Laws of Thermodynamics and Properties
Key Takeaways
- The First Law of Thermodynamics: energy is conserved — Q - W = ΔU (closed system) or Q̇ - Ẇ = ṁΔh + ... (open system).
- The Second Law: entropy of an isolated system always increases; heat flows spontaneously from hot to cold.
- Thermodynamic properties: temperature (T), pressure (P), volume (V), internal energy (U), enthalpy (H = U + PV), entropy (S).
- Specific heats: cv (constant volume) and cp (constant pressure); for ideal gases, cp - cv = R and k = cp/cv.
- Property diagrams (T-s, P-v, P-h) show process paths and are essential for solving thermodynamics problems.
- Phase changes occur at constant temperature and pressure; latent heat is absorbed/released during phase change.
Laws of Thermodynamics and Properties
FE Exam Weight: Thermodynamics and Heat Transfer accounts for 9-14 questions (~10% of the exam). This section combines thermodynamic analysis with heat transfer modes.
The Laws of Thermodynamics
Zeroth Law
If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A is in thermal equilibrium with C. (This establishes the concept of temperature.)
First Law (Conservation of Energy)
Closed System (no mass flow):
Sign convention: Q positive into system, W positive out of system.
Open System (steady-state, steady-flow):
where:
- Q̇ = rate of heat transfer
- Ẇs = shaft power (work per unit time)
- h = specific enthalpy
- V = velocity
- z = elevation
Second Law
Kelvin-Planck: No heat engine can convert ALL heat into work (there must be waste heat).
Clausius: Heat cannot spontaneously flow from cold to hot without external work.
Entropy inequality:
- = 0 for reversible processes
-
0 for irreversible (real) processes
Third Law
The entropy of a perfect crystalline substance approaches zero as temperature approaches absolute zero.
Thermodynamic Properties
| Property | Symbol | Intensive/Extensive |
|---|---|---|
| Temperature | T | Intensive |
| Pressure | P | Intensive |
| Volume | V (total) / v (specific) | Extensive / Intensive |
| Internal Energy | U / u | Extensive / Intensive |
| Enthalpy | H = U + PV / h = u + Pv | Extensive / Intensive |
| Entropy | S / s | Extensive / Intensive |
| Specific Heats | cp, cv | Intensive |
Ideal Gas Relations
| Property | Formula |
|---|---|
| Internal energy change | Δu = cv ΔT |
| Enthalpy change | Δh = cp ΔT |
| Entropy change | Δs = cv ln(T₂/T₁) + R ln(v₂/v₁) |
| Δs = cp ln(T₂/T₁) - R ln(P₂/P₁) | |
| cp - cv | = R (specific gas constant) |
| Specific heat ratio | k = cp/cv |
For air (standard values): cp = 1.005 kJ/(kg·K), cv = 0.718 kJ/(kg·K), k = 1.4, R = 0.287 kJ/(kg·K)
Thermodynamic Processes (Ideal Gas)
| Process | Condition | Work (W) | Key Relation |
|---|---|---|---|
| Isothermal | T = constant | W = nRT ln(V₂/V₁) | PV = constant |
| Isobaric | P = constant | W = PΔV | V/T = constant |
| Isochoric | V = constant | W = 0 | P/T = constant |
| Adiabatic | Q = 0 | W = -ΔU | PVᵏ = constant |
| Polytropic | PVⁿ = constant | W = (P₂V₂-P₁V₁)/(1-n) | General case |
| Isentropic | s = constant (reversible adiabatic) | — | PVᵏ = const, T₂/T₁ = (P₂/P₁)^((k-1)/k) |
Phase Diagrams
T-s Diagram (Temperature vs. Entropy)
- Area under the curve = heat transferred (Q = ∫T ds)
- Horizontal line = isothermal process
- Vertical line = isentropic process
P-v Diagram (Pressure vs. Specific Volume)
- Area under the curve = work done (W = ∫P dv)
- Shows the two-phase dome (vapor dome)
P-h Diagram (Pressure vs. Enthalpy)
- Most useful for refrigeration cycle analysis
- Phase boundaries clearly shown
An ideal gas undergoes an isothermal expansion. Which statement is true?
Air (cp = 1.005 kJ/kg·K) is heated at constant pressure from 300 K to 500 K. The specific enthalpy change is:
According to the Second Law of Thermodynamics, which is impossible?