Non-isothermal reactors: energy balances and adiabatic operation

Why it matters

Real reactors are rarely isothermal. An exothermic reaction heats its own contents, which raises the rate, which releases heat faster — the basis of both efficient adiabatic reactors and thermal runaway. Coupling the energy balance to the mole balance tells you the temperature profile, the cooling duty, and whether equilibrium will cap conversion.

Key ideas

Energy balance for a flow reactor (steady state). Heat added − shaft work + enthalpy in − enthalpy out = 0. With conversion as the variable and ΔCp ≈ 0, the enthalpy change splits into sensible heat of the stream and heat released by reaction. Per mole of A fed, the "lumped" heat capacity is Cp' = Σθi·Cpi, where θi = Fi0/FA0 (inerts and excess reactants included).

Adiabatic operation. No heat exchange (Q = 0) and negligible shaft work. The energy balance becomes a straight line linking conversion and temperature — the adiabatic operating line:

• T = T0 + (−ΔHr)·XA/Cp'.
• Exothermic (ΔHr < 0): temperature rises with conversion. Endothermic: it falls.
• The slope dXA/dT = Cp'/(−ΔHr). Adding inerts increases Cp' and flattens the temperature rise.
• The adiabatic temperature rise ΔTad = (−ΔHr)/Cp' is the rise at complete conversion. For liquids it is often written (−ΔHr)·CA0/(ρ·cp). The line holds for batch, CSTR and PFR alike, because it expresses only the first law; what differs between reactors is the mole balance it is combined with.

Combining with the mole balance.

• Adiabatic PFR: step along the reactor; at each XA, read T from the operating line, compute k(T) and −rA(XA, T), and integrate τ = CA0·∫dXA/(−rA).
• Adiabatic CSTR: the exit X and T must satisfy both the operating line and the mole balance simultaneously. Graphically, intersect the energy-balance line with the mole-balance curve X(T); more than one intersection means multiple steady states (next topic).
• Non-adiabatic reactors: heat exchange UA(T − Ta) appears in the balance, tilting the operating line and allowing temperature control.

Reversible reactions and equilibrium. For a reversible exothermic reaction the equilibrium conversion falls with temperature (van 't Hoff). An adiabatic exothermic reactor climbs its operating line until it meets the equilibrium curve, which caps conversion. That is why industrial reactors for SO₂ oxidation or ammonia synthesis use several adiabatic beds with inter-stage cooling.

Heat duty for isothermal operation. To hold T = T0 the exchanger must remove the reaction heat: Q = FA0·XA·(−ΔHr).

Formulas

Cp' = Σ θi·Cpi, θi = Fi0/FA0

• Cpi: molar heat capacity of species i (J/mol·K); Cp' in J/(mol A fed)·K.

XA = Cp'·(T − T0)/(−ΔHr) or T = T0 + (−ΔHr)·XA/Cp' (adiabatic, ΔCp ≈ 0)

• ΔHr: heat of reaction per mole of A (J/mol), negative if exothermic; T0: feed temperature (K).

ΔTad = (−ΔHr)/Cp' = (−ΔHr)·CA0/(ρ·cp)

• ρ: density (kg/m³); cp: specific heat (J/kg·K); CA0 in mol/m³.

UA·(Ta − T) − FA0·Cp'·(T − T0) + (−ΔHr)·FA0·XA = 0 (steady CSTR with heat exchange)

• U: overall heat-transfer coefficient (W/m²·K); A: exchange area (m²); Ta: coolant temperature (K).

dT/dV = [(−ΔHr)·(−rA) − U·a·(T − Ta)] / (FA0·Cp') (PFR with heat exchange)

• a: heat-exchange area per unit reactor volume (m²/m³).

Q = FA0·XA·(−ΔHr) (heat removal for isothermal operation, feed at reactor temperature)

d ln K/dT = ΔHr/(R·T²) (van 't Hoff)

Worked examples

Example 1 (standard, liquid adiabatic rise). A liquid feed with CA0 = 2 kmol/m³ (2000 mol/m³), ρ·cp = 4.2 × 10⁶ J/m³·K, reacts with ΔHr = −80 kJ/mol A in an adiabatic reactor. Find ΔTad and the rise at 75 % conversion.

1. ΔTad = (−ΔHr)·CA0/(ρ·cp) = 80 000 × 2000/(4.2 × 10⁶) = 38.1 K.
2. At XA = 0.75: ΔT = 0.75 × 38.1 = 28.6 K.

ΔTad = 38.1 K; rise at 75 % conversion = 28.6 K.

Example 2 (GATE level, gas with inerts). Gaseous A → R, ΔHr = −60 kJ/mol, Cp(A) = Cp(R) = 150 J/mol·K, is fed with nitrogen (Cp = 30 J/mol·K) in a mole ratio N₂ : A = 4 : 1 at 400 K to an adiabatic PFR. Find the exit temperature at XA = 0.6.

1. Cp' = 1 × 150 + 4 × 30 = 270 J/(mol A)·K (ΔCp = 0 since Cp(A) = Cp(R)).
2. ΔTad = 60 000/270 = 222.2 K.
3. T = 400 + 222.2 × 0.6 = 533.3 K.

T ≈ 533 K. Without the nitrogen, ΔTad would be 400 K and T = 640 K — inerts are a simple way to limit hot spots.

Example 3 (adiabatic CSTR and cooling duty). An adiabatic CSTR is fed at 300 K with a first-order reactant; ΔTad = 100 K. The exit is measured at 350 K, where k = 0.1 s⁻¹.

1. XA = (350 − 300)/100 = 0.5.
2. τ = XA/[k·(1 − XA)] = 0.5/(0.1 × 0.5) = 10 s. For comparison, if FA0 = 10 mol/s, ΔHr = −80 kJ/mol and the reactor were held isothermal at the feed temperature with XA = 0.8, the cooling duty would be Q = 10 × 0.8 × 80 000 = 640 kW.

Common mistakes

• Using the heat capacity of A alone and ignoring inerts or solvent in Cp'.
• Sign errors: for exothermic reactions (−ΔHr) is positive and T rises with X.
• Mixing per-mole and per-kilogram heat capacities, or J and kJ.
• Assuming an adiabatic exothermic reversible reaction can reach high conversion — the equilibrium curve stops it.
• Using the adiabatic line for a reactor that exchanges heat.

For GATE CH

Expect NAT questions on adiabatic temperature rise, exit temperature or conversion from the adiabatic line, the conversion-temperature intersection for a CSTR, cooling duty for isothermal operation, and the effect of inerts. Conceptual questions test the interplay of the adiabatic line with the equilibrium curve. Practise setting up Cp' per mole of A fed with excess reactants and inerts.

Quick check

1. What is the slope dX/dT of the adiabatic operating line?
2. Doubling the inert flow does what to ΔTad?
3. For an endothermic reaction in an adiabatic reactor, does T rise or fall?
4. Why do SO₂ converters use inter-stage cooling?

Answers: 1. Cp'/(−ΔHr); 2. lowers it (Cp' increases); 3. it falls; 4. because the adiabatic line meets the falling equilibrium curve at low conversion, so cooling between beds lets conversion climb further.