Thermodynamic Cycles and Applications

Heat Engines and the Carnot Limit

A heat engine absorbs heat Q_H from a hot reservoir, produces net work W_net, and rejects heat Q_L to a cold reservoir; by the first law W_net = Q_H − Q_L, and thermal efficiency is η = W_net/Q_H = 1 − Q_L/Q_H. The Carnot cycle — two reversible isothermal and two reversible adiabatic (isentropic) steps — gives the highest efficiency any engine can achieve between two fixed temperatures: $\eta_{Carnot} = 1 - \frac{T_L}{T_H}$ with both temperatures in kelvin. This is a ceiling, not a design target: a real Rankine or Otto engine always falls short because of irreversibilities (friction, finite-rate heat transfer, throttling).

Worked Carnot example. A power plant takes in heat at T_H = 800 K and rejects to a river at T_L = 300 K. The Carnot limit is η = 1 − 300/800 = 1 − 0.375 = 0.625 (62.5%). If the boiler supplies Q_H = 1000 kW of heat, the maximum possible work output is W_net = η·Q_H = 0.625 × 1000 = 625 kW, and the minimum heat that must be rejected is Q_L = Q_H − W_net = 375 kW. Note that lowering T_L (better condenser cooling) raises η more cheaply than raising T_H, which is limited by turbine metallurgy.

Vapor Power Cycle: Rankine

The Rankine cycle is the workhorse of fossil and nuclear steam plants. Four steady-flow devices each handle the working fluid (water/steam):

Because each device is open and steady-flow, apply Q̇ − Ẇ = ṁΔh per device. Net work per unit mass = turbine work − pump work = (h₃ − h₄) − (h₂ − h₁), heat added = (h₃ − h₂), and: $\eta = \frac{(h_3 - h_4) - (h_2 - h_1)}{h_3 - h_2}$ Pump work is tiny compared with turbine work (liquid is nearly incompressible: w_pump ≈ v_f·ΔP), so it is sometimes neglected. Superheating, reheating, and regeneration raise the average temperature of heat addition and thus efficiency.

Gas Power Cycles: Otto, Diesel, Brayton

Otto cycle (spark-ignition gasoline engines) — two isentropic and two constant-volume steps: $\eta_{Otto} = 1 - \frac{1}{r^{k-1}}, \quad r = \frac{V_1}{V_2}\ (\text{compression ratio})$ Worked example: r = 9, k = 1.4 → η = 1 − 1/9^0.4 = 1 − 1/2.408 = 1 − 0.415 = 0.585 (58.5%). Efficiency rises with compression ratio, but in real gasoline engines r is limited by knock (auto-ignition) to about 8–11.

Diesel cycle — heat is added at constant pressure (during injection) rather than constant volume: $\eta_{Diesel} = 1 - \frac{1}{r^{k-1}}\cdot\frac{r_c^{\,k} - 1}{k(r_c - 1)}$ where r_c = V₃/V₂ is the cutoff ratio. At the same compression ratio the bracketed term makes Diesel slightly less efficient than Otto, but diesel engines run at much higher r (15–22), so in practice they are more efficient.

Brayton cycle (gas turbines/jet engines) — compressor → combustor → turbine, with η = 1 − 1/r_p^((k−1)/k), where r_p = P₂/P₁ is the pressure ratio.

Refrigeration and Heat Pumps

Run a power cycle backward and it MOVES heat from cold to hot using work input. The vapor-compression cycle:

Coefficient of Performance measures "useful effect per work input": $COP_R = \frac{Q_L}{W_{net}} = \frac{h_1 - h_4}{h_2 - h_1}, \qquad COP_{HP} = \frac{Q_H}{W_{net}} = COP_R + 1$ The Carnot maxima are COP_R = T_L/(T_H − T_L) and COP_HP = T_H/(T_H − T_L). COP routinely exceeds 1 because the device transports heat rather than generating it — this surprises students expecting an "efficiency" below 1.

Combustion and Psychrometrics

Combustion. Complete combustion of a hydrocarbon: C_xH_y + (x + y/4)O₂ → x CO₂ + (y/2) H₂O. The air–fuel ratio AFR = mass of air / mass of fuel; stoichiometric AFR for gasoline ≈ 14.7:1. A lean mixture has excess air (AFR > 14.7), a rich mixture has excess fuel. Incomplete combustion produces CO and unburned hydrocarbons; high flame temperatures form NOₓ.

Worked COP example. A Carnot refrigerator keeps a freezer at 263 K while rejecting heat to a 303 K kitchen. COP_R = T_L/(T_H − T_L) = 263/(303 − 263) = 263/40 = 6.58. To remove 2 kW of heat from the freezer, the compressor needs W = Q_L/COP = 2/6.58 = 0.30 kW. A real unit, with irreversibilities, needs more.

Psychrometrics treats moist air as a mixture of dry air and water vapor:

Worked humidity-ratio example. Air at 25 °C and φ = 50% at 101.3 kPa total pressure. The saturation pressure of water at 25 °C is P_sat ≈ 3.17 kPa, so the vapor partial pressure is P_v = φ·P_sat = 0.50 × 3.17 = 1.585 kPa. The humidity ratio is ω = 0.622·P_v/(P − P_v) = 0.622 × 1.585/(101.3 − 1.585) = 0.00989 kg water/kg dry air.

On the psychrometric chart, dry-bulb is the horizontal axis and humidity ratio the vertical axis; curves of constant relative humidity, diagonal constant-wet-bulb (≈ constant-enthalpy) lines, and constant-volume lines let you read any state from two known properties. Air conditioning cools and dehumidifies; the latent load (removing moisture) often exceeds the sensible load (lowering temperature).

Cycle Summary

The table below contrasts the major cycles, their working fluid, and the governing efficiency or performance relation:

For the reversed (refrigeration) cycles, remember the identity COP_HP = COP_R + 1: the heat delivered to the hot space Q_H equals the heat absorbed Q_L plus the work W added by the compressor, so dividing through by W gives COP_HP = COP_R + 1 directly.