Thermodynamics Principles

Thermodynamic System

  • Definition: A definite quantity of matter contained within a closed surface for analysis.
    • System: Also termed as thermodynamics system or control system.
    • Surroundings: Mass or region outside the system.
    • Boundary: The surface separating the system from its surroundings.
    • Control Boundary: Specifically defines the separation.
    • Control Volume: Volume enclosed by the boundary.
    • Control Space: Space enclosed within the boundary.
    • Universe: Combination of the system and surroundings.

Types of Thermodynamic Systems

  1. Open System
    • Characteristics: Exchanges both energy and mass across boundaries.
    • Examples: Automobile engine, human body.
  2. Closed System
    • Characteristics: Exchanges energy, but not mass across boundaries.
    • Examples: Earth, beaker of water.
  3. Isolated System
    • Characteristics: No exchange of heat, work, or mass with surroundings.
    • Examples: Closed thermos bottle.
    • Observable Properties: Pressure, temperature, and volume that can be observed directly or indirectly.

Thermodynamic Properties

  • Intensive Property: Properties that have the same value in any part of the system; independent of the system mass (e.g., pressure, temperature).
  • Extensive Property: Properties that depend on the mass of the system and are not uniform throughout (e.g., mass, volume, energy, enthalpy).

Thermodynamic State

  • Definition: Specific condition of the system determined by its properties.
  • A change in properties signifies a “change of state.”

Thermodynamic Process, Path, and Cycle

  • Thermodynamic Process: The manner in which a system's state changes (e.g., constant pressure, constant volume).
  • Thermodynamic Path: Series of state changes during a process.
  • Thermodynamic Cycle: Sequence of processes where the initial and final states are the same, forming a closed loop.

Processes: Reversible and Irreversible

  • Reversible Process: System returns to its original state by reversing the factors that caused the change (e.g., compression and expansion of a gas).
  • Irreversible Process: Cannot return to its original state through the same path; lacks equilibrium during changes (e.g., mixing paint).

Thermodynamic Equilibrium

  • Definition: A state where the system’s state does not change over time without external interference.
  • Conditions of Equilibrium:
    • Mechanical: Net force and moment are zero.
    • Thermal: Temperature is uniform throughout.
    • Chemical: Chemical potentials are equal.
    • Electrical: Electrical potentials are uniform.

Density and Specific Volume

  • Density ($\rho$): Mass per unit volume, $\rho = \frac{m}{V}$ (SI unit: kg/m³).
  • Specific Volume ($v$): Volume per unit mass, $v = \frac{V}{m}$ (SI unit: m³/kg).

Specific Gravity and Specific Weight

  • Specific Gravity: Ratio of a substance's density to the density of a reference material (no unit).
  • Specific Weight ($\gamma$): Weight per unit volume, $\gamma = \frac{F_g}{V}$ (SI unit: N/m³).

Pressure

  • Definition: Force exerted per unit area (SI unit: Pa or N/m²).
  • Types of Pressure:
    • Absolute Pressure: Total pressure measured from perfect vacuum.
    • Gauge Pressure: Pressure difference between system and atmospheric pressure.
    • Vacuum Pressure: Pressure below atmospheric levels.

Temperature Measurements

  • Definition: Measure of hotness or coldness of a substance.
  • Key Points:
    • Melting Point: Temperature at which a solid becomes a liquid.
    • Boiling Point: Temperature at which a liquid turns into gas.
    • Triple Point: Condition at which solid, liquid, and gas phases coexist.
    • Critical Point: Beyond which substances exist only as gases.

Sample Problems on Density and Temperature

  • Density Problem: Given $m = 2.27$ kg and $V = 0.1$ m³, find $\rho$:
    • $\rho = \frac{2.27}{0.1} = 22.7$ kg/m³.
  • Specific Volume Problem: Given $V = 0.037$ m³ and $m = 5$ kg, find $v$:
    • $v = \frac{0.037}{5} = 0.0074$ m³/kg.
  • Temperature Conversion: Convert from Fahrenheit to Celsius:
    • Example: $130^{\circ}F = 54.44^{\circ}C$; then to Kelvin: $K = 54.44 + 273.15 = 327.59 K$.