Optimum temperature progression and multiple steady states
Best temperature policies on the conversion-temperature chart, the optimum temperature progression for reversible exothermic reactions, and CSTR multiple steady states, stability, ignition and extinction.
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Why it matters
For reversible exothermic reactions — SO₂ oxidation, ammonia and methanol synthesis, water-gas shift — temperature pulls in two directions: hotter is faster, cooler gives a higher equilibrium conversion. Following the optimum temperature progression minimises catalyst or reactor volume. In stirred tanks, the same heat-generation feedback creates multiple steady states, ignition and extinction, which every operator of an exothermic CSTR must understand.
Key ideas
Conversion–temperature (X–T) chart. Plot lines of constant rate on X vs T. For an irreversible reaction the rate always increases with T at fixed X, so the best policy is simply the highest temperature allowed by materials, catalyst life or side reactions. For a reversible endothermic reaction both rate and equilibrium improve with T, so again use the highest allowable temperature.
Reversible exothermic reactions. At fixed X, the rate rises with T at first, passes through a maximum, and falls to zero at the equilibrium temperature for that X. The locus of these maxima is the optimum temperature progression (OTP). It lies below the equilibrium curve and falls as conversion rises: start hot (rate limited, far from equilibrium), finish cooler (equilibrium limited). A PFR following the OTP needs the minimum volume. In practice the OTP is approximated by adiabatic beds with inter-stage cooling, by cold-shot (quench) injection between beds, or by cooled multitubular reactors.
Finding the OTP. At fixed X, set ∂(−rA)/∂T = 0. For A ⇌ R with −rA = k1·CA0·(1 − X) − k2·CA0·X this gives k1·E1·(1 − X) = k2·E2·X, which can be solved for T at each X.
Steady states of a CSTR. The exit conditions must satisfy both:
- the mole balance, giving X as a function of T — for first order X_MB = kτ/(1 + kτ), an S-shaped curve; multiplied by (−ΔHr) it becomes the heat-generation curve G(T);
- the energy balance, a straight line — the heat-removal line R(T) = Cp'·(1 + κ)·(T − Tc). Each intersection is a steady state. With an S-shaped G and a straight R there can be one or three intersections.
Stability. At an intersection, if a small rise in T increases heat removal more than heat generation (slope of R > slope of G), the reactor returns: stable. If the slope of G exceeds that of R, a small upset grows: unstable. With three steady states, the lower (extinguished, low X) and upper (ignited, high X) are stable; the middle one is unstable and cannot be held without control.
Ignition and extinction. Slowly raising the feed temperature shifts R to the right; at a tangency the low state disappears and the reactor jumps to the high state (ignition). Lowering the feed temperature again, the reactor stays ignited until a lower tangency (extinction). This hysteresis is why start-up and shut-down procedures matter.
Formulas
−rA = k1·CA0·(1 − X) − k2·CA0·X (A ⇌ R, first order both ways)
k1·E1·(1 − X) = k2·E2·X (condition for maximum rate at fixed X)
T_opt = (E2 − E1) / {R·ln[(k20·E2·X)/(k10·E1·(1 − X))]}
- k10, k20: frequency factors (s⁻¹); E1, E2: activation energies (J/mol), E2 − E1 = −ΔHr for an exothermic reaction; R = 8.314 J/mol·K.
T_eq = (E2 − E1) / {R·ln[(k20·X)/(k10·(1 − X))]} (equilibrium temperature for conversion X)
G(T) = (−ΔHr)·X_MB(T), X_MB = kτ/(1 + kτ) (first-order CSTR)
R(T) = Cp'·(1 + κ)·(T − Tc), κ = UA/(FA0·Cp'), Tc = (T0 + κ·Ta)/(1 + κ)
- Cp': heat capacity per mole of A fed (J/mol·K); UA: exchanger conductance (W/K); T0: feed temperature; Ta: coolant temperature (K).
Stable if dR/dT > dG/dT at the steady state
dX_MB/dT = X·(1 − X)·E/(R·T²) (first-order CSTR, slope of the S-curve)
Worked examples
Example 1 (standard, OTP). For A ⇌ R: k1 = 10⁶·exp(−50 000/RT) s⁻¹, k2 = 10¹³·exp(−100 000/RT) s⁻¹ (so ΔHr = −50 kJ/mol). Find the equilibrium temperature and the optimum temperature at X = 0.5.
- Equilibrium at X = 0.5 requires k1 = k2: T_eq = 50 000/[8.314 × ln(10¹³/10⁶)] = 50 000/(8.314 × 16.12) = 373.1 K.
- Optimum: T_opt = 50 000/{8.314 × ln[(10¹³ × 100 000 × 0.5)/(10⁶ × 50 000 × 0.5)]} = 50 000/[8.314 × ln(2 × 10⁷)] = 50 000/(8.314 × 16.81) = 357.7 K.
- Repeating at X = 0.2 gives 389.9 K and at X = 0.8 gives 330.5 K.
T_eq = 373 K; T_opt = 358 K at X = 0.5. The OTP falls from about 390 K to 330 K as X goes from 0.2 to 0.8.
Example 2 (GATE level, stability). An adiabatic CSTR (feed 300 K, ΔTad = 200 K) runs a first-order reaction with E/R = 10 000 K. At T = 400 K, kτ = 1. Show that 400 K is a steady state and test its stability.
- Mole balance: X_MB = kτ/(1 + kτ) = 1/2 = 0.5.
- Energy balance (adiabatic line): X_EB = (T − T0)/ΔTad = (400 − 300)/200 = 0.5. Both agree, so 400 K is a steady state.
- Slope of the S-curve: dX_MB/dT = X(1 − X)·(E/R)/T² = 0.25 × 10 000/400² = 0.0156 K⁻¹.
- Slope of the energy line: 1/ΔTad = 0.005 K⁻¹.
- The generation slope exceeds the removal slope, so the state is unstable.
400 K is the unstable middle steady state. Solving numerically, the other two steady states are near 300 K (X ≈ 0) and 498.6 K (X ≈ 0.993).
Common mistakes
- Running a reversible exothermic reactor isothermally at high temperature and expecting high conversion — equilibrium falls with T.
- Placing the OTP on or above the equilibrium curve; the rate is zero there.
- Judging stability by temperature level instead of comparing the slopes of G(T) and R(T).
- Forgetting coolant terms: κ and Tc shift and tilt the heat-removal line.
- Assuming the steady state reached depends only on design values; with hysteresis it depends on the start-up path.
For GATE CH
Typical questions: qualitative shape of the OTP and why it falls with conversion; best temperature policy for irreversible, endothermic and exothermic reversible reactions; counting and classifying CSTR steady states from G–R plots; computing Tc or κ; checking whether a given T is a steady state from mole and energy balances. Practise drawing the S-curve with straight removal lines for different feed temperatures and coolant flows.
Quick check
- For an irreversible exothermic reaction, what temperature policy minimises volume?
- Does the OTP for a reversible exothermic reaction rise or fall with X?
- Of three CSTR steady states, which are stable?
- With κ = 1, T0 = 300 K, Ta = 290 K, find Tc.
Answers: 1. the highest allowable temperature; 2. it falls; 3. the lowest and the highest; 4. Tc = (300 + 290)/2 = 295 K.
Interview questions
All Chemical Reaction Engineering interview questionsTry answering each one aloud before you open it.
1.What is meant by 'optimum temperature progression' in chemical reaction engineering?Concept
It is the temperature at each conversion that gives the maximum reaction rate, so a reactor that follows it reaches a given conversion in the smallest volume. For irreversible reactions and reversible endothermic reactions this is simply the highest allowable temperature. For reversible exothermic reactions the rate at fixed conversion passes through a maximum below the equilibrium temperature, so the optimum progression starts hot and falls as conversion rises, staying below the equilibrium curve. It is approximated industrially with staged adiabatic beds and inter-cooling or quenching.
2.Explain the concept of multiple steady states in a chemical reactor.Concept
Multiple steady states occur when a chemical reactor can operate at different stable conditions (temperature, pressure, concentration) for the same set of input parameters. This phenomenon is often observed in non-linear systems, such as those involving exothermic reactions, where the reactor can have more than one stable operating point.
3.Why is it important to consider optimum temperature progression in exothermic reactions?Application
In exothermic reactions, heat is released, which can lead to temperature increases that may cause runaway reactions or undesired side reactions. By considering optimum temperature progression, engineers can control the temperature to maintain reaction stability, improve yield, and ensure safety.
4.What could happen if a reactor operating under multiple steady states is disturbed?Application
If a reactor operating under multiple steady states is disturbed, it may shift from one steady state to another. This can lead to significant changes in reaction conditions, such as temperature and concentration, potentially affecting product quality and safety. It is crucial to design control systems to manage such transitions.
5.How can the presence of multiple steady states affect the design of a chemical reactor?Application
The presence of multiple steady states requires careful design and control strategies to ensure the reactor operates at the desired steady state. Engineers must consider factors like heat removal, feed composition, and catalyst activity to avoid unintended shifts between steady states, which could impact efficiency and safety.
6.Describe a method to determine the optimum temperature progression for a given reaction.Application
Write the rate as a function of conversion and temperature, including the reverse reaction, and at each fixed conversion find the temperature where ∂(−rA)/∂T = 0. For A ⇌ R with first-order steps this gives k1·E1·(1 − X) = k2·E2·X, which can be solved for T at each X. Plotting these points on the conversion–temperature chart gives the locus of maximum rates. The reactor design then follows that locus, in practice as closely as staged beds with cooling or a cooled tube allow.
7.What role does catalyst activity play in achieving optimum temperature progression?Application
Catalyst activity sets how fast the reaction goes at a given temperature. A more active catalyst lets the reactor reach high rates at lower temperatures, which for reversible exothermic reactions is valuable because the equilibrium conversion is higher there, so the final beds can run cooler. Catalyst limits — sintering above a maximum temperature and too low activity below an ignition temperature — bound the usable part of the optimum progression. As the catalyst deactivates, operators raise the bed temperatures to keep the rate up.
8.Calculate the heat removal rate required to maintain a steady state in a reactor with an exothermic reaction releasing 500 kJ/mol at a rate of 2 mol/s.Numerical
To maintain a steady state, the heat removal rate must equal the heat generated by the reaction. Heat generated = 500 kJ/mol × 2 mol/s = 1000 kJ/s. Therefore, the heat removal rate required is 1000 kJ/s.
9.A CSTR has two possible stable steady states, at 300 K and at 350 K. How do you decide whether a steady state is stable, and which one would you operate at?Application
Stability is judged from the slopes of the heat-generation curve G(T) and the heat-removal line R(T) at the steady state: if dR/dT > dG/dT, a small temperature rise removes more heat than it generates and the reactor returns, so the state is stable. With an S-shaped G curve, the lowest and highest intersections are stable and the middle one is unstable. The choice between the low and high stable states is not about stability but economics and safety: the low state usually has negligible conversion, so reactors are normally run at the upper (ignited) state with enough cooling margin to avoid runaway.
10.What is the impact of feed composition on the occurrence of multiple steady states in a reactor?Application
Feed composition can significantly impact the occurrence of multiple steady states. Variations in reactant concentrations can alter reaction kinetics and thermodynamics, potentially leading to different steady states. Engineers must carefully control feed composition to ensure the reactor operates at the desired steady state.
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