

and air phases, respectively, where air containing certain amount vapor
(moisture). If we ignore the small amount of air in the water and let
x H 2O,u  1, what is the water vapor concentrat ion along the interface in
the air - vapor phase?? The temperatures, Tu and Ts are 310 K and the
total pressure, Ps , is 105 Pa. It is also assumed that the thermodynamic
equilibrum exists at the interface.

Concentration
at Interface
At the equilibriu m conditon, the water vapor concentrat ion along the
water - air interface in the s - surface should be in the saturation
condition. Equilibri um saturation data for water - air systemsis available
in a conventional steam table. At Ts  310 K, the saturation vapor
pressure is 6,224 Pa. If the total pressure is 105 Pa, x H 2O can be found as
x H 2O,s

ci pi 6,224
 

 0.06224
5
c
p
10

and
(0.06224) 18
m H 2O,s 
 0.0396
(0.06224) 18  (1 - 0.06224) 29
where the molecular weight of air was assumed to be 29 u.

Concentration at Interface in a
Salt-Water System
In many processes, solubility data can be found in a chemical handbook.
For an example, consider a slab of salt dissolving in water at 30 oC and 1
bar (105 Pa), the corresponding solibility of salt in water is 36.3 g/100g
(usually in cgs unit). If the salt is pure, m NaCl, u  1, the NaCl concentrat ion
along the salt - water interface in the s - side at the equilibriu m conditon
can be found as
36.3
 0.266
100  36.3
On average, seawater in the oceans has a salinity of about 3.5% (35 g/kg),
m NaCl,s 

which means that every kilogram (roughly one litre by volume) of seawater
has approximately 35 grams of dissolved salts (predominantly sodium (Na  )
and chloride (Cl - ) ions).

Concentrations at Interface & Henry Number
In many processes, particularly thoseinvolving the absorption of
a gas into a liquid , the gas is frequently sparingly soluble. For such
a dilute solution, solubility data can be convenient ly represented by
x A,s  HeA  x A,u
where He A is the Henry number for species A. He is inversely
proportional to total pressure (p) and is proportional to
temperature (T). The product of He number and p for a given
species is a function of T only for a specific range of applications :
He A p  C HeA (T).
where C HeA is the Henry constant for species A. In general, the Henry
constant is dependent on the solute, the solvent and the temperature.

Concentrations at Interface & Henry Number
Let' s considerin g the absorption of carbon dioxide in water from a stream
of pure CO 2 at 3 bar and 300 K. The corresponding C He is 1710 bar; Thus,
HeCO2  C Hei (T)/p  1710/3  570, and
x CO2 ,u  x CO2 ,s / HeCO2  1 / 570  0.00175,
where a small partial pressure of water vapor at the s - surface may affect the
solubility of the gas in water, since, following Henry's law, the solubility of
a gas in a liquid is directly proportional to the partial pressure the gas above
the liquid. Advantage can be taken for increasing the solubility with pressure
in bottled or canned soft drinks or carbonated water. Note that, for highly
soluble gases, such as sulfur dioxide and ammonia, the above linear solubility
equations may not hold well.

Solubility of Carbonated Soft Drinks
An everyday example of Henry's law is given by
carbonated soft drinks. Before the bottle or can
of carbonated drink is opened, the gas above the
drink is almost pure carbon dioxide at a pressure higher than
atmospheric pressure. The drink itself contains dissolved
carbon dioxide. When the bottle or can is opened, some of
this gas escapes, giving the characteristic hiss (or "pop" in
the case of a sparkling wine bottle). Because the partial
pressure of carbon dioxide above the liquid is now lower,
some of the dissolved carbon dioxide comes out of solution
as bubbles. If a glass of the drink is left in the open, the
concentration of carbon dioxide in solution will come into
equilibrium with the carbon dioxide in the air, and the drink
will go "flat".

One-dimensional Transient Diffusion Formulation
Let' s considerin g a simplified one - dimensiona l transient diffusion problem.
Referring to the figue on the left, the principle of conservation of species A
applied to the elemental control volume V located between z and z  z
requires that the changes of species A on the control volume, V equals the
net infow of species A across the volume boundary :
 A V t  t -  A V t  (J A z  z  J A z )t
Substituting V  Az and J A  jA A, and dividing through
by Azt, , the above equation becomes
 A t  t -  A t
( jA z  z  jA z )

t
z
Letting z and t approaching to infinitesmally small, one
 A
j
has
 A
t
z
Introducing Fick' s law, i.e., jAz  ρD A-Bdm A /dz, gives
 A  ( m A )  
dm A 

  ρD A -B

t
t
z 
dz 

One-dimensional Transient Diffusion Formulation
In case of a dilute solution,  can be assumed to be constant and if the
process is isothermal, D A -Bcan also be taken to be a constant. Then the
above equation can be simplified as
m A
 2mA
 D A B
t
z 2
the above equation is analog to the one - dimensiona l transient heat
conduction equation, i.e.,
T
 2T
 2
t
z
where  is the thermal diffusivit y.
Diffusion in solids is characteri zed by its dilute solution and low - diffusion
coefficien t (~ 10-9 - 10-12 m 2 /s) and has many industrial applications, such as
hardening of steels by carbonization, and coloring of a clear saphire by
packing it with titanium oxide powders in a furnace to 2000C for 600 hrs.

HW#3 Hardening of Steel by Carbonization 1
In hardening of steels by heating steel components in a
carbonaceo us material in a furnace. By considerin g a
a slab of steel in a furnace at 800C with an initial
concentrat ion of species A equal to m A,0 (t  0), where
species A is carbon and B is iron. At carbonizat ion, the
concentrat ion of apecies A along the boundary (z  0) is
suddenly raised to m A,u , please describe the concentrat ion
profile of species A as a function of time, t, if D A-B is
constant. Also determine the rate of dissolution of species
A as a function of time.

Hardening of Steel by Carbonization 2
Carbonization involves small amount
of C 2 in iron, which is a typical dilute
solution case. The governing equation
becomes
m A
2mA
 D A B
t
z 2

with the initial condition : m A ( z,0)  m A,0 ,
and boundary conditions : m A (0, t)  m A,u & m A ( , t)  m A,0 .
The solution for the above problem can be found as


m A - m A,0
z

 erfc 
 4D t 
m A,u - m A,0
A- B 

where erfc is the complement ary error function.

HW#3 Hardening of Steel by Carbonization 3
[erfc(η )]
2  η2 η
Since

e
z
z
π
m A,u  m A,0 (z 2 /4DA  Bt)
m A
One has

e
z
πD A-B t
From the Fick' s law, jAz  ρD A
jAz 

m A
z

ρ
(z 2 /4DA  B t)
(m A,u  m A,0 )e
πt/DA-B

The concentrat ion gradient at z  0 can be
used to definite the penetration depth, δ p :
m A

z


z 0

m A,u  m A,0
πD A-B t



m A,u  m A,0
δp

Then δ p  πD A-B t  1.77( D A-B t)1/2  ( D A-B t)1/2
δ p is proportional to the square root of both the
diffusion coefficien t and time.

HW#3b Hardening of Steel by Carbonization 4
The diffusion coeffieien t of C 2 at 800 C is approximately
5.6  10 -10 m 2 /s. Also the metallugic al data indicates that the
maximum concentrat ion of C 2 in iron at 800 C at the phase
interface is 1.5%, so that m A,u can be assumed to be 0.015.
If the initial C 2 concentrat ion of the iron ( an mild steel) is
0.2%, please find for the location 1 mm below the iron surface
to have a C 2 concentrat ion of 0.8%.

HW#3b Hardening of Steel by Carbonization 5


z

.
The solution for the above problem is
 erfc


m A,u - m A,0
 4D A-B t 
Since m A,0  0.002, m A,u  0.015, m A  0.008, and z  0.001m,
m A - m A,0


 0.008 - 0.002
z


erfc
 0.462
 4D t  0.015  0.002
A- B 

Using a table for the error function, one has

z
 0.52
4D A-B t

z2
(0.001) 2
t

 1650s  0.46 h.
4D A-B (0.52) 2 4(5.6  10 10 )(0.52) 2
Comments
1. The diffusion coefficien t of C 2 in iron increases exponentially with temperature,
D A  D o e -Q/R uT , where Do is a proportionconstan, and Q is the activation energy. The
process normally is carried out at relatively high temperatureto the time required.
2. The boundary condition of m(0, t) was specified using the maximum value. To
have a more accurate result, m A,u requires the solubility data by considerin g the
diffusion in the specific carbonaceo us material.

HW#3b Hardening of Steel by Carbonization6
m A
2mA
The governing equation is
 D A B
t
z 2
with m A ( z,0)  m A,0 , and m A (0, t)  m A,b & m A ( , t)  m A,0 .
The solution for the above problem can be found as




m A - m A,0
m - m A,b
z
z
 or A

 erfc 
 erf 
 4D t  m A,0 - m A,b
 4D t 
m A,b - m A,0
A- B 
A- B 


where erf and erfc is the error and the error complement ary functions, respectively.
a. The above error function conversion can be proved as follows


m A - m A,b - m A  m A,b  m A,0  m A,0 m A,b - m A,0 m A  m A,0
z





 erf


m A,0 - m A,b
m A,b - m A,0
m A,b - m A,0 m A,b - m A,0
 4D A-B t 






m A  m A,0
m A  m A,0
z
z
z






1
 erf

 1 - erf
 erfc






m A,b - m A,0
m A,b - m A,0
 4D A-B t 
 4D A-B t 
 4D A-B t 
where erfc( )  1 - erf( ).
b. Let   z/ 4D A-B t and erf( ) 

2  2 
One has
[erf( )] 
e
,
z
z


2 η  η2
e dη & erf(0)  0, erf( )  1.
π 0
m A m A,0  m A,b z 2/(4DA-Bt)

e
z
D A B t

Mass Convection and Mass Transfer Rate
• Convective transport occurs when a constituent of the fluid (mass or
heat of a component in a mixture) is carried along with the fluid. The
amount carried past a plane of unit area perpendicular to the velocity
(the mass flux) is the product of the velocity and the concentration.
• It is generally used to describe the mass transfer between a surface
and a moving fluid.
• When mass transfer rates are low, there is a simple analogy between
heat convection and mass convection. Similar to heat convection, the
rate of mass convection is also a complicated function of surface
geometry, fluid velocity, composition, and properties.
• The mass transfer rate (in mass per second) is used to measure the
rate at which liquids and solids are moving. The density of the liquid
mixture can be assumed to be constant at low mass transfer rate
conditions, in which the amount of mass transfer of one species or the
mass fraction involved is relatively low (less than a few % of the total
liquid density).

Low Mass Transfer Rate & Convective Boundary Condition
A typical problem of low mass transfer rates is the evaporation of water into
air. The water-air interface might be the surface of a water reservoir or the
surface of a falling water droplet in cooling tower. In such situations, the mass
fraction of water vapor in the air is relatively small. The highest concentration
is at the interface, but even if the water temperature is as high as 50 ºC, the
corresponding mH2O,s at 1 atm, the saturated partial pressure of is only 0.077.
A typicalconvective boundary (or interface) condition for a water droplet in air can
be expressed as
ρ
m
h
jAr r  R  -D A-B A
 h m (ρ A,b - ρ A, )  If ρ is a constant, A
 m (m A,b - m A, )
r r  R
r r  R D A-B
where h m is the convective mass transfer coefficien t and is in [m/s]; R is the radius
of the droplet; the subscripts b represents the value at the boundary or interface while
 indicates the value at the ambient condition or the location away from the boundary.
The above convective boundary condition can also be expressed as
c
x
- D A- B A
 h m ( cA,b - cA, )  If c is a constant, D A-B A
 h m (x A,b - x A, )
r r  R
r r  R
Here, the mass transfer coefficien t, h m , is analog to the heat transfer coefficien t, h.

Heat and Mass Transfer Analogy
From the analogous forms of several dimensionless heat transfer and mass transfer
equations, it follows that, for a prescribed geometry and equivalent boundary
conditions, the functional dependencies of the Nusselt number, Nu, and the Sherwood
number, Sh, are equivalent. This means that the mass transfer coefficients, hm, might
be obtained from the correlations developed for heat transfer where the Prandtl
number (Pr = cpm/k) is replaced by Schmidt number (Sc = m/(DA-B)) and the Nusselt
number (Nu = hL/k) is replaced by Sherwood number (Sh = hmL/ DA-B).

Nu  f (Re, Pr)  Sh  f (Re, Sc)
where Re  ρUL/μ. Here, Sh is the ratio of the convective mass transfer
to diffusive mass transfer while Sc is the ratio of the momentum
(viscosity) diffusion rate to mass (molecular ) diffusion rate.
For example, in the case of fully developed turbulent flow in a smooth pipe, based
on heat transfer analyses, the correlation for Nud can be found as
0.4
Nu d  0.023Re 0.8
, for Pr  0.5
d Pr
0.4
which for mass transfer becomes Sh d  0.023Re 0.8
S
c
, for Sc  0.5
d

Many analogies, such as Chilton and Colburn analogy, Reynolds analogy, Prandtl–
Taylor analogy had been developed to directly relate heat transfer coefficients,
mass transfer coefficients, with different degrees of accuracy.

Mass Convection Correlations
The table below lists some other correlations to determine the mass transfer
coefficient for external forced convection flow. The expressions for the flat
plate are obtained from the solutions of the boundary layer equations, while
the other formulas are experimental correlations

Evaporative Cooling
If qadd  0, qcon  qevap

• The term evaporative cooling originates from association of the latent
energy created by evaporation at a liquid interface with a reduction in the
thermal energy of the liquid. If evaporation occurs in the absence of other
energy transfer processes, the thermal energy and temperature of the
liquid, must decrease.
• If the liquid is to be maintained at a fixed temperature, energy loss due to
evaporation must be replenished by other means (to compensate the
required latent heat). If no specific energy input, such as radiation from
the sun, it is most like that the convection heat transfer at the interface
can provide the only means of energy inflow to the liquid, an energy
balance yields qcon  qevap

Evaporative Interface Condition & Wet- & Dry-Bulb Psychrometer
If qadd  0, qcon  qevap , which means the convective heat flux equal to the
evaporation rate times the enthalpy of vaporization (h fg ) :
qcon   h(Tb  T )  qevap  h m [  A,b (Tb )   A, (T )]h fg or
 h(Tb  T )  h m [  (Tb )mA,b (Tb )   (T )mA, (T )]h fg
where m A,b (  A,b ) is frequently equal to its saturated concentrat ion, m A,sat (  A,sat ).
By considerin g radiation to the ambient (qrad ) and additional energy input,
qadd , the energy balance becomes : qcon  qadd  qevap  qrad .
Let’s considering a wet- & dry-bulb
Psychrometer, which is widely used for
measuring the moisture content of air.
In its simplest form, the air is made to
flow over a pair of thermometers, where
one of which has its bulb made wet by a
wick. Evaporation of water from the
wick causes the wet bulb to
cool from its steady-sate temperature, which is a function of the air temperature
measured by the dry bulb and the associated air humidity.

Wet- & Dry-Bulb Psychrometer 1
Use of wet and dry bulb
temperature measurements to
determine the relative humidity
of an air stream
Assumption:
1. Vapor follows the idea gas law, 2. Qusi - steady sate is assumed,
3. Radiation is negligible , and 4. Conduction along the thermometer is negligible .
M aterial Properties :
Air at 308K and 1 atm :   1.135 kg/m 3 , c p  1007 J / kg  K ,   23.7  106 m 2 / s.
Saturated vapor at 298K, v g  44.25m3/kg & h fg  2443kJ/kg, and
v g  15.52 m3/kg at 318K
Air - vapor at 1 atm, D A-B  0.26  104 m 2 /s at 298K,
D A-B  0.26  104 m 2 /s  (308 / 298)3 / 2  27  106 m 2 /s and
Le (Lewis number  Sc / Pr)   / DA B  0.878 at 308K.

Wet- & Dry-Bulb Psychrometer 2
Relative Humidity and Ideal Gas
In an ideal gas, the gas law can be expressed as p/ρ 

R uT
, where m*m is molar mass in
*
mm

[kg/kmol], and R u is the ideal gas constant equal to 8.314 J/K - mol . The relative humidity
is defined as, rh 

p H 2O, (T )
p H 2O,sat (T )

, where p H 2O, (T ) is the partial pressure of water vapor in

the mixture (air) and p H 2O,sat (T ) is the saturated vapor pressure of water at T .
Based on the ideal gas law,
R u T
R u T
&
p
(
T
)


(T
)
.
H 2O, sat

H 2O, sat

m*m
m*m

p H 2O, (T )   H 2O, (T )

substituting the above two equations into the defination rh , one has

 H O, (T )R u T / m*m  H O, (T )
rh 


*
p H O,sat (T )  H O,sat (T )R u T / m m  H O,sat (T )
p H 2O, (T )
2

2

2

2

2

Wet- & Dry-Bulb Psychrometer 3
Since the boundary condition for evaporating cooling is
 h(Tb  T )  h m [  A,b (Tb )   A, (T )]h fg , Dividing by  A,sat (T ) , one has
(T  Tb )
hm   A,sat (Tb )  A, (T ) 
 h fg


.
 A,sat (T )
h   A,sat (T )  A,sat (T ) 
 (T )
Since A,   RH , the relative humidity,
 A,sat (T )
  A,sat (Tb )

 A,sat (Tb )
(T  Tb )  h 
(T  Tb )  h 
 





,





RH 
RH


 A,sat (T )  A,sat (T )h fg  h m 
  A,sat (T )
  A,sat (T )h fg  hm 
where the horizontal line over the variable denotes the mean of the variable.
The coefficien t ratio, h / h m is evaluated using a more simple correlatio n reported
by Knudsen & Katz (1958) :
Nu d  hd / k  CRedm Pr1/ 3 and Sh d  h m d / D A B  CRedm Sc1/ 3 ,
where the constants C and m depend on Re d and all propertiesare evaluated at film
temperature, (Tb  T )/2  308K.
* J.D. Knudsen, D. L. Katz, Fluid Dynamics and Hat Transfer McGraw - Hill,
New York, 1958.

Wet- & Dry-Bulb Psychrometer 4
Then the ratio (h/h m ) becomes
h
k

h m D A B

1/ 3

 Pr 
 
 Sc 

k
1.135  1007  23.7 x10 6
1/ 3
1/3
Le 



0.878
D A B
27 x10 6

h
J

 960.67 3
,
m K
hm
where Le is the Lewis number and k  c p .

 A,sat (Tb )
 A,sat (T ) 15.52
The term
can be found to be

 0.351 and
 A,sat (T )
 A,sat (Tb ) 44.25
 A,sat (T )  1 /  A,sat (T ).
Then RH

 A,sat (Tb )
(T  Tb )  k
1/ 3 
Le 



 A,sat (T )  A,sat (T )h fg  D A B

 0.351 

( 45  25)
 960.67  0.351  0.122  0.229
3
(1 / 15.52) x 2443x10

Wet- & Dry-Bulb Psychrometer: Discussions
a) The wet - and dry - bulb psychrometer is widely used for measuring the moisture
content of air. The above caculation is accurate only for humidity measuremen t in
water vapor - air mixtures at low temperature, when the saturate vapor concentrat ion
is relatively low, that is, the usual conditions encountere d in air - conditioning practice.
b) Amore accurate empirical correlatio n developed by Churchill & Bernstein (1997)
for a flow over a circular cylinder in overall average condition can also be used
4/5

  Red 5 / 8 
0.62 Re Pr
Nud  hd / k  0.3 
  , for Red Pr  0.2
1  
[1  (0.4 / Pr)2 / 3 ]1 / 4   282,000  
where all properties are evaluated at the film or mean temperature(  Tb  T )/2
1/ 2
d

1/ 3

The corresponding mass transfer correlatio n becomes
Sh d  h m d / DA B

1/ 2
d

1/ 3

0.62 Re Sc
 0.3 
[1  (0.4 / Sc 2 / 3 ]1 / 4

  Red 

1  
  282,000 

5/8 4 /5





, for RedSc  0.2

To apply theabove correlatio ns for the h/ h m ratio, a numerical scheme is needed.
* *S. W. Churchill, M . Bernstein, ASM E J. Heat Transfer, 99, 300, 1977.

Wet- & Dry-Bulb Psychrometer: Discussions
c. The effect of ignoring radiation heat transfer is worth
further study.
d. Also the wet - bulb temperature recorded by a
psychrmeter is in general not the same as the adiabatic
saturation temperature or thermodynamic wet - bulb
temperature that is the temperature at which water, by
evaporating into air in a constant pressure - mixing
process, can bring the air to saturation adiabatica lly
at the same temperature.

Evaporation of a Water Droplet 1
A water droplet with a diameter of 50mm initially at 315K is sprayed into air stream at
315 K, 1.05x10 5 Pa, and 50.5% of relative humidity. Please estimate the droplet life time.
Assumptions or simplifica tions :
a) After a short initial transient, the droplet temperature reaches a steady state value. A
quasi - steady state condition is assumed in this problem.
b) Radiation heat transfer is negligible , since the associated temperature is relatively low.
c) the droplet is entrained or slowly falling into a thick air, the Sherwood number (Sh) is in
its lower bond  5 and the ratio of convective coefficien ts is assumed to be constant,
i.e., h/h m  k / D A B . All propertiesare evaluated at the film temperature.
Analysis and Results :
Using the convective boundary condition, the evaporation rate of the droplet is

jA,b A  -D A-B A
A  h m (  A,b -  A, ) A  h m  (m A,b - m A, )A
r r  R
where A is the surface of the droplet equal tod 2 . A mass balance on a droplet requires
that the rate of mass loss equal to the evaporation rate. Since the droplet volume 
d π 3 
dd
2
 2 h m (m A,b - m A, )/ w
 d ρ w    jA,b A   h mρ(m A,b - m A, )d 
dt  6
dt

where  w is the water density.

π 3
d ,
6

Evaporation of a Water Droplet 2
Since Shd  h md/ D A-B  5  h m  5D A-B / d
0

dd


 10D A-B  air (m A,b - m A, )/(  w d)   ddd  10D A-B
(m - m A, )  dt
do
0
dt
 w A,b

d 02

 w d 02
 10D A-B
(m - m A, )   
2
 w A,b
20D A-B  (m A,b - m A, )
The ambient H 2 0 concentrat ion at T  315K and 1 atm, m A, can be obtained as follows :
The pressure at the saturate condition can be found as p A,sat  0.08135 x 105 Pa (from Text)
 p A,  ( RH )  p A,sat  0.505 x 0.8135 x105  0.0411x10 5 Pa,
x A,  p A, / p  0.0411 / 1.050  0.0391, with M H 2O  18.02kg/kmol; M air  28.97kg/kmol
m A, 

0.0391M H 2O
0.0391M H 2O  (1  0.0391) M air



0.0391x18
 0.0246
0.0391x18  (1  0.0391) x 29

To obtain Tb or m A,b , the boundary condition for the evaporating cooling can be applied :
 h(Tb  T )  hm [  (Tb )mA,b (Tb )   (T )mA, (T )]h fg . Since h/h m  k / DA B was given
DA B h fg
hm
Tb  T 
h fg [  (T )mA, (T ) -  (Tb )mA,b (Tb )] 
[  (T )mA, (T ) -  (Tb )mA,b (Tb )]
h
k

Evaporation of a Water Droplet 3
Iteration is required to obtain m A,b , since the Tb is also unknown. In the first
iteration, Tb  T1  305 K. From the table in the text :
p A,sat  0.04714x10 5 Pa at 305K, x A,sat  p A,sat /p  0.04714/1.05  0.0449 
m A,b 

0.04497(18)
0.8081

 0.02835.
0.0449(18)  (1  0.0449)(29) 28.5061

Data : m A,b  0.02835 at 305K, m A,  0.0246 at 315K, ρ at 315K  1.12kg/m 3 ,
ρ at 305K  1.15kg/m 3 , the film temperature, Tm  (315  305) / 2  310K,
h fg  2.4x10 6 J/kg at 310K, D A-B  25.6 x10 4 (310 / 298) 3 / 2  27 x10 6 m 2 /s,
and k  0.027W/m - K at 310K.
Tb  T 

D A B h fg

[  (Tb )mA,b (Tb ) -  (T )mA, (T )]

k
27x10 6  2.4x10 6
 315 
(1.15  0.02835 - 1.12  0.0246)  315 - 11  304K
0.027
Since Tb  304K is close to the assumed T1 , 305K, no further iteration is needed.

Evaporation of a Water Droplet 4
Tm  (315  304)/2  310K,  air (310 K )  1.13kg / m 3 ,  w (310 K )  995.7kg / m 3 ,
At 304 K, p A,sat  0.04454x10 5 Pa, x A,sat  p A,sat /p  0.04454/1.05  0.04242 
m A,b

0.04242(18)
0.8082


 0.02676
0.04242(18)  (1  0.04242)(29) 28.5061

 w d 02

995.7 x (50 x10 6 ) 2


20D A-B  air (m A,b - m A, ) 20  27 x10 6  1.13(0.02676  0.0246)
2.489 x10 6

 1.89s for d o  50x10 -6 m
4
6.102  x10  0.00216
995.7 x (100 x10 6 ) 2
-6
If d o  100x10 m,  
 7.55s
6
1.318 x10
HW 4a : If the ratio of convective coefficien ts, h/h m is changed to 2k/D A B , please
repeat the example problem. Please compare your results with the above results
and discuss your findings.
4b : If the Sh number is changed to 10 (h/h m is still k/D A B ), please repeat the example
problem presented in the class and discuss your finding.

Evaporation of a Water Droplet: Discussions 1
a) Significan t errors can be caused by theassumption of Sh  5 and the ratio of
convective coefficien ts, h/h m ,  k / D A B and these parameters should be studied.
For example, to obtain a more reliable solution for a prblem of a freely falling
droplet,a heat transfer correlatio n related to freely falling droplets reported by
Ranz and M arshall (1952) could be used for Sh number and h/h m ratio estimations :
1/3
1/3
Nu d  2  0.6Re1/2
 Shd  2  0.6Re1/2
d Pr
d Sc

Since Re d is a function of the droplet velocity, which can be accelerate d by the
gravity force and affected by air drag, and frequently numerical schemes are needed
estimate both Sh number and the h/h m ratio. If Re d is a constant and equal to1000,
 Sh d  hm d / D A B  2  0.6(1000)1/ 2 ( 4.18)1/ 3  32.564.  hm  32.564 D A B / 50 x10 6.
Since Shd  h m d/ D A-B  32.564  h m  32.564D A-B /d,
0
τ
dd
ρ
 65.128D A-Bρ air (m A,b - m A, )/(ρ w d)   ddd  65.128D A-B
(m A,b - m A, )  dt
do
0
dt
ρw

d 02

 w d 02
 65.128D A-B
(m A,b - m A, )   
2
w
130.256D A-B  (m A,b - m A, )
* W. Ranz, W. M arshall, Evaporation from drops, Chem. Eng. Prog., 48, 141 - 146, 1952.

Species Continuity Equation: Mass Conservation 1
The figure in the left shows the conservation of mass for
species A in a small element V(  xyz ) in Cartesian
coordinates. The mass balance for species A for the
volume element ΔVcan be expressed by
 Rate of
  Rate of species   Rate of species
species A
   A in by mass    A out by mass 

 
 

accumulati on  inflow
 outflow

 Rate of species   Rate of species A transfer 

  A generation by    to V by diffusion

 

chemical reaction  from surroundings


The mass of component A flowing into and out from V through three pairs of
parallel surface element perpendicular to the x-, y-, and z - coordinates. One has
 A u A 
j Ax 


  A 



V


u


u


x

y

z

r

V

j

j

x  yz


 A A

 Ax
A A
A
Ax


x
x
 t 






where rA is the rate of species A generation by chemical reaction per unit of volume.

Species Continuity Equation: Mass Conservation 2
Let Δx, Δy,and Δz become infinitesi mal small, i.e.,  0 and divided by V, one has
 A
 u
 
 
  A A  rA    DA A 
t
x
x 
x 
Using vector forms to include the components in y and z directions , the above
equation becomes
 A
   (  AU A )    ( DA A )  rA (general) *
t
Let' s consider t he case of incompressible fluids, From the continuity equation, one
has   U A  0 and so   (  AU A )  U A A   A  U A  U A A
Since ρ A  ρm A , for constant  and D A , the above mass - balance equation can
be simplified as
 A
 U A   A  DA2  A  rA (constant  and D A ) *
t
The above equations with * can be called the species differenti al mass - balance
equations or the species continuity equations.

Species Continuity Equation: Mass Conservation 3
Similar equations can also obtained based molar density, c A :
c A
   ( c AU A )    ( DAc A )  rA*
(general) *
t
c A
 U A  c A  DA2c A  rA*
(constant  and D A ) *
t
The species continuity equation in terms of mass concentation can be
expressed in Cartesian coordinates :
 u A v A wA 
 A   A
 A
 A 
  u x
 uy
 uz


   A 
 
t 
x
y
z 
y
z 
 x
  2  A  2  A  2  A   DA  A DA  A DA  A 
  
 DA 




  rA
2
2
2 
y
z   x x
x y
x z 
 x
For constant  and D A , the above equation becomes
 2 A 2 A 2 A 
 A   A
 A
 A 
  rA
  u x
 uy
 uz


  DA 
2
2
2 
t 
x
y
z 
y
z 
 x

Species Continuity Equation: Mass Conservation 4
M olar concentrat ion based on Cartesian coordinates for the case of constant  and D A :
  2cA  2cA  2cA  *
c A  c A
c A
c A 
  u x
 uy
 uz
  D A  2  2  2   rA
t  x
y
z 
y
z 
 x
To solve the convective mass trasfer problems, either boudary theory or numerical schemes
are needed. The following two papers dealing with the evaporation of a droplet and
convection effects are selected here to illustrate the approaches and procedures involved in
the numerical analysis :
1. A. M . Briones, J. S. Ervin, M icrometer - sized water droplet impingemen t dynamics and
evaporation on a flat dry surface, Langmuir, 26, 13272 - 13286, 2011.
2. S. Torfi, S. M . H. Nejad, Absorptionand evaporation mass transfer simulation of a
droplet by finite volume method and SIM PLEC algorithm, Applied M echanics &
M aterials,110 - 116, 4315 - 4323, 2012.
****
H.W. #5 Find the species continuity equation based on cylindrical coordinates in terms of
mass concentrat ion for the cases of constant  and D A .

Carbon Hollow Fiber Application
Carbon Dioxide (and/or H2S) is separated from natural gas (CH4) by selectively
permeation through a hollow fiber membrane (see the figure below). The driving
force is the partial pressure difference across the membrane for CO2 (and/or
H2S) , CH4 and other gas components. CO2 (and/or H2S) is the “fast” gas
whereas CH4 is the “slow” gas. This membrane technology is based on
polymeric hollow fiber. The pressurized feed gas enters the bundle from the
shell side, the methane stays under pressure, and the CO2 (and/or H2S) is
collected at a lower pressure. High selectivity
makes 95%+ methane recovery available. A
typical membrane system consists of a pretreatment skid and a series of membrane
modules

Air Liquide (MEDAL)
Newport, DE USA

Technical Writing Improvement
Technical writing essentially is the communication abilities
with technical data and language. I do not have any shortcut
for the improvement of these abilities, since, in most of time, I
also have difficulties to write a technical paper with the quality
I feel comfortable with. This should be a long term effort. I
have few suggestions:
• I should be able to have lunches with any of your graduate
students, so that they have to talk to me one-to-one in English
while I have a better understanding of your students as well as
the Czech culture, as a whole.
• To set-up a PC workstation with an English operating system
with the software of Photoshop, MathLab, PDF professional
version, and ...... (immersing in a globalized world!)
• Encouraging students to have a gmail account to read its
English news at least 20 mins everyday. To have n gmail
account, you can have a face-to-face talk to anyone in the world
FREE!

A Personal Note

Acknowledgements
& Thank You!!
The generous funding provided by Czech Ministry of
Education, Youth and Sports through Project No.
HEATEAM - CZ.1.07/2.3.00/20.0188 and professorship by
Brno University of Technology (BUT) for me to stay in Brno
are acknowledged. I am thankful to my host, Professor
Miroslav Raudenský, for providing the wonderful
opportunity for me to work with his talented group. Special
thanks are to Messrs Ondrej Kristof and Tomáš Veselý for
their technical support in preparing this presentation. I also
like to express my gratitude and appreciation to Miss Hana
Hladilova and Mrs Zuzana Vetešníková for their enormous
patience and hospitality for making my stay in Brno a lovely
experience.

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