First law for open systems: steady flow energy equation

Mass conservation, flow work and the steady flow energy equation applied to nozzles, diffusers, turbines, compressors, throttles, heat exchangers and mixing, with stagnation temperature for high-speed flow.

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Why it matters

Almost every component of an aircraft engine — intake diffuser, compressor, combustor, turbine, nozzle — is an open system with fluid flowing steadily through it. The steady flow energy equation (SFEE) is the tool that gives compressor power, turbine output, nozzle exit velocity and combustor heat release from inlet and exit states. It is the most used equation in propulsion.

Key ideas

Control volume. For flow devices we fix a region in space (control volume, CV) and track mass and energy crossing its control surface at inlets and exits, plus heat and shaft work crossing its walls.

Steady flow means nothing at any point inside the CV changes with time: the mass and energy stored in the CV are constant, mass flow in equals mass flow out, and properties at each inlet and exit are steady (they may differ from point to point, but not in time). Engines at constant throttle and turbines at constant load are steady-flow devices; filling a tank is not.

Mass conservation (continuity). ṁ = ρ·A·V = A·V / v. For one inlet and one exit, ṁ₁ = ṁ₂.

Flow work. Pushing a fluid element into the CV against pressure p needs work p·v per unit mass. This flow work is why enthalpy h = u + p·v, not u, appears in the SFEE: each kg of flowing fluid carries h + V²/2 + g·z.

SFEE. Per unit mass, with q heat added and w shaft work done by the fluid: q − w = (h₂ − h₁) + (V₂² − V₁²)/2 + g·(z₂ − z₁)

Typical simplifications by device.

  • Nozzle (accelerates flow): w = 0, usually q ≈ 0, Δz ≈ 0 → h₁ + V₁²/2 = h₂ + V₂²/2. Enthalpy drop becomes kinetic energy.
  • Diffuser (aircraft intake): the reverse — kinetic energy recovered as enthalpy (pressure and temperature rise).
  • Turbine: usually adiabatic, KE change small → w = h₁ − h₂.
  • Compressor/pump: adiabatic → w_in = h₂ − h₁.
  • Throttling valve: q = 0, w = 0, ΔKE ≈ 0 → h₁ = h₂. For an ideal gas the temperature is then unchanged.
  • Heat exchanger/combustor: w = 0, → q = h₂ − h₁ per kg of each stream.
  • Adiabatic mixing chamber: Σṁ·h (in) = Σṁ·h (out).

Stagnation enthalpy. h₀ = h + V²/2. In an adiabatic, no-work flow (nozzle, diffuser, duct), h₀ is constant. For an ideal gas this gives the stagnation temperature T₀ = T + V²/(2·cp) — the temperature measured by a probe that brings the flow to rest, important in high-speed flight.

Units. h is in kJ/kg but V²/2 comes out in J/kg (m²/s²). Divide V²/2 by 1000 before adding it to h. At 45 m/s the kinetic energy is only about 1 kJ/kg, which is why it is often neglected for turbines and compressors but never for nozzles and diffusers.

Formulas

ṁ = ρ·A·V = A·V / v Q̇ − Ẇ = ṁ·[(h₂ − h₁) + (V₂² − V₁²)/2 + g·(z₂ − z₁)] h = u + p·v Δh = cp·ΔT (ideal gas) V₂ = √(V₁² + 2·(h₁ − h₂)) (adiabatic nozzle, h in J/kg) T₀ = T + V²/(2·cp) (ideal gas, cp in J/(kg·K))

  • ṁ: mass flow rate, kg/s; ρ: density, kg/m³; A: flow area, m²; V: velocity, m/s
  • Q̇: heat transfer rate into the CV, kW; Ẇ: shaft power out of the CV, kW
  • h: specific enthalpy, kJ/kg (convert to J/kg when combining with V²/2 in m²/s²)
  • z: elevation, m; g = 9.81 m/s²
  • T₀: stagnation temperature, K

Applies to steady flow with one inlet and one exit; sum ṁ·(h + V²/2 + g·z) over all ports for several streams.

Worked examples

Example 1 (standard). Air enters an adiabatic nozzle at 600 K with 30 m/s and leaves at 450 K and 150 kPa. The mass flow is 2 kg/s. Find the exit velocity and exit area. cp = 1.005 kJ/(kg·K), R = 0.287 kJ/(kg·K).

  1. V₂ = √(V₁² + 2·cp·(T₁ − T₂)) with cp in J/(kg·K)
  2. V₂ = √(30² + 2 × 1005 × 150) = √(900 + 301 500) = 549.9 m/s
  3. Exit density: ρ₂ = p₂ / (R·T₂) = 150 / (0.287 × 450) = 1.161 kg/m³
  4. A₂ = ṁ / (ρ₂·V₂) = 2 / (1.161 × 549.9) = 3.13 × 10⁻³ m²

Answer: V₂ ≈ 550 m/s, A₂ ≈ 31.3 cm²

Example 2 (GATE level). Hot gas (treat as air, cp = 1.005 kJ/(kg·K)) flows through a turbine at 5 kg/s. Inlet: 1200 K, 50 m/s. Exit: 750 K, 200 m/s. Heat lost from the casing is 50 kW. Find the power output.

  1. SFEE solved for power: Ẇ = Q̇ + ṁ·[(h₁ − h₂) + (V₁² − V₂²)/2]
  2. Enthalpy term: ṁ·cp·(T₁ − T₂) = 5 × 1.005 × 450 = 2261.25 kW
  3. KE term: ṁ·(V₁² − V₂²)/2 = 5 × (2500 − 40 000)/2 / 1000 = −93.75 kW
  4. Q̇ = −50 kW (heat leaves)
  5. Ẇ = −50 + 2261.25 − 93.75 = 2117.5 kW ≈ 2.12 MW

The higher exit velocity and the casing loss both reduce the shaft power.

Common mistakes

  • Adding V²/2 in J/kg to h in kJ/kg without dividing by 1000.
  • Using u instead of h for a flowing stream (forgetting flow work).
  • Getting the sign of the compressor work wrong: in the Q̇ − Ẇ form, compressor Ẇ is negative.
  • Assuming a throttle changes the temperature of an ideal gas (it does not; real gases and vapours may cool).
  • Neglecting KE in a nozzle or diffuser problem, where it is the whole point.
  • Applying the SFEE to unsteady problems such as tank filling.

For GATE AE

Expect SFEE numericals on nozzles and diffusers (exit velocity, exit temperature, exit area), turbine and compressor power with or without KE and heat loss, throttling, and stagnation temperature for high-speed flow — a direct link to the propulsion syllabus. Conceptual questions test which terms vanish for each device. Practise writing the SFEE in full first, then striking out terms with a reason.

Quick check

  1. What is conserved across an adiabatic throttling valve?
  2. Turbine: ṁ = 2 kg/s, h₁ = 3000 kJ/kg, h₂ = 2500 kJ/kg, adiabatic, KE and PE negligible. Find the power.
  3. Air at 200 m/s, 250 K: find the stagnation temperature (cp = 1005 J/(kg·K)).
  4. Why does enthalpy rather than internal energy appear in the SFEE?
  5. In a diffuser, does the static pressure rise or fall?

Answers: 1. Enthalpy. 2. 1000 kW. 3. About 269.9 K. 4. Each kg of flowing fluid carries flow work p·v, and u + p·v = h. 5. It rises.

Interview questions

All Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is the steady flow energy equation in the context of open systems?Concept

    The SFEE is the first law applied to a control volume in steady flow, where nothing inside the control volume changes with time, so energy in equals energy out. For one inlet and one exit: Q̇ − Ẇ = ṁ[(h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)], with Q̇ the heat added, Ẇ the shaft power delivered, h specific enthalpy, V velocity and z elevation. Enthalpy appears instead of internal energy because each kilogram of fluid also carries flow work p·v across the boundary.

  2. 2.Explain the significance of each term in the steady flow energy equation.Concept

    In the steady flow energy equation, Q̇ represents the rate of heat transfer into the system, which can increase the system's energy. Ẇ is the rate of work done by the system, which decreases the system's energy. The term ṁ(h2 - h1) represents the change in enthalpy, accounting for energy changes due to temperature and pressure differences. The kinetic energy change is represented by 0.5ṁ(v2² - v1²), which accounts for changes in velocity. Finally, ṁ(gz2 - gz1) represents the potential energy change due to elevation differences.

  3. 3.Why is the steady flow energy equation important in aerospace engineering?Application

    The steady flow energy equation is crucial in aerospace engineering because it helps analyze and design systems where fluids flow continuously, such as jet engines, turbines, and compressors. It allows engineers to calculate the energy transformations and efficiencies of these systems, ensuring they operate safely and effectively. Understanding these energy changes is essential for optimizing performance and fuel efficiency in aerospace applications.

  4. 4.What assumptions are typically made when applying the steady flow energy equation?Concept

    When applying the steady flow energy equation, it is typically assumed that the flow is steady, meaning the properties at any given point do not change with time. The system is also assumed to be open, allowing mass to enter and exit. Additionally, it is often assumed that the process is adiabatic (no heat transfer) or that heat transfer is known, and that changes in kinetic and potential energy are negligible unless specified otherwise.

  5. 5.How does the steady flow energy equation apply to a jet engine?Application

    Each component of a turbojet is treated as its own steady-flow control volume. In the intake (diffuser) kinetic energy is converted into enthalpy; the compressor absorbs work w = h₂ − h₁; the combustor adds heat q = h₃ − h₂ at roughly constant pressure; the turbine delivers w = h₃ − h₄, which drives the compressor; and the nozzle converts the remaining enthalpy drop into jet kinetic energy. The SFEE gives the temperatures and the jet velocity, and the thrust then follows from the momentum equation, F ≈ ṁ(V_jet − V_flight).

  6. 6.Why is it important to consider changes in kinetic and potential energy in the steady flow energy equation?Application

    Considering changes in kinetic and potential energy in the steady flow energy equation is important because these changes can significantly impact the energy balance in certain systems. For example, in high-speed flows or systems with significant elevation differences, ignoring these terms could lead to inaccurate calculations of energy transformations and system performance. Including these terms ensures a more comprehensive analysis and accurate predictions of system behavior.

  7. 7.Calculate the work done by a turbine if the mass flow rate is 2 kg/s, the specific enthalpy at the inlet is 3000 kJ/kg, and at the outlet is 2500 kJ/kg. Assume no heat transfer and negligible changes in kinetic and potential energy.Numerical

    To calculate the work done by the turbine, use the steady flow energy equation: Ẇ = ṁ(h1 - h2). Here, ṁ = 2 kg/s, h1 = 3000 kJ/kg, and h2 = 2500 kJ/kg. Substituting these values, Ẇ = 2 kg/s * (3000 kJ/kg - 2500 kJ/kg) = 2 kg/s * 500 kJ/kg = 1000 kJ/s or 1000 kW. Therefore, the work done by the turbine is 1000 kW.

  8. 8.A compressor increases the pressure of air from 100 kPa to 500 kPa. If the mass flow rate is 1.5 kg/s and the specific enthalpy increases by 200 kJ/kg, calculate the power input to the compressor. Assume no heat transfer and negligible changes in kinetic and potential energy.Numerical

    To calculate the power input to the compressor, use the steady flow energy equation: Ẇ = ṁ(h2 - h1). Here, ṁ = 1.5 kg/s and the change in specific enthalpy (h2 - h1) = 200 kJ/kg. Substituting these values, Ẇ = 1.5 kg/s * 200 kJ/kg = 300 kJ/s or 300 kW. Therefore, the power input to the compressor is 300 kW.

  9. 9.Explain how the steady flow energy equation can be simplified for an adiabatic nozzle.Application

    For an adiabatic nozzle, the steady flow energy equation can be simplified by assuming no heat transfer (Q̇ = 0) and no work done (Ẇ = 0). The equation then focuses on the changes in kinetic energy and enthalpy: 0 = ṁ(h2 - h1) + 0.5ṁ(v2² - v1²). This simplifies to h1 + 0.5v1² = h2 + 0.5v2², indicating that the decrease in enthalpy is converted into an increase in kinetic energy, resulting in a higher exit velocity.

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