Heat Transfer
Main ref: Chemical Engineering, Volume 1: Fluid Flow, Heat Transfer
and Mass Transfer, Coulson and Richardson, 6th ed, Chapter 9
Chapter 9
Thermodynamics and Heat Transfer
• Heat: The form of energy that can be transferred from one
system to another as a result of temperature difference.
• Thermodynamics is concerned with the amount of heat
transfer as a system undergoes a process from one
equilibrium state to another.
• Heat Transfer deals with the determination of the rates of
such energy transfers as well as variation of temperature.
• The transfer of energy as heat is always from the highertemperature medium to the lower-temperature one.
• Heat transfer stops when the two mediums reach the same
temperature.
• Heat can be transferred in three different modes:
conduction, convection, radiation
3
Heat Transfer in Chemical Processes
• In most chemical processes, heat is given out or
absorbed
• Fluids must often be heated or cooled, e.g. distillation
units, reaction vessels
• Important that heat is transferred at suitable rate
• It can also be important to minimize heat loss e.g.
from a pipe
• Control of flow of heat important in chemical
engineering
Conduction, Convection and Thermal Radiation
Conduction refers to the transport of energy in a medium due to a temperature gradient.
Solid: Transfer of vibrational energy between molecules. Liquid: Also transfer of kinetic
energy between molecules. Metals: Also via movement of free electrons (hence metals
have high thermal conductivity).
Convection refers to heat transfer between a surface and a fluid (at rest or in motion), or
between two fluids. This process involves mixing of fluid elements – not governed by
temperature difference alone.
Two types:
Natural convection: Due to density difference, e.g. when a pool is heated from below
Forced convection: Eddy movement in a fluid, e.g. when a pipe is heated from the outside
Thermal Radiation All materials
radiate thermal energy in the form of
electromagnetic
waves.
When
radiation falls on a second body, it is
reflected, transmitted or absorbed –
if absorbed, the temperature (heat)
of the second body increases.
Individual and Overall Heat Transfer Coefficients
Consider heat transfer between flames heating a pipe and the liquid
inside:
Radiation from flames, conduction through the pipe wall, and convection
to/through the liquid.
Q = UADT
Q = heat transferred per unit time (J/s)
A = area available for flow of heat (m2)
DT = temperature difference between flame and liquid in pipe (K)
U = overall heat transfer coefficient (W/(m2 K))
UUdepends
(heaton:
transfer coeff) is the constant of
between the “rate of heat
•proportionality
Mechanism of heat transfer
•transfer
Fluid dynamics
(Q)”of fluid
and “the driving force for the heat
• Properties of wall of pipe (thermal conductivity) and its thickness
(DT) multiplied
byvary
thewith
area”.
•transfer
U not necessarily
a constant – may
temperature and DT (eqn
above not necessarily a linear relationship)
Heat Transfer by Conduction
Consider heat transfer through a wall of thickness x:
Rate of heat flow Q over the area A and a small
distance dx:
The negative sign indicates temperature gradient is
in the opposite direction of the flow of heat.
k = thermal conductivity of the material (W/(m K))
If wall thickness is x, and boundary temperatures are T1 and T2:
Overall heat transfer coefficient (U) is here equal to k/x
Thermal Conductivities (room temperature)
Think of it as transfer of kinetic/vibrational energy
between adjacent atoms/molecules – if they are far
away from one another, as in gases, very low
thermal conductivity. The exact opposite case is a
metal, where atoms are organized close to one
another in the solid state.
•
•
•
•
Metals: very high
Non-metallic solids: lower
Non-metallic liquids: even lower
Gases: very low
Thermal Diffusivity
How fast does heat diffuse through a material?
cp
Specific heat, J/kg · °C: Heat capacity
per unit mass
c p

Heat capacity (storage), J/m3·°C: Heat
capacity per unit volume
Thermal diffusivity, m2/s: Represents
how fast heat diffuses through a material
A material that has a high thermal
conductivity or a low heat capacity will
obviously have a large thermal diffusivity.
The larger the thermal diffusivity, the faster
the propagation of heat through the
medium.
A small value of thermal diffusivity means
that heat is mostly absorbed by the
material and a small amount of heat is
conducted further.
Heat Transfer by Convection
Heat transfer by convection is a result of movement of fluid on a macroscopic scale in the form
of eddies and circulating currents.
Natural Convection: Currents arise from the heating process itself. Consider heating of a
beaker containing a liquid using a heat source under the beaker. The density of liquid near the
bottom decreases as the temperature increases, and it thus rises, being replaced by colder
liquid from above.
Forced Convection: Currents are provided by an agitator or turbulent flow in a pipe.
If convective heat transfer is taking place from a heated surface to a fluid, there are no
circulating currents in the immediate vicinity of the surface – a film of fluid (free of turbulence)
covers the surface. Heat transfer by thermal conduction occurs through this film. More agitation
 Thinner film  Higher rate of heat transfer.
If resistance mainly lies in this film:
The film thickness is typically unknown, and the eqn is written in terms of h, the heat transfer
coefficient of the film:
EXAMPLE 1
A warm object of initial temperature 60 C is cooling in air of
temperature 22 C. The temperature of the object is measured as a
function of time. Given the data below, what is the value of the heat
transfer coefficient?
Air temperature
Newton’s Law of Cooling:
The rate of heat loss of a body is directly proportional to the difference in
the temperatures between the body and its surroundings provided the
temperature difference is small and the nature of radiating surface remains same.
Consider a warm object cooling in air:
y (t )  Tt  Tair
dy d
 Tt  Tair    hTt  Tair    hy
dt dt
y  y0 e  ht
Tt  Tair  T0  Tair e
DT
 e  ht
DT0
 DT 
   ht
ln
 DT0 
 ht
Because the rate of change in
temperature difference, i.e. differential with
respect to time, is proportional to the
temperature difference as stated in
Newton’s Law of Cooling (constant of
proportionality = h)
First order kinetics!
Same type of eqn as in reaction kinetics.
A plot of ln(DT/DT0) vs t should give
a straight line with slope –h.
EXAMPLE 1
A warm object of initial temperature 60 C is cooling in air of
temperature 22 C. The temperature is measured as a function of
time. What is the value of the heat transfer coefficient?
Approximately linear – follows Newton’s
Law of Cooling.
h = 0.0607 Jmin-1K-1 = 1.01 x 10-3 W K-1
Note: Here h also incorporates the area
term.
EXAMPLE 2
The Cost of Heat Loss through a Roof
The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m
thick, and is made of a flat layer of concrete with thermal conductivity k =
0.8 W/m °C . The temperatures of the inner and the outer surfaces of the
roof one night are measured to be 25°C and 0°C, respectively, for a
period of 10 hours.
(a) What is the rate of heat loss through the roof?
(b) What is the cost of that heat loss if the cost of electricity is $0.2/kWh.
The kilowatt hour (symbol kWh) is a derived unit of
energy equal to 3.6 megajoules. If the energy is being
transmitted at a constant rate over a period of time,
the total energy in kilowatt hours is the power in
kilowatts multiplied by the time in hours. The
kilowatt hour is commonly used as a billing unit for
energy delivered to consumers by electric utilities.
SOLUTION
(a) Heat transfer through the roof is by conduction and the area of the roof is:
A = 6 m × 8 m = 48 m2
The steady rate of heat transfer through the roof is determined to be
Q = kA(T1-T2)/x = (0.8)(48)(25-0)/0.25 = 3840 W = 3.84 kW
(b) The amount of heat lost through the roof during 10 h and its cost:
Q x Dt =(3.84 kW)(10 h) = 38.4 kWh
Cost of that heat loss (i.e. per 10 h) =
= (Amount of energy) x (Unit cost of energy) = (38.4 kWh)($0.2/kWh) = $7.68
Cost/month = (cost/day)×(30 days/month)= $7.68×30 = $230.4
EXAMPLE 3
Estimate the heat loss per square meter of surface through a
brick wall 0.5 m thick when the inner surface is at 400 K and the
outer surface is at 300 K. The thermal conductivity of the brick
may be taken as 0.7 W/mK.
SOLUTION
EXAMPLE 3
Estimate the heat loss per square meter of surface through a
brick wall 0.5 m thick when the inner surface is at 400 K and the
outside surface is at 300 K. The thermal conductivity of the brick
may be taken as 0.7 W/mK.
Note that here the area of the
wall is set as 1, thus giving us the
answer per square meter.
Thermal resistances (x/k) in series
Chemical Engineering, Volume 1: Fluid Flow, Heat Transfer and Mass
Transfer, Coulson and Richardson, 6th ed, p. 390
EXAMPLE 4
EXAMPLE 4
Temperature drop is
proportional to thermal
resistance

Material Balances
Main ref: Elementary Principles of Chemical Processes, Felder/Rousseau, 1st ed, Chapter 2
Chemical Plant
Extremely complex !!
• In industry, chemical processes typically involve several
steps
• An industrial process is normally treated as a series of unit
operations – one unit operation refers to a basic step
involving a chemical transformation (chemical reaction) or
physical transformation (e.g. filtration, crystallization)
• Materials are transferred between the different steps (unit
operations)
• Essential to know amount/composition/condition of material
entering/leaving process unit
• We need means of representing complex processes
(qualitatively and quantitatively)
General material balance equation
qin and qout denote flow rates into and out of the process unit.
Methane flows in and out of the unit.
If qin ≠ qout
Possible explanations:
1. Methane is leaking
2. Methane is consumed or generated (chemical reaction)
3. Methane is accumulating in the unit
4. The measurement(s) of qin and/or qout are wrong
Input – Output + Generation – Consumption = Accumulation
This balance can refer to total mass of material
or to any molecular or atomic species involved
Main ref: Elementary Principles of Chemical Processes, Felder/Rousseau, 1st ed; p. 85
Material balances on reactive systems
• Chemical reaction(s) occur, i.e. compounds are
consumed/generated
• Must include generation and consumption terms in balance
equation
• Stoichiometric equation imposes constraints on the relative
amounts of compounds
(e.g. if A → B, cannot start with 1 mol of A and end up with
2 mol of B)
If no accumulation:
Input + Generation = Output + Consumption
Main ref: Elementary Principles of Chemical Processes, Felder/Rousseau, 1st ed; p. 118
Limiting reagents
Consider the chemical reaction:
2SO2 + O2 → 2SO3
The reactants are in stoichiometric proportions if the molar ratio equals
that of the balanced reaction equation, i.e. 2 mol SO2 react with 1 mol O2.
If reactants are fed into a reactor in stoichiometric proportions, and if the
reaction proceeds to completion, all reactants are consumed.
But, consider you have 2 mol SO2 and only 0.8 mol O2 – in this case
reaction will stop once all O2 is consumed, and 0.4 mol SO2 (= 20% of 2)
remain unreacted. In this case, O2 is the limiting reagent.
Chemical reactions take time to proceed – it is often not practical to design
processes such that complete conversion of the limiting reactant is reached.
Instead, the output stream contains both reactants and products, and the
separated reactant is then recycled back into the reactor. Consider shape of
conversion-time plot (slope decreases with time).
Limiting reagents
Consider the chemical reaction:
2A + 3B → 2C
We conduct this reaction in batch, loading the reactor with
3.21 moles of A and 4.03 moles of B. The reaction is run to
maximum conversion.
1. Which is the limiting reagent?
2. What is the number of moles of C generated?
3. How much more of the limiting reagent would have to be
added in order for there to be no reactants remaining at full
conversion?
Limiting reagents
2A + 3B → 2C
We conduct this reaction in batch, loading the reactor with
3.21 moles of A and 4.03 moles of B. The reaction is run to
maximum conversion.
1. Which is the limiting reagent?
For all of A to react, we need 3/2 molecules of B for each A.
3.21 moles x 3/2 = 4.815 moles
We only have 4.03 moles of B
 B is the limiting reagent
Limiting reagents
Consider the chemical reaction:
2A + 3B → 2C
We conduct this reaction in batch, loading the reactor with
3.21 moles of A and 4.03 moles of B. The reaction is run to
maximum conversion.
2. What is the number of moles of C generated?
• Given that B is the limiting reagent, the reaction will stop when
all B is gone.
• 1 mole of B gives 2/3 mole of C based on the stoichiometric
equation.
• The number of moles of C obtained will equal the initial number
of moles of B multiplied by 2/3, i.e. 2/3 x 4.03 = 2.69 moles
Limiting reagents
Consider the chemical reaction:
2A + 3B → 2C
We conduct this reaction in batch, loading the reactor with
3.21 moles of A and 4.03 moles of B. The reaction is run to
maximum conversion.
3. How much more of the limiting reagent would have to be
added in order for there to be no reactants remaining at full
conversion?
For all of A to react, we need 3/2 molecules of B for each A.
3.21 moles x 3/2 = 4.815 moles
We only have 4.03 moles of B
 We need to add 4.815 – 4.03 = 0.785 moles of B
Example of a material balance involving chemical reaction
Production of acrylonitrile
Acrylonitrile
Reactor
Mol per 1 mol air
Composition per 1 mol total
Information provided:
• Feed contains 10% propylene, 12% ammonia, and 78% air (all mol%)
• 30% conversion of the limiting reagent is achieved
• 100 mol (total) feed as basis (this is what is fed to reactor)
Questions:
• Which is the limiting reagent?
• By how much are the other two reagents present in excess?
• Determine molar amounts of all product gases.
Main ref: Elementary Principles of Chemical Processes, Felder/Rousseau, 1st ed; p. 123
Reactor
Composition per 1 mol
Limiting reagent:
(NH3/C3H6) = 1.2
(O2/C3H6) = 1.64
Stoichiometric ratio = 1
Stoichiometric ratio = 1.5
O2 being fed: 0.78 x 100 x 0.21 = 16.4 mol
(mol fraction O2 in air = 0.21)
C3H6 being fed: 0.100 x 100 = 10 mol
Both NH3 and O2 are fed in more than stoichiometric ratio relative to C3H6 – this is
the limiting reagent.
Excess reagents:
% excess NH3 = (12 – 10)/10 = 0.20 = 20%
% excess O2 = (16.4 – 15)/15 = 0.093 = 9.3%
Reactor
Composition
per 1 mol
Molar amounts of product gases:
Conversion of C3H6 = 30%  0.7 x 10 = 7 mol C3H6
From balanced reaction equation (3 mol of C3H6 have reacted)
NH3: 12 – 3 = 9 mol
O2: 16.4 – (3 x 1.5) = 11.9 mol
C3H3N: 3 mol
N2: 61.6 mol
(not involved in reaction)
H2O: 3 x 3 = 9 mol
Input - Consumption
Input - Consumption
Generation
Input = Output
Generation
The same principles apply to more
complex systems
The task is typically to determine unknown process parameters for a system based on the
information given. The way forward is to set up equations based on what is known in the form of
material balances, then solve these equations for the unknown(s). Note that to solve for X
unknown parameters, you need at least X independent equations (material balances).
The more complex case below illustrates the process for production of non-alcoholic beer.
From CEIC1000 Lecture Notes
Multiphase systems
• Many industrial processes do not involve any chemical reactions.
• Many process units are simply mixers that blend streams of
different compositions.
• Anther common type of process is separation, e.g. isolate A from
mixture of A and B.
• Separation processes typically include two phases: gas/liquid,
solid/liquid, gas/solid, immiscible liquids.
• Much easier to separate two different phases than two
compounds in the same phase.
Examples of separation processes in two-phase systems:
• Distillation - difference in boiling point between two liquids
• Crystallization - difference in melting point
• Liquid extraction - contact a liquid mixture with a solvent that is
miscible with one component but immiscible with the other, then
separate the two phases
Example of liquid extraction
We wish to separate a mixture of
benzene and methanol (miscible
liquids).
Water and methanol are miscible, but
water and benzene are not. If water is
added to a mixture of benzene and
methanol, most of the methanol will
end up in the aqueous phase. Once
the mixture has settled (into two
immiscible layers), separation is
straightforward.
Benzene
Methanol
Add
water
Benzene
Water + Methanol