Energy Balances

Energy is conserved too. Learn the steady-flow energy balance with enthalpy, sensible and latent heat, heats of reaction and formation, Hess's law, heat duties for heaters, coolers and reactors, adiabatic temperature, and steam and cooling-water requirements.

Heating, cooling, evaporating and reacting all involve energy. An energy balance tells you the heat duty of every exchanger and reactor jacket, the steam and cooling water a plant needs, and whether a reactor might overheat.

Enthalpy is the key property: for flowing streams, heat added equals the enthalpy change (when work and kinetic/potential energy are negligible).

An analogy

An energy balance is the plant's utility bill: every joule of steam you buy must go somewhere — into products, cooling water or losses.

Enthalpy and reference states

Enthalpy has no absolute value; choose reference states (e.g. elements at 25 °C, or components at a convenient temperature/phase). Only differences matter, so be consistent.

Sensible, latent and reaction heat

Heating without phase change uses c_p ΔT; phase changes use latent heats; reactions release (exothermic, ΔH < 0) or absorb (endothermic) heat computed from heats of formation or combustion.

Utilities

Steam supplies heat by condensing (≈ 2000–2100 kJ/kg at typical pressures); cooling water removes heat with a limited temperature rise (≈ 10 K); refrigeration for below-ambient cooling. Energy balances size these utilities.

Adiabatic reactors and heat integration

With no heat removal, reaction heat raises the temperature: adiabatic temperature rise ΔT_ad = (−ΔH_r) X C_A0/(ρ c_p). Runaway risk arises when heat generation exceeds removal. Pinch analysis integrates hot and cold streams to minimise utilities.

Formulas

• Steady-flow energy balance: Q̇ − Ẇ_s = ΣṁĤ_out − ΣṁĤ_in
• Sensible heat: Q̇ = ṁ c_p ΔT
• Latent heat: Q̇ = ṁ λ
• Heat of reaction (Hess): ΔH_r° = Σν ΔH_f°(products) − Σν ΔH_f°(reactants)
• Reactor heat removal: Q̇ = ξ̇ · ΔH_r

Worked example: Reactor cooling duty

Given: 50 kmol/h reacts, ΔH_r = −80 kJ/mol; Cooling water 30 → 40 °C

Find: Heat removal and water flow.

• Q = 50 000 mol/h × 80 kJ/mol = 4.0 × 10⁶ kJ/h = 1111 kW
• ṁ_w = 1111/(4.18 × 10) = 26.6 kg/s

Answer: ≈ 1.11 MW, ≈ 26.6 kg/s water

Worked example: Heat of combustion

Given: 6.23 kmol CH₄ (100 kg); LHV 802 kJ/mol

Find: Heat released.

• Q = 6234 mol × 802 kJ/mol = 5.0 × 10⁶ kJ

Answer: 5.0 GJ

Common mistakes

• Wrong: Using inconsistent reference states. — Right: Define references for every component and stick to them.
• Wrong: Forgetting the sign of ΔH_r. — Right: Exothermic ΔH_r < 0 releases heat — heat must be removed.

Applications

• Exchanger & utility sizing: Heat duties set exchanger areas and steam/cooling demand.
• Reactor safety: Cooling capacity vs reaction heat prevents runaways.
• Energy efficiency: Pinch analysis and heat recovery cut utility costs.

Key takeaways

• Q̇ = ΔH for steady flow (no work).
• Sensible c_pΔT, latent λ, reaction ξΔH_r.
• Size steam and cooling water from duties.
• Exothermic reactors need reliable cooling.

Going further

• Temperature-dependent ΔH_r: Kirchhoff's law adjusts heats of reaction using Δc_p.
• Pinch analysis: Composite curves find minimum hot/cold utility targets.
• Thermal runaway: Semenov diagrams compare heat generation and removal curves.