Power, Refrigeration Cycles, Psychrometrics, and Second Law

The Second Law and Carnot Limits

The second law sets the ceiling on every cycle. A heat engine takes heat Q_H from a hot reservoir, produces net work W_net, and rejects Q_L to a cold reservoir. By the first law over a complete cycle, W_net = Q_H − Q_L. Its thermal efficiency is:

η = W_net / Q_in = 1 − Q_L/Q_H

The second law forbids converting all heat to work (Kelvin-Planck statement), so η < 1 always. The maximum possible efficiency between two fixed-temperature reservoirs is the Carnot efficiency:

η_Carnot = 1 − T_L/T_H (absolute temperatures)

No real or ideal cycle between those two reservoirs can exceed it, and the Carnot value depends only on the reservoir temperatures, not the working fluid.

Worked example. An engine operates between 800 K and 300 K. η_Carnot = 1 − 300/800 = 0.625, or 62.5%. A claimed engine doing better than this between the same reservoirs violates the second law and is impossible. The killer trap: using °C (e.g., 527 and 27) in the ratio gives a different, wrong answer — Carnot relations demand kelvin or rankine. Entropy ties to the second law: for any real process s_gen ≥ 0, and a reversible adiabatic process is isentropic (Δs = 0).

Refrigerators and Heat Pumps: COP

Refrigerators and heat pumps move heat against the temperature gradient using work input. They are rated by coefficient of performance, not efficiency:

• Refrigerator: COP_R = Q_L / W_net (useful effect = cooling).
• Heat pump: COP_HP = Q_H / W_net (useful effect = heating).
• Note COP_HP = COP_R + 1.

Carnot limits: COP_R,Carnot = T_L/(T_H − T_L) and COP_HP,Carnot = T_H/(T_H − T_L), again with absolute temperatures.

Worked example. A refrigerator keeps a space at 268 K while rejecting to 298 K. Max COP_R = 268/(298 − 268) = 268/30 = 8.9. A real unit will be lower. Reporting an efficiency for a refrigerator (or a COP for an engine) is a guaranteed wrong answer.

Power and Refrigeration Cycles

Rankine cycle (steam power): pump → boiler (heat in) → turbine (work out) → condenser (heat out). Use property tables for each state. η = w_net / q_in = (w_turbine − w_pump)/q_boiler. Pump work w_p = v_f(p₂ − p₁) is usually small.

Vapor-compression refrigeration: compressor → condenser → expansion valve (throttle, h constant) → evaporator. The throttle is isenthalpic (h₃ = h₄). COP_R = (h₁ − h₄)/(h₂ − h₁).

Air-standard cycles use ideal-gas relations, not tables:

• Otto (gasoline): η = 1 − r^(1−k), with compression ratio r and k = 1.4.
• Diesel: η depends on r and the cutoff ratio r_c.
• Brayton (gas turbine): η = 1 − (p₂/p₁)^((1−k)/k) = 1 − r_p^((1−k)/k).

A useful efficiency ranking the FE expects: for the same compression ratio, Otto > Diesel, because Diesel's heat addition occurs at higher entropy. But real diesels run much higher compression ratios, so in practice they beat gasoline engines. Isentropic efficiency rates real rotating devices against the ideal isentropic process and differs by device: a turbine is η = w_actual/w_ideal = (h₁ − h₂a)/(h₁ − h₂s) (actual work is less than ideal), while a compressor/pump is η = w_ideal/w_actual = (h₂s − h₁)/(h₂a − h₁) (actual work is more than ideal). Flipping these two formulas is a frequent trap.

Worked example. An Otto engine with r = 9, k = 1.4: η = 1 − r^(1−k) = 1 − 9^(−0.4) = 1 − 0.415 = 0.585 (58.5%). Raising r to 11 lifts η to 1 − 11^(−0.4) ≈ 0.617, which is why higher compression improves efficiency until knock limits it.

Psychrometrics

Psychrometrics analyzes moist air (air–water-vapor mixtures) for HVAC problems, read off the psychrometric chart in the handbook. The key properties:

• Dry-bulb temperature T_db — ordinary thermometer reading.
• Wet-bulb temperature T_wb — evaporative cooling reading.
• Humidity ratio (specific humidity) ω = mass of water vapor per mass of dry air (kg/kg).
• Relative humidity φ = p_vapor/p_sat (0–100%).
• Dew point — temperature at which condensation begins (φ = 100%).
• Enthalpy h of the mixture (per kg dry air).

Given any two independent properties, the chart fixes the rest. Sensible heating/cooling moves horizontally (ω constant), because no moisture is added or removed. Cooling below the dew point removes moisture (dehumidification), dropping ω and following the saturation curve. Evaporative cooling follows a constant wet-bulb (≈constant enthalpy) line as water evaporates into the stream. Humidification raises ω. A typical FE item: air is cooled below its dew point and the condensate rate is ṁ_water = ṁ_dry air(ω₁ − ω₂).

Always track whether moisture is added (humidification) or removed (dehumidification) to pick the right ω direction.