Cycles, Gas & Combustion — Cheat Sheet
Gas processes, cycle efficiencies, compression work and combustion stoichiometry.
Ideal gas
| Equation of state | pV = mRT, R = R_u/M, R_u = 8314.46 J/kmol·K |
| Mayer's relation | c_p − c_v = R |
| Ratio | k = c_p/c_v (1.4 diatomic, 1.67 monatomic) |
| Air | R = 287, c_p = 1005, c_v = 718 J/kg·K |
Processes (pVⁿ = const)
| Isobaric | n = 0; W = p·ΔV, Q = m·c_p·ΔT |
| Isothermal | n = 1; W = mRT·ln(V₂/V₁), Q = W, ΔU = 0 |
| Isentropic | n = k; Q = 0, ΔS = 0, T₂/T₁ = (p₂/p₁)^((k−1)/k) |
| Isochoric | n = ∞; W = 0, Q = m·c_v·ΔT |
| Polytropic work | W = (p₁V₁ − p₂V₂)/(n − 1) |
| Entropy | ΔS = m[c_v·ln(T₂/T₁) + R·ln(V₂/V₁)] |
Closed vs flow work
| Closed system | W = ∫p dV — a piston |
| Steady flow | W = ∫V dp — a compressor, larger by n |
| Ordering | Isothermal takes MOST p·dV work, LEAST V·dp work |
| Consequence | Intercooling helps flow machines, not pistons |
Compression
| Isothermal (floor) | P = p₁V̇·ln(r_p) |
| Isentropic | P = [k/(k−1)]·p₁V̇·[r_p^((k−1)/k) − 1] |
| Optimum staging | Equal ratio per stage, r_stage = r_p^(1/N) |
| Discharge temp | T₂ₛ = T₁·r_p^((k−1)/k) |
| Receiver | V = FAD × t × p_atm / (p_max − p_min) |
Isentropic efficiency
| Compressor | η = (h₂ₛ − h₁)/(h₂ − h₁) — actual is MORE work |
| Turbine | η = (h₁ − h₂)/(h₁ − h₂ₛ) — actual is LESS work |
| Typical | recip/screw 0.70–0.85, centrifugal 0.80–0.90, large turbine 0.85–0.92 |
| Note | Polytropic efficiency > isentropic for a compressor (reheat) |
Cycle efficiency
| Carnot | η = 1 − T_c/T_h (kelvin) |
| Otto | η = 1 − r^(1−k) |
| Diesel | η = 1 − r^(1−k)·[(r_c^k − 1)/(k(r_c − 1))] |
| Brayton | η = 1 − r_p^((1−k)/k) |
| Rankine | η = (w_turbine − w_pump)/q_in |
| Reality | Real engines reach 60–75% of the air-standard figure |
Combustion
| Stoichiometry | CxHy + (x + y/4)(O₂ + 3.76N₂) |
| Air/fuel (mass) | methane 17.2, petrol 15.1, propane 15.7 |
| Air ratio | λ = actual A/F ÷ stoichiometric A/F |
| Max dry CO₂ | natural gas ~11.7%, oil ~15.5% |
| Rule of thumb | 3% O₂ ≈ 15% excess air on gas |
| Siegert loss | q_A = (T_flue − T_air)(A/(21 − O₂) + B) |
| LHV vs HHV | gas LHV/HHV ≈ 0.90 — the net/gross gap |
Transient heating
| Time constant | τ = m·c_p/(h·A) |
| Response | T(t) = T∞ + (T₀ − T∞)e^(−t/τ) |
| Landmarks | 63% at 1τ, 95% at 3τ, 99% at 5τ |
| Validity | Bi = hL_c/k < 0.1, L_c = V/A |
| If Bi > 0.1 | Surface leads the core — true soak time is LONGER |
Formula reference — verify against the governing standard. Not a substitute for engineering judgment.