Question

An aircraft with a turbojet engine is flying with \(V_0=250\ \text{m s}^{-1}\) at an altitude where \(\rho=1\ \text{kg m}^{-3}\). The inlet area is \(A_0=1\ \text{m}^2\). The average exhaust-gas speed at the nozzle exit, with respect to the aircraft, is \(V_{e,\text{rel}}=550\ \text{m s}^{-1}\). The exit pressure equals ambient and fuel–air ratio is negligible. Find the uninstalled thrust (rounded to the nearest integer).

Hint:

If an exit speed is given relative to the aircraft and directed aft, convert to the inertial frame as \(V_e=V_0 - V_{e,\text{rel}}\). With \(p_e=p_a\), the thrust reduces to \(\dot m\,(V_0 - V_e) = \dot m\,(V_{e,\text{rel}}+V_0)\).

Answer 1

Step 1: Mass flow rate through the intake.
With negligible fuel mass, \(\dot m\) is the air mass flow captured by the intake. In the aircraft frame, the incoming air speed magnitude is \(V_0\), so \[ \dot m \;=\; \rho\,A_0\,V_0 \;=\; (1)\,(1)\,(250)\;=\; 250\ \text{kg s}^{-1}. \] 

Step 2: Momentum (velocity) term for thrust.
Uninstalled turbojet thrust (pressure term zero since \(p_e=p_a\)) is the momentum flux difference: \[ T \;=\; \dot m\,(V_e - V_0)\ +\ (p_e-p_a)A_e. \] Because the given exhaust speed \(V_{e,\text{rel}}\) is relative to the aircraft and directed aft, the absolute (earth-frame) exhaust speed is \[ V_e \;=\; V_0 - V_{e,\text{rel}} \;=\; 250 - 550 \;=\; -300\ \text{m s}^{-1} \] (negative sign means aft, opposite to the aircraft's forward \(+x\) direction). Hence \[ V_e - V_0 = (-300) - 250 = -550\ \text{m s}^{-1}. \] Taking the sign into account, the thrust acts forward with magnitude \[ T \;=\; \dot m\,(V_0 - V_e)\;=\; 250 \times (250 - (-300)) \;=\; 250 \times 550 \;=\; 137{,}500\ \text{N}. \] (Pressure term is zero.)
 \[\boxed{T=137{,}500\ \text{N}}\]