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Aerospace

Net thrust of a jet engine

Thrust from the momentum change of the air and fuel through an engine.

Ref
AER-07
Rev
v1
Field
Aerospace
Result
F [kN]

F in newtons. ṁ_a the intake air mass flow, ṁ_e = ṁ_a(1 + f) the exit mass flow including fuel, Ve exit velocity, V0 flight speed, pe and p0 exit and ambient pressure, Ae exit area. The calculator takes a pressure-matched nozzle, so the last term is zero.

A jet engine pushes the aircraft by throwing air backwards faster than it arrived. Thrust is the extra momentum given to each second’s worth of air and fuel.

Calculator

Solve for
kN
kg/s
5500
0.010.04
m/s
3501,000
m/s
0300
  • Specific thrust F/ṁa362 N·s/kg
  • Thrust power F·V04.525 MW

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Net thrust (kN) current value 18.1 kN
ṁaIntake air flowKey driver
1.81181
VeExit velocity
5.3538.5
V0Flight speed
15.630.6
fFuel-air ratio
17.818.7
  • Thrust falls with flight speed because the intake momentum is subtracted: a static engine gives the most thrust.
  • Fuel flow adds a small bonus to the exit mass; ignoring it typically underestimates thrust by 1–3%.
  • Efficiency is best when exhaust speed is close to flight speed, which is why turbofans move a lot of air slowly.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kN)Local stepChange in result
ṁa Intake air flow5 to 500 kg/s1.81181179.2+10%+10 %
Ve Exit velocity350 to 1,000 m/s5.3538.533.15+10%+16.9 %
V0 Flight speed0 to 300 m/s15.630.615+10%−6.91 %
f Fuel-air ratio0.01 to 0.04 17.818.70.9+undefined%—

Sweep of Exit velocity

Exit velocity (m/s)Net thrust (kN)
3505.35
431.39.494
512.513.64
593.817.78
67521.93
756.326.07
837.530.21
918.834.36
1,00038.5

Sweep one variable

Net thrust (kN)Hover, touch, or use the arrow keys on the chart
10203040 4005006007008009001000 Exit velocity (m/s) now 18.1

Variables and how each one contributes

F

Net thrust

Result · kN

What the equation gives you. Everything else feeds it.

ṁa

Intake air flow

kg/s · typical 5 to 500

Air swallowed by the engine each second. Directly proportional.

f

Fuel-air ratio

· typical 0.01 to 0.04

Fuel mass per unit of air mass, typically 0.01–0.04. Adds fuel mass to the exhaust, a small positive effect.

Ve

Exit velocity

m/s · typical 350 to 1,000

Speed of the exhaust relative to the engine. The most direct lever: thrust rises almost one-for-one with exit velocity.

V0

Flight speed

m/s · typical 0 to 300

Speed of the air entering the engine relative to it. Reduces thrust: the faster the aircraft flies, the less extra speed the engine can add.

Worked example with the values above

  1. Exit mass flow: ṁe = 50 × (1 + 0.02) = 51 kg/s.
  2. Momentum terms: 51 × 600 − 50 × 250 = 18,100 N.
  3. Net thrust 18.1 kN (pressure term zero).

Assumptions and limits

  • Steady one-dimensional flow, one uniform exit.
  • Pressure-matched nozzle (pe = p0).
  • Speeds relative to the engine; ignores intake drag and installation losses.

Common mistakes

  • Forgetting the intake momentum.
  • Using exit velocity relative to the ground.
  • Using this for a propeller engine without conversion.

Statistics and uncertainty

Thrust is a difference of two similar large quantities, so errors are magnified. With the exit term about 2.4 times the intake term, a 2% error in Ve becomes roughly a 3.4% error in thrust.

Where it is used

  • Engine selection and thrust-lapse estimates.
  • Missile and aircraft performance.
  • Comparing turbojet and turbofan layouts.

Maths you need

  • Momentum: mass × velocity
  • Newton’s second law: force is the rate of change of momentum

Related formulas

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