First law for closed systems and steady-flow devices

Why it matters

The first law is the energy bookkeeping behind every engine, compressor, turbocharger, radiator and nozzle. Closed-system analysis describes what happens inside an engine cylinder with the valves shut; the steady-flow energy equation (SFEE) describes the turbocharger compressor, the intercooler and the exhaust nozzle. Almost every thermodynamics numerical begins with one of these two balances.

Key ideas

First law for a cycle. For any closed system undergoing a cycle, the net heat transfer equals the net work transfer: ∮δQ = ∮δW. A machine that produces work continuously without any energy input (a perpetual-motion machine of the first kind, PMM1) is therefore impossible.

First law for a process (closed system). Q − W = ΔE, where E = U + KE + PE. For a stationary system ΔE = ΔU. Internal energy is a property (point function), even though Q and W individually are path functions. Sign convention: Q into the system positive, W done by the system positive.

Enthalpy. H = U + PV, a property. For a closed system at constant pressure with only boundary work, Q = ΔH; for a constant-volume process, Q = ΔU.

Ideal-gas specific heats. For an ideal gas u and h depend on temperature only: du = c_v dT and dh = c_p dT, with c_p − c_v = R and γ = c_p/c_v. For air take c_p ≈ 1.005 kJ/kg·K, c_v ≈ 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, γ = 1.4.

Control volumes and steady flow. In an open system, mass crosses the boundary. Pushing mass in or out needs flow work Pv per unit mass, which is why enthalpy h = u + Pv appears in every flow equation. Steady flow means nothing inside the control volume changes with time: mass flow in equals mass flow out, and the energy stored is constant.

Common steady-flow devices and usual assumptions:

• Nozzle / diffuser: no work, usually adiabatic, ΔPE ≈ 0. Enthalpy converts to kinetic energy (nozzle) or back (diffuser).
• Turbine / compressor / pump: usually adiabatic, ΔKE and ΔPE small, so w = h₁ − h₂ (turbine output) or w_in = h₂ − h₁ (compressor input).
• Throttling valve: no work, adiabatic, ΔKE negligible → h₁ = h₂ (isenthalpic). For an ideal gas T stays constant; for a real fluid T usually drops (refrigeration expansion valve).
• Heat exchanger: no work; heat lost by one stream = heat gained by the other when the outer shell is insulated.
• Mixing chamber: Σṁh in = Σṁh out for adiabatic mixing.

Unsteady (transient) flow. Filling or emptying a tank is not steady flow; use the uniform-flow form, e.g. charging an evacuated rigid tank from a line gives u₂ = h_line, so for an ideal gas T₂ = γ T_line.

Formulas

Q − W = ΔU (closed, stationary system) — Q, W, ΔU in J or kJ.

ΔU = m c_v (T₂ − T₁), ΔH = m c_p (T₂ − T₁) — ideal gas, any process. m: kg; c_v, c_p: kJ/kg·K; T: K.

c_p − c_v = R, γ = c_p / c_v — ideal gas.

ṁ = ρ A V = A V / v — mass flow rate, kg/s; ρ: kg/m³; A: m²; V: velocity, m/s; v: specific volume, m³/kg.

Q̇ − Ẇ = ṁ [(h₂ − h₁) + (V₂² − V₁²)/2 + g (z₂ − z₁)] — SFEE, single inlet (1) and outlet (2).

• Q̇: heat rate into the CV, W (or kW); Ẇ: shaft power out of the CV, W; h: J/kg; V: m/s; z: m.
• Per unit mass: q − w = Δh + ΔV²/2 + g Δz. Remember V²/2 is in J/kg; divide by 1000 for kJ/kg.

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

h₁ = h₂ — throttling.

u₂ = h_line → T₂ = γ T_line — charging an evacuated, insulated rigid tank with an ideal gas.

Worked examples

Example 1 (standard, closed system). 0.5 kg of air in a piston–cylinder is heated at a constant pressure of 200 kPa from 300 K to 500 K. Find W, ΔU and Q. Take R = 0.287, c_v = 0.718, c_p = 1.005 kJ/kg·K.

1. Boundary work at constant pressure: W = P (V₂ − V₁) = m R (T₂ − T₁) = 0.5 × 0.287 × 200 = 28.7 kJ (by the gas).
2. ΔU = m c_v ΔT = 0.5 × 0.718 × 200 = 71.8 kJ.
3. Q = ΔU + W = 71.8 + 28.7 = 100.5 kJ.
4. Check: at constant pressure Q = ΔH = m c_p ΔT = 0.5 × 1.005 × 200 = 100.5 kJ. Consistent.

Example 2 (GATE level, steady flow). A turbocharger compressor takes in 0.5 kg/s of air at 100 kPa, 300 K and 10 m/s, and delivers it at 500 K and 50 m/s. Heat lost from the casing is 10 kJ/kg. Changes in elevation are negligible. Find the power required. Take c_p = 1.005 kJ/kg·K.

1. SFEE per unit mass: q − w = (h₂ − h₁) + (V₂² − V₁²)/2.
2. Δh = c_p (T₂ − T₁) = 1.005 × 200 = 201.0 kJ/kg.
3. ΔKE = (50² − 10²)/2 = 1200 J/kg = 1.2 kJ/kg.
4. Heat is lost, so q = −10 kJ/kg.
5. w = q − Δh − ΔKE = −10 − 201.0 − 1.2 = −212.2 kJ/kg (negative: work is done on the air).
6. Power: Ẇ = ṁ w = 0.5 × (−212.2) = −106.1 kW, so the compressor needs 106.1 kW of input power.

Note how small the kinetic-energy term is compared with Δh; that is why it is often neglected for compressors and turbines, but never for nozzles and diffusers.

Common mistakes

• Using ΔU = m c_v ΔT only for constant-volume processes. For an ideal gas it holds for every process; it is Q = m c_v ΔT that needs constant volume.
• Using u instead of h for flowing streams; the flow work Pv must be included.
• Adding V²/2 in J/kg to h in kJ/kg without dividing by 1000.
• Losing the sign of heat loss or compressor work. Write the balance with the convention, then let the sign come out of the algebra.
• Treating throttling as isothermal for real fluids; it is isenthalpic.
• Applying the steady-flow equation to tank filling or emptying.

For GATE ME

Typical questions: Q, W and ΔU for an ideal gas through constant-pressure, constant-volume, isothermal and polytropic processes, often around a cycle where ∮δQ = ∮δW is the check; SFEE applied to nozzles (exit velocity), turbines and compressors (power), throttling (h₁ = h₂, finding quality with a throttling calorimeter) and tank charging (T₂ = γ T_line). Practise identifying which terms vanish for each device before writing numbers.

Quick check

1. A closed system gains 800 J of internal energy while doing 300 J of work. What heat was added?
2. Steam enters an adiabatic nozzle at very low velocity and its enthalpy drops by 125 kJ/kg. What is the exit velocity?
3. What property stays constant across a throttling valve?
4. An insulated, evacuated rigid tank is filled with air from a line at 300 K. What is the final air temperature?

Answers: 1. Q = 800 + 300 = 1100 J. 2. V = √(2 × 125 000) = 500 m/s. 3. Enthalpy. 4. T₂ = γ T_line = 1.4 × 300 = 420 K.