11.2 Thermodynamics

11.2 Thermodynamics

Quick Answer: Thermodynamics describes heat, work, and temperature. On the PAF GD Pilot initial test you must know the three temperature scales, the three modes of heat transfer, Q = m c delta-T, the first law delta-U = Q - W, the gas laws (Boyle, Charles, Gay-Lussac), and the Carnot efficiency eta = 1 - Tc/Th.

Thermodynamics is the physics of energy in transit. A jet engine is a heat engine: it takes in chemical energy, burns it to raise the temperature and pressure of air, extracts work by expanding the gas through a turbine, and ejects the cooler exhaust. The same laws that govern a Carnot engine govern every power plant on an airframe.

Temperature Scales

Temperature is measured on three common scales:

The Kelvin scale is absolute: 0 K is absolute zero, the temperature at which all classical molecular motion ceases. Conversions: T(K) = T(deg C) + 273.15 and T(deg F) = (9/5) T(deg C) + 32. All gas-law and Carnot-efficiency calculations must use kelvin.

Heat Transfer

Heat flows from a hotter body to a cooler one by three mechanisms:

1. Conduction — energy transfers through a material by particle collisions without bulk motion. Metals conduct well because free electrons carry energy; air is a poor conductor, which is why double-glazed cabin windows insulate.
2. Convection — energy moves by the bulk motion of a fluid. Hot air rises because it is less dense; this drives circulation in a cabin and cooling in a radiator.
3. Radiation — energy travels as electromagnetic waves and needs no medium. A dark-painted aircraft skin absorbs solar radiation and heats up; a polished skin reflects it.

Specific Heat Capacity

The specific heat capacity c of a substance is the heat needed to raise 1 kg of it by 1 K:

Q = m * c * delta-T

with Q in joules, m in kg, c in J/(kg K), and delta-T in K (or deg C, since the interval is the same). Water has a remarkably high c = 4200 J/(kg K), which is why it is used as a coolant.

Worked Example: Cooling an Engine Cylinder

A copper engine valve of mass 0.20 kg at 350 deg C is quenched in oil to 50 deg C. With c_copper = 390 J/(kg K), the heat released is Q = 0.20 x 390 x (350 - 50) = 0.20 x 390 x 300 = 23,400 J = 23.4 kJ. This is the energy the oil must absorb.

First Law of Thermodynamics

The first law is conservation of energy for a thermodynamic system:

delta-U = Q - W

where delta-U is the change in internal energy, Q is heat added to the system, and W is work done by the system. If the gas expands, it does work on the surroundings, so W is positive and delta-U falls. If heat is added while volume is fixed (isochoric), W = 0 and all Q becomes internal energy, raising the temperature.

Worked Example: Isothermal Expansion

One mole of ideal gas expands isothermally at 300 K from 1.0 x 10^-3 m^3 to 2.0 x 10^-3 m^3 against a constant external pressure of 1.0 x 10^5 Pa. The work done by the gas is W = P delta-V = 1.0 x 10^5 x (2.0 - 1.0) x 10^-3 = 100 J. Because temperature is constant, internal energy does not change, so delta-U = 0 and the first law gives Q = W = 100 J: 100 J of heat must flow in to keep the gas at 300 K while it expands.

Gas Laws

For a fixed mass of an ideal gas, the three named gas laws combine into the ideal gas equation PV = nRT:

Temperatures in these equations must be in kelvin. A common PAF error is to plug in Celsius values.

Worked Example: Cabin Pressurization

A cabin contains air at 1.0 x 10^5 Pa and 300 K. If the pressure drops to 2.5 x 10^4 Pa at high altitude while the volume is fixed, the new temperature (Gay-Lussac) is T2 = T1 x (P2/P1) = 300 x (2.5 x 10^4 / 1.0 x 10^5) = 300 x 0.25 = 75 K. Such a temperature is uninhabitable, which is why cabin air is heated and pressurized.

Heat Engines and Carnot Efficiency

A heat engine takes heat Qh from a hot reservoir at Th, converts some to work W, and rejects the rest Qc to a cold reservoir at Tc. The Carnot engine is the most efficient possible engine operating between those two reservoirs:

eta_Carnot = 1 - Tc / Th        (T in kelvin)

No real engine can exceed this limit. Raising Th (hotter combustion) or lowering Tc (colder sink) improves the ceiling. A jet engine's high turbine inlet temperature is why materials science is so important in aviation.

Worked Example: Turbojet Upper Limit

A turbojet operates between a turbine inlet temperature of 1200 K and an exhaust temperature of 300 K. The Carnot efficiency is eta = 1 - 300/1200 = 1 - 0.25 = 0.75, that is, 75%. The real engine's thermal efficiency is lower because of friction, blade leakage, and non-ideal combustion, but 75% is the theoretical ceiling for those reservoir temperatures.

Entropy and the Second Law

The second law of thermodynamics states that the total entropy of an isolated system never decreases. Entropy S measures the dispersal of energy: for a reversible process dS = dQ_rev / T. Real processes are irreversible, so entropy always increases. In aviation terms, the exhaust from a jet engine carries entropy away with it; that energy can never be fully recovered as work.