heat and mass transfer

1. 1
COMBINED HEAT AND MASS TRANSFER
In MEG 310.2, the basic principles of heat and mass transfer were extensively introduced. Thus, in
this section of the course, the fundamental equations are only highlighted. Although some salient
aspects of mass transfer are briefly expounded, the main focus here is the study of combined heat
and mass transfer.
References and Bibliography (reading list)
1. Cengel Y., Ghajar, A. J.: Heat and Mass Transfer Fundamentals and Applications, 4th
Ed., McGraw
Hill, Boston, 2011.
1. Rogers Y., Mayhew Y.: Engineering thermodynamics Work and Heat Transfer, 4th
Ed., Addison-
Wesley Longman, Singapore, 2001.
These notes are built mainly from Ref. 1.
Heat Transfer vs Mass Transfer
From Einstein’s 𝐸 = 𝑚𝑐2
, it is clear that heat and mass transfer can be reasonably conflated as
forms of energy.
Table 1: Summary of Heat and Mass Transfer Equations.
Mode
Form
Conduction* Convection Radiation
Heat Transfer
𝑄̇ 𝑐𝑐𝑐𝑐 = −𝑘𝑘
𝑑𝑑
𝑑𝑑
𝑄̇ 𝑐𝑐𝑐𝑐 = ℎ 𝑐𝑐𝑐𝑐 𝐴 𝑠(𝑇𝑠 − 𝑇∞) 𝑄̇ 𝑟𝑟𝑟
= 𝜎𝐴 𝑠 𝑇4
Mass Transfer
𝑚̇ 𝑑𝑑𝑑𝑑 = −𝐷𝐴𝐴 𝐴
𝑑𝐶𝐴
𝑑𝑑
𝑚̇ 𝑐𝑐𝑐𝑐 = ℎ 𝑚𝑚𝑚𝑚 𝐴 𝑠(𝐶𝑠
− 𝐶∞)
**N/A
*Mass conduction is also called mass diffusion.
**Mass is not transferred by radiation! (Sublimation and evaporation are not radiation processes)
𝑄̇ is the rate of heat transfer 𝐶 is the concentration
𝑚̇ is the rate of mass transfer 𝑇 is the absolute temperature
𝑘 is the thermal conductivity 𝐶𝐴 is the concentration of species A at the
location
ℎ is the transfer coefficient (𝑇𝑠 − 𝑇∞) : temperature difference across the
thermal boundary layer
𝐷𝐴𝐴 is the diffusion coefficient (𝐶𝑠 − 𝐶∞) : concentration difference across the
concentration boundary layer
𝐴 𝑎𝑎𝑎 𝐴 𝑠 are the cross sectional and surface
area respectively
From the equations highlighted in Table 1, Heat transfer is seen to be driven by a temperature
difference, while mass transfer is driven by a difference in concentration.

2. 2
Q.
1. What is the difference between mass transfer and bulk fluid motion (say fluid flow in an
inclined pipe)?
2. Why is conduction expressed in terms of cross sectional area but convection in terms of
surface area?
Mass Transfer (A review)
It has been seen that the driving force behind mass transfer is a concentration difference. Now we
see how the concentration is defined.
i. In terms of mass
Here, concentration is defined as mass per unit volume (density). Considering a small volume V
within a mixture of species:
concentration of each species (partial concentration): 𝜌𝑖 =
𝑚 𝑖
𝑉
total concentration of the mixture:
𝜌 =
𝑚
𝑉
= � 𝑚𝑖/𝑉 = � 𝜌𝑖 1
i represents each species in the mixture.
Thus the concentration of the mixture is a simple sum of the concentration of its constituent species.
Note that, in mass terms, concentration is defined as density.
ii. In terms of moles
In terms of moles, concentration is defined as amount of matter in kmol per unit volume. This is
known as molar concentration or molar density. As before,
concentration of each species (partial concentration): 𝐶𝑖 = 𝑁𝑖/𝑉
total concentration of the mixture:
𝜌 =
𝑁
𝑉
= � 𝑁𝑖/𝑉 = � 𝐶𝑖 2
Hence the concentration of the mixture is a simple sum of the molar concentration of its constituent
species.
Relations between mole fraction, mass fraction and pressure fraction
Mole fraction of each species is defined as:
𝑦𝑖 =
𝑁𝑖
𝑁
=
𝐶𝑖
𝐶
Now,

3. 3
𝑚 = 𝑁𝑀
M is the molar mass. Dividing both sides by V gives
𝜌 = 𝐶𝐶
The mass fraction is defined as:
𝑤𝑖 =
𝜌𝑖
𝜌
3
𝑤𝑖 =
𝜌𝑖
𝜌
=
𝐶𝑖 𝑀𝑖
𝐶𝐶
= 𝑦𝑖
𝑀𝑖
𝑀
4
For an ideal gas
𝑃𝑃 = 𝑁𝑁𝑁, 𝑃 = � 𝑃𝑖
The pressure fraction is defined as:
𝑃𝑖
𝑃
. Hence,
𝑃𝑖
𝑃
=
𝑁𝑖 𝑅𝑅
𝑉
𝑁𝑁𝑁
𝑉
= 𝑦𝑖 5
Mass Transfer across an Interface between Boundaries at Different States
Gas/ Liquid Interface
At the interface of a gas and liquid, say at the surface of a river or between a fizzy drink and the gas
(Co2) stored between its surface and the container’s top/seal, there is mass transfer between the gas
and the liquid, if the gas dissolves in the liquid.
Two cases are studied here:
Case 1: Dilute solutions i.e. the gas is weakly soluble in the liquid
For dilute solutions, Henry’s law applies at the interface i.e. just below and above the surface of the
liquid:
𝑦𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠 ∝ 𝑦𝑖 𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠 6
𝑦𝑖 𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠𝑑𝑒 refers to the mole fraction of the gas species in the mixture just below the surface of
the liquid (interface).
𝑦𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠 refers to the mole fraction of the gas species above the surface of the liquid (interface)
But for an ideal gas, from Eq. 5,
𝑃𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠
𝑃
= 𝑦𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠. Thus,
𝑃𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠 ∝ 𝑃𝑦𝑖 𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠

4. 4
Fig. 1: A representation of the gas/liquid interface
Hence at the interface,
𝑦𝑖 𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠 =
𝑃𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠
𝑘1 𝑃
7
𝑘1 is a constant of proportionality
P is the total pressure of the gas mixture above the liquid
𝑘1 𝑃 is known as the Henry constant (see Table 2)
Table 2 [1]: Henry constant for various gases dissolved in water under varying temperatures at low
to medium pressures.
This means that the mole fraction of the gas just below the surface of the liquid is proportional to
the partial pressure of the gas in the gas mixture above the liquid. This mole fraction will be
homogeneous throughout the liquid only if the liquid is in thermodynamic phase equilibrium
(mainly for pure liquids).
Case 2: Gas is strongly soluble in the liquid e.g. ammonia and water.
In this case, Raoult’s law applies:
𝑃𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠 = 𝑦𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠 𝑃 = 𝑦𝑖 𝑙𝑖𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠 𝑃𝑖 𝑠𝑠𝑠(𝑇) 8
𝑃𝑖 𝑠𝑠𝑠(𝑇) is the saturation pressure of the gas species at the interface temperature (from tables)

5. 5
P is the total pressure.
Q:
How do the two previous relations account for surface area? (Surely, the surface area of the contact
surface must affect mass transfer!)
Gas/Solid Interface
Diffusion of a gas into a solid is a very important consideration in metallurgy. An example is
hydrogen embrittlement. It is modelled as
𝐶𝑖 𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠 ∝ 𝑃𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠 9
and
𝐶𝑖 𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠 = 𝔍𝑃𝑖 𝑔𝑔𝑔 𝑠𝑠𝑠𝑠 10
𝔍 is the solubility of the gas in the solid
Note that: 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 𝔍𝐷𝐴𝐴 11
Recall that 𝐷𝐴𝐴 is the diffusion coefficient.
This process can be reversible i.e. the gas diffuses in the solid and dynamic equilibrium ensues. In the
irreversible case, the gas dissolves in the solid and is ‘locked in’.
Q1. A polymer plate is in contact with hydrogen gas at 298 K and 250 kPa. Determine the molar
densities of hydrogen in the polymer at the interface
Q2. For a bottle of fizzy soda water (carbonated water) at 300 K and 1.3 bar, assuming that the CO2 is
trapped between the seal and pure water, determine (a.) the mole fraction of the water vapour in
the CO2 gas (b.) the mass of CO2 in a 1 litre drink.
Mass Diffusion
Mass conduction is referred to as mass diffusion.
Steady State Mass Diffusion
Recall that for heat transfer, the rate of heat transfer through a solid plane wall of thickness L with
inlet temperature 𝑇1 and outlet temperature 𝑇2 under steady conditions (i.e. the temperature
distribution of the wall does not change with time) is given by
𝑄̇ 𝑐𝑐𝑐𝑐 = −𝑘𝑘
𝑇1 − 𝑇2
𝐿
Similarly, the rate of transfer of mass of a species A through a plane wall of thickness L with mass
fraction of the species at the inlet 𝑤 𝐴1 and at the outlet 𝑤 𝐴2 under steady conditions is given by
𝑚̇ 𝑑𝑑𝑑𝑑 = 𝜌𝐷𝐴𝐴 𝐴
𝑤 𝐴1 − 𝑤 𝐴2
𝐿
12

6. 6
Proof:
Recall that mass diffusion is given by Fick’s law (in mass terms)
𝑚̇ 𝑑𝑑𝑑𝑑
𝐴
= −𝜌𝐷𝐴𝐴
𝑑𝑑 𝐴
𝑑𝑑
= 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 13
*Constant because steady state is considered.
Now integrating over the inlet and outlet surface of a solid plane wall of thickness L measured along
the x axis from the origin
𝑚̇ 𝑑𝑑𝑑𝑑
𝐴
� 𝑑𝑑
𝐿
0
= −𝜌𝐷𝐴𝐴 �
𝑑𝑑 𝐴
𝑑𝑑
𝑤 𝐴2
𝑤 𝐴1
giving Eq. 12
𝑚̇ 𝑑𝑑𝑑𝑑 = 𝜌𝐷𝐴𝐴 𝐴
𝑤 𝐴1 − 𝑤 𝐴2
𝐿
Similarly in mole terms, Eq. 12 is expressed as
𝑁̇ 𝑑𝑑𝑓𝑓 = 𝐶𝐷𝐴𝐴 𝐴
𝑦𝐴1 − 𝑦𝐴2
𝐿
14
Note that this analysis only valid for a small concentration of species for which 𝜌 𝑎𝑎𝑎 𝐷𝐴𝐴 are
constant.
Q.
Recall that steady state heat transfer through a cylindrical section is given by
𝑄̇ 𝑐𝑐𝑐𝑐 = 2𝜋𝜋𝜋
𝑇1 − 𝑇2
𝐼𝐼(𝑟2/𝑟1)
Now as done in the previous section, show that steady state mass diffusion through a cylindrical
section is given by (in mole terms)
𝑁̇ 𝑑𝑑𝑑𝑑 = 2𝜋𝜋𝜋𝐷𝐴𝐴
𝑦𝐴1 − 𝑦𝐴2
𝐼𝐼(𝑟2/𝑟1)
15
and in mass terms,
𝑚̇ 𝑑𝑑𝑑𝑑 = 2𝜋𝜋𝜋𝐷𝐴𝐴
𝑤 𝐴1 − 𝑤 𝐴2
𝐼𝐼(𝑟2/𝑟1)
16
𝑟2 𝑎𝑎𝑎 𝑟1are the outer and inner radii of the cylinder respectively. L is its length.
Note that in terms of partial pressure, for a plane wall, Eqs. 12 and 13 become:
𝑁̇ 𝑑𝑑𝑑𝑑 = 𝔍 𝐴𝐴 𝐷 𝐴𝐴 𝐴
𝑃𝐴1 − 𝑃𝐴2
𝐿
17

7. 7
Eq. 17 is obtained by substituting Eq. 10 in Eq. 14 (i.e. in terms of the concentrations of the species
on the solid side of both side of interface).
Q. Express the molar transfer rate across a cylinder in terms of partial pressures (i.e. Eq. 15 in terms
of partial pressures – similar to Eq. 17).
It can now be seen that the analogy between heat and mass transfer can be expressed as
Table 2: Analogy between heat and mass transfer
Mass Diffusion
Heat Conduction In mass terms In mole terms
T wi yi
k 𝜌𝐷𝐴𝐴 𝐶𝐷𝐴𝐴
𝑞̇
𝑗𝑖 =
𝑚̇ 𝑑𝑑𝑑𝑑
𝐴 𝚥𝚤� =
𝑁̇ 𝑑𝑑𝑑𝑑
𝐴
𝛼 𝐷𝐴𝐴 𝐷𝐴𝐴
* for species i
Q3. A thin polymer membrane of thickness 2mm separates nitrogen from the atmosphere. The
molar concentrations of nitrogen inside and outside the membrane are 0.065 and 0.003 kmol/m3
respectively. The diffusion coefficient in the polymer is 5.3 x 10-10
m2
/s. What is the mass flux
through the membrane under steady conditions?
Q4. The solubility of hydrogen gas in steel in terms of its mass fraction is given in terms of its mass
fraction as 𝑤ℎ2
= 2.09 𝑥 10−4
exp �−
3950
𝑇
� 𝑃ℎ2
0.5
; where 𝑃ℎ2
is the partial pressure of hydrogen in
bars and T is the temperature in K. if natural gas is transported in a 1 cm thick and 3 m internal
diameter steel pipe at 500kPa pressure and the mole fraction of hydrogen in the natural gas is 8%,
determine the highest rate of hydrogen loss through a 100 m long section of pipe at steady
conditions at a temperature of 293 K if the pipe is exposed to air. The diffusivity of hydrogen in steel
is 2.9 x 10-13
m2
/s. [1]
Transient Mass Diffusion
Transient diffusion occurs before steady conditions form. Transient mass diffusion is important in
case hardening heat treatment processes e.g. carburisation, boriding and nitriding which involves
hardening the outer surface (layers) of a metallic material while allowing the inner section to stay
ductile and malleable (soft). Transient diffusion occurs in relatively short time periods usually at very
high temperatures (relatively).
Recall that transient heat transfer is given by (proof beyond the scope of this course)
𝑇(𝑥, 𝑡) − 𝑇𝑖
𝑇𝑠 − 𝑇𝑖
= 𝑒𝑒𝑒𝑐�
𝑥
2√ 𝛼𝛼
�
𝑇𝑠 is the fluid temperature and𝑇𝑖 is the temperature of the solid
erfc is called the complementary error function. It is a very complex integral but can easily be read
off tables see Table 3 [1].

8. 8
Table 3 [1]: The complementary error function
Thus, transient mass diffusion can be written as
𝐶𝐴(𝑥, 𝑡) − 𝐶𝐴𝐴
𝐶𝐴 𝑠 − 𝐶𝐴 𝑖
= 𝑒𝑒𝑒𝑒 �
𝑥
2� 𝐷𝐴𝐴 𝑡
� 18
𝐶𝐴𝐴 is the initial concentration of species A at t=0
𝐶𝐴𝐴 is its concentration at the inner side of the exposed surface
Now, the depth of penetration is (proof not required)
� 𝜋𝜋𝐴𝐴 𝑡 19
Q5. In a carburising process, the top surface of a steel part of initial carbon content 0.15% is to be
hardened in a furnace at 1200 K. The mass fraction of the exposed surface of the part is maintained
at 1.25% by the carbon-rich environment. The mass diffusivity for carbon in steel at 1200 K is 8 x 10-
12
m2
/s. If the process is to continue until the mass fraction of carbon at a depth of 0.80mm is raised
to 35%, how long must the process last? [1]
Mass Convection
Like heat convection which is a combination of heat conduction and convective heat transfer due to
bulk fluid motion, mass convection is a combination of mass diffusion and mass convection due to
bulk fluid motion.
In basic fluid mechanics, the thickness of the (velocity) boundary layer was measured from the
surface (which the fluid is in contact with) to the boundary where the fluid velocity is 0.99 of the free
stream velocity. Similar definitions to this velocity gradient have been made for a thermal boundary
layer for heat gradients, and the principle can be extended to define a concentration boundary layer
for species concentration gradients (see Fig. 1 [1]).

9. 9
Now, back to our definition of heat convection as
ℎ𝑒𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ℎ𝑒𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + ℎ𝑒𝑒𝑒 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑑𝑑𝑑 𝑡𝑡 𝑏𝑏𝑏𝑏 𝑓𝑓𝑓𝑓𝑓 𝑚𝑚𝑚𝑚𝑚𝑚 20
At the surface, Prandtl’s no slip condition means that the velocity of the fluid with respect to the
surface is zero. Thus, heat transfer is wholly by heat conduction at the surface. In the same way,
mass transfer at the surface is wholly by mass conduction.
Fig 2: a. An illustration of the velocity, thermal and concentration boundary layers in internal flow
(b) concentration boundary layers in external flow
Like fluid dynamics, heat convection is an overwhelmingly empirical subject. Thus, dimensionless
numbers are also used to classify heat transfer regimes. Using the afore-mentioned analogy
between heat and mass transfer, the dimensionless numbers used in heat transfer can be extended
to mass transfer.
In Fig. 1, the relative magnitudes of molecular momentum diffusion (in the velocity boundary layer)
and heat diffusion (in the thermal boundary layer) are related by the Prandtl number:
𝑃𝑃 =
𝜈
𝛼
=
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑇ℎ𝑒𝑒𝑒𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
21
The Schmidt number relates momentum diffusion to mass diffusion (in the concentration boundary
layer):
𝑆𝑆 =
𝜈
𝐷𝐴𝐵
=
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
22
And the Lewis number relates thermal and mass diffusion:
𝐿𝐿 =
𝑆𝑆
𝑃𝑃
=
𝛼
𝐷𝐴𝐴
=
𝑇ℎ𝑒𝑒𝑒𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑖𝑖𝑖
23
Recall that, from Table 1, heat convection in external flows is given by:
𝑄̇ 𝑐𝑐𝑐𝑐 = ℎ 𝑐𝑐𝑐𝑐 𝐴 𝑠(𝑇𝑠 − 𝑇∞)
ℎ 𝑐𝑐𝑐𝑐 is the average heat transfer coefficient
(𝑇𝑠 − 𝑇∞) is the temperature difference across the thermal boundary layer
And convective mass transfer:

10. 10
𝑚̇ 𝑐𝑐𝑐𝑐 = ℎ 𝑚𝑚𝑚𝑚 𝐴 𝑠(𝜌 𝐴 𝑠 − 𝜌 𝐴 ∞) 24
ℎ 𝑚𝑚𝑚𝑚, 𝜌 is the average mass transfer coefficient and average density of the fluid
(𝜌 𝐴 𝑠 − 𝜌 𝐴 ∞) is the mass concentration difference of species A across the concentration boundary
layer.
In your previous study of heat transfer, ℎ 𝑐𝑐𝑐𝑐 was nondimensinalised using the Nusselt number as
𝑁𝑁 =
ℎ 𝑐𝑐𝑐𝑐 𝐿
𝑘
25
Similarly, the Sherwood number is used to nondimensionalise ℎ 𝑚𝑚𝑚𝑚 as
𝑆ℎ =
ℎ 𝑚𝑚𝑚𝑚 𝐿
𝐷𝐴𝐴
26
The Stanton number is important because it is used to relate the above numbers.
The Stanton number for heat transfer is
𝑆𝑡ℎ𝑒𝑒𝑒 = 𝑁𝑢
1
𝑅𝑅 𝑃𝑃
=
ℎ 𝑐𝑐𝑐𝑐
𝜌𝜌𝐶 𝑝
27
V is the velocity outside the boundary layer or mean velocity
𝐶 𝑝 is the specific heat of the fluid
For mass transfer,
𝑆𝑡 𝑚𝑚𝑚𝑚 = 𝑆ℎ
1
𝑅𝑅 𝑆𝑆
=
ℎ 𝑚𝑚𝑚𝑚
𝑉
28
Other important points to note include:
In forced convection
𝑁𝑁 = 𝑓(𝑅𝑅, Pr) 29
𝑆ℎ = 𝑓(𝑅𝑅, 𝑆𝑆) 30
In natural convection transfer
𝑁𝑁 = 𝑓(𝐺𝐺, Pr) 31
𝑆ℎ = 𝑓(𝐺𝐺, 𝑆𝑆) 32
Recall that
𝐺𝑟ℎ𝑒𝑒𝑒 =
(𝑇𝑠−𝑇∞)𝛽𝛽𝐿 𝑐
3
𝜈2
33
and

11. 11
𝐺𝑟 𝑚𝑚𝑚𝑚 =
(𝜌∞ − 𝜌𝑠)𝑔𝐿 𝑐
3
𝜌𝜈2
34
Gr is the Grashof number
𝐿 𝑐 is the concentration entry length of the boundary layer (Fig 1)
Analogy between friction, heat and mass transfer coefficients
Fig. 1 illustrates the simultaneous concepts of velocity, heat and mass transfer in fluid dynamics.
Thus, analogies relating these seemingly unrelated concepts are necessary. These analogies will
emphasise the relevance of the dimensionless numbers highlighted in the previous section.
Note that the corresponding coefficients in velocity, heat and mass transfer are the friction factor 𝑓,
ℎ 𝑐𝑐𝑐𝑐 and ℎ 𝑚𝑚𝑚𝑚 respectively.
Reynold’s Analogy
It deals with the special case where:
Pr ≈ Sc ≈ 1
which makes it possible to write (proof not necessary)
𝑓
2
𝑅𝑅 = 𝑁𝑁 = 𝑆ℎ 35
i.e.
𝑓
2
𝑉∞ 𝐿
𝜈
=
ℎ 𝑐𝑐𝑐𝑐 𝐿
𝑘
=
ℎ 𝑚𝑚𝑚𝑚 𝐿
𝐷𝐴𝐴
36
Since P = Sc = 1
𝑓
2
=
𝑁𝑁
𝑅𝑅 𝑃𝑃
=
𝑆ℎ
𝑅𝑅 𝑆𝑆
which is
𝑓
2
= 𝑆𝑡ℎ𝑒𝑒𝑒 = 𝑆𝑡 𝑚𝑚𝑚𝑚 37
By measuring one of these coefficients, it becomes possible to determine the other two. But only if
Pr ≈ Sc ≈ 1.
Chilton-Colburn Analogy
This analogy extends the relationship beyond Pr ≈ Sc ≈ 1. It is written as
f
2
= Stheat 𝑃𝑟
2
3 = Stmass 𝑆𝑐
2
3 38
For

12. 12
0.6 < Pr, Sc < 60
These analogies are only applicable to regimes where the rate of mass transfer of the concerned
species is small relative to its flow rate.
Rewriting Eq. 38 as
Stheat
Stmass
= �
𝑆𝑆
𝑃𝑃
�
2
3
giving
hheat
hmass
= 𝜌𝐶 𝑝 �
𝑆𝑆
𝑃𝑟
�
2
3
= 𝜌𝐶 𝑝 �
𝛼
𝐷𝐴𝐴
�
2
3
= 𝜌𝐶 𝑝 𝐿𝑒
2
3 39
Eq. 39 is a very useful form of the Chilton-Colburn analogy.
Determining the Mass Transfer Coefficient
The mass transfer coefficient can be determined:
1. From the friction coefficient or the heat transfer coefficient using the Chilton-Colburn
analogy. This can only be done if the friction or heat transfer coefficient is known.
2. By choosing the same boundary conditions and replacing the Nusselt and Prandtl numbers
by the Sherwood and Schmidt numbers respectively (as shown in Table 3). See also Eqs. 29-
32.
Table 4 [1]: Sherwood number relations in mass convection corresponding to the Nusselt number in
heat convection for different flow regimes

13. 13
Q6. In an experiment to determine the average heat transfer coefficient for air flow of a part, air at 1
atm at a free stream velocity of 3 m/s is blown over the part covered with a layer of naphthalene.
The part has surface area of 0.8 m2
, and results show that 110 g of naphthalene sublimated in 45
mins. The temperature of the part and naphthalene was kept at 298 K during the experiment, at
which the mass diffusivity of naphthalene is 0.61 x 10-5
m2
/s and its vapour pressure is 11 Pa.
Determine the heat transfer coefficient using the Chilton-Colburn analogy. [1]
Simultaneous Heat and Mass Transfer