Balances on multiple-unit processes

Real processes are chains of units. In the back end of the ethanol plant, whole stillage from the beer column goes to decanter centrifuges, which split it into wet cake and thin stillage. Part of the thin stillage returns to the slurry tank as backset; the rest goes to a multiple-effect evaporator that concentrates it to syrup. The syrup is mixed with the wet cake and dried in rotary dryers to make distillers dried grains with solubles, sold as animal feed. The dryers are the largest energy users in the plant after distillation, so Ingrid asks Theo to work out how much water each unit removes. That means a balance on four connected units, which is the subject of this lesson.

Subsystems and boundaries

In a multiple-unit process, a balance can be written around any subsystem: a single unit, a mixing point, a splitting point, a group of units, or the whole process. Each choice of boundary gives a different set of equations. The flows that cross the boundary appear in its balances; flows that stay inside cancel out. Drawing several boundaries on one flowchart and counting the degrees of freedom for each is the key to solving these problems efficiently.

The overall boundary around the whole stillage train has as inputs only the whole stillage, and as outputs the backset, the evaporator vapour, the dryer vapour and the dried grains. The wet cake, thin stillage, evaporator feed and syrup are internal and do not appear. An overall balance is often the quickest way to find product flows, but it cannot tell you anything about internal streams.

Degrees of freedom for multiple units

Count degrees of freedom for each subsystem separately, using only the unknowns and equations that belong to it. Then look for a subsystem with zero degrees of freedom, solve it, and mark its streams as known. That usually reduces the degrees of freedom of a neighbouring subsystem to zero, and the solution proceeds unit by unit. For the whole process, the total degrees of freedom equal the sum of the unknowns in all streams minus all independent equations on all units. Balances on every unit plus an overall balance are not all independent: the overall balance is the sum of the unit balances.

A useful bookkeeping rule: for a process of \(k\) units handling \(N\) species with no reaction, there are at most \(kN\) independent material balances, whichever \(k\) boundaries you choose, as long as together they include every unit. Choosing which \(k\) boundaries to use is a matter of convenience.

Splitters and mixers

A stream splitter divides a stream into two or more streams of identical composition. Because the outlets have the same composition as the inlet, the splitter gives only one independent balance, a total balance, no matter how many components there are. Writing species balances on a splitter adds nothing new, and counting them as extra equations is a classic mistake. The composition restrictions take their place: they are already used when every outlet is labelled with the inlet's fractions.

A mixing point combines streams of different composition into one. It behaves like any other unit, with as many independent balances as there are species. Mixers are where recycle streams rejoin the process, so they appear in nearly every recycle problem in Lesson 5.

The stillage train

The order of solution followed the degrees of freedom. The centrifuge had two unknowns and two balances, so it went first. Its outputs made the splitter solvable, then the evaporator, then the mixer and dryer. An overall balance at the end served as a check rather than as a step. In a different problem, for example one where the grains production is given and the stillage flow is unknown, the overall balance might be the place to start.

# Stillage train: centrifuge -> evaporator -> dryer, with backset split off the thin stillage
S = 215_000.0          # kg/h whole stillage from the beer column
s_S = 0.130            # solids mass fraction in whole stillage
s_cake, s_thin = 0.35, 0.070   # wet cake and thin stillage solids fractions
backset = 68_110.0     # kg/h of thin stillage returned to the slurry tank (Lesson 3)
s_syrup = 0.30         # syrup solids leaving the evaporator
s_ddgs = 0.90          # dried grains solids

# Centrifuge: total and solids balances in two unknowns (cake C, thin stillage T)
T = S * (s_cake - s_S) / (s_cake - s_thin)
C = S - T
print(f"Centrifuge: wet cake {C:9.0f} kg/h, thin stillage {T:9.0f} kg/h")

# Splitter: same composition in both outlets
E_feed = T - backset
print(f"Splitter: backset {backset:9.0f} kg/h, evaporator feed {E_feed:9.0f} kg/h")

# Evaporator: solids are a tie component
syrup = E_feed * s_thin / s_syrup
vap_evap = E_feed - syrup
print(f"Evaporator: syrup {syrup:9.0f} kg/h, water evaporated {vap_evap:9.0f} kg/h")

# Mixer + dryer: wet cake and syrup combine, dryer removes water to 90 % solids
solids_in = C * s_cake + syrup * s_syrup
dryer_feed = C + syrup
ddgs = solids_in / s_ddgs
vap_dryer = dryer_feed - ddgs
print(f"Dryer: feed {dryer_feed:9.0f} kg/h at {solids_in / dryer_feed:.3f} solids, DDGS {ddgs:9.0f} kg/h, water out {vap_dryer:9.0f} kg/h")

# Overall check on the train (boundary around all four units)
out = backset + vap_evap + ddgs + vap_dryer
print(f"Overall: in {S:.0f} kg/h, out {out:.0f} kg/h")
solids_out = backset * s_thin + ddgs * s_ddgs
print(f"Solids: in {S * s_S:.0f} kg/h, out {solids_out:.0f} kg/h")

Centrifuge: wet cake     46071 kg/h, thin stillage    168929 kg/h
Splitter: backset     68110 kg/h, evaporator feed    100819 kg/h
Evaporator: syrup     23524 kg/h, water evaporated     77294 kg/h
Dryer: feed     69596 kg/h at 0.333 solids, DDGS     25758 kg/h, water out     43838 kg/h
Overall: in 215000 kg/h, out 215000 kg/h
Solids: in 27950 kg/h, out 27950 kg/h

A second example: two columns in series

The ethanol from the beer column is still wet. The rectifier concentrates the beer column vapour toward the azeotrope, the composition at which a boiling ethanol and water mixture gives a vapour of the same composition, about 95.6 percent ethanol by mass at atmospheric pressure. Ordinary distillation cannot pass that point, so the rectifier product goes to molecular sieves, beds of zeolite pellets that adsorb water molecules but not the larger ethanol molecules. Treating the two separations in series is another multiple-unit problem.

Wet basis and dry basis

Moisture in solids can be reported two ways, and both appear in the stillage train. On a wet basis, moisture is the mass of water divided by the total mass of wet material; grains at 10 percent moisture contain 10 kg of water in every 100 kg. On a dry basis, moisture is the mass of water divided by the mass of dry solids; the same grains hold 11.1 kg of water per 100 kg of dry solids. Feed buyers quote wet-basis moisture, while dryer designers often use dry basis, because the dry solids flow is constant through a dryer and the water removed is then simply the dry solids flow times the change in dry-basis moisture.

For the dryer in the example, the feed at 33.3 percent solids has a dry-basis moisture of 2.00 kg of water per kg of solids, and the product at 90 percent solids has 0.111. With 23 180 kg/h of dry solids, the water removed is \(23\,180 \times (2.00 - 0.111) = 43\,790\ \text{kg/h}\), matching the earlier result within rounding. When a problem gives moistures, always check which basis is meant before you start.

Strategy for larger problems

As problems grow, the hand procedure becomes a strategy for ordering calculations. Write the degrees of freedom of every unit and of the overall process in a small table. Start where the count is zero. After each unit is solved, update the table. If no unit has zero degrees of freedom, look for a combination of units, or set up the equations for several units together and solve them as a linear system. Process simulators use the same two approaches: sequential modular simulators solve one unit at a time in a chosen order, while equation-oriented simulators solve all the equations at once.