Simultaneous heat and mass transfer processes involve the transfer of both heat and mass, such as in evaporation. Evaporation from a water body requires latent heat from the surface and surroundings to vaporize the water molecules (mass transfer). The rate of heat transfer is equal to the rate of mass transfer multiplied by the latent heat of vaporization. Several approximations are provided to calculate the rate of mass transfer. The rate of heat transfer by convection is set equal to the rate of mass transfer multiplied by the latent heat, resulting in an expression relating the temperature difference between the surface and surroundings to the rates of heat and mass transfer.

1. 1
Simultaneous (Unified) Heat and Mass Transfer
In the previous sections, we studied heat and mass transfer separately. However, processes like
phase change processes which involve mass transfer – such as evaporation, sublimation, ablation
and sweating, there is simultaneous heat and mass transfer. Processes like these underline the
relevance of the Chilton-Colburn analogy.
Consider evaporation from the surface of a water body. If the air above the water body is saturated
i.e. relative humidity is 100%, no evaporation occurs (using static equilibrium analysis). Thus no heat
or mass transfer occurs. But if the air is not saturated, water evaporates from the surface and mixes
with the air just above the surface as water vapour. Vapourisation requires latent heat which comes
from the surface water, deeper parts of the water, and radiation from the surroundings. Thus
𝑄̇ = 𝑚̇ 𝑣ℎ 𝑓𝑓 1
𝑚̇ 𝑣 is the rate of mass evaporation
ℎ 𝑓𝑓 is the latent heat absorbed
𝑄̇ represents all heat sources (convection from the fluid, radiation from the environment,
conduction from within the fluid, artificial heating or chemical reactions)
Table 1 [1]:
Now, in Table 1 [1], several approximations for 𝑚̇ 𝑣 are presented.
Assuming all sources of heat but convection are negligible, Eq. 1 can be written as
𝑄̇ 𝑐𝑐𝑐𝑐 = 𝑚̇ 𝑣ℎ𝑓𝑓 2
From Table 1,
ℎ 𝑐𝑐𝑐𝑐 𝐴 𝑠(𝑇∞ − 𝑇𝑠) =
ℎ 𝑐𝑐𝑐𝑐 𝐴 𝑠ℎ 𝑓𝑓
𝐶 𝑝 𝐿𝑒
2
3
𝑀𝑣
𝑀
𝑃𝑣,𝑠 − 𝑃𝑣,∞
𝑃
3

2. 2
Q: In Eq. 3, compared to the expression in Table [1], what happened to the density?
giving
𝑇𝑠 = 𝑇∞ −
ℎ 𝑓𝑓
𝐶 𝑝 𝐿𝑒
2
3
𝑀𝑣
𝑀
𝑃𝑣,𝑠 − 𝑃𝑣,∞
𝑃
4
Eq. 4 gives a relationship between the temperature difference between the boundary (represented
by s) and the surrounding gas (represented by ∞) under steady conditions.
Q1. A canned drink is to be cooled by wrapping a wet cloth around it and blowing air to it with a fan.
If the environmental conditions are 1 atm, 33 o
C and 50% humidity, how cool can the drink get when
steady state conditions are reached? [1]
Q2. In a blending plant, pressurised cans of product are tested in a well agitated hot water bath
heated to, and maintained at, 55 o
C by an electric heater, and open to ambient air. The dimensions
of the bath are 3 x 1.5 x 0.3 m and the cans are to be tested to determine if they can withstand
temperatures up to the bath’s temperature of 55 o
C during transportation. If the average conditions
in the plant are 0.9 bar, 30 o
C, 45% relative humidity, and the plant surfaces are at an average
temperature of 25 o
C; determine the rate of heat loss from the top surface of the water bath by
a. radiation
b. natural convection
c. evaporation [1]

3. 3
Exergy
References and Bibliography (reading list)
1. Rogers Y., Mayhew Y.: Engineering thermodynamics Work and Heat Transfer, 4th
Ed., Addison-
Wesley Longman, Singapore, 2001.
1. Boxer G.,: Theory, Worked Examples and Problems, Macmillan, London, 1978.
*Theory from Ref. 1, examples from Ref. 2 (old but very reliable).
Exergy (availability) is the maximum useful work that can be done by a system interacting with its
environment which is at constant pressure p0 and temperature T0 [1]. Since all practical processes
are irreversible, availability presents a means for practically evaluating the irreversibilities of a
process or cycle [2].
As the system at initial state p1, T1 interacts with the surroundings, its final state approaches that of
its surroundings p0, T0. The interaction can happen in two ways:
• The system does work on its surroundings while degenerating to the p0, T0
• The surroundings do work on the system which approaches p0, T0
The first type of system is simpler and thus used in the following analysis.
Closed Systems
For a closed system made of a compressed fluid at state p1,T1 behind a closed piston, the useful work
done when it expands to the state of the surroundings is derived from the First Law as
𝑄 + 𝑊 = 𝑈1 − 𝑈0 1
Since the system does work on the surroundings,
|𝑊| = 𝑄 + (𝑈1 − 𝑈0) 2
Now as the piston also does work on the surroundings at constant pressure po the useful work is less.
Recall that for a constant pressure process 𝑊 = −𝑝𝑝𝑝, in this case 𝑊𝑠𝑠𝑠𝑠 = −𝑝0(𝑉0 − 𝑉1)
Thus the useful work done by the system is
|𝑊| = 𝑄 + (𝑈1 − 𝑈0) − 𝑝0(𝑉0 − 𝑉1)
which can be rewritten as
|𝑊| = 𝑄 − (𝑈0 − 𝑈1) − 𝑝0(𝑉0 − 𝑉1) 4
In order to express Q in terms of temperature, we use the principle of increasing entropy which
states that: the change of entropy of the surroundings and of the system is greater than or equal to
zero.

4. 4
When heat is transferred to the system from the surroundings which is at a constant temperature T0
∆𝑆 + �−
𝑄
𝑇0
� ≥ 0 5
𝑄 ≤ 𝑇0∆𝑆 6
Substituting Eq. 6 in 4 gives
|𝑊| ≤ 𝑇0(𝑆0 − 𝑆1) − (𝑈0 − 𝑈1) − 𝑝0(𝑉0 − 𝑉1) 7
rearranged to give
|𝑊| ≤ (𝑈1 + 𝑝0 𝑉1 − 𝑇0 𝑆1) − (𝑈0 + 𝑝0 𝑉0 − 𝑇0 𝑆0) 8
The maximum value of Eq. 8 corresponding to the maximum work is obtained when the process is
reversible. Therefore
|𝑊|max = (𝑈1 + 𝑝0 𝑉1 − 𝑇0 𝑆1) − (𝑈0 + 𝑝0 𝑉0 − 𝑇0 𝑆0) 9
written as
|𝑊|max = 𝐴1 − 𝐴0 10
where 𝐴 = (𝑈 + 𝑝 𝑜 𝑉 − 𝑇0 𝑆) is the Non Flow Exergy Function.
Stead-Flow Open Systems
Consider an open system, with an infinite source reservoir supplying fluid at p1,T1 at velocity C1 at
height z1 at a steady rate; discharging to the surroundings at p0,T0 and zero velocity at datum z0.
From the steady flow energy equation,
𝑄̇ + 𝑊̇ = 𝑚̇ [(ℎ0 − ℎ1) +
1
2
(𝐶0
2
− 𝐶1
2) + 𝑔(𝑧0 − 𝑧1)] 11
rearranging
�𝑊̇ � = �𝐻̇1 + 𝑚̇ �
1
2
𝐶1
2
+ 𝑔𝑧1�� − �𝐻̇0 + 𝑚̇ 𝑔𝑧0� + 𝑄̇ 12
Note that the enthalpy terms encompass the work done against the surroundings at entry and exit.
As in the non-flow case, 𝑄̇ is expressed in terms of temperature using the principle of increasing
entropy.
∆𝑆̇ + �−
𝑄̇
𝑇0
� ≥ 0
𝑄̇ ≤ 𝑇0∆𝑆̇
Now neglecting the comparatively small K.E and P.E terms, Eq. 12 can be written as

5. 5
�𝑊̇ � ≤ −(𝐻̇0 − 𝐻̇1) + 𝑇0�𝑆̇0 − 𝑆̇1� 13
Again, the maximum work is done when the process is reversible i.e.
�𝑊̇ � 𝑚𝑚𝑚
= −�𝐻̇0 − 𝐻̇1� + 𝑇0�𝑆̇0 − 𝑆̇1� = 𝐵̇1 − 𝐵̇0 14
𝐵 = (𝐻 − 𝑇0 𝑆) = (𝑈 + 𝑝𝑝 − 𝑇0 𝑆) is the Steady Flow Exergy Function.
Irreversibility (i)
In expansion, irreversibility is defined as
𝑖 = |𝑊| 𝑚𝑚𝑚 − |𝑊| 𝑎𝑎𝑎𝑎𝑎𝑎 15
The reverse is the case for compression
𝑖 = 𝑇0�∆𝑆𝑠𝑠𝑠𝑠𝑠𝑠 + ∆𝑆𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠� 16
Effectiveness (ε)
For the case where W or Q is being transferred to the surroundings from the system
𝜀 =
𝑔𝑔𝑔𝑔 𝑖𝑖 𝑏 𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑙𝑙𝑙𝑙 𝑜𝑜 𝑏 𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠
17
When W or Q is being transferred to the system from the surroundings,
𝜀 =
𝑔𝑔𝑔𝑔 𝑖𝑖 𝑏 𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠
𝑙𝑙𝑙𝑙 𝑜𝑜 𝑏 𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
18
where 𝑏 =
𝐵
𝑚
Tutorials
1. Consider a gas expanding in a turbine from state p1,T1 to a pressure p0 as shown in Fig. 1.
Fig 1: expansion in a turbine
Show that the maximum useful work per unit mass flow is given by
|𝑊| 𝑚𝑚𝑚 = 𝑐 𝑝(𝑇1 − 𝑇0) − 𝑐 𝑝 𝑇0 𝐼𝐼
𝑇1
𝑇0
+ 𝑅𝑇0 𝐼𝐼
𝑝1
𝑝0

6. 6
where 𝑏 =
𝐵
𝑚
2. A jet of water at 85 o
C running at a rate 30 kg/s mixes adiabatically with another jet of water at
40 o
C running at 40 kg/s. calculate the rate of loss of available energy.
3. Two vessels A and B with volumes 4 m3
and 3 m3
respectively are connected by a pipe containing a
valve. With the valve shut, A contains air at 6 bar, 450 K whilst there is a vacuum in B. With the valve
opened, calculate the loss in available energy if the flow is adiabatic, and the volume of the valve is
comparatively small. [2]
4. A rigid, perfectly insulated vessel is divided into two parts by a non-permeable membrane. One
part contains 0.06 m3
of nitrogen at 6.5 bar and 80 o
C, and the other part contains 0.07 m3
of oxygen
at 13 bar and 40 o
C. If the membrane suddenly fails and ruptures completely, calculate the
irreversibility when both gases mix.
Assume that both gases are perfect with 𝐶 𝑝, 𝑁2
= 1.04
𝑘𝑘
𝑘𝑘
𝐾 and 𝐶 𝑝, 𝑂2
= 0.92
𝑘𝑘
𝑘𝑘
𝐾. [2]