Thermodynamics and Its Applications

Syllabus

• Thermodynamics: systems and processes, properties of pure substances, ideal and real gases, zeroth, first and second laws, work and heat, property tables and charts, availability and irreversibility, thermodynamic relations.
• Applications: air and gas compressors, vapour and gas power cycles (regeneration, reheat), IC engines (Otto, Diesel, dual cycles), refrigeration and air-conditioning (vapour and gas cycles, heat pumps, psychrometry), turbomachinery (impulse and reaction, velocity diagrams, Pelton, Francis and Kaplan turbines, steam and gas turbines).

Key concepts and formulas

First law. Closed system: Q − W = ΔU. Steady flow: q − w = Δh + ΔV²/2 + gΔz. For an ideal gas, Δu = c_vΔT, Δh = c_pΔT, c_p − c_v = R, γ = c_p/c_v.

Second law and entropy. Carnot efficiency η = 1 − T_L/T_H. COP of a Carnot refrigerator T_L/(T_H − T_L) and heat pump T_H/(T_H − T_L), so COP_HP = COP_R + 1. Clausius inequality ∮ δQ/T ≤ 0. Entropy change of an ideal gas: Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁). Availability (exergy) of heat Q at T: Q(1 − T₀/T).

Air-standard cycles.

• Otto: η = 1 − 1/r^γ−1 (r = compression ratio).
• Diesel: η = 1 − (1/r^γ−1) · (ρᵞ − 1)/(γ(ρ − 1)) (ρ = cut-off ratio). For the same compression ratio, η_Otto > η_Dual > η_Diesel.
• Brayton: η = 1 − 1/r_p^(γ−1)/γ (r_p = pressure ratio). Maximum net work at r_p = (T_max/T_min)^{γ/(2(γ−1))}.
• Rankine: η = (turbine work − pump work)/heat added, found from steam tables. Reheat raises dryness at exit; regeneration raises efficiency.

Refrigeration. 1 TR = 3.5167 kW (often 3.5 kW). COP = refrigerating effect/work.

Psychrometry. Specific humidity ω = 0.622 p_v/(p − p_v); relative humidity φ = p_v/p_sat.

Turbomachines. Specific speed of a turbine N_s = N√P/H^5/4; of a pump N_s = N√Q/H^3/4. Pelton (impulse) wheels suit high heads and low N_s; Francis medium; Kaplan low heads and high N_s. The maximum efficiency of a Pelton wheel occurs at blade speed u = V/2.

Worked example

Find the efficiency of an Otto cycle with r = 8 and of a Brayton cycle with r_p = 10 (γ = 1.4), and the COP of a Carnot refrigerator between −10 °C and 30 °C. Then find the work done in compressing 1 kg of air isothermally at 300 K from 1 bar to 5 bar.

import math

g = 1.4
print(f"Otto, r = 8: eta = {1 - 1 / 8 ** (g - 1):.4f}")
print(f"Brayton, rp = 10: eta = {1 - 1 / 10 ** ((g - 1) / g):.4f}")
TL, TH = 263.15, 303.15
cop = TL / (TH - TL)
print(f"Carnot COP_R = {cop:.4f}, COP_HP = {cop + 1:.4f}")
R, T = 0.287, 300
print(f"isothermal work = {R * T * math.log(5):.3f} kJ/kg (done on the gas)")

Otto, r = 8: eta = 0.5647
Brayton, rp = 10: eta = 0.4821
Carnot COP_R = 6.5787, COP_HP = 7.5787
isothermal work = 138.573 kJ/kg (done on the gas)

Practice questions