Thermodynamics fundamentals for the PE Mechanical exam

Thermodynamics is the foundation under every cycle, turbine, compressor, and refrigeration problem on the PE Mechanical exam. The exam tests application, not theory: a first-law balance, a Carnot ceiling, the right property state, and the discipline to use absolute temperature. Get those four habits right and most of the area falls out.

First law — energy conservation

Closed system: ΔU = Q − W (heat in positive, work by the system positive). Open system, steady flow (SFEE), per unit mass: q − w = (h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁).

For most equipment two terms dominate and the rest vanish: a turbine/compressor → w ≈ −(h₂ − h₁); a nozzle → (V₂² − V₁²)/2 ≈ −(h₂ − h₁); a throttle → h₂ ≈ h₁. Recognizing which device you have tells you which terms to keep.

Heat engine: Q_in = 900 kJ at 60% thermal efficiency → W_net = 0.60 × 900 = 540 kJ, Q_out = 360 kJ.

Second law, entropy, and the Carnot bound

No engine between two reservoirs beats Carnot: η_Carnot = 1 − T_cold/T_hot (absolute T). For the reverse cycle, COP_ref = T_c/(T_h − T_c) and COP_hp = T_h/(T_h − T_c), with COP_hp = COP_ref + 1 for the same cycle. Real devices always fall short — entropy generation is positive.

Carnot ceiling: between 750 K and 300 K, η = 1 − 300/750 = 0.60 (60%). Treat it as the ceiling — a real cycle at those temperatures must come in lower.

Properties and states

A pure substance is fixed by two independent properties; the trap is the state:

• Saturated mixture (under the dome): interpolate on quality x — h = h_f + x·h_fg (same form for s, v).
• Superheated (right of the dome): read the superheat table at (P, T) directly — there is no x.
• Compressed/subcooled liquid: approximate with saturated-liquid values at the temperature.

Quality: where h_f = 420 and h_fg = 2,257 kJ/kg, a mixture at x = 0.90 has h = 420 + 0.90 × 2,257 ≈ 2,451 kJ/kg.

Ideal gas

P·V = m·R·T (absolute T and P; R is the specific gas constant). Process relations follow — constant volume → P/T constant; constant pressure → V/T constant; isentropic → P·Vᵏ constant.

Mass in a tank: air at 500 kPa, 2.0 m³, 300 K, R = 0.287 kJ/kg·K → m = PV/(RT) = (500 × 2)/(0.287 × 300) ≈ 11.6 kg. Rigid tank heated: 200 kPa at 300 K → 450 K gives P₂ = 200 × (450/300) = 300 kPa (kelvin throughout).

Where people lose points

• Gauge vs. absolute — the ideal-gas law and every T-ratio/efficiency formula need absolute T (K or °R) and absolute P. Using °C/°F or gauge pressure is the single most common thermo error.
• Sign convention on work/heat — closed-system ΔU = Q − W; flipping W's sign inverts the answer.
• Wrong property state — applying a quality formula to superheated vapor, or saturated values when the state is superheated. Always check which side of the dome you're on.
• Treating Carnot as achievable — it's an upper bound; a real cycle is always lower.
• Closed vs. open balance — using ΔU = Q − W on a flow device (it needs the SFEE with enthalpy), or vice versa.

Drill it

Practice these on real problems — the tutor walks any you miss, grounded in the worked solution. Where this lives: TFS Module 1 — Thermal/Fluid Principles (this area); the laws become cycles in Power cycles — Rankine & Brayton and the shared Refrigeration cycle & COP. Pair it with Heat transfer fundamentals.