Exam Revision Notes: Heat and Mass Transfer

Course Contents 1. Heat Transfer: Conduction

2. Mass Transport: Diffusion

3. Transient Conduction

4. Transient Diffusion

5. Convective Heat and Mass Transport

6. Combined Transport and the Overall Transport Coefficient

7. Applications of Transport Phenomena

a. Heat Exchangers

b. Mass Transport in Physiological/Biological Systems

Note Objectives Revision should allow you to:

1. Solve steady state 1-D conduction problems in planar, cylindrical, spherical geometries including composite

systems, systems with thermal conductivity dependent on temperature and systems with heat generation

2. Calculate Mass Transfer rates for simple systems using Fick’s Law, in different forms

3. Calculate mass transfer rates in simple binary systems in which molecular diffusion is superimposed on bulk

flow, e.g. Diffusion controlled reaction at a catalyst surface

4. Solve transient conduction/diffusion problems in finite media

5. Estimate heat and mass transfer coefficients for convective transport applications, by use of appropriate

correlations

6. Estimate overall coefficients of heat and mass transport from individual coefficients

7. Analyze and apply these concepts to heat exchanger problems and problems in physiological mass transfer

8. Formulate the complete governing equations with suitable boundary conditions for mass and heat transport

in representative settings and verify if a particular solution is valid.

Underlying Mathematics: A basic language

Functions A function is a relation which associates a real number , with another real number . The real numbers are

said to lie in a special set called the domain. It should be noted that functions can be described by their graph (a plot

of vs ) and can be generalized to include many different types of variables ( ). The physical

significance of functions is that functions re the basic way of describing continuous processes and systems. Such

functions usually possess a domain between two given constants ([ ] [ ]) based on the process or system

being considered.

Derivatives A derivative is a special kind of continuous function. These functions have the condition that at any point , the

limit ( ) exists and is equal to the value of the function . Be careful when finding the derivative of

a function and don’t jump around. The derivative of a function is defined as:

Assuming ofcourse that the above limit exists. It should be noted that for a function to be differentiable at any point

it is necessary (but not sufficient) that it is continuous.

Derivatives are very important in applied science subjects like engineering as they allow for the mapping of objects

that are varying with time. That is a derivative represents the rate of change of a function with respect to .

Integration Integration is essentially a continuous summation. Hence we may use it to calculate the area under a given curve.

Additionally, you should recall that there are two major types of integration: definite and indefinite. Definite

integration involves the presence of lower and upper limits to the summation, while indefinite integration does not

involve such limits. Thus it should be noted that indefinite integration requires the presence of an additional

constant of integration to be added in. This is extremely important.

Integration is also considered to be the opposite of mathematical operation to differentiation, and hence in order to

retrieve the original function from a differential function, integration must be conducted. This works both ways and

in order to retrieve the original function from an integral, the expression must be differentiated.

Functions of many variables

If we were given a function , we can define its partial derivatives as

. These expressions are simply

the derivative of a function with respect to one of the two variables keeping the other variable fixed. We can repeat

this process to obtain second order derivatives such as:

. For most functions we can say that

.

Maxima and Minima This is once again a very basic concept of maths but you should refresh yourself of these basics before things get

tough.

Recall the condition for a maximum of at is:

Recall the condition for a minima of at is:

Gradient Usually when dealing with real world situations, everything is in 3D space. Hence even functions are based upon the

three most basic components of space. Hence while calculating the gradient function, you must be aware of the fact

that the following notation is commonly used. In this course you will only be attempting 1D problems so don’t worry.

[

]

is a vector quantity where the component contains information of the rate of change of with respect to and

the and components are similarly related to the rates of change of in those directions.

Conduction Conduction heat transfer is the process by which heat flows through a solid. In the conduction mode, heat is

transferred through a complex submicroscopic mechanism in which atoms interact by elastic and inelastic collisions

to propagate the energy from regions of higher temperature to regions of lower temperature. From an engineering

point of view there is no need to delve into the complexities of the mechanisms because the rate of heat

propagation can be predicted by Fourier’s law, which incorporates the mechanistic features of the process into a

physical property known as the thermal conductivity. Although conduction also occurs in liquids and gases it is rarely

the predominant transport mechanism in fluids – once heat begins to flow in a fluid, even if no external force is

applied, density gradients are set up and convective currents are set in motion. In convection, thermal energy is thus

transported on a macroscopic scale as well as on a microscopic scale, and convection currents are generally more

effective in transporting heat than conduction alone, where the motion is limited to submicroscopic transport of

energy.

Conduction heat transfer has been a fertile field for applied mathematicians for the past 200 years. The governing

physical relations are partial differential equations, which are susceptible to solution by classical methods. Famous

mathematicians, including Laplace and Fourier, spent part of their lives seeking and tabulating useful solutions to

heat conduction problems. However, the analytic approach to conduction is limited to relatively simple geometric

shapes and to boundary conditions that can only approximate the situation in realistic engineering problems.

Although today the advent of extremely fast computers have allowed all types of conduction problems to be solved,

we must first understand the basic analytic approach to conduction using the basic geometric shapes and boundary

conditions.

Definitions Temperature is defined as a measure of average microscopic energy of particles. It allows us to determine how hot

or cold a given material is. It is an intensive property and hence is not dependent on the mass or amount of a given

substance.

Conduction is defined as the transport of heat through a material medium which relies on a temperature difference

for heat transfer without material transport.

Fourier’s Law It has been empirically observed that the heat flow rate (flux) through a substance is proportional to the

temperature difference between the boundaries of the substance divided by the distance between the boundaries.

Thus we have:

Here temperature must be in Kelvin and the heat flux ( ) is the heat flow per unit area for a given unit time (W/m2).

This empirical deduction led Fourier to determine his fundamental law governing conduction. Fourier’s Law states:

Two important points that should be noted about this formula. Firstly, the minus sign is a consequence of the second

law of thermodynamics which requires that heat must flow in the direction from higher to lower temperature (Look

at OP1). Secondly, is a material property called the ‘Conductivity’ and can be calculated as a function of

temperature and pressure. As a rough indicator,

.

The expression for Fourier’s Law given above, however, only takes into account the heat flow in the direction.

Hence if we were to write out Fourier’s Law in full we can say that in general:

As can be seen the conductivity for each expression above is the conductivity associated with each direction given as

a subscript. That is is the conductivity associated with the direction etc. Usually the conductivity for a given

substance does not change in the 3 different directions but if given a complex material or a peculiar situation this

can occur. However, for most cases conductivity in each direction is the same and hence Fourier’s Law can be

simplified even further into:

Notice here that the heat flux is a vector as you have a , and a component. If we were to re-write this above

equation in cylindrical co-ordinates we have:

This may seem complicated as we are now introducing new variables.

For knowledge ( on the diagram) represents the radial direction, (

on the diagram) represents the angle of rotation in a given plane, and

represents which plane you are operating in (i.e. the depth). In this

course, we shall only be dealing with heat transfer in the radial direction

so the only heat flux equation in cylindrical co-ordinates that you will

need to know is the first one.

Example

An insulated metallic sphere of density , specific heat and radius

is heated by electric coils imbedded in it (occupying a negligible fraction

of the volume). The current is passed through the coils (resistance R0)

staring from is as follows: a sinusoidal fluctuating current:

, for periods of the sinusoidal variation, followed by a steady current for another units of

time.

a) Starting from first principles, and considering the heat energy transferred to the sphere between time and

, derive the equation governing the variation of the temperature of the sphere (assumed uniform

throughout the sphere) with time. Explain your derivation

b) Assuming that the sphere starts at a temperature T0, what is its temperature at the end of the process?

The diagram for this question is drawn on the attached sheet of paper OP2.

(a)

Let us commence this problem with the expression for heat loss for a given electrical component. We know that heat

loss is given by:

We also know that:

Hence we have the following expression:

Now let us consider the time interval to , where is extremely small. We can safely say that in this

extremely small interval, the current remains for all intents and purposes constant. Hence we have:

[ ]

This is the equation that governs the variation of the temperature of the sphere with time.

(b)

Using the expression derived above we must integrate each side:

∫

[

∫

∫

]

The reason for the two integrals in this expression is to accommodate for the varying current being supplied. The

first integral takes into account the current while it is being applied as a sinusoidal function, whilst the second

integral takes into account the other constant period of current application. Now the way in which the upper limits

of the integrals on the right hand side of the expression above have been derived is shown below:

We want to know what is when , or 1 revolution. Hence we have:

Hence for N periods we have:

Note this is the time in seconds taken for the periods to be completed. Now we can solve the problem.

∫

[

∫

∫

]

[

∫

(

)

]

[

∫

]

Remember that:

Hence:

[[

]

]

[

]

[

]

Hence the temperature of the sphere at the end of the entire process is given by the above expression.

From Empirical Laws to Governing Equations So far we have only been looking at expressions derived through empirical data. It’s time to start deriving governing

equations for heat transfer by conduction, along with possible heat generation. For now let us use Cartesian co-

ordinates as it provides the simplest example. Let’s perform an energy balance using a control volume that consists

of a cuboid of dimensions , and . Hence we have:

Now in a time we can say that the rate of inflow of heat in the direction is given

by:

Additionally, the outflow of heat in the direction is given by:

Similarly such contributions are present for the other 2 directions as well. We can write them as a large expression as

shown here:

This expression accounts for the net inflow and outflow in each direction. Now we must account for the generation

and accumulation terms. The generation term is given by:

Where is the heat generated per unit volume per unit time. Finally the accumulation term is simply the heat

equation that we are most used to. That is the expression:

This expression may be written as the following in this case.

Hence if we put all of these terms together we have:

Now let us divide each term by and . This gives us:

( )

Now if we take the limit as , , and tend to 0 we have:

[

]

The Conduction Equation The expression we have derived above is in essence the conduction equation we are after. However, we need an

expression in terms of measurable quantities. Hence, for each of the heat fluxes in the equation above, we must

substitute for Fourier’s Law. Hence we get the following expression:

[

[

]

[

]

[

]]

[

[

]

[

]

[

]]

[

]

[

]

[

]

If we assume that the conductivities in each direction are the same and are constant, and that the heat generation is

0 we have:

[

]

[

]

[

]

[

]

[

]

[

]

Where:

Analysis of Conduction

The Conduction Equation: Another Look So far the conduction equation that we have derived is a partial differential equation that looks like:

[

]

[

]

[

]

Now if we need to find the temperature variation in a material with both spatial location and time we need 2 pieces

of additional information.

1. What was the initial temperature distribution? (Required for unsteady state problems)

2. What happens at the boundary? (Boundary conditions)

As we shall most of the time dealing with steady state problems, let us ignore the first of these conditions and look

instead at the second one. At the boundary we can say there are 3 important conditions that must be considered.

1. Fixed temperature (Dirichlet Conditions)

2. No – Flux Condition (Neuman)

3. Newton’s Law of Cooling

The Dirichlet Condition is easy to understand as all it means is that the temperature at the boundary is simply

constant. The second condition is a little bit more difficult to explain.

Consider an insulated slab (drawn on attached sheet of paper OP3). Let us consider surface 1 and 3. At surface 1 or 3

we can say that:

is never 0. Hence we can say that:

Now let us consider surfaces 2 and 4. At these surfaces we can say:

is never 0. Hence we can say that:

Hence these expressions explain the No-Flux condition. This condition can be represented as:

Where is the normal vector.

The third and final boundary condition simply means that the flux at the boundary is governed by Newton’s Law of

Cooling. This law states:

Where is the heat flux, is the heat transfer coefficient (unique for a given substance), is the temperature of

the boundary and is the temperature of the environment.

Having discussed the 3 boundary conditions completely, there are a few different versions of the conduction

equation that you should be aware of. Firstly the conduction equation in cylindrical co-ordinates is:

[

]

[

]

[

]

Also the conduction equation in spherical co-ordinates is:

[

]

[

]

[

]

Conduction in a Slab Lets say we have steady state conduction in a uniform solid slab with constant conductivity. Lets say the ends of the

slab are at fixed temperatures and . (Diagram is give on attached piece of paper OP4). Now lets say we would

like to find out how temperature varies within the slab. In essence we want a temperature profile. We must use the

conduction equation and work from first principles. As here we have conduction occurring only in one direction, our

conduction equation simplifies to give simply the direction term. Additionally, as there is no heat generation in this

case and the slab is at steady state, both the heat generation and accumulation terms are 0. Hence we have the

following expression for the conduction equation (using the Cartesian version here as it fits question best).

[

]

[

]

[

]

[

]

Now we know that the conductivity is 0, hence, we can say that:

If we now integrate both sides we get:

Separating variables and integrating again gives us:

∫ ∫

This is the key point at any problem involving conduction between a set of two boundaries. You must check that you

have 2 constants and ensure that you have information for the temperatures and positions of the boundaries. Here,

we know that when , . Hence we can say that:

Substituting back into the equation gives:

Now if we substitute in the other boundary condition (i.e. when ), we have:

(

)

Therefore substituting gives:

This equation gives us the temperature profile within a slab whose boundaries are at different temperatures. Two

things must be noticed here. Firstly, the profile is linear and secondly the profile is independent of the conductivity.

That is no matter what the material the slab is made of, the temperature profile at steady state will always look like

this. It should be noted that the conductivity only affects the heat flux.

Example

Consider a slab 1 m in length, with , and . What is the flux and the

temperature profile?

Diagram is drawn on attached sheet of paper. OP5.

Let us start by calculating the temperature profile. Starting from the conduction equation we can work our way

carefully down to the temperature profile already calculated above:

Substituting in the values we have been provided, we have:

This is the temperature profile for this slab. Now to calculate the heat flux, we must first have an expression for

.

Hence we have:

Now using Fourier’s Law we have:

Note that while in this case we could have calculated the heat flux first (by substituting

) it is better to find

the temperature profile and then differentiate to get to the flux. This is because while this method works for one

slab with a linear profile, it may not work for more complicated scenarios.

Conduction in Composite Media: Series of Slabs Let us consider a series of slabs stacked next to each other. (Diagram on attached sheet. OP6). A few new things

have been added to the diagram. Firstly, the variable represents the length of the slab that lies from co-

ordinate to co-ordinate . Secondly it must be noted that the heat flux travelling from one slab to the next is the

same, i.e.:

In questions involving such a large number of slabs stacked together, we are often only given the initial temperature

( ) and the final temperature ( ). Now lets say if we were asked to find the temperature profile in this case.

To do this we must first find the heat flux transferred from 1 slab to the next. Hence, let us attempt this question by

first considering the heat flux being transferred from each slab. Consider slab 1:

Now consider slab 2:

Now consider slab 3:

Hence, for the slab we have:

Now it should be noticed that these equations for the heat flux all form part of a telescoping series, and if we simply

add all of them together we get:

[∑

]

From this equation we are usually able to calculate the heat flux, as we are often given the initial and final

temperatures as well as the lengths and conductivities of each of the slabs. Thus once we have calculated the heat

flux, in order to find the temperature at a certain location , we must first find which slab that location lies in. Once

we have determined which slab this position is contained within, we must go back to the earlier heat flux

equations and solve for the temperatures at either end of the particular slab we are concerned about. This can be

tedious as you may have to work your way down solving for all the temperatures until you finally arrive at the slab

you require. Once you know the end temperatures of that slab, you can use the profile calculated earlier for a slab to

determine the temperature at a given location. Thus the temperature variation in a series of slabs is simply a series

of linear profiles each with a differing slope.

It should be noted here that often we are also asked to calculate the heat energy being transferred between slabs.

To do this it should be realized that:

Where is the heat energy, and is the area. Hence the above equation may be written as:

[ ]

[∑

]

Conduction in Slabs where Conductivity is not Constant Sometimes the conductivity may vary with temperature in which case the calculations become slightly more

challenging. For example lets say if we were asked to calculate the steady state temperature variation in a slab with

length , which had boundary temperatures of and and were told that the conductivity varied according

to the expression .

Diagram for this problem is given on the sheet attached. OP7.

We must begin from first principles and hence we should start from the conduction equation again. We have:

[

]

[

]

[

]

However, we know that the system is at steady state and hence there is no accumulation. Additionally we are not

told about the presence of any heat generation and know that heat is only being transferred in the direction.

Hence we can simplify the equation to:

[

]

Now at this stage we cannot pull the conductivity out of the differential as it is not a constant but rather is varying

with position. Hence we must integrate this expression:

Now by separating variables, substituting for the conductivity and integrating we have:

∫ ∫

Substituting for boundary conditions we have:

Substituting back into the original equation we have:

Using the other boundary condition we have:

Substituting back into the original equation we have:

This expression gives us the temperature profile when the conductivity is not constant. A few points should be noted

about this type of question. Firstly, whilst performing this question, we stumbled upon an expression that was

identical to Fourier’s Law. This is shown below:

From this expression we can see that:

Hence this allows for a very easy way in which to calculate the heat flux. However, in this case the conductivity is

varying with temperature, so one should ask the question what value of the conductivity should be used. The answer

may be derived as follows.

∫

∫

∫

Now if we multiply by

, we get the following expression.

[

∫

]

Which in turn becomes:

[

∫

]

This expression shed light on exactly which conductivity value to use whilst calculating the heat flux for a varying

conductivity. As we can see the first term of the expression above is simply

. The second term in square brackets

is essentially an average value of the conductivity. In essence this term represents the sum of all the conductivity

terms between two intervals divided by the total number of these terms: an average. Hence this expression may be

written as:

Hence when the conductivity is varying with temperature, this is the formula that must be used to calculate the heat

flux.

r2,T

2

r1,T

1

Analogy between Heat Transfer and Current Flow, and interpretation as Resistance An interesting analogy may be made between Heat Transfer and Current Flow. This may not be directly related to

the course but it helps to understand some of the fundamental equations derived above, particularly the expression

for the heat flux for several slabs stacked together. Now we know from Ohm’s Law that:

Where is the current, is the voltage and is the resistance. Now if we were to relate each one of these

components into the corresponding component in heat transfer, we find that the current is extremely similar to

the heat flux, the potential difference is extremely similar to the temperature difference (i.e. across the entire

medium ), and the total resistance for the given medium is ∑

. Hence the relationship we form by

simply linking each of these components according to Ohm’s law is exactly the same as the one derived earlier for a

multi-slab medium.

[ ]

[∑

]

Conduction in Cylinders Due to the shape of most industrial pipes being cylindrical, we must

understand carefully how conduction occurs in cylinders. In order to

determine the temperature profile in a cylinder, let us start by considering

the situation at steady state and the material possessing a constant

conductivity. From the diagram on the left we know that at the

temperature is , while at the temperature is . Hence in essence we are

trying to determine the amount of heat that is transferred out of a pipe via

conduction. We must start with the conduction equation, however, instead

of using Cartesian co-ordinates let us use the equation adapted for cylindrical co-ordinates. Hence we have:

[

]

[

]

[

]

Due to the system being at steady state the accumulation term is 0. Additionally, let us assume that there is no heat

generation and that all heat flow occurs only in the radial direction. Hence we have:

[

]

As the conductivity is constant we can say that:

[

]

[

]

Integrating gives:

If we now separate variables and integrate again we have:

∫

∫

Now substituting for the boundary conditions and solving we have:

Substituting back into the original equation we have:

(

)

Substituting for the other boundary condition we have:

( )

( )

Substituting back into the original equation we have:

[

( )] (

)

This is the temperature profile for through the annulus of two cylinders. Now if we were asked to determine the

heat flux being transferred through the cylinder, we could do this simply by differentiating this function and plugging

it into Fourier’s as is shown below.

[

( )]

[

( )]

From this expression we can clearly see that the heat flux varies depending on the radial position being considered.

This leads heat flux to being a slightly more complex function to deal with in cylinders. However, it should be noted

that if we consider the heat energy being transferred in a cylinder, we obtain the following expression.

[

( )]

[

( )]

[

( )]

As you can see from the expression for the heat energy being transferred in a cylinder, the heat being transferred is

not dependent on the radial position, but is rather dependent on the overall length of the pipe. Hence this is the

more commonly used in cylindrical circumstances. You must carefully remember that the total heat flow through any

cylindrical surface is a constant. The expression derived above may be rewritten as:

Where is the log mean area and is given by the expression:

( )

In essence the log mean area is the difference in surface area between the outer and inner cylinders divided by the

logarithm of the ratio of the radii. It is a purely geometric entity and is in effect the average area of heat transfer.

The problem of pipe lagging and heat loss

The expressions we have derived above have a great many industrial applications. One of the more commonly used

applications is to determine exactly how does the thickness of the annulus between the cylinders (i.e. ), affect

heat transfer. This is an important relationship to determine to ensure that all

pipes in an industrial plant have a certain specified ‘thickness’ to prevent the

product from losing all of its heat and hence cooling down significantly from

the original temperature. In this case let us once again consider the same

diagram as given on the right, but in this case we do not know what is.

Instead we are trying to determine the exact relationship between the heat

transferred per unit length of the pipe

, and the radius of the outer cylinder

. A few assumptions need to be made before the calculations commence,

and these are:

1. Neglect all radiative heat loss from the outer surface of the outer cylinder

2. is known and all the heat on the surface of the outer cylinder is transferred primarily through Newton’s

Law of Cooling to the atmosphere

3. The temperature of the atmosphere (“Far Field” temperature) is

4. The surface temperature shall be called , but it is an unknown and varies with

5. The temperature profile through a cylinder is: [

(

)] (

)

Having made these assumptions let us begin the calculations. First of all we know that the heat flux leaving the outer

cylinder is equal the heat flux being used to heat the environment. Hence we get the following expression:

Now using the temperature profile to find the derivative with respect to we have:

[

( )]

[

( )]

Now at , we know the temperature is hence we have:

r2,T

2

r1,T

1

[

( )]

Rearranging we have:

( )

Now for the sake of this algebraic manipulation let us assume:

( )

Hence we have:

Now we can solve for . We get:

Now in order to solve for the heat released per unit length at the outer radius we must use the following

expression:

Substituting for we have:

(

)

(

)

(

)

Now substituting for we have:

(

( ))

(

)

(

)

Hence this is the expression that relates the heat loss per unit length to the outer radius . Now we would like to

find out what happens to the heat loss as varies. To do this we must first check if this expression possesses any

maximum or minimum points to ensure that the heat loss is not maximum or minimum at a certain point. Thus we

have:

(

)

Using the quotient rule we have:

( )

Hence we have:

(

)

(

) (

)

[

( )

]

When the gradient function is 0 we have:

(

) (

)

[

( )

]

(

) (

)

(

)

Surprisingly as we can see there appears to be a local minima at the point where

. This implies that as you

increase the thickness between the cylinders, the heat loss initially increases until the point where the outer radius is

after which any further increments in the outer radius will lead to the heat loss decreasing. This is an extremely

intriguing point and should be noted.

Conduction through Array’s of Cylinders As with the slabs before, let us consider how we would approach a question which involved the transfer of heat

through a series of cylinders. (Diagram is given on attached sheet of paper as OP8). Now the first point that should

be noted is that while with slabs we can work with heat flux (due to the area of heat transfer between slabs

remaining constant throughout the slabs) with cylinders we must only deal with heat energy. This is because as

shown earlier, the heat flux in a cylinder varies for every radial position. This may be understood better by thinking

that that in a cylinder, if you move in the radial direction, the area of heat transfer must increase as the very radius is

increasing. The heat energy being transferred, however, remains constant despite the area of heat transfer

changing. Hence the expression derived for the heat being transferred must be used for each and every cylinder.

However, it should be remembered that as before we are only told the temperatures of the first surface and final

surface (i.e. and ). Hence we have:

Where:

( )

Rearranging and substituting for (where is the length of the annulus) we have:

Similarly for each of the other cylinders we have:

As we can see this is a telescopic series and hence by adding them together we result in the following expression:

∑

Rearranging we have:

∑

Having found the heat flow rate, we can now easily find the interface temperatures step by step, and thus

temperature profiles in each cylinder.

Conduction with Heat Generation In many conduction and heat transfer problems there are auxiliary heat generation effects. For example:

1. Absorption of neutrons in a fuel element of a reactor

2. Exothermic chemical reactions

3. Absorption of radiation

Let us first consider the effect of such heat generation in a rectangular slab. Hence lets say we have a rectangular

slab (with constant conductivity) whose ends are maintained at fixed known temperatures. (Diagram is given on

attached sheet OP9). Lets start off by considering uniform heat generation throughout the slab i.e. let be the heat

generation per unit volume. Assuming we have heat transfer in the direction, using the conduction equation in

Cartesian co-ordinates we gain the following expression.

[

]

[

]

[

]

Integrating with respect to we have:

Integrating again we have:

Now using the boundary condition we have:

Substituting back into the equation we have:

Now using the other boundary condition we have:

Substituting back into the equation we have:

[

]

This is the generic profile for a slab when there is constant heat generation. Now if the length of the slab is (where

and ) then we have:

[

]

[

]

Hence it should be noted that when there is heat generation, the conductivity does affect the temperature profile.

From this it follows that the heat flux at a given point on the slab is given by:

[

]

Now let us try and analyze this system by considering a few special cases. Lets consider where . (Diagram on

attached sheet OP10). This leads to the following temperature profile:

Now lets see what the heat flux of this special case is at the boundaries, i.e. where and . The expressions

we obtain are:

When , we have:

And when , we have:

Hence as it can be seen, the heat flux at the two ends of the slab with heat generation is equal and opposite to each

other. Now let us consider the consider the case where . Now let us try and find the maximum point of the

temperature distribution. For now let us assume that giving us the diagram in OP11. The temperature

profile for this slab with heat generation is:

[

]

Now as we want the maximum point of the temperature distribution, we must solve for

. Hence we have:

So hence this calculated value of is where the maximum temperature is. However, it should be noticed that this

value can either lie within the slab, or outside the slab. If the value is outside the slab the temperature at one end

becomes the maximum value. Hence in order to ensure that the maximum value of the temperature occurs within

the slab we can impose the following inequality:

We know that if:

Then the maximum temperature will occur at the very edge of the slab. Hence, when:

The maximum temperature lies either on the boundary or within the slab. This can be simplified to:

Now it should be noticed that if , the heat generation per unit volume, is extremely small then

will be

huge. This would result in the maximum not being within the slab. The opposite is true if the heat generation per

unit volume is extremely large. We may calculate the minimum value for the heat generation for the maximum to lie

within the domain by using the above expression:

Hence the minimum value for the heat generation to ensure that the maximum temperature remains within the slab

is:

Summary of Conduction We have completed conduction. So far it should be noted that we have done:

1. Fourier’s Law

2. Conduction Equation

3. Conduction in Slabs

4. Conduction in Cylinders

5. Conduction in Slabs with Heat generation

Diffusion

Mass Transport Mass transport is the transport of material involving concentration differences. Pure transport of material (e.g. fluid

flow) without concentration differences is not counted as mass transfer. There are two major modes of mass

transfer:

1. Diffusion

2. Convective Mass Transfer

Diffusion is analogous to conduction, convective mass transfer is analogous to convection. However, it should be

note that there is no analogue of radiation in mass transfer.

Concepts of Mass Transfer and Terminology The difficulty encountered whilst dealing with mass transfer is due to the presence of mixtures of different species.

In order to develop equations involving flux, and eventually concentration terminology, we label each of the

different types of species. That is if there were species then we have the species . Now another

aspect of notation that must be remembered is that the local molar concentration of species is denoted by . This

simply represents the moles of species per unit volume (i.e. ) and can vary not only with position but also

with time.

Now in a given mixture with species, the total concentration is given by:

The total concentration can also depend upon spatial location and time. Now let us consider the velocities of each of

the species. As before the velocity for a species is denoted by . It follows then that the average velocity ( ) of

the mixture is given by the following expression.

∑

∑

(

)

That is the sum of the product of each concentration with its respective velocity divided by the total concentration.

Also it may be seen that the mole fraction of species is:

So now we have sorted out a significant portion of the notation. We must now find some way of defining what

diffusion actually is. Hence, let us define the diffusion of species to be , or the velocity of species relative

to the molar average velocity of the mixture.

Now as we had the heat flux for conduction (which was the heat energy transferred per unit area) we would like to

similarly determine an expression for the molar flux which similarly is the molar flow rate of a given species ,

transferred per unit area. Hence the units we require here are:

It should be noticed that these units are given by the product of concentration per unit volume and velocity. Hence

we can define the molar flux of species to be:

(Note: With this definition, the average velocity can be considered the total molar flux divided by the total

concentration).

Now the problem with this quantity is that we have defined diffusion relative to the average velocity. Hence, in order

to use the molar flux and then somehow link it to the average velocity to come up with an expression for the

diffusion of a certain species is extremely difficult. Instead let us consider a moving frame of reference. In fact, let us

consider a frame of reference that is travelling at exactly . Now in this frame of reference, the molar flux of species

may be defined by the following expression:

It is exactly the same definition but is just taking into account the fact that the frame of reference being used is

moving. Note the different letters being used for the molar flux. Both and represent exactly the same thing but

is in a moving frame of reference while is in a stationary frame of reference. Now it should be noted that if we

consider the sum of all molar fluxes for species 1 to in the moving frame of reference, we get 0. This is because

some species will have a velocity that is less than the average velocity, while some will have a velocity that is higher

than the average velocity. Hence these will add and subtract to give an overall value of 0. This is not the case with

the molar flux in a stationary frame of reference.

∑

Example

For a ternary mixture of A, B,C, at a particular location, ca =0.5M, cb=0.2M, cc=0.1M, va = 2cm/s,vb=4cm/s,vc =8cm/s

(all in the z-direction) calculate the absolute and diffusive molar fluxes of A,B,C at this location.

Let us start with the absolute flux. First we must make sure that we convert the concentrations to and

velocities to .

Now in order to calculate the diffusive molar flux we must first calculate the average velocity of the mixture. The

average velocity is given by the sum of the absolute molar fluxes for each species divided by the total concentration.

Hence we have:

Hence we can now calculate the diffusive flux for each species.

( )

(

)

( )

(

)

( )

(

)

Check that the sum of the diffusive flux for every species tallies to 0.

Hence everything works out nicely.

Empirical Law of Mass Diffusion Through various experiments Adolf Fick came up with the following empirical law for mass diffusion.

This is known as Fick’s Law and it relates the diffusive flux to the gradient of mole fraction in the direction the

material is diffusing. Now here as before relates to the total concentration and is a constant known as the

molecular diffusivity. Now lets consider a binary system. We have for species 1:

Here notice the two numbers in the subscript for the molecular diffusivity. This indicates that the constant

represents the molecular diffusivity in a binary system of species 1 through species 2. For the other species, Fick’s

Law becomes:

Now we know that:

And:

Thus it follows that:

Now it should be noted that

. Hence:

Remember this is only valid for a binary mixture.

Now it should be noted that despite the diffusive flux being much easier to work with than the absolute flux, it is

difficult to directly measure, especially when there are a significant number of species in the mixture. However,

contrastingly, the absolute flux is directly measurable. Hence, we need an expression that can somehow relate these

two types of flux.

Let us start off by considering a binary system. The absolute flux for species 1 would be:

This may be written as:

Now at this stage although we have a link between the absolute and diffusive flux, we don’t know what the second

term here is. Hence let us try and determine exactly what this second term means.

∑

For a binary system this becomes:

Hence if we consider the product above we have:

(

)

Hence substituting back into the above equation, we have:

Finally using Fick’s Law:

It should be noted that a similar equation be written for . From this equation we can measure all terms and

effectively calculate what the diffusive flux is. Now in the case that the total concentration is independent of

position, i.e. the total concentration is constant throughout the mixture, then this expression may written as:

Note the total concentration can be considered to be constant if you have a mixture of ideal gases at constant

pressure and temperature. This is because we know that:

Hence the total concentration will be a constant.

If the system however is not binary, then this equation simply becomes:

( ∑

)

From Empirical Law to Governing Equations As we did with Fourier’s Empirical law of conduction, we must similarly use Fick’s Law of empirical law of diffusion to

come up with an overall governing law for mass transfer by mass flux along with possible generation. Our starting

point should be using the conservation of mass.

Lets consider a cube of dimensions , and to be our control volume. Let us

consider this control volume in the time interval to . For every species we

have the following general mass balance:

The inflow through the face with normal in direction in the time interval specified is:

( )

The corresponding outflow is:

( )

Similar expressions may be written for the inflow and outflow in every other direction for species in the specified

time interval.

Now let us consider the generation term. Let the mass generation, rate per unit volume for species be . Hence

the mass of species generated is given by the following expression:

Now let us consider the accumulation term. This term is simply given by:

The reason for the multiplication of the volume to the difference of the total concentrations is due to the fact that

when concentration is multiplied by volume you get the answer in moles. Hence if we substitute each of these terms

back into the original expression we have:

[( ) ( ) ] [( )

( ) ]

[( ) ( ) ]

Now if we multiply by -1 divide each term by the volume and we have:

[( ) ( ) ]

[( ) ( ) ]

[( ) ( ) ]

Now if we take the limit as , , and tend to 0 we have:

If we now multiply both sides by we have:

[

]

Hence this is the equation for the mass balance for every species . This expression can be considered to be

analogous to the conduction equation in the conduction part of the course.

A Bit of History: Adolf Fick (1829 – 1901) Adolf Fick was a German Physiologist. He initially studied maths and physics. He received his

doctorate in Marburg in 1851, which involved investigating visual errors due to astigmatism. He

later became a professor of physiology in the University of Zurich and the in the University of

Wurzburg. He came up with his famous Fick’s law of diffusion in 1855. He conducted several

studies of muscle contraction, and how the air mixed into the lungs. Additionally he also

carefully understood the working of the heart and the hydrodynamics of circulation.

He also constructed the first contact lens which he made from heavy brown glasses. These were first tested on

rabbits, then himself and subsequently upon various volunteers. These contact lenses involved grape sugar solution

between the glass and the cornea.

Analysis of Diffusion There are many different types of scenarios that arise with diffusion. We shall consider all of then slowly.

Diffusion of Species A through a static Species B

A good example of such a case would be when a liquid (A) evaporates into a vertical column of stagnant gas (B).

Ofcourse this would only work if B is insoluble in A. Now let us begin.

Let us call the liquid mole fraction at the surface of the liquid , and that at the upper

surface ( ) . Now lets say we were asked to calculate how concentration changes with

position i.e. a concentration profile.

This is a question analogous to calculating the temperature profile in the conduction part of

the course. The difference here is that we must use two equations simultaneously and work

towards a solution, rather than use 1 equation solitarily. Hence from Fick’s Law we have:

Now we know that is stationary hence . Thus Fick’s Law can be reduced to:

Now we cannot just integrate here because we are not sure whether varies with or with . Hence at this point

we must use the mass balance. At steady state, without generation and with diffusion only in the direction we

have:

[

]

By integrating we get:

Hence through the mass balance we can see that the absolute flux of the liquid is a constant. Hence if we use this

information in the earlier expression we have:

∫ ∫

Now if we input the two boundary conditions we have we end up with the following expression:

Substituting back we have:

z1

z

2

[

]

Using the other boundary condition we have:

[

]

[

]

Substituting back gives:

[

] [

]

[

] [

]

[

]

Thus this is the concentration profile for through the container. Additionally note that we have also determined a

method to calculate the flux of . This is shown below.

[

]

As here we know the mole fractions of substance at the boundaries we can find the absolute flux of . Additionally

as we know that the absolute flux is a constant, we know that it must remain the same throughout the container.

Thus this question has been answered quite easily and through it we managed to come up with an expression for the

absolute flux. However, if we were specifically asked by the question to determine the absolute flux, then instead of

going through all of this tedious algebra, you should use a definite integration process, i.e. integrate definitely across

the entire domain. For example consider the expression below:

∫

∫

[

]

All of these quantities are known and hence the absolute flux is calculable. Now this expression can be further

simplified as follows:

[

]

Remember this is only true because in this case this system is a binary system and has only 2 species present within

it. This allows for further manipulations, as are shown below.

Lets first multiply and divide by . It must be noted that due to the nature of this binary system

. Hence we have:

[

]

Remembering the definition of a log mean to be

(

), we have:

Where:

[

]

FIND OUT WHAT IS THE SIGNIFICANCE OF ?

Now we have obtained the concentration profile for above, as well as the flux. However, it should be remembered

that since changes with position, must also change to ensure the total mole fraction is equal to 1. Due to this

variation in mole fraction through the container there must be a diffusive flux for (as recall diffusive flux measures

the rate of change of mole fraction of a species with direction). Hence in order to calculate this diffusive flux for

we must start from Fick’s Law.

Now we know that , and hence:

Now it must be remembered from above that we had derived the absolute flux to be:

This ofcourse can be expressed to be:

Hence if we substitute this expression into the earlier expression we have:

Thus is stationary but has a diffusive flux sufficient to ensure that it cancels exactly with the diffusive flux of to

ensure that:

∑

Accounting for a Moving Interface

So far we have been assuming that when the liquid evaporated, the amount that evaporated was so small that the

volume of liquid left at the bottom of the cylinder was unaffected. However, now we must consider the possibility

that the very surface of the liquid could decrease with time. Hence consider exactly the same situation as before.

Before we progress, we must make one extremely important assumption. We must assume

that the interface motion is low enough that the computation of the mass transport based

on the previous considerations is valid. Hence we can use all of the equations we have

derived before. The equation we have for the absolute flux of is:

Now this ofcourse can be simply written as:

Now assuming we have a perfect gas mixture we know that:

Hence through simple substitution we arrive at:

Let us call the partial pressure for component , given by the expression:

Where is the total pressure. Hence we have:

In order to ensure that everything is in terms of partial pressures we can multiply top and bottom by total pressure

resulting in:

Now let us consider to simply be equal to the top of the container and call it 0. Additionally, let us call the

position of the interface and lets call it . Hence we have:

( )

Now it should be noted that in essence represents the evaporation rate, so in questions which ask for you to

calculate the evaporation rate, it means to calculate . Now in order to complete the above expression we must

use some information from the mass balance. However, the above mass balance only possesses a term for the total

concentration of as can be seen below. We would like to only take into account the concentration of in the

liquid phase. Hence as we know that is the evaporation rate, we can safely say that in a time the number of

moles of the will be leaving the liquid are given by the expression:

Now it should also be noted that in this exact time the interface position changes from to . Hence

another expression which denotes the number of moles of leaving the liquid over a time interval of is:

z

Here it should be noted that represents the concentration of in the liquid phase (which we are assuming to be

uniform and constant throughout the liquid). Hence equating these two expressions we have:

Hence we can see that:

Equating the two expressions for the absolute flux of or the evaporation rate of and we have:

( )

( )

Separating variables and integrating gives:

∫ ( )

∫

( )

√ ( )

Hence we have found an expression relating the interface level to the time. It may be noted that this expression may

also have been derived without using partial pressures. That is we could have used the fact that:

[

]

To derive exactly the same expression above. Interestingly, it should be noted that the experimental method of

calculating molecular diffusive constants is through the use of this equation. First of all the level of the interface

should be noted at , call it . Then it should be noted again at , call it . Then we have:

( )∫

∫

( )

If a graph is plotted of vs this expression can be confirmed and may be calculated.

Equimolar Counter Diffusion

Equimolar counter diffusion involves once again a binary system. However, here both species are non-stationary.

Instead, each of these species are moving in opposite directions with exactly the same absolute flux. (Diagram drawn

on attached sheet OP12). In order to find the concentration here we must start from Fick’s Law. In such a case Fick’s

Law is:

As we know:

Substituting gives:

Once again we must now make sure that does not vary with either or . Hence using the mass balance we

have:

[

]

Integrating gives:

Hence using this information back into the original equation we have:

∫ ∫

Using boundary conditions we have:

Using other boundary condition gives:

Hence rearranging gives the concentration profile to be:

Example

To maintain a pressure close to 1atm, an industrial pipeline containing ammonia gas is vented to the ambient air.

Venting is achieved by tapping the pipe and inserting a 5mm diameter tube which extends for 10m into the

atmosphere. With the entire system at 298K, determine the mass rate of ammonia leaking to the atmosphere and

the mass rate of contamination of the pipe with air.

The diagram for this question is on the attached sheet of paper OP 13. Before we commence a few assumptions

must be made. Firstly we must assume that both the gases involved behave as ideal gases. Additionally, a further

assumption that must be made is that and . Let us start by calculating the mass rate of

Ammonia leaking to the atmosphere.

Integrating across the whole domain gives:

( )

(Check this answer). Now as this is equimolar counter diffusion we can say that:

Hence we have:

Diffusion Mass Transfer in Biology

Diffusion plays a critical role in cell biology, both inside a single cell, as well as a means of communication between

cells. For example diffusion is integral in the following processes:

1. Transport of molecules inside a cell, from the cytoplasm to membrane

2. Transport of molecules between cells serving as a means of cell to cell communication

3. Transport of chemical in a tissue playing an important role in development

Diffusion Mass Transfer with Reaction

This is a slightly different process than the rest. Let us consider the case where a reactant diffuses towards a

surface which contains a catalyst, then having reacted according to the equation , the product diffuses

away. Let us try and find the concentration profiles and absolute flux for the both reactant and product. The diagram

for this is given on the sheet OP14. Let us assume that the reaction occurs extremely fast on the surface and that all

the molecules of arriving on the surface are immediately converted to ( ). Hence this reaction may be

considered to be a diffusion controlled reaction. The mole fraction of on the surface is thus and the mole

fraction of is 1. Additionally for every two molecules of reaching the surface, one molecule of is generated and

returns. Hence we can say that on the surface:

Now let us begin answering this question by first considering a mass balance for each of the two reactants

separately. As there is no reaction occurring between the bulk and the catalyst surfaces the generation term is 0 for

both. Hence at steady state for the reactant we have:

For the product we have:

Hence both and are constants. But remember that at the catalyst surface we are already aware that

. Hence as both the absolute fluxes are constants, this expression must hold for every point all over the

system. Now lets consider Fick’s Law for the product . We have:

We know that . Hence:

Rearranging gives:

We know that is an unknown constant, hence by integrating this equation and then using the two constants to

include the boundary conditions we can generate a concentration profile. This has been done below. It should be

noted that .

∫

∫

(

)

(

)

(

)

(

)

[

](

)

[

](

)

[

](

)

Additionally it should be noted that the absolute flux for the product is given by the expression:

(

)

In this scenario the absolute flux of the product is also the rate at which it is produced and is hence an extremely

useful value. Similar expressions for the concentration profile of and the absolute flux of may be determined.

Example

Consider a solid waste treatment system using aerobic fermentation to decompose organic matter into chemical

constituents. Oxygen concentration at the surface is known. Oxygen is taken up in the organic matter at a rate which

is proportional to its local concentration. At steady state, determine how the concentration of oxygen varies in the

organic matter.

It should be noted that Fick’s Law for a solid/liquid is:

Notice that the first term may be dropped. Additionally, as the

total concentration may be considered to be constant this

expression may be written as:

Now let us begin this question by conducting a mass balance on a

slice of width in this system. The mass balance for species

(Oxygen) at steady state is:

[

]

Now notice that the generation term is proportional to the local concentration of oxygen at that given location.

Hence we may say that . The minus sign here indicates the fact that the concentration is being used up.

This gives us the following expression:

In an effort now to get the entire expression in terms of concentrations, we must substitute for . Hence we have:

(

)

Now we know that the boundary conditions we have here are, at , , and at ,

.

Hence we should technically solve as per usual, introducing these boundary conditions when we have our two

constants. However, how do we solve this second order differential equation?

In order to do this at this stage we must guess a possible function that could be. Lets say . Now if this

function is differentiated twice we have:

Impermeable surface

oxygen c

A=c

A,0

z

Hence we can say that

. This function works however, we immediately encounter a problem. The second

boundary condition requires the first differential of this function to be equal to be 0. However, this cannot occur as

an exponential function will never be equal to 0 unless it is multiplied by 0. Hence we must redefine our function.

Instead of the expression above let us say:

This expression not only allows for 2 boundary conditions, but also has the additional negative sign in the power of

the second exponential to ensure that the function can in fact be equal to 0. Using this function we can find the

boundary conditions. We have:

Hence substituting back we have:

Now if we differentiate this function we have:

Using the second boundary condition we have:

[

]

Substituting back into the original equation we have:

(

)

(

)

Transient Conduction and Diffusion So far in all our work in conduction and diffusion, we have been assuming that everything takes place at steady

state. This assumption is very useful as it allows us to neglect the accumulation term in the mass and conduction

balances and helps us to solve the question. However, the stark reality is that in real life majority of the processes

take place at unsteady state and even if they do, their behaviour before they reach steady state is critically

important. For example when you heat something in the microwave or if a pollutant spreads in a river. Additionally it

must be noted that just studying steady state gives you no idea how long it takes to reach steady state. For example,

the steady state temperature profile in a slab for constant conductivity is independent of the conductivity, but the

time taken to essentially reach steady state depends upon the conductivity (the term contains the conductivity).

Hence we must determine not only how a temperature or concentration profiles varies in space but also how it

varies with time.

In order to factor in both special and temporal information that is required at unsteady state we must utilize the

complex mathematical tool of Fourier’s mathematics.

Fourier’s Mathematics Let us start off by considering a function between 0 and , which has a value of 0 at both boundaries. One possible

solution for this type of function is simply where . Just to make this a little bit more generic, let us

consider a function between 0 and that must be equal to 0 at each of the boundaries. Then we get a function like

(

) where .

Now this is the point where you will need to take a leap of faith as this point is well beyond your mathematical

capability. Any suitable function which takes values of 0 at 0 and , can be written as a superposition of these sine

wave functions. This is a very surprising and non trivial result. This statement implies that:

∑ (

)

Note: are numbers and that there are an infinite number of them. Also you should realize that as tends towards

, tends towards . This prevents the series from blowing up. Additionally, the term ‘suitable function’ above

implies a function which is smooth and continuous in the domain specified. However, functions which have a finite

number of discontinuities may also be considered suitable. Now it should also be realized that when a function is

decomposed as a sum of sine functions, there is only one way in which this may be done. Hence if two Fourier series

are equal everywhere, then necessarily, the corresponding coefficients are equal. This is shown mathematically

below:

∑ (

)

∑ (

)

Then this expression implies that for all :

Now the next question that must be considered is how exactly do we determine the value of each of these

coefficients . There is a formula which does this for us successfully.

∫ (

)

Hence given any function or profile, you can find the Fourier coefficients, and vice versa.

Back to transient conduction and diffusion Let us consider conduction for the time being. Lets say we have a temperature profile whose values at the boundary

are , not zero. Now in order to ensure that we have a function which is 0 at both boundaries we must manipulate

the temperature profile a bit. Hence if we take our temperature profile and subtract from it, then we are sure

that at the boundaries the apparent temperature is 0. Hence the function can be considered to be mapped

by Fourier’s mathematics. Hence we have:

∑ (

)

Now it should be noted that all the spatial variation of is contained in the sine part of this summation, and all of

the temporal information is located in the coefficient which is a function of time. In essence by writing the

function as a Fourier series, we have factored out the temporal and spatial information into two different

components. Hence now if we would like to determine how the temperature profile varies with time, we must

simply find the Fourier coefficients and how they vary with time.

In order to determine the Fourier coefficients we must use some information from the conduction equation. The

conduction equation states:

[

]

[

]

[

]

With conduction in only the direction and no generation we have:

Where

. Now if we substitute for here we have:

∑ (

)

[ ∑ (

)

]

[ ∑ (

)

]

∑

(

)

∑ (

) (

)

Now as we saw before, if two series are equal then the corresponding Fourier coefficients must be equal to one

another. Hence we can say that for all values of :

We have almost found the temporal variation of every coefficient . Now the only remaining piece in this puzzle is

the initial condition. Usually we are given information regarding the initial condition of the temperature profile.

Hence we have:

∑ (

)

You should note that sometimes we are given information like . Then using this function of we

can use the expression below to solve for .

∫ (

)

Given this piece of information we can easily determine all the Fourier coefficients. Let us assume for now that

at the boundaries, i.e. where . Hence at this point let us call the Fourier coefficient to

be . Now using the expression below and separating variables and integrating definitely gives:

∫

∫

(

)

Hence here is an expression mapping exactly how any Fourier coefficient varies with time. Now knowing how this

coefficient varies with time allows us to reconstruct the full spatial profile and also know how it varies with time. This

shall be discussed in further detail.

Summary of Procedure so far...

1. We had fixed, equal temperatures on the boundaries

2. We expanded the temperature profile into a suitable series (Fourier series in this case). The coefficients of

the sine functions are where the temporal information is contained

3. Plug this into the governing conduction equation. This results in ordinary differential equation in time for

each of the coefficients in the series

4. These ordinary differential equations are all decoupled and can be solved easily. In order to find the solution

initial conditions for these coefficients are required

5. Expanding the initial condition profile in the same series, gives the values of the initial values of the

coefficients

6. Knowing as a function of time, we can reconstruct the full temperature profile as a function of time

A few important points to note:

If the temperatures at the boundary are fixed but not equal, this procedure can be adapted easily

If the boundary conditions are different, the expansion in the sine series will not work. However expansion in

a different series (e.g. cosine) will work

If the initial condition contains only a few sine functions, only those sine components are present

subsequently

Example

Suppose the temperature at both boundaries is maintained at a fixed temperature and the initial

condition of the temperature profile is:

(

)

If and when , how does the temperature profile evolve?

Looking at the initial condition, we can see that there is only 1 sine mode i.e. the (

) mode.

Hence we have:

∑ (

)

Substituting gives:

(

) ∑ (

)

(

) ∑ (

)

As a result of this initial condition, all Fourier coefficients are 0 apart from . Hence we have:

(

) (

) (

)

(

) (

)

Now at :

Now we know that the time variation of any Fourier coefficient is given by:

In this case we have:

Separating variables and integrating gives:

Now we know that:

∑ (

)

Which in this scenario is:

(

)

Substituting in our expression for we have:

[

] (

)

Hence we have determined exactly how the temperature profile varies with time. It should be noted that if you take

the limit as tends to infinity, you can get the steady state temperature profile for this slab.

Suppose the temperature at both boundaries is maintained at a fixed temperature and the initial condition of the

temperature profile is:

(

)

If , , and when , what are these coefficient values at time ?

To perform this question we must write (

) as a sum of sine functions. In order to do this let us consider

as this is most likely the function that will have some sort of relationship to . Hence we have:

Hence we can see that:

Using this expression we can say that:

(

)

(

)

(

)

Hence we can rewrite the initial condition as:

(

)

(

)

(

)

Now substituting this into the Fourier series temperature profile, we have for :

(

)

(

) ∑ (

)

(

)

(

) ∑ (

)

We can see now that there are only 2 sine modes that are active and these are when and . Hence we

have:

(

)

(

) (

) (

)

By equating coefficients we can say that:

And:

Now we know that the expression describing the way in which each of these Fourier coefficients varies with time is:

Hence for the first Fourier coefficient we have:

And for the third Fourier coefficient we have:

Hence substituting these coefficients back into the original Fourier series profile we have:

(

) (

)

(

) (

)

Thus at the end of this question we have established a complete temperature profile as a function of space and time.

It should be noted above that the 3rd sine mode decays 9 times as fast as the first mode, i.e. as time progresses, the

dominant transient behaviour is that of the first mode. The steady state temperature profile is once again simply .

It should be noted that all of these type of problems may be done for diffusion and mass transfer as well.

Transient Conduction and Generation This is not much of a change really. In fact it is exactly the same setting as before, except for an extra term that

accounts for the position dependent generation. For example, if the position dependent generation were:

(

)

Now, if we were asked how the Fourier coefficient changed with time we would perform the question as follows.

The conduction equation tells us that:

(

)

Now as we are assuming the same situation as before, we know that the temperature at the boundaries of the slab

are both . Hence, we can say that:

∑ (

)

∑ (

)

Substituting the profile into the conduction equation expression we have above, we can say that:

[ ∑ (

)

]

[ ∑ (

)

] (

)

∑

(

)

(

)

∑ (

) (

)

∑

(

)

(

)

∑ (

) (

)

Now from this expression we can see that when :

And we can see for the remaining values of :

Now we know that when we integrate for the remaining values we get the expression:

However, the expression for is slightly more difficult to solve. When it is integrated however, it looks like:

ASK HIM HOW THIS COMES ABOUT!

Transient Diffusion with Reaction This is pretty much identical to the scenario with the heat generation.

Example

Consider a slab of length L where oxygen is diffusing, but is also

consumed everywhere via a first order reaction. Both ends are exposed

to an ambient concentration of . If concentration of oxygen is initially

non-uniform, how does it change with time?

We know from the mass balance that:

[

]

However, it should be remembered that for transient problems, where we are dealing mostly with solids there is a

constant total concentration. Additionally in a solid the first term of Fick’s law may drop out as . Hence

from Fick’s Law we can say that:

Hence substituting for this into the mass balance we have:

Now the reaction term is – . Minus because the reactant is being consumed, is the rate at which it is being

consumed, and is the concentration of the oxygen. Hence substituting gives:

oxygen

oxygen

–

Now using Fourier’s series we have:

∑ (

)

∑ (

)

Substituting for we have:

[ ∑ (

)

]

[ ∑ (

)

] – [ ∑ (

)

]

∑

(

)

∑ (

) (

)

∑ (

)

Now if we divide by ∑ (

)

we have:

[

]

Separating variables and integrating definitely gives:

∫

[

]∫

(

) [

]

[

]

Hence the way in which the concentration profile varies with time is:

∑ (

)

∑ [

]

(

)

Convection Convection is a mode of heat/mass transport which involves a flow of matter. There are two main types of

convection:

1. Forced Convection

In this type of convection, fluid flow is driven by some process. This aids in heat and mass transfer owing to a

temperature and concentration differential being setup

2. Free Convection

In this type of convection fluid flow arises as a result of temperature differences in the fluid, due to a

significant density variation with temperature (buoyancy effects) or concentration

Forced Convection A good example of a forced convection problem is when there is flow of a fluid in a

pipe, driven by a pressure differential. The calculations involving the pressure

differentials would be considered to be a fluid mechanics problem. Now the pipe could

be at a different temperature than the fluid that is entering. Hence, the temperature of

the fluid could change as it proceeds through the pipe. The temperature profile that

provides this information is complicated and varies with position. This part of the

problem would be considered a heat transport problem. Think of this as if it were

talking about fluid flow past a heated cylinder. You should also note that there could be an analogous situation for

convective mass transport.

Free Convection In free convection, there is no fluid flow without the temperature

differential. Hence there is induced fluid flow and convective heat

transfer. A good example of this would be the condensation of gas to

liquid when exposed to a cold plate and corresponding flow. There may be

similar situations in mass transfer.

Forced Convection The extent of heat transfer in forced convection is contained in the heat transfer coefficient, .

In essence this is value tells you how good the energy removal is. If you recall in the formula

, is the local heat transfer coefficient. The heat transfer coefficient is

not a fundamental material property, but it rather depends upon the material, flow

properties and flow configuration (i.e. shape etc).

Non-Dimensionalization

Consider the following example:

Assume steady state conditions and determine how the

concentration of oxygen varies in the organic matter.

From the mass balance we know that:

We know the values of the boundary conditions. That is at ,

and at (the length of the slab), the absolute flux

. Now before we continue let us

have a small dimensional analysis. By looking at all of the variables we are using here we can see that we have:

1. The rate constant which has units of [ ]

2. The diffusion constant which has units of [

]

3. The initial concentration which has units of [

]

4. The concentration profile which has units of [

]

5. The length of the slab which has units of [ ]

T0

Ambient T

1

T

a

Impermeable surface

oxygen c

A=c

A

,0

z

6. The direction along the slab which has units of [ ]

Now by rearranging these variables we can come up with one dimensionless parameter (lets call it ). This is:

(

)

Now let us try to make almost all of the variables that we will be dealing with dimensionless. Thus we have:

And:

Now using these dimensionless variables we can rewrite our above boundary conditions. The first boundary

condition which is , may be rewritten as when . Additionally the second boundary

condition which is

when can be rewritten as

when . In fact not only can our

boundary conditions be rewritten in such a way, but also our fundamental mass balance which was

Can be rewritten as:

So why do we want everything dimensionless? The reason is that if we have a dimensionless equation it acts almost

like a ratio. Hence for any particular reactor the dimensionless profile will be the same and the actual profile can be

obtained easily from the dimensionless profile. Now this equation above is a universal dimensionless equation with a

fixed solution which depends upon this parameter .

Now to find the concentration profile we must simply separate variables and integrate. Thus we have:

Let us guess

. Now we know the . Hence . This leads to the following profile:

Now we use the boundary conditions to determine exactly what and are. We have:

Hence we have:

Hence substituting back into the profile we have:

(

)

As can be seen the solution depends upon the parameter .

Back to Forced Convection... In order to proceed with forced convection problems we must have some way of determining what exactly the heat

transfer coefficient is. However, as we already know the heat transfer coefficient is not a fundamental material

property that one can simply test. Rather it depends upon the material, flow properties and flow configuration!

Hence information about the heat transfer coefficients is only obtained after comprehensive and careful

experimental investigations.

The next question that must be asked is how exactly the empirical information is reported. It is not easy to attribute

a quantity to the type of material, flow properties or flow configurations! Additionally how can one simply compare

the empirical data between two completely different situations?

The answer to the first question is through the use of the Nusselt, Reynold and Prandtl Numbers. The answer to the

second question is to ensure that all data presented in dimensionless. This allows for an easy comparison to different

situations.

The Nusselt Number is given by the following expression:

Here is the much required heat transfer coefficient, is a suitable length scale and is the conductivity of the

fluid. The Nusselt Number can be considered to be a measure of the convective transport. If we have a local Nusselt

Number, then we get a local heat transfer coefficient. However, if we have an average Nusselt Number then we get

an average heat transfer coefficient. Additionally if an average heat transfer coefficient is used, then an average

Nusselt Number shall result. Also it should be noted that normally length scale is considered a constant rather than a

variable.(The Nusselt Number gives the ratio of convective to conductive heat transfer across a boundary).

The Reynold’s Number is given by the following expression:

Here, is the density, is the maximum velocity, is the diameter or length of the pipe or surface and is the

dynamic viscosity. This number allows us to specify the flow configuration. It allows us to determine whether the

flow is laminar or turbulent. (The Reynolds Number gives the ratio of the inertial forces to the viscous forces in a

fluid).

The Prandtl Number is given by the following expression:

Here you know all of the terms. Just remember

. (The Prandtl Number gives the ratio of momentum

diffusivity to thermal diffusivity).

Now it should be remembered that the empirical information that we get from investigations is written as an

equation relating the Nusselt number to the Reynolds and Prandtl numbers. Hence, once we have a specified

Reynolds and Prandtl number we can calculate the heat transfer coefficient for that system, given ofcourse that the

conductivity of the fluid and the length is known! These type of expressions that relate the 3 types of numbers are

called correlations.

Thus in a heat transfer problem, an experimental/computation/theoretical investigation is used to form a

relationship involving the Nusselt number (containing the heat transfer coefficient) and the Reynolds and Prandtl

numbers for forced convection.

Here are examples of some correlations (you do not have to learn any):

1. For convection involving a sphere

2. Laminar Flow past a flat plate of fixed temperature ( represents the distance from leading edge)

3. Turbulent Flow past a flat plate of fixed temperature ( represents the distance from leading edge)

4. Cylinder in Cross Flow ( is the diameter and the values of and are determined empirically, depending

on the Re, after curve fitting (values are given in lecture 12))

Flow Past A Heated Sphere

Using these correlations, we can determine the value of the Nusselt Number, and hence the value of heat transfer

coefficient. Once this has been determined we can easily determine exactly what the heat flux or heat energy being

transferred is. For example consider a fluid flowing past a sphere at a velocity of . Let us assume that the

temperature of the sphere is and the average temperature of the fluid is . Using the correlation for the sphere

we have:

Substituting all of the required information into this equation we find:

(

)

(

)

It should be noted here that as the velocity of the fluid increases, it in effect increases the Nusselt number which in

turn increases the heat transfer coefficient. When the velocity of the fluid is 0, the Nusselt number is simply 2 in this

case, which signifies that there is no convection occurring but only conduction in the external field. It should be

remembered that in effect the Nusselt number is giving us the ratio of convective to conductive heat transfer. In this

case we can see that heat transfer coefficient is simply given by:

(

)

Substituting this value into the equation:

( )

(

) ( )

This equation gives you the flux transferred. If you would like to determine the heat energy transferred, then we

must multiply by the area of heat transfer, which in this case is the surface area of a sphere.

(

) ( )

(

) ( )

Flow Past a Flat Plate

Let us now consider the situation where we have a fluid flowing past a flat plate at a velocity of . We know from

fluid mechanics that the flat plate will exert some frictional forces on the fluid hence cause a change in the velocity

very close to the boundary. In fact, if we assume the no-slip condition, then the fluid in contact with the flat plat will

have a velocity of absolutely 0. (A diagram of this is on attached sheet. OP15.) If you look at the curve drawn on this

diagram, the area under the curve is all the area where the velocity is considered to significantly deviate from ,

and all the area above the curve is where the velocity may be considered to be the same as . Now as the value

being considered increases, so does the length of the boundary layer. However, as you can see from the curve,

although the boundary layer initially increases sharply, its rate of increase with respect to decreases extremely

quickly and it begins to level off. Hence, at a certain point we can conclusively say that any further increase in will

not affect the length of the boundary layer . Hence at this point we can say that has become independent of . It

was found that this curve displaying the boundary layer could be mapped by the expression

. Hence it can be said

that the velocity increases monotonically from at the plate boundary, to far away. The exact distance this ‘far

away’ relates to is given by the expression:

√

The boundary layer is defined as the period between , as at , the velocity may be considered for all

practical purposes o be exactly .

Now the correlation that may be used to depict the way in which the flow affects the heat transfer coefficient in this

scenario is:

It should be remembered that this correlation holds only if the flow is laminar. In experimental observations it was

found that for this scenario, the flow was laminar only when the Reynolds Number .

A similar correlation was derived for the case of turbulent flow, and this is given by the following expression:

So far everything discussed should make sense. However, an important point should be remembered that applies for

all the correlations discussed above. All of these correlations are based on flow around the outside of an object. One

may now logically ask, what about flow within another object, such as flow through a pipe. The answer is there are

lots of correlations for these scenarios as well. However, once the heat transfer coefficient has been determined,

and the heat flux transferred is required we face an inherent problem. This problem occurs due to the equation

below:

( )

The problem is that the temperature of the fluid within a pipe or within an object varies based on a certain profile.

So what value of should we use? The answer is we must use the average temperature. The average temperature

within a pipe/object is given by the following expression:

∫

Here is the mass flow rate, is the velocity (as a function of ), is the temperature (as a function of ), is the

heat capacity and is the area.

Example

Air at pressure of 6 kN/m2 and temperature of 300 flows with a velocity of 10m/s over a flat plate 0.5m long.

Estimate the cooling rate per unit width of the plate needed to maintain it at a surface temperature of 27 .

Properties of air at mean temp. of 437K, given pressure are:

k=0.0364W/m, Pr=0.687, =0.000521m2/s

In order to perform this question we must first determine how much heat is being supplied to the plate. You should

also note that here, there is flow around the outside of the object. Hence we do not need to calculate the average

temperature. Let us first find out what type of flow we are faced with. Hence, in order to do this, let us calculate the

Reynolds number for the air at the very end of the plate. This value of the Reynolds number shall be the maximum

value, and if this is within a certain flow type, then we can be rest assured that all of the flow is within this flow type.

Hence we have:

This is well within the laminar region which goes till . Hence using the laminar correlation for a flat plate we

have:

Substituting gives:

( (

)

)

Now by using Newton’s law we have the heat transferred from the fluid to the plate is (think about where the

energy is transferred from when deciding which temperature to put first in the equation):

( )

Substituting gives:

Now let us consider a small slice, of the slab. The heat transferred through this slab is only . As we know the

width of this slab is unit, we can say:

Hence by integrating across the whole length, we have:

∫ ∫

(

)

If this is the amount of heat being transferred to the slab, then an equivalent value of heat must be removed from

the slab to maintain it at the initial temperature of .

Thus when faced with a problem in forced convection where the use of correlations becomes necessary, adopt the

following approach:

1. Note the flow geometry and flow conditions

2. Select appropriate temperature and obtain fluid properties at that temperature

3. Obtain the Reynolds number which indicates flow regime

4. Choose an appropriate correlation and obtain the heat transfer coefficient

5. In some cases iteration may be necessary for example where an unknown temperature has to be

determined

The Mass Transfer Side of Forced Convection So far we have only looked at the heat transferred due to convection. However, in addition to just heat transfer,

there is also mass transfer that is occurring simultaneously. Convective mass transport works extremely similarly to

heat transfer. In fact one of the only differences is the names of the dimensionless coefficients used in the mass

transport correlations. Instead of the Nusselt Number, in mass transfer we have the Sherwood Number. This is

defined by the following expression:

Here is the mass transfer coefficient and is the reason behind all of these correlations. is the length, could be

the diameter or length based on the geometry. Finally is the diffusive constant. During this course we shall only

be considering the mass transfer associated when a certain species diffuses through another species , when is

in extremely dilute concentrations. Next in mass transfer, instead of the Prantl number we have the Schmidt number

which is given by the following expression:

The Reynolds number is used in mass transfer as well so there is no issue with that. Now instead of using Newton’s

Law of cooling, we use the expression for molar flux to calculate the amount of mass transfer occurring. This

expression is given by:

Here is the molar flux, and the units for the mass transfer coefficient are .

Correlations for Convective Mass Transport

Correlations for convective mass transport can be found in the case that physical properties are constant, mass

transfer rates are small, no chemical reaction, viscous dissipation can be neglected (which it can for many fluids

because it turns out to be very small) and no radiation.

When these conditions are met, it is found that the correlations established for heat transfer exactly parallel those of

mass transfer for the same flow geometry (and commensurate boundary conditions). Hence, we may simply replace

the dimensionless numbers with the ones used in mass transfer and these use the correlations to obtain the mass

transfer coefficient.

For example, for flow past a flat plate containing material at a different concentration from the fluid the correlation

is:

Laminar flow past a sphere:

Transport Resistance and Overall Transport Coefficients Often, due to a need for insulation or prevention of mass transfer, an understanding of heat and mass transfer

resistance is required. In order to achieve this, we must compare heat and mass transport to electricity.

It may be understood that the equivalent of a potential difference in electricity is the temperature/concentration

difference in heat and mass transport. Additionally, the current in electricity may be equated to the heat and mass

flux. If this is the case then the resistance may be simply represented as the potential difference over the current.

Hence in the case of heat transfer, the resistance can be equated to:

Now if we consider the resistance in slabs we can see that:

Hence the resistance for a given slab is

, and just like a series of resistors, if we have slabs then the resistance for

each of them can be summed together to calculate a total resistance. Now what about a slab which contains a fluid

in it, or a gap of air between slabs? In this location, convection governs the amount of heat transfer, hence we can

say that:

So in this case, the resistance is given by the inverse of the heat transfer coefficient. Once again this may be summed

for a series of slabs. It should be noted that the higher the heat transfer coefficient, the lower the resistance. Now

consider a series of slabs of fluid next to each other. The total resistance is given by:

It should be noted that the total resistance is equivalent to the inverse of the total heat transfer coefficient. So you

may simply add up all of the heat transfer coefficients and take the reciprocal of this value to determine the total

resistance.

For mass transfer, there is an exact parallel to these situations. Just quickly running through the expressions we

have:

Now if we consider the resistance in slabs we can see that:

For convective mass transport we have:

Also as before you should remember that the total mass transfer coefficient is the reciprocal of the total resistance.

Let us now try an example question. Consider the following question:

Consider flow through pipe, as well as flow outside pipe. We have convection outside the pipe, convection inside the

pipe, as well as conduction through the pipe walls. Far away temperature shall be called Bulk Mean Outside

Temperature denoted as . Similarly the temperature within the pipe shall be called the bulk mean internal

temperature and shall be denoted as . The internal radius is , the external radius is and the conductivity

through the pipe is .

How can we write the steady state heat transfer rate in terms of far away temperature and the average bulk

temperature in the pipe?

In order to answer this question let us consider first the heat transferred from the bulk outside conditions to the

outer pipe surface. Hence we have:

Here is the heat transfer coefficient of the outside conditions. Hence we have:

Here is the outer pipe surface area. If we now perform a similar procedure for the heat transferred from the inner

surface of the pipe to the inner bulk conditions we have:

Now in order to determine the heat transfer through the wall of the pipe we must recall that the temperature profile

for conduction through the annulus of a cylinder is:

[

( )

] (

)

Now if we plug this into the flux equation we have:

[

( )

]

[

(

)]

[

(

)]

Now at this point this expression is all well and good, however, we must add an extra negative sign now to ensure

that the heat is being transferred from the outer surface of the cylinder to the inner surface of the cylinder. Hence

the expression becomes:

[

(

)]

Now let us write an expression for the overall heat transfer through the pipe based on the outer surface. Thus we

have:

Here is the overall heat transfer coefficient, and would be the only unknown in this question if it were posed to

us. So how would we calculate this overall heat transfer coefficient? It is at this stage where we must use our

knowledge of resistances. We know that the reciprocal of the overall heat transfer coefficient is the total resistance.

Hence the resistance for the first convective layer is:

Now the resistance for the inner convective layer we have:

Now we need the resistance for this layer based on the outer surface, hence we must somehow get the outer

surface area into this equation. Hence if we multiply both sides by we have:

Now we must perform a similar procedure for the conductive layer in the pipe wall. Hence we have:

[

(

)]

Multiplying by we have:

[

(

)]

(

)

(

)

Therefore by adding the resistances together we get the total resistance, which is also equal to the inverse of the

total heat transfer coefficient. Hence we have:

(

)

Through this formula we may determine the overall heat transfer coefficient and hence determine the overall heat

being transferred. It should be noted that in the cylinder, flux is not constant, hence the resistance depends upon

where it is taken relative to. In this case we considered the outer surface, but a similar equation may be derived for

the overall heat transfer coefficient based around the inner surface.

Interaction of fluid flow and heat/mass transport: Forced Convection Sometimes we are given a particular fluid flow profile – from Fluid Mechanics – and then asked to determine the

temperature or concentration profiles and heat or molar fluxes. If this can be achieved, we can also determine

exactly how the fluid flow affects the temperature and concentration of the fluid.

In order to achieve this let us consider the following type of fluid flow:

Consider steady state incompressible and two dimensional flow where the velocity is and the velocity is .

Now as we know that the flow is incompressible we can see that:

At any given position the velocity gradients are constant. In essence this is a mass balance. Now we know from the

derivation earlier that the mass balance for a system which is experiencing convection, diffusion and generation is:

[

]

Now let us consider dilute amounts of in medium flowing at a velocity of and in the and directions

respectively. We have:

( )

Similarly we can say that:

( )

Now it should be noted here that earlier we had said that the velocity is and the velocity is . What this

actually means is that the average velocities in each of these directions are and respectively. Hence this leads to:

And

Now we also know that the diffusive flux is given by the equation:

Where is the total concentration. Now it should be noted that as we are considering incompressible flow here

(which in essence means a liquid), we can assume that the total concentration is constant. Hence we have:

Hence we have:

And

Now if we substitute these values for the absolute flux into the overall mass balance we have:

[

]

[

(

)

(

)]

(

)

(

)

(

)

(

)

Now by using the product rule we can differentiate the following terms of the above expression:

And

Substituting back into the earlier equation we have:

(

)

(

)

However, now it should be noticed that from the earlier statement we know:

Hence:

(

)

(

)

(

)

(

)

(

)

At steady state we can see that this expression becomes:

(

)

(

)

Hence we have found a relationship between the concentration profiles and the average velocities. If we now use a

similar approach with an energy balance we get:

(

)

[

]

[

]

Here represents the viscous dissipation and may be calculated using the following formula:

(

)

[(

)

(

)

]

(

)

This expression basically allows us to calculate the amount of heating that occurs in the fluid due to viscous shearing.

So let us now consider an example where we can start with a fluid mechanics problem and link it to heat and mass

transfer.

Example

Consider a couette flow in a channel. The upper plate is at a temperature of and the lower plate is at a

temperature of . Now lets say if the upper plate moves at a velocity of and the plates are a distance apart,

what is the temperature profile between the two plates.

You may assume that the fluid properties are weakly dependent on temperature.

Now from fluid mechanics it is known that the velocity profile through the channel is:

Now at steady state, velocity is only in the direction and depends only on . At any given value of there is no

variation of any quantity in the direction. Hence using this information we can say that the temperature varies only

in the direction. Hence now if we go back to our convection energy balance we have:

(

)

[

]

[

]

(

)

[ ]

[

]

Hence we have:

[

]

Now in order to solve this equation we must go to the equation for the viscous shearing. Hence we have:

(

)

[(

)

(

)

]

(

)

(

)

[ ]

Hence we have:

(

)

Now in order to solve this differential we must use the velocity profile above. Thus we have:

Substituting gives:

(

)

Substituting this back into the convection energy balance we have:

[

]

[

] (

)

Now if we have constant conductivity, we have:

(

)

Thus solving this equation by integrating gives us the temperature profile for the channel being considered. To solve

it we must integrate twice. As can be seen the right hand side does not consist of a or a . Hence we don’t need an

exponential function as was used when integrating a second order differential earlier on. In this case the profile must

simply be quadratic. Hence we have:

(

)

Integrating again gives:

(

)

Now by substituting the boundary conditions which are: when , and when , , we have:

Substituting gives:

(

)

The second boundary condition gives:

(

)

Substituting gives the overall temperature profile to be:

(

)

It should be noticed from this example that fluid flow has the effect of making the temperature profile parabolic

instead of linear. Now if we were to calculate the heat flux in the above example at and , (as clearly

heat flux varies with ) we would get:

(

(

)

)

This heat flux could be either into or out of the fluid depending on the temperatures. At the heat flux is:

(

)

And once again this could be into or out of the fluid depending on the temperatures. An interesting couple of

observations should be made at this stage. Firstly consider the scenario where the temperatures at either end of the

channel were equal. The temperature profile would read:

(

)

And would still remain parabolic! This is slightly counter intuitive but is a direct result of the fluid flow. Additionally

now if we go back to the general case, the maximum temperature in the channel can be calculated to be present at:

(

)

(

)

(

)

(

)

(

)

And in the case that the temperatures are equal, the maximum temperature becomes present at:

(

)

Viscous heating is a concept that is of great importance to all types of engineers particularly when it comes to space

travel. For example, the air flow next to a rocket results in huge amounts of viscous heating and hence must be

taken into account.

Mass Transport in Couette Flow Just as we just considered heat transport in Couette flow, we must also now consider the concentration profile in

couette flow. Hence imagine the following example:

Consider a couette flow in a channel. The temperature of the plates are the same. The fluid contains dissolved solute

and both the upper and lower plates contain pores to allow diffusion of the solute through them. Assuming the

concentration of the upper plate is and that of the lower plate is , calculate the concentration profile

between the two plates.

You may assume that the diffusion constant is essentially independent of temperature.

Now once again it is known from fluid mechanics that at steady state, the velocity profile through the channel is:

Where is the velocity in the direction. The velocity ( ) is 0 ofcourse. Now from the momentum balance

conducted above we have:

(

)

(

)

Now at steady state and for fully developed flow, we can clearly see that there will be no variation of velocity or any

other quantities in the direction. Additionally we know that the velocity is and there is no mass generation.

Thus we have:

( )

(

)

Hence, assuming is constant, our overall equation becomes:

Integrating we get:

Separating variables and integrating again we have:

∫ ∫

Substituting boundary conditions which are , when and when , we have:

Substituting back we have:

From the other boundary condition we have:

Substituting back gives us the overall concentration profile as:

(

)

Thus it can be seen that the concentration in such a scenario is clearly linear. However, care should be taken when

attempting such problems due to the assumptions made here. Remember, steady state was assumed, also the plates

were assumed to be infinitely long and the flow was assumed to be fully developed resulting in no changes in

concentration with . Thus be careful when attempting a question.

So what if we relax the assumption regarding the infinitely long plate? In industry the process of falling film is used

very commonly, allow impurities to dissolve into or out of a flowing fluid. The film usually flows down an inclined

surface. The velocity of the topmost layer of fluid is . Lets say the flow is fully developed and at steady state. Over

a length the fluid comes into contact with a solute in the atmosphere. Now lets say we were asked to calculate

how much solute was absorbed by the liquid assuming it was carrying no solute initially.

So let us start with the fact that the fluid flow is fully developed. Hence

from fluid mechanics it is known that:

( [

]

)

This is clearly a parabolic fluid flow profile. Now when the flow is initially

exposed to the solute, the system is not at steady state. However, over

time, the system slowly reaches steady state. Now in this scenario the

difficulty arises due to the fact that we cannot justify the claim that the

concentration profile of the solute does not vary with . As we are now

dealing with plates of a finite length and are particularly focusing at

the point where pure liquid enters between the plates, we can

understand that as the liquid travels further through the pipe, there is a larger amount of solute that can diffuse into

it. This is because the liquid is in contact with the plates for a longer amount of time. Hence, we cannot now assume

that the concentration of the fluid is independent of the direction, as we could have done in the earlier scenario

where the plates were infinitely long. This leads to the following balance:

(

)

(

)

Now this simplifies to:

(

)

(

)

(

)

(

)

Now unfortunately at this stage we cannot proceed due to the additional terms present. On the left hand side we

have a convective term and the right hand side we have the diffusion terms. In order to proceed and actually

generate an answer, we must make some sort of approximations. The first assumption we make is that the fluid is

flowing so fast that we can neglect diffusion in the direction of the flow. In this case that would mean diffusion in the

direction would go to 0. This leads to the following expression:

(

)

In essence this expression is telling us that the diffusion in the direction is balanced by convective transport of the

solute in the direction. Now unfortunately, the solution to this problem is extremely difficult and requires much

more complex mathematics than is currently known. Hence, we must approach this problem from a different

viewpoint. Remember the objective of this question is to determine the number of moles of solute being dissolved in

the fluid per unit time. Hence we know from earlier knowledge that the absolute flux for a liquid is given by the

expression:

Now in this particular scenario we know the velocity . Hence we can say that the absolute flux of the solute is

simply given by the expression:

L

Solid surface

x : direction of flow

y: perpendicular

Now usually the solution to this problem is difficult, however for some special cases it may be solved relatively

easily. If the contact time is small, it may be assumed that the problem is almost identical to the case where diffusion

occurs through an infinite medium. In essence due to the short contact time, the effect of the solid surface is not felt

by the solute. Now when such problems are given to you, you will be given a concentration profile that will look like

the following:

(

√ )

2 things must be noted about this concentration profile. Firstly, this concentration profile is given to us under the

assumption that . Secondly, the error function of a given variable is defined as:

√ ∫

It should be noted that and . Having established this we can now substitute the concentration

profile into our equation for the absolute flux expressed above:

[

(

√ )

]

{

[ ]

[

(

√ )

]

}

[

(

√ )

]

Now one of the results that we must simply learn but which has its origins in the chain rule is:

(

√ ∫

)

√

So now let

√

. Hence we have:

√

(∫

)

√

√

(

√ )

(

√

)

(

)√

Therefore we conclude by saying that:

(

)√

(

)√

Interaction of Fluid Flow and Conduction Suppose you are posed with the following scenario. You have two plates

which are porous and large. The upper plate is at a temperature of and

the lower plate is to be maintained at a temperature of . To ensure that

the lower plate is kept at the specified temperature a coolant gas is blown

upwards at a velocity of . Now lets say we were asked to determine how

the flow would affect the temperature variation and heat transfer.

The fluid mechanics part of this problem is very easy as the velocity is

simply 0 in the direction and is in the direction. Now from the

energy balance we can gather that:

(

)

[

]

[

]

[

]

When the conductivity is constant we have:

Solving this differential equation gives:

Now in order to determine the constants we use the boundary conditions:

Substituting back we have:

Using the second boundary condition we have:

(

)

T

1

T0

(

)

(

)

Substituting back into the original equation gives:

(

)

[

(

)

]

(

)

(

)

(

)

(

)

Notice now that when is 0 we get back the original temperature profile! Thus we can clearly examine exactly the

relationship between the temperature profile and the speed of the fluid.

A2

A2

A0 kc

dy

cdD

dy

dcV

Flow and diffusion ,consumption of A (first order reaction)

A convection reaction diffusion equation

Boundary conditions: cA=0 at y=0,

cA=c

A,0, at y=L

1

C=cA/c

A0, Y=y/L

0CD

Lk

dY

dC

D

LV

dY

Cd 20

2

2

C=0, Y=0

C=1, Y=1

Peclet number (Pe)

Damkohler number (Da)

membrane

Heat Exchangers A heat exchanger is a device used to implement an exchange of heat energy between two fluids at different

temperatures separated by a solid wall. It is a basic component of engineering. There are 3 major classes of heat

exchangers: counter current, co-current and cross flow. Additionally there are 3 major ways in which a heat

exchanger may be set up. These are shell and tube heat exchangers, double pipe heat exchangers and plate heat

exchangers.

A typical heat exchanger problem involves:

1. A fluid at a particular temperature being used to heat up a fluid flowing at a particular flow rate to a

specified temperature. Usually we assume no heat loss and hence we can calculate the exit temperature of

the heating fluid

It should be noted at this stage, however, that if we are given a

particular heat exchanger, we will unfortunately not be able to

determine the desired temperature change at steady state. This is

because the amount of heat transfer depends on the time of heat

transfer, the rate of transfer which in turn depends on the material

and flow properties and configuration.

This is where knowledge of transport rates comes in. So the

question is how do we find what the exiting temperatures are and

what is the size of a heat exchanger to obtain the desired

temperature change?

To answer these questions let us consider the following situation.

How does the concentration vary with position Y? Second order equation with constant coefficients Exponential solution will work

m2

–mPe –Da=0

Two roots:

exp(m1Y) &

exp(m2Y)

Two solutions: The general solution of such an equation can be written as C(Y) =C

1 exp(m

1Y) +C

2

exp(m2Y) Use boundary conditions

Let us say we have a double pipe heat exchanger. Note that position 1

always indicates the cold stream entry. Now let us consider the cold fluid

which is in this scenario flowing in the inner tube. Observe the region

between and . An energy balance for this fluid reveals that:

But it is known that:

By substitution, this leads us to the following expression:

Therefore the total heat transferred to the cold stream is simply:

This assumes is constant but if it is not do an integration using the required expression. Now the heat transferred

by the hot stream is given by:

Now it should be noticed that here as there is no heat loss, the two expressions for total heat transferred are the

same. Hence we have:

As we know that this equality is true, we can deduce a number of things by simply looking at the relationship

between the specific heat capacities of the two streams. Hence if:

Then logically, the temperature change in the hotter stream is greater than the temperature change in the cooler

stream. Now lets say if we wanted to determine exactly the required surface area for a heat transfer. In order to

calculate this value we must do some heat transport analysis. As before we know:

Where is the heat flux transferred and is the overall heat transfer coefficient obtained from a given correlation.

Now from this equation we get:

Where is the surface area between and and is the heat transferred across this distance. It should be

noticed that both and depend upon . Hence we must be careful when dealing with them. Rearranging we

have:

Now recall from above that we have an expression that reads:

M

c

T

c

1

T

c

2

T

c

T

h

Substituting this equation into the equation for the area and we have:

Now simply by integrating across the entire domain we have the following expression:

∫

This expression is obviously obtained assuming the heat transfer coefficient does not vary with position. Now this

expression may appear to be all well and good, however, it should be seen that we are asked to integrate

with

respect to and yet have no relationship for and throughout the heat exchanger. At this point there are a

number of cases that arise. Firstly and the most simple case that arises is the case where is constant throughout

the heat exchanger. This may only occur where the hot fluid is a saturated vapour and through the heat exchanger is

condensed to a saturated liquid isothermally. In this case this integral can be easily done:

∫

(

)

(

)

Using this expression the overall area of the heat exchanger may be easily determined. It should be noticed that in

the case that is constant, this expression works for both co – current and counter current heat exchangers. This is

due to the fact that since is constant it makes no difference where the hot stream inlet is. Now in order to write

this expression more neatly, let us multiply and divide by the cold temperature difference to obtain:

(

)

This expression may also be written as:

[( ) ( )] (

)

The reason for this multiplication is simply due to the logarithm term divided by the difference of the two terms

present within the logarithm gives us the reciprocal of the log mean temperature. This leads to the following

expression:

Where:

( ) ( )

(

)

The above expression may be rearranged neatly to give:

Since the temperature difference driving the heat transfer is changing from position to position, this expression says

that the effective temperature difference for the heat transfer is the log mean temperature difference.