Thermodynamic Properties, Energy Balances, and Ideal Gases

System Definition and Properties

The first decision in any thermodynamics problem is the system boundary. A closed system has a fixed mass with no mass crossing the boundary (a sealed piston-cylinder, a rigid tank). A control volume / open system allows mass to flow across the boundary (a turbine, nozzle, compressor, throttle, or heat exchanger). The energy balance differs for each, so misclassifying the system is the costliest early error in a thermo problem.

Key properties: internal energy u, enthalpy h = u + pv, entropy s, specific volume v = 1/ρ, plus pressure p and temperature T. Intensive (per-unit-mass) properties carry lowercase symbols; multiply by mass m for the extensive total. Distinguish path functions (heat Q and work W — depend on the process route, written with δ) from point/state functions (u, h, s — depend only on the endpoints, written with d). The state postulate says two independent intensive properties fix the state of a simple compressible substance, so any two of {p, T, v, u, h, s, x} pin down all the rest from the tables. )

Reading Property Tables and Quality

For a pure substance like water/steam, identify the region before reading a table:

• Compressed (subcooled) liquid: T < T_sat at the given p.
• Saturated mixture: T = T_sat; liquid and vapor coexist. Use quality x.
• Superheated vapor: T > T_sat at the given p.

In the saturated mixture region, quality x = m_vapor / m_total, and any property is interpolated linearly between the saturated-liquid and saturated-vapor values:

y = y_f + x·y_fg = y_f + x(y_g − y_f)

where f denotes saturated liquid, g saturated vapor, and fg the difference. 8(v_g − v_f), and similarly for u, h, and s. A classic trap is assigning a quality to a compressed-liquid (x would be < 0) or superheated state (x would be > 1), where x has no meaning — x is only defined 0–1 inside the dome. x = 0 is exactly saturated liquid; x = 1 is exactly saturated vapor. A second trap is double-interpolation in superheated tables: when both p and T fall between table entries you must interpolate twice.

When in doubt about which region you are in, compare the given T to T_sat at the given p: below it means compressed liquid, equal means a mixture, above it means superheated.

The First Law

Closed system (per unit mass): Q − W = ΔU, or q − w = u₂ − u₁. Sign convention in the FE Handbook: Q into the system is positive, W done BY the system is positive. Boundary (moving-boundary) work for a quasi-static process is w = ∫p dv — the area under the process curve on a p–v diagram. Special cases: constant-volume process does zero boundary work; constant-pressure process gives w = pΔv; and for an ideal gas an isothermal process gives w = RT ln(v₂/v₁).

Steady-flow energy equation (SFEE) for a single-stream control volume:

q − w_s = (h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)

Use enthalpy here, not internal energy, because the flow work pv is already bundled into h. Kinetic and potential terms are usually negligible except in nozzles and diffusers, where velocity change is the whole point. Common device simplifications: a throttle is h₁ = h₂ (isenthalpic); a nozzle converts enthalpy to kinetic energy with w_s = 0; a turbine/compressor is often adiabatic (q = 0).

Worked example — adiabatic steam turbine. Steam enters at h₁ = 3450 kJ/kg and exits at h₂ = 2300 kJ/kg, adiabatic (q = 0), KE/PE negligible. Specific work w_s = h₁ − h₂ = 1150 kJ/kg. At ṁ = 5 kg/s, power = ṁ w_s = 5 × 1150 = 5750 kW delivered to the shaft.

Ideal-Gas Relations

When the substance is a gas far from its critical point, use the ideal-gas law:

pv = RT or pV = mRT

with absolute pressure and absolute temperature (K or R), and R the specific gas constant (air: R = 287 J/(kg·K) = 0.287 kJ/(kg·K)). The combined gas law p₁V₁/T₁ = p₂V₂/T₂ requires absolute T — using °C produces nonsense ratios.

For ideal gases: Δu = cv ΔT, Δh = cp ΔT, R = cp − cv, and k = cp/cv (air k ≈ 1.4). Isentropic (reversible adiabatic) processes obey pv^k = constant and T₂/T₁ = (p₂/p₁)^((k−1)/k).

For air, the standard values are cp = 1.005 kJ/(kg·K), cv = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K), k = 1.40 — note R = cp − cv (1.005 − 0.718 ≈ 0.287) holds. The universal gas constant is R_u = 8.314 kJ/(kmol·K), and R = R_u/M for a gas of molar mass M (air M ≈ 29 kg/kmol). Process exponents: isobaric (n=0), isothermal (n=1), isentropic (n=k), isochoric (n=∞) are special cases of the polytropic pv^n = constant.

Worked example. Air is compressed isentropically from 100 kPa, 300 K to 800 kPa. T₂ = T₁(p₂/p₁)^((k−1)/k) = 300 × (800/100)^(0.4/1.4) = 300 × 8^0.286 = 300 × 1.811 = 543 K. The compressor work per unit mass is then Δh = cp(T₂ − T₁) = 1.005 × (543 − 300) ≈ 244 kJ/kg.