First law for closed systems and steady-flow devices

Why it matters

The first law is the energy bookkeeping behind every thermal machine: how much heat a boiler must supply, how much power a turbine delivers, how hot a compressor discharge gets, how fast a nozzle jet leaves. For a mechatronics engineer the same balance sizes a motor's cooling, a pneumatic compressor or a hydraulic power pack. Writing the right form of the equation for a closed system or a steady-flow device, with consistent signs and units, solves most thermodynamics problems.

Key ideas

Statement. Energy cannot be created or destroyed; it only changes form. For any system, net energy transferred in as heat and work equals the change in stored energy.

Closed system (control mass). Stored energy E = U + KE + PE. With heat added to the system taken as positive and work done by the system taken as positive:

• ΔE = Q − W. For a stationary system (no change in KE or PE), ΔU = Q − W.
• Over a cycle ΔU = 0, so ∮δQ = ∮δW: net heat in equals net work out. Historically this is how the first law was stated (Joule's experiments).
• Internal energy is a property (a point function); Q and W are path functions, but their difference Q − W is the same for every path between two states.
• For an ideal gas, u depends only on T: Δu = c_v ΔT for any process. Similarly Δh = c_p ΔT.
• The first law holds for real (irreversible) processes too; the work just cannot then be computed as ∫p dV.

Control volume and steady flow. In an open system, mass crossing the boundary carries energy h + V²/2 + gz per kg. The term pv inside h = u + pv is the flow work needed to push the mass across the boundary. Steady flow means nothing inside the control volume changes with time:

• mass flow in = mass flow out (ṁ₁ = ṁ₂ for one inlet and one outlet),
• the rates of heat and work are constant,
• properties at each inlet and outlet do not change with time (they may differ from point to point).

Typical steady-flow devices and their usual simplifications.

• Turbine / compressor / pump: usually adiabatic, with ΔKE and ΔPE small. Then w = h₁ − h₂ (turbine output) or w_in = h₂ − h₁ (compressor input). For a pump handling liquid, w_in ≈ v Δp.
• Nozzle / diffuser: no work, nearly adiabatic, so the enthalpy drop becomes kinetic energy: V₂² − V₁² = 2(h₁ − h₂).
• Throttling valve: no work, no heat, negligible ΔKE, so h₂ = h₁ (isenthalpic). For an ideal gas the temperature then stays constant; for real fluids it usually drops (used in refrigeration).
• Heat exchanger / boiler / condenser: no work. Heat lost by the hot stream equals heat gained by the cold stream if the outer shell is insulated.
• Mixing chamber: Σṁh in = Σṁh out, with Σṁ in = Σṁ out.

Unsteady flow (filling or emptying a tank) needs the transient form of the balance; a classic result is that an evacuated insulated tank filled from a line ends with u₂ = h_line, so for an ideal gas T₂ = γT_line.

Formulas

Q − W = ΔU + ΔKE + ΔPE — closed system, any process.

• Q: heat added (kJ), W: work done by the system (kJ), U: internal energy (kJ).

ΔU = m c_v (T₂ − T₁), ΔH = m c_p (T₂ − T₁) — ideal gas with constant specific heats.

• m: mass (kg), c_v, c_p: specific heats (kJ/kg·K; air 0.718 and 1.005), T: temperature (K or °C, since only a difference appears).

∮δQ = ∮δW — any closed system undergoing a cycle.

ṁ (h₁ + V₁²/2 + g z₁) + Q̇ = ṁ (h₂ + V₂²/2 + g z₂) + Ẇ — steady-flow energy equation (SFEE), one inlet (1) and one outlet (2).

• ṁ: mass flow rate (kg/s), h: specific enthalpy (J/kg), V: velocity (m/s), g = 9.81 m/s², z: elevation (m), Q̇: heat transfer rate into the CV (W), Ẇ: shaft power out of the CV (W). If h is in kJ/kg, divide V²/2 and gz by 1000.

ṁ = ρ A V = A V / v — continuity. ρ: density (kg/m³), A: flow area (m²), v: specific volume (m³/kg).

V₂ = √(2 (h₁ − h₂) + V₁²) — adiabatic nozzle, h in J/kg.

w_pump ≈ v (p₂ − p₁) — reversible adiabatic pump with an incompressible liquid (kJ/kg when p is in kPa).

Worked examples

Example 1 (closed system). A rigid tank holds 2 kg of air. A paddle wheel does 50 kJ of work on the air while 10 kJ of heat leaks out. Take c_v = 0.718 kJ/kg·K. Find ΔU and the temperature rise.

1. Signs: Q = −10 kJ (heat leaves); W = −50 kJ (work done on the system).
2. ΔU = Q − W = −10 − (−50) = +40 kJ.
3. ΔU = m c_v ΔT → ΔT = 40 / (2 × 0.718) = 27.86 K.

ΔU = 40 kJ, ΔT ≈ 27.9 K

Example 2 (GATE level, turbine with KE and heat loss). Steam enters a turbine at 5 kg/s with h₁ = 3230 kJ/kg and V₁ = 50 m/s, and leaves with h₂ = 2550 kJ/kg and V₂ = 150 m/s. Heat loss from the casing is 40 kW; neglect elevation change. Find the power output.

1. SFEE solved for power: Ẇ = ṁ [(h₁ − h₂) + (V₁² − V₂²)/2] + Q̇.
2. Enthalpy drop: 3230 − 2550 = 680 kJ/kg.
3. KE term: (50² − 150²)/2 = (2500 − 22 500)/2 = −10 000 J/kg = −10 kJ/kg.
4. Q̇ = −40 kW (heat leaves the CV).
5. Ẇ = 5 × (680 − 10) − 40 = 3350 − 40 = 3310 kW.

Ẇ ≈ 3.31 MW

Example 3 (nozzle). Steam enters an adiabatic nozzle at 30 m/s with h₁ = 3000 kJ/kg and leaves with h₂ = 2760 kJ/kg. Find the exit velocity.

1. V₂ = √(2 (h₁ − h₂) + V₁²) with h in J/kg.
2. V₂ = √(2 × 240 000 + 900) = √480 900 = 693.5 m/s.

V₂ ≈ 693 m/s

Common mistakes

• Adding V²/2 in J/kg to h in kJ/kg. Divide the kinetic and potential terms by 1000 or convert h to J/kg.
• Sign slips: writing W = +50 kJ when the paddle wheel does work on the gas, or treating a heat loss as positive.
• Using ΔU = m c_v ΔT only for constant-volume processes. For an ideal gas it holds for every process; c_v is just the name of the property.
• Applying the closed-system form ΔU = Q − W to a turbine. Flowing systems need the enthalpy (SFEE) form.
• Assuming a throttling process is isothermal for every fluid. It is isenthalpic; only an ideal gas keeps the same temperature.
• Neglecting the kinetic-energy term for nozzles and diffusers, where it is the whole point of the device.

For GATE ME

Expect straightforward SFEE applications — turbine or compressor power, nozzle exit velocity, throttling, mixing and heat-exchanger balances — plus closed-system problems using ΔU = Q − W with c_v ΔT and boundary work from the previous topic. Tank filling and emptying (unsteady flow) appears occasionally. Practise drawing the control volume first, listing which terms vanish, and keeping kJ/kg and J/kg apart.

Quick check

1. A gas receives 120 kJ of heat and does 80 kJ of work. What is ΔU?
2. What property stays constant across a throttling valve?
3. Write the work per kg of an adiabatic turbine with negligible changes in KE and PE.
4. An adiabatic nozzle has an enthalpy drop of 50 kJ/kg and negligible inlet velocity. Find the exit velocity.
5. For an ideal gas, does Δu = c_v ΔT apply to a constant-pressure process?

Answers: 1. 40 kJ. 2. Specific enthalpy. 3. w = h₁ − h₂. 4. √(2 × 50 000) ≈ 316 m/s. 5. Yes; u of an ideal gas depends only on T.