Thermodynamics

Thermodynamics studies energy, heat, and work. It governs engines, power plants, refrigerators, and any system that converts energy.

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.

In practice: temperature is well defined, and 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. The entropy of an isolated system always increases.

Implications:

• No heat engine can be 100% efficient.
• Processes have a natural direction (irreversibility).
• Not all heat can be converted into work.

Entropy (S): a 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.

In practice: absolute zero (0 K) cannot be reached.

Work and Heat

Work

Mechanical work in thermodynamics:

W = ∫ P dV

For constant pressure: W = P ΔV

Example: a 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)
      v
   [Engine] -> W_out (work)
      v
    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 reaches 100% efficiency.

Carnot Efficiency (Maximum Possible)

η_Carnot = 1 - T_cold/T_hot

Where temperatures are in Kelvin (absolute)

Example: an engine operating between 600 K and 300 K.

η_max = 1 - 300/600 = 0.5 = 50%

Real engines achieve roughly 30 to 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 to 10 for gasoline engines, giving theoretical η of 50 to 60%.

Diesel Cycle

Similar to Otto but combustion at constant pressure. Higher compression ratios (15 to 20), so 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 to 45%.

Refrigeration Cycles

A refrigerator or AC moves heat from cold to hot, the opposite of an 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: an AC removes 5 kW from a room while the 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, and 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 the 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 and produces 300 J work. What is the 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 the Carnot limit.
• Heat engines: convert thermal energy to work.
• Refrigeration: move heat from cold to hot (requires work input).

Next Steps

Continue to 06-fluid-mechanics.md for liquid and gas flow.