Jet Engine & Gas Turbine Calculator

Brayton-cycle performance of turbojets (with afterburner), turbofans and turboshaft / turboprop engines at any altitude and speed: thrust or shaft power, fuel consumption, exhaust speed and temperature, station by station.

TorqueMath.com · Jet Engine & Gas Turbine Calculator
Engine type
Typical engines
m
M
kg/s
: 1
K
K reheat temp.

Component efficiency

%
%

Station by station

Total (stagnation) temperature and pressure along the engine. Station numbers follow the usual SAE ARP 755 convention.

StationWhereTemperaturePressure

Effect of turbine temperature

Same engine and flight condition, only the turbine inlet temperature changes. Hotter means more thrust from the same airflow, at some cost in fuel per unit of thrust.

How it is calculated

Air is compressed, heated at constant pressure in the combustor and expanded through the turbine, which drives the compressor (and fan, or an output shaft). What is left speeds the gas up in the nozzle. This is the Brayton cycle with real component efficiencies.

Atmosphere (ISA) T = 288.15 − 0.0065·h (to 11 km), then 216.65 K; V0 = M · √(1.4 · 287 · T) Intake (ram) Tt2 = T0 (1 + 0.2 M²) pt2 = p0 (1 + 0.2 M²)^3.5 × recovery Compressor Tt3 = Tt2 [1 + (π^0.286 − 1) / ηc] Combustor fuel/air f = (cpg·Tt4 − cpa·Tt3) / (ηb·LHV − cpg·Tt4) LHV = 43 MJ/kg, 5 % pressure loss Turbine cpg (1+f)(Tt4 − Tt5) = cpa (Tt3 − Tt2) + BPR · cpa (Tt13 − Tt2) Nozzle V9 = √(2·cpg·Tt·η (1 − (p0/pt)^0.25)) expanded to ambient Thrust F = ṁcore [(1+f)·V9 − V0 + BPR·(V19 − V0)] TSFC = fuel flow / F Turboshaft P = ṁ · cpg (1+f)(Tt5 − Tt6) · ηm free power turbine down to 1.05 · p0

Air: cp = 1005 J/kg·K, γ = 1.4. Hot gas: cp = 1148 J/kg·K, γ = 1.333. Results are design-point estimates of the kind used in preliminary design (Saravanamuttoo, Gas Turbine Theory). Real engines add bleed, cooling air and variable geometry, so expect real thrust and fuel figures within about ±10 %.

This calculator estimates the design-point performance of a gas turbine from the Brayton cycle with real component efficiencies. Choose a turbojet, turbofan or turboshaft and set flight altitude and Mach number, air mass flow, pressure ratio, turbine inlet temperature and compressor and turbine efficiency. It returns thrust or shaft power, fuel flow, specific fuel consumption, thermal efficiency and the temperature and pressure at each station, using the ISA standard atmosphere. Aerospace students, engineers, model jet builders and aviation enthusiasts use it to see how engines trade thrust, fuel and temperature.

Worked example: a J79-class afterburning turbojet at sea level

  1. Inlet: at sea level in the ISA atmosphere, T0 = 288.15 K and p0 = 101.3 kPa. The engine is static (Mach 0), and 98 % intake pressure recovery gives pt2 = 99.3 kPa.
  2. Compressor: pressure ratio 13.5 at 85 % efficiency gives Tt3 = 288.15 × [1 + (13.5^0.286 − 1) / 0.85] = 662 K (389 °C) and pt3 = 13.4 bar.
  3. Combustor: heating 77 kg/s (170 lb/s) of air to 1250 K takes a fuel/air ratio f = (1148 × 1250 − 1005 × 662) / (0.99 × 43 000 000 − 1148 × 1250) = 0.0187, or 1.44 kg/s of fuel.
  4. Turbine: driving the compressor takes about 380 kJ per kg of air, so the gas cools to 925 K and expands to 314 kPa at 88 % turbine efficiency.
  5. Nozzle and thrust: expanding to ambient pressure gives a jet velocity of 707 m/s, so F = 77 × 1.0187 × 707 = 55.5 kN (12 470 lbf). TSFC = 1.44 / 55.5 = 26.0 g/(kN·s), or 0.92 lb/(lbf·h).
  6. Afterburner: reheating the exhaust to 1950 K raises jet velocity to 1014 m/s and thrust to 82.0 kN (18 420 lbf), but total fuel flow rises to 3.83 kg/s and TSFC to 46.7 g/(kN·s), or 1.65 lb/(lbf·h).

The model gives 55.5 kN dry and 82.0 kN with afterburner, within a few percent of the published J79-GE-17 ratings of roughly 53 kN and 79 kN. The afterburner adds about 48 % thrust for 2.7 times the fuel flow, which is why it is used only for takeoff and short dashes.

FAQ

How does a jet engine produce thrust?

A jet engine takes in air, compresses it, burns fuel in it at nearly constant pressure, and lets the hot gas expand through a turbine and then a nozzle. The turbine extracts just enough energy to drive the compressor, and the rest accelerates the gas out of the nozzle. By Newton's second and third laws, thrust equals the rate of momentum change: F = ṁ × (V_jet − V_flight), plus a pressure term if the nozzle does not expand fully to ambient pressure. More air or a faster jet means more thrust.

How does a turbofan produce thrust, and what is bypass ratio?

A turbofan uses its core, a small turbojet, to drive a large fan. Most of the air passes the fan and goes around the core, and the bypass ratio is that bypass mass flow divided by the core flow: about 5–6 : 1 for the CFM56 and 9–12 : 1 for the newest airliner engines. Accelerating a large mass of air a little is more efficient than accelerating a small mass a lot, because propulsive efficiency ≈ 2 / (1 + V_jet / V_flight). On a high-bypass engine the fan produces roughly 80 % of takeoff thrust.

What is TSFC (thrust specific fuel consumption)?

TSFC is the fuel mass flow divided by thrust: how much fuel the engine burns per unit of thrust per unit of time. It is quoted in g/(kN·s), kg/(daN·h) or lb/(lbf·h); 1 lb/(lbf·h) = 28.3 g/(kN·s). A 1950s turbojet burns about 0.9 lb/(lbf·h) dry, a modern high-bypass turbofan about 0.30–0.35 at takeoff and 0.5–0.55 at cruise, and an afterburner pushes it to 1.6–2.0. For shaft engines the equivalent is specific fuel consumption per kW or hp of shaft power.

What is the Brayton cycle?

The Brayton cycle is the thermodynamic cycle of every gas turbine: isentropic compression, heat addition at constant pressure, isentropic expansion and heat rejection at constant pressure. The ideal efficiency depends only on pressure ratio: η = 1 − 1 / π^((γ−1)/γ). For a pressure ratio of 30 and γ = 1.4 that is 62 %. Real engines lose some of this to compressor and turbine inefficiency, combustor pressure loss and cooling air, so their actual thermal efficiency is lower and also depends strongly on turbine inlet temperature.

Why is turbine inlet temperature so important?

Turbine inlet temperature sets how much energy each kilogram of air can carry. The turbine must first return the compressor work, and whatever is left becomes jet velocity or shaft power, so raising the temperature from 1250 K to 1650 K can increase specific thrust or power by half or more. It also allows a higher pressure ratio and therefore better efficiency. The limit is the turbine blades: modern single-crystal nickel blades with internal air cooling and thermal barrier coatings let engines run gas temperatures well above the metal's melting point.

What is the difference between a turbojet, turbofan, turboprop and turboshaft?

All four share the same gas generator: compressor, combustor and turbine. A turbojet sends all its air through the core and produces thrust only from the hot jet, which suits supersonic flight. A turbofan adds a fan driven by the turbine and moves extra air around the core, which is quieter and more efficient at subsonic speeds. A turboshaft expands the gas through a free power turbine that drives an output shaft, as in helicopters. A turboprop is a turboshaft geared to a propeller, with a small residual jet thrust.