Process Design and Analysis is a module that looks at networks of unit operations and unit processes and seeks to be able to completely understand them in terms of material and energy flows. This means that:

We should be able to calculate the total amounts of all chemicals present at any part of any chemical process.
We should be able to evaluate temperatures and/or pressures at any part of a chemical process.
We should be able to calculate the total energy required to operate any chemical process (albeit sometimes under ideal conditions, with several assumptions).

Most of the time, this module is just concerned with knowing these things under 'steady-state' conditions, meaning that if we took a snapshot of all material flowrates, all stream compositions, all temperatures at all points, etc, at one time, the values should be the same as if we did it again at another time. Often, this then allows us to work out (very) roughly how big some process units may need to be, how much the overall process might cost, what some of the major hazards of such a plant might be, etc.

Sometimes we need to know what's occurring in unsteady-state conditions which can be a little more difficult to work out. Other modules in subsequent years deal with these scenarios in a lot more detail - they're critical for the control of processes (i.e. how computers can be programmed to maintain temperatures, pressures, etc, or allow the establishment of a newly-desired state) and for the design of particular process hardware (e.g. heat exchangers, distillation columns, etc).

In order to start to do this analysis, we need to be familiar with concepts surrounding chemical reactions, thermodynamics, linear algebra (mathematical concepts appear all over the place as soon as we want quantitative results), and many more, however, there are just 2 major concepts that you should really bear in mind

Conservation of mass

This is the foundation of material balancing.

In all circumstances, mass is neither created nor destroyed. This means that for any system that we can define, any mass that enters the system must either leave the system or remain inside it.

LaTeX: input=output+accumulation (1a)

Given that we are mostly interested in steady state systems here, the accumulation term will equal zero, leaving:

LaTeX: input=output (1b)

Thus if we sum the masses of all material entering a particular system per unit time, it should equal the sum of all masses of material leaving the system [er unit time.

Such a concept can also apply to specific chemical components, however it's possible in a reactor unit for specific chemicals to be used up or created (note, the overall mass of material will remain unchanged). Thus, for a specific chemical at steady state:

LaTeX: input+generation=output+consumption (2)

The system would be anything that we decide to put a system boundary around, as indicated in the diagram below. Everything outsideof that boundary is referred to as the surroundings.

We can put anything we like, however complex, inside here with the important thing being that anything that crosses the boundary from the surroundings to the system is an 'input', and anything crossing the boundary from the system to the surroundings is an 'output'. We can only analyse a system in terms of its inputs and outputs, and so anything that sits inside the system is ignored for the purposes of that specific bit of analysis.

A material balance example

An example could be a furnace, into which we inject pure methane and pure molecular oxygen. Let's say that there are 2 input streams to the furnace:

Methane feed: 200 kg h^-1.
Oxygen feed: 1200 kg h^-1.

There is a single chimney from which all outputs products exit the furnace. We may be interested to calculate the composition of this output product stream. We don't quite have enough information to do this, but maybe we can make some assumptions so that we can get a sensible response. Let's assume that:

all methane is consumed, and
all methane undergoes complete combustion, that is it combines with oxygen in such proportions that it only forms carbon dioxide and water as products.

We can now conduct some material balances over this furnace unit to calculate the quantities of all components in the input and output streams.

The input was originally defined:

Component	Methane Feed (kg h^-1)	Oxygen Feed (kg h^-1)	Total (kg h^-1)
Methane	200	0	200
Oxygen	0	1200	1200
Total	200	1200	1400

The output needs to be calculated, and given that a chemical reaction has occurred, we should use equation (2) above to do a mass balance on the specific components. Methane is trivial (our first assumption stated that it was all consumed) so let's start with the oxygen.We'll rearrange that equation slightly to take account of the fact that it's the output flowrate that we'd like.

LaTeX: ouput=input+generation-consumption

LaTeX: output=1200+0-consumption

It's clearly the consumption term that we need to evaluate. Chemical reactions are always considered in molar terms (whole numbers of molecules collide with other whole numbers of discrete molecules, and so considering the pure masses of components doesn't help.

The second assumption told us that:

$LaTeX: CH_4+2O_2\longrightarrow\:CO_2+2H_2O$

And so we find that for every mole of methane consumed, 2 moles of oxygen were required. It's easy to calculate the number of moles of methane that were used up; it's the entirety of the methane that entered the furnace, in molar terms. The relative molecular mass of methane is 16 kg kmol^-1, and so we know that 12.5 kmol h^-1 of methane entered the furnace. Hence, 25 kmol h^-1 of oxygen was consumed. We can input this into equation 2 like so, where M_r,O2 stands for the RMM of oxygen:

$LaTeX: output=1200+0-M_{r,O_2}\left(2\left(\frac{200}{16}\right)\right)=1200-32\left(2\left(12.5\right)\right)=1200-800=400$

Thus 400 kg h^-1 remain in the output stream from the furnace. This looks a little convoluted, because we needed to make sure that each term has units of kg h^-1. If we like, we could do this in molar terms instead, but we always need to be really careful that when we do molar balances, we only ever consider a single component at once. The total number of moles across a chemical reactor or chemical process are never required to be conserved.

It has now become apparent that the oxygen was fed in excess. This is very common in these types of units. We want the combustion to be clean and we want all of our fuel to be consumed in such a case. It's much better to provide an excess of oxygen than to risk wasting the fuel and to have 'dirty' flue gas contaminated with carbon monoxide, coke, etc.

We should now be able to complete this analysis for the carbon dioxide and water. We know 12.5 kmol h^-1 of methane is consumed, and from the stoichiometric relationship, we can see that 12.5 kmol h^-1 of CO₂ and 25 kmol h^-1 of water should be formed. These components have different Mr values that should be applied so that we can know the mass flowrates:

$LaTeX: mass\:CO_2=12.5\times44=550$

$LaTeX: mass\:water=25\times18=450$

We can summarise all of this:

Component	Total input (kg h^-1)	Total output (kg h^-1)
Methane	200	0
Oxygen	1200	400
Carbon Dioxide	0	550
Water	0	450
Total	1400	1400

We can clearly see that the total masses of material entering and leaving the furnace are equal, thus mass has been conserved. In other scenarios, we may find that the conversion of neither reactant is complete, or that there are multiple competing reactions, or that reactions occur in series (where the product of one reaction becomes the reactant in other - in reality, this is what actually occurs in combustion reactions, where transitions to CO₂ occur via intermediates).

Regardless of the complexity of the scenario, mass will always be conserved, and so we apply the data we have and the laws of chemistry appropriately and sequentially until we are able to define every stream that exists n terms of its chemical composition.

Conservation of energy

This is the foundation of energy balancing.

In all circumstances, energy is neither created nor destroyed, however its form can change, and we occasionally need to consider where the energy is coming from and going to. This means that for any system that we can define, any energy that enters the system must either leave it or remain inside it. Thus we could write:

$LaTeX: initial\:state\:+energy\:in=final\:state\:+energy\:out\:+accumulation$

For steady state systems, the accumulation term would equal zero (nothing is changing at any specific point as time passes), and we could define a net transfer of energy as the difference between the energy in and energy out terms. Hence we could rewrite this as:

$LaTeX: final\:state-initial\:state=net\:energy\:transferred$

This will be the basis of an energy balance.

There are two types of system that we should consider: (1) closed, and (2) open.

In a closed system, no matter crosses the system boundary. An example of this would be a closed saucepan of water, from which no vapour escapes and into which nothing can enter. We could conduct an energy balance to calculate how much energy we may have to add to this system (e.g. via the hob) to heat the contents from one temperature to another.

In an open system, matter can cross the system boundary. An example might be a distillation column into which material is continuously fed, and from which different product fractions are continuously withdrawn.

Closed system

The state of the system at any point can be described by three different terms:

Potential energy (E_P): due to the position that a system has inside a potential field (gravitational, electromagnetic, etc)

Kinetic energy (E_K): due to the motion of a system relative to some frame of reference.

Internal energy (U): due to the motion of the system’s molecules, e.g. random, translational, vibrational, rotational motion. This is a function of the system’s temperature.

Given that these can be evaluated, the difference between these values at different states can also be evaluated. As such, we can evaluate the difference between the final and initial states by evaluating and summing the differences in each of these terms.

$LaTeX: \Delta E_P+\Delta E_K+\Delta U=net\:energy\:transferred$

There are two terms that describe the net energy passing across the system boundary:

Heat (Q) – energy exchange between system and surroundings caused by temperature driving force. A positive value means the net heat transfer is into the system.
Work (W) – any energy transfer that is not heat. Often achieved by compressing, expanding, pumping, stirring, etc. A positive value means a net work done by the system on the surroundings.

The sign convention associated with these terms means that the following generalised steady state energy balance is found for closed systems:

$LaTeX: \Delta E_P+\Delta E_K+\Delta U=Q-W$

Given that on most scales, the change in internal energy and heat transferred will dominate, we simplify this to:

$LaTeX: \Delta U=Q$

Open system

This system is set up in a similar way to that of a closed system, except that matter can now cross a system boundary, and such material flows as well as having energy associated with them (inflow and outflow terms are analogous to initial and final state terms in closed systems) these flows can do ‘flow work’ on the system as they cross the system boundary entering, and do flow work on the surroundings as they leave the system. This work term is related to the volume and pressure of the matter crossing a system boundary.

Thus, the work is now described as the sum of ‘shaft work’ and ‘flow work’, which is the net difference between flow work in and flow work out:

$LaTeX: W=W_S+W_{fl}=W_S+\left(P_oV_o-P_iV_i\right)$

Where ‘i’ and ‘o’ represent inflow and outflow respectively.

We can once again state that the overall steady state energy balance would take the form:

$LaTeX: input\:state+transfer\:in=output\:state+transfer\:out$

Rearranging, incorporating different types of energy and the definitions of flow work:

$LaTeX: E_{Pi}+E_{Ki}+U_i+Q=E_{Po}+E_{Ko}+U_o+W_S+\left(P_oV_o-P_iV_i\right)$

We can rearrange this:

$LaTeX: \left[\left(U_o-P_oV_o\right)-\left(U_i-P_iV_i\right)\right]+\left[E_{Po}-E_{Pi}\right]+\left[E_{Ko}-E_{Ki}\right]=Q-W_S$

This may seem a little bit of a strange thing to do, however there is one last definition that is useful. The enthalpy (H) of a system is defined as:

LaTeX: H=U+PV

And so the change in enthalpy ( $LaTeX: \Delta H$ ) can also be incorporated here:

$LaTeX: \Delta H+\Delta E_P+\Delta E_K=Q-W_S$

And once again, in almost all scenarios that we’ll consider, the changes in potential energy and kinetic energy and the shaft work term are all negligible compared to the magnitude of the other terms, we can simplify this to:

$LaTeX: \Delta H=Q$

Essentially we’re saying that the change in enthalpy between 2 states (say, the input and output of a process unit) is equal to the net energy transfer to/from the system.

The use of energy to achieve a change in state is defined by latent heat changes sensible heat changes.

Latent heat changes look at changes in phase (solid to liquid, liquid to vapour, etc). Latent heats of specific phase changes for specific chemicals are available in many data sources.

Sensible heat changes look at changes in material temperature. The specific heat capacity of a particular chemical in a particular state governs this change and tell you exactly how much energy you need to give a unit quantity of that material to raise its temperature by 1°C.

If you struggle with relevant questions in the quiz, you should be able to use this information to remind you how to proceed.

An energy balance example

Water flows through a pipe at a rate of 8 kg s^-1. A heater is employed to heat the pipe such that the water, which is initially at 15°C is changed into steam at 130°C at 1 atm. Calculate the heat duty required.

Specific heat capacity of water at constant pressure: 4.18 kJ kg^-1 K^-1

Saturation temperature at 1atm: 99.97°C

Latent heat of vaporisation of water: 2256 kJ kg^-1

Specific heat capacity of steam at constant pressure: 2.05 kJ kg^-1 K^-1

We can consider the heated section of the pipe as our system. The input is water flowing at 8 kg s^-1 at a temperature of 15°C and the output is steam flowing at 8 kg s^-1 at a temperature of 130°C. Note that we are not told about any other material flows and so although we technically weren’t told ‘input’ and ‘output’ flowrates, we know that conservation of mass must apply and so the flowrate of 8 kg s^-1 must be valid everywhere.

We know that water at 15°C will have a certain enthalpy,and similarly steam at 130°C will have an enthalpy value. This difference in enthalpy must be provided by the heater, and if we assume the system is 100% efficient (adiabatic – i.e. no heat is lost to the surroundings) we can say:

$LaTeX: Q=\Delta H$

The total enthalpy change between the 2 states is the same, regardless of the route used to calculate it, so let’s break this problem into 3 processes:

Heating the water from 15°C to the saturation temperature (boiling point), a sensible heat change
Vaporising the water so that it turns from liquid to gas, a latent heat change
Heating the steam from the saturation temperature to 130°C, a sensible heat change

As such, 3 calculations are straight forward. Let's consider a time basis of 1 second (i.e. we'll calculate the energy used in 1 second).

$LaTeX: \Delta H_1=mc_{p,w}\left(T_s-T_i\right)=8\times4.2\times\left(99.97-15\right)=2855\:kJ$

$LaTeX: \Delta H_2=ml_v=8\times2256=18048\:kJ$

$LaTeX: \Delta H_3=mc_{p,s}\left(T_o-T_s\right)=8\times2.05\times\left(130-99.97\right)=492.5\:kJ$

We can sum these to get Q

$LaTeX: Q=\Delta H=\sum_{i=1}^{3} \Delta H_{i}=2855+18048+492.5=21396 kJ$

We know that this quantity of heat is delivered per second, so we can actually say that 21396 kW of power are required.

Note how much more energy was required to vaporise the water, as opposed to just changing its temperature. This is because the latent heat value is much higher than specific heat values. We know that this is because changing temperature just makes molecules vibrate with more energy but vaporising something means that we have to physically overcome intermolecular bonding.

How this heat is actually delivered is another matter! For example, if we use electric heating, how much power do we need to use to heat a metal tube to the correct temperature to do this, and how long does the heated section need to be? If we used a heat exchanger and used a hot fluid to do the heating (a very common example is to use hot reactor products to ‘preheat’ cold reactants) how big does this need to be? What shape will it be? How will we maximise the possibility to transfer heat given that the 2 streams will never be in physical contact? How do we ensure the most efficient process (for example, if we start to have layers of grime build up on tube walls (fouling) the heat transfer is greatly reduced)? How do we ensure that all of this is done safely? These are all things that a chemical engineer needs to consider when designing process units.

As long as you consider this, your current levels of physical chemistry and reaction chemistry should be fine...

The questions in this quiz look at some scenarios that we may expect our first years to be able to analyse. The first few questions are thoroughly explained and are set up in such a way that as long as you remember the principles of conservation of mass and energy, you should be able to answer - we'd certainly expect first year students to be able to answer these at the end of semester 1 (by the end of December). The last couple of questions are a little more in-depth, and involve multiple stages of calculation. These are the sorts of things that we'd expect first years to be confident with at the end of the module.

For the first few questions, you'll see some auto-generated feedback. If you've got something wring, we may suggest what you might have done to cause this. For the remaining ones, see if you can go back over your working and find what might have gone wrong. We assume that your teachers may also be interested in seeing if they can solve these...

Best of luck!