4: Thrust of a Jet Propulsion system

Assume that we are in a reference frame moving with the engine, under steady-flow conditions. A control surface encloses the engine, with area A on the upstream and downstream faces. The flight velocity is u₁, and the ambient pressure is P₁. Because of flow acceleration (or deceleration) at the inlet, the "capture area) of air A_i is different from the area of the inlet. The mass flow rate of air entering the inlet is ma, and the mass flow rate leaving the nozzle is me, with the exit nozzle area being A_e. The exit velocity is ue, and the pressure at the nozzle exit is P_e. The fuel flow rate into the engine is m_f, and the mass flow rate of air leaving the control volume through its sides is m_s. The reaction to the engine thrust is felt by the engine at the attachment point, and is equal to the magnitude of the thrust, T.

Momentum Conservation gives:

Steady:

x-momentum:

Assume that pressue and velocity are constant over the entire control surface except at the exhaust area A_e. Thus,

Using continuity,

, or

. Now,

Thus,

Net momentum outflow from the control volume is

Thus,

, or

Define a fuel-air ratio

Thus,

Note:

Thrust comes from:

a) Increase in momentum of the air + fuel (momentum thrust)

b) Pressure at the exit plane being higher than the outside pressure (pressure thrust).

When the exit velocity is subsonic,

P_e = P₁

_{because a pressure difference cannot be maintained across a subsonic jet.}

_{Thus, for subsonic jet exhausts,}

For engines with bypass flow (some flow going through a fan or propeller, but not going through the burner), the thrust equation is

where

is the bypass ratio