Continuous high-temperature short-time sterilisation causes less loss of heat-labile nutrients than batch sterilisation to the same Del factor mainly because

spore inactivation has a higher activation energy than nutrient degradation, so raising temperature accelerates killing more than damage

nutrient degradation has a higher activation energy than spore inactivation

the Del factor required falls as temperature rises

spores are killed by pressure rather than by heat

Which of the following statements about membrane separations are correct?

Ultrafiltration membranes are rated by molecular-weight cut-off.

Cross-flow operation reduces cake build-up compared with dead-end operation.

Concentration polarisation raises the solute concentration at the membrane surface above that in the bulk.

Ultrafiltration membranes are rated by molecular-weight cut-off.

Cross-flow operation reduces cake build-up compared with dead-end operation.

Concentration polarisation raises the solute concentration at the membrane surface above that in the bulk.

Reverse osmosis membranes retain only particles larger than 0.1 µm.

A protein with pI 5.0 is to be captured from a buffer at pH 7.5. It will bind to

an anion exchanger such as DEAE

neither, because it carries no net charge

a gel-filtration matrix, by charge

Which of the following statements about protein chromatography are correct?

In hydrophobic interaction chromatography, proteins are loaded at high salt concentration.

His-tagged proteins bound to Ni-NTA are eluted with imidazole.

Protein A binds the Fc region of IgG.

In hydrophobic interaction chromatography, proteins are loaded at high salt concentration.

His-tagged proteins bound to Ni-NTA are eluted with imidazole.

In gel filtration, the smallest molecules elute first.

Protein A binds the Fc region of IgG.

A loop under proportional-only control oscillates with constant amplitude at an ultimate gain Ku = 10 and an ultimate period Pu = 4 min. The Ziegler–Nichols PID settings are

K_c = 4.5, τ_I = 3.33 min, τ_D = 0

K_c = 5, τ_I = 4 min, τ_D = 1 min

K_c = 10, τ_I = 2 min, τ_D = 0.5 min

A proportional controller on a first-order process leaves a steady-state offset after a set-point change. Adding which control action eliminates the offset?

Feed-forward action on the set point

The cooling-water valve on the jacket of a large exothermic fermenter should be chosen so that, on failure of the instrument air supply, it

fails open, so it is an air-to-close valve

fails closed, so it is an air-to-open valve

holds its last position, whatever the actuator type

fails closed, so it is an air-to-close valve

Upstream and Downstream Processing, Measurement and Control

The second chapter of Section 4 of the Biotechnology paper takes the syllabus’s second sub-heading, Upstream and Downstream Processing, which in this paper also carries process instrumentation and control. Upstream: media formulation and optimisation, and sterilisation of air and media with the Del factor. Downstream: membrane filtration and ultrafiltration, high-speed and ultracentrifugation, cell disruption, the principles of ion-exchange, gel-filtration, hydrophobic-interaction and affinity chromatography, GC, HPLC and FPLC, and extraction, adsorption and drying. Then measurement devices and valves, first- and second-order systems, feedback and feed-forward control, P, I and D control, and controller tuning.

1. Media formulation and optimisation

A medium supplies carbon and energy (glucose, sucrose, molasses, starch), nitrogen (ammonium salts, urea, corn steep liquor, yeast extract, peptones), phosphorus, sulfur, trace metals, vitamins and growth factors, and for particular processes precursors (phenylacetic acid for penicillin G), inducers (IPTG, methanol for the Pichia AOX1 promoter), buffers and antifoams. Defined media give reproducibility and easy downstream processing; complex media are cheap and give high yields but vary between batches. Formulation starts from an elemental balance: to make 30 g/L of cells containing 12% nitrogen needs 3.6 g/L of N, and since (NH₄)₂SO₄ is 28/132 = 21.2% N, about 17 g/L of the salt.

🎯 Why one-factor-at-a-time is replaced by statistical design

Varying one component at a time misses interactions and needs many runs. A Plackett–Burman design screens many factors in few runs (N runs for up to N − 1 factors) to find the few that matter; response surface methodology — central composite or Box–Behnken designs — then fits a quadratic model around the optimum of those few, which captures curvature and interaction.

2. Sterilisation of media and air — thermal death kinetics and the Del factor

Heat kills organisms by first-order kinetics, dN/dt = −k_d·N, so N = N₀e^(−k_d·t), and the decimal reduction time D = 2.303/k_d is the time for a tenfold kill. k_d follows Arrhenius, k_d = A·e^(−E/RT). The design quantity is the Del factor, ∇ = ln(N₀/N) = ∫k_d dt. N₀ is the total count in the whole batch and N the acceptable final count — usually 10⁻³, a one-in-a-thousand chance that a batch is contaminated. A 10 m³ batch with 10⁵ spores/mL has N₀ = 10⁵ × 10⁷ = 10¹², so ∇ = ln(10¹²/10⁻³) = ln 10¹⁵ = 34.54. In batch sterilisation heating, holding and cooling all contribute, ∇_total = ∇_heat + ∇_hold + ∇_cool; if heating and cooling supply 14.54, the holding stage at k_d = 2.5 min⁻¹ must supply 20.0, so t_hold = 20/2.5 = 8 min. Because N₀ grows with volume, a larger vessel needs a larger ∇ for the same risk.

🎯 Why high-temperature short-time sterilisation spares the medium

Spore inactivation has a much higher activation energy than the thermal degradation of vitamins, sugars and proteins, so raising the temperature speeds killing far more than it speeds damage. Continuous sterilisation at a higher temperature for seconds therefore reaches the required ∇ with less nutrient loss than a long batch hold — and it also removes the slow heating and cooling ramps that batch sterilisation cannot avoid.

Air for aeration is sterilised by filtration, not heat: depth filters of fibre or glass wool capture particles by inertial impaction, interception and diffusion, and cartridge membrane filters of 0.2 µm rating act as absolute barriers; they must be kept dry, since a wetted filter passes organisms. Heat-labile media components (vitamins, serum, some antibiotics) are likewise filter-sterilised through 0.2 µm membranes and added after the bulk medium has been autoclaved.

3. Filtration, centrifugation and cell disruption

Pressure-driven membrane processes
Process	Retains	Bioprocess use
Microfiltration	Particles of about 0.1–10 µm: cells, debris	Cell harvesting and clarification; 0.2 µm sterile filtration
Ultrafiltration	Macromolecules, rated by molecular-weight cut-off (MWCO) in kDa	Concentrating proteins; diafiltration for buffer exchange
Nanofiltration and reverse osmosis	Small organics and multivalent ions (NF); essentially all solutes and salts (RO)	Water purification; concentrating small products

Dead-end filtration builds a cake whose resistance grows with the volume filtered (the Ruth equation; filter aids such as diatomaceous earth keep compressible cakes porous, and rotary vacuum drum filters handle fungal mycelia). Cross-flow (tangential) filtration sweeps the membrane surface, limiting cake growth. Flux still falls because retained solute accumulates at the surface — concentration polarisation, which can progress to a gel layer — and because of fouling. Centrifugation accelerates settling: Stokes gives the terminal velocity v = d²(ρₚ − ρ)g/18µ, and in a centrifuge g is replaced by ω²r, so the relative centrifugal force is ω²r/g. Disc-stack and tubular-bowl machines harvest cells continuously; ultracentrifuges separate macromolecules and organelles either by rate-zonal sedimentation through a sucrose gradient (by sedimentation coefficient, in Svedberg units) or isopycnically in CsCl (by buoyant density).

Cell disruption releases intracellular products. Mechanical methods — the high-pressure homogeniser, the bead mill, sonication at bench scale — dominate at large scale. Protein release from a homogeniser is first order in the number of passes, ln[R_m/(R_m − R)] = kN at a fixed pressure: if one pass releases 40%, k = −ln 0.6 = 0.511, and 90% release needs N = ln 10/0.511 = 4.5, so 5 passes. Non-mechanical methods include lysozyme (with EDTA to open the Gram-negative outer membrane), osmotic shock (which releases periplasmic proteins), detergents, alkali, solvents and freeze–thaw. More disruption releases more product but also finer debris that is harder to remove, so the degree of disruption is itself an optimisation.

4. Principles of chromatography — ion exchange, gel filtration, HIC, affinity, GC, HPLC and FPLC

Separation principle and elution for each mode
Mode	Separates by	Binding and elution
Ion exchange	Net charge	Above its pI a protein is negative and binds an anion exchanger (DEAE, Q); below its pI it binds a cation exchanger (CM, SP); elute with a salt gradient or pH shift
Gel filtration (size exclusion)	Hydrodynamic size	No binding; large molecules are excluded from the pores and elute first; K_av = (V_e − V₀)/(V_t − V₀)
Hydrophobic interaction	Surface hydrophobicity	Load at high salt (ammonium sulfate), which strengthens hydrophobic contact; elute by decreasing salt
Affinity	Specific biological binding	His-tag on Ni-NTA eluted with imidazole; IgG Fc on protein A eluted at low pH; GST tag on glutathione eluted with free glutathione
GC	Volatility and partition into a stationary liquid	Inert carrier gas; volatile, thermally stable analytes (or derivatives); flame-ionisation or thermal-conductivity detection
HPLC and FPLC	Any of the above chemistries on small particles	HPLC runs at high pressure, often reverse phase (C18) for small molecules and peptides; FPLC runs at moderate pressure with biocompatible materials for native proteins

Column performance is judged by resolution, Rs = 2(t_R2 − t_R1)/(w₁ + w₂), with Rs ≥ 1.5 giving baseline separation; peaks at 8 and 10 min with base widths 0.8 and 1.2 min give Rs = 4/2 = 2.0. Efficiency is the plate number N = 16(t_R/w)² and the plate height HETP = L/N, which the van Deemter equation H = A + B/u + Cu splits into eddy diffusion, longitudinal diffusion and mass-transfer resistance, with an optimum flow velocity. In gel filtration a protein eluting at 70 mL from a column with V₀ = 40 mL and V_t = 120 mL has K_av = 30/80 = 0.375, and K_av falls roughly linearly with log(molecular weight) for calibration.

5. Extraction, adsorption and drying

In liquid–liquid extraction the partition coefficient K = C_solvent/C_aqueous and the extraction factor E = K × (solvent volume/feed volume) decide recovery: a single equilibrium stage leaves a fraction 1/(1 + E) in the feed and extracts E/(1 + E). With K = 4 and a solvent-to-feed ratio of 0.5, E = 2 and two-thirds (0.667) is extracted; splitting the same solvent into two cross-current stages with E = 1 each leaves (1/2)² = 0.25, so staging beats one large contact. Penicillin is extracted into butyl or amyl acetate from acidified broth, where the uncharged acid partitions into the solvent, then back into a neutral aqueous buffer. Aqueous two-phase systems (PEG–dextran, PEG–salt) partition proteins gently without organic solvents.

Adsorption is described by isotherms: Langmuir, q = q_max·KC/(1 + KC), for monolayer coverage of identical sites (q_max = 50 mg/g, K = 0.2 L/mg, C = 20 mg/L gives q = 50 × 4/5 = 40 mg/g), and Freundlich, q = K·C^(1/n), an empirical form for heterogeneous surfaces. Adsorption on activated carbon or resins captures small products and removes colour and impurities. Drying has a constant-rate period, while the surface stays wet and evaporation is limited by heat and mass transfer to it, then, below the critical moisture content, a falling-rate period controlled by moisture diffusing out from inside. Heat-sensitive proteins are freeze-dried (frozen, then ice sublimed under vacuum in primary drying and bound water desorbed in secondary drying); spray drying dries fine droplets in seconds for enzymes and food products.

6. Measurement devices, valves, system dynamics, controllers and tuning

Measurement: temperature by resistance thermometers (Pt100) or thermocouples; pH by a steam-sterilisable glass electrode; dissolved oxygen by a polarographic (Clark) or galvanic membrane electrode, or an optical sensor based on luminescence quenching; pressure by diaphragm gauges; gas flow by rotameters or thermal mass-flow meters; foam by conductance probes; and exit gas by paramagnetic O₂ and infrared CO₂ analysers, which give the oxygen uptake rate, the CO₂ evolution rate and their ratio, the respiratory quotient.
Valves: a control valve is an actuator on a valve body. Air-to-open valves fail closed and air-to-close valves fail open on loss of air, and the choice is made for safety — a steam valve should fail closed, a cooling-water valve on an exothermic fermenter should fail open. Trim characteristics are linear, equal-percentage or quick-opening. Aseptic service favours diaphragm valves, which have no crevices and can be steam-sterilised.

A first-order system G(s) = K/(τs + 1) responds to a step of size M as y = KM(1 − e^(−t/τ)), reaching 63.2% of its final change at t = τ: a thermometer with τ = 2 min plunged from 25 °C into 75 °C reads 25 + 50 × 0.632 = 56.6 °C after 2 min. A second-order system G(s) = K/(τ²s² + 2ζτs + 1) is overdamped for ζ > 1, critically damped at ζ = 1 and underdamped for ζ < 1, when its overshoot is exp(−πζ/√(1 − ζ²)) — 16.3% at ζ = 0.5 — and the decay ratio is the overshoot squared. Feedback control measures the controlled variable and corrects after an error appears; feed-forward measures a disturbance and acts before it upsets the process, which needs a process model and is usually combined with feedback.

Controller modes, Gc(s) = Kc[1 + 1/(τ_I s) + τ_D s], and Ziegler–Nichols settings from the ultimate gain Ku and period Pu
Mode	Effect	Ziegler–Nichols setting
P	Fast, but leaves an offset: for a set-point step on a process of gain Kp, offset = 1/(1 + KcKp) of the step	K_c = 0.5 K_u
PI	Integral action removes offset but slows the response and can make it oscillate	K_c = 0.45 K_u, τ_I = P_u/1.2
PID	Derivative action anticipates from the rate of change of error and damps oscillation, but amplifies noise	K_c = 0.6 K_u, τ_I = P_u/2, τ_D = P_u/8

🧠 Tuning in one line

Ziegler–Nichols closed-loop tuning raises a P-only gain until the loop oscillates steadily (Ku, Pu) and reads the settings from the table: Ku = 10, Pu = 4 min gives PID Kc = 6, τ_I = 2 min, τ_D = 0.5 min. The open-loop alternative fits a first-order-plus-dead-time model to a process reaction curve and uses Cohen–Coon or IMC rules.

Key takeaways

Del factor ∇ = ln(N₀/N) = ∫k_d dt with N₀ for the whole batch and N typically 10⁻³; D = 2.303/k_d; HTST works because spore kill has the higher activation energy.
MF harvests cells, UF concentrates proteins by MWCO, cross-flow limits cake but not concentration polarisation; homogeniser release is first order in passes.
Above pI, bind an anion exchanger; gel filtration elutes large first; HIC binds at high salt; His-tags elute with imidazole; Rs = 2Δt_R/(w₁ + w₂).
Single-stage extraction recovers E/(1 + E); several smaller stages beat one large one; Langmuir q = q_maxKC/(1 + KC); freeze-drying sublimes ice under vacuum.
First order reaches 63.2% at t = τ; overshoot exp(−πζ/√(1 − ζ²)); P leaves offset 1/(1 + KcKp), I removes it, D anticipates; Z–N PID is 0.6Ku, Pu/2, Pu/8.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

To screen eleven medium components for significant effects on titre using only twelve runs, the appropriate design is
1. a Plackett–Burman design
2. a central composite design
3. a full factorial design
4. one-factor-at-a-time variation
Show answer
Answer: A — a Plackett–Burman design
Plackett–Burman designs screen up to N − 1 factors in N runs, here 11 in 12, estimating main effects only, to pick the few that matter. A central composite design is a response-surface tool for optimising those few; a full factorial of 11 two-level factors needs 2048 runs; one factor at a time needs many runs and misses interactions.
A process must produce 30 g/L of biomass that contains 12% nitrogen by dry weight, with (NH₄)₂SO₄ (molar mass 132 g/mol) as the sole nitrogen source. The minimum (NH₄)₂SO₄ concentration required is ______ g/L (round off to the nearest integer).
Numerical answer — type the value.
Show answer
Answer: 17
Nitrogen needed = 0.12 × 30 = 3.6 g/L. (NH₄)₂SO₄ carries two N atoms, 28 g per 132 g, so it is 21.2% N and 3.6/0.212 = 17 g/L. Counting one nitrogen per formula unit (14/132) doubles the answer to about 34.
A 10 m³ batch of medium contains 10⁵ viable spores per mL. For a probability of contamination of 1 in 1000 batches (N = 10⁻³), the required Del factor is ______ (round off to two decimal places).
Numerical answer — type the value.
Show answer
Answer: 34.54
N₀ = 10⁵ spores/mL × 10⁷ mL = 10¹², so ∇ = ln(N₀/N) = ln(10¹²/10⁻³) = ln 10¹⁵ = 15 × 2.3026 = 34.54. Using the concentration per mL instead of the total count, or log₁₀ instead of ln (giving 15), are the two usual errors.
For the batch above (required ∇ = ln 10¹⁵), the heating and cooling stages together contribute a Del factor of 14.54. If the specific death rate at the holding temperature is 2.5 min⁻¹, the holding time required is ______ min.
Numerical answer — type the value.
Show answer
Answer: 8
∇_hold = ∇_total − (∇_heat + ∇_cool) = 34.54 − 14.54 = 20.0; at constant temperature ∇_hold = k_d·t, so t = 20.0/2.5 = 8 min. Ignoring the heating and cooling contributions gives 13.8 min, an over-sterilisation that damages the medium unnecessarily.
Continuous high-temperature short-time sterilisation causes less loss of heat-labile nutrients than batch sterilisation to the same Del factor mainly because
1. spore inactivation has a higher activation energy than nutrient degradation, so raising temperature accelerates killing more than damage
2. nutrient degradation has a higher activation energy than spore inactivation
3. the Del factor required falls as temperature rises
4. spores are killed by pressure rather than by heat
Show answer
Answer: A — spore inactivation has a higher activation energy than nutrient degradation, so raising temperature accelerates killing more than damage
With k = Ae^(−E/RT), the reaction with the larger E is the more temperature-sensitive. Spore death has the larger activation energy, so at a higher temperature the required ∇ is reached in a time short enough that the lower-E nutrient reactions barely proceed. Reversing the activation energies would make HTST worse, and the required ∇ is set by N₀ and N, not by temperature.
In a high-pressure homogeniser, protein release follows ln[R_m/(R_m − R)] = kN. If one pass releases 40% of the releasable protein, the minimum whole number of passes needed to release at least 90% is ______.
Numerical answer — type the value.
Show answer
Answer: 5
One pass: k = ln[1/(1 − 0.4)] = ln(1/0.6) = 0.511. For 90%: N = ln[1/(1 − 0.9)]/k = ln 10/0.511 = 2.303/0.511 = 4.5, so 5 whole passes. Assuming each pass adds 40 percentage points gives 3 passes, which ignores that each pass acts only on what remains.
Which of the following statements about membrane separations are correct?
1. Ultrafiltration membranes are rated by molecular-weight cut-off.
2. Cross-flow operation reduces cake build-up compared with dead-end operation.
3. Concentration polarisation raises the solute concentration at the membrane surface above that in the bulk.
4. Reverse osmosis membranes retain only particles larger than 0.1 µm.
Show answer
Answer: A — Ultrafiltration membranes are rated by molecular-weight cut-off.; B — Cross-flow operation reduces cake build-up compared with dead-end operation.; C — Concentration polarisation raises the solute concentration at the membrane surface above that in the bulk.
UF is specified by MWCO in kDa, cross-flow sweeps retained material off the surface, and polarisation is the accumulation of rejected solute in the boundary layer that lowers flux. Reverse osmosis is the tightest process, rejecting dissolved salts; a 0.1 µm cut-off describes microfiltration.
A protein with pI 5.0 is to be captured from a buffer at pH 7.5. It will bind to
1. an anion exchanger such as DEAE
2. a cation exchanger such as CM
3. neither, because it carries no net charge
4. a gel-filtration matrix, by charge
Show answer
Answer: A — an anion exchanger such as DEAE
At a pH above its pI a protein has lost protons and carries a net negative charge, so it binds the positively charged DEAE groups of an anion exchanger and is eluted by rising salt. It would bind a cation exchanger only below pH 5; at the pI it carries no net charge; gel filtration separates by size, not charge.
Two peaks elute at 8.0 and 10.0 min with base widths of 0.8 and 1.2 min respectively. The chromatographic resolution is ______.
Numerical answer — type the value.
Show answer
Answer: 2
Rs = 2(t_R2 − t_R1)/(w₁ + w₂) = 2 × 2.0/(0.8 + 1.2) = 4.0/2.0 = 2.0, above the 1.5 needed for baseline separation. Omitting the factor of 2 — dividing the gap by the sum of the widths — gives 1.0 and wrongly suggests overlapping peaks.
A gel-filtration column has void volume V₀ = 40 mL and total bed volume V_t = 120 mL. A protein elutes at V_e = 70 mL. Its partition coefficient K_av is ______ (round off to three decimal places).
Numerical answer — type the value.
Show answer
Answer: 0.375
K_av = (V_e − V₀)/(V_t − V₀) = (70 − 40)/(120 − 40) = 30/80 = 0.375. K_av = 0 means complete exclusion (elution at V₀) and K_av near 1 means free access to the pores; dividing by V_t instead of (V_t − V₀) gives 0.25.
Which of the following statements about protein chromatography are correct?
1. In hydrophobic interaction chromatography, proteins are loaded at high salt concentration.
2. His-tagged proteins bound to Ni-NTA are eluted with imidazole.
3. In gel filtration, the smallest molecules elute first.
4. Protein A binds the Fc region of IgG.
Show answer
Answer: A — In hydrophobic interaction chromatography, proteins are loaded at high salt concentration.; B — His-tagged proteins bound to Ni-NTA are eluted with imidazole.; D — Protein A binds the Fc region of IgG.
High salt strengthens hydrophobic contacts, so HIC loads at high and elutes at low salt; imidazole competes with histidine for the nickel; protein A captures antibodies by their Fc and releases them at low pH. In gel filtration small molecules enter the pores and take the longer path, so the largest elute first.
An antibiotic has a partition coefficient K = 4 (solvent/aqueous). A single equilibrium extraction uses solvent equal to half the volume of the aqueous feed. The fraction of antibiotic extracted into the solvent is ______ (round off to three decimal places).
Numerical answer — type the value.
Show answer
Answer: 0.667
The extraction factor E = K × (V_s/V_f) = 4 × 0.5 = 2, and a single stage extracts E/(1 + E) = 2/3 = 0.667, leaving 1/3 in the feed. Taking K itself as the ratio of amounts (giving 0.8) ignores the unequal volumes.
A thermometer behaves as a first-order system with time constant 2 min. It is moved suddenly from a bath at 25 °C to one at 75 °C. Its reading after 2 min is ______ °C (round off to one decimal place).
Numerical answer — type the value.
Show answer
Answer: 56.6
y = y₀ + ΔT(1 − e^(−t/τ)) = 25 + 50(1 − e^(−1)) = 25 + 50 × 0.632 = 56.6 °C. Applying 63.2% to 75 °C instead of to the 50 °C change gives 47.4, and forgetting the initial 25 °C gives 31.6.
An underdamped second-order process has damping coefficient ζ = 0.5. The percentage overshoot of its response to a step input is ______ (round off to one decimal place).
Numerical answer — type the value.
Show answer
Answer: 16.3
Overshoot = exp(−πζ/√(1 − ζ²)) = exp(−π × 0.5/0.866) = exp(−1.814) = 0.163, so 16.3%. The decay ratio would be its square, 2.7%. Omitting the square root in the denominator gives exp(−2.09) = 12.3%, a common slip.
A loop under proportional-only control oscillates with constant amplitude at an ultimate gain Ku = 10 and an ultimate period Pu = 4 min. The Ziegler–Nichols PID settings are
1. K_c = 6, τ_I = 2 min, τ_D = 0.5 min
2. K_c = 4.5, τ_I = 3.33 min, τ_D = 0
3. K_c = 5, τ_I = 4 min, τ_D = 1 min
4. K_c = 10, τ_I = 2 min, τ_D = 0.5 min
Show answer
Answer: A — K_c = 6, τ_I = 2 min, τ_D = 0.5 min
Ziegler–Nichols PID: Kc = 0.6Ku = 6, τ_I = Pu/2 = 2 min, τ_D = Pu/8 = 0.5 min. The 4.5 and 3.33 min pair is the PI setting (0.45Ku, Pu/1.2), and running at Kc = Ku would leave the loop on the edge of instability.
A proportional controller on a first-order process leaves a steady-state offset after a set-point change. Adding which control action eliminates the offset?
1. Integral action
2. Derivative action
3. A larger proportional band
4. Feed-forward action on the set point
Show answer
Answer: A — Integral action
Integral action keeps changing the output for as long as any error persists, so the loop can come to rest only at zero error. Raising Kc shrinks the offset, 1/(1 + KcKp) of the step, but never removes it, and a larger proportional band means a smaller Kc and a bigger offset; derivative action responds only to changing error.
The cooling-water valve on the jacket of a large exothermic fermenter should be chosen so that, on failure of the instrument air supply, it
1. fails open, so it is an air-to-close valve
2. fails closed, so it is an air-to-open valve
3. holds its last position, whatever the actuator type
4. fails closed, so it is an air-to-close valve
Show answer
Answer: A — fails open, so it is an air-to-close valve
If cooling is lost, heat released by the culture can drive the temperature up and kill it, so the safe failure state is full cooling: the valve must fail open, which means air holds it closed — air-to-close. A steam valve is the opposite case and should fail closed. The last option pairs failing closed with the wrong actuator type.