Chemistry · Lesson

Chemical processes, mass balances and scale-up

Use independent component balances and declared unit boundaries to distinguish separation, recycle, accumulation and geometric scale effects.

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By the end, you can

Close solute and carrier balances separately.
Distinguish internal circulation from fresh external input.
Integrate accumulation in an ideal nonreactive mixed tank.
Compare area and volume scaling without treating geometry as an operating design.

A process balance starts with a chosen boundary

A chemical process connects incoming streams, transformations, separations and outgoing streams. Draw the boundary before counting material. An internal recycle crosses a unit boundary but may remain inside the overall process boundary.

Specify whether a stream rate measures total mass, a species amount or an elemental inventory. A reaction can change the amounts of species while conserving the appropriate elements and total mass within its complete account.

A stream split and an internal recycle need different boundaries

Take an original hypothetical steady two-component separator. Feed is 150 kg h⁻¹ with solute mass fraction 0.240. One dilute output is 80.0 kg h⁻¹ with solute fraction 0.100. No chemical identity, equilibrium phase relation or separator technology is assigned. Total mass balance gives the other output 70.0 kg h⁻¹.

Feed solute is 36.0 kg h⁻¹ and dilute-output solute 8.00 kg h⁻¹, leaving rich-output solute 28.0 kg h⁻¹. Its mass fraction is 28.0/70.0=0.400. Carrier balance independently gives 114 kg h⁻¹ incoming and 72.0+42.0=114 kg h⁻¹ outgoing. The total equation is the sum of the two component equations; it does not add a third independent component constraint.

Changing only dilute-output flow to 85.0 kg h⁻¹, retaining its mass fraction and the same feed, gives rich flow 65.0 kg h⁻¹, solute 27.5 kg h⁻¹ and fraction 0.423076923. Both component inventories still close. These stipulated mass-fraction observations do not prove that a particular separation achieves them.

For a separate symbolic steady conversion A→P, let each entity retain one TrackedUnit. Fresh A is 37.0 kmol TrackedUnit h⁻¹. Assign single-pass conversion X=0.300, complete P separation and purge fraction p=0.100 of unconverted A; the rest returns to the mixer. No impurity, additional source or actual molecular reaction is included.

The mixer recurrence is F_reactor=F_fresh+(1−p)(1−X)F_reactor. Thus F_reactor=37.0/[1−0.900×0.700]=100 kmol TrackedUnit h⁻¹. Product is 30.0, unconverted A 70.0, purge 7.00 and recycle 63.0 on that basis. Mixer closure is 100=37.0+63.0; whole-process external closure is 37.0=30.0+7.00.

Overall fresh-feed conversion is product/fresh=30.0/37.0=0.810810811, different from single-pass 0.300. If assigned purge becomes 0.200 at the same stipulated conversion, reactor/product/purge/recycle become 84.0909091/25.2272727/11.7727273/47.0909091 kmol TrackedUnit h⁻¹, and overall conversion 0.681818182. Actual recycling can change composition and rate; holding single-pass conversion fixed is a teaching assumption.

Conversion, yield and selectivity need their denominators

Conversion describes how much of a chosen reactant is consumed relative to the specified incoming amount. Yield relates a desired product to a stated theoretical or feed basis. Selectivity compares products or pathways using an explicit convention.

A large conversion can coexist with low desired-product yield when other reactions consume the reactant. State the reaction stoichiometry and basis before combining percentages. A feed’s limiting component and an actual reaction extent are different quantities.

A transient inventory and a scale ratio have their own assumptions

For a separate ideal nonreactive mixed liquid tank, assign fixed volume V=84.0 L, equal inlet and outlet q=7.00 L min⁻¹, inlet tracked-solute concentration C_in=0.600 mol L⁻¹ and initial concentration C₀=0.100 mol L⁻¹. Volumes are additive and outlet concentration equals the tank concentration. The component balance is dC/dt=(q/V)(C_in−C), with C(t)=C_in+(C₀−C_in)exp(−qt/V).

Hydraulic space time V/q=12.0 min is a flow-model parameter, not every molecule’s residence time. At 18.0 min, concentration is 0.488434920 mol L⁻¹ and tank amount 41.0285333 mol. Initial amount 8.40 mol plus integrated input 75.6 mol equals final amount plus integrated output 42.9714667 mol. Accumulation matters; the finite-time state is not exactly the limiting steady concentration.

Halving both flows to 3.50 L min⁻¹ gives space time 24.0 min. At the same 18.0 min, concentration is 0.363816724 mol L⁻¹, tank amount 30.5606048 mol, integrated input 37.8 mol and output 15.6393952 mol. The component balance again closes. Neither version provides a complete solvent or heat balance.

For another independent geometric illustration, compare spheres of radii 0.120 and 0.360 m. V=4πr³/3 gives volumes 0.00723822947 and 0.195432196 m³; A=4πr² gives areas 0.180955737 and 1.62860163 m². Tripling radius scales volume by 27 and area by nine.

Assign uniform volumetric heat-generation rate 4.80 kW m⁻³ and an independent surface removal-capacity rate 0.300 kW m⁻², solely to illustrate compatible scaling. Small-sphere generation/capacity are 0.0347435015/0.0542867211 kW, ratio 0.640; large-sphere values are 0.938074540/0.488580489 kW, ratio 1.920. Volume grows faster than area under these fixed assignments.

Twenty-seven small spheres have the same total volume as the large one, with total area 4.88580489 m², three times the large surface. Their assigned total generation is 0.938074540 kW and capacity 1.46574147 kW, retaining ratio 0.640. This is geometry, without a specified reaction, heat-transfer coefficient, material, coolant or engineering design. It establishes no feasible or safe operating arrangement.

Changing the flow changes replacement time in an ideal mixed tank

Tank concentration / mol L⁻¹

Elapsed time / min

Assigned q=7.00 L min⁻¹
Halved q=3.50 L min⁻¹
— First mixed-tank model

Underlying data

Elapsed time / min	Tank concentration / mol L⁻¹
0	0.09999999999999998 · Assigned q=7.00 L min⁻¹
18	0.4884349199257851 · Assigned q=7.00 L min⁻¹
36	0.5751064658160681 · Assigned q=7.00 L min⁻¹
0	0.09999999999999998 · Halved q=3.50 L min⁻¹
18	0.36381672362949263 · Halved q=3.50 L min⁻¹
36	0.4884349199257851 · Halved q=3.50 L min⁻¹

Original constant-volume nonreactive component model C(t)=0.600−0.500exp(−qt/84.0), using q in L min⁻¹ and t in min. Circles show q=7.00; squares q=3.50 at 0,18,36 min. Dashed curve shows only the first model. These assigned ideal mixtures approach the same limiting concentration at different rates; they are not reaction-rate or equipment-performance measurements.

A recycle can return unconverted material

A separator can return part of a stream to an earlier unit. A recycle may increase overall utilization even when single-pass conversion remains limited. Material separation and chemical conversion must remain separate terms in the ledger.

A component that is not removed or consumed can accumulate in a closed recycle. A purge creates an output that can control its inventory within the specified model. The overall balance includes that output; repeated circulation does not create additional fresh feed.

Scale-up changes more than the amount processed

Geometry, surface area, mixing, heat transfer and transport can change as a system grows. A reaction model fitted under one set of conditions does not automatically predict a larger process. Residence-time and temperature distributions may also change.

Use the original balances below to understand boundaries and units. They do not specify manufacturing equipment, operating conditions or a practical procedure for running a chemical process.

Worked examples

Close both component balances

Use the stated 150 kg h⁻¹ feed and 80.0 kg h⁻¹ dilute output.

The rich stream is 70.0 kg h⁻¹.
Its solute is 36.0−8.00=28.0 kg h⁻¹, mass fraction 0.400.
Carrier closes 114=72.0+42.0 kg h⁻¹.

Total mass is consistent with the two independent component balances.

Count fresh input at the external boundary

Use fresh 37.0 kmol TrackedUnit h⁻¹, single-pass conversion 0.300 and purge 0.100.

The steady mixer recurrence gives reactor A feed 100.
Product/purge/recycle are 30.0/7.00/63.0 kmol TrackedUnit h⁻¹.
Overall fresh-feed conversion is 30.0/37.0=0.810810811.

The internal recycle is neither fresh feed nor new creation of TrackedUnits.

Include the changing tank inventory

Use the assigned 84.0 L ideal tank with q=7.00 L min⁻¹ at 18.0 min.

Space time is 12.0 min and C=0.488434920 mol L⁻¹.
Final tank amount is 41.0285333 mol.
Initial 8.40 plus input 75.6 equals final plus output 42.9714667 mol.

Input and output differ because the tank inventory changes.

Apply different powers of the radius ratio

Compare radii 0.120/0.360 m and the stated fixed volumetric/surface rates.

Volume scales 27-fold and area ninefold.
Generation/capacity ratios are 0.640 and 1.920.
The ratio scales with radius under these assignments.

The geometric comparison does not provide an operating design.

Try it yourself

Change only dilute output to 85.0 kg h⁻¹ at mass fraction 0.100. Find rich flow, solute flow and mass fraction.
Show the answer and reasoning
65.0 kg h⁻¹; 27.5 kg h⁻¹ solute; fraction 0.423076923.
Total mass gives the stream flow; component balance gives 36.0−8.50=27.5 kg h⁻¹.
Set assigned purge fraction to 0.200, retaining fresh 37.0 and single-pass conversion 0.300. Find the steady flow account.
Show the answer and reasoning
Reactor/product/purge/recycle=84.0909091/25.2272727/11.7727273/47.0909091 kmol TrackedUnit h⁻¹; overall conversion 0.681818182.
Product plus external purge stays 37.0; the mixer counts recycle internally.
Halve both tank flows to 3.50 L min⁻¹. Compare at 18.0 min.
Show the answer and reasoning
Space time 24.0 min; C=0.363816724 mol L⁻¹; tank 30.5606048 mol; integrated in/out 37.8/15.6393952 mol.
The exponential argument halves while the initial and inlet concentrations remain fixed.
Use twenty-seven original small spheres to match the large sphere’s total volume under the same assigned rates. Find total area and generation/capacity ratio.
Show the answer and reasoning
Area 4.88580489 m², three times the large area; generation/capacity 0.938074540/1.46574147 kW, ratio 0.640.
Matching total volume does not match surface area. Engineering feasibility requires additional evidence.

What to carry forward

Independent component balances determine compatible stream inventories.
Internal recycle and external fresh feed belong to different boundaries.
A transient component balance must include accumulation.
Area and volume scale differently, while real scale-up also needs transport and process evidence.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.