Introduction to Convection:
Mass Transfer
Chapter Six and Appendix E
Sections 6.1 to 6.8 and E.4
Concentration Boundary
Layer
• Features
The Concentration Boundary Layer
– A consequence of evaporation or
sublimation of species A from a
liquid or solid surface across
which a second fluid species B
is flowing.
– A region of the flow characterized
by species fluxes and concentration
gradients.
– A region between the surface and
free stream whose thickness  c
increases in the flow direction.
– Why does  c increase in the flow
direction?
– Manifested by a surface species
flux N A, s and a convection mass
transfer coefficient hm .
– What is the heat transfer analog
to Fick’s law?
c 
C A, s  C A  y 
C A, s  C A , 
N A   DAB
 0.99
C A
y
Fick ' s Law
N A, s   DAB
C A
y
y 0
N A, s  hm  CA, s  CA, 
hm 
 DAB C A / y y  0
C A, s  C A, 
Concentration Boundary (cont.)
• Definitions
Term
Species molar flux
Variable
N A, s
kmol/s·m2
Species molar rate
N A, s
kmol/s
Species mass flux
nA, s
kg/s·m2
Species mass rate
nA
kg/s
Species molar concentration
CA
kmol/m3
Species mass concentration (density)
A
kg/m3
Species molecular weight
MA
kg/kmol
Convection mass transfer coefficient
hm
m/s
Binary diffusion coefficient1
DAB
m2/s
1
Table A.8
Units
Concentration Boundary
Layer (cont.)
• Convection Calculations
Species Molar Flux:
N A, s  hm  CA, s  CA, 
Species Mass Flux:
nA  M A N A ,  A  M ACA

nA, s  hm  A, s   A, 

Total Transfer Rates:
N A, s  hm As  CA, s  CA, 

nA, s  hm As  A, s   A, 

Average Mass Transfer Coefficient:
1
hm   hm dx
As As
Vapor Concentration
Species Vapor Concentration or Density
• At a Vapor/Liquid or Vapor/Solid Interface
The vapor concentration/density corresponds to saturated conditions at
the interface temperature Ts .
C A, s  C A, sat Ts 
 A, s   A, sat Ts   M ACA, sat Ts 
Assuming perfect gas behavior, the concentration/density may be estimated from
knowledge of the saturation pressure.
p A, sat Ts   A, s
C A, s 

RTs
MA
The concentration may also be directly determined from saturation tables.
E.g., from Table A.6 for saturated water,
 A, s  vg1 Ts   M ACA, s
Vapor Concentration (cont.)
• Free Stream Conditions
– The free stream concentration/density may be determined from knowledge
of the vapor pressure, p A,  , assuming perfect gas behavior.
C A,  
p A, 
RT

 A, 
MA
– For water vapor in air, the free stream concentration/density may be
determined from knowledge of the relative humidity,  .
 
p A, 
p A, sat T 
For dry air,
  0

C A, 
C A, sat T 

 A, 
 A, sat T 
Similarity
Species Boundary Layer Equation and Similarity
• Species boundary layer approximation:
C A
C A
y
x
• Species equation for a non-reacting boundary layer:
C A
C A
 2C A
u
v
 DAB
x
y
y 2
What is the physical significance of each term?
Is this equation analogous to another boundary layer equation?
Similarity (cont.)
• Dimensionless form of the species boundary layer equation:
C  C A, s
C A*  A
C A,   C A, s
*
C A*
1  2C A*
* C A
u
v

x*
y* Re L Sc y *2
*
Sc 
v
 the Schmidt number
DAB
How may the Schmidt number be interpreted?
• Functional dependence for a prescribed geometry:

C A*  f x* , y* ,Re L , Sc
hm 
 DAB C A / y
C A, s  C A, 
y 0

DAB  C A,   C A, s  C A*

L  C A, s  C A,   y*
y*  0
DAB C A*

L y*
y*  0
Similarity (cont.)
The dimensionless local convection mass transfer coefficient is then
hm L C A*
Sh 

DAB y*

 f x* , Re L Sc

y*  0
Sh  local Sherwood number
What is the functional dependence of the average Sherwood number?
Analogies
Analogies
• Heat and Mass Transfer Analogy
From analogous forms of the dimensionless boundary layer energy and
species equations, it follows that, for a prescribed geometry and equivalent
boundary conditions, the functional dependencies of Nu and Sh are equivalent.


Sh  f  x , Re, Sc 
Nu  f x* , Re, Pr
*
Since the Pr and Sc dependence of Nu and Sh, respectively, is typically
of the form Prn and Scn, where n is a positive exponent (0.30 ≤ n ≤ 0.40),


Nu
Sh
*

f
x
,
Re

Pr n
Sc n
hL / k hm L / DAB

n
Pr
Sc n
h
k Pr n
k


Le n   c p Le1 n
n
hm DAB Sc
DAB
Le 

DAB
 the Lewis number
Analogies (cont.)
• Reynolds Analogy
For dp* / dx*  0
Re
Cf
 Nu
2
and Pr  Sc  1,
Nu  Sh
it follows that
Cf Sh

 Stm
2 Re
where
h
Sh
 m  the mass transfer Stanton number
Re Sc V
• Modified Reynolds Analogy
Cf
 Stm Sc 2 / 3  jm
0.6  Sc  300
2
Stm 
Colburn j factor for mass transfer
– Applicable to laminar flow if dp*/dx* ~ 0.
– Generally applicable to turbulent flow without restriction on dp*/dx*.
Evaporative Cooling
Evaporative Cooling
• 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 hence the 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. Assuming convection
heat transfer at the interface to provide the only means of energy inflow to
the liquid, an energy balance yields
  qevap

qconv
Evaporation Cooling (cont.)
h T  Ts   nAh fg  hm   A, sat Ts    A,  h fg
T  Ts    hm / h    A, sat Ts    A,  h fg
Steady-state
Cooling
Obtained from heat/mass
transfer analogy
• With radiation from the interface and heat addition by other means,
  qadd
  qevap
  qrad

qconv
Problem: Naphthalene Sublimation Method
Problem 6.60: Use of the naphthalene sublimation technique to obtain the
average convection heat transfer coefficient for a gas
turbine blade.
KNOWN: Surface area and temperature of a coated turbine blade. Temperature and pressure of air
flow over the blade. Molecular weight  M A  128.16 kg/kmol  and saturation vapor pressure of the
naphthalene coating. Duration of air flow and corresponding mass loss of naphthalene due to
sublimation.
FIND: Average convection heat transfer coefficient.
SCHEMATIC:
Naphthalene coating
Air flow
p = 1 atm
Too = 27oC
Ts = 27oC, As = 0.05 m2
pA,sat = 1.33x10-4 bar
m = 8 g for t = 30 min
Problem: Naphthalene (cont.)
ASSUMPTIONS: (1) Applicability of heat and mass transfer analogy, (2) Negligible change in As
due to mass loss, (3) Naphthalene vapor behaves as an ideal gas, (4) Solid/vapor equilibrium at surface
of coating, (5) Negligible vapor density in freestream of air flow.
3
-6
PROPERTIES: Table A-4, Air (T = 300K):  = 1.161 kg/m , cp = 1007 J/kgK,  = 22.5  10
2
-5 2
m /s. Table A-8, Naphthalene vapor/air (T = 300K): DAB = 0.62  10 m /s.
ANALYSIS: From the rate equation for convection mass transfer, the average convection mass
transfer coefficient may be expressed as
hm 

nA
As A,s  A, 


m / t
As  A,s
where
A,s  A,sat  Ts  
MA pA,sat
 Ts
128.16 kg / kmol 1.33  104 bar

 6.83  104 kg / m3
0.08314 m3  bar / kmol  K  300K 
Hence,
hm 
0.008 kg /  30 min 60s / min 

0.05m2 6.83 104 kg / m3

 0.13m / s
Problem: Naphthalene (cont.)
Using the heat and mass transfer analogy with n = 1/3,
2/3
  
2
/
3
 0.130 m / s 1.161kg / m3 
h  h m  cp Le
 h m  cp 

 DAB 




2/3

6

5
1007 J / kg  K 22.5 10 / 0.62 10
 359 W / m 2  K
COMMENTS: The naphthalene sublimation technique is used extensively to determine convection
coefficients associated with complex flows and geometries.
Problem: Wet-bulb Thermometer
Problem 6.74: Use of wet and dry bulb temperature measurements to determine
the relative humidity of an air stream.
SCHEMATIC:
ASSUMPTIONS: (1) Perfect gas behavior for vapor, (2) Steady-state conditions, (3)
Negligible radiation, (4) Negligible conduction along thermometer.
3
PROPERTIES: Table A-4, Air (308K, 1 atm):  = 1.135 kg/m , cp = 1007 J/kgK,  = 23.7
-6 2
 10 m /s;
3
Table A-6, Saturated water vapor (298K): vg = 44.25 m /kg, hfg = 2443 kJ/kg; (318K): vg = 15.52 m3/kg;
-4 2
-4 2
Table A-8, Air-vapor (1 atm, 298K): DAB = 0.26  10 m /s, DAB (308K) = 0.26  10 m /s
3/2
-4 2
 (308/298) = 0.27  10 m /s, Le = /DAB = 0.88.
Problem: Wet-bulb Thermometer
ANALYSIS: Dividing the energy balance on the wick by  A,sat T  ,
  A,sat  Ts 
A, 


.

T

T




 A,sat 
A,sat  
With   A, /  A,sat  T     for a perfect gas and h/hm obtained from the heat and mass
transfer analogy,
T  Ts
h 
 h fg  m 
 A,sat  T 
 h 
 
 cp
 A,sat  Ts 

 T  Ts  .
 A,sat  T  Le2/3  A,sat  T  h fg
where
A,sat  Ts 
A,sat  T 

vg  T 

vg  Ts 

15.52
 0.351
44.25
A,sat  T   15.52 m3 / kg

1
 0.064 kg/m3 .
Hence,
  0.351 
1.135 kg/m3 1007 J/kg  K 
 0.88
2/3

0.064 kg/m3 2.443 106 J/kg
  0.351  0.159  0.192.

 45  25  K