Enthalpy

Enthalpy h = u + pv is the energy carried by a flowing fluid. Learn why it's used for open systems, sensible and latent heat, c_p Δt calculations, steam enthalpies and heating curves.

When fluid flows into a device, it brings its internal energy plus the 'flow work' pv needed to push it in. Adding them gives enthalpy, which is why every steady-flow calculation — boilers, turbines, air heaters — is written in terms of h.

Enthalpy also captures latent heat: boiling water absorbs 2257 kJ/kg at 100 °C with no temperature rise.

An analogy

Enthalpy is like a delivery charge: the fluid's own energy (u) plus the cost of pushing it through the door (pv).

Sensible vs latent heat

Sensible heat changes temperature (Q = mcΔT). Latent heat changes phase at constant temperature: melting ice needs 334 kJ/kg, boiling water 2257 kJ/kg at 1 atm — over five times the energy to heat it from 0 to 100 °C.

Why enthalpy for open systems

In the steady-flow energy equation the flow work pv appears naturally with u, so engineers tabulate h. At constant pressure, heat added equals Δh — which is why c_p (not c_v) is used for flowing gases.

Reference states

Only differences in h matter. Steam tables set u = 0 for saturated liquid at 0.01 °C; refrigerant tables use other references (IIR: h = 200 kJ/kg at 0 °C liquid). Never mix values from tables with different references.

Enthalpy of reaction and psychrometrics

Combustion releases enthalpy of formation differences (heating values: LHV/HHV). Moist-air enthalpy (per kg dry air) combines air and vapour: h ≈ 1.005T + ω(2501 + 1.86T) — the basis of HVAC psychrometric charts.

Formulas

• Definition: h = u + p·v
• Ideal gas / liquid sensible heat: Δh = c_p ΔT
• Heater / boiler: Q̇ = ṁ (h₂ − h₁)
• Latent heat: Q = m · h_fg

Worked example: Air heater duty

Given: ṁ = 1 kg/s air, 20 → 200 °C; c_p = 1.005 kJ/kg·K

Find: Heater power.

• Q̇ = 1 × 1.005 × 180 = 180.9 kW

Answer: ≈ 181 kW

Worked example: Boiler heat input

Given: Feedwater h₁ = 195 kJ/kg; Steam out 3 MPa 400 °C, h₂ = 3232 kJ/kg; ṁ = 10 kg/s

Find: Boiler heat rate.

• Q̇ = 10 × (3232 − 195) = 30 370 kW

Answer: ≈ 30.4 MW

Common mistakes

• Wrong: Forgetting latent heat when a phase change occurs. — Right: Check the phase at each state; add h_fg when boiling/condensing.
• Wrong: Using c_v for flowing gases. — Right: Use c_p for constant-pressure/flow processes.

Applications

• Boilers & HRSGs: Heat duty = ṁ Δh across economiser, evaporator and superheater sections.
• HVAC: Enthalpy of moist air drives cooling-coil capacity including dehumidification.
• Process heating: Steam is used for heating because condensing latent heat delivers huge energy at constant temperature.

Key takeaways

• h = u + pv — the energy of flowing fluid.
• Sensible: c_pΔT; latent: h_fg.
• Q̇ = ṁΔh for heaters, boilers, coolers.
• Only differences matter; keep one reference.

Going further

• Joule–Thomson effect: Throttling a real gas changes its temperature; below the inversion temperature it cools — the basis of gas liquefaction.
• Stagnation enthalpy: h₀ = h + V²/2 is conserved in adiabatic flow — key for high-speed flows and turbomachinery.
• Heating values: HHV includes condensing the water in combustion products; condensing boilers recover part of that difference.