First law of thermodynamics for closed and open systems

First law for cycles and closed-system processes, enthalpy and flow work, the steady-flow energy equation for turbines, compressors, nozzles and throttles.

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Why it matters

The first law is conservation of energy written for engineering devices. It tells you how much power a compressor needs, how much heat a furnace or quench tank must handle, what a turbine delivers and how fast steam leaves a nozzle. In production engineering it is used for compressed-air plants, heat-treatment and casting energy balances, coolant systems and every heat exchanger.

Key ideas

First law for a cycle. For any closed system undergoing a cycle, the net heat supplied equals the net work done: ∮δQ = ∮δW. Joule's paddle-wheel experiments established this equivalence.

First law for a process in a closed system. Energy E (= internal energy U + kinetic + potential) is a property. For a process, Q − W = ΔE, and for a stationary system Q − W = ΔU. Here Q is heat added to the system and W is work done by the system. Internal energy is the energy stored at the molecular level; for an ideal gas it depends only on temperature, so dU = m·c_v·dT for any process, not only constant-volume ones.

Enthalpy. H = U + pV (specific h = u + pv). For an ideal gas dh = c_p·dT. In a constant-pressure quasi-static process of a closed system, Q = ΔH. Enthalpy matters most for open systems, because the pv part represents flow work.

Special closed-system processes (ideal gas).

  • Constant volume: W = 0, so Q = ΔU = m·c_v·ΔT.
  • Constant pressure: W = mRΔT and Q = m·c_p·ΔT.
  • Isothermal: ΔU = 0, so Q = W.
  • Adiabatic: Q = 0, so W = −ΔU; expansion cools the gas, compression heats it.
  • Free (unresisted) expansion of a gas in an insulated vessel: Q = 0, W = 0, so ΔU = 0 and an ideal gas keeps its temperature.

Open systems and flow work. For a control volume, mass crossing the boundary carries energy with it. Pushing a unit mass into or out of the control volume requires flow work pv, so each kilogram of flowing fluid carries h + V²/2 + gz.

Steady-flow energy equation (SFEE). At steady state (no change of mass or energy inside the control volume), for one inlet (1) and one outlet (2): Q̇ − Ẇ = ṁ[(h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)] Typical simplifications:

  • Turbine: adiabatic, KE and PE changes small → Ẇ = ṁ(h₁ − h₂).
  • Compressor or pump: Ẇ (negative, power input) = Q̇ − ṁ(h₂ − h₁); for liquids, Ẇ_in ≈ ṁ·v·Δp.
  • Nozzle: no work, adiabatic → V₂ = √(2(h₁ − h₂) + V₁²). A diffuser does the reverse.
  • Throttling valve or porous plug: no work, adiabatic, negligible KE change → h₂ = h₁ (isenthalpic).
  • Heat exchanger: no work; the heat lost by one stream equals the heat gained by the other.
  • Boiler and condenser: no work → Q̇ = ṁ(h₂ − h₁).

Unsteady (transient) flow. Filling or emptying a tank requires an energy balance that includes the change of energy inside; for example, charging an evacuated rigid tank from a line at constant state gives u_tank = h_line, so the gas in the tank ends hotter than the supply line (T = γT_line for an ideal gas).

PMM-1. A perpetual-motion machine of the first kind, delivering work continuously without any energy input, would violate the first law and is impossible.

Formulas

Q − W = ΔU (closed, stationary)

  • Q: heat added (kJ); W: work done by system (kJ); ΔU: internal energy change (kJ).

ΔU = m·c_v·ΔT, ΔH = m·c_p·ΔT

  • Ideal gas, any process; c_v, c_p (kJ/kg·K); air c_p = 1.005, c_v = 0.718, R = 0.287 kJ/kg·K.

H = U + p·V

  • Enthalpy (kJ).

Q̇ − Ẇ = ṁ·[(h₂ − h₁) + (V₂² − V₁²)/2 + g·(z₂ − z₁)]

  • SFEE; Q̇, Ẇ (kW); ṁ (kg/s); h (kJ/kg); V (m/s); z (m). Divide the V² and gz terms by 1000 to get kJ/kg.

V₂ = √(2·(h₁ − h₂) + V₁²)

  • Adiabatic nozzle; h in J/kg (not kJ/kg).

h₁ = h₂

  • Throttling.

Worked examples

Example 1 (standard): constant-pressure heating of air. Given: 2 kg of air at 200 kPa is heated at constant pressure from 300 K to 500 K in a piston–cylinder. c_p = 1.005, c_v = 0.718, R = 0.287 kJ/kg·K. Find W, ΔU and Q.

  1. W = p·ΔV = m·R·ΔT = 2 × 0.287 × 200 = 114.8 kJ
  2. ΔU = m·c_v·ΔT = 2 × 0.718 × 200 = 287.2 kJ
  3. Q = ΔU + W = 287.2 + 114.8 = 402.0 kJ
  4. Check: m·c_p·ΔT = 2 × 1.005 × 200 = 402.0 kJ ✓ Answer: W = 114.8 kJ, ΔU = 287.2 kJ, Q = 402 kJ

Example 2 (GATE level): air compressor. Given: a compressor takes in 0.5 kg/s of air at 300 K and delivers it at 480 K. Heat lost to the surroundings is 10 kW. Neglect KE and PE changes. Take c_p = 1.005 kJ/kg·K. Find the power input.

  1. SFEE: Q̇ − Ẇ = ṁ·c_p·(T₂ − T₁).
  2. Q̇ = −10 kW (heat leaves the system).
  3. Ẇ = Q̇ − ṁ·c_p·(T₂ − T₁) = −10 − 0.5 × 1.005 × 180 = −10 − 90.45 = −100.45 kW
  4. Negative W means work is done on the air. Answer: power input ≈ 100.5 kW

Example 3 (GATE level): steam nozzle. Given: steam enters an adiabatic nozzle with h₁ = 3000 kJ/kg and V₁ = 50 m/s, and leaves with h₂ = 2800 kJ/kg. Find the exit velocity.

  1. V₂ = √(2·(h₁ − h₂) + V₁²), with h in J/kg.
  2. V₂ = √(2 × 200 000 + 50²) = √402 500 = 634.4 m/s Answer: V₂ ≈ 634 m/s

Common mistakes

  • Mixing kJ/kg with m²/s²: V²/2 is in J/kg, so divide by 1000 before adding to h in kJ/kg.
  • Losing the sign of heat lost or work input.
  • Using c_p for ΔU, or thinking ΔU = m·c_v·ΔT applies only to constant-volume processes.
  • Treating throttling as isothermal; it is isenthalpic (isothermal only for an ideal gas).
  • Writing ΔH = Q − W for an open system; the correct steady-flow form uses ṁ·Δh plus KE and PE.
  • Forgetting that the inlet velocity term matters in a nozzle when V₁ is not small.

For GATE PI

Expect closed-system numericals with ideal-gas processes, SFEE problems on compressors, turbines, nozzles, throttling and heat exchangers, and conceptual questions on flow work, enthalpy, free expansion and PMM-1. A frequent trap is the sign of heat loss in a compressor or turbine, or a unit slip between kJ/kg and m²/s². Practise writing the SFEE, crossing out terms with a reason, and then solving.

Quick check

  1. In an insulated rigid tank, gas expands freely into vacuum. What are Q, W and ΔU?
  2. Write the first law for a cycle.
  3. What property stays constant in a throttling process?
  4. A turbine receives steam at 3200 kJ/kg and discharges at 2600 kJ/kg (adiabatic, negligible KE). Work per kg?

Answers: 1. All zero 2. ∮δQ = ∮δW 3. Enthalpy 4. 600 kJ/kg

Interview questions

All Thermal and Fluids Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is the first law of thermodynamics for a closed system?Concept

    The first law of thermodynamics for a closed system states that the change in internal energy of the system is equal to the heat added to the system minus the work done by the system. Mathematically, it is expressed as ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added, and W is the work done by the system.

  2. 2.Explain the first law of thermodynamics for an open system.Concept

    For a control volume, energy also crosses the boundary with the mass, and each kilogram carries h + V²/2 + gz, where the pv part of h = u + pv is the flow work needed to push it in or out. At steady state with one inlet and one outlet the steady-flow energy equation is Q̇ − Ẇ = ṁ[(h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)]. It simplifies device by device: an adiabatic turbine gives Ẇ = ṁ(h₁ − h₂), a throttle gives h₁ = h₂, a nozzle converts enthalpy drop into kinetic energy.

  3. 3.How does the first law of thermodynamics apply to a heat engine?Application

    In a heat engine, the first law of thermodynamics is used to relate the heat input, work output, and change in internal energy. The engine absorbs heat from a high-temperature reservoir, converts part of it into work, and rejects the remaining heat to a low-temperature reservoir. The first law ensures that the energy balance is maintained, meaning the heat input equals the sum of work output and heat rejected.

  4. 4.Why is enthalpy used in the first law of thermodynamics for open systems?Application

    Enthalpy is used in the first law of thermodynamics for open systems because it accounts for both the internal energy and the flow work associated with mass entering or leaving the system. In open systems, mass flow is a significant factor, and enthalpy conveniently combines internal energy and pressure-volume work, simplifying the energy balance equations.

  5. 5.What happens if a closed system undergoes an adiabatic process?Application

    In an adiabatic process, no heat is exchanged with the surroundings (Q = 0). For a closed system, the first law of thermodynamics simplifies to ΔU = -W, meaning the change in internal energy is equal to the negative of the work done by the system. If the system does work, its internal energy decreases, and if work is done on the system, its internal energy increases.

  6. 6.Describe a real-world example where the first law of thermodynamics is applied in an open system.Application

    A common real-world example is a steam turbine in a power plant. Steam enters the turbine at high pressure and temperature, does work on the turbine blades, and exits at a lower pressure and temperature. The first law of thermodynamics is used to calculate the work output of the turbine by considering the enthalpy change of the steam and any heat losses.

  7. 7.How does the first law of thermodynamics explain the operation of a refrigerator?Application

    Over one cycle the refrigerant's internal energy returns to its starting value, so the net energy in must equal the net energy out. The refrigerant absorbs Q_L from the cold space and receives compressor work W, and rejects Q_H to the surroundings, so Q_H = Q_L + W. That is why the condenser always rejects more heat than the evaporator absorbs. The first law alone does not say heat cannot flow from cold to hot without work; that is the second law.

  8. 8.Calculate the work done by a closed system if 500 J of heat is added and the internal energy increases by 200 J.Numerical

    Using the first law of thermodynamics for a closed system, ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added, and W is the work done by the system. Here, ΔU = 200 J and Q = 500 J. Rearranging the equation gives W = Q - ΔU = 500 J - 200 J = 300 J. Therefore, the work done by the system is 300 J.

  9. 9.What is the significance of the first law of thermodynamics in energy conservation?Concept

    The first law is the principle of conservation of energy: energy can change form and cross boundaries as heat, work or with mass flow, but it cannot be created or destroyed. The total energy of an isolated system is constant; for a closed system the change in stored energy equals the net heat in minus the net work out. In practice it is the basis of every energy balance — finding compressor power, furnace heat input or cooling loads — and it rules out perpetual-motion machines of the first kind.

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