Chemical Engineering Principles

The big picture

Here is a small experiment you could run this afternoon. A chemist puts 5 grams of salicylic acid in a flask, adds acetic anhydride and a drop of acid catalyst, warms it in a water bath, cools it, filters the crystals, and holds up about 5 grams of aspirin. Elapsed time: an hour. Now I hand you a different assignment. Make 40,000 tonnes of aspirin a year, every year, at a cost of a few cents a tablet, with no worker injured, no river polluted, and no batch out of specification. Nothing in the flask experiment tells you how to do that. The chemistry is identical. Almost everything else is different.

That gap is where chemical engineering lives. It is not chemistry practiced by people who like machines, and it is not mechanical engineering applied to pipes. It is a distinct discipline built to answer a distinct question: given a chemical or physical transformation that works, how do you carry it out continuously, safely, economically, and at industrial scale? Answering that question turns out to require a specific and surprisingly small toolkit, and this course is that toolkit.

Here is the plan for today. First, what actually changes when you scale a process up, because the difficulties are not the ones most people expect. Then the single organizing insight of the profession, the unit operations idea, which lets you learn a few dozen pieces of equipment and thereby understand thousands of plants. Then how engineers draw processes, from the block flow diagram you can sketch in ten seconds to the process flow diagram that carries real numbers. We finish with a flowsheet you will meet again and again in this course, so that the rest of the material has somewhere concrete to land.

Why scale-up is hard

The instinct is that making a thousand times more of something means buying a vessel a thousand times bigger. The instinct is wrong, and the reason is geometry. Double the linear dimensions of a tank and its volume goes up by a factor of 2^3 = 8, while its surface area goes up by only 2^2 = 4. Volume grows faster than area. Every quantity that scales with volume, the amount of material, the heat released by a reaction, the inventory of anything hazardous, races ahead of every quantity that scales with area, the heat you can remove through the wall, the oxygen you can transfer across a surface, the light you can shine in.

Put numbers on it. Suppose a reaction releases 500 kJ per kilogram of product and you run it in a jacketed vessel. In a 1 litre flask holding 1 kg of contents, the heat load is modest and the flask has roughly 0.05 m^2 of wall per kilogram of contents. Scale to a 10 m^3 reactor holding 10,000 kg, and the surface-to-volume ratio has collapsed to roughly 0.0025 m^2 per kilogram, a factor of about twenty worse. The reaction that gently warmed a flask now has twenty times less wall per kilogram through which to dump the same heat per kilogram. This is not a hypothetical annoyance. Runaway reactions, in which heat generation outruns heat removal and the temperature climbs until something ruptures, are among the most common causes of serious industrial accidents, and they are fundamentally a scale-up problem.

Mixing tells the same story. In a flask you swirl and everything is uniform in a second. In a 50 m^3 stirred tank, the time for the impeller to blend the contents can be tens of seconds, and during those seconds part of the vessel is at a different concentration and temperature from the rest. A reaction sensitive to local concentration will give you a different product distribution in the big tank than in the flask, with identical reagents. Add to this that real feedstocks contain impurities the lab used a purified bottle to avoid, that continuous operation must run for years rather than an hour, and that the economics now turn on pennies per kilogram, and you can see why the plant is not a big flask.

Key idea: Scale-up is hard because volume grows faster than area, so heat removal, mixing, and mass transfer all get harder as the equipment gets bigger, and these limits, not the chemistry, usually decide how a process is built.

The unit operations idea

Faced with that, an industry could have developed into thousands of unrelated crafts: the sulfuric acid trade, the soap trade, the dye trade, each with its own apprenticeship. For a while it did. The change came from a simple observation, articulated in Britain by George E. Davis, whose 1887 lectures in Manchester became the first handbook of the field, and named in the United States by Arthur D. Little in 1915. Little's insight was this: any chemical process, no matter what it makes, can be resolved into a series of the same basic physical steps. He called them unit operations.

Distillation is distillation whether you are separating alcohol from a fermentation broth or naphtha from crude oil. Filtration is filtration whether the solids are pharmaceutical crystals or mine tailings. Heat exchange, evaporation, drying, absorption, extraction, crystallization, size reduction, fluid transport: a modest list of operations, each governed by the same equations everywhere it appears. Learn the list, learn the equations, and you can walk into an unfamiliar plant and read it. That is why chemical engineers move so freely between industries, and why a degree earned studying petroleum equations lands people in pharmaceuticals, semiconductors, food, and batteries.

Here is the working inventory. Notice how each entry answers a question about what is being changed.

Unit operation	What it does	Governing idea	Everyday example
Reaction	Changes chemical identity	Kinetics and equilibrium	Catalytic converter
Distillation	Separates by volatility	Vapor-liquid equilibrium	Whiskey still
Absorption	Washes a gas with a liquid	Gas-liquid mass transfer	Scrubber on a stack
Extraction	Separates by solubility in two liquids	Liquid-liquid partition	Decaffeinating coffee
Filtration	Separates solids from fluids	Flow through a porous cake	Coffee filter
Evaporation and drying	Removes a volatile liquid	Latent heat and mass transfer	Making powdered milk
Heat exchange	Moves heat between streams	Conduction and convection	Car radiator
Fluid transport	Moves material through the plant	Pressure drop and pump work	Municipal water main
Crystallization	Forms a pure solid from solution	Solubility and supersaturation	Rock candy
Size reduction	Breaks solids into smaller pieces	Energy per new surface created	Coffee grinder

Two later refinements are worth knowing. In the 1960s the field reorganized much of this teaching around transport phenomena, the recognition that momentum, heat, and mass all obey mathematically parallel transport laws, so that one set of equations covers pressure drop, heating, and drying at once. That is the subject of Module 5. More recently, the same reasoning has been pushed downward in scale into biology and electronics, so that fermenters, bioreactors, and semiconductor deposition chambers are analyzed with the tools you are about to learn.

Key idea: Every chemical process resolves into a small set of reusable unit operations, each governed by the same physics everywhere it appears, which is why a chemical engineer can read a plant in an industry they have never worked in.

How a process is drawn

Chemical engineers think on paper in a specific way, and learning the notation is learning to think in it. Drawings come in three levels of detail.

A block flow diagram (BFD) is the ten-second sketch: boxes for functions, arrows for streams. Crude oil in, a box labeled distillation, arrows out labeled gasoline, kerosene, diesel, residue. It shows the logic of the process and nothing else. Engineers draw these constantly, on whiteboards and napkins, because most process arguments are settled at this level.

A process flow diagram (PFD) is the working document. Now the boxes become recognizable equipment symbols, the streams are numbered, and a table beneath the drawing carries the numbers: flow rate, composition, temperature, and pressure for every numbered stream. Major equipment is tagged with a letter and number by convention, so R-101 is a reactor, C-101 a column, E-101 an exchanger, P-101 a pump, V-101 a vessel. When someone says the plant is designed, they usually mean the PFD is finished and its stream table closes.

A piping and instrumentation diagram (P&ID) adds everything needed to build and operate: every valve, every line size, every instrument, every control loop, every relief device. A P&ID for a modest plant runs to dozens of sheets and is the document that safety reviews are conducted on. We will not draw P&IDs in this course, but you should know that the safety analysis in Module 6 happens over these drawings, line by line.

The habit worth building right now is stream labeling. Every arrow on your diagram gets a number, and every numbered stream gets a total flow and a composition. A diagram with unlabeled arrows is a picture; a diagram with labeled streams is a calculation waiting to be done. Nearly every problem in Modules 2 and 3 begins with the instruction: draw the flowsheet and label every stream.

Key idea: Draw first, label every stream with a flow rate and composition, and the algebra will follow; a labeled flowsheet is the setup for essentially every calculation in this course.

A flowsheet you will meet again

Let us build one now, because this exact process returns in five later lessons. A refinery side stream delivers 100 kmol/h of a liquid mixture that is 50 mole percent benzene and 50 mole percent toluene. Both are valuable, and both are worth far more pure than mixed. They boil at different temperatures at atmospheric pressure, benzene at 80.1 C and toluene at 110.6 C, so distillation is the obvious tool.

The flowsheet is short. Stream 1, the feed, enters a distillation column, C-101, partway up. Heat is added at the bottom by a reboiler, E-102, which boils liquid and sends vapor up the column. Vapor leaving the top is condensed in E-101; part of that condensate is drawn off as the overhead product, stream 2, and the rest is returned to the top of the column as reflux. Liquid collecting at the bottom leaves as stream 3, the bottoms product. Two products, two heat exchangers, one column, one recycle loop inside the column itself.

Now label it. Suppose the design calls for the overhead to be 95 mole percent benzene and the bottoms to be 5 mole percent benzene. What are the two product flows? You do not need any thermodynamics to answer, only bookkeeping: whatever benzene comes in must go out. Total moles: D + B = 100 kmol/h. Benzene moles: 0.95 D + 0.05 B = 0.50 x 100 = 50 kmol/h. Substituting B = 100 - D gives 0.95 D + 5 - 0.05 D = 50, so 0.90 D = 45 and D = 50 kmol/h, with B = 50 kmol/h. Half the feed leaves at the top, half at the bottom, and we have not yet said a word about temperature, pressure, or column height.

That calculation is a material balance, and it is the subject of the next module. Later you will compute the bubble point of this feed (92.1 C), its relative volatility (about 2.5), the number of equilibrium stages the column needs (13, including the reboiler), and the reboiler duty in megawatts (about 1.1 MW, costing roughly 1 GJ of steam per tonne of product). One flowsheet, six lessons, numbers that agree with each other throughout. That consistency is not a stylistic flourish. It is what a real design package looks like.

Key idea: A process is a network of unit operations connected by streams, and the first thing you can always compute about it, before any physics, is the bookkeeping of what goes in and what comes out.

Common misconceptions

• Chemical engineering is chemistry with more math. Chemists discover and characterize transformations; chemical engineers design the equipment, flows, heat, and controls that carry them out continuously at scale. Many chemical engineers take no chemistry course after the second year and spend their careers on transport, control, and economics.
• Scaling up just means buying bigger equipment. Volume outruns area, so heat removal, mixing, and mass transfer all degrade with size. The scale-up problem is usually a transport problem, not a chemistry problem.
• Chemical engineers only work in oil and chemicals. Refining and petrochemicals are the historical core, but the same unit operations run pharmaceutical plants, semiconductor fabs, breweries, water treatment works, battery factories, and carbon capture units.
• The flowsheet is a drawing, so it is not really quantitative. A finished process flow diagram carries a stream table in which every stream has a flow, composition, temperature, and pressure, and every balance closes. It is the most quantitative document in the project.

Recap

• Chemical engineering exists to carry chemical and physical transformations out continuously, safely, and economically at scale.
• Scale-up is hard because volume grows as the cube of size while area grows as the square, so heat removal and mixing get relatively worse in bigger equipment.
• The unit operations idea, named by Arthur D. Little in 1915, resolves any process into a small set of reusable physical steps, each governed by the same equations everywhere.
• Processes are drawn as block flow diagrams (logic), process flow diagrams (equipment plus a stream table), and piping and instrumentation diagrams (everything needed to build, operate, and review for safety).
• Label every stream with a flow rate and composition, because that is the setup for the material and energy balances that follow.
• Our running example, a 100 kmol/h column splitting a 50/50 benzene-toluene feed into 95 percent and 5 percent products, gives D = 50 kmol/h and B = 50 kmol/h from bookkeeping alone.

Sources

1. Britannica. (n.d.). Chemical engineering. Encyclopaedia Britannica. britannica.com
2. American Institute of Chemical Engineers. (n.d.). About AIChE. AIChE. aiche.org
3. Wikipedia contributors. (n.d.). Unit operation. Wikipedia. en.wikipedia.org
4. Engineering LibreTexts. (n.d.). Chemical engineering bookshelf. LibreTexts. eng.libretexts.org

The big picture

On 23 September 1999, after a journey of 669 million kilometres, the Mars Climate Orbiter fired its engine to enter orbit and was never heard from again. The investigation found the cause quickly and it was humiliating: one software module supplied thruster impulse in pound-force seconds while the navigation software expected newton seconds. Nobody made an arithmetic mistake. A number crossed an interface without its unit attached, and a 193 million dollar spacecraft flew into the Martian atmosphere.

You will not lose a spacecraft in this course, but you will lose marks, and later you may size a pump three times too small. Chemical engineering is unusually exposed to unit errors because our numbers travel between disciplines constantly: a chemist reports concentration in moles per litre, a supplier quotes flow in cubic metres per hour, a pump curve is drawn in metres of head, a steam bill arrives in gigajoules, and a control system reports pressure in bar gauge. All of these describe the same plant, and you are the person who must make them agree.

Here is the plan for today. First, dimensions and units, and the single mechanical technique, the conversion factor, that makes unit changes foolproof. Then dimensional homogeneity, which is the cheapest error check ever invented, plus the dimensionless groups that come out of it. Then the four process variables that describe every stream in this course: flow rate, composition, pressure, and temperature, each with the conversions that trip people up. We will work our running benzene-toluene stream through all four so that Module 2 can start with numbers you already trust.

Dimensions, units, and the conversion factor

A dimension is a kind of physical quantity: length, mass, time, temperature, amount of substance. A unit is a specific measure of that dimension: the metre, the kilogram, the second, the kelvin, the mole. Dimensions are what nature has; units are what we chose. This course uses SI throughout, with the derived units built from the base ones.

Quantity	SI unit	Built from	Useful equivalents
Force	newton, N	kg m/s^2	1 N = 0.2248 lbf
Pressure	pascal, Pa	N/m^2	1 atm = 101.325 kPa = 1.01325 bar = 760 mmHg = 14.696 psi
Energy	joule, J	N m	1 kJ = 0.2388 kcal; 1 GJ = 277.8 kWh
Power	watt, W	J/s	1 kW = 3600 kJ/h = 1.341 hp
Viscosity	pascal second, Pa s	kg/(m s)	1 cP = 0.001 Pa s
Amount	mole, mol	base unit	1 kmol = 1000 mol

Every conversion you will ever do rests on one trick. Any true equality between two quantities can be written as a fraction equal to one. Since 1 h = 3600 s, the fraction (1 h / 3600 s) equals 1, and so does its reciprocal. Multiplying by one changes nothing physically, so you may insert as many of these as you like and cancel units like algebra. Convert our feed of 100 kmol/h to mol/s: 100 kmol/h x (1000 mol / 1 kmol) x (1 h / 3600 s) = 100,000 / 3600 = 27.8 mol/s. The kmol cancels, the hour cancels, and mol/s survives. Nothing was memorized.

Do a harder one. A heat transfer coefficient is 500 W/(m^2 K) and you want it in kJ/(h m^2 K), because your energy balance is in kJ/h. Write it out: 500 J/(s m^2 K) x (3600 s / 1 h) x (1 kJ / 1000 J) = 500 x 3.6 = 1800 kJ/(h m^2 K). Notice that the temperature unit did not need converting: a kelvin and a Celsius degree are the same size, so any per-degree coefficient is numerically identical in K and in C. That is true for intervals and differences, never for temperature values themselves, which is the distinction most people get wrong first.

Key idea: Write every conversion as a multiplication by a fraction equal to one, cancel units like algebra, and never carry a number without its unit attached.

Dimensional homogeneity and dimensionless groups

Every valid physical equation is dimensionally homogeneous: each additive term has the same dimensions, and both sides match. This gives you a free audit of any equation you write or read. If someone hands you Q = U A and claims it gives heat duty in watts, check: U is W/(m^2 K), A is m^2, so U A is W/K, not W. Something is missing, and it is the temperature difference. The correct form is Q = U A dT, which gives (W/(m^2 K))(m^2)(K) = W. The check took five seconds and caught a real error.

Sometimes a combination of variables cancels out completely, and the result is a dimensionless group. The most famous in this field is the Reynolds number, Re = rho v D / mu, where rho is density, v velocity, D pipe diameter, and mu viscosity. Check it. Viscosity in Pa s is kg/(m s), since Pa = N/m^2 = kg/(m s^2) and multiplying by seconds gives kg/(m s). So Re has units (kg/m^3)(m/s)(m) divided by kg/(m s), which is (kg/(m s)) divided by (kg/(m s)), equal to 1. Dimensionless.

Why care? Because dimensionless groups are the reason small experiments predict large equipment. Two flows with the same Reynolds number behave the same way regardless of their absolute size, so a laboratory rig and a full-scale pipeline can be compared honestly. Module 5 uses Re for pressure drop, the Nusselt and Prandtl numbers for heat transfer, and the Sherwood and Schmidt numbers for mass transfer. All of them come from this same habit of asking what cancels.

Key idea: Terms that are added must share dimensions, both sides of an equation must match, and combinations that come out dimensionless let small-scale experiments predict full-scale behavior.

Process variable 1: flow rate, on three different bases

A stream can be measured three ways, and you must be fluent in all three. Mass flow rate (kg/s, kg/h, t/h) is what conservation of mass applies to directly. Molar flow rate (mol/s, kmol/h) is what stoichiometry applies to. Volumetric flow rate (m^3/h, L/min) is what a flowmeter or a pump actually sees, and it is the treacherous one, because volume depends on temperature and pressure, especially for gases.

Convert our feed all three ways. It is 100 kmol/h of 50 mole percent benzene (molar mass 78.11 kg/kmol) and 50 mole percent toluene (92.14 kg/kmol). The average molar mass is the mole-fraction-weighted sum: M = 0.50 x 78.11 + 0.50 x 92.14 = 39.06 + 46.07 = 85.13 kg/kmol. Mass flow = 100 kmol/h x 85.13 kg/kmol = 8513 kg/h, which is 8513/3600 = 2.365 kg/s, or 8.5 tonnes per hour. For volume, take the liquid mixture density as roughly 870 kg/m^3 at ambient temperature: volumetric flow = 8513 / 870 = 9.79 m^3/h, about 163 litres per minute. That last number is what you would specify to a pump vendor.

For gases the volumetric number is nearly meaningless without stating conditions, so engineers quote gas flows at defined standard conditions. Using 0 C and 1 atm, one kmol of ideal gas occupies 22.41 m^3, so 100 kmol/h of any gas is 2241 standard cubic metres per hour. Warm that same gas to 200 C at constant pressure and the actual volumetric flow becomes 2241 x (473.15/273.15) = 3882 m^3/h, a 73 percent increase in the pipe with no change whatsoever in the mass or moles flowing. This is why gas ducts are enormous and why the phrase standard cubic metres, written Sm^3, must never be dropped.

Key idea: Mass conserves, moles react, and volume merely flows; always know which basis you are on, and never quote a gas volumetric flow without the temperature and pressure it refers to.

Process variable 2: composition, and why 50 percent is ambiguous

Composition can be given as a mass fraction or a mole fraction, and they are not the same number. Convert between them by taking a convenient basis. Suppose a tank is labeled 50 percent benzene by mass. Take 100 kg: that is 50 kg of benzene and 50 kg of toluene. Moles of benzene = 50 / 78.11 = 0.640 kmol; moles of toluene = 50 / 92.14 = 0.543 kmol; total = 1.183 kmol. Mole fraction of benzene = 0.640 / 1.183 = 0.541. So 50 percent by mass is 54.1 percent by mole. The lighter component always has the larger mole fraction, because a kilogram of it contains more molecules.

Going the other way, our 50 mole percent feed is, per 100 kmol, 39.06 kg of benzene per 85.13 kg total, so 45.9 percent benzene by mass. Whenever a problem says percent, find out which percent, because a 4 point discrepancy will ruin a balance.

Two more composition measures appear constantly. Molarity is moles of solute per litre of solution, which is convenient in the laboratory but temperature dependent, since the solution expands when heated. Parts per million is a mass ratio for liquids and solids (1 ppm = 1 mg per kg) but conventionally a mole or volume ratio for gases (1 ppm = 1 mole of species per million moles of gas). A permitted emission of 25 ppm of a solvent in a stack gas means 25 moles per million moles of flue gas, and reading it as a mass ratio would misstate the limit by the ratio of the molar masses.

Key idea: Mass fractions and mole fractions differ whenever the components have different molar masses; convert by taking a 100 kg or 100 mol basis, and always ask which percent is meant.

Process variable 3: pressure, absolute and gauge

Pressure is force per unit area, one pascal being one newton per square metre. The pascal is inconveniently small, so plants run on kPa, bar, or, in the United States, psi. Standard atmospheric pressure is 101.325 kPa.

The trap is the reference point. Most gauges read zero when open to the room, so they report gauge pressure, the amount above atmospheric. Thermodynamics and the ideal gas law require absolute pressure, measured from perfect vacuum. The relation is simple: P(absolute) = P(gauge) + P(atmospheric). A reactor gauge reading 250 kPa is at 250 + 101.3 = 351.3 kPa absolute, and using 250 in a gas law calculation would understate the moles present by nearly 30 percent. Below atmospheric pressure the same logic gives vacuum readings: a condenser at 80 kPa vacuum is at 101.3 - 80 = 21.3 kPa absolute.

Pressure also gets expressed as a height of fluid, because a static column of liquid produces P = rho g h. Ten metres of water produces 997 kg/m^3 x 9.81 m/s^2 x 10 m = 97,806 Pa, about 97.8 kPa, which is why a suction lift of ten metres is roughly the practical limit for a pump drawing water at atmospheric pressure. Mercury, at 13,600 kg/m^3, produces the same pressure in only 0.734 m, which is why barometers use it. When a pump curve is quoted as 40 m of head, that is a pressure rise of rho g h, and for water it is 997 x 9.81 x 40 = 391 kPa.

Key idea: Gauge pressure is measured from the local atmosphere and absolute pressure from vacuum; every thermodynamic formula wants absolute, and a static liquid column of height h produces P = rho g h.

Process variable 4: temperature, values versus intervals

Temperature scales were treated carefully in Engineering Thermodynamics (ENGR 230); the essentials here are T(K) = T(C) + 273.15 and T(F) = 1.8 T(C) + 32. Absolute temperature in kelvins is required by the ideal gas law, by Arrhenius kinetics, and by every efficiency expression.

The distinction that causes real errors is between a temperature value and a temperature interval. A stream at 80 C is at 353.15 K, a conversion that adds 273.15. But a stream heated by 80 degrees Celsius has been heated by 80 kelvin, because the two units have the same size step. So a heat capacity of 4.184 kJ/(kg C) is numerically identical to 4.184 kJ/(kg K), and you never convert it. In Fahrenheit-based work the interval conversion is 1 F interval = 5/9 K, which is not the same as the 1.8 T + 32 formula for values. Get in the habit of asking whether a temperature in a formula is a level or a difference.

Finally, a word about significant figures. Plant data are rarely better than three figures: a flowmeter good to one percent, a composition from a gas chromatograph good to perhaps half a percent. Reporting a reboiler duty as 1,111,247 W is not precision, it is noise. Round to what your worst input supports, usually three figures, and carry extra digits only inside the calculation. This course quotes three or four figures and rounds honestly at the end.

Key idea: Temperature values need the 273.15 offset while temperature intervals do not, so per-degree properties are the same number in C and in K.

Common misconceptions

• Percent is percent. Fifty percent by mass of a benzene-toluene mixture is 54.1 percent by mole. Always establish the basis before using a composition in a balance.
• Gauge pressure can be used in the ideal gas law if the number is large. No. The gas law requires absolute pressure regardless of magnitude; at 250 kPa gauge the error is nearly 30 percent.
• A heat capacity in kJ/(kg C) must be converted before use with kelvins. It must not. Degrees Celsius and kelvins are the same size, so per-degree quantities are numerically identical.
• Volumetric flow rate describes a gas stream adequately. Only with the temperature and pressure attached. The same 100 kmol/h occupies 2241 m^3/h at 0 C and 3882 m^3/h at 200 C.
• More decimal places mean a better answer. The answer is only as good as the weakest measurement feeding it; three significant figures is usually the honest limit for plant data.

Recap

• Dimensions are kinds of quantity, units are the measures we chose; convert by multiplying by fractions equal to one and cancelling.
• Every equation must be dimensionally homogeneous, and combinations that cancel to dimensionless, such as Re = rho v D / mu, let small experiments predict large equipment.
• Flows come on three bases: mass (conserved), molar (reacts), and volumetric (what meters see, and temperature and pressure dependent for gases).
• Our running feed, 100 kmol/h at 50 mole percent benzene, is 8513 kg/h, 2.365 kg/s, about 9.79 m^3/h of liquid, and 45.9 percent benzene by mass.
• Absolute pressure equals gauge plus atmospheric, and a static column gives P = rho g h, so 10 m of water is 97.8 kPa.
• Temperature values need the offset to kelvins; temperature intervals do not, so per-degree properties are the same number in C and K.

Sources

1. National Institute of Standards and Technology. (n.d.). NIST Chemistry WebBook. U.S. Department of Commerce. webbook.nist.gov
2. National Institute of Standards and Technology. (n.d.). The International System of Units (SI). NIST. nist.gov
3. OpenStax. (2019). Measurements. In Chemistry 2e. Rice University. openstax.org
4. National Aeronautics and Space Administration. (1999). Mars Climate Orbiter mishap investigation board phase I report. NASA. nasa.gov

The big picture

If you learn one thing from this course and forget everything else, learn this: matter does not vanish. A plant is a box with pipes going in and pipes coming out, and every kilogram that enters either leaves or piles up inside. That sentence, written as an equation and applied ruthlessly, is called a material balance, and it is the single most used tool in chemical engineering. Process design starts with it. Plant troubleshooting starts with it. Environmental permitting, yield accounting, and most of the arguments in a control room are settled by it.

It is also, honestly, a bookkeeping exercise, and that is its charm. You do not need to know how a reactor works to write a balance around it. You do not need thermodynamics, kinetics, or transport. You need to draw a boundary, count what crosses it, and do algebra. In Lesson 1 we already solved a distillation column this way and got D = 50 kmol/h without knowing a single thing about vapor-liquid equilibrium.

Here is the plan for today. First the general balance equation in its full form, with all four terms, and the simplifications that collapse it for most problems. Then the choice of what to balance: total mass, individual species, moles, or atoms, and when each is legitimate. Then a procedure you should follow every single time, including the degree-of-freedom count that tells you in advance whether the problem can be solved. Then three worked units: a mixer, an evaporator, and our distillation column, done properly.

The general balance equation

Choose a system, meaning a region with a boundary, exactly as in thermodynamics. For any quantity you care to track, over any period of time you care to choose, the following is always true:

Accumulation = Input - Output + Generation - Consumption

Read it slowly. Input and output are amounts crossing the boundary in the streams. Generation and consumption are amounts created or destroyed inside by chemical reaction. Accumulation is the change in the amount held inside the system. Nothing else can happen to matter, so nothing else appears.

Two simplifications do most of the work. If the process is at steady state, nothing inside is changing with time, so accumulation is zero. If the balanced quantity is not created or destroyed, generation and consumption are zero. Together they give the workhorse form for this lesson: Input = Output. Simple as that looks, applying it correctly to the right quantity around the right boundary is the whole art.

Which quantities have zero generation? Total mass always does, in any chemical process, because mass is conserved by chemical reaction to a precision far beyond anything we measure. Atoms of each element always do, for the same reason. Individual chemical species do not, if they react: burning methane consumes methane and generates carbon dioxide. Total moles do not, unless the reaction happens to have equal moles on both sides. So for non-reactive processes you may balance total mass, each species, and total moles freely. For reactive ones you may still balance total mass and each element, and this is why the atom balance is the reliable fallback when reactions confuse you.

Key idea: Accumulation = Input - Output + Generation - Consumption is always true; at steady state with no reaction it becomes Input = Output, and total mass and atoms of each element never have generation terms.

Batch, continuous, and the meaning of steady state

Processes come in three modes. A batch process is charged, run, and emptied: a pharmaceutical crystallizer, a brewery fermenter, a laboratory flask. There is no input or output during the run, so the balance reduces to Accumulation = Generation - Consumption, and you write it over the whole batch. A continuous process has material flowing in and out constantly: a refinery, an ammonia plant, a wastewater works. If nothing changes with time it is at steady state and accumulation vanishes. A semibatch process has flow in or out but not both, such as slowly feeding a reagent into a stirred vessel to keep a reaction gentle.

Continuous plants are preferred for large tonnages because they run unattended, use equipment continuously rather than in cycles, and are easier to control at fixed conditions. Batch dominates where volumes are small, products change often, or contamination between products must be prevented, which is why most pharmaceutical and specialty chemical manufacture is still batch. Note that a batch plant can be at steady state in an averaged sense over many identical batches, which is how batch plants are given annual capacity figures.

Key idea: Continuous processes at steady state have zero accumulation, batch processes accumulate everything and are balanced over the run, and semibatch has flow in one direction only.

A procedure worth following every time

Undisciplined balance solving turns easy problems into hours of frustration. Use this sequence.

1. Draw the flowsheet. Boxes and arrows are enough.
2. Label every stream with a symbol for its total flow and symbols or values for its composition. Put known values on the drawing, not in your head.
3. Choose a basis. If a flow is given, use it. If only compositions are given, invent one, typically 100 kg/h or 100 mol/h, solve, and scale at the end.
4. Count unknowns and independent equations. This is the degree-of-freedom analysis.
5. Write the balances, starting with the one that involves the fewest unknowns.
6. Solve, then check by an independent balance you did not use.

Step 4 deserves emphasis because it saves the most time. Degrees of freedom equals the number of unknowns minus the number of independent equations. If it is zero, the problem has exactly one solution and you can proceed. If positive, information is missing and you must find another specification. If negative, the problem is over-specified, which usually means the data are inconsistent and you have discovered an error before wasting an hour.

How many independent equations does a unit give you? For a non-reactive unit handling n species, you get n independent balances, not n + 1. You may write a balance on each species and one on total mass, but the total is just the sum of the species balances, so only n of those n + 1 equations are independent. Compositions summing to one supplies one relation per stream. A splitter, which merely divides a stream without changing it, is the sharpest example: all outlet compositions are identical to the inlet, so no matter how many species flow through it, a splitter provides only one independent balance, the total.

Key idea: Count unknowns against independent equations before solving; a non-reactive unit with n species gives n independent balances, and a splitter gives exactly one.

Worked example 1: a mixer

A plant receives 100 kg/h of 20 weight percent sodium hydroxide solution and must dilute it with pure water to 8 weight percent. How much water, and what is the product flow?

Draw it: two inlets (concentrated solution at 100 kg/h and water at W kg/h), one outlet (product at M kg/h). Unknowns: W and M, so two. Species: NaOH and water, so two independent balances. Degrees of freedom zero. Proceed.

Balance on NaOH, which appears in only two streams and is therefore the smart first choice: 0.20 x 100 = 0.08 x M, so 20 = 0.08 M and M = 250 kg/h. Total mass balance: 100 + W = M = 250, so W = 150 kg/h. Check with a water balance, which we did not use: water in = 0.80 x 100 + 150 = 80 + 150 = 230 kg/h; water out = 0.92 x 250 = 230 kg/h. It closes.

Notice the tactic: pick the species that appears in the fewest streams, because that balance has the fewest unknowns. Here NaOH appears in two streams while water appears in three, so the NaOH balance solved immediately for M.

Worked example 2: an evaporator

A juice concentrator takes 1000 kg/h of feed containing 10 weight percent dissolved solids and boils off water until the product contains 40 weight percent solids. What are the product and vapor flows?

The vapor is pure water, which makes solids a perfect choice for the first balance because solids appear in only two streams. Solids in = 0.10 x 1000 = 100 kg/h. Solids out = 0.40 x P. Setting them equal gives P = 100 / 0.40 = 250 kg/h. Total mass: 1000 = P + V, so V = 750 kg/h of water evaporated.

Now look at what that number means physically. Removing 750 kg/h of water requires supplying its latent heat, roughly 2257 kJ/kg at atmospheric pressure, so about 1,693,000 kJ/h, or 470 kW, just to concentrate a tonne per hour of juice. Evaporation is expensive, which is why real plants use multiple-effect evaporators that reuse the vapor from one stage to heat the next, and why the energy balances of Module 3 matter as much as the material balances of this one. The species called solids here is a tie component: something that passes through the process into exactly one outlet, letting you solve for a flow in one line.

Key idea: Choose the balance on the species appearing in the fewest streams; a tie component that goes to only one outlet often solves the problem in a single equation.

Worked example 3: the benzene-toluene column, done formally

Return to the column from Lesson 1. Feed F = 100 kmol/h at mole fraction 0.50 benzene. Distillate D at 0.95 benzene, bottoms B at 0.05 benzene. Note that we are balancing moles here rather than mass. That is legitimate because no reaction occurs, so moles are conserved as surely as mass.

Unknowns: D and B. Species: benzene and toluene, so two independent balances. Degrees of freedom zero.

Total moles: D + B = 100. Benzene: 0.95 D + 0.05 B = 0.50 x 100 = 50. Substitute B = 100 - D into the second: 0.95 D + 0.05(100 - D) = 50, so 0.95 D + 5 - 0.05 D = 50, giving 0.90 D = 45 and D = 50 kmol/h, B = 50 kmol/h.

Check with the toluene balance: in = 0.50 x 100 = 50 kmol/h; out = 0.05 x 50 + 0.95 x 50 = 2.5 + 47.5 = 50 kmol/h. Closed.

It is worth converting these to mass to see the sizes involved. Distillate: 50 kmol/h at an average molar mass of 0.95 x 78.11 + 0.05 x 92.14 = 74.20 + 4.61 = 78.81 kg/kmol, giving 3941 kg/h. Bottoms: 0.05 x 78.11 + 0.95 x 92.14 = 3.91 + 87.53 = 91.44 kg/kmol, giving 4572 kg/h. Total out = 8513 kg/h, which is exactly the mass flow we computed for the feed in Lesson 2. The molar balance and the mass balance agree, as they must.

What have we learned about the column? Its product flows, and nothing else. We do not know how tall it is, how much heat the reboiler needs, or whether the separation is even physically possible with this feed. Those questions need equilibrium and energy, which come later. But observe how much a balance gives for free: if a plant manager tells you the column feeds 100 kmol/h at 50 percent and produces 60 kmol/h of 95 percent distillate, you can reply immediately that the numbers are impossible, because 0.95 x 60 = 57 kmol/h of benzene would be leaving overhead when only 50 kmol/h enters. Balances catch lies.

Key idea: A material balance is a hard constraint on reality, so it both solves design problems and exposes impossible or mismeasured plant data.

When measurements do not close

On a real plant, balances never close exactly. Meters drift, samples are unrepresentative, and small streams go unmeasured. If a balance is out by half a percent, suspect instrumentation. If it is out by ten percent, suspect a stream you have not drawn: a vent, a drain, a leak, or an accumulation in a tank you assumed was steady. This process is called data reconciliation, and it is one of the most valuable things a young engineer can do in a plant, because the missing stream is often worth real money or is a safety problem in disguise.

Common misconceptions

• Moles are always conserved. Moles are conserved only when no reaction occurs or when the reaction has equal moles on each side. Mass and atoms are always conserved.
• A unit handling n species gives n + 1 independent balances. It gives n. The total balance is the sum of the species balances and adds nothing new.
• A splitter with three components gives three balances. It gives one. Splitting does not change composition, so the species balances are all multiples of the total.
• If you have as many equations as unknowns you can always solve. Only if the equations are independent. Writing the total balance plus every species balance for a two-species unit looks like three equations but is only two.
• A balance that does not close means the arithmetic is wrong. On a real plant it usually means a stream is missing from your drawing or a meter is wrong, and finding out which is genuinely useful work.

Recap

• The general balance equation is Accumulation = Input - Output + Generation - Consumption, applied to any quantity around any boundary.
• At steady state accumulation is zero; for species that do not react, generation and consumption are zero, leaving Input = Output.
• Total mass and elemental atoms never have generation terms; individual species and total moles may.
• Follow the procedure: draw, label, choose a basis, count degrees of freedom, balance the species appearing in the fewest streams, solve, and check with an unused balance.
• A non-reactive unit with n species supplies n independent balances; a splitter supplies exactly one.
• Worked results to remember: the diluter needs 150 kg/h of water for 250 kg/h of product; the evaporator makes 250 kg/h of concentrate and boils off 750 kg/h of water; the column gives D = 50 kmol/h (3941 kg/h) and B = 50 kmol/h (4572 kg/h).

Sources

1. Engineering LibreTexts. (n.d.). Material balances. LibreTexts Chemical Engineering. eng.libretexts.org
2. OpenStax. (2019). Reaction stoichiometry. In Chemistry 2e. Rice University. openstax.org
3. Wikipedia contributors. (n.d.). Mass balance. Wikipedia. en.wikipedia.org
4. American Institute of Chemical Engineers. (n.d.). Resources for chemical engineers. AIChE. aiche.org

The big picture

A single unit is a warm-up. Real flowsheets are networks, and the interesting ones eat their own tail: a stream leaves a unit, gets separated, and part of it goes back to where it came from. That loop, the recycle, is one of the most powerful ideas in process design, and it is also where students first feel genuinely stuck, because every unknown seems to depend on every other one.

The stuck feeling has a specific cure. You are allowed to draw the boundary anywhere. Draw it around the whole plant and the recycle stream is invisible, because it never crosses your boundary. Suddenly a problem with six unknowns has two. Then draw a second, smaller boundary that captures one more unknown, and a third. Solving recycle problems is not harder algebra; it is choosing boundaries in the right order.

Here is the plan for today. First, boundary strategy on multi-unit flowsheets, including the two boundaries beginners forget: the mixing point and the splitting point. Then a recycle problem worked from end to end, with the crucial distinction between single-pass and overall conversion. Then purge streams, with the ammonia loop worked in full numbers, because purge is where recycle stops being free. We finish with bypass, the recycle's mirror image.

Boundary strategy

On any flowsheet you may draw a balance envelope around any region you like, and each choice gives a valid, independent set of equations. Four choices repay attention.

The overall envelope encloses the entire process. Only the fresh feeds and the final products cross it. Every internal stream, including all recycles, is invisible. This is almost always the right first move, because it is the boundary with the fewest unknowns.

An individual unit envelope encloses one piece of equipment. Use it once the overall balance has fixed the outer streams.

A mixing point is the junction where fresh feed meets recycle. It is not a piece of equipment, just a tee in a pipe, but it is a perfectly legitimate system and its balance is often the one that finally connects the recycle to everything else.

A splitting point is the tee where a stream divides into product and recycle. Remember from the last lesson that it gives only one independent balance because composition is unchanged in every branch.

The strategy is a sequence: overall envelope first, then whichever internal boundary now has only one unknown, then the next. If you find yourself with four simultaneous equations in four unknowns, you probably skipped the overall balance.

Key idea: Draw the overall envelope first, because recycle streams never cross it, then work inward to boundaries that each add one unknown at a time.

Why recycle at all

Recycle costs money: extra piping, an extra compressor or pump, extra separation duty. Engineers do it anyway, for four reasons. To recover unconsumed reactant, which is the dominant reason and the subject of the next section. To recover a catalyst or a solvent, because throwing away a solvent is usually both expensive and an environmental problem. To control a reactor, since recycling cool product back into a hot reactor is one of the standard ways of moderating an exothermic reaction. And to conserve utilities, as when condensate is returned to a boiler house rather than discarded.

A recycle problem worked completely

Consider a liquid-phase reaction A goes to B. The reactor achieves only 30 percent conversion of the A fed to it, which is typical when equilibrium or selectivity limits how far you can push. Downstream, a separator recovers essentially all the B as product and returns essentially all the unreacted A to the reactor inlet. Fresh feed is 100 kmol/h of pure A. Find the product rate, the reactor feed rate, the recycle rate, and both conversions.

Start with the overall envelope. Crossing it: 100 kmol/h of A entering, and product P leaving. Since the separator returns all unreacted A to the loop, no A leaves the plant at all. Every atom of A that enters must depart as B, so P = 100 kmol/h of B. The overall conversion, defined on the fresh feed, is 100 percent.

Now the reactor. Let its combined feed be F1. The reactor consumes 30 percent of the A fed to it, and at steady state that consumption must equal the B production, which we now know is 100 kmol/h. So 0.30 F1 = 100 and F1 = 333.3 kmol/h. The reactor outlet is therefore 0.70 x 333.3 = 233.3 kmol/h of A plus 100 kmol/h of B.

Now the mixing point, where fresh feed and recycle combine to make the reactor feed: 100 + R = 333.3, so R = 233.3 kmol/h. The recycle ratio, recycle divided by fresh feed, is 2.33. Check at the separator: in comes 233.3 A plus 100 B; out goes 100 B as product and 233.3 A as recycle. Closed.

Now stare at the two conversion numbers, because this is the lesson. The single-pass conversion, measured on the reactor feed, is 30 percent. The overall conversion, measured on the fresh feed, is 100 percent. A reactor that looks feeble in isolation delivers complete utilization of raw material once wrapped in a recycle loop. That single fact explains an enormous amount of industrial practice.

What does a better reactor buy you? Repeat with 60 percent single-pass conversion: F1 = 100 / 0.60 = 166.7 kmol/h, R = 66.7 kmol/h, recycle ratio 0.67. The product is identical at 100 kmol/h, but the reactor, the separator, and the recycle pump all handle roughly half as much material. Recycle converts a chemistry problem into an equipment-size problem, and the design question becomes economic: is it cheaper to build a better reactor or to build bigger separation and recycle equipment? Ammonia converters run at only 10 to 20 percent conversion per pass and get away with it precisely because the recycle answer wins there.

Key idea: Single-pass conversion is measured on the reactor feed and overall conversion on the fresh feed; recycle lets a low single-pass conversion still yield near-complete overall use of raw material, at the cost of larger equipment.

The trouble with loops: inerts accumulate

Now break the story. Suppose the fresh feed contains something that does not react and is not removed by the separator, for instance argon in the hydrogen and nitrogen fed to an ammonia plant. Trace it. Argon enters the loop, passes through the reactor unchanged, passes through the separator into the recycle, and comes back. More argon enters with the next kilogram of fresh feed. Nothing removes it. There is no steady state: the argon concentration in the loop climbs without bound, diluting the reactants, dropping their partial pressures, and eventually shutting the reaction down.

The cure is a purge: a small stream deliberately bled from the recycle loop and thrown away or burned. It seems wasteful, and it is, because the purge carries away valuable reactant along with the inert. But it is the only way to give the inert an exit, and at steady state the inert balance sets the purge rate exactly.

Purge worked: the ammonia loop

Take an ammonia synthesis loop, N2 + 3 H2 gives 2 NH3. Fresh feed is 100 mol/s consisting of 74.25 mol/s hydrogen, 24.75 mol/s nitrogen (a clean 3 to 1 ratio), and 1.00 mol/s argon, a typical trace from the air separation and reforming steps upstream. The converter is followed by a chiller that condenses out essentially pure liquid ammonia; the uncondensed gas splits into recycle and purge. Suppose the process is designed to hold the loop gas at 10 mole percent argon.

Start with the argon balance over the overall envelope. Argon enters at 1.00 mol/s and can leave only in the purge, since it does not react and does not condense with the ammonia. The purge has the same composition as the recycle, because they come from a splitter. So argon out = 0.10 x P, and setting that equal to 1.00 gives P = 10 mol/s.

That purge of 10 mol/s is 1.0 mol/s of argon and 9.0 mol/s of hydrogen plus nitrogen. Since the non-inert part of the loop stays near the 3 to 1 stoichiometric ratio, that is 6.75 mol/s of hydrogen and 2.25 mol/s of nitrogen thrown away every second.

Now the nitrogen balance, again overall. Nitrogen in = 24.75 mol/s. Nitrogen out in the purge = 2.25 mol/s. The difference, 22.50 mol/s, was consumed by reaction. Stoichiometry gives ammonia produced = 2 x 22.50 = 45.0 mol/s, and hydrogen consumed = 3 x 22.50 = 67.5 mol/s, which checks against the hydrogen balance: 74.25 in, 6.75 out in the purge, difference 67.5. Overall nitrogen conversion is 22.50 / 24.75 = 90.9 percent.

Confirm with total mass, which must balance even though moles do not. In: 74.25 x 2.016 + 24.75 x 28.02 + 1.00 x 39.95 = 149.7 + 693.5 + 40.0 = 883.2 g/s. Out: 45.0 x 17.03 + 6.75 x 2.016 + 2.25 x 28.02 + 1.00 x 39.95 = 766.4 + 13.6 + 63.0 + 40.0 = 883.0 g/s. Agreement to within rounding.

Now run the trade-off, which is the real design question. Suppose you tolerate 20 mole percent argon in the loop instead of 10. The argon balance gives P = 1.00 / 0.20 = 5 mol/s, of which 1.0 is argon and 4.0 is reactant (3.0 hydrogen, 1.0 nitrogen). Nitrogen consumed rises to 24.75 - 1.00 = 23.75 mol/s, ammonia production rises to 47.5 mol/s, and overall nitrogen conversion improves to 96.0 percent. Hydrogen lost in the purge falls from 9.1 percent of the fresh feed to 4.0 percent.

Loop argon	Purge rate	Hydrogen lost	Ammonia made	Overall N2 conversion
10 mole percent	10 mol/s	6.75 mol/s (9.1 percent)	45.0 mol/s	90.9 percent
20 mole percent	5 mol/s	3.00 mol/s (4.0 percent)	47.5 mol/s	96.0 percent

So why not run at 40 percent argon and purge almost nothing? Because argon takes up space. At fixed loop pressure, raising the inert fraction lowers the partial pressures of nitrogen and hydrogen, which lowers the reaction rate and the equilibrium ammonia content per pass. You then need a larger converter, a larger recycle compressor, and more compression power for the same production. The purge rate is a genuine optimum between reactant thrown away and equipment plus power bought, and real plants land somewhere in this range. In practice the purge is not simply vented: it is burned as fuel in the reformer, or its hydrogen is recovered by membrane or pressure-swing adsorption and returned, which shifts the optimum again.

Key idea: An inert entering a recycle loop must have an exit, and the inert balance alone fixes the purge rate; the choice of loop inert concentration trades reactant loss against reactor and compressor size.

Bypass, the mirror image

A bypass is a stream deliberately routed around a unit and recombined with its outlet. Its usual purpose is control by blending. A dryer that can only produce bone-dry solid can deliver a specified 3 percent moisture if part of the wet feed bypasses it and is blended back. A heat exchanger sized for the worst case can hold a constant outlet temperature year round if a controlled fraction of the cold stream bypasses the tubes. Balance-wise, a bypass is easy: it involves a splitting point (one independent balance) and a mixing point, and the overall envelope again ignores it entirely.

Common misconceptions

• Recycle streams must be included in the overall balance. They must not. An internal recycle never crosses the overall envelope, which is exactly what makes that envelope so useful.
• Conversion is a single number. There are two, and confusing them is the most common error in this topic. Single-pass conversion uses the reactor feed as the basis; overall conversion uses the fresh feed.
• A purge is just waste caused by poor design. A purge is the mathematical necessity that gives inerts an exit. Without it there is no steady state at all.
• Increasing the purge always improves the process. A larger purge lowers inert concentration and helps the reaction, but throws away more reactant. There is an optimum, not a direction.
• Purge composition can be chosen independently of the recycle. Purge and recycle leave a splitting point, so they have identical compositions.

Recap

• Choose boundaries deliberately: overall envelope first, then units, mixing points, and splitting points, adding one unknown at a time.
• Recycle recovers unreacted feed, solvents, and catalyst, and can moderate reactor temperature, at the cost of larger separation and pumping equipment.
• In the worked case, 30 percent single-pass conversion with total recovery gives 100 percent overall conversion, a reactor feed of 333.3 kmol/h and a recycle of 233.3 kmol/h on a fresh feed of 100 kmol/h.
• Doubling single-pass conversion to 60 percent halves the recycle to 66.7 kmol/h without changing production.
• Inerts accumulate in loops unless purged; the inert balance alone fixes the purge rate, giving P = 10 mol/s at 10 percent loop argon and 45.0 mol/s of ammonia from a 100 mol/s fresh feed.
• Raising loop inert concentration cuts purge losses but slows the reaction and enlarges the converter and compressor, so the purge rate is an economic optimum.

Sources

1. Britannica. (n.d.). Haber-Bosch process. Encyclopaedia Britannica. britannica.com
2. Wikipedia contributors. (n.d.). Plastic recycling. Wikipedia. en.wikipedia.org
3. Engineering LibreTexts. (n.d.). Material balances on multiple-unit processes. LibreTexts Chemical Engineering. eng.libretexts.org
4. U.S. Department of Energy. (n.d.). Industrial energy and decarbonization. DOE. energy.gov

The big picture

Everything so far assumed molecules keep their identity. Now they do not. Put methane and oxygen into a burner and methane comes out as carbon dioxide and water; the methane balance has a consumption term, the carbon dioxide balance has a generation term, and Input = Output is simply false for both. Only three things survive a reaction untouched: total mass, atoms of each element, and, if you are careful about what you mean, energy.

This is not a complication so much as a promotion. Reactive balances are where chemical engineering stops being general bookkeeping and starts being chemistry at scale. They are also where the vocabulary that industry actually argues about lives. When a plant manager says the unit is running at 92 percent conversion but selectivity has drifted to 88, both numbers are precise, both are computed from a reactive balance, and mixing them up will get you laughed out of a meeting. Yield, conversion, and selectivity are different quantities that all sound like efficiency, and today we pin each of them down.

Here is the plan for today. First stoichiometry as a set of ratios and the limiting reagent that falls out of it. Then the three methods for solving reactive balances: molecular species balances with generation terms, extent of reaction, and atomic balances. Then a fully worked two-reaction example, the partial oxidation of ethylene, which forces you to distinguish conversion, yield, and selectivity because all three differ. We finish with combustion calculations and the excess air convention, which sets up the energy balances in Module 3.

Stoichiometry and the limiting reagent

A balanced equation is a statement of ratios. For ammonia synthesis, N2 + 3 H2 gives 2 NH3, the ratios say that one mole of nitrogen consumed always accompanies three moles of hydrogen consumed and two moles of ammonia produced. It says nothing about how fast, how far, or under what conditions. Stoichiometry is a constraint, not a prediction.

We write coefficients with a sign convention that pays off immediately: the stoichiometric coefficient nu is negative for reactants and positive for products. So for ammonia synthesis, nu(N2) = -1, nu(H2) = -3, nu(NH3) = +2.

Feed reactants in exactly the stoichiometric ratio and both run out together. Feed anything else and one runs out first: the limiting reagent. To find it, divide the moles fed of each reactant by its stoichiometric coefficient magnitude and take the smallest. Suppose you feed 50 mol of nitrogen and 120 mol of hydrogen. For nitrogen, 50 / 1 = 50. For hydrogen, 120 / 3 = 40. Hydrogen is limiting. The nitrogen needed to consume all the hydrogen is 40 mol, so 10 mol of nitrogen is left over: nitrogen is present in 25 percent excess, since (50 - 40)/40 = 0.25.

Excess is normal and deliberate. You feed excess air to a burner to make sure the fuel burns completely, excess of a cheap reagent to drive an expensive one to full conversion, and excess of one reactant to suppress a side reaction. The convention to memorize is that percentage excess is always computed against the theoretical requirement for complete consumption of the limiting reagent, not against the total fed.

Key idea: Divide moles fed by the stoichiometric coefficient for each reactant; the smallest ratio identifies the limiting reagent, and percentage excess is measured against the theoretical requirement for that limiting reagent.

Three ways to solve a reactive balance

The first method is molecular species balances. Write the general balance for each compound and keep the generation and consumption terms explicitly. It is the most transparent method and the most tedious, because every reaction contributes a term to several equations.

The second method is the extent of reaction, and it is the one to reach for by default. Define a single variable xi (Greek letter xi, in units of moles or moles per hour) that measures how far the reaction has proceeded. Then, for every species at once, n(out) = n(in) + nu x xi. One variable replaces all the generation and consumption terms in the whole reaction. With ammonia synthesis at xi = 22.5 mol/s, nitrogen falls by 22.5, hydrogen by 67.5, and ammonia rises by 45.0, exactly the numbers from the ammonia loop in the last lesson. For a system with several reactions, use one extent per independent reaction: n(out) = n(in) + sum over reactions of nu(i,j) x xi(j).

The third method is atomic species balances. Since atoms are never created or destroyed, write Input = Output for each element and forget the reactions entirely. This is the right choice when the reactions are unknown, numerous, or messy, which describes most combustion and much biological processing. Its weakness is that it does not tell you anything about individual reactions, only about elements.

Key idea: Extent of reaction replaces every generation and consumption term with one variable per independent reaction, while atomic balances sidestep reaction detail entirely by counting elements.

Conversion, yield, and selectivity

Three words, three meanings, endless confusion. Fix them now.

Conversion answers: how much of the reactant disappeared? It equals (moles of reactant fed minus moles left) divided by moles fed. It says nothing about what the reactant turned into.

Yield answers: how much desired product did I get, compared with the maximum theoretically possible from the feed? A common definition is moles of desired product formed divided by the moles that would have formed if all the limiting reagent had gone to that product.

Selectivity answers: of the reactant that did react, what fraction went where I wanted? It is often quoted as moles of desired product formed divided by moles of reactant consumed, expressed as a percentage of the stoichiometric maximum, and sometimes as the ratio of desired to undesired product.

The relationship that ties them together is worth memorizing: yield is approximately conversion multiplied by selectivity. A reactor with 90 percent conversion and 50 percent selectivity wastes half of what it converts and yields 45 percent. A reactor with 25 percent conversion and 95 percent selectivity yields only about 24 percent per pass, but with recycle it can turn nearly all the feed into product, which is why industry frequently chooses low conversion and high selectivity. Selectivity is usually worth more than conversion, because unconverted reactant can be recycled while a byproduct is often unrecoverable waste.

Worked example: partial oxidation of ethylene

Ethylene oxide is made by passing ethylene and oxygen over a silver catalyst. Two reactions compete:

1. C2H4 + 0.5 O2 gives C2H4O, the desired reaction, extent xi1
2. C2H4 + 3 O2 gives 2 CO2 + 2 H2O, complete combustion of the feedstock, extent xi2

Feed 100 mol of ethylene and 100 mol of oxygen. Measurements give an ethylene conversion of 25 percent and a selectivity of 80 percent to ethylene oxide. Find the full outlet composition, the yield, and the oxygen left over.

Ethylene reacted = 0.25 x 100 = 25 mol. Of that, 80 percent went to ethylene oxide, so 20 mol of ethylene became ethylene oxide, and the remaining 5 mol burned. Since each reaction consumes one ethylene per unit extent, xi1 = 20 mol and xi2 = 5 mol.

Now apply n(out) = n(in) + sum of nu x xi to every species.

Species	In (mol)	nu in reaction 1	nu in reaction 2	Out (mol)
C2H4	100	-1	-1	100 - 20 - 5 = 75
O2	100	-0.5	-3	100 - 10 - 15 = 75
C2H4O	0	+1	0	0 + 20 = 20
CO2	0	0	+2	0 + 10 = 10
H2O	0	0	+2	0 + 10 = 10

Total outlet = 75 + 75 + 20 + 10 + 10 = 190 mol, compared with 200 mol in. Moles are not conserved, and there is no reason they should be.

Check with atomic balances, which must close regardless. Carbon in = 100 x 2 = 200. Carbon out = 75 x 2 + 20 x 2 + 10 x 1 = 150 + 40 + 10 = 200. Hydrogen in = 100 x 4 = 400. Out = 75 x 4 + 20 x 4 + 10 x 2 = 300 + 80 + 20 = 400. Oxygen atoms in = 100 x 2 = 200. Out = 75 x 2 + 20 x 1 + 10 x 2 + 10 x 1 = 150 + 20 + 20 + 10 = 200. All three close exactly, which is the strongest check available on a reactive balance.

Now the three efficiency numbers. Conversion of ethylene = 25 / 100 = 25 percent. Selectivity = 20 mol of ethylene oxide per 25 mol of ethylene consumed = 80 percent, or equivalently 4 moles of ethylene oxide per mole of ethylene burned. Yield = 20 mol of ethylene oxide out of the 100 mol theoretically possible from 100 mol of ethylene = 20 percent, and note 0.25 x 0.80 = 0.20, confirming the relationship.

Is that a bad reactor? Not at all. The 75 mol of unreacted ethylene and 75 mol of unused oxygen are recycled, so overall the plant converts nearly all its ethylene. The 5 mol that burned, however, is gone forever, and it also released a great deal of heat that the reactor must remove, which is why real ethylene oxide reactors are enormous multitubular heat exchangers full of catalyst. Selectivity, not conversion, is what a process like this is optimized for.

Key idea: Conversion counts reactant that disappeared, selectivity counts where it went, yield counts product against the theoretical maximum, and yield is approximately conversion times selectivity.

Combustion and the excess air convention

Combustion problems are common enough to have their own conventions. Air is taken as 21 mole percent oxygen and 79 mole percent nitrogen, and the nitrogen is treated as inert, which is close enough for balances although not for pollution, since a little of it makes nitrogen oxides at flame temperature.

Theoretical air is the amount exactly required for complete combustion of all the fuel to carbon dioxide and water. Real burners use excess air to ensure complete burnout despite imperfect mixing.

Work an example we will reuse in Module 3. Burn 100 mol/s of methane with 20 percent excess air. Methane burns as CH4 + 2 O2 gives CO2 + 2 H2O, so theoretical oxygen is 200 mol/s. With 20 percent excess, oxygen fed is 240 mol/s, and air fed is 240 / 0.21 = 1142.9 mol/s, carrying 902.9 mol/s of nitrogen. Assuming complete combustion, the extent is 100 mol/s and the flue gas is: carbon dioxide 100, water 200, leftover oxygen 240 - 200 = 40, nitrogen 902.9, for a total of 1242.9 mol/s.

Two ways of reporting that composition matter in practice. On a wet basis, carbon dioxide is 100 / 1242.9 = 8.05 percent, water 16.1 percent, oxygen 3.2 percent, nitrogen 72.6 percent. On a dry basis, which is what a stack analyzer reports because the sample is cooled and the water condenses out, the total is 1042.9 mol/s and carbon dioxide is 100 / 1042.9 = 9.59 percent, oxygen 3.84 percent, nitrogen 86.6 percent. A historical dry-basis stack analysis is still called an Orsat analysis. If a problem gives you a flue gas composition without saying which basis, assume dry and say so.

Notice how useful that leftover oxygen figure is. Measuring 3.8 percent oxygen in a dry stack gas tells a plant engineer immediately that the burner is running around 20 percent excess air. Too little and you get unburned fuel, soot, and carbon monoxide; too much and you are heating nitrogen up the chimney and wasting fuel. Combustion control is largely the management of that one number.

Key idea: Theoretical air is what stoichiometry demands and excess air is what burners actually use; measuring leftover oxygen in the dry stack gas is the standard way of inferring how much excess is present.

Common misconceptions

• Total moles are conserved in a reaction. Only if the stoichiometric coefficients happen to sum equally. Mass and atoms are conserved; moles generally are not, as the ethylene example shows with 200 mol in and 190 mol out.
• High conversion means a good process. A reactor at 95 percent conversion and 60 percent selectivity destroys more feedstock than one at 25 percent conversion and 95 percent selectivity operating with recycle.
• Yield and conversion are the same thing. Yield accounts for where the reactant went; conversion only says it left. They coincide only when selectivity is 100 percent.
• Percentage excess is computed on the total feed. It is computed against the theoretical requirement for complete consumption of the limiting reagent.
• Stack gas analyses are on a wet basis. Analyzers condense the water out, so quoted flue gas compositions are normally dry, and confusing the two shifts every percentage.

Recap

• Stoichiometric coefficients are negative for reactants and positive for products, and the limiting reagent is found from the smallest ratio of moles fed to coefficient.
• Reactive balances can be solved by molecular balances with generation terms, by extent of reaction (one variable per independent reaction, n(out) = n(in) + nu x xi), or by atomic balances.
• Conversion measures reactant consumed, selectivity measures where it went, yield measures product against the theoretical maximum, and yield is roughly conversion times selectivity.
• The worked ethylene oxidation with 25 percent conversion and 80 percent selectivity gives xi1 = 20, xi2 = 5, an outlet of 75 C2H4, 75 O2, 20 C2H4O, 10 CO2, and 10 H2O, and a yield of 20 percent.
• Atom balances on carbon, hydrogen, and oxygen all closed at 200, 400, and 200 respectively, which is the strongest available check.
• Burning 100 mol/s of methane with 20 percent excess air needs 1142.9 mol/s of air and gives a flue gas of 1242.9 mol/s containing 3.84 percent oxygen on a dry basis.

Sources

1. OpenStax. (2019). Reaction stoichiometry. In Chemistry 2e. Rice University. openstax.org
2. Chemistry LibreTexts. (n.d.). Limiting reagents. LibreTexts. chem.libretexts.org
3. Wikipedia contributors. (n.d.). Extent of reaction. Wikipedia. en.wikipedia.org
4. U.S. Energy Information Administration. (n.d.). Natural gas explained. EIA. eia.gov

The big picture

Ask a plant manager what keeps them awake and the answer is rarely chemistry. It is the energy bill. A large chemical site can spend more on steam, fuel, and electricity than on everything else except raw materials, and the chemical industry as a whole is the largest industrial consumer of energy on the planet. Every one of those costs was set on a drawing board by an engineer computing an enthalpy change. Material balances tell you how big the equipment must be; energy balances tell you what it will cost to run for the next twenty years.

There is a second reason to care, and it is the more urgent one. Energy that you fail to account for does not disappear; it shows up as temperature. A reaction whose heat release you underestimated will heat its own contents, speed itself up, release heat faster still, and run away. Module 6 examines what that looks like when it happens to a real plant. Energy balances are a safety tool before they are an economic one.

Here is the plan for today. First, why flowing streams are described by enthalpy rather than internal energy, and why enthalpy is always a difference from an arbitrary reference. Then heat capacity, and the sensible heat calculation that follows from it. Then latent heat, the energy of phase change, which is usually much larger than people expect. We finish by putting real numbers on both for the same stream, so you can see the ratio for yourself. Engineering Thermodynamics (ENGR 230) develops the first and second laws in full; this lesson takes the results and puts them to work on process equipment.

Why enthalpy, and why it is always relative

A closed system stores internal energy U, the sum of the molecular kinetic and potential energies of its contents. For flowing streams a second term appears. To push a stream of specific volume v into a unit held at pressure P, the upstream fluid must do work of P times v per kilogram, called flow work, and the downstream fluid receives the same on the way out. Because that term appears every single time material crosses a boundary, engineers fold it into a combined property once and for all: enthalpy, H = U + P V, or per kilogram, h = u + P v.

The payoff is that for a steady-flow unit with negligible changes in kinetic and potential energy, the energy balance reduces to a very clean statement: Q - Ws = delta H, where Q is heat added to the unit, Ws is shaft work done by it, and delta H is the enthalpy of everything leaving minus the enthalpy of everything entering. Most process units have no shaft work at all, so for exchangers, heaters, condensers, and reactors the balance is simply Q = delta H. That is the equation this module runs on.

Now a point that confuses almost everyone once. There is no such thing as the absolute enthalpy of a stream. Enthalpy is defined only up to an additive constant, so every table of enthalpy data is quoted relative to some arbitrary reference state: steam tables usually take saturated liquid water at 0.01 C as zero, while thermochemical tables take the elements in their standard states at 25 C and 1 bar as zero. This is not a defect, because every calculation you will ever do involves a difference, and the constant cancels. The rule that follows is absolute: within one calculation, all enthalpies must come from the same reference state. Mixing two tables with different references is a classic and expensive mistake.

Key idea: Enthalpy bundles internal energy with flow work so that steady-flow units obey Q - Ws = delta H, and enthalpies are always differences from an arbitrary reference that must be kept consistent throughout a calculation.

Heat capacity and sensible heat

Sensible heat is the energy that changes a substance's temperature without changing its phase, and it is sensible in the old sense of being detectable by a thermometer. The property that governs it is the specific heat capacity, the energy required to raise one kilogram (or one mole) by one degree.

Two versions exist. At constant volume, energy goes only into internal energy, giving Cv. At constant pressure, some energy also does expansion work, so Cp is larger. Flow processes happen at nearly constant pressure, so Cp is the one this course uses almost exclusively. For liquids and solids the two are nearly identical; for ideal gases they differ by the gas constant, Cp - Cv = R, which is 8.314 J/(mol K).

For a stream heated from T1 to T2 with no phase change, Q = m x Cp x (T2 - T1) on a mass basis, or Q = n x Cp x (T2 - T1) on a molar basis. Recall from Lesson 2 that a Celsius degree equals a kelvin, so Cp values are numerically the same in either. Heat capacity itself varies with temperature, which real design work handles with a polynomial such as Cp = a + bT + cT^2, integrated over the range. For the temperature spans in this course an average Cp is close enough, and I will say when it is not.

Substance	Cp, kJ/(kg K)	Boiling point at 1 atm, C	Latent heat of vaporization, kJ/kg
Water (liquid)	4.18	100.0	2257
Steam (vapor, near 1 atm)	2.0	-	-
Air (gas)	1.005	-	-
Benzene (liquid)	1.74	80.1	394
Toluene (liquid)	1.70	110.6	360
Ethanol (liquid)	2.44	78.4	841
Carbon steel	0.49	-	-

Water's heat capacity of 4.18 kJ/(kg K) is remarkably large, several times that of most organics and eight times that of steel. That single number explains why water is the universal industrial coolant, why oceans moderate climate, and why a water-cooled exchanger can absorb a large duty with a modest flow.

Key idea: Sensible heat is Q = m Cp dT, with Cp at constant pressure the relevant property for flow processes, and water's unusually high Cp is why it cools nearly everything in industry.

Latent heat, and why it dominates

Now boil something. Hold water at 100 C and 1 atm and keep adding heat: the temperature does not move, yet energy pours in. That energy is going into pulling molecules apart against their attractions, converting liquid into vapor at constant temperature. It is latent heat, and for vaporization it carries the symbol delta H(vap).

The numbers are startling. Water's latent heat of vaporization at 100 C is 2257 kJ/kg. Its latent heat of fusion at 0 C is 334 kJ/kg. Compare those with the 4.18 kJ/kg needed to raise a kilogram by one degree, and you see that boiling a kilogram of water costs the same as heating it by 540 degrees, if only it could stay liquid that long.

Let us make it concrete with a duty you might actually be handed. Take 1000 kg/h of water at 20 C and produce saturated steam at 100 C and 1 atm. There are two steps.

Sensible: Q = 1000 kg/h x 4.18 kJ/(kg K) x (100 - 20) K = 1000 x 4.18 x 80 = 334,400 kJ/h. Dividing by 3600 gives 92.9 kW.

Latent: Q = 1000 kg/h x 2257 kJ/kg = 2,257,000 kJ/h, which is 627.0 kW.

Total: 2,591,400 kJ/h, or 719.8 kW. The latent portion is 2257 / 2591 = 87 percent of the whole duty. Heating the water through eighty degrees, which sounds like the main event, is a seventh of the job.

Everything downstream in this course follows from that ratio. It is why evaporators and dryers are the energy hogs of a plant, why distillation is expensive (a column boils its contents over and over), why multiple-effect evaporators that reuse vapor to heat the next stage exist at all, and why the single most effective energy-saving move in process design is usually to avoid a phase change rather than to make one more efficient. It is also why steam is the industrial heating medium of choice: condensing steam gives back 2257 kJ for every kilogram, at a constant temperature you can set by choosing the pressure.

Key idea: Latent heat is typically several hundred to several thousand kJ/kg, dwarfing sensible heat over ordinary temperature spans, so phase changes dominate the energy bill of most processes.

Reading property data

Where do these numbers come from? For water, from steam tables, which list enthalpy of saturated liquid, saturated vapor, and their difference at every pressure. For everything else, from compilations such as the NIST Chemistry WebBook, which is free, authoritative, and worth learning to navigate now. Two habits will keep you out of trouble. First, check the temperature at which a property is quoted, since a heat of vaporization falls as temperature rises and reaches zero at the critical point. Second, check whether a value is per mole or per kilogram; benzene's heat of vaporization is 30.8 kJ/mol, which divided by its molar mass of 78.11 kg/kmol gives 394 kJ/kg, and the two numbers differ by a factor of 78.

One useful simplification: for ideal mixtures, enthalpies are additive. The enthalpy of a stream is the sum over components of the mass (or moles) of each times its specific enthalpy. Real mixtures have a heat of mixing, but for the hydrocarbon and similar mixtures in this course it is small enough to ignore, and I will flag the cases where it is not, such as sulfuric acid in water, where mixing releases enough heat to boil the solution.

A worked preheat on the running example

Return to our benzene-toluene feed: 8513 kg/h at 25 C, to be preheated to its bubble point of 92.1 C before entering the column. Take the liquid mixture heat capacity as roughly 1.72 kJ/(kg K), the mass-weighted blend of the two values in the table.

Q = 8513 kg/h x 1.72 kJ/(kg K) x (92.1 - 25) K = 8513 x 1.72 x 67.1 = 982,500 kJ/h, which is 272.9 kW, call it 273 kW.

Hold that number. In Lesson 14 we will find that the reboiler of the same column needs about 1.11 MW, roughly four times as much, and the reason is exactly the ratio you just met: the preheater only raises temperature, while the reboiler vaporizes. Two hundred and seventy-three kilowatts is also a real cost. At a steam price of about 10 US dollars per gigajoule, 982,500 kJ/h is roughly 9.8 dollars an hour, or 86,000 dollars a year for a preheater most people would never look at twice. This is the sense in which energy balances are the economics of a plant.

Key idea: Enthalpies of ideal mixtures add, property data must be checked for temperature and for a mass or mole basis, and a preheat duty that looks trivial is worth tens of thousands of dollars a year.

Common misconceptions

• Enthalpy has an absolute value you could look up. It does not. Every table is relative to a chosen reference state, and all references must match within one calculation.
• Adding heat always raises temperature. Not during a phase change. Boiling water at 1 atm stays at 100 C no matter how hard you heat it; the energy goes into latent heat.
• Sensible heat is the main energy cost of a process. Usually the reverse. In the worked example, latent heat was 87 percent of the total duty.
• Cp and Cv are interchangeable. For liquids and solids nearly so, but for gases they differ by R, and flow processes at constant pressure require Cp.
• Heat capacity is constant. It varies with temperature, and over wide ranges you must integrate a polynomial rather than use a single value.

Recap

• Enthalpy H = U + PV folds flow work into a single property, giving the steady-flow balance Q - Ws = delta H, or Q = delta H when there is no shaft work.
• Enthalpy values are always relative to a reference state, and references must be consistent throughout a calculation.
• Sensible heat is Q = m Cp dT; water's Cp of 4.18 kJ/(kg K) is exceptionally high, which is why water cools industry.
• Latent heat changes phase at constant temperature: 2257 kJ/kg to vaporize water at 100 C and 334 kJ/kg to melt ice.
• Making 1000 kg/h of steam from 20 C water needs 92.9 kW sensible plus 627.0 kW latent, a total of 719.8 kW, of which the phase change is 87 percent.
• Preheating our 8513 kg/h benzene-toluene feed from 25 C to its 92.1 C bubble point requires 273 kW, worth roughly 86,000 dollars a year in steam.

Sources

1. National Institute of Standards and Technology. (n.d.). NIST Chemistry WebBook: Thermophysical properties. U.S. Department of Commerce. webbook.nist.gov
2. OpenStax. (2016). Heat transfer, specific heat, and calorimetry. In University physics volume 2. Rice University. openstax.org
3. OpenStax. (2016). Phase changes. In University physics volume 2. Rice University. openstax.org
4. U.S. Department of Energy. (n.d.). Steam systems. Office of Energy Efficiency and Renewable Energy. energy.gov

The big picture

Last lesson gave you the property, enthalpy, and the two ways it changes, sensible and latent. Today you use them on equipment. The good news is that the method never varies. Draw the boundary, list every stream in and out, evaluate each stream's enthalpy relative to a common reference, and set Q minus Ws equal to the difference. Do that carefully on a heater, a mixer, a valve, and an exchanger, and you can do it on anything.

The bad news is that beginners lose marks on the same three things every time: choosing different reference states for different streams, forgetting that a phase change happened somewhere in the middle, and dropping a unit conversion between kJ/h and kW. We will deal with all three explicitly.

Here is the plan for today. First a quick justification for ignoring kinetic and potential energy, since I do not want you taking it on faith. Then the inlet-outlet enthalpy table, which is the bookkeeping device that makes these problems mechanical. Then five worked units in increasing order of interest: a heater, an adiabatic mixer, a throttling valve, a two-stream exchanger, and a pump. We finish by converting a duty into a utility bill, which is what the calculation is usually for.

Why kinetic and potential energy usually drop out

The full steady-flow energy balance carries terms for kinetic and potential energy: Q - Ws = delta H + delta(KE) + delta(PE). We routinely throw the last two away, and you should know why rather than merely being told.

Take a liquid moving at a typical process velocity of 3 m/s. Its kinetic energy per kilogram is v^2 / 2 = 9 / 2 = 4.5 J/kg. Now lift it ten metres: potential energy per kilogram is g x h = 9.81 x 10 = 98.1 J/kg. Compare those with the enthalpy change of heating that same kilogram of water by 25 degrees: 4180 x 25 = 104,500 J/kg. The kinetic term is 0.004 percent of it and the potential term 0.09 percent. Both are far below the accuracy of any plant measurement.

So the terms drop out, but not always. They matter in nozzles and diffusers, where velocity change is the entire point; in tall columns handling low-enthalpy changes; and in compressible high-velocity gas flow. The habit worth forming is to estimate the terms once, in your head, rather than to assume.

Key idea: Kinetic and potential energy per kilogram are typically a few to a hundred joules while process enthalpy changes are tens to hundreds of kilojoules, so both terms drop out except in nozzles and high-velocity flow.

The inlet-outlet enthalpy table

Here is the device that turns energy balances into arithmetic. Make a table with one row per species and four columns: moles or mass in, specific enthalpy in, moles or mass out, specific enthalpy out. Fill the enthalpies relative to a reference state you choose and write at the top of the page. Then Q = sum of (n x h) out minus sum of (n x h) in.

Choosing the reference cleverly saves work. Pick the inlet condition of the largest stream as your zero, and that whole column becomes zeroes. Pick 25 C and 1 atm if the problem involves reactions, because that is where heats of formation are tabulated. There is no wrong choice, only inconsistent ones.

Worked unit 1: a heater

Preheat 8513 kg/h of our benzene-toluene feed from 25 C to 92.1 C, exactly as in the last lesson. Reference: liquid feed at 25 C, so h(in) = 0. Then h(out) = Cp x dT = 1.72 x 67.1 = 115.4 kJ/kg. Q = 8513 x 115.4 = 982,500 kJ/h = 273 kW. Positive Q means heat is added, which is the sign convention throughout: Q into the system is positive, work out of it is positive.

Worked unit 2: an adiabatic mixer

Two water streams meet in an insulated tee: 3000 kg/h at 20 C and 1000 kg/h at 80 C. What comes out?

Adiabatic means Q = 0, and there is no shaft work, so delta H = 0. Take 20 C as reference. Then h(in) = 3000 x 0 + 1000 x 4.18 x 60 = 250,800 kJ/h. Out, 4000 kg/h at unknown T: h(out) = 4000 x 4.18 x (T - 20). Setting them equal, 4000 x 4.18 x (T - 20) = 250,800, so T - 20 = 15.0 and T = 35.0 C.

You could have written that as a weighted average, (3000 x 20 + 1000 x 80) / 4000 = 35 C, and for equal heat capacities that shortcut is exact. It fails the moment the streams differ in composition or phase, so learn the balance and treat the shortcut as a check.

Worked unit 3: a throttling valve

A valve has no shaft work and, being small and fast, essentially no heat transfer. So Q = 0, Ws = 0, and the balance says delta H = 0. A throttle is isenthalpic.

That single fact does a lot of work in process engineering. For an ideal gas, constant enthalpy means constant temperature, so throttling an ideal gas does nothing thermally. For a real gas the Joule-Thomson effect appears and most gases cool on throttling, which is the basis of refrigeration and of air liquefaction. And for a saturated liquid, throttling to a lower pressure produces flash: some of the liquid vaporizes, and because vaporization needs latent heat that can only come from the liquid itself, the remaining liquid cools to the new saturation temperature. Every flash drum, every refrigeration expansion valve, and every steam trap in a plant runs on this one line of algebra.

Key idea: A throttling valve is isenthalpic, which is why real gases cool on expansion and why a saturated liquid flashes and self-cools when its pressure is dropped.

Worked unit 4: a two-stream heat exchanger

This is the unit you will meet most often, so let us do it carefully and keep the numbers for Lesson 12.

A process stream of 10,000 kg/h of an organic liquid with Cp = 2.5 kJ/(kg K) must be cooled from 90 C to 45 C. Cooling water is available at 25 C and, by site policy, may not be returned warmer than 40 C, because above that scaling and biological growth become problems in the cooling tower.

Take the hot side first. Mass flow = 10,000 / 3600 = 2.778 kg/s. Q = 2.778 x 2.5 x (90 - 45) = 2.778 x 112.5 = 312.5 kW. That is the duty: the exchanger must move 312.5 kW out of the hot stream.

Now the cold side. If the exchanger loses no heat to the surroundings, all 312.5 kW enters the water. So m(water) = Q / (Cp x dT) = 312.5 / (4.184 x 15) = 312.5 / 62.76 = 4.979 kg/s, which is 17,900 kg/h, or about 17.9 tonnes per hour.

Pause on that. Cooling one stream of ten tonnes an hour takes nearly eighteen tonnes an hour of water, and that is with water's exceptionally high heat capacity working in our favor. Cooling water is one of the largest mass flows on any chemical site, which is why cooling towers are the tallest structures on many of them and why raising the allowed return temperature by even a few degrees is a real economic lever.

Notice what this balance has not told you: how big the exchanger must be. Energy says how much heat must move; it takes the rate equation Q = U A dT(lm) to say through how much steel, and that is Lesson 12.

Worked unit 5: a pump

Pumps are the case where shaft work matters. For a liquid, which is essentially incompressible, the ideal work of raising pressure is Ws = V x delta P per unit mass, where V is specific volume. Pumping water (specific volume 0.001 m^3/kg) through a pressure rise of 400 kPa requires 0.001 x 400,000 = 400 J/kg of ideal work. At 20 kg/s, that is 8.0 kW of hydraulic power. A real pump at 70 percent efficiency draws 8.0 / 0.70 = 11.4 kW at the shaft, and the missing 3.4 kW ends up as heat in the fluid and the bearings.

Note the sign convention in action. Work done by the system is positive Ws, so a pump, which has work done on it, carries a negative Ws and therefore raises the enthalpy of the stream. Turbines are the reverse. Get this wrong and your energy balance will be off by exactly twice the work term, which is a distinctive and recognizable error.

Key idea: An energy balance gives the duty, not the equipment size; heat is positive into the system, shaft work is positive out of it, and pump work raises stream enthalpy.

From duty to utility bill

Process duties are paid for in utilities, and the conversion is short. Suppose our 273 kW preheater is served by low-pressure steam that condenses at about 2100 kJ/kg. Steam required = 982,500 kJ/h divided by 2100 kJ/kg = 468 kg/h. At a typical industrial steam cost of roughly 20 US dollars per tonne, that is about 9.40 dollars an hour, or 82,000 dollars a year on continuous operation.

The 312.5 kW cooler is paid for differently. Cooling water costs little per tonne, perhaps 0.05 dollars, but 17.9 tonnes an hour still comes to about 0.90 dollars an hour, plus the fan and pump power of the cooling tower. Heating is expensive and cooling is cheap, which is a large part of why heat integration, matching hot streams that need cooling against cold streams that need heating, is one of the highest-return activities in process design. We return to it in Lesson 16.

Key idea: Convert every duty into a utility flow and a cost; heating is typically an order of magnitude more expensive than cooling, which is what makes heat integration worth pursuing.

Common misconceptions

• Adiabatic means the temperature does not change. Adiabatic means no heat crosses the boundary. Temperatures can change a great deal, as in the adiabatic mixer or an adiabatic reactor.
• A throttling valve cools every fluid. It is isenthalpic, not isothermal. An ideal gas does not change temperature at all; real gases usually cool; a saturated liquid flashes and cools.
• An energy balance tells you the size of the exchanger. It gives the duty only. Area requires the rate equation with a heat transfer coefficient and a temperature driving force.
• Kinetic and potential energy can always be neglected. Usually, but not in nozzles, diffusers, or high-velocity compressible flow. Estimate them once rather than assuming.
• Pump work can be ignored because pumps are small. For liquids it is often small compared with thermal duties, but it is never zero, and mistaking its sign doubles the error.

Recap

• Use an inlet-outlet enthalpy table with a single stated reference state, then Q - Ws equals enthalpy out minus enthalpy in.
• Kinetic energy at 3 m/s is 4.5 J/kg and potential energy over 10 m is 98 J/kg, both negligible against enthalpy changes of order 100 kJ/kg.
• Adiabatic mixing of 3000 kg/h at 20 C with 1000 kg/h at 80 C gives 35.0 C.
• Throttling valves are isenthalpic, giving Joule-Thomson cooling in real gases and flash vaporization with self-cooling in saturated liquids.
• Cooling 10,000 kg/h of an organic from 90 C to 45 C is a duty of 312.5 kW and needs 17,900 kg/h of cooling water heated from 25 C to 40 C.
• The 273 kW preheater consumes about 468 kg/h of low-pressure steam, roughly 82,000 dollars a year, while the cooling duty costs under a dollar an hour.

Sources

1. Engineering LibreTexts. (n.d.). Energy balances on process units. LibreTexts Chemical Engineering. eng.libretexts.org
2. OpenStax. (2016). The first law of thermodynamics. In University physics volume 2. Rice University. openstax.org
3. National Institute of Standards and Technology. (n.d.). NIST Chemistry WebBook. U.S. Department of Commerce. webbook.nist.gov
4. U.S. Department of Energy. (n.d.). Energy efficiency in industry. Office of Energy Efficiency and Renewable Energy. energy.gov

The big picture

Every reaction moves energy. Rearranging atoms means breaking bonds, which costs energy, and forming new ones, which releases it, and the two rarely cancel. When formation wins, the reaction is exothermic and the vessel gets hot. When breaking wins, it is endothermic and the vessel gets cold unless you feed it heat. Ammonia synthesis is exothermic and its converters need cooling. Steam reforming of methane is strongly endothermic and its reformers sit inside enormous fired furnaces. Neither could be designed without the number we compute today.

The stakes here are higher than efficiency. An exothermic reaction in a badly cooled vessel raises its own temperature, and rising temperature raises reaction rate, which raises heat release, which raises temperature. That feedback loop is the thermal runaway, and it has destroyed plants and killed people. When you compute a heat of reaction you are computing, among other things, how much energy is available to be released if control is lost. Module 6 will return to this with real cases.

Here is the plan for today. First heats of reaction and the elegant trick, Hess's law, that lets us tabulate a modest number of formation enthalpies and compute any reaction from them. Then combustion and heating values, including the higher and lower distinction that quietly appears on every fuel invoice. Then a fully worked reactor balance and an adiabatic flame temperature. We finish with simultaneous balances, where composition and temperature depend on each other and the problem must be iterated.

Heat of reaction and Hess's law

The standard heat of reaction, written delta H(rxn) at 25 C and 1 bar, is the enthalpy change when stoichiometric amounts of reactants at 25 C become products at 25 C. By the sign convention, exothermic reactions have negative delta H, because the system loses enthalpy to the surroundings.

Measuring one for every reaction would be hopeless, so we exploit the fact that enthalpy is a state property: the change depends only on the endpoints, not the route. That is Hess's law, and it lets us route every reaction through a common waypoint, the elements. Define the standard heat of formation delta Hf as the enthalpy change when one mole of a compound forms from its elements in their standard states, and set delta Hf of every element in its standard state to zero. Then for any reaction:

delta H(rxn) = sum over products of (nu x delta Hf) minus sum over reactants of (nu x delta Hf)

Work one. Methane combustion, CH4 + 2 O2 gives CO2 + 2 H2O. Using delta Hf values of -74.9 kJ/mol for methane, -393.5 for carbon dioxide, -241.8 for water vapor, and zero for oxygen:

delta H(rxn) = [(-393.5) + 2 x (-241.8)] - [(-74.9) + 2 x 0] = (-393.5 - 483.6) + 74.9 = -877.1 + 74.9 = -802.2 kJ/mol.

Strongly exothermic, as anyone who has lit a gas ring would expect. Now repeat with liquid water, delta Hf = -285.8 kJ/mol: delta H(rxn) = [(-393.5) + 2 x (-285.8)] + 74.9 = -965.1 + 74.9 = -890.2 kJ/mol. The two answers differ by 88 kJ/mol, which is exactly twice water's latent heat of vaporization at 25 C, 44 kJ/mol, because the second case additionally condenses two moles of water.

Key idea: Enthalpy is a state property, so Hess's law lets any heat of reaction be assembled from tabulated heats of formation as products minus reactants.

Heating values, higher and lower

That 88 kJ/mol difference has a commercial name. The higher heating value (HHV, or gross calorific value) assumes the product water ends up liquid, so its latent heat is counted as recovered. The lower heating value (LHV, or net) assumes the water leaves as vapor up the stack, so that latent heat is lost.

For methane, dividing by the molar mass of 16.04 kg/kmol: LHV = 802.2 / 16.04 = 50.0 MJ/kg and HHV = 890.2 / 16.04 = 55.5 MJ/kg. The gap is 10 percent, which is more than enough to start an argument about who met an efficiency guarantee. European and American practice differ in which value is quoted by default, and equipment efficiencies above 100 percent, sometimes advertised for condensing boilers, are simply LHV-based numbers for a device that actually condenses the water and recovers the latent heat. Whenever you see a heating value or a thermal efficiency, ask which basis.

Key idea: Higher heating value counts the latent heat of the product water and lower heating value does not, a 10 percent difference for methane that must be stated whenever efficiency is quoted.

A reactor energy balance, worked

Take the furnace from Lesson 5: 100 mol/s of methane burned with 20 percent excess air, giving a flue gas of 100 mol/s carbon dioxide, 200 water, 40 leftover oxygen, and 902.9 nitrogen, a total of 1242.9 mol/s.

The clean way to organize a reactive energy balance is the heat of reaction method: imagine the process in three hypothetical steps, since enthalpy does not care about the route. Step one, cool the reactants from their inlet temperature to 25 C. Step two, react completely at 25 C, releasing extent times delta H(rxn). Step three, heat the products from 25 C to their outlet temperature. Sum the three enthalpy changes and set the total equal to Q.

With reactants entering at 25 C, step one is zero. Step two releases 100 mol/s x 802.2 kJ/mol = 80,220 kJ/s, or 80.2 MW, using the lower heating value because the flue gas leaves hot and the water stays as vapor. Step three is whatever it takes to heat 1242.9 mol/s of flue gas to the stack temperature.

Suppose the furnace raises steam and the flue gas leaves at 200 C. Using an average flue gas heat capacity of about 34 J/(mol K) over that modest range, step three costs 1242.9 x 0.034 x (200 - 25) = 1242.9 x 5.95 = 7395 kJ/s, or 7.4 MW carried up the stack. The heat available to the process is therefore 80.2 - 7.4 = 72.8 MW, a thermal efficiency of 90.8 percent on an LHV basis. Lower the stack temperature and efficiency rises, which is exactly what economizers do, and the limit is set by acid dew point corrosion rather than by thermodynamics.

Adiabatic flame temperature

Now ask a different question. What if no heat is removed at all? Then Q = 0, the reaction enthalpy has nowhere to go except into the product gases, and the temperature they reach is the adiabatic flame temperature.

Set the sum of the three steps to zero: 0 = -80,220 + 1242.9 x Cp x (T - 25). Flue gas heat capacity rises with temperature, so use a high-temperature average of about 38 J/(mol K), that is 0.038 kJ/(mol K). Then T - 25 = 80,220 / (1242.9 x 0.038) = 80,220 / 47.23 = 1698 K, giving T = about 1720 C.

Is that right? It is the right order of magnitude and the right lesson: with 20 percent excess air the flame runs some hundreds of degrees below the roughly 1950 C obtained at exactly stoichiometric air, because the extra nitrogen and oxygen must be heated too. Two effects make the true value lower still. Heat capacity keeps climbing with temperature, and above about 1700 C carbon dioxide and water begin to dissociate, absorbing energy. Real measured flame temperatures for methane in air are near 1950 C stoichiometric and lower with excess air, so our estimate is honest but approximate, and a design would use temperature-dependent heat capacities and an equilibrium calculation.

The design consequence is immediate: excess air is a cheap and effective way to control flame temperature, which is how furnace designers keep tube metal below its limit and how nitrogen oxide formation, which is extremely temperature sensitive, is suppressed.

Key idea: With no heat removal, all the reaction enthalpy heats the products, and the resulting adiabatic flame temperature falls as excess air rises because more inert mass must be heated.

Where the heat actually goes: the ethylene oxide reactor

Return to the two-reaction example from Lesson 5, where 100 mol of ethylene fed gave xi1 = 20 mol to ethylene oxide and xi2 = 5 mol to complete combustion. Compute the heat of each reaction from formation data: ethylene +52.4 kJ/mol, ethylene oxide -52.6, carbon dioxide -393.5, water vapor -241.8.

Reaction 1: delta H = (-52.6) - (52.4) = -105.0 kJ/mol. Reaction 2: delta H = [2 x (-393.5) + 2 x (-241.8)] - 52.4 = (-787.0 - 483.6) - 52.4 = -1323.0 kJ/mol.

Total heat released per 100 mol of ethylene fed = 20 x 105.0 + 5 x 1323.0 = 2100 + 6615 = 8715 kJ.

Look hard at that split. The desired reaction, which made all the product, contributed 2100 kJ. The side reaction, which consumed only 5 mol of ethylene and made nothing of value, contributed 6615 kJ, or 76 percent of the total heat load. A selectivity loss of one part in five imposes three quarters of the cooling duty, and if cooling falls behind, the temperature rises, and combustion, which has the higher activation energy, accelerates faster than the desired reaction, destroying selectivity further and releasing yet more heat. That is a runaway waiting for an excuse, and it is why industrial ethylene oxide reactors are thousands of narrow tubes packed with catalyst and immersed in boiling coolant. Chemistry, energy, and safety are the same problem.

Key idea: A small selectivity loss to a strongly exothermic side reaction can dominate the reactor heat load and set up a runaway, which is why selectivity, cooling design, and safety are inseparable.

Simultaneous material and energy balances

So far material balances were solved first and energy balances afterward, because composition was given. Often it is not. In an adiabatic reactor, the extent of reaction determines the temperature rise, while the temperature determines the equilibrium extent. Neither can be found without the other, and the two balances must be solved together.

The standard method is a controlled iteration. Guess an outlet temperature. Use it to obtain the equilibrium or kinetically achievable conversion. Do the material balance for that conversion. Do the energy balance to compute the temperature that conversion would actually produce. Compare with your guess, adjust, and repeat. The loop converges quickly because the two relationships pull in opposite directions for an exothermic reaction: more conversion means more heat means higher temperature means, by Le Chatelier, less equilibrium conversion.

That opposition is not just a numerical convenience; it is why ammonia converters are built as several catalyst beds with cooling between them. Each bed runs until the rising temperature stalls the equilibrium, then the gas is cooled and enters the next bed with its equilibrium restored. The staged converter is a picture of a simultaneous material and energy balance rendered in steel. Process simulators solve these systems automatically now, but they solve exactly the loop described here, and knowing that is what lets you tell a converged answer from a plausible-looking one.

Key idea: When conversion and temperature depend on each other, guess a temperature, close the material balance, close the energy balance, and iterate; staged reactors with interstage cooling are the physical expression of that trade-off.

Common misconceptions

• Exothermic reactions have positive delta H. The opposite. The system loses enthalpy, so delta H is negative; methane combustion is -802.2 kJ/mol on an LHV basis.
• Higher and lower heating values are alternative measurements of the same thing. They are different quantities differing by the latent heat of the product water, 10 percent for methane, and quoting the wrong one misstates every efficiency.
• The adiabatic flame temperature is what a real burner reaches. Real flames lose heat by radiation, and dissociation absorbs energy above about 1700 C, so measured temperatures fall below the simple calculation.
• Adding excess air always improves combustion. It ensures burnout and lowers flame temperature, but every extra mole of nitrogen is heated and sent up the stack, so too much excess air simply wastes fuel.
• Side reactions matter only for yield. They frequently dominate the heat load, as the ethylene case shows, and therefore determine the cooling design and the runaway risk.

Recap

• Standard heat of reaction equals the sum of products' heats of formation minus reactants', by Hess's law, since enthalpy is a state property.
• Methane combustion is -802.2 kJ/mol with water as vapor and -890.2 kJ/mol with water as liquid, giving LHV 50.0 MJ/kg and HHV 55.5 MJ/kg.
• The heat of reaction method routes the calculation through 25 C: cool reactants, react at 25 C, heat products.
• Burning 100 mol/s of methane releases 80.2 MW; with flue gas leaving at 200 C, 7.4 MW goes up the stack and 72.8 MW is available, an LHV efficiency of 90.8 percent.
• With no heat removal the same combustion at 20 percent excess air gives an adiabatic flame temperature of roughly 1720 C by simple calculation, lower than the stoichiometric value because extra air must be heated.
• In the ethylene oxide reactor, the 5 mol of ethylene that burned released 6615 of the 8715 kJ, showing how a small selectivity loss can dominate the cooling duty and the runaway risk.
• When conversion and temperature are coupled, iterate: guess T, close the material balance, close the energy balance, adjust.

Sources

1. National Institute of Standards and Technology. (n.d.). NIST Chemistry WebBook: Thermochemical data. U.S. Department of Commerce. webbook.nist.gov
2. OpenStax. (2019). Hess's law. In Chemistry 2e. Rice University. openstax.org
3. U.S. Energy Information Administration. (n.d.). Units and calculators explained: Heat content of fuels. EIA. eia.gov
4. Wikipedia contributors. (n.d.). Adiabatic flame temperature. Wikipedia. en.wikipedia.org

The big picture

Here is the fact that the entire separations half of chemical engineering rests on. Put a mixture of two liquids in a closed flask, warm it until it just begins to boil, and sample the tiny bubble of vapor above the surface. That vapor is not the same mixture as the liquid. It is richer in whichever component is more volatile. Do nothing clever at all, simply let nature reach equilibrium, and you have performed a separation.

Not a very good one, admittedly. Our benzene-toluene feed is 50 percent benzene as a liquid, and the first vapor off it will be about 71 percent benzene. One step of boiling and condensing takes you from 50 to 71, not to 95. But do it again on the condensate, and again, and again, and each step climbs. A distillation column is a machine for performing that step dozens of times in a single vessel, continuously. Everything about column design comes back to the number 71.

Here is the plan for today. First vapor pressure, the property that makes one component more volatile than another, and the Antoine equation that gives it as a function of temperature. Then Raoult's law and Dalton's law, which together let you compute the composition of the vapor in equilibrium with any liquid. Then two calculations you should be able to do by hand: the bubble point, where a liquid begins to boil, and the dew point, where a vapor begins to condense. Then relative volatility, the single number that summarizes how easy a separation will be. We finish with Henry's law, the version of all this that applies to a dilute gas dissolved in a liquid, which is what absorption columns run on. Engineering Thermodynamics (ENGR 230) covers cycles and the second law; here we want only what process design needs.

Vapor pressure and the Antoine equation

Leave a puddle of any liquid in a closed vessel and molecules escape from the surface until the space above is saturated. The pressure exerted by that saturated vapor is the liquid's vapor pressure, written P star, and it depends only on temperature. It rises steeply with temperature, roughly exponentially, because escape requires molecules to have enough energy to break free of their neighbors and the fraction with that energy climbs sharply. When the vapor pressure reaches the pressure pushing down on the liquid, boiling begins throughout the bulk; that is what a boiling point is.

For calculation, vapor pressure is tabulated as the Antoine equation, log10 P star = A - B / (T + C), with constants fitted for each substance over a stated temperature range. With P in mmHg and T in degrees Celsius, benzene has A = 6.90565, B = 1211.033, C = 220.79, and toluene has A = 6.95464, B = 1344.8, C = 219.482.

Try them at 92.1 C. For benzene: 1211.033 / (92.1 + 220.79) = 1211.033 / 312.89 = 3.8705, so log10 P star = 6.90565 - 3.8705 = 3.0352 and P star = 1084 mmHg. For toluene: 1344.8 / (92.1 + 219.482) = 1344.8 / 311.582 = 4.3160, so log10 P star = 6.95464 - 4.3160 = 2.6386 and P star = 435 mmHg. In kPa, multiplying by 0.13332, these are 144.6 kPa and 58.0 kPa. Benzene, the lighter molecule, exerts about two and a half times the vapor pressure of toluene at the same temperature, and that ratio is the whole reason the separation works.

Key idea: Vapor pressure measures a substance's tendency to escape into the gas phase, rises steeply with temperature, and is computed from the Antoine equation.

Raoult's law and Dalton's law

Now put two liquids together. If the molecules are chemically similar, so that a benzene molecule surrounded by toluene feels much the same as when surrounded by benzene, then each component escapes in proportion to how much of the surface it occupies. That is Raoult's law:

p(i) = x(i) x P star(i)

where p(i) is the partial pressure of component i above the mixture, x(i) is its liquid mole fraction, and P star(i) is its pure-component vapor pressure at that temperature. Meanwhile Dalton's law says the vapor is described by y(i) = p(i) / P, where P is the total pressure. Put them together and you have the working equation of vapor-liquid equilibrium:

y(i) x P = x(i) x P star(i)

Two lines of algebra, and you can compute the composition of vapor in equilibrium with any liquid, at any temperature, for any ideal mixture. Everything in Lesson 14 follows from it.

Worked bubble point

The bubble point is the temperature at which a liquid of given composition, held at a given pressure, forms its first bubble of vapor. The condition is that the partial pressures must add up to the total pressure: sum of x(i) P star(i) = P.

Take our feed: 50 mole percent benzene, 50 mole percent toluene, at 101.3 kPa, which is 760 mmHg. Since vapor pressures depend on temperature and we do not know the temperature, we must iterate. That sounds worse than it is.

Guess 90 C. Benzene: 1211.033 / 310.79 = 3.8967, log10 P star = 3.0090, P star = 1021 mmHg. Toluene: 1344.8 / 309.482 = 4.3454, log10 P star = 2.6092, P star = 407 mmHg. Sum: 0.5 x 1021 + 0.5 x 407 = 510 + 204 = 714 mmHg. Too low, so we must go hotter.

Guess 92 C. Benzene: 1211.033 / 312.79 = 3.8717, log10 P star = 3.0339, P star = 1081 mmHg. Toluene: 1344.8 / 311.482 = 4.3174, log10 P star = 2.6372, P star = 434 mmHg. Sum: 540.5 + 217 = 757.5 mmHg. Very close, still slightly low.

Guess 92.1 C, using the vapor pressures computed above, 1084 and 435 mmHg. Sum: 542 + 217.5 = 759.5 mmHg, which is 760 to within our precision. The bubble point is 92.1 C.

Now the payoff. The vapor composition follows immediately from Dalton: y(benzene) = x P star / P = (0.50 x 1084) / 760 = 542 / 760 = 0.714. The liquid is half benzene; the first vapor off it is 71.4 percent benzene. In kPa the same arithmetic reads 72.3 / 101.3 = 0.714, as it must.

Note also that the bubble point of 92.1 C lies between the boiling points of the pure components, 80.1 C for benzene and 110.6 C for toluene, but not at the midpoint. Mixtures boil over a range, not at a point, which is the next calculation.

Key idea: At the bubble point the partial pressures sum to the total pressure; for our 50/50 feed at 101.3 kPa that gives 92.1 C, and the vapor formed is 71.4 percent benzene.

Worked dew point

The dew point is the mirror image: the temperature at which a vapor of given composition, cooled at constant pressure, forms its first drop of liquid. The condition is that the liquid mole fractions must sum to one: sum of y(i) P / P star(i) = 1.

Take a vapor that is 50 mole percent benzene at 760 mmHg. Guess 98 C: benzene P star = 1279 mmHg, toluene P star = 523 mmHg. Then x(benzene) = 380 / 1279 = 0.297 and x(toluene) = 380 / 523 = 0.726, summing to 1.023. Above one, so we must go hotter. Guess 99 C: P star values 1315 and 540 mmHg, giving 0.289 + 0.704 = 0.993. Just below one. Interpolating gives a dew point of about 98.8 C, at which the first liquid formed is only 29 percent benzene.

So a 50/50 mixture starts boiling at 92.1 C and is not fully vaporized until 98.8 C. Between those temperatures liquid and vapor coexist, with the vapor always richer in benzene and the liquid always richer in toluene. That temperature gap is the boiling range, and it is the region in which every distillation column operates.

Relative volatility

Doing Antoine arithmetic at every stage of a column would be tedious, and for a well-behaved pair there is a shortcut. Define the relative volatility alpha as the ratio of the pure-component vapor pressures:

alpha = P star(light) / P star(heavy) = 1084 / 435 = 2.49, call it 2.5.

Check how well it holds at the other end of the range: at 98.8 C the values are about 1308 and 536, giving alpha = 2.44. Over a seven degree span the relative volatility moved by two percent, which for hand calculation is constant. Assuming constant alpha lets us write the equilibrium relation in closed form, with no temperature in it at all:

y = alpha x / (1 + (alpha - 1) x)

Test it at x = 0.5 with alpha = 2.5: y = 1.25 / (1 + 1.5 x 0.5) = 1.25 / 1.75 = 0.714, matching the bubble point calculation exactly. This one equation carries us through the entire stage-by-stage column calculation in Lesson 14.

Relative volatility is also the single best predictor of how hard a separation will be. Above about 2, separation is easy and a modest column will do. Between 1.2 and 1.5, expect a tall column with a large reflux and a serious energy bill. As alpha approaches 1, the vapor and liquid have the same composition, no separation occurs no matter how many stages you build, and you must find another method. Propylene and propane, with alpha near 1.1, need columns of well over a hundred trays and are among the most energy-intensive separations practiced anywhere.

Key idea: Relative volatility is the ratio of pure-component vapor pressures, roughly constant over a modest range, and it both simplifies the equilibrium equation to y = alpha x / (1 + (alpha - 1) x) and predicts how difficult the separation will be.

Henry's law for dilute solutes

Raoult's law describes a component that makes up much of the liquid. For a sparingly soluble gas dissolved in a liquid, the environment around each solute molecule is entirely solvent, and a different but equally simple proportionality holds: Henry's law, p = k(H) x, where k(H) is a constant for that solute, solvent, and temperature.

Work an example we will reuse. Air above water at 25 C has an oxygen partial pressure of 0.21 atm, and the Henry constant for oxygen in water is about 4.3 x 10^4 atm on a mole fraction basis. Then x = 0.21 / 43,000 = 4.9 x 10^-6. One litre of water contains 55.5 mol, so the dissolved oxygen is 55.5 x 4.9 x 10^-6 = 2.7 x 10^-4 mol, which at 32 g/mol is about 8.7 mg. The measured value at 25 C is 8.3 mg/L, so the law is good to roughly five percent here. Note the constant is large, meaning the gas prefers the gas phase overwhelmingly, which is why fish need gills with enormous surface area and why aerating a fermenter is a serious engineering problem.

Two practical consequences follow. Solubility of a gas is proportional to its partial pressure, which is why a carbonated drink fizzes when opened and why divers must decompress slowly. And solubility of gases falls as temperature rises, opposite to most solids, which is why warm river water holds less oxygen and why thermal discharges are regulated.

Key idea: Henry's law, p = k(H) x, governs dilute dissolved gases, making solubility proportional to partial pressure, and it is the equilibrium relation on which absorption column design rests.

Common misconceptions

• A mixture boils at one temperature. Only pure substances and azeotropes do. Our 50/50 benzene-toluene mixture boils from 92.1 C to 98.8 C, and the compositions change continuously in between.
• The vapor above a boiling mixture has the same composition as the liquid. If it did, distillation would be impossible. The vapor is enriched in the more volatile component: 71.4 percent benzene above a 50 percent liquid.
• Raoult's law applies to everything. It applies well to chemically similar molecules. Ethanol and water, or acetone and chloroform, deviate strongly, which is the subject of the next lesson.
• Relative volatility can be increased by adding more stages. Alpha is a thermodynamic property of the mixture at that temperature and pressure. Stages exploit it; they do not change it. Changing pressure or adding a third component does.
• Henry's law and Raoult's law are rivals. They are the dilute and concentrated limits of the same physical picture, and each is accurate in its own regime.

Recap

• Vapor pressure rises steeply with temperature and is computed from the Antoine equation; at 92.1 C benzene is at 1084 mmHg and toluene at 435 mmHg.
• Raoult's law gives p(i) = x(i) P star(i) and Dalton's law gives y(i) = p(i) / P, which combine into y(i) P = x(i) P star(i).
• Bubble point: sum of x(i) P star(i) = P. For the 50/50 feed at 101.3 kPa this gives 92.1 C, with the first vapor 71.4 percent benzene.
• Dew point: sum of y(i) P / P star(i) = 1. For a 50 percent benzene vapor this gives 98.8 C, with the first liquid 29 percent benzene.
• Relative volatility alpha is about 2.5 for benzene and toluene and nearly constant across the range, giving y = alpha x / (1 + (alpha - 1) x).
• Alpha predicts difficulty: above 2 is easy, near 1.1 requires over a hundred trays, and alpha = 1 makes separation by distillation impossible.
• Henry's law, p = k(H) x, governs dilute dissolved gases and predicts about 8.7 mg/L of oxygen in water in equilibrium with air at 25 C against a measured 8.3 mg/L.

Sources

1. National Institute of Standards and Technology. (n.d.). NIST Chemistry WebBook: Antoine equation parameters. U.S. Department of Commerce. webbook.nist.gov
2. Wikipedia contributors. (n.d.). Raoult's law. Wikipedia. en.wikipedia.org
3. Chemistry LibreTexts. (n.d.). Henry's law. LibreTexts. chem.libretexts.org
4. OpenStax. (2019). Solubility. In Chemistry 2e. Rice University. openstax.org

The big picture

Last lesson gave you equations. Today we turn them into pictures, because separation engineers think in diagrams, and a diagram shows at a glance what a column of numbers hides. Two plots do almost all the work: the temperature-composition diagram, which shows where a mixture boils, and the x-y diagram, which shows how much each equilibrium stage buys you.

Then we break the nice assumption. Raoult's law works beautifully for benzene and toluene, molecules so alike that neither notices the substitution. It fails badly for ethanol and water, and the failure is not a small correction: it produces the azeotrope, a composition at which distillation simply stops working, and which is the reason you cannot buy 100 percent ethanol from a still at any price.

Here is the plan for today. First the two diagrams and the lever rule, with a worked flash calculation on our running mixture. Then non-ideality, activity coefficients, and azeotropes, with the industrial workarounds. Then the harder and more interesting question: what does thermodynamics say is the absolute best any separation could ever do, and how close do real columns get? The answer, which is honest and slightly humbling, is not very close at all.

The temperature-composition diagram

Plot temperature on the vertical axis against benzene mole fraction on the horizontal, at a fixed pressure of 101.3 kPa. The left edge is pure toluene, boiling at 110.6 C; the right edge is pure benzene, boiling at 80.1 C. Two curves connect them. The lower one is the bubble point curve, giving the temperature at which a liquid of each composition starts to boil. The upper one is the dew point curve, giving the temperature at which a vapor of each composition starts to condense. Below both, everything is liquid. Above both, everything is vapor. Between them lies a lens-shaped two-phase region where liquid and vapor coexist.

We have already computed two points on it: for a composition of 0.50 benzene, the bubble point is 92.1 C on the lower curve and the dew point is 98.8 C on the upper. A horizontal line drawn at any temperature inside the lens is a tie line: it meets the bubble curve at the composition of the liquid and the dew curve at the composition of the vapor, and those two phases are the ones actually in equilibrium at that temperature.

The lever rule and a worked flash

The tie line does more than give compositions; it gives amounts. If the overall mixture composition is z and the tie line ends are x (liquid) and y (vapor), then a mass balance on the light component requires z = (V/F) y + (1 - V/F) x, which rearranges to the lever rule:

V / F = (z - x) / (y - x)

The vapor fraction is the distance from the liquid end to the overall composition, divided by the length of the whole tie line. Work it on our mixture.

Heat the 50/50 feed to 95 C at 101.3 kPa, partway into the two-phase lens. This is a flash: one equilibrium stage, and one of the most common units in a plant. From the Antoine equations at 95 C, benzene has P star = 1176 mmHg and toluene 477 mmHg.

Find the liquid composition from the bubble point condition at this temperature: x(1176) + (1 - x)(477) = 760, so 699 x = 283 and x = 0.405. Then the vapor composition from Dalton: y = x P star / P = 0.405 x 1176 / 760 = 0.626.

Now the lever rule: V/F = (0.500 - 0.405) / (0.626 - 0.405) = 0.095 / 0.221 = 0.43. Forty-three percent of the feed vaporizes. Check it: 0.43 x 0.626 + 0.57 x 0.405 = 0.269 + 0.231 = 0.500, the feed composition, as required.

Sanity-check the limits too. At 92.1 C, the bubble point, the liquid composition equals the feed and V/F is zero. At 98.8 C, the dew point, the vapor composition equals the feed and V/F is one. The flash calculation interpolates between those, and the whole two-phase lens is now quantitative.

Key idea: A tie line across the two-phase region gives the compositions of the coexisting phases, and the lever rule turns their positions into the fraction vaporized; one flash on our feed at 95 C vaporizes 43 percent.

The x-y diagram

The second picture drops temperature entirely and plots vapor composition y against liquid composition x. The diagonal line y = x is the do-nothing line: if the equilibrium curve lay on it, vapor and liquid would be identical and no separation would be possible. The real equilibrium curve for benzene and toluene bows above the diagonal, and the vertical distance between curve and diagonal at any x is exactly how much enrichment one equilibrium stage delivers.

Read the curve at x = 0.5 and you find y = 0.714, a gain of 0.214. Read it at x = 0.9 and you find y = 2.5 x 0.9 / (1 + 1.5 x 0.9) = 2.25 / 2.35 = 0.957, a gain of only 0.057. Separation gets harder as you approach purity, which is why the last few percent of a specification often costs more than the first eighty. That single observation explains a great deal about product pricing.

Key idea: On the x-y diagram, the gap between the equilibrium curve and the diagonal is the enrichment one stage gives; it shrinks near both pure ends, which is why high purity is expensive.

When molecules do notice each other

Raoult's law assumed a benzene molecule is indifferent to whether its neighbors are benzene or toluene. Now take ethanol and water. An ethanol molecule in pure ethanol sits among hydrogen-bonding hydroxyl groups; in water it is surrounded by a strongly hydrogen-bonded network that would rather bond to itself than to ethanol's hydrocarbon tail. The result is that ethanol is effectively squeezed out, escaping to the vapor far more readily than its mole fraction would suggest.

We handle this with an activity coefficient gamma, a correction factor multiplying Raoult's law:

p(i) = gamma(i) x(i) P star(i)

An ideal mixture has gamma = 1 everywhere. When unlike molecules dislike each other, gamma is greater than one and the mixture shows a positive deviation: higher vapor pressure than Raoult predicts, and a boiling point lower than expected. When unlike molecules attract each other strongly, as acetone and chloroform do through hydrogen bonding, gamma is less than one, giving a negative deviation and a higher boiling point.

Azeotropes

Push a positive deviation far enough and something remarkable happens. Recall that relative volatility with activity coefficients is alpha = gamma(1) P star(1) / (gamma(2) P star(2)). As ethanol becomes concentrated, its activity coefficient falls toward one while water's rises, and at some composition the whole ratio passes through exactly 1. At that point y = x: the vapor has the same composition as the liquid, and further distillation achieves nothing at all. That composition is an azeotrope.

For ethanol and water at atmospheric pressure the azeotrope sits at about 95.6 percent ethanol by mass (89.5 mole percent) and boils at 78.2 C, below the boiling point of either pure component. This is why rectified spirit tops out near 96 percent and why absolute ethanol cannot be made by ordinary distillation. Negative deviations produce maximum-boiling azeotropes instead: nitric acid and water form one at about 68 percent acid, boiling at 122 C, which is why concentrated nitric acid from a still stops there.

Industry gets around azeotropes in four ways, and all four are in commercial use. Pressure swing exploits the fact that azeotropic composition shifts with pressure, so two columns at different pressures can step past it. Azeotropic distillation adds an entrainer that forms a new, more convenient azeotrope; benzene was used for ethanol historically and cyclohexane later. Extractive distillation adds a high-boiling solvent that changes the activity coefficients without vaporizing, which is how ethylbenzene and styrene are handled. And hybrid processes hand the last step to a different physics entirely: modern fuel ethanol plants distil to about 93 percent and then remove the remaining water with molecular sieve adsorption, which does not care about vapor-liquid equilibrium at all.

Key idea: Activity coefficients correct Raoult's law for non-ideality, and when they drive the relative volatility to exactly one an azeotrope forms, at which distillation stops and a different mechanism must take over.

Which model to use

Real design uses correlations fitted to data: Wilson, NRTL, and UNIQUAC for liquid activity coefficients, UNIFAC for estimating them from molecular groups when data are missing, and equations of state such as Soave-Redlich-Kwong or Peng-Robinson for hydrocarbons and high pressures. Every process simulator asks you to choose one, and choosing badly is among the most common serious errors in industrial simulation: a column designed with an ideal model for a strongly non-ideal system can be wrong by many stages, or can miss an azeotrope entirely and predict a separation that is thermodynamically impossible. The rule of thumb is that hydrocarbons and similar molecules tolerate simple models, while anything with hydrogen bonding, ions, or a large size difference needs a fitted activity coefficient model and real data behind it.

What thermodynamics allows

Now the deeper question. Mixing is spontaneous: pour benzene into toluene and they mix by themselves, increasing entropy. Separation therefore runs uphill and must be paid for. How much, at minimum?

For an ideal binary, the minimum work to unmix one mole of feed completely is W(min) = -R T (x1 ln x1 + x2 ln x2). For our 50/50 feed at 298 K: W(min) = -8.314 x 298 x (0.5 ln 0.5 + 0.5 ln 0.5) = 8.314 x 298 x 0.693 = 1717 J/mol, about 1.7 kJ per mole of feed.

Now compare with reality. In Lesson 14 our column will boil 125 kmol/h of vapor for a feed of 100 kmol/h, at roughly 32 kJ/mol of vaporization enthalpy, so it consumes about 40 kJ of reboiler heat per mole of feed. But heat is not work, and the comparison must be fair: that heat is delivered at about 383 K and rejected at about 298 K, so its work equivalent is at most the Carnot fraction, 1 - 298/383 = 0.222, giving about 8.9 kJ/mol of work equivalent. Against a thermodynamic minimum of 1.7 kJ/mol, our column is roughly 19 percent efficient in second-law terms.

That figure is typical. Distillation is a thermodynamically wasteful separation, and it dominates industrial practice anyway, because it is simple, continuous, scalable, and needs no exotic materials. Commonly cited estimates put separations at a large fraction of industrial energy use, with distillation the biggest single share. It is one of the great standing engineering opportunities: anything that separates at closer to the thermodynamic minimum, such as well-designed membranes or adsorption, is attacking a genuinely enormous prize. Lesson 15 looks at the alternatives.

Equilibrium also sets two more specific limits that Lesson 14 will use. At infinite reflux, when no product is withdrawn and everything is returned, you need the fewest possible stages, given by the Fenske equation; for our column that minimum is about 6.4 stages. At infinite stages, you need the least possible reflux; for our column that minimum reflux ratio is about 1.10. A real column must exceed both, and choosing where to sit between them is the classic trade of capital against energy.

Key idea: Separation costs work because mixing is spontaneous; the minimum for our feed is 1.7 kJ/mol while the column actually spends about 8.9 kJ/mol of work equivalent, roughly 19 percent efficiency, and real designs must exceed both the minimum stages and the minimum reflux.

Common misconceptions

• An azeotrope is a compound. It is not. It is a composition at which the vapor and liquid happen to match, and it shifts with pressure, which is exactly what pressure-swing distillation exploits.
• Adding more trays can beat an azeotrope. At the azeotrope alpha equals one, so a stage achieves nothing. Infinite trays achieve nothing infinitely often.
• Distillation is efficient because it is so widely used. It is used because it is robust and cheap to build, not because it is thermodynamically good; our worked column runs near 19 percent second-law efficiency.
• The lever rule needs equal molar flows or ideal behavior. It is just a mass balance on a tie line and holds whenever the two end compositions are the phases in equilibrium.
• Any simulator will get the answer right if the input data are correct. The choice of thermodynamic model is itself input, and an ideal model on a hydrogen-bonding system can predict separations that cannot physically occur.

Recap

• The temperature-composition diagram has a bubble curve, a dew curve, and a two-phase lens between them crossed by tie lines.
• The lever rule gives V/F = (z - x)/(y - x); flashing our 50/50 feed at 95 C gives x = 0.405, y = 0.626, and 43 percent vaporized.
• On the x-y diagram the gap between the equilibrium curve and the diagonal is the enrichment per stage: 0.214 at x = 0.5 but only 0.057 at x = 0.9.
• Activity coefficients correct Raoult's law; positive deviations give minimum-boiling azeotropes such as ethanol and water at 95.6 percent by mass and 78.2 C.
• Azeotropes are beaten by pressure swing, entrainers, extractive distillation, or a different physics such as molecular sieve adsorption.
• Minimum separation work for our feed is 1.7 kJ/mol, against roughly 8.9 kJ/mol of work equivalent actually spent, so distillation runs near 19 percent second-law efficiency.
• Equilibrium also fixes a minimum stage count (about 6.4 here) and a minimum reflux ratio (about 1.10), and real columns must exceed both.

Sources

1. Wikipedia contributors. (n.d.). Azeotrope. Wikipedia. en.wikipedia.org
2. Britannica. (n.d.). Distillation. Encyclopaedia Britannica. britannica.com
3. Chemistry LibreTexts. (n.d.). Raoult's law and ideal mixtures of liquids. LibreTexts. chem.libretexts.org
4. U.S. Department of Energy. (n.d.). Industrial process heat and separations. Office of Energy Efficiency and Renewable Energy. energy.gov

The big picture

Before anything can react, be heated, or be separated, it has to get there. Moving fluid through pipe is the most ordinary thing a chemical plant does and it consumes a startling share of the plant's electricity. Pumps and compressors are among the largest single categories of industrial motor load, so a design decision as dull as pipe diameter shows up on the electricity bill for the next thirty years.

It is also the part of the subject where a small number does something dramatic. Pressure drop in a pipe scales as roughly the fifth power of the inverse diameter. Choose a pipe one size too small and you may need a pump three times larger. Choose a suction line badly and the pump will cavitate, sounding like it is pumping gravel, and destroy its own impeller within weeks. Both failures are entirely predictable from equations you can work by hand.

Here is the plan for today. First viscosity and the Reynolds number, and the distinction between laminar and turbulent flow that governs everything else. Then the mechanical energy balance and the Darcy-Weisbach equation, worked completely on a real pipe. Then fittings, and how pressure drop scales with diameter. Then pumps: how they are characterized, how the operating point is set, and the cavitation calculation that catches more design errors than any other single check.

Viscosity, Reynolds number, and two kinds of flow

Viscosity is a fluid's resistance to shearing. Formally, for a Newtonian fluid, shear stress equals viscosity times velocity gradient. Water at 25 C has a viscosity of about 0.00089 Pa s, air about 0.000018 Pa s, and heavy oils a thousand times more than water. Many industrial fluids are not Newtonian at all: polymer melts, slurries, ketchup, and blood have viscosities that depend on how hard you shear them, which is a substantial subject of its own.

What matters most is not viscosity alone but its balance against inertia, and that ratio is the Reynolds number we met in Lesson 2: Re = rho v D / mu. Below about 2100, viscous forces dominate and the flow is laminar, moving in orderly parallel layers with a parabolic velocity profile. Above about 4000 it is turbulent, full of chaotic eddies, with a much flatter velocity profile. Between the two lies a transition region nobody designs in deliberately.

The distinction is not academic. Turbulent flow mixes vigorously, which makes heat and mass transfer far better, at the price of much higher friction. Laminar flow is quiet and cheap to push but transfers heat poorly. Almost all industrial pipe flow of low-viscosity liquids is turbulent; flow of heavy oils and polymers is often laminar, and their heat exchangers must be designed accordingly.

Key idea: Reynolds number compares inertia with viscosity; below about 2100 flow is laminar and orderly, above about 4000 it is turbulent, which greatly improves mixing and transfer while raising friction.

The mechanical energy balance and friction

For flow between two points in a piping system, the mechanical energy balance, essentially Bernoulli's equation with a friction term added, says that pressure energy, kinetic energy, and potential energy can trade among themselves, plus whatever a pump adds, minus what friction destroys. In head units, metres of fluid:

P1/(rho g) + v1^2/(2g) + z1 + h(pump) = P2/(rho g) + v2^2/(2g) + z2 + h(friction)

Friction head is given by the Darcy-Weisbach equation:

h(f) = f x (L/D) x v^2 / (2g)

where f is the Darcy friction factor, L the pipe length, D the inside diameter, and v the average velocity. For laminar flow, f = 64/Re exactly. For turbulent flow, f depends on Reynolds number and on relative roughness, the ratio of surface roughness to diameter, through the implicit Colebrook equation, whose graphical form is the famous Moody chart. Commercial steel has a roughness of about 0.045 mm; drawn tubing is smoother, concrete much rougher.

A pipe worked from end to end

Pump 0.020 m^3/s of water at 25 C through 100 m of 4 inch schedule 40 steel pipe, whose inside diameter is 0.1023 m. Water density is 997 kg/m^3 and viscosity 0.00089 Pa s.

Velocity. Cross-sectional area A = pi D^2 / 4 = 3.1416 x 0.1023^2 / 4 = 0.008219 m^2. So v = Q/A = 0.020 / 0.008219 = 2.43 m/s, comfortably inside the usual 1 to 3 m/s rule of thumb for liquid lines.

Reynolds number. Re = rho v D / mu = 997 x 2.43 x 0.1023 / 0.00089 = 248 / 0.00089 = 2.8 x 10^5. Firmly turbulent.

Friction factor. Relative roughness = 0.000045 / 0.1023 = 0.00044. Reading the Moody chart, or solving the Colebrook equation, gives f = 0.018.

Friction head. h(f) = 0.018 x (100 / 0.1023) x (2.43^2) / (2 x 9.81) = 0.018 x 977.5 x 5.92 / 19.62 = 0.018 x 977.5 x 0.3017 = 5.31 m of water.

Pressure drop. delta P = rho g h(f) = 997 x 9.81 x 5.31 = 51,900 Pa, about 52 kPa.

Power. Hydraulic power = Q x delta P = 0.020 x 51,900 = 1038 W, about 1.04 kW. A centrifugal pump at 70 percent efficiency therefore draws 1.04 / 0.70 = 1.48 kW at the shaft, and rather more at the motor terminals.

Fittings, and why the pipe is the easy part

Real lines are not bare pipe. Every elbow, tee, valve, and sudden change of area destroys energy, and these losses are handled with resistance coefficients: h = K v^2 / (2g). Typical values are K = 0.75 for a standard 90 degree elbow, 0.17 for a fully open gate valve, and 6 or more for a fully open globe valve, which is why globe valves are used for throttling and gate valves for isolation.

Add to our line four elbows and one globe valve: K(total) = 4 x 0.75 + 6 = 9.0. The velocity head v^2/(2g) is 5.92 / 19.62 = 0.302 m, so fitting losses are 9.0 x 0.302 = 2.72 m, on top of the 5.31 m of straight pipe. Total friction head is 8.03 m, so the fittings added more than half again as much as 100 metres of pipe. Beginners size the pipe carefully and forget the valves; the valves often win.

Key idea: Frictional head loss is f (L/D) v^2/(2g) for pipe and K v^2/(2g) for each fitting, and in a compact plant the fittings can dominate the straight pipe.

Why diameter matters so much

Hold the flow rate fixed and change the pipe size. Velocity varies as 1/D^2, so v^2 varies as 1/D^4, and the L/D term contributes another 1/D. Pressure drop therefore scales as roughly 1/D^5.

Double our pipe from 0.1023 m to 0.2045 m and the pressure drop falls by a factor of about 2^5 = 32, from 52 kPa to under 2 kPa, and the pumping power falls with it. The steel, however, costs perhaps twice as much per metre, plus larger valves, flanges, and supports.

That is the classic economic pipe size problem: capital cost rises with diameter while operating cost falls steeply. The optimum for ordinary liquid service lands near 1 to 3 m/s, and for gases nearer 15 to 30 m/s because gas density is low and the pressure drop penalty per unit velocity is smaller. Those velocity rules of thumb are not arbitrary tradition; they are the memorized answer to an optimization you have just done in one line.

Key idea: Pressure drop scales as about the inverse fifth power of diameter, so one pipe size up cuts pumping cost dramatically while raising capital cost, and the standard velocity rules of thumb encode that optimum.

Pumps and the operating point

Two families of pump cover nearly everything. Centrifugal pumps spin a fluid outward and convert velocity into pressure. They are simple, cheap, have no valves, handle large flows smoothly, and dominate industrial service. Their head falls as flow rises, so they are described by a pump curve of head against flow. Positive displacement pumps trap a fixed volume and push it along, delivering nearly constant flow regardless of pressure. They handle viscous fluids and high pressures and give accurate metering, but they must never be run against a closed discharge, since the pressure will rise until something bursts, which is why they always have a relief valve.

The flow a centrifugal pump actually delivers is set by intersecting its pump curve with the system curve, which is the head the piping demands, equal to static lift plus friction that grows roughly as the square of flow. Where the two cross is the operating point. Throttling a valve steepens the system curve and moves the point to lower flow, which works but wastes energy as heat in the valve; a variable speed drive lowers the pump curve instead and saves that energy, which is why variable speed drives have spread so widely.

Cavitation and net positive suction head

Here is the calculation that saves pumps. If the pressure anywhere at the pump inlet falls below the liquid's vapor pressure, the liquid boils, forming bubbles that are carried into the high pressure region of the impeller and collapse violently. This is cavitation, it sounds like pumping gravel, it erodes metal, and it will wreck an impeller in weeks.

The guard is net positive suction head. The available value is NPSH(a) = (P(atmospheric) - P(vapor)) / (rho g) + z - h(friction, suction), where z is the liquid level height above the pump (negative for a lift). The pump manufacturer states a required value, NPSH(r), and the design rule is that available must exceed required with a margin, commonly a metre or more.

Work it for water at 25 C, where vapor pressure is 3.17 kPa. NPSH(a) = (101,300 - 3,170) / (997 x 9.81) = 98,130 / 9781 = 10.03 m. Suppose the pump sits 3 m above the liquid level and the suction line costs 1 m of friction: NPSH(a) = 10.03 - 3 - 1 = 6.0 m. Against a required 4 m, that is a comfortable 2 m margin.

Now pump the same water at 80 C, where vapor pressure has risen to 47.4 kPa. NPSH(a) = (101,300 - 47,400) / 9781 = 53,900 / 9781 = 5.51 m, minus 3 m of lift and 1 m of friction gives 1.5 m, well below the 4 m required. This pump will cavitate. Nothing changed except the temperature.

This is why hot liquids, boiling liquids, and anything drawn from a vacuum column are pumped from vessels mounted high above the pump, giving a flooded suction, and why the distance between a column sump and its bottoms pump is a real design constraint rather than a matter of convenience.

Key idea: A pump cavitates when inlet pressure falls to the liquid's vapor pressure; available net positive suction head must exceed the manufacturer's requirement, and it collapses as the liquid gets hotter.

Common misconceptions

• Pressure drop is proportional to velocity. In turbulent flow it goes roughly as velocity squared, and at fixed flow rate as the inverse fifth power of diameter.
• Fittings are a minor correction. In our worked line, four elbows and one globe valve added 2.72 m against 5.31 m for 100 metres of pipe.
• A bigger pump fixes cavitation. Cavitation is a suction-side problem. Raise the liquid level, lower the pump, cool the liquid, or reduce suction line friction; a bigger pump usually requires more net positive suction head, not less.
• Throttling a valve to reduce flow saves energy. It reduces flow, but the pump still works against a steeper system curve and the excess energy is destroyed in the valve; a variable speed drive saves it.
• Turbulence should be avoided. It costs pumping power but is enormously beneficial for mixing and for heat and mass transfer, which is why exchangers are deliberately designed for turbulent flow.

Recap

• Reynolds number classifies flow: laminar below about 2100, turbulent above about 4000, with turbulence improving transfer and increasing friction.
• The mechanical energy balance trades pressure, velocity, and elevation head against pump head and friction head.
• Worked line: 0.020 m^3/s of water in 4 inch pipe gives v = 2.43 m/s, Re = 2.8 x 10^5, f = 0.018, h(f) = 5.31 m, delta P = 52 kPa, hydraulic power 1.04 kW, shaft power 1.48 kW at 70 percent efficiency.
• Four elbows and a globe valve add 2.72 m of head, more than half the straight-pipe loss.
• Pressure drop scales as about 1/D^5, so doubling diameter cuts it by roughly 32 times; velocity rules of thumb of 1 to 3 m/s for liquids encode the economic optimum.
• Centrifugal pumps operate where their curve crosses the system curve; positive displacement pumps need relief valves.
• Available net positive suction head fell from 6.0 m to 1.5 m simply by heating the water from 25 C to 80 C, which would cavitate a pump requiring 4 m.

Sources

1. OpenStax. (2016). Bernoulli's equation. In University physics volume 1. Rice University. openstax.org
2. OpenStax. (2016). Viscosity and turbulence. In University physics volume 1. Rice University. openstax.org
3. Wikipedia contributors. (n.d.). Darcy-Weisbach equation. Wikipedia. en.wikipedia.org
4. U.S. Department of Energy. (n.d.). Pumping systems. Office of Energy Efficiency and Renewable Energy. energy.gov

The big picture

Module 3 told you how much heat must move. It said nothing about how fast, or through how much steel. That is the difference between a balance and a rate equation, and it is the difference between knowing a duty is 312.5 kW and knowing you must buy a 19 square metre exchanger. Today we get the rate equations, first for heat and then for mass, and you will find they are the same equation twice.

That repetition is not a coincidence. Heat flows down a temperature gradient, mass flows down a concentration gradient, and momentum flows down a velocity gradient, all with the same mathematical form: flux equals a conductivity times a gradient. Recognizing this is what lets a single chapter of theory serve exchangers, dryers, absorbers, and reactors alike.

Here is the plan for today. First the three mechanisms of heat transfer and the numbers that go with them. Then the overall heat transfer coefficient, which is a statement about resistances in series and about which one is actually limiting. Then the log mean temperature difference and a full exchanger sizing, including the counter-current versus co-current comparison that is the single most valuable thing to know about exchangers. Then mass transfer: diffusion, two-film theory, and a worked rate, with the analogy to heat made explicit.

Three ways heat moves

Conduction is transfer through a stationary material by molecular contact, described by Fourier's law: q = k A dT / dx, where k is the thermal conductivity. Copper is about 400 W/(m K), carbon steel 45, stainless steel 16, water 0.6, mineral wool insulation 0.04, and still air 0.026. That six-order range is why heat exchangers are made of metal and insulation is made of trapped gas.

Put a number on it. A 10 mm carbon steel wall with a 20 degree difference across it passes q/A = 45 x 20 / 0.010 = 90,000 W/m^2, or 90 kW per square metre. Compare that with the few kW/m^2 an exchanger actually achieves and the lesson is immediate: the metal wall is almost never the limiting resistance. What limits transfer is the fluid clinging to each side of it.

Convection is transfer between a surface and a moving fluid, written as Newton's law of cooling: q = h A (T(surface) - T(fluid)), where h is the film or convective heat transfer coefficient. Unlike k, h is not a material property; it depends on velocity, geometry, and fluid properties, and it is obtained from correlations in dimensionless groups, typically Nusselt number as a function of Reynolds and Prandtl numbers. Typical magnitudes in W/(m^2 K): natural convection in gases 5 to 25, forced convection in gases 25 to 250, forced convection in liquids 100 to 15,000, and boiling or condensing 2500 to 100,000.

Radiation is transfer by electromagnetic waves, needing no medium, and it goes as the fourth power of absolute temperature through the Stefan-Boltzmann law. Below about 400 C it is usually a minor term in process equipment, but in a fired furnace it is the dominant mode, which is why furnace design is a distinct specialty.

Key idea: Conduction through metal is fast, convection through the fluid films is slow, and radiation matters mainly in fired equipment, so exchanger performance is set by the fluid films rather than by the wall.

The overall heat transfer coefficient

In a real exchanger, heat passes through a fluid film, then the wall, then another film. Resistances in series add, exactly as in an electrical circuit:

1/U = 1/h(inside) + x/k + 1/h(outside) + fouling resistances

Work an example. Cooling water inside gives h = 3000 W/(m^2 K); an organic liquid outside gives h = 800; the wall is 2 mm of steel, so x/k = 0.002/45 = 0.0000444.

1/U = 0.000333 + 0.0000444 + 0.00125 = 0.001628, giving U = 614 W/(m^2 K).

Look at the three terms. The organic film contributes 0.00125, which is 77 percent of the total resistance. The water film contributes 20 percent. The steel wall contributes under 3 percent. If you want a better exchanger, improving the organic side is the only move that matters, and polishing the steel or making the wall thinner is a waste of effort. That principle, find the controlling resistance and attack it, applies far beyond heat transfer.

Now add fouling: scale, corrosion products, biological growth, and polymer deposits build up on both surfaces during service. Typical fouling resistances of 0.0002 m^2 K/W on each side add 0.0004 to the total, giving 1/U = 0.002028 and U = 493 W/(m^2 K), a 20 percent loss. Exchangers are therefore designed with fouling allowances and cleaned on a schedule, and a fouled exchanger that no longer meets its duty is one of the most common causes of a plant being unable to make rate.

Our design value of U = 500 W/(m^2 K) in the next section is exactly this fouled figure, which is what a designer would actually use.

Key idea: Overall resistance is the sum of film, wall, and fouling resistances, the largest one controls, and fouling can easily cost twenty percent of performance.

Log mean temperature difference and a worked sizing

The rate equation for an exchanger is Q = U A dT, but which dT? The temperature difference between the streams changes continuously along the exchanger. The correct average, for constant U and constant heat capacities, is the log mean temperature difference:

dT(lm) = (dT1 - dT2) / ln(dT1 / dT2)

where dT1 and dT2 are the differences at the two ends.

Take our exchanger from Lesson 7: cool 10,000 kg/h of an organic from 90 C to 45 C, duty 312.5 kW, with cooling water going from 25 C to 40 C.

Counter-current, meaning the streams flow in opposite directions. At the hot end the organic enters at 90 C and meets water leaving at 40 C, so dT1 = 50 C. At the cold end the organic leaves at 45 C and meets water entering at 25 C, so dT2 = 20 C. Then dT(lm) = (50 - 20) / ln(50/20) = 30 / 0.9163 = 32.7 C. Area A = Q / (U dT(lm)) = 312,500 / (500 x 32.7) = 312,500 / 16,370 = 19.1 m^2.

Co-current, meaning both streams enter at the same end. Now the hot end has the 90 C organic meeting 25 C water, so dT1 = 65 C, and at the far end the 45 C organic meets 40 C water, so dT2 = 5 C. Then dT(lm) = (65 - 5) / ln(13) = 60 / 2.565 = 23.4 C, and A = 312,500 / (500 x 23.4) = 26.7 m^2.

Arrangement	dT1	dT2	LMTD	Area required
Counter-current	50 C	20 C	32.7 C	19.1 m^2
Co-current	65 C	5 C	23.4 C	26.7 m^2

Counter-current needs 40 percent less area for identical duty and identical terminal temperatures. That is why essentially every process exchanger is counter-current. There is a second and sharper reason: co-current flow can never heat the cold stream above the hot stream's outlet temperature, because they approach each other and meet. Counter-current can, an arrangement called a temperature cross, and it is often exactly what a heat integration scheme requires.

Two practical caveats. Shell-and-tube exchangers with multiple tube passes are partly co-current, so their LMTD is multiplied by a correction factor F, typically 0.8 to 1.0; a design with F below about 0.75 is usually redone with more shells. And when one side is boiling or condensing, its temperature is constant, so the arrangement hardly matters.

Key idea: Exchangers are sized by Q = U A dT(lm), and counter-current flow gives a larger log mean difference, less area, and the possibility of a temperature cross, which is why it is nearly universal.

Exchanger types in one paragraph

The double pipe exchanger is one pipe inside another, cheap and used for small duties. The shell and tube exchanger, a bundle of tubes inside a cylindrical shell, is the industrial workhorse: robust, high pressure capable, and mechanically cleanable. The plate exchanger stacks corrugated plates, giving very high coefficients and compactness with easy cleaning, and it dominates food and dairy service, though gaskets limit temperature and pressure. Air-cooled exchangers use finned tubes and fans where cooling water is scarce. And a fired heater is used when process temperatures exceed what steam can supply.

Mass transfer and the analogy

Now change one word. Instead of heat flowing down a temperature gradient, consider a species flowing down a concentration gradient. Fick's law says J = -D dC/dx, where J is molar flux and D the diffusivity. The parallel with Fourier's law is exact, and it extends further: Newton's law of viscosity gives momentum flux down a velocity gradient. Three transports, one mathematical form.

Transport	Law	Driving force	Property	Dimensionless groups
Momentum	Newton	Velocity gradient	Kinematic viscosity	Reynolds
Heat	Fourier	Temperature gradient	Thermal diffusivity	Nusselt, Prandtl
Mass	Fick	Concentration gradient	Diffusivity	Sherwood, Schmidt

The numbers, however, are humbling. Diffusivities in gases are around 10^-5 m^2/s, while in liquids they are around 10^-9 m^2/s, ten thousand times smaller. Molecular diffusion through a liquid is desperately slow: it would take hours for a dye to spread across a beaker unaided. Every practical liquid process therefore relies on bulk motion, stirring, and turbulence to carry material close to where it is needed, leaving only a thin film for diffusion to cross.

That picture is formalized as two-film theory. At a gas-liquid interface, the two bulk phases are assumed well mixed and at equilibrium right at the interface, with all the resistance concentrated in a thin stagnant film on each side. The rate is then written N = k (C(bulk) - C(interface)) with a mass transfer coefficient k for each film, and the two resistances add in series exactly as heat film resistances did. Whichever film has the larger resistance controls, and a soluble gas like ammonia in water is gas-film controlled while a sparingly soluble one like oxygen is liquid-film controlled.

A worked mass transfer rate

Aerate a 10 m^3 fermenter. The liquid-side coefficient is k(L) = 2 x 10^-5 m/s and the sparger produces an interfacial area of a = 50 m^2 per m^3 of liquid. The saturation concentration of oxygen in water in equilibrium with air at 25 C is C star = 8.3 g/m^3, from the Henry's law calculation in Lesson 9, and the organisms hold the bulk concentration down to C = 2.0 g/m^3.

Rate = k(L) x a x V x (C star - C) = (2 x 10^-5) x 50 x 10 x (8.3 - 2.0) = 0.01 x 6.3 = 0.063 g/s, or 227 g of oxygen per hour.

Notice that the coefficient and the area appear only as their product, k(L)a, which is what is actually measured and reported for a fermenter. That is a general lesson: in gas-liquid contacting, creating interfacial area is usually easier and more effective than improving the coefficient, which is why absorbers are filled with packing, fermenters have spargers and impellers, and distillation trays are designed to make bubbles. Notice also the small driving force: the maximum possible is 8.3 g/m^3, so no amount of engineering can push oxygen in faster than that gradient allows, and oxygen transfer is the classic limit on how dense an aerobic culture can be grown.

Key idea: Mass transfer obeys the same flux-equals-coefficient-times-driving-force form as heat, resistances in the gas and liquid films add, and because interfacial area and coefficient appear as the product k(L)a, creating area is usually the effective design lever.

Common misconceptions

• A thicker or thinner exchanger wall changes performance much. The metal contributes under 3 percent of the resistance in our worked case; the fluid films dominate.
• The overall coefficient is an average of the two film coefficients. It is a harmonic combination of resistances, so it is always smaller than the smaller of the two coefficients.
• Co-current and counter-current differ only in plumbing. Counter-current gave 19.1 m^2 against 26.7 m^2 for the same duty, and only counter-current permits a temperature cross.
• Diffusion will mix a liquid tank given time. Liquid diffusivities near 10^-9 m^2/s make unaided mixing hopeless at industrial scale; bulk flow does the work and diffusion only crosses a thin film.
• Fouling is a maintenance problem, not a design problem. It is designed for explicitly with fouling resistances, and ignoring it produced a 20 percent optimistic coefficient in our example.

Recap

• Conduction follows Fourier's law with k from 400 W/(m K) for copper to 0.026 for air; convection follows q = h A dT with h from 5 to 100,000 W/(m^2 K); radiation matters mainly above about 400 C.
• Resistances add: 1/U = 1/h(in) + x/k + 1/h(out) + fouling, giving U = 614 W/(m^2 K) clean and 493 fouled in the worked case, with the organic film controlling.
• dT(lm) = (dT1 - dT2)/ln(dT1/dT2); for our exchanger it is 32.7 C counter-current and 23.4 C co-current.
• Areas are 19.1 m^2 counter-current against 26.7 m^2 co-current for the same 312.5 kW duty, and only counter-current allows a temperature cross.
• Heat, mass, and momentum transfer share one mathematical form, with Fick's law paralleling Fourier's.
• Liquid diffusivities near 10^-9 m^2/s are ten thousand times smaller than gas values, so bulk motion does the transport and films do the diffusing.
• Aerating a 10 m^3 fermenter with k(L)a = 0.001 per second and a 6.3 g/m^3 driving force transfers 227 g of oxygen per hour.

Sources

1. OpenStax. (2016). Mechanisms of heat transfer. In University physics volume 2. Rice University. openstax.org
2. Wikipedia contributors. (n.d.). Logarithmic mean temperature difference. Wikipedia. en.wikipedia.org
3. Engineering LibreTexts. (n.d.). Heat and mass transfer. LibreTexts Chemical Engineering. eng.libretexts.org
4. American Institute of Chemical Engineers. (n.d.). Heat transfer resources. AIChE. aiche.org

The big picture

Everything else in a plant exists to serve the reactor. Feed preparation gets material to it, separations clean up what comes out, and recycle sends back what it failed to convert. Get the reactor wrong and no amount of downstream cleverness rescues the process. Chemical reaction engineering is the part of the discipline that decides what kind of vessel, how big, at what temperature, and with what catalyst.

The subject begins with a distinction worth stating plainly. Thermodynamics tells you whether and how far; kinetics tells you how fast. Diamond is thermodynamically unstable with respect to graphite at room conditions, and converts at a rate so slow that engagement rings are safe for geological time. Ammonia synthesis is thermodynamically favorable at moderate temperature and, without a catalyst, so slow at that temperature as to be worthless. Thermodynamics sets the ceiling; kinetics decides whether you can reach it in a vessel you can afford.

Here is the plan for today. First rate laws and the Arrhenius temperature dependence, with a worked number for how much a ten degree rise buys you. Then the three ideal reactors, batch, stirred tank, and plug flow, with their design equations derived from balances you already know, and a full sizing comparison for the same duty. Then why anyone would choose the larger reactor anyway, and how stirred tanks in series close the gap. Then catalysis, and finally how reactor choice controls selectivity, which is where the last three modules come together.

Rate laws

The rate of reaction of a species A, written -r(A), is the moles of A consumed per unit volume per unit time, with SI units of mol/(m^3 s). The negative sign is convention so that consumption rates come out positive. For most reactions the rate depends on concentration through a rate law, commonly of the power-law form -r(A) = k C(A)^n, where n is the reaction order and k the rate constant.

Order is an experimental fact, not a consequence of the balanced equation. Only for elementary reactions, those that occur in a single molecular event, do the orders match the stoichiometric coefficients. Most industrial reactions proceed through several steps and have orders that may be fractional, may be negative in a product, and may change with conditions. The units of k follow from the order: for first order, per second; for second order, m^3/(mol s).

Temperature: the Arrhenius equation

Rate constants depend strongly on temperature through the Arrhenius equation, k = A exp(-Ea / (R T)), where Ea is the activation energy, R the gas constant, and A the pre-exponential factor. Physically, only molecules with energy above Ea can react, and the fraction that do rises exponentially with temperature.

Work a number. For a typical activation energy of 80 kJ/mol, what does raising the temperature from 300 K to 310 K do? The ratio is exp[(Ea/R)(1/T1 - 1/T2)] = exp[(80,000/8.314) x (1/300 - 1/310)] = exp[9622 x 1.075 x 10^-4] = exp(1.035) = 2.8. A ten degree rise nearly triples the rate. This is the origin of the old rule of thumb that reaction rates roughly double for every ten degrees, which corresponds to an activation energy near 53 kJ/mol.

Two consequences follow. Positively, raising temperature is by far the cheapest way to speed up a reaction. Negatively, this exponential sensitivity is the engine of thermal runaway: an exothermic reaction that heats itself by ten degrees roughly triples its own heat release, which heats it further. Cooling capacity, meanwhile, grows only linearly with temperature difference. When an exponential races a straight line, the exponential wins, and that is the whole mechanism behind the accidents in Lesson 16.

Key idea: Rate constants follow k = A exp(-Ea/RT), so a ten degree rise typically doubles to triples the rate; heat generation is exponential in temperature while heat removal is only linear, which is why runaways happen.

Three ideal reactors

Each ideal reactor is just the general balance equation applied with a particular assumption about mixing.

A batch reactor is charged, sealed, and run. No flow in or out, so accumulation equals generation: V dC(A)/dt = r(A) V. For a first-order reaction this integrates to C(A) = C(A0) exp(-k t), or in terms of conversion, t = -ln(1 - X) / k.

A continuous stirred tank reactor (CSTR) is perfectly mixed, so the composition everywhere inside equals the composition at the outlet. That is the crucial and slightly shocking assumption: the entire vessel sits at the exit concentration. At steady state, in minus out equals consumption, giving tau = C(A0) X / (-r(A)), where tau = V / v0 is the space time. For first order this becomes tau = X / (k (1 - X)).

A plug flow reactor (PFR) is a tube in which fluid moves as coherent plugs with no mixing along the flow direction. Each plug behaves exactly like a small batch reactor travelling down the pipe, so the design equation integrates the same way: tau = -ln(1 - X) / k for first order. Note that the PFR space time and the batch reaction time are identical expressions, which is the deepest thing to notice in this lesson.

A sizing comparison worked out

Take a first-order liquid-phase reaction with k = 0.05 per minute, a feed of 100 litres per minute, and a target conversion of 90 percent.

CSTR. tau = X / (k (1 - X)) = 0.90 / (0.05 x 0.10) = 0.90 / 0.005 = 180 minutes. Volume V = tau x v0 = 180 x 100 = 18,000 L = 18.0 m^3.

PFR. tau = -ln(1 - 0.90) / 0.05 = ln(10) / 0.05 = 2.303 / 0.05 = 46.1 minutes. Volume = 46.1 x 100 = 4610 L = 4.61 m^3.

The stirred tank needs 3.9 times the volume for identical duty. Now repeat at 50 percent conversion: CSTR tau = 0.50/(0.05 x 0.50) = 20 min, V = 2.00 m^3; PFR tau = ln(2)/0.05 = 13.9 min, V = 1.39 m^3, a ratio of only 1.44.

Conversion	CSTR volume	PFR volume	Volume ratio
50 percent	2.00 m^3	1.39 m^3	1.44
90 percent	18.0 m^3	4.61 m^3	3.90

The reason is the perfect mixing assumption. In a CSTR the whole vessel sits at the exit concentration, which is the lowest concentration in the process, and therefore reacts at the slowest possible rate. In a PFR the concentration falls gradually from inlet to outlet, so most of the reactor runs at a higher concentration and a faster rate. The penalty grows with conversion because the exit concentration keeps falling.

Key idea: A CSTR operates entirely at its outlet concentration and therefore at its slowest rate, so it needs more volume than a plug flow reactor for the same conversion, and the penalty grows sharply as conversion rises.

So why build a stirred tank?

Because volume is not the only cost. Stirred tanks are cheap to build, easy to clean, and simple to instrument. They handle slurries and solids that would plug a tube. Above all they are superb at temperature control: the contents are uniform, a jacket or internal coil sees the whole volume, and there is no hot spot to find. A plug flow reactor running an exothermic reaction develops a temperature peak somewhere down its length that must be located and cooled, which is why industrial fixed-bed reactors are built as thousands of narrow tubes surrounded by boiling coolant.

There is also a middle path. Put stirred tanks in series and the concentration steps down in stages, approaching the smooth profile of a plug flow reactor. For our first-order case with three equal tanks reaching 90 percent overall, each needs a space time of 23.1 minutes, giving 69.3 minutes total and a combined volume of 6.93 m^3, against 18.0 for a single tank and 4.61 for a plug flow reactor. Three tanks recover most of the difference, and this is exactly why continuous polymerization and fermentation trains are built as cascades.

Catalysis

A catalyst increases the rate of a reaction by providing a lower-energy pathway, without being consumed. Lower Ea in the Arrhenius equation means an enormously larger k at the same temperature, which usually means the reaction can be run cooler, which for an exothermic equilibrium reaction also means a better equilibrium yield. Something close to nine in ten industrial chemical processes involve a catalyst somewhere.

State the limit clearly, because it is regularly forgotten: a catalyst cannot change the equilibrium position. It lowers the barrier equally in both directions, so it speeds the approach to equilibrium without moving it. Any claim that a catalyst improves yield beyond equilibrium is wrong. What a catalyst can do is improve selectivity, by accelerating one pathway more than another, and in practice that is often its most valuable property.

Homogeneous catalysts are dissolved in the same phase as the reactants, giving excellent contact and selectivity but posing the problem of separating a dissolved catalyst from the product. Heterogeneous catalysts are solids in contact with fluid reactants, usually a metal dispersed on a porous support, and dominate industry because separation is trivial: the fluid flows past and the catalyst stays.

A heterogeneous catalytic reaction has seven steps in series: diffusion of reactant from the bulk to the particle surface, diffusion into the pores, adsorption on an active site, reaction on the surface, desorption, diffusion out of the pores, and diffusion back into the bulk. Any of these can be the slowest step. When pore diffusion limits, only the outer shell of each pellet is doing useful work, described by an effectiveness factor below one, and the cure is smaller pellets or thinner catalyst layers, traded against the pressure drop that smaller pellets cause.

Catalysts also die, in three ways. Sintering is the loss of active surface as metal crystallites grow at high temperature. Coking is the deposition of carbon that blocks pores, usually reversible by burning it off. Poisoning is the chemisorption of an impurity onto active sites, often irreversible; sulfur is the classic poison for nickel and platinum catalysts, which is why feed desulfurization units sit ahead of reformers, and why leaded petrol had to be eliminated before catalytic converters could work.

Key idea: Catalysts lower the activation barrier in both directions, so they speed the approach to equilibrium and can improve selectivity but can never shift equilibrium; heterogeneous catalysts dominate industry and fail by sintering, coking, and poisoning.

Reactor choice as a selectivity tool

Here is where reaction engineering earns its keep. Suppose the desired product is an intermediate in a series, A goes to B goes to C, and B is what you want. Let B build up too long and it turns into C. You want a reactor in which every molecule spends the same time inside, so you can stop at the optimum: that is a plug flow or batch reactor. A stirred tank, in which some molecules leave immediately and some linger for hours, is precisely wrong.

Now suppose two reactions compete in parallel for the same reactant, and the desired one has the higher order in A. High concentration favors the desired reaction, so you want a reactor that keeps concentration high: plug flow again, or a batch reactor early in its run. If instead the undesired reaction has the higher order, you want concentration kept low everywhere, and a stirred tank, which holds the entire volume at the lowest concentration in the system, becomes the right answer. The same feature that made the CSTR large makes it selective.

Finally, a caution about the ideal models. Real vessels have channeling, dead zones, and bypassing, characterized by measuring a residence time distribution with a tracer pulse. A tank that looks well mixed but has a stagnant corner will underperform its design, and tracer testing is one of the standard diagnostic tools when a reactor fails to meet its conversion.

Key idea: Choose plug flow or batch when you need uniform residence time or high concentration, and a stirred tank when low concentration or excellent temperature control is what protects selectivity.

Common misconceptions

• Reaction order can be read from the balanced equation. Only for elementary reactions. Industrial kinetics are usually multistep and may show fractional or negative orders.
• A catalyst improves the yield beyond equilibrium. It cannot. It lowers the barrier both ways, speeding the approach to equilibrium and often improving selectivity, but never moving the equilibrium point.
• The bigger reactor is always the worse design. The CSTR is 3.9 times larger here, yet it may still win on temperature control, solids handling, cost, and selectivity.
• Plug flow reactors are safer because they are small. They develop hot spots that stirred tanks do not, which is why fixed-bed exothermic reactors are built as multitubular exchangers full of coolant.
• Raising temperature is a free way to increase output. It raises the rate of every reaction, usually the undesired ones faster, and it raises heat release exponentially against a linearly growing cooling capacity.

Recap

• Thermodynamics decides whether and how far a reaction goes; kinetics decides how fast, and rate laws are experimental facts rather than consequences of stoichiometry.
• Arrhenius behavior means a 10 K rise multiplies the rate by about 2.8 for an activation energy of 80 kJ/mol, and this exponential dependence is the mechanism of thermal runaway.
• Design equations: batch and plug flow give t or tau = -ln(1 - X)/k for first order, while a CSTR gives tau = X/(k(1 - X)).
• For k = 0.05 per minute, 100 L/min, and 90 percent conversion, the CSTR needs 18.0 m^3 and the PFR 4.61 m^3, a ratio of 3.9 that falls to 1.44 at 50 percent conversion.
• Three equal stirred tanks in series need 6.93 m^3 in total, recovering most of the gap, which is why cascades are common.
• Catalysts lower activation energy in both directions, cannot change equilibrium, and deactivate by sintering, coking, and poisoning.
• Reactor choice controls selectivity: plug flow or batch for uniform residence time and high concentration, stirred tank for low concentration and superior temperature control.

Sources

1. Wikipedia contributors. (n.d.). Chemical reactor. Wikipedia. en.wikipedia.org
2. Chemistry LibreTexts. (n.d.). The Arrhenius equation. LibreTexts. chem.libretexts.org
3. OpenStax. (2019). Catalysis. In Chemistry 2e. Rice University. openstax.org
4. American Institute of Chemical Engineers. (n.d.). Catalysis and reaction engineering. AIChE. aiche.org

The big picture

Walk onto any refinery or large chemical site and the tall silver towers you see are distillation columns. There are tens of thousands of them in operation worldwide, and they perform the large majority of industrial separations. They are also, as Lesson 10 showed, thermodynamically wasteful, running near twenty percent second-law efficiency. Industry uses them anyway because they are simple, continuous, made of ordinary steel, scale to enormous throughput, and have no moving parts inside. Understanding a column properly is the single most transferable piece of separations knowledge you can acquire.

Today we finish the job we started in Lesson 1. Our column has been carrying numbers through this whole course: a feed of 100 kmol/h at 50 mole percent benzene, products at 95 and 5 percent, D and B both 50 kmol/h, a bubble point of 92.1 C, and a relative volatility of 2.5. By the end of this lesson you will know how many trays it needs, how tall and wide it is, and what it costs to run.

Here is the plan for today. First the anatomy of a column and the crucial question of what reflux is actually for. Then the equilibrium stage idea and how real trays relate to it. Then the McCabe-Thiele construction, which we will work numerically stage by stage so you can follow it without a chart. Then the two limiting cases, minimum stages and minimum reflux, and the economic optimum between them. We finish with duties, diameter, and height.

What a column is

A distillation column is a vertical vessel containing trays or packing, with a reboiler at the bottom that boils liquid to generate rising vapor, and a condenser at the top that condenses the vapor leaving. Feed enters partway up. Vapor rises, liquid flows down, and on every tray the two are brought into intimate contact so that they approach equilibrium. Because vapor is always enriched in the light component relative to the liquid it touches, each contact step moves the vapor up the composition scale and the liquid down it.

The section above the feed is the rectifying section, whose job is to purify the vapor going overhead. The section below is the stripping section, whose job is to strip the last light component out of the liquid going to the bottom. You need both, which is why the feed enters in the middle rather than at either end.

Now the question that separates people who have understood distillation from people who have memorized it: what is reflux for? Part of the condensed overhead is returned to the top of the column instead of being taken as product. Why give back product you have just made?

Because without it, the rectifying section has no liquid at all. The contacting that enriches the vapor requires a descending liquid stream for the ascending vapor to exchange with, and above the feed the only possible source of that liquid is returned condensate. With zero reflux, vapor would rise through the top section untouched and leave at whatever composition the feed tray produced. Reflux is not a refinement; it is the thing that makes the top of the column exist. The reflux ratio R = L/D, liquid returned divided by product withdrawn, is therefore the single most important operating variable a column has.

Key idea: Reflux supplies the descending liquid that rising vapor exchanges with, so without it the rectifying section cannot separate at all, and the reflux ratio is the master operating variable of any column.

Equilibrium stages and real trays

The design abstraction is the theoretical stage, also called an equilibrium stage: a contact at which vapor and liquid leave in exact equilibrium with each other. Real trays do not achieve equilibrium, because contact time is finite and mixing imperfect. The gap is bridged by an efficiency. Overall column efficiency, the ratio of theoretical stages to actual trays, runs from about 40 percent for difficult, viscous, or low-relative-volatility systems to 80 percent or more for easy ones; 60 percent is a reasonable default for hydrocarbons, and we will use it.

Physically, contact happens either on trays, perforated plates holding a layer of liquid that vapor bubbles through, or in packing, either dumped rings or structured corrugated sheets that spread liquid as a film. Trays tolerate fouling and wide flow ranges; packing gives lower pressure drop, which matters enormously in vacuum service where a high pressure drop would raise the bottom temperature and cook the product. Packing performance is quoted as HETP, the height equivalent to a theoretical plate.

The McCabe-Thiele construction

McCabe and Thiele showed in 1925 that a binary column can be solved graphically on the x-y diagram, and the method remains the best way to understand what a column does. It rests on constant molar overflow: the assumption that molar liquid and vapor flows are constant within each section. That holds when the two components have similar molar latent heats, which for benzene (30.8 kJ/mol) and toluene (33.2 kJ/mol) is an excellent approximation.

Three lines go on the diagram alongside the equilibrium curve y = 2.5x/(1 + 1.5x).

The rectifying operating line is a mass balance around the top of the column: y = (R/(R+1)) x + x(D)/(R+1). Choose a reflux ratio of R = 1.5. Then y = (1.5/2.5) x + 0.95/2.5 = 0.6 x + 0.38. It passes through the point (0.95, 0.95) on the diagonal, as every rectifying line must.

The feed line, or q-line, describes the thermal condition of the feed. Our feed arrives as a saturated liquid at its bubble point of 92.1 C, so q = 1 and the q-line is vertical at x = 0.50.

The stripping operating line is the balance around the bottom. It runs from (0.05, 0.05) on the diagonal to the point where the rectifying line meets the q-line. At x = 0.50 the rectifying line gives y = 0.6(0.50) + 0.38 = 0.68, so the intersection is (0.50, 0.68). The slope is (0.68 - 0.05)/(0.50 - 0.05) = 0.63/0.45 = 1.4, giving y = 1.4 x - 0.02.

Stepping off the stages

Now the construction. Start at the top with x = 0.95. With a total condenser, the vapor leaving stage 1 has y = 0.95. Alternate between the equilibrium relation, which gives the liquid on a stage from the vapor leaving it, and the operating line, which gives the vapor entering a stage from the liquid leaving the one above. Inverting the equilibrium relation gives x = y / (2.5 - 1.5 y).

Stage	y entering equilibrium	x leaving stage	Operating line used
1	0.950	0.884	Rectifying
2	0.910	0.802	Rectifying
3	0.861	0.713	Rectifying
4	0.808	0.627	Rectifying
5	0.756	0.554	Rectifying
6	0.712	0.497	Rectifying (feed stage next)
7	0.676	0.455	Stripping
8	0.617	0.392	Stripping
9	0.529	0.310	Stripping
10	0.414	0.221	Stripping
11	0.289	0.140	Stripping
12	0.176	0.079	Stripping
13	0.090	0.038	Reboiler

Follow one line to see the pattern. At stage 3 the vapor entering equilibrium is y = 0.861, so the liquid in equilibrium with it is x = 0.861/(2.5 - 1.5 x 0.861) = 0.861/1.208 = 0.713. Feeding that x into the rectifying line gives the next vapor: y = 0.6(0.713) + 0.38 = 0.808, which is the entry for stage 4.

At stage 6 the liquid composition falls to 0.497, just below the feed composition of 0.50, so from stage 7 onward we switch to the stripping line. That switch point is the feed stage. Stepping continues until stage 13 produces x = 0.038, which has passed the bottoms specification of 0.05, and we stop.

The answer: 13 theoretical stages. The last one is the reboiler, which is itself an equilibrium stage because it boils liquid and returns vapor in equilibrium with it, so the column proper needs 12 theoretical trays with the feed on tray 7. At 60 percent overall efficiency that is 12/0.6 = 20 actual trays.

Key idea: Alternating between the equilibrium curve and the operating lines steps off theoretical stages; our column needs 13 including the reboiler, with the feed at stage 7 and about 20 real trays at 60 percent efficiency.

The two limits, and the optimum between them

Two extreme cases bracket every design.

Total reflux means returning all the condensate and withdrawing no product. Both operating lines collapse onto the diagonal, the driving force is as large as it can be, and the stage count is the smallest possible. For a constant-alpha binary this minimum number of stages is given by the Fenske equation: N(min) = ln[(x(D)/(1 - x(D))) x ((1 - x(B))/x(B))] / ln(alpha) = ln[19 x 19] / ln(2.5) = ln(361)/0.9163 = 5.889/0.9163 = 6.4 stages. Real columns are started up at total reflux, which is a good way to check that the trays are working before product is drawn.

Minimum reflux is the other limit: reduce R until the operating line just touches the equilibrium curve, at which point the driving force vanishes there and an infinite number of stages would be needed. For our saturated-liquid feed, the pinch occurs at the feed composition, where the equilibrium vapor is y = 0.714. Then R(min) = (x(D) - y)/(y - x(F)) = (0.95 - 0.714)/(0.714 - 0.50) = 0.236/0.214 = 1.10.

So our chosen R = 1.5 is 1.36 times the minimum, squarely inside the usual industrial range of 1.1 to 1.5 times R(min). The reason that range exists is a genuine optimum. Raising reflux boils more liquid, which costs energy in the reboiler and cooling in the condenser, but it reduces the stage count, which reduces the height and cost of the column. Total annual cost falls as you move off the infinite-stage limit at R(min), then rises again as energy dominates, and the shallow minimum sits at a modest multiple of R(min). Because energy prices have risen relative to steel over the decades, modern designs sit lower in that range than older ones did.

Key idea: Minimum stages occur at total reflux (6.4 here by Fenske) and minimum reflux at infinite stages (1.10 here), and the economic design sits at 1.1 to 1.5 times the minimum reflux, trading energy against column height.

Duties, diameter, and height

Now size the equipment. With constant molar overflow, the vapor rate in the rectifying section is V = D(R + 1) = 50 x 2.5 = 125 kmol/h. That is the vapor the reboiler must generate and the condenser must condense.

Condenser and reboiler duties. Using an average molar heat of vaporization of about 32 kJ/mol, the condenser duty is 125,000 mol/h x 32 kJ/mol = 4,000,000 kJ/h = 4.0 GJ/h, which is 1.11 MW. The reboiler duty is essentially the same for a saturated liquid feed. Compare that with the 273 kW feed preheater from Lesson 6, and you see the reboiler is four times larger, exactly as latent heat predicted it would be.

Utilities. Steam at 2100 kJ/kg of condensing enthalpy: 4,000,000/2100 = 1905 kg/h, about 1.9 tonnes an hour. Cooling water for the condenser, heated 15 degrees: 1,111,000/(4184 x 15) = 17.7 kg/s, which is nearly 64 tonnes an hour. The distillate is 3941 kg/h, so the energy intensity is 4.0 GJ per 3.94 tonnes, about 1.0 GJ per tonne of product, which is a realistic figure for a benzene-toluene split and a useful benchmark to carry around.

Diameter. The vapor at the top is about 125 kmol/h at 1 atm and roughly 92 C, so its volumetric flow is 125 x 22.41 x (365/273) = 3745 m^3/h, or 1.04 m^3/s. Trays flood if vapor velocity gets too high, and a typical design superficial velocity is around 1 m/s, giving a cross-sectional area of about 1.04 m^2 and a diameter of about 1.2 m.

Height. Twenty trays at a typical 0.5 m spacing is 10 m, plus disengagement space at the top and a liquid sump at the bottom, so the vessel is roughly 14 to 15 m tall. A 1.2 m by 15 m tower processing 8.5 tonnes an hour: that is the whole design, and every number in it traces back to a balance or an equilibrium calculation you have now done by hand.

Key idea: With V = D(R + 1) = 125 kmol/h, the column needs 1.11 MW of reboiler duty, 1.9 t/h of steam, 64 t/h of cooling water, and a vessel about 1.2 m in diameter and 15 m tall.

Common misconceptions

• Reflux is wasted product. Without reflux there is no liquid in the rectifying section and no rectification at all; it is the mechanism, not an overhead.
• More trays always give better separation. Only up to the pinch. Below minimum reflux no number of trays reaches the specification, and above minimum stages the extra trays do nothing.
• A theoretical stage is a tray. Real trays fall short of equilibrium; at 60 percent efficiency our 12 theoretical trays became 20 actual ones.
• The reboiler is just a heater. It is a full equilibrium stage, which is why our 13 stages became 12 trays plus the reboiler.
• The condenser duty can be ignored. It is comparable to the reboiler duty and drives a cooling water flow of 64 tonnes an hour in this modest column.

Recap

• A column has a rectifying section above the feed and a stripping section below, with a reboiler generating vapor and a condenser supplying reflux.
• Reflux provides the descending liquid without which the rectifying section cannot separate; R = L/D is the master operating variable.
• McCabe-Thiele assumes constant molar overflow and uses a rectifying line y = 0.6x + 0.38, a vertical q-line at x = 0.50, and a stripping line y = 1.4x - 0.02.
• Stepping between the equilibrium curve and the operating lines gives 13 theoretical stages including the reboiler, feed at stage 7, or about 20 real trays at 60 percent efficiency.
• Fenske gives 6.4 minimum stages at total reflux, and minimum reflux is 1.10, so the chosen R = 1.5 sits at 1.36 times the minimum.
• Vapor rate V = D(R + 1) = 125 kmol/h gives a 1.11 MW reboiler, 1.9 t/h of steam, 64 t/h of cooling water, and about 1.0 GJ per tonne of product.
• The finished column is roughly 1.2 m in diameter and 15 m tall.

Sources

1. Britannica. (n.d.). Distillation. Encyclopaedia Britannica. britannica.com
2. Wikipedia contributors. (n.d.). Continuous distillation. Wikipedia. en.wikipedia.org
3. Engineering LibreTexts. (n.d.). Distillation and separation processes. LibreTexts Chemical Engineering. eng.libretexts.org
4. U.S. Department of Energy. (n.d.). Industrial efficiency and decarbonization. Office of Energy Efficiency and Renewable Energy. energy.gov

The big picture

Distillation is the default, not the answer. It fails outright at azeotropes. It destroys heat-sensitive products, which rules it out for most pharmaceuticals and biologicals. It is hopeless for dilute streams, because you would boil the entire solvent to recover a trace. It cannot touch permanent gases that will not condense at any sensible temperature. And when relative volatility is near one, it becomes so tall and so energy-hungry that alternatives win on cost.

For each of those failures there is a separation built on a different physical property. Absorption uses solubility. Extraction uses differential solubility in two immiscible liquids. Membranes use differences in how fast molecules pass through a barrier. Adsorption uses affinity for a solid surface. Today we take each in turn, work numbers on the first two, and then look at the economics that decides among them, because in practice the choice is almost always financial.

Here is the plan for today. First absorption, with a worked solvent rate. Then extraction, with a worked recovery that demonstrates the most important principle in all staged contacting. Then membranes and adsorption, with the energy comparison that shows why they matter. Then a compact decision framework. Then process economics: scaling laws, capital against operating cost, and the payback calculation that gets projects approved or rejected.

Absorption

Absorption, or scrubbing, contacts a gas with a liquid solvent that dissolves the component you want to remove. It is the standard tool for cleaning gas streams: removing carbon dioxide and hydrogen sulfide from natural gas with amine solutions, removing sulfur dioxide from flue gas with limestone slurry, recovering ammonia or solvent vapors, and capturing carbon dioxide from power plant exhaust.

The equilibrium relation is Henry's law from Lesson 9, usually written in mole fractions as y = m x, where m is the slope. The design equation is a mass balance, an operating line, exactly analogous to distillation, and the same staged or packed contactors are used.

Work an example. A gas stream of 100 kmol/h contains 2 mole percent of a solute; 95 percent of it must be removed. The equilibrium slope is m = 1.5, and the solvent enters clean.

Solute entering = 2.0 kmol/h; leaving in the gas = 0.1 kmol/h; therefore absorbed = 1.9 kmol/h. The carrier gas is 98 kmol/h and passes through unchanged.

What is the least solvent that could possibly do this? The limit occurs when the leaving liquid is in equilibrium with the entering gas, which is the richest it could ever get: x(max) = y(in)/m = 0.02/1.5 = 0.01333. Then L(min) = 1.9/0.01333 = 142.5 kmol/h. At exactly that rate the driving force vanishes at the bottom and you would need an infinitely tall column, precisely as at minimum reflux in distillation.

Practice uses 1.2 to 2 times the minimum. Take 1.5 times: L = 214 kmol/h, giving an actual exit liquid composition of 1.9/214 = 0.0089, comfortably below saturation. A useful check is the absorption factor A = L/(m G) = 214/(1.5 x 98) = 1.46; values above about 1.25 give efficient absorption, and values below 1 mean the solvent will saturate before the gas is clean no matter how tall the column.

Absorption is nearly always paired with stripping, the reverse operation that regenerates the solvent so it can be recycled, typically by heating it or contacting it with steam. In amine systems the absorber runs cold and the stripper hot, and the energy to regenerate the solvent, commonly cited at roughly 3 to 4 GJ per tonne of carbon dioxide for conventional amine capture, is the dominant cost of the whole process. That is a striking illustration of a general truth: in a solvent process, the separation is usually cheap and the regeneration is where the money goes.

Key idea: Absorption is sized from a mass balance against a Henry's law equilibrium, with a minimum solvent rate set by saturation and practice at 1.2 to 2 times that; the paired stripping step that regenerates the solvent usually dominates the cost.

Extraction, and why staging wins

Liquid-liquid extraction contacts a feed solution with a solvent that is immiscible with it but dissolves the target better. It is the tool of choice when the components boil too closely, when the product would decompose on boiling, or when the solute is dilute. Real examples: recovering antibiotics from fermentation broth, extracting aromatics from reformate with sulfolane, decaffeinating coffee with supercritical carbon dioxide, and recovering copper, uranium, and rare earths from leach liquors, where solvent extraction is the backbone of modern hydrometallurgy.

The equilibrium is described by a distribution coefficient K, the ratio of solute concentration in the solvent phase to that in the feed phase at equilibrium.

Now the calculation that teaches the key lesson. Feed: 1000 kg/h containing 50 kg/h of solute in 950 kg/h of water. Solvent available: 500 kg/h. Distribution coefficient K = 4.

One contact using all 500 kg/h. Let E be the solute extracted. Then the solvent-phase fraction is E/500 and the water-phase fraction is (50 - E)/950, and K = 4 requires 950 E = 4 x 500 x (50 - E) = 100,000 - 2000 E. So 2950 E = 100,000 and E = 33.9 kg/h, a recovery of 67.8 percent.

Two contacts using 250 kg/h each. First stage: 950 E1 = 4 x 250 x (50 - E1) = 50,000 - 1000 E1, so 1950 E1 = 50,000 and E1 = 25.6 kg/h, leaving 24.4 kg/h in the water. Second stage on that raffinate: 1950 E2 = 1000 x 24.4, so E2 = 12.5 kg/h. Total extracted = 38.1 kg/h, a recovery of 76.3 percent.

Arrangement	Solvent used	Solute recovered	Recovery
One contact	500 kg/h	33.9 kg/h	67.8 percent
Two contacts of 250 kg/h	500 kg/h	38.1 kg/h	76.3 percent

Identical solvent, eight and a half points more recovery, purely by splitting the contact in two. Arranging the same stages counter-current, so fresh solvent meets the most depleted feed, does better still. This is the deepest principle in separations, and it is the same one that makes a distillation column work: many small contacts, arranged counter-current, beat one large contact every time. It is why columns have trays, why extractors are built as mixer-settler trains, and why absorbers are packed.

Key idea: Dividing a fixed amount of solvent among several counter-current stages recovers substantially more solute than a single contact, which is the organizing principle behind every staged separation.

Membranes and adsorption

Membranes separate by letting some molecules through a thin barrier faster than others, driven by a pressure or concentration difference. Pressure-driven liquid processes form a family by pore size: microfiltration for particles and bacteria, ultrafiltration for proteins and macromolecules, nanofiltration for divalent ions, and reverse osmosis for salt. Gas membranes separate nitrogen from air, recover hydrogen from refinery purge streams (which is exactly the option mentioned for the ammonia loop purge in Lesson 4), and remove carbon dioxide from natural gas. Pervaporation, where the permeate evaporates on the far side, is used commercially to dehydrate ethanol past its azeotrope.

Two properties characterize a membrane, and they fight each other: permeance, how fast material passes, and selectivity, how much better one species passes than another. Materials that are highly permeable tend to be poorly selective, a trade-off so consistent it is drawn as an empirical upper bound on performance plots, and improving it is a major research field.

Membranes matter because they can be far more energy-efficient than thermal separations. Modern seawater reverse osmosis desalinates at roughly 3 kWh per cubic metre of fresh water, against a thermodynamic minimum near 1 kWh per cubic metre, which is a second-law efficiency around 35 percent. Compare our distillation column at about 19 percent and you can see why membranes displace distillation wherever a suitable material exists. Their limitations are real: fouling, limited chemical and thermal resistance, modest achievable purity in a single pass, and the fact that no membrane exists for many separations you would like to make.

Adsorption uses a solid with high affinity and enormous internal surface area, such as activated carbon, silica gel, or a zeolite molecular sieve. It shines on dilute streams and on final polishing, exactly where distillation is worst. Pressure-swing adsorption produces most of the world's industrial-purity hydrogen and much of its on-site oxygen and nitrogen, and molecular sieves are what finish fuel ethanol past its azeotrope.

Choosing a separation

Situation	Preferred method	Why
Volatility difference, thermally stable, not dilute	Distillation	Cheap, continuous, scalable
Component to remove from a gas	Absorption	Solubility difference, no condensation needed
Close boilers or heat-sensitive product	Extraction	Operates cold, uses solubility not volatility
Dilute solute or final polishing	Adsorption	High affinity at very low concentration
Gas separations, desalination, azeotrope breaking	Membranes	No phase change, much lower energy
Solids from liquids	Filtration or centrifugation	Mechanical, cheap, no phase change

In practice hybrids dominate: distil to 93 percent ethanol and finish with molecular sieves; absorb carbon dioxide into amine and strip it with steam; extract an antibiotic and crystallize it. Each step is used where it is strongest.

Process economics

Now the discipline that decides. Costs split into capital expenditure, the one-time cost of building, and operating expenditure, the recurring cost of running: raw materials, utilities, labor, and maintenance. For commodity chemicals raw materials usually dominate operating cost, often over half of it, which is why the selectivity lesson in Module 2 was worth so much attention. A yield improvement of one percentage point on a large plant can be worth more than an entire utility optimization.

Capital costs scale non-linearly, captured by the six-tenths rule: Cost2 = Cost1 x (Capacity2 / Capacity1)^0.6. Suppose a 100,000 tonne per year plant costs 50 million dollars. Doubling capacity gives 50 x 2^0.6 = 50 x 1.516 = 75.8 million dollars, not 100 million. Capital per annual tonne falls from 500 dollars to 379. That exponent, typically 0.6 to 0.7, is the mathematical statement of economy of scale, and it is why commodity chemical plants keep getting bigger and why small plants struggle to compete on price. It works in reverse too, which is why modular small-scale production has to justify itself on flexibility, transport savings, or feedstock access rather than on unit cost.

To get from equipment to plant cost, engineers use Lang factors: total installed cost is roughly 4 to 5 times purchased equipment cost for a fluid-processing plant, once piping, instrumentation, electrical work, structures, engineering, and contingency are included. A vessel quoted at 100,000 dollars becomes something near half a million installed, which regularly surprises people costing a project for the first time. Published cost data are updated to the present with a cost index such as the Chemical Engineering Plant Cost Index.

Projects are then judged. The simplest screen is simple payback: capital divided by annual saving. A heat recovery project costing 400,000 dollars that saves 150,000 dollars a year of steam has a payback of 400,000/150,000 = 2.7 years. Many industrial firms want under two or three years for a retrofit, which is a demanding hurdle and the reason many thermodynamically sensible energy projects are never built. More rigorous measures, net present value and discounted cash flow rate of return, account for the time value of money and are what a real investment decision uses; payback remains the screening tool because it takes ten seconds.

Key idea: Capital scales with roughly the 0.6 power of capacity and multiplies by a Lang factor of 4 to 5 from equipment to installed cost, while raw materials usually dominate operating cost, so yield improvements and scale are the two biggest economic levers.

Common misconceptions

• Distillation is always the cheapest separation. It is the default, but it fails at azeotropes, on heat-sensitive materials, on dilute streams, and on permanent gases, and membranes beat it substantially on energy where a suitable material exists.
• Using more solvent in one contact is the way to improve extraction. Splitting the same solvent into counter-current stages recovers more: 76.3 percent in two contacts against 67.8 percent in one.
• Absorption cost is the absorber. The stripping step that regenerates the solvent usually dominates, at roughly 3 to 4 GJ per tonne of carbon dioxide for conventional amine capture.
• Doubling plant capacity doubles the capital cost. It multiplies it by about 2^0.6, or 1.52, which is the whole basis of economy of scale.
• Installed cost is close to equipment cost. Lang factors of 4 to 5 apply for fluid-processing plants once piping, instruments, structures, and engineering are counted.

Recap

• Absorption removes a component from a gas into a solvent; the worked case absorbs 1.9 kmol/h with a minimum solvent of 142.5 kmol/h and a design rate of 214 kmol/h, giving an absorption factor of 1.46.
• Solvent regeneration by stripping is usually the dominant cost of an absorption process.
• Extraction with K = 4 recovers 67.8 percent in one contact of 500 kg/h of solvent but 76.3 percent when the same solvent is split into two stages.
• Many small counter-current contacts beat one large contact, which is the principle behind trays, packing, and mixer-settler trains alike.
• Membranes trade permeance against selectivity and reach about 35 percent second-law efficiency in seawater desalination, against roughly 19 percent for our distillation column.
• Adsorption excels on dilute streams and final polishing, producing most industrial-purity hydrogen and finishing fuel ethanol past its azeotrope.
• Economics: capital scales as capacity to the 0.6 power, Lang factors of 4 to 5 convert equipment to installed cost, and a 400,000 dollar project saving 150,000 dollars a year pays back in 2.7 years.

Sources

1. Wikipedia contributors. (n.d.). Liquid-liquid extraction. Wikipedia. en.wikipedia.org
2. Britannica. (n.d.). Reverse osmosis. Encyclopaedia Britannica. britannica.com
3. American Institute of Chemical Engineers. (n.d.). Separations technology resources. AIChE. aiche.org
4. U.S. Department of Energy. (n.d.). Carbon capture and separations. Office of Fossil Energy and Carbon Management. energy.gov

The big picture

We end where the profession's obligations are heaviest. Everything in this course exists so that engineers can build things that work. Process safety is the discipline of making sure that when they do not work, nobody dies.

Start with a distinction that took the industry decades and several catastrophes to internalize. Personal safety is about slips, cuts, and falls, the things that injure one worker at a time, measured by recordable injury rates and improved by handrails and training. Process safety is about releases of large amounts of energy or hazardous material: fires, explosions, and toxic clouds that can kill many people at once. The two are barely related, and a site can have an excellent injury record while sitting one instrument failure from a disaster.

The plan: hazards and how they are identified, inherently safer design, two real cases treated factually, sustainability, and finally what chemical engineers actually do for a living.

Hazards, risk, and identification

A hazard is an inherent property with the potential to cause harm: flammable, toxic, reactive, or under pressure. Risk combines the likelihood of a harmful event with its consequence. Being careful manages risk but never removes a hazard; only changing what you have does that.

The common process hazards are worth naming with numbers. Flammability is bounded by the explosive limits: methane burns in air only between about 5 and 15 mole percent, a narrow window but an easy one to pass through. Flash point is where a liquid gives off enough vapor to ignite, about -43 C for gasoline and 52 C or higher for diesel. Toxicity is set by exposure limits, reactivity covers substances that decompose or react violently with water or air, and stored energy covers pressure and temperature: a vessel of hot liquid above its atmospheric boiling point holds enough energy to destroy itself if containment is lost.

Hazards are found systematically, not by intuition. The dominant method is HAZOP, hazard and operability study, developed at ICI in Britain in the 1960s. A multidisciplinary team works through the piping and instrumentation diagrams node by node, applying deliberately mechanical guide words to each parameter: NO or NONE, MORE, LESS, AS WELL AS, PART OF, REVERSE, OTHER THAN, applied to flow, temperature, pressure, level, and composition. No flow in this line? Reverse flow through this pump? The method is exhaustive and does not depend on anyone having a good idea on the day. Complementary tools include what-if analysis, failure modes and effects analysis, quantitative fault tree analysis, and layer of protection analysis, which counts the independent safeguards between a cause and a consequence to judge whether residual risk is tolerable.

Safety is built as layers of protection, from the inside out: inherently safer design, process control, alarms and operator response, safety instrumented systems, pressure relief, containment such as dikes and blast walls, and emergency response. Each layer has holes, and an accident happens when the holes line up. That image, the Swiss cheese model, is why investigations of major accidents never find a single cause: they find a chain in which every link failed.

Key idea: Hazards are inherent properties and risk is likelihood times consequence; hazards are found by systematic methods such as HAZOP, and protection is built as independent layers whose simultaneous failure produces a disaster.

Inherently safer design

The most important idea in process safety came from Trevor Kletz, an ICI engineer who summarized it in one sentence: what you do not have cannot leak. Rather than adding safeguards around a hazard, remove or shrink the hazard. Four strategies follow.

Minimize the quantity held: make a dangerous intermediate on demand rather than storing tonnes, or use a small continuous reactor instead of a large batch vessel. Substitute a less hazardous material, such as a water-based solvent for a flammable one. Moderate the conditions: dilute, refrigerate, lower the pressure, or handle a substance as a solution rather than a gas. Simplify, because every added trip and interlock is one more thing that can fail or be bypassed.

This hierarchy sits above the others because added safeguards must be maintained, powered, tested, and not defeated over decades, whereas a hazard that is not present requires nothing at all.

Bhopal, 1984

On the night of 2 to 3 December 1984, at a Union Carbide India Limited pesticide plant in Bhopal, Madhya Pradesh, water entered a storage tank containing methyl isocyanate, a highly toxic and highly reactive intermediate. The resulting violent exothermic reaction raised the tank's temperature and pressure, and roughly thirty to forty tonnes of methyl isocyanate vapor were released over a densely populated area while people slept. The Madhya Pradesh government has recorded 3,787 deaths from the release; other estimates run several times higher, over half a million people were exposed, and chronic injury and litigation continued for decades. It remains the deadliest industrial accident in history, and each of its lessons maps onto a concept above.

Inventory. The plant stored many tonnes of a lethal intermediate that is made and consumed within the process and need not be stockpiled at all. After Bhopal the industry moved decisively toward making such intermediates on demand. This is minimization, the single change that would have most reduced the consequence.

Layers that were not there. The refrigeration system that would have kept the tank cool had been shut down, the flare tower meant to burn escaping vapor was not in service, and the vent gas scrubber was not sized for a release of this magnitude. Layer after layer was absent or inadequate on the night it mattered: the Swiss cheese model in its most literal form.

Siting, planning, and decline. Dense informal settlement had grown up immediately around the plant, there was no effective community warning or evacuation plan, and local physicians did not know what the gas was or how to treat exposure; process safety does not stop at the fence line. The plant was also operating well below capacity in a period of cost reduction, with reduced staffing and deferred maintenance, a gradual degradation visible in advance to anyone looking for it.

Texas City, 2005

On 23 March 2005, at the BP refinery in Texas City, Texas, an isomerization unit was being restarted. A raffinate splitter tower was overfilled with hydrocarbon liquid, which rose through the overhead system and discharged from a blowdown drum venting directly to the atmosphere. A vapor cloud formed at ground level and found an ignition source, believed to be a running vehicle engine nearby. The explosion killed 15 people and injured about 180. The U.S. Chemical Safety Board investigated and published a detailed report, and its findings are, if anything, more instructive than Bhopal's, because nothing exotic was involved.

Process safety is not personal safety. BP's recordable injury rate at the site was good and was cited internally as evidence of a safe operation. It measured slips and cuts, and said nothing about the likelihood of a vapor cloud explosion. The investigation made this distinction central, and leading process safety indicators such as overdue inspections, bypassed alarms, and loss-of-containment events came into wide use afterward.

Siting and obsolete equipment. All 15 people who died were in or immediately around portable trailers placed close to the unit, and nothing about their work required them to be there; facility siting studies using blast and toxic exposure modeling became a standard requirement afterward. The atmospheric blowdown stack, meanwhile, should already have been replaced by a flare system, a change known and recommended before the event.

Instrumentation and fatigue. The tower level indication was unreliable and a high-level alarm did not function, so operators had no accurate picture of what was inside, and decisions cannot be better than the data they rest on. Crews had also worked long consecutive shifts, while startups are the most hazardous phase of operation and demand the most alert people. An independent panel later examined the site's safety culture and cost pressures, and the case is now taught in every process safety course in the world.

Key idea: Bhopal teaches that the safest inventory is the one you do not hold and that safeguards must be maintained and sized for the real hazard; Texas City teaches that a good injury rate says nothing about process safety, and that siting, instrumentation, obsolete equipment, and fatigue are engineering issues.

The regulatory response

These accidents built the modern framework. In the United States, the Occupational Safety and Health Administration's Process Safety Management standard requires covered facilities to compile process safety information, conduct hazard analyses, write procedures, manage change, investigate incidents, and audit compliance, while the Environmental Protection Agency's Risk Management Program addresses offsite consequences. The U.S. Chemical Safety Board investigates major incidents independently and deliberately has no enforcement power, so its recommendations carry weight because they are technical rather than punitive. In Europe the Seveso Directives, named after a 1976 dioxin release in Italy, impose comparable duties.

One requirement deserves emphasis. Management of change means any modification, however small, gets the same engineering scrutiny as the original design. Many major accidents, including the 1974 Flixborough explosion that killed 28 people, trace to a temporary modification installed without review. If you take one procedural habit from this course, take that one.

Sustainability and green engineering

The same analytical habits apply to waste and energy. Green chemistry, set out as twelve principles by Anastas and Warner in 1998, argues that preventing waste beats treating it, that syntheses should maximize incorporation of feedstock into product, that safer solvents should be chosen, that catalysis beats stoichiometric reagents, and that products should be designed to degrade.

Atom economy makes the second principle quantitative: mass of desired product divided by total mass of reactants. Direct oxidation of ethylene, C2H4 plus 0.5 O2 giving C2H4O, has reactant mass 28.05 + 16.00 = 44.05 and product mass 44.05, an atom economy of 100 percent. The older chlorohydrin route to the same product co-produced calcium chloride in bulk and scored far lower. That comparison, not the reaction yield, tells you which route is intrinsically cleaner.

The E-factor, introduced by Roger Sheldon, measures the reality: kilograms of waste per kilogram of product. Bulk chemicals typically run about 1 to 5, fine chemicals 5 to 50, and pharmaceuticals often 25 to over 100. That last figure is not incompetence; it reflects multistep syntheses, solvent-intensive purification, and regulatory constraints on changing a validated process. It does show where the opportunity is.

Heat integration, developed as pinch analysis by Linnhoff and colleagues in the late 1970s, systematizes the idea from Lesson 7: plot all hot streams needing cooling and all cold streams needing heating as composite curves, slide them together to a chosen minimum approach temperature, and read off the true minimum external heating and cooling the process can ever require. Setting that target before designing the exchanger network routinely finds double-digit energy savings.

At the largest scale, the chemical sector is the biggest industrial energy consumer and a major source of carbon dioxide, with ammonia synthesis and steam cracking the heaviest loads. Decarbonizing it means electrifying heat, replacing fossil-derived hydrogen with hydrogen from water and low-carbon electricity, capturing unavoidable process emissions, and closing loops through chemical recycling of plastics. These are material balances, energy balances, separations, and reactor design, which is to say they are this course.

Key idea: Atom economy and the E-factor measure how much feedstock becomes product rather than waste, pinch analysis targets the true minimum energy a process needs, and decarbonizing the chemical industry is a material and energy balance problem at global scale.

What chemical engineers actually do

The industries are broader than the name suggests. Petroleum refining and petrochemicals remain the historical core, followed by polymers, specialty and agricultural chemicals, and industrial gases. Pharmaceuticals and biotechnology employ large numbers, as do food, brewing, and consumer products. Semiconductor manufacturing depends on chemical engineers for deposition, etching, and ultrapure chemicals. Pulp and paper, mining, water treatment, and environmental consulting draw on the same toolkit, and energy is the fastest-changing area: batteries, hydrogen, biofuels, and carbon capture all need people who can close a balance.

The roles vary more than the industries. A process engineer at an operating plant troubleshoots, optimizes, and supports production, and is where most graduates start. A design engineer at an engineering contractor develops flowsheets, sizes equipment, and produces the drawings a plant is built from. Others run production, scale up new chemistry in research and development, design automation in process control, or lead hazard studies in process safety. Many move into technical service, regulatory affairs, consulting, finance, and management, where the degree is well regarded for certifying quantitative discipline under real constraints.

On credentials: in the United States licensure runs through the Fundamentals of Engineering examination, supervised experience, and the Professional Engineer examination administered through NCEES, and matters most for consulting and work affecting public safety. AIChE, founded in 1908, is the professional body. The U.S. Bureau of Labor Statistics reported a median annual wage for chemical engineers of about 112,000 dollars in May 2023, near the top of engineering disciplines, with growth projected around the average for all occupations.

One honest closing note. Many graduates never size a distillation column professionally. What endures is the habit of mind this course has trained: draw the boundary, count what crosses it, insist the balance closes, find the controlling resistance, separate what is thermodynamically possible from what is kinetically achievable, and put a number on everything before forming an opinion. That habit is the real qualification you have earned.

Key idea: Chemical engineers work across refining, pharmaceuticals, semiconductors, food, water, and energy, and the durable skill is the quantitative habit of closing balances rather than any single unit operation.

Common misconceptions

• A good injury rate means a safe plant. Texas City had a good recordable injury rate and suffered a catastrophic explosion. Personal safety metrics do not measure process safety.
• Adding more safeguards is the best way to manage a hazard. Inherently safer design ranks above added protection, because a hazard you do not have requires no maintenance, testing, or vigilance.
• Major accidents have a single cause. They have chains. Every layer had a hole and the holes lined up, which is why small modifications still need full engineering review under management of change.
• A high reaction yield means a clean process. Yield says nothing about how much feedstock mass ends up as product; atom economy and the E-factor do. And chemical engineers do not only work with chemicals: the same balances run semiconductor fabs, breweries, water treatment works, and battery plants.

Recap

• Process safety concerns fires, explosions, and toxic releases and is distinct from personal safety; hazards are found systematically by HAZOP, protection is built as independent layers, and inherently safer design ranks first: minimize, substitute, moderate, simplify, because what you do not have cannot leak.
• Bhopal in 1984 released some thirty to forty tonnes of methyl isocyanate with more than 3,787 recorded deaths, teaching inventory minimization, maintained and adequately sized safeguards, siting, and emergency planning.
• Texas City in 2005 killed 15 and injured about 180, teaching that injury rates do not measure process safety, and that siting, instrumentation, obsolete blowdown stacks, and fatigue are engineering issues.
• Regulation followed in OSHA Process Safety Management, the EPA Risk Management Program, the Chemical Safety Board, and the Seveso Directives, with management of change as the key procedural discipline.
• Green engineering measures performance by atom economy (100 percent for direct ethylene oxidation) and E-factor (1 to 5 for bulk chemicals, often over 25 for pharmaceuticals), while pinch analysis targets minimum energy.
• Chemical engineers work across refining, pharmaceuticals, semiconductors, food, water, and energy, with a U.S. median wage near 112,000 dollars in May 2023, and the durable skill is the habit of closing balances.

Sources

1. U.S. Chemical Safety and Hazard Investigation Board. (2007). Investigation report: Refinery explosion and fire, BP Texas City. CSB. csb.gov
2. Britannica. (n.d.). Bhopal disaster. Encyclopaedia Britannica. britannica.com
3. Occupational Safety and Health Administration. (n.d.). Process safety management. U.S. Department of Labor. osha.gov
4. U.S. Environmental Protection Agency. (n.d.). Green chemistry. epa.gov
5. U.S. Bureau of Labor Statistics. (n.d.). Chemical engineers. Occupational Outlook Handbook. bls.gov