Thermodynamics

Thermodynamics is the study of energy, heat, and work. It governs engines, power plants, refrigerators, and any system involving energy conversion.

Basic Concepts

System, Surroundings, and Boundary

┌─────────────────────────┐
│     Surroundings        │
│   ┌───────────────┐     │
│   │   System      │     │  ← Boundary
│   │ (what we're   │     │
│   │  analyzing)   │     │
│   └───────────────┘     │
└─────────────────────────┘

Types of Systems:

• Closed: Fixed mass, energy can cross boundary (piston-cylinder)
• Open: Mass and energy can cross boundary (turbine, compressor)
• Isolated: No mass or energy exchange (ideal thermos)

Properties and State

Intensive Properties (independent of mass):

• Temperature (T)
• Pressure (P)
• Density (ρ)

Extensive Properties (depend on mass):

• Mass (m)
• Volume (V)
• Energy (E)

State: Condition defined by properties (P, T, V, etc.)

Temperature Scales

Conversions:

K = °C + 273.15
°F = (9/5)°C + 32

Pressure

P = F/A

Common units:
- Pascal (Pa): 1 Pa = 1 N/m²
- Bar: 1 bar = 100,000 Pa
- Atmosphere (atm): 1 atm = 101,325 Pa
- PSI: 1 psi = 6,895 Pa

Absolute vs Gauge Pressure:

P_absolute = P_gauge + P_atmospheric

The Four Laws of Thermodynamics

Zeroth Law

If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.

Practical meaning: Temperature is well-defined. Thermometers work!

First Law (Energy Conservation)

Energy cannot be created or destroyed, only converted from one form to another.

ΔE = Q - W

Where:
- ΔE = change in internal energy (J)
- Q = heat added to system (J)
- W = work done by system (J)

Sign convention:

• Q positive: heat added TO system
• W positive: work done BY system

For a closed system:

Q - W = ΔU + ΔKE + ΔPE

Usually for stationary systems: Q - W = ΔU

Second Law

Heat naturally flows from hot to cold. Entropy of an isolated system always increases.

Implications:

• No heat engine can be 100% efficient
• Processes have a natural direction (irreversibility)
• Can't convert all heat into work

Entropy (S): Measure of disorder or energy unavailability

ΔS ≥ Q/T  (equality for reversible processes)

Third Law

As temperature approaches absolute zero, entropy approaches a constant minimum.

Practical meaning: Can't reach absolute zero (0 K).

Work and Heat

Work

Mechanical work in thermodynamics:

W = ∫ P dV

For constant pressure: W = P ΔV

Example: Piston expands gas at constant P = 200 kPa, volume increases from 0.1 m³ to 0.3 m³.

W = P ΔV = 200,000 × (0.3 - 0.1) = 40,000 J = 40 kJ

Heat Transfer Mechanisms

1. Conduction (through solid material)

q = kA(T₁ - T₂)/L

Where:
- q = heat transfer rate (W)
- k = thermal conductivity (W/m⋅K)
- A = area (m²)
- L = thickness (m)

2. Convection (by fluid motion)

q = hA(T_surface - T_fluid)

Where h = convection coefficient (W/m²⋅K)

3. Radiation (electromagnetic waves)

q = εσA(T₁⁴ - T₂⁴)

Where:
- ε = emissivity (0-1)
- σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²⋅K⁴

Ideal Gas Law

For ideal gases (air, many gases at room conditions):

PV = nRT  or  PV = mRT/M

Where:
- P = absolute pressure (Pa)
- V = volume (m³)
- n = number of moles
- R = universal gas constant = 8.314 J/(mol⋅K)
- m = mass (kg)
- T = absolute temperature (K)
- M = molecular weight (kg/mol)

Specific gas constant:

R_specific = R/M

For air: R_air ≈ 287 J/(kg⋅K)

Example: Air in a Tank

A 2 m³ tank contains air at 25°C and 300 kPa. Find the mass of air.

P = 300,000 Pa
V = 2 m³
T = 25 + 273 = 298 K
R = 287 J/(kg⋅K)

PV = mRT
m = PV/(RT) = (300,000 × 2)/(287 × 298) = 7.0 kg

Thermodynamic Processes

Common Processes

Adiabatic Process

For ideal gas:

PV^γ = constant
TV^(γ-1) = constant

Where γ = c_p/c_v (specific heat ratio)
For air: γ ≈ 1.4

Heat Engines and Cycles

A heat engine converts heat into work.

Basic Engine Cycle

    Q_in (from hot reservoir)
      ↓
   [Engine] → W_out (work)
      ↓
    Q_out (to cold reservoir)

Thermal Efficiency

η = W_out / Q_in = (Q_in - Q_out) / Q_in = 1 - Q_out/Q_in

Where:
- η = efficiency (dimensionless, often expressed as %)
- W_out = net work output
- Q_in = heat input

No real engine can be 100% efficient!

Carnot Efficiency (Maximum Possible)

η_Carnot = 1 - T_cold/T_hot

Where temperatures are in Kelvin (absolute)

Example: Engine operates between 600 K and 300 K.

η_max = 1 - 300/600 = 0.5 = 50%

Real engines achieve 30-40% of Carnot efficiency.

Otto Cycle (Gasoline Engines)

1. Intake stroke (isobaric)
2. Compression (adiabatic)
3. Combustion (isochoric, heat added)
4. Power stroke (adiabatic expansion)
5. Exhaust (isochoric, heat rejected)

Otto efficiency:

η = 1 - 1/r^(γ-1)

Where r = compression ratio = V_max/V_min

Typical r = 8-10 for gasoline engines → η ≈ 50-60% (theoretical)

Diesel Cycle

Similar to Otto but combustion at constant pressure. Higher compression ratios (15-20) → higher efficiency.

Rankine Cycle (Steam Power Plants)

1. Pump (liquid water, increase pressure)
2. Boiler (add heat, create steam)
3. Turbine (steam expands, produce work)
4. Condenser (reject heat, condense to liquid)

Typical efficiency: 35-45%

Refrigeration Cycles

A refrigerator/AC moves heat from cold to hot (opposite of engine).

Coefficient of Performance (COP)

For refrigeration:

COP_R = Q_in / W_in = Q_in / (Q_out - Q_in)

Where:
- Q_in = heat removed from cold space
- W_in = work input (compressor)

For heat pump:

COP_HP = Q_out / W_in

COP_HP = COP_R + 1

Example: AC removes 5 kW from room, compressor uses 1.5 kW.

COP = 5/1.5 = 3.33

Vapor-Compression Cycle

1. Evaporator (absorb heat, refrigerant evaporates)
2. Compressor (increase pressure and temperature)
3. Condenser (reject heat, refrigerant condenses)
4. Expansion valve (reduce pressure)

Used in: Refrigerators, air conditioners, heat pumps.

Practice Problems

Problem 1: First Law

A system absorbs 500 J of heat and does 200 J of work. What is the change in internal energy?

<details> <summary>Solution</summary>

ΔU = Q - W = 500 - 200 = 300 J

</details>

Problem 2: Ideal Gas

Air (R = 287 J/kg⋅K) at 100 kPa and 300 K occupies 1 m³. Find the mass.

<details> <summary>Solution</summary>

PV = mRT
m = PV/(RT) = (100,000 × 1)/(287 × 300) = 1.16 kg

</details>

Problem 3: Carnot Efficiency

An engine operates between 800 K and 300 K. What is maximum efficiency?

<details> <summary>Solution</summary>

η_max = 1 - T_cold/T_hot = 1 - 300/800 = 0.625 = 62.5%

</details>

Problem 4: Engine Efficiency

An engine receives 1000 J heat, produces 300 J work. What is efficiency?

<details> <summary>Solution</summary>

η = W/Q_in = 300/1000 = 0.3 = 30%

</details>

Key Takeaways

✓ First Law: Energy is conserved (Q - W = ΔU)
✓ Second Law: Heat flows hot to cold, entropy increases
✓ Ideal Gas: PV = mRT
✓ Efficiency: η = W_out/Q_in, always less than Carnot limit
✓ Heat engines: Convert thermal energy to work
✓ Refrigeration: Move heat from cold to hot (requires work)

Next Steps

Thermal energy often involves fluids in motion:

• Chapter 06: Fluid Mechanics: liquid and gas flow
• Study specific thermodynamic cycles in detail
• Learn about combustion and chemical thermodynamics

Pro Tip: Always use absolute temperature (Kelvin) in thermodynamic equations!