Thermodynamics Cheatsheet Cheatsheet

🔥

Basics & System Types

FUNDAMENTAL

System Type	Definition	Mass Transfer	Example
Open (Control Volume)	Fixed region in space; mass crosses boundary	Yes	Turbine, nozzle, compressor
Closed (Control Mass)	Fixed mass; energy crosses boundary	No	Piston-cylinder, rigid tank
Isolated	No energy or mass crosses boundary	No	Thermos flask (ideal)

Property Type	Definition	Examples
Intensive	Independent of system mass	Temperature (T), Pressure (P), Density (ρ), Specific volume (v)
Extensive	Depends on system mass	Volume (V), Mass (m), Energy (E), Enthalpy (H), Entropy (S)
Specific (derived)	Extensive property per unit mass	v = V/m, u = U/m, h = H/m, s = S/m

Thermodynamic Properties

TemperatureT (K or °C)

PressureP (Pa, bar, atm, MPa)

Specific Volumev = V/m (m³/kg)

Densityρ = m/V = 1/v (kg/m³)

Internal Energyu = U/m (kJ/kg)

Enthalpyh = u + Pv (kJ/kg)

Entropys = S/m (kJ/kg·K)

Unit Conversions

unit_conversions.py

# Temperature
T(K)    = T(°C) + 273.15
T(°F)   = T(°C) × 9/5 + 32
T(R)    = T(°F) + 459.67

# Pressure
1 bar   = 10^5 Pa = 100 kPa
1 atm   = 101.325 kPa
1 MPa   = 10^6 Pa = 10 bar

# Energy & Power
1 kJ    = 1000 J
1 kWh   = 3600 kJ
1 kW    = 1 kJ/s
1 HP    = 745.7 W

Equation of State — Ideal Gas

Form	Equation	Notes
General	PV = nRT	n = moles, R = 8.314 J/(mol·K)
Per unit mass	Pv = RT	R = R_univ / M (specific gas constant)
Combined	P₁V₁/T₁ = P₂V₂/T₂	For fixed mass of ideal gas
Density form	P = ρRT	ρ = 1/v

ideal_gas_calcs.py

# Specific gas constants for common gases
R_air  = 287.0   # J/(kg·K)
R_N2   = 296.8   # J/(kg·K)
R_O2   = 259.8   # J/(kg·K)
R_CO2  = 188.9   # J/(kg·K)
R_H2O  = 461.5   # J/(kg·K)
R_He   = 2077.0  # J/(kg·K)

# Example: Find density of air at 300 K, 1 atm
P = 101325   # Pa
T = 300      # K
rho = P / (R_air * T)   # 1.177 kg/m³

💡

State postulate: The equilibrium state of a simple compressible system is completely specified by two independent, intensive properties. All other properties can be derived from these two.

⚖️

Laws of Thermodynamics

CORE

Law	Statement	Equation	Significance
0th Law	If A is in thermal equilibrium with B and C, then B and C are in equilibrium with each other	T_A = T_B = T_C	Basis for temperature measurement
1st Law (Closed)	Energy cannot be created or destroyed, only converted	ΔU = Q − W	Energy conservation for a process
1st Law (Open)	Energy balance for a control volume with mass flow	Q̇ − Ẇ = ṁ(h₂ + V₂²/2 + gz₂) − ṁ(h₁ + V₁²/2 + gz₁)	Steady-flow energy equation (SFEE)
2nd Law (Kelvin-Planck)	No cyclic device converts heat entirely into work	W_cycle ≤ Q_H	Impossibility of 100% efficient heat engine
2nd Law (Clausius)	Heat cannot flow spontaneously from cold to hot	Q_C / T_C ≤ Q_H / T_H	Refrigeration/heat pump limitation
3rd Law	Entropy of a perfect crystal approaches zero at 0 K	S → 0 as T → 0 K	Absolute reference for entropy

First Law — Closed System

first_law_closed.txt

─── First Law for Closed Systems ───

General form (per unit mass):
  q − w = Δu = u₂ − u₁

  q  = heat transfer per unit mass (kJ/kg)
  w  = work per unit mass (kJ/kg)
  Δu = change in internal energy

For a process 1→2:
  Q₁₋₂ − W₁₋₂ = U₂ − U₁

Types of work:
  • Boundary work (expansion/compression):
    W = ∫P dV  (for quasi-equilibrium process)

  • Shaft work: W_shaft (turbine, compressor)
  • Electrical work: W_elec = VI·Δt

Cyclic process:
  ∮δQ = ∮δW   (Q_net = W_net)

First Law — Open System (SFEE)

steady_flow_energy.txt

─── Steady Flow Energy Equation (SFEE) ───

  Q̇ − Ẇ = ṁ[(h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)]

Simplified forms:
  • Turbine:     Ẇ = ṁ(h₁ − h₂)          [Q≈0, ΔKE≈0, ΔPE≈0]
  • Compressor:  Ẇ = ṁ(h₂ − h₁)          [W_in]
  • Nozzle:      V₂²/2 = h₁ − h₂ + V₁²/2  [W=0, Q=0]
  • Throttling:  h₁ = h₂                  [Q=0, W=0, ΔKE=0, ΔPE=0]

Boundary Work Calculations

Process	Work (W)	Notes
Isobaric (P = const)	W = P(V₂ − V₁)	Area under P-V curve (rectangle)
Isothermal (T = const, ideal gas)	W = P₁V₁ ln(V₂/V₁) = mRT ln(V₂/V₁)	Hyperbolic curve
Isentropic (s = const)	W = (P₂V₂ − P₁V₁)/(1−γ) = (P₁V₁ − P₂V₂)/(γ−1)	PV^γ = const
Polytropic (PVⁿ = const)	W = (P₂V₂ − P₁V₁)/(1−n) = (P₁V₁ − P₂V₂)/(n−1)	General case, n ≠ 1

boundary_work.py

# Example: Isothermal expansion of ideal gas
P1 = 500   # kPa
V1 = 0.1   # m³
V2 = 0.3   # m³
T  = 300   # K (constant)

# W = P1*V1 * ln(V2/V1)
W = P1 * V1 * 1000 * __import__('math').log(V2/V1)
# W = 50,000 × ln(3) = 54,931 J = 54.93 kJ

⚠️

Sign convention: Q > 0 = heat added TO the system, W > 0 = work done BY the system. This is the IUPAC convention. Some textbooks (especially older ones) use the opposite sign for W. Always check which convention is being used!

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Thermodynamic Processes

PROCESSES

Process	Condition	P-V Relation	T-V Relation	Heat (Q)	Work (W)
Isobaric	P = const	P = C	V/T = const	Q = m·cp·ΔT	W = P(V₂−V₁)
Isochoric (Isovolumetric)	V = const	—	P/T = const	Q = m·cv·ΔT	W = 0
Isothermal	T = const (ideal gas)	PV = const	T = const	Q = W	W = P₁V₁ ln(V₂/V₁)
Adiabatic	Q = 0 (no heat transfer)	PV^γ = const	TV^(γ−1) = const	Q = 0	W = (P₁V₁−P₂V₂)/(γ−1)
Polytropic	PVⁿ = const	PVⁿ = const	TV^(n−1) = const	Q = ΔU + W	W = (P₁V₁−P₂V₂)/(n−1)
Isentropic	s = const, reversible + adiabatic	PV^γ = const	TV^(γ−1) = const	Q = 0	W = (P₁V₁−P₂V₂)/(γ−1)

Specific Heats of Ideal Gases

Gas	cp (kJ/kg·K)	cv (kJ/kg·K)	γ = cp/cv	M (g/mol)
Air	1.005	0.718	1.400	28.97
N₂	1.039	0.743	1.400	28.01
O₂	0.918	0.658	1.395	32.00
CO₂	0.846	0.657	1.289	44.01
H₂	14.307	10.183	1.405	2.016
He	5.193	3.116	1.667	4.003
Ar	0.520	0.312	1.667	39.95
CH₄	2.254	1.736	1.299	16.04

Mayer's relation: cp − cv = R (specific gas constant).
Monatomic gases: γ ≈ 5/3 = 1.667. Diatomic: γ ≈ 7/5 = 1.4. Polyatomic: γ ≈ 4/3 = 1.33.

Polytropic Process Relations

polytropic_relations.txt

─── Polytropic Process: PVⁿ = const ───

Relations between states:
  P₁V₁ⁿ = P₂V₂ⁿ
  T₁V₁^(n-1) = T₂V₂^(n-1)
  T₁^(n) / P₁^(n-1) = T₂^(n) / P₂^(n-1)

  i.e., T₂/T₁ = (V₁/V₂)^(n-1) = (P₂/P₁)^((n-1)/n)

Special cases of polytropic index n:
  n = 0 → Isobaric     (P = const)
  n = 1 → Isothermal   (PV = const)
  n = γ → Isentropic   (PV^γ = const)
  n → ∞ → Isochoric    (V = const)

Heat transfer:
  Q = m·cv·(n − γ)/(n − 1) · (T₂ − T₁)
  Q = m·cn·(T₂ − T₁)
  where cn = cv·(n − γ)/(n − 1) is the polytropic specific heat

process_calculations.py

import math

# Adiabatic compression of air
gamma = 1.4
P1 = 100    # kPa
T1 = 300    # K
V1 = 0.5    # m³
P2 = 800    # kPa

# Temperature after adiabatic compression
T2 = T1 * (P2/P1)**((gamma-1)/gamma)
# T2 = 300 × 8^(0.286) = 300 × 1.811 = 543.3 K

# Volume ratio
V2 = V1 * (P1/P2)**(1/gamma)
# V2 = 0.5 × (1/8)^(0.714) = 0.5 × 0.2527 = 0.126 m³

# Work done on the gas (compression, negative work out)
W = (P1*V1 - P2*V2) / (gamma - 1)
# W = (50 - 101.1) / 0.4 = -127.7 kJ

# Internal energy change (Q=0 for adiabatic)
m = 0.6    # kg
cv = 0.718 # kJ/(kg·K)
dU = m * cv * (T2 - T1)  # = 0.6 × 0.718 × 243.3 = 104.8 kJ
# Note: dU = -W (first law: Q = dU + W, Q=0 → dU = -W)

🚫

Common mistake: In a throttling process (valve, capillary tube), enthalpy is constant (h₁ = h₂), NOT temperature. For an ideal gas, throttling results in zero temperature change. For real gases, it can cause cooling or heating depending on the Joule-Thomson coefficient.

🔁

Thermodynamic Cycles

CYCLES

Cycle	Type	Working Fluid	Applications	Efficiency
Carnot	Reversible, External Combustion	Any	Theoretical benchmark	η = 1 − T_C/T_H
Otto	Reversible, Spark Ignition	Air (ideal)	Petrol/Gasoline engines	η = 1 − 1/r^(γ−1)
Diesel	Reversible, Compression Ign.	Air (ideal)	Diesel engines	η = 1 − (1/r^(γ−1))·[(rₖ^γ−1)/(γ(rₖ−1))]
Rankine	Vapor power cycle	Steam/Water	Steam power plants	η_th ≈ 33–45%
Brayton	Gas power cycle (open)	Air/Combustion gas	Gas turbines, jet engines	η = 1 − 1/r_p^((γ−1)/γ)

Carnot Cycle (Most Efficient)

carnot_cycle.txt

─── Carnot Cycle (Theoretical Maximum Efficiency) ───

Four reversible processes:
  1→2 : Isothermal expansion at T_H (absorb heat Q_H)
  2→3 : Isentropic expansion (T_H → T_C)
  3→4 : Isothermal compression at T_C (reject heat Q_C)
  4→1 : Isentropic compression (T_C → T_H)

Thermal efficiency:
  η_Carnot = 1 − T_C/T_H = 1 − Q_C/Q_H

  • T_H = source (hot reservoir) temperature [K]
  • T_C = sink (cold reservoir) temperature [K]
  • Maximum possible efficiency between two reservoirs

Carnot COP (Refrigerator):
  COP_R = T_C / (T_H − T_C)

Carnot COP (Heat Pump):
  COP_HP = T_H / (T_H − T_C)

carnot_example.py

# Carnot engine between 600 K and 300 K
T_H = 600  # K
T_C = 300  # K

eta = 1 - T_C/T_H  # 0.50 = 50%

Q_H = 1000  # kJ heat input
W_net = eta * Q_H  # 500 kJ
Q_C = Q_H - W_net  # 500 kJ rejected

# As refrigerator: COP_R = T_C/(T_H - T_C) = 300/300 = 1.0
COP_R = T_C / (T_H - T_C)
COP_HP = T_H / (T_H - T_C)  # 600/300 = 2.0

Otto Cycle (Spark Ignition)

otto_cycle.txt

─── Otto Cycle (Petrol/SI Engine) ───

Four processes:
  1→2 : Isentropic compression
  2→3 : Constant volume heat addition (combustion)
  3→4 : Isentropic expansion (power stroke)
  4→1 : Constant volume heat rejection

Thermal efficiency:
  η_Otto = 1 − 1/r^(γ−1)

  r = V₁/V₂ = compression ratio (typically 8–12 for SI)
  γ = cp/cv = 1.4 for air

Mean effective pressure:
  MEP = W_net / (V₁ − V₂)

Key: Efficiency depends ONLY on compression ratio, not on
peak temperature or pressure. Higher r → higher η (but
auto-ignition limits r in practice).

Diesel Cycle (Compression Ignition)

diesel_cycle.txt

─── Diesel Cycle (CI Engine) ───

Four processes:
  1→2 : Isentropic compression
  2→3 : Constant pressure heat addition (fuel injection)
  3→4 : Isentropic expansion (power stroke)
  4→1 : Constant volume heat rejection

Thermal efficiency:
  η_Diesel = 1 − (1/r^(γ−1)) × [(rₖ^γ − 1)/(γ(rₖ − 1))]

  r  = compression ratio = V₁/V₂  (typically 14–25 for CI)
  rₖ = cutoff ratio = V₃/V₂

Note: For rₖ = 1 (no constant-P heat addition), the Diesel
cycle reduces to the Otto cycle.

Higher cutoff ratio → lower efficiency (opposite trend to
compression ratio).

Brayton Cycle (Gas Turbine)

brayton_cycle.txt

─── Brayton Cycle (Gas Turbine / Jet Engine) ───

Open cycle (ideal):
  1→2 : Isentropic compression (compressor)
  2→3 : Constant pressure heat addition (combustor)
  3→4 : Isentropic expansion (turbine)
  4→1 : Constant pressure heat rejection (exhaust)

Thermal efficiency:
  η_Brayton = 1 − 1/r_p^((γ−1)/γ)

  r_p = P₂/P₁ = pressure ratio (typically 10–30)
  γ = 1.4 for air

Back work ratio:
  BWR = Ẇ_compressor / Ẇ_turbine

  Gas turbines have high BWR (~40–80%), meaning a large
  fraction of turbine work goes back to drive the compressor.

With regenerator:
  η increases; heat from exhaust pre-heats compressed air

Rankine Cycle (Steam Power Plant)

rankine_cycle.txt

─── Ideal Rankine Cycle ───

Four processes:
  1→2 : Isentropic compression (pump, liquid)
  2→3 : Constant pressure heat addition (boiler)
  3→4 : Isentropic expansion (turbine, vapor)
  4→1 : Constant pressure heat rejection (condenser)

Efficiency:
  η = W_net / Q_in = (Ẇ_turbine − Ẇ_pump) / Q_boiler

  Ẇ_turbine = ṁ(h₃ − h₄)
  Ẇ_pump   = ṁ(h₂ − h₁) = ṁ·v₁·(P₂ − P₁)  [liquid approx]
  Q_boiler  = ṁ(h₃ − h₂)
  Q_cond    = ṁ(h₄ − h₁)

Improvements:
  • Reheat: Expand in stages, reheat between turbines
  • Regeneration: Extract steam for feedwater heating
  • Supercritical: P > 22.064 MPa (no distinct boiling)
  • Binary: Two cycles for different temp ranges

Cycle Comparison	Carnot	Otto	Diesel	Brayton	Rankine
Heat addition	Isothermal	Const. Volume	Const. Pressure	Const. Pressure	Const. Pressure
Heat rejection	Isothermal	Const. Volume	Const. Volume	Const. Pressure	Const. Pressure
Compression	Isentropic	Isentropic	Isentropic	Isentropic	Isentropic (liquid)
Expansion	Isentropic	Isentropic	Isentropic	Isentropic	Isentropic (vapor)
Working fluid	Any (ideal)	Air	Air	Air/combustion gas	Steam/water
Practical η	Theoretical max	25–35%	35–45%	20–40%	33–45%

💡

Carnot theorem: No heat engine operating between two reservoirs can be more efficient than a Carnot engine. All reversible engines operating between the same two reservoirs have the same efficiency. This is the ultimate benchmark for all real cycles.

🌀

Entropy & Second Law Analysis

ENTROPY

entropy_definitions.txt

─── Entropy (S) — Measure of Disorder ───

Definition (Clausius):
  dS = δQ_rev / T      [kJ/K]

For a process 1→2:
  ΔS = S₂ − S₁ = ∫₁² (δQ/T)_rev

For ideal gas (per unit mass):
  Δs = cv·ln(T₂/T₁) + R·ln(V₂/V₁)
     = cp·ln(T₂/T₁) − R·ln(P₂/P₁)

Entropy generation (S_gen):
  ΔS_system + ΔS_surroundings = S_gen ≥ 0

  S_gen > 0 : Irreversible process
  S_gen = 0 : Reversible (internally) process
  S_gen < 0 : Impossible (violates 2nd law)

Principle of Entropy Increase:
  Isolated system: ΔS_isolated = S_gen ≥ 0

Clausius Inequality

clausius_inequality.txt

─── Clausius Inequality ───

  ∮ (δQ / T) ≤ 0

For a cycle:
  If the cycle is reversible:   ∮ δQ/T = 0
  If the cycle is irreversible: ∮ δQ/T < 0

For a process:
  ∫₁² (δQ/T) ≤ ΔS = s₂ − s₁

  Equality holds for reversible (internally) processes.
  Inequality holds for irreversible processes.

Physical meaning: In any real (irreversible) process,
some energy is "degraded" — less available to do work.

Entropy Change Calculations

entropy_changes.py

import math

# 1. Solid/Liquid (approx constant specific heat)
#    ΔS = m·c·ln(T2/T1)
m = 2.0       # kg
c = 0.45      # kJ/(kg·K) (steel)
T1, T2 = 300, 600  # K
dS = m * c * math.log(T2/T1)
# dS = 2 × 0.45 × ln(2) = 0.624 kJ/K

# 2. Ideal gas, isothermal expansion
R = 287       # J/(kg·K)
T = 400       # K (constant)
V1, V2 = 0.1, 0.3  # m³
dS = m * R/1000 * math.log(V2/V1)
# Uses: dS = m·R·ln(V2/V1)

# 3. Heat reservoir at T_H transfers Q to system
T_H = 500    # K
Q = 100      # kJ
dS_surroundings = -Q / T_H  # -0.20 kJ/K

Irreversibilities & Exergy

Source of Irreversibility	Description	Examples
Friction	Mechanical energy converted to heat	Bearing friction, pipe flow friction
Heat transfer across ΔT	Finite temperature difference	Heat exchanger, cooling fin
Unrestrained expansion	Gas expanding into vacuum	Burst balloon, free expansion
Mixing	Mixing of different substances	Mixing hot and cold fluids
Chemical reaction	Spontaneous reactions	Combustion, corrosion
Electrical resistance	I²R losses	Wires, motor windings
Inelastic deformation	Permanent shape change	Plastic deformation, hysteresis

exergy_availability.txt

─── Exergy (Availability) ───

Maximum useful work obtainable as system comes to
equilibrium with the environment (dead state: P₀, T₀):

Closed system exergy:
  X = (U − U₀) + P₀(V − V₀) − T₀(S − S₀)

Flow exergy (open system):
  x = (h − h₀) − T₀(s − s₀) + V²/2 + gz

Exergy destruction (irreversibility):
  I = T₀ · S_gen

Second law efficiency:
  η_II = (Exergy recovered) / (Exergy supplied)

  • Heat engine:     η_II = W_net / X_heat_in
  • Refrigerator:    η_II = X_removed / W_net
  • Heat pump:       η_II = X_heat_delivered / W_net

🚫

Entropy never decreases for an isolated system. The universe itself can be treated as an isolated system, so ΔS_universe = S_gen ≥ 0. This is the most general statement of the second law. Every real process increases the total entropy of the universe.

💧

Properties & Steam Tables

PROPERTIES

Phase Diagram & Terminology

phase_diagram.txt

─── P-v and T-v Diagram for Pure Substance ───

Regions:
  • Compressed (subcooled) liquid: T < T_sat at given P
  • Saturated liquid-vapor mixture: T = T_sat (two-phase)
  • Superheated vapor: T > T_sat at given P
  • Supercritical fluid: P > P_critical, T > T_critical

Key properties on saturation line:
  h_fg = h_g − h_f   (latent heat of vaporization)
  s_fg = s_g − s_f
  v_fg = v_g − v_f
  u_fg = u_g − u_f

For a mixture with quality x:
  h  = h_f  + x · h_fg
  s  = s_f  + x · s_fg
  v  = v_f  + x · v_fg
  u  = u_f  + x · u_fg

  x = m_vapor / m_total    (quality, 0 ≤ x ≤ 1)

Critical point of water:
  T_c = 373.95 °C  (647.1 K)
  P_c = 22.064 MPa (220.64 bar)
  v_c = 0.003155 m³/kg

Triple point of water:
  T_t = 0.01 °C (273.16 K)
  P_t = 0.6117 kPa

Steam Tables — Saturated Water

T (°C)	P_sat (kPa)	h_f (kJ/kg)	h_fg (kJ/kg)	h_g (kJ/kg)	s_f	s_g
25	3.17	104.8	2442.3	2547.2	0.367	8.558
50	12.35	209.3	2382.7	2592.1	0.704	8.075
100	101.3	419.0	2257.0	2676.0	1.307	7.355
150	475.8	632.2	2114.3	2746.5	1.842	6.838
200	1554	852.4	1940.7	2793.1	2.331	6.432
250	3973	1085.3	1715.7	2801.0	2.794	6.073
300	8581	1344.0	1404.7	2748.7	3.254	5.705

At 100 °C (1 atm), latent heat h_fg = 2257 kJ/kg. This decreases as pressure increases and becomes zero at the critical point.

Steam Tables — Superheated Vapor

P (MPa)	T (°C)	v (m³/kg)	h (kJ/kg)	s (kJ/kg·K)
0.1	150	1.9367	2776.4	7.6134
0.1	300	2.6388	3074.3	8.2158
1.0	200	0.2060	2827.9	6.6940
1.0	500	0.3541	3478.5	7.7622
5.0	400	0.0578	3195.7	6.6459
10.0	500	0.0328	3374.6	6.5995

Steam Table Interpolation

steam_interpolation.py

import math

# Linear interpolation between two states
# Given: At T=100°C, h_f=419.0 and at T=150°C, h_f=632.2
# Find: h_f at T=120°C

T1, T2 = 100, 150   # °C
h_f1, h_f2 = 419.0, 632.2
T = 120  # °C

h_f = h_f1 + (h_f2 - h_f1) * (T - T1) / (T2 - T1)
# h_f = 419.0 + (632.2 - 419.0) × (120 - 100)/(150 - 100)
# h_f = 419.0 + 213.2 × 0.4 = 419.0 + 85.28 = 504.28 kJ/kg

# For a wet mixture with quality x:
# Example: P = 200 kPa, x = 0.6
# From tables: h_f = 504.7, h_fg = 2201.9 kJ/kg
x = 0.6
h_f_val = 504.7
h_fg_val = 2201.9
h = h_f_val + x * h_fg_val
# h = 504.7 + 0.6 × 2201.9 = 504.7 + 1321.1 = 1825.8 kJ/kg

💡

Quick state determination: Compare given properties with saturation data. If T < T_sat(P) → compressed liquid. If T = T_sat(P) and x is between 0 and 1 → wet mixture. If T > T_sat(P) → superheated vapor. For compressed liquid, properties ≈ saturated liquid at same T.

🚗

IC Engines (SI & CI)

ENGINES

Feature	SI (Spark Ignition)	CI (Compression Ignition)
Fuel	Gasoline (petrol)	Diesel
Ignition	Spark plug	Self-ignition by compression
Air-fuel mixing	Carburetor / fuel injection (in manifold)	Direct fuel injection into cylinder
Compression ratio	6–12 (lower to avoid knock)	14–25 (higher for auto-ignition)
Load control	Throttling (quantity governing)	Fuel control (quality governing)
Speed	High (3000–6000 RPM typical)	Lower (1500–3000 RPM typical)
Thermal efficiency	25–35%	35–45%
Weight/power ratio	Lower (lighter)	Higher (heavier)
Fuel cost	Higher	Lower
Applications	Cars, motorcycles, small equipment	Trucks, buses, ships, generators

SI Engine — Otto Cycle Analysis

otto_engine.py

import math

# Otto cycle parameters
gamma = 1.4
r = 10.0     # compression ratio
T1 = 300     # K (intake temp)
P1 = 100     # kPa
cv = 0.718   # kJ/(kg·K)

# Process 1→2: Isentropic compression
T2 = T1 * r**(gamma - 1)
P2 = P1 * r**gamma
# T2 = 300 × 10^0.4 = 753.6 K
# P2 = 100 × 10^1.4 = 2511.9 kPa

# Process 2→3: Constant volume heat addition
# Q_in = m * cv * (T3 - T2)
T3 = 2000    # K (peak temperature)
q_in = cv * (T3 - T2)  # kJ/kg
# q_in = 0.718 × (2000 - 753.6) = 894.5 kJ/kg

# Process 3→4: Isentropic expansion
T4 = T3 / r**(gamma - 1)
P4 = P1 * (T4/T1)
# T4 = 2000 / 10^0.4 = 796.2 K

# Heat rejected (4→1, constant volume)
q_out = cv * (T4 - T1)
# q_out = 0.718 × (796.2 - 300) = 356.3 kJ/kg

# Efficiency
eta = 1 - 1 / r**(gamma - 1)
# eta = 1 - 1/10^0.4 = 0.602 = 60.2%
# (Ideal; actual ~30-35% due to losses)

# Mean Effective Pressure
W_net = q_in - q_out
v1 = 0.861  # m³/kg (from ideal gas)
v2 = v1 / r
MEP = W_net / (v1 - v2)
print(f"Efficiency: {eta*100:.1f}%, MEP: {MEP:.1f} kPa")

CI Engine — Diesel Cycle Analysis

diesel_engine.py

import math

# Diesel cycle parameters
gamma = 1.4
r = 18.0     # compression ratio (higher than SI)
rk = 2.5     # cutoff ratio (V3/V2)
T1 = 300     # K
cv = 0.718   # kJ/(kg·K)
cp = 1.005   # kJ/(kg·K)

# Isentropic compression (1→2)
T2 = T1 * r**(gamma - 1)
# T2 = 300 × 18^0.4 = 936.7 K

# Constant pressure heat addition (2→3)
T3 = T2 * rk
# T3 = 936.7 × 2.5 = 2341.8 K

# q_in = cp * (T3 - T2)
q_in = cp * (T3 - T2)
# q_in = 1.005 × (2341.8 - 936.7) = 1412.2 kJ/kg

# Isentropic expansion (3→4)
T4 = T3 * (1/r)**(gamma-1) * rk**gamma
# Or: V4/V3 = V1/V2 × V2/V3 = r/rk
T4 = T3 * (rk / r)**(gamma - 1)

# Heat rejected (4→1, constant volume)
q_out = cv * (T4 - T1)

# Diesel efficiency
eta = 1 - (1/(r**(gamma-1))) * ((rk**gamma - 1)/(gamma*(rk - 1)))
# For r=18, rk=2.5, gamma=1.4:
# eta = 1 - (1/18^0.4) × ((2.5^1.4 - 1)/(1.4×1.5))
# eta ≈ 0.615 = 61.5%

Engine Performance Parameters

Parameter	Formula	Typical Values
Indicated thermal eff.	η_it = IP / (ṁ_f × CV)	SI: 30%, CI: 40%
Brake thermal eff.	η_bt = BP / (ṁ_f × CV)	SI: 25%, CI: 35%
Mechanical eff.	η_mech = BP / IP	80–90%
Specific fuel cons. (brake)	BSFC = ṁ_f / BP	SI: 0.25–0.35 kg/kWh
Mean effective pressure	MEP = W / V_d	SI: 800–1000 kPa, CI: 700–900 kPa
Volumetric efficiency	η_vol = (m_actual / m_theoretical)	SI: 80–85%, CI: 85–90%

Abnormal Combustion

Issue	Cause	Effect	Prevention
Knocking (SI)	Auto-ignition of end gas before spark flame arrives	Noise, vibration, engine damage	Reduce compression ratio, use higher octane fuel, advance timing
Diesel knock	Too rapid combustion after ignition delay	Rough running, noise	Use higher cetane fuel, increase compression temp
Pre-ignition	Hot spots ignite mixture before spark	Loss of power, possible damage	Clean deposits, reduce spark plug heat range

Octane ↑ → Anti-knock ↑Cetane ↑ → Ignition quality ↑Stoichiometric: A/F = 14.7:1

⚠️

Octane vs Cetane: Octane number measures anti-knock quality (SI engines) — higher is better. Cetane number measures ignition quality (CI engines) — higher means easier ignition. n-Heptane has octane = 0; Iso-octane has octane = 100. Methanol has octane ~110. Diesel fuel cetane typically 40–55.

✈️

Gas Turbines & Jet Propulsion

PROPULSION

Brayton Cycle — Detailed Analysis

brayton_analysis.py

import math

# Brayton cycle parameters
gamma = 1.4
cp = 1.005   # kJ/(kg·K)
rp = 12.0    # pressure ratio
T1 = 300     # K (inlet)
T3 = 1400    # K (turbine inlet, TIT)

# Isentropic compression (1→2)
T2 = T1 * rp**((gamma - 1) / gamma)
# T2 = 300 × 12^0.286 = 610.2 K
w_comp = cp * (T2 - T1)
# w_comp = 1.005 × (610.2 - 300) = 311.7 kJ/kg

# Isentropic expansion (3→4)
T4 = T3 / rp**((gamma - 1) / gamma)
# T4 = 1400 / 12^0.286 = 688.1 K
w_turb = cp * (T3 - T4)
# w_turb = 1.005 × (1400 - 688.1) = 714.5 kJ/kg

# Net work and efficiency
w_net = w_turb - w_comp
q_in = cp * (T3 - T2)
eta = w_net / q_in
# eta = 1 - 1/rp^0.286 = 1 - 1/12^0.286 = 0.508 = 50.8%

# Back work ratio
bwr = w_comp / w_turb
# BWR = 311.7 / 714.5 = 0.436 = 43.6%

print(f"eta = {eta*100:.1f}%, BWR = {bwr*100:.1f}%")
print(f"W_net = {w_net:.1f} kJ/kg")

Jet Propulsion — Thrust & Propulsion

jet_propulsion.txt

─── Turbojet Engine ───

Thrust:
  F = ṁ_a × (V_exit − V_inlet) + (P_exit − P_amb) × A_exit

  Simplified (P_exit ≈ P_amb):
  F = ṁ_a × (Vₑ − V₀)

Thrust specific fuel consumption:
  TSFC = ṁ_f / F   [kg/(N·s)]

Propulsive efficiency:
  η_p = 2 × V₀ / (V₀ + Vₑ)
  V₀ = flight velocity, Vₑ = exhaust velocity

Overall efficiency:
  η_o = η_th × η_p = (Thrust power) / (ṁ_f × CV)

  Thrust power = F × V₀
  η_th = (Vₑ² − V₀²) / (2 × q_in)

Turbine Types & Comparison

Type	Application	Bypass Ratio	Efficiency	Speed
Turbojet	Fighter jets	0	High speed, low η_p	Supersonic
Low-bypass Turbofan	Military, bizjets	0.5–2	Moderate	Subsonic/Supersonic
High-bypass Turbofan	Airliners	5–12	High η_p, quiet	Subsonic
Turboprop	Regional, cargo	Very high	Best at low speed	< 0.6 Mach
Ramjet	Missiles, hypersonic	0	No compressor needed	> Mach 3
Scramjet	Hypersonic vehicles	0	Supersonic combustion	> Mach 5

Brayton Cycle Improvements

Improvement	Description	Effect on η	Effect on W_net
Increase pressure ratio	Higher rp	Increases (up to a point)	Increases then decreases
Increase TIT	Higher T₃ (turbine inlet temp)	Increases	Increases significantly
Regeneration	Use exhaust heat to preheat compressed air	Increases at low rp	No change (heat recovery)
Intercooling	Cool between compressor stages	Increases (with reheat)	Increases
Reheat	Heat between turbine stages	Increases (with intercooling)	Increases

Ericsson cycle = Brayton with perfect intercooling + perfect regeneration → achieves Carnot efficiency. Modern gas turbines use TIT up to 1700 °C with blade cooling techniques.

💡

Regeneration benefit: A regenerator is most effective when the exhaust temperature (T₄) is much higher than the compressed air temperature (T₂). At high pressure ratios, T₂ approaches T₄, and regeneration provides diminishing returns. Therefore, regeneration is most beneficial at moderate pressure ratios.

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