Enthalpy
Definition, Formula, Applications & Why It Matters — Explained for Engineering Students
Last Updated: March 2026
📌 Key Takeaways
- Enthalpy (H) = U + PV — internal energy plus pressure-volume product.
- At constant pressure: QP = ΔH. Heat added at constant pressure equals the enthalpy change.
- For ideal gases: ΔH = nCpΔT. Enthalpy depends only on temperature.
- Enthalpy is essential for open system analysis — turbines, compressors, nozzles, and heat exchangers all use enthalpy in energy equations.
- In steam tables, enthalpy values (hf, hfg, hg) are the primary data used for Rankine cycle calculations.
- SI unit: J or kJ. Specific enthalpy: kJ/kg.
1. Definition and Formula
Enthalpy
H = U + PV
Where: H = enthalpy (J or kJ), U = internal energy (J or kJ), P = pressure (Pa or kPa), V = volume (m³)
Specific enthalpy: h = u + Pv (per unit mass, kJ/kg)
Enthalpy is a state function — its value depends only on the current state of the system (P, T, composition), not on how the system reached that state. It is also an extensive property (proportional to the amount of substance); specific enthalpy h = H/m is intensive.
Since U, P, and V are all state functions, H is automatically a state function as well. We cannot measure H in absolute terms (just like U), but we can always measure changes in enthalpy (ΔH), which is what matters for engineering calculations.
2. Why Enthalpy Exists — The Physical Meaning
Enthalpy was not invented arbitrarily — it was created to simplify two extremely common situations in engineering:
Reason 1: Constant-Pressure Processes
Most chemical and engineering processes occur at constant pressure (atmospheric or controlled). At constant pressure, the First Law of Thermodynamics gives Q = ΔU + PΔV = Δ(U + PV) = ΔH. So the heat added at constant pressure is simply the enthalpy change — no need to calculate work separately.
Reason 2: Open Systems (Flow Processes)
When fluid flows into an open system (like a turbine), it carries two types of energy: its internal energy (U), and the work needed to push it through the boundary against pressure (PV, called flow work). Enthalpy H = U + PV combines both into a single quantity, enormously simplifying the energy equation for open systems.
Without enthalpy, every open system energy balance would require separately tracking internal energy and flow work. With enthalpy, the Steady Flow Energy Equation uses just h values from steam tables — clean and practical.
3. Enthalpy and Constant-Pressure Processes
Heat at Constant Pressure
QP = ΔH = H&sub2; − H&sub1;
The heat added to (or removed from) a system at constant pressure equals the change in enthalpy.
This is why enthalpy appears everywhere in chemistry (reaction enthalpies, heat of formation, calorimetry) and in engineering processes that occur at roughly constant pressure: boilers, heat exchangers, chemical reactors, and atmospheric processes.
Compare with constant-volume processes, where QV = ΔU. The choice between using ΔH or ΔU depends entirely on whether the process is at constant pressure or constant volume.
4. Enthalpy of Ideal Gases
For ideal gases, enthalpy depends only on temperature — not on pressure or volume:
Enthalpy Change — Ideal Gas
ΔH = nCpΔT = nCp(T&sub2; − T&sub1;)
Per unit mass: Δh = cp(T&sub2; − T&sub1;)
Where Cp = molar specific heat at constant pressure
This is because for ideal gases: H = U + PV = U + nRT. Since U depends only on T (for ideal gas) and nRT depends only on T, H also depends only on T.
The relationship between Cp and Cv for ideal gases:
Cp − Cv = R (Mayer’s relation)
γ = Cp / Cv (ratio of specific heats)
For air: Cp = 1.005 kJ/kg·K, Cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, γ = 1.4
5. Enthalpy in Open Systems
Enthalpy is the key property for analysing every open system device:
| Device | Energy Equation (simplified SFEE) | What Changes |
|---|---|---|
| Turbine | Wout = h&sub1; − h&sub2; | Enthalpy drops → work produced |
| Compressor | Win = h&sub2; − h&sub1; | Enthalpy rises → work consumed |
| Heat Exchanger | Q = ˙m(h&sub2; − h&sub1;) | Enthalpy changes → heat transferred |
| Nozzle | V&sub2;² = V&sub1;² + 2(h&sub1; − h&sub2;) | Enthalpy drops → velocity increases |
| Throttling Valve | h&sub1; = h&sub2; (isenthalpic) | Enthalpy unchanged → pressure drops, temperature may change |
In every case, the analysis comes down to finding enthalpy values at the inlet and outlet — usually from steam tables, Mollier diagrams, or ideal gas relations.
6. Enthalpy in Steam Tables
For water/steam (the working fluid in Rankine cycles), enthalpy values are obtained from steam tables:
| Symbol | Meaning | Where to Find It |
|---|---|---|
| hf | Specific enthalpy of saturated liquid | Saturation table (at given P or T) |
| hg | Specific enthalpy of saturated vapour | Saturation table |
| hfg | Latent heat of vaporisation = hg − hf | Saturation table |
| h (wet mixture) | h = hf + x·hfg, where x = dryness fraction | Calculated using quality x |
| h (superheated) | Specific enthalpy of superheated steam | Superheated steam table (at given P and T) |
Steam tables are indispensable for power plant analysis. Every Rankine cycle problem requires looking up h values at each state point.
7. Enthalpy vs Internal Energy — When to Use Which
| Use Enthalpy (H) when… | Use Internal Energy (U) when… |
|---|---|
| Process is at constant pressure | Process is at constant volume |
| Analysing open systems (flow processes) | Analysing closed systems (no mass flow) |
| Using SFEE for turbines, compressors, nozzles | Using Q = ΔU + W for piston-cylinder systems |
| Working with steam tables (h values are primary) | Working with ideal gas at constant V |
8. Worked Numerical Examples
Example 1: Enthalpy Change of Air
Problem: 3 kg of air (cp = 1.005 kJ/kg·K) is heated from 25°C to 400°C at constant pressure. Find ΔH.
Solution
ΔH = mcp(T&sub2; − T&sub1;) = 3 × 1.005 × (400 − 25) = 3 × 1.005 × 375
ΔH = 1,130.6 kJ
Since this is a constant-pressure process, the heat added is also 1,130.6 kJ.
Example 2: Turbine Work Using Enthalpy
Problem: Steam enters a turbine at h&sub1; = 3,230 kJ/kg and exits at h&sub2; = 2,540 kJ/kg. Mass flow rate is 10 kg/s. Find the power output.
Solution
Power = ˙m × (h&sub1; − h&sub2;) = 10 × (3230 − 2540) = 10 × 690
Power = 6,900 kW = 6.9 MW
Example 3: Enthalpy of Wet Steam
Problem: Steam at 200 kPa has a dryness fraction of 0.85. Using steam table values hf = 504.7 kJ/kg and hfg = 2,201.6 kJ/kg, find the specific enthalpy.
Solution
h = hf + x·hfg = 504.7 + 0.85 × 2201.6 = 504.7 + 1871.4
h = 2,376.1 kJ/kg
9. Common Mistakes Students Make
- Confusing enthalpy with entropy: Enthalpy (H, units: kJ) measures energy content. Entropy (S, units: kJ/K) measures disorder. They sound similar but are completely different properties.
- Using Cv instead of Cp for enthalpy: ΔH = nCpΔT, not nCvΔT. Cv gives ΔU. Cp gives ΔH. Mixing them up is a common error.
- Thinking enthalpy is only for open systems: While enthalpy is most commonly used in open systems, it is equally valid for any constant-pressure process in a closed system. QP = ΔH applies to both.
- Forgetting flow work in open system analysis: When fluid enters an open system, it carries internal energy AND flow work (PV). Enthalpy accounts for both. Using internal energy alone in SFEE gives wrong results.
- Not interpolating in steam tables: If the given pressure or temperature is between two table entries, you must interpolate linearly to find the correct h value. Using the nearest value without interpolation introduces unnecessary error.
10. Frequently Asked Questions
What is enthalpy?
Enthalpy (H) is a thermodynamic property defined as H = U + PV. It represents the total energy of a system including both its internal energy and the energy associated with its pressure and volume. For constant-pressure processes, the heat transferred equals the enthalpy change. For open systems, enthalpy replaces internal energy in the energy equation because it automatically includes flow work.
What is the difference between enthalpy and internal energy?
Internal energy (U) accounts only for the energy stored within the system from molecular motion and forces. Enthalpy (H = U + PV) adds the pressure-volume product, which represents flow work in open systems or boundary work at constant pressure. Use U for constant-volume closed systems; use H for constant-pressure processes and open systems.
Why is enthalpy important for open systems?
When fluid flows through a turbine, compressor, or heat exchanger, it carries energy in two forms: internal energy (U) and flow work (PV, the work needed to push the fluid through the boundary). Enthalpy combines both (H = U + PV) into a single property, allowing the Steady Flow Energy Equation to be written compactly using just h values. Without enthalpy, every flow problem would require separate internal energy and flow work calculations.